optimal control of investment-reinsurance problem for an

19
Research Article Optimal Control of Investment-Reinsurance Problem for an Insurer with Jump-Diffusion Risk Process: Independence of Brownian Motions De-Lei Sheng, 1 Ximin Rong, 1,2 and Hui Zhao 1 1 Department of Mathematics, Tianjin University, Tianjin 300072, China 2 Center for Applied Mathematics, Tianjin University, Tianjin 300072, China Correspondence should be addressed to De-Lei Sheng; [email protected] Received 18 March 2014; Revised 20 June 2014; Accepted 25 June 2014; Published 24 July 2014 Academic Editor: Simone Marsiglio Copyright © 2014 De-Lei Sheng et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. is paper investigates the excess-of-loss reinsurance and investment problem for a compound Poisson jump-diffusion risk process, with the risk asset price modeled by a constant elasticity of variance (CEV) model. It aims at obtaining the explicit optimal control strategy and the optimal value function. Applying stochastic control technique of jump diffusion, a Hamilton-Jacobi-Bellman (HJB) equation is established. Moreover, we show that a closed-form solution for the HJB equation can be found by maximizing the insurer’s exponential utility of terminal wealth with the independence of two Brownian motions () and 1 (). A verification theorem is also proved to verify that the solution of HJB equation is indeed a solution of this optimal control problem. en, we quantitatively analyze the effect of different parameter impacts on optimal control strategy and the optimal value function, which show that optimal control strategy is decreasing with the initial wealth and decreasing with the volatility rate of risk asset price. However, the optimal value function (; ; ) is increasing with the appreciation rate of risk asset. 1. Introduction By means of investment and reinsurance, insurers can protect themselves against potentially large losses or ensure their earnings remain relatively stable. erefore, many optimal investment and reinsurance problems have arisen in insur- ance risk management and have been extensively studied in the literature. In the older forms, reinsurance was oſten referred to as “proportional” reinsurance; few studies pay attention to reinsurance. Since Borch [1] studied the safety load- ing of reinsurance premiums, a vast amount of literature is particularly concerned about reinsurance. However, the excess-of-loss reinsurance is a tool commonly employed in risk management in the recent thirty years. Tapiero and Zuckerman [2] gave the optimum excess-loss reinsurance under a dynamic framework. Taylor [3] studied reserving consecutive layers of inwards excess-of-loss reinsurance. Cao and Xu [4] investigated both proportional and excess-of-loss reinsurance under investment gains. Gu et al. [5] investigated optimal control of excess-of-loss reinsurance and investment for insurers under a constant elasticity of variance model but without compound Poisson jump in their research. Zhao et al. [6] studied optimal excess-of-loss reinsurance and investment problem for an insurer with jump-diffusion risk process under the Heston model. It is well known that the compound Poisson process is the most popular and useful model to describe claims process ever since the classical Cram´ er-Lundberg model in risk theory. However, as a compound Poisson process perturbed by a standard Brownian motion, jump diffusion has been researched extensively in the recent ten years. Jump diffusion can give more practical description of claims than continuous models, which widely used to describe dynamics of surplus process. Yang and Zhang [7] investigated the problem of optimal investment for insurer with jump-diffusion risk process. Li et al. [8] studied the threshold dividend strategy for renewal jump-diffusion process. Ruan et al. [9] studied the optimal portfolio and consumption with habit formation in a jump-diffusion market. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 194962, 19 pages http://dx.doi.org/10.1155/2014/194962

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Page 1: Optimal Control of Investment-Reinsurance Problem for an

Research ArticleOptimal Control of Investment-ReinsuranceProblem for an Insurer with Jump-Diffusion Risk ProcessIndependence of Brownian Motions

De-Lei Sheng1 Ximin Rong12 and Hui Zhao1

1 Department of Mathematics Tianjin University Tianjin 300072 China2 Center for Applied Mathematics Tianjin University Tianjin 300072 China

Correspondence should be addressed to De-Lei Sheng tjhbsdl126com

Received 18 March 2014 Revised 20 June 2014 Accepted 25 June 2014 Published 24 July 2014

Academic Editor Simone Marsiglio

Copyright copy 2014 De-Lei Sheng et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper investigates the excess-of-loss reinsurance and investment problem for a compound Poisson jump-diffusion risk processwith the risk asset price modeled by a constant elasticity of variance (CEV) model It aims at obtaining the explicit optimal controlstrategy and the optimal value function Applying stochastic control technique of jump diffusion aHamilton-Jacobi-Bellman (HJB)equation is established Moreover we show that a closed-form solution for the HJB equation can be found by maximizing theinsurerrsquos exponential utility of terminal wealth with the independence of two Brownian motions 119882(119905) and 119882

1(119905) A verification

theorem is also proved to verify that the solution of HJB equation is indeed a solution of this optimal control problem Then wequantitatively analyze the effect of different parameter impacts on optimal control strategy and the optimal value function whichshow that optimal control strategy is decreasing with the initial wealth 119909 and decreasing with the volatility rate of risk asset priceHowever the optimal value function 119881(119905 119909 119904) is increasing with the appreciation rate 120583 of risk asset

1 Introduction

Bymeans of investment and reinsurance insurers can protectthemselves against potentially large losses or ensure theirearnings remain relatively stable Therefore many optimalinvestment and reinsurance problems have arisen in insur-ance risk management and have been extensively studied inthe literature

In the older forms reinsurance was often referred toas ldquoproportionalrdquo reinsurance few studies pay attentionto reinsurance Since Borch [1] studied the safety load-ing of reinsurance premiums a vast amount of literatureis particularly concerned about reinsurance However theexcess-of-loss reinsurance is a tool commonly employed inrisk management in the recent thirty years Tapiero andZuckerman [2] gave the optimum excess-loss reinsuranceunder a dynamic framework Taylor [3] studied reservingconsecutive layers of inwards excess-of-loss reinsurance Caoand Xu [4] investigated both proportional and excess-of-lossreinsurance under investment gains Gu et al [5] investigated

optimal control of excess-of-loss reinsurance and investmentfor insurers under a constant elasticity of variance modelbut without compound Poisson jump in their research Zhaoet al [6] studied optimal excess-of-loss reinsurance andinvestment problem for an insurer with jump-diffusion riskprocess under the Heston model

It is well known that the compound Poisson process is themost popular and useful model to describe claims processever since the classical Cramer-Lundberg model in risktheory However as a compound Poisson process perturbedby a standard Brownian motion jump diffusion has beenresearched extensively in the recent ten years Jump diffusioncan givemore practical description of claims than continuousmodels which widely used to describe dynamics of surplusprocess Yang and Zhang [7] investigated the problem ofoptimal investment for insurer with jump-diffusion riskprocess Li et al [8] studied the threshold dividend strategyfor renewal jump-diffusion process Ruan et al [9] studiedthe optimal portfolio and consumption with habit formationin a jump-diffusion market

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 194962 19 pageshttpdxdoiorg1011552014194962

2 Abstract and Applied Analysis

Not only does the insurer cede part of premiums forreinsurance but the insurer also invests in a financial marketconsisting of one risk-free asset and a risk asset The constantelasticity of variance (CEV) model describes the volatilityof risk asset that received widespread interests after beingproposed by Cox and Ross [10] for European option pricingpartly because the CEV model can capture the impliedvolatility skew and can explain the volatility smile as wellHsu et al [11] gave the integration and detailed derivationfor constant elasticity of variance (CEV) option pricingmodel Campi et al [12] investigated systematic equity-based credit risk for a CEV model with jump to defaultApplying Legendre transform and dual theory Xiao et al [13]investigated analytical strategies for annuity contracts underconstant elasticity of variance model The CEV model wasapplied to the optimal investment and reinsurance problemsin Gu et al [14] By maximizing the expected exponentialutility of terminal wealth Lin and Li [15] studied the optimalreinsurance and investment for a jump-diffusion risk processunder the CEV model

The goal of this paper is therefore to investigate theoptimal control strategy of excess-of-loss reinsurance andinvestment problem for an insurer with compound Poissonjump-diffusion risk process from the stochastic control pointof view Under the hypothesis that two Brownian motions119882(119905) and 119882

1(119905) are mutually independent we present

three main results We first give the optimal excess-of-lossreinsurance strategy and investment strategy respectivelyaccording to the first-order necessary condition ofmaximumWe also obtain the optimal value function by plugging areasonable conjecture into the HJB equation and solvinglinear second-order partial differential equations In the thirdpart we specialize in a verification theorem We also provethe verification theorem in detail inspired by the results ofTaksar and Zeng [16] Gu et al [5] and Zhao et al [6] At lastwe present some numerical examples to illustrate the impactsof model parameters

The paper is organized as follows Section 2 introducesthe risk asset price model under the CEV model and thewealth process with jump diffusion In Section 3 we con-sider the optimal excess-of-loss reinsurance and investmentstrategy and a verification theorem as well Section 4 containssome numerical simulations and theoretical results Section 5brings this paper to an end

2 The Model

Throughout this paper (ΩF 119875 F1199050le119905le119879

) denotes a com-plete probability space satisfying the usual condition wherea finite constant 119879 gt 0 represents the investment timehorizon two standard Brownianmotions119882

1(119905) and119882(119905) are

independent of each other 120574 is a Poisson random measureF

119905= F119882

119905or F

1198821

119905or F

120574

119905represents the minimal 120590-field

generated by F119882

119905 F119882

1

119905 and F

120574

119905 which contains all the

information available until time 119905 All stochastic processesinvolved in this paper are supposed to be F

119905adapted In

addition let 119909 and 119910 = min119909 119910 below

Following the same formulation of Zhao et al [6] thesurplus process of an insurer is described by the followingjump-diffusion model

d119877 (119905) = 119888119889119905 + 120590d119882(119905) minus d119862 (119905) (1)

where 119888 and 120590 are positive constants 119882(119905) is a standardBrownian motion and 120590119882(119905) describes the uncertaintyassociated with the surplus of the insurer at time 119905 Assume119862(119905) = sum

119873(119905)

119894=1119885

119894which represents the cumulative claims until

time 119905 where 119873(119905) is a homogeneous Poisson process withintensity 120582 and the claim sizes 119885

119894 119894 ge 1 are independent

and identically distributed positive random variables withcommon distribution 119865(119911) Denote the mean value 119864[119885

119894] =

120583infin

and

119863 = sup 119911 119865 (119911) le 1 lt +infin (2)

Suppose that 119865(0) = 0 0 lt 119865(119911) lt 1 if 119911 isin (0 119863) and 119865(119911) =1 if 119911 ge 119863 The premium according to the expected valueprinciple is 119888 = (1 + 120578)120582120583

infin where 120578 gt 0 is the safety loading

of the insurerAccording to the theory of Poisson random measure

(referring to Oslashksendal and Sulem [17]) we can rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894 120574 denotes the

Poisson random measure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(3)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (4)

so the compensated Poisson random measure of 120574 is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (5)

Insurance company can purchase excess-of-loss rein-surance to reduce the risk Supposing the insurerrsquos (fixed)retention level denoted by 119886 the corresponding reserveprocess is

d119877 (119905) = 119888(119886)119889119905 + 120590d119882(119905) minus d119862(119886)

(119905) (6)

and the reinsurerrsquos premium rate is

119888(119886)

= (1 + 120578) 120582120583infinminus (1 + 120579) 120582 120583

infinminus 119864 [119885

119894and 119886]

= (120578 minus 120579) 120582120583infin+ (1 + 120579) 120582119864 [119885

119894and 119886]

119864 [119885119894and 119886] = int

119886

0

119911d119865 (119911) + int

infin

119886

119886d119865 (119911)

= int

119886

0

(1 minus 119865 (119911)) d119911 ≜ int

119886

0

119865 (119911)d119911

(7)

Abstract and Applied Analysis 3

and 119862(119886)= sum

119873(119905)

119894=1(119885

119894and 119886) satisfying

119890int119905

0119890minus119903119904d119862(119886)(119904)

lt infin forall119905 lt infin (8)

In particular if the retention level 119886 = 119863 then

119888(119886)

= (1 + 120578) 120582120583infin= 119888

(119863)= 119888 (9)

Without loss of generality we always assume the reinsur-ance is not cheap that is 120579 gt 120578 and 120579 denotes the safetyloading of the reinsurer

Moreover the insurer also can invest in a financial marketconsisting of one risk-free asset and one risky assetThe priceprocess 119878

0(119905) of the risk-free asset is given by

d1198780(119905) = 119903119878

0(119905) d119905 (10)

where 119903 gt 0 is the risk-free interest rate The price process119878(119905) of the risky asset is described by the constant elasticity ofvariance (CEV) model

d119878 (119905) = 119878 (119905) [120583d119905 + 119896119878120573(119905) d119882

1(119905)] (11)

where 120583 and 119896 are positive constants 120583 gt 119903 is the expectedinstantaneous return rate of the risky asset 119896119878120573

(119905) is theinstantaneous volatility 120573 is the elasticity parameter and119882

1(119905) is a standard Brownian motion When 120573 = 0 a CEV

model degenerates into a GBM In this paper we assume thestandard Brownian motion119882

1(119905) is independent of119882(119905)

The control 120572 = (119886(119905) 120587(119905)) 119905 isin [0 119879] is a two-dimen-sional F(119905) adapted stochastic process where 119886(119905) isin [0 119863]

is the retention level of the excess-of-loss at time 119905 in which119886(119905) equiv 119863 means ldquono reinsurancerdquo and 119886(119905) equiv 0 means ldquofullreinsurancerdquo and 120587(119905) represents the proportion invested inthe risky asset Here short-selling is not allowed that is120587(119905) ge 0The amount invested in the risk-free asset is119883(119905)(1minus120587(119905)) where 119883(119905) is the wealth process for the insurer underthe strategy 120572 and the dynamics of119883(119905) is given by

d119883 (119905) = [(120578 minus 120579) 120582120583infin+ (1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120583120587 (119905)119883 (119905) + 119903 (1 minus 120587 (119905))119883 (119905) ] d119905

+ 120590d119882(119905) + 120587 (119905)119883 (119905) 119896119878120573(119905) d119882

1(119905) minus d119862(119886)

(119905)

119883 (0) = 1199090

(12)

A strategy 120572 is said to be admissible if forall119905 isin

[0 119879] (119886(119905) 120587(119905)) is F(119905) progressively measurable and119864[int

infin

0120587

2(119905)119883

2(119905)119878

2120573(119905)] lt infin 119886(119905) isin [0 119863] and 120587(119905) isin

[0 +infin) The set of all admissible strategies denoted byΛ and (12) has a unique (strong) solution Suppose theinsurer has a utility function 119880(119909) which is strictly concaveand continuously differentiable on (minusinfin +infin) and aims tomaximize the expected utility of hisher terminal wealth thatis

max120572isinΛ

119864 [119880 (119883 (119879))] (13)

3 Main Results

In this section we solve the excess-of-loss reinsurance andinvestment problem with independence of two Brownianmotions119882(119905) and119882

1(119905) By maximizing the expected utility

of terminal wealth the optimal strategy and value functionare given At the beginning let us give the following expo-nential utility function of an risk aversion insurer

119880 (119909) = minus1

119902119890minus119902119909

119902 gt 0 (14)

This utility function has a constant absolute risk aversionparameter 119902 and is the only utility function under the princi-ple of ldquozero utilityrdquo giving a fair premium that is independentof the level of reserves of insurers For an admissible strategy120572 = (119886(119905) 120587(119905)) we define the value function as

119867120572(119905 119909 119904) = 119864 [119880 (119883 (119879)) | 119883 (119905) = 119909 119878 (119905) = 119904] (15)

and the optimal value function is

119867(119905 119909 119904) = sup120572isinΛ

119867120572(119905 119909 119904) (16)

with the boundary condition

119867(119879 119909 119904) = minus1

119902119890minus119902119909

(17)

The objective of the insurer is to find an optimal strategy120572

lowast= (119886

lowast(119905) 120587

lowast(119905)) such that 119867(119905 119909 119904) = 119867

120572lowast

(119905 119909 119904) where119886

lowast(119905) is called the optimal reinsurance strategy and 120587

lowast(119905) is

called the optimal investment strategyFor any119867120572

isin 119862122

([0 119879]times119877+times119877

+) define the generator

A119867120572(119905 119909 119904)

= 119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909

+ 120583119904119867120572

119904+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+ (120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909+ 119896

2120587119909119904

2120573+1119867

120572

119909119904

+ 119867120572

119909(1 + 120579) 120582int

119886

0

119865(119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

(18)

where 119867120572

119905 119867120572

119909 119867120572

119904 119867120572

119909119909 and 119867

120572

119904119904denote the corresponding

first- and second-order partial derivatives of 119867120572(119905 119909 119904) with

respect to (wrt) the corresponding variables respectivelyApplying the classical tools of stochastic optimal control

we can derive the following Hamilton-Jacobi-Bellman (HJB)equation for problem (16)

sup120572isinΛ

A119867120572(119905 119909 119904) = 0 (19)

with the boundary condition (17)

4 Abstract and Applied Analysis

Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem

Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast

= (119886lowast 120587

lowast) in feedback form where

one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the

whole interval [0 119879] always is

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (20)

and the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(21)

The optimal value function is119881 (119905 119909 119904)

= minus1

119902exp minus 119902

times [119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909) d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

times 119865 (119909) d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

]

(22)

(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(23)

And the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(24)

The optimal value function is given by

119881 (119905 119909 119904)

= minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

=

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]

119892 (119905 119904)

= [(minus(120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573) 119890

2120573119903119905+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573

+ ((120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573)1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ (2120573 + 1) (120583 minus 119903)2

times (4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573) 119905 isin [0 119905

0)

119892 (119905 119904)

=

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

119905 isin [1199050 119879]

119891 (119905)

=(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [119905

0 119879]

Abstract and Applied Analysis 5

119891 (119905)

=(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

minus(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905

0)

+ (1 + 120579) 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [0 119905

0)

(25)Proof See Appendix A

Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= (120583 minus 119903) [1

119902minus

(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)

)

2119902119903]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587

lowast(119905) = ((120583 minus 119903)

1199091199021198962)119890

minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]

A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy

Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem

Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions

(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt

(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)

2minus 1205832120583)

then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast

= (119886lowast(119905) 120587

lowast(119905)) given in Theorem 1

Proof See Appendix B

Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump

4 Numerical Results

In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge

ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar

Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890

minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583

infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1

120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the

optimal strategy when 120573 ge 0

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= [

[

120583 minus 119903

119902minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

2119902119903

]

]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(27)

where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price

6 Abstract and Applied Analysis

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of x on the optimal investment strategy 120587

x = 5

x = 3

x = 1

Figure 1 The optimal investment strategy 120587 decreases with respectto 119909

Deriving 120587lowast with respect to 119909 we have

120597120587lowast

120597119909= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(28)

because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Remark 5 Obviously the optimal investment strategy 120587lowast

decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1

However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset

Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890

2119903119905120573 and 120583 gt (120583minus119903)we have

120597120587lowast

120597120583=

119890119903(119905minus119879(1+2120573))

119904minus2120573

[1198902119903119879120573

120583 minus 1198902119903119905120573

(120583 minus 119903)]

1198962119902119903119909gt 0 (29)

Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2

010 011 012 013 014 015 016

06

08

10

12

14

120583

120587

The effect of 120583 on the optimal investment strategy 120587

t = 8

t = 5

t = 1

Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset

Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt

119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)

Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions

120597120587lowast

120597120573= minus (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [1198902119903119905120573

(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

= (119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

(30)

Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573

gt 1198902119903119905120573

gt 0 and ln[119904] gt 0If 1199042 gt 119890

(120583minus119903)(119879minus119905) take logarithm on both sides

ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)

We have

1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905)

gt 1198902119903119905120573

119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0

(32)

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 2: Optimal Control of Investment-Reinsurance Problem for an

2 Abstract and Applied Analysis

Not only does the insurer cede part of premiums forreinsurance but the insurer also invests in a financial marketconsisting of one risk-free asset and a risk asset The constantelasticity of variance (CEV) model describes the volatilityof risk asset that received widespread interests after beingproposed by Cox and Ross [10] for European option pricingpartly because the CEV model can capture the impliedvolatility skew and can explain the volatility smile as wellHsu et al [11] gave the integration and detailed derivationfor constant elasticity of variance (CEV) option pricingmodel Campi et al [12] investigated systematic equity-based credit risk for a CEV model with jump to defaultApplying Legendre transform and dual theory Xiao et al [13]investigated analytical strategies for annuity contracts underconstant elasticity of variance model The CEV model wasapplied to the optimal investment and reinsurance problemsin Gu et al [14] By maximizing the expected exponentialutility of terminal wealth Lin and Li [15] studied the optimalreinsurance and investment for a jump-diffusion risk processunder the CEV model

The goal of this paper is therefore to investigate theoptimal control strategy of excess-of-loss reinsurance andinvestment problem for an insurer with compound Poissonjump-diffusion risk process from the stochastic control pointof view Under the hypothesis that two Brownian motions119882(119905) and 119882

1(119905) are mutually independent we present

three main results We first give the optimal excess-of-lossreinsurance strategy and investment strategy respectivelyaccording to the first-order necessary condition ofmaximumWe also obtain the optimal value function by plugging areasonable conjecture into the HJB equation and solvinglinear second-order partial differential equations In the thirdpart we specialize in a verification theorem We also provethe verification theorem in detail inspired by the results ofTaksar and Zeng [16] Gu et al [5] and Zhao et al [6] At lastwe present some numerical examples to illustrate the impactsof model parameters

The paper is organized as follows Section 2 introducesthe risk asset price model under the CEV model and thewealth process with jump diffusion In Section 3 we con-sider the optimal excess-of-loss reinsurance and investmentstrategy and a verification theorem as well Section 4 containssome numerical simulations and theoretical results Section 5brings this paper to an end

2 The Model

Throughout this paper (ΩF 119875 F1199050le119905le119879

) denotes a com-plete probability space satisfying the usual condition wherea finite constant 119879 gt 0 represents the investment timehorizon two standard Brownianmotions119882

1(119905) and119882(119905) are

independent of each other 120574 is a Poisson random measureF

119905= F119882

119905or F

1198821

119905or F

120574

119905represents the minimal 120590-field

generated by F119882

119905 F119882

1

119905 and F

120574

119905 which contains all the

information available until time 119905 All stochastic processesinvolved in this paper are supposed to be F

119905adapted In

addition let 119909 and 119910 = min119909 119910 below

Following the same formulation of Zhao et al [6] thesurplus process of an insurer is described by the followingjump-diffusion model

d119877 (119905) = 119888119889119905 + 120590d119882(119905) minus d119862 (119905) (1)

where 119888 and 120590 are positive constants 119882(119905) is a standardBrownian motion and 120590119882(119905) describes the uncertaintyassociated with the surplus of the insurer at time 119905 Assume119862(119905) = sum

119873(119905)

119894=1119885

119894which represents the cumulative claims until

time 119905 where 119873(119905) is a homogeneous Poisson process withintensity 120582 and the claim sizes 119885

119894 119894 ge 1 are independent

and identically distributed positive random variables withcommon distribution 119865(119911) Denote the mean value 119864[119885

119894] =

120583infin

and

119863 = sup 119911 119865 (119911) le 1 lt +infin (2)

Suppose that 119865(0) = 0 0 lt 119865(119911) lt 1 if 119911 isin (0 119863) and 119865(119911) =1 if 119911 ge 119863 The premium according to the expected valueprinciple is 119888 = (1 + 120578)120582120583

infin where 120578 gt 0 is the safety loading

of the insurerAccording to the theory of Poisson random measure

(referring to Oslashksendal and Sulem [17]) we can rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894 120574 denotes the

Poisson random measure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(3)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (4)

so the compensated Poisson random measure of 120574 is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (5)

Insurance company can purchase excess-of-loss rein-surance to reduce the risk Supposing the insurerrsquos (fixed)retention level denoted by 119886 the corresponding reserveprocess is

d119877 (119905) = 119888(119886)119889119905 + 120590d119882(119905) minus d119862(119886)

(119905) (6)

and the reinsurerrsquos premium rate is

119888(119886)

= (1 + 120578) 120582120583infinminus (1 + 120579) 120582 120583

infinminus 119864 [119885

119894and 119886]

= (120578 minus 120579) 120582120583infin+ (1 + 120579) 120582119864 [119885

119894and 119886]

119864 [119885119894and 119886] = int

119886

0

119911d119865 (119911) + int

infin

119886

119886d119865 (119911)

= int

119886

0

(1 minus 119865 (119911)) d119911 ≜ int

119886

0

119865 (119911)d119911

(7)

Abstract and Applied Analysis 3

and 119862(119886)= sum

119873(119905)

119894=1(119885

119894and 119886) satisfying

119890int119905

0119890minus119903119904d119862(119886)(119904)

lt infin forall119905 lt infin (8)

In particular if the retention level 119886 = 119863 then

119888(119886)

= (1 + 120578) 120582120583infin= 119888

(119863)= 119888 (9)

Without loss of generality we always assume the reinsur-ance is not cheap that is 120579 gt 120578 and 120579 denotes the safetyloading of the reinsurer

Moreover the insurer also can invest in a financial marketconsisting of one risk-free asset and one risky assetThe priceprocess 119878

0(119905) of the risk-free asset is given by

d1198780(119905) = 119903119878

0(119905) d119905 (10)

where 119903 gt 0 is the risk-free interest rate The price process119878(119905) of the risky asset is described by the constant elasticity ofvariance (CEV) model

d119878 (119905) = 119878 (119905) [120583d119905 + 119896119878120573(119905) d119882

1(119905)] (11)

where 120583 and 119896 are positive constants 120583 gt 119903 is the expectedinstantaneous return rate of the risky asset 119896119878120573

(119905) is theinstantaneous volatility 120573 is the elasticity parameter and119882

1(119905) is a standard Brownian motion When 120573 = 0 a CEV

model degenerates into a GBM In this paper we assume thestandard Brownian motion119882

1(119905) is independent of119882(119905)

The control 120572 = (119886(119905) 120587(119905)) 119905 isin [0 119879] is a two-dimen-sional F(119905) adapted stochastic process where 119886(119905) isin [0 119863]

is the retention level of the excess-of-loss at time 119905 in which119886(119905) equiv 119863 means ldquono reinsurancerdquo and 119886(119905) equiv 0 means ldquofullreinsurancerdquo and 120587(119905) represents the proportion invested inthe risky asset Here short-selling is not allowed that is120587(119905) ge 0The amount invested in the risk-free asset is119883(119905)(1minus120587(119905)) where 119883(119905) is the wealth process for the insurer underthe strategy 120572 and the dynamics of119883(119905) is given by

d119883 (119905) = [(120578 minus 120579) 120582120583infin+ (1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120583120587 (119905)119883 (119905) + 119903 (1 minus 120587 (119905))119883 (119905) ] d119905

+ 120590d119882(119905) + 120587 (119905)119883 (119905) 119896119878120573(119905) d119882

1(119905) minus d119862(119886)

(119905)

119883 (0) = 1199090

(12)

A strategy 120572 is said to be admissible if forall119905 isin

[0 119879] (119886(119905) 120587(119905)) is F(119905) progressively measurable and119864[int

infin

0120587

2(119905)119883

2(119905)119878

2120573(119905)] lt infin 119886(119905) isin [0 119863] and 120587(119905) isin

[0 +infin) The set of all admissible strategies denoted byΛ and (12) has a unique (strong) solution Suppose theinsurer has a utility function 119880(119909) which is strictly concaveand continuously differentiable on (minusinfin +infin) and aims tomaximize the expected utility of hisher terminal wealth thatis

max120572isinΛ

119864 [119880 (119883 (119879))] (13)

3 Main Results

In this section we solve the excess-of-loss reinsurance andinvestment problem with independence of two Brownianmotions119882(119905) and119882

1(119905) By maximizing the expected utility

of terminal wealth the optimal strategy and value functionare given At the beginning let us give the following expo-nential utility function of an risk aversion insurer

119880 (119909) = minus1

119902119890minus119902119909

119902 gt 0 (14)

This utility function has a constant absolute risk aversionparameter 119902 and is the only utility function under the princi-ple of ldquozero utilityrdquo giving a fair premium that is independentof the level of reserves of insurers For an admissible strategy120572 = (119886(119905) 120587(119905)) we define the value function as

119867120572(119905 119909 119904) = 119864 [119880 (119883 (119879)) | 119883 (119905) = 119909 119878 (119905) = 119904] (15)

and the optimal value function is

119867(119905 119909 119904) = sup120572isinΛ

119867120572(119905 119909 119904) (16)

with the boundary condition

119867(119879 119909 119904) = minus1

119902119890minus119902119909

(17)

The objective of the insurer is to find an optimal strategy120572

lowast= (119886

lowast(119905) 120587

lowast(119905)) such that 119867(119905 119909 119904) = 119867

120572lowast

(119905 119909 119904) where119886

lowast(119905) is called the optimal reinsurance strategy and 120587

lowast(119905) is

called the optimal investment strategyFor any119867120572

isin 119862122

([0 119879]times119877+times119877

+) define the generator

A119867120572(119905 119909 119904)

= 119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909

+ 120583119904119867120572

119904+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+ (120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909+ 119896

2120587119909119904

2120573+1119867

120572

119909119904

+ 119867120572

119909(1 + 120579) 120582int

119886

0

119865(119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

(18)

where 119867120572

119905 119867120572

119909 119867120572

119904 119867120572

119909119909 and 119867

120572

119904119904denote the corresponding

first- and second-order partial derivatives of 119867120572(119905 119909 119904) with

respect to (wrt) the corresponding variables respectivelyApplying the classical tools of stochastic optimal control

we can derive the following Hamilton-Jacobi-Bellman (HJB)equation for problem (16)

sup120572isinΛ

A119867120572(119905 119909 119904) = 0 (19)

with the boundary condition (17)

4 Abstract and Applied Analysis

Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem

Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast

= (119886lowast 120587

lowast) in feedback form where

one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the

whole interval [0 119879] always is

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (20)

and the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(21)

The optimal value function is119881 (119905 119909 119904)

= minus1

119902exp minus 119902

times [119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909) d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

times 119865 (119909) d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

]

(22)

(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(23)

And the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(24)

The optimal value function is given by

119881 (119905 119909 119904)

= minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

=

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]

119892 (119905 119904)

= [(minus(120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573) 119890

2120573119903119905+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573

+ ((120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573)1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ (2120573 + 1) (120583 minus 119903)2

times (4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573) 119905 isin [0 119905

0)

119892 (119905 119904)

=

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

119905 isin [1199050 119879]

119891 (119905)

=(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [119905

0 119879]

Abstract and Applied Analysis 5

119891 (119905)

=(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

minus(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905

0)

+ (1 + 120579) 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [0 119905

0)

(25)Proof See Appendix A

Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= (120583 minus 119903) [1

119902minus

(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)

)

2119902119903]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587

lowast(119905) = ((120583 minus 119903)

1199091199021198962)119890

minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]

A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy

Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem

Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions

(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt

(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)

2minus 1205832120583)

then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast

= (119886lowast(119905) 120587

lowast(119905)) given in Theorem 1

Proof See Appendix B

Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump

4 Numerical Results

In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge

ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar

Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890

minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583

infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1

120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the

optimal strategy when 120573 ge 0

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= [

[

120583 minus 119903

119902minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

2119902119903

]

]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(27)

where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price

6 Abstract and Applied Analysis

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of x on the optimal investment strategy 120587

x = 5

x = 3

x = 1

Figure 1 The optimal investment strategy 120587 decreases with respectto 119909

Deriving 120587lowast with respect to 119909 we have

120597120587lowast

120597119909= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(28)

because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Remark 5 Obviously the optimal investment strategy 120587lowast

decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1

However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset

Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890

2119903119905120573 and 120583 gt (120583minus119903)we have

120597120587lowast

120597120583=

119890119903(119905minus119879(1+2120573))

119904minus2120573

[1198902119903119879120573

120583 minus 1198902119903119905120573

(120583 minus 119903)]

1198962119902119903119909gt 0 (29)

Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2

010 011 012 013 014 015 016

06

08

10

12

14

120583

120587

The effect of 120583 on the optimal investment strategy 120587

t = 8

t = 5

t = 1

Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset

Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt

119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)

Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions

120597120587lowast

120597120573= minus (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [1198902119903119905120573

(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

= (119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

(30)

Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573

gt 1198902119903119905120573

gt 0 and ln[119904] gt 0If 1199042 gt 119890

(120583minus119903)(119879minus119905) take logarithm on both sides

ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)

We have

1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905)

gt 1198902119903119905120573

119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0

(32)

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 3: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 3

and 119862(119886)= sum

119873(119905)

119894=1(119885

119894and 119886) satisfying

119890int119905

0119890minus119903119904d119862(119886)(119904)

lt infin forall119905 lt infin (8)

In particular if the retention level 119886 = 119863 then

119888(119886)

= (1 + 120578) 120582120583infin= 119888

(119863)= 119888 (9)

Without loss of generality we always assume the reinsur-ance is not cheap that is 120579 gt 120578 and 120579 denotes the safetyloading of the reinsurer

Moreover the insurer also can invest in a financial marketconsisting of one risk-free asset and one risky assetThe priceprocess 119878

0(119905) of the risk-free asset is given by

d1198780(119905) = 119903119878

0(119905) d119905 (10)

where 119903 gt 0 is the risk-free interest rate The price process119878(119905) of the risky asset is described by the constant elasticity ofvariance (CEV) model

d119878 (119905) = 119878 (119905) [120583d119905 + 119896119878120573(119905) d119882

1(119905)] (11)

where 120583 and 119896 are positive constants 120583 gt 119903 is the expectedinstantaneous return rate of the risky asset 119896119878120573

(119905) is theinstantaneous volatility 120573 is the elasticity parameter and119882

1(119905) is a standard Brownian motion When 120573 = 0 a CEV

model degenerates into a GBM In this paper we assume thestandard Brownian motion119882

1(119905) is independent of119882(119905)

The control 120572 = (119886(119905) 120587(119905)) 119905 isin [0 119879] is a two-dimen-sional F(119905) adapted stochastic process where 119886(119905) isin [0 119863]

is the retention level of the excess-of-loss at time 119905 in which119886(119905) equiv 119863 means ldquono reinsurancerdquo and 119886(119905) equiv 0 means ldquofullreinsurancerdquo and 120587(119905) represents the proportion invested inthe risky asset Here short-selling is not allowed that is120587(119905) ge 0The amount invested in the risk-free asset is119883(119905)(1minus120587(119905)) where 119883(119905) is the wealth process for the insurer underthe strategy 120572 and the dynamics of119883(119905) is given by

d119883 (119905) = [(120578 minus 120579) 120582120583infin+ (1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120583120587 (119905)119883 (119905) + 119903 (1 minus 120587 (119905))119883 (119905) ] d119905

+ 120590d119882(119905) + 120587 (119905)119883 (119905) 119896119878120573(119905) d119882

1(119905) minus d119862(119886)

(119905)

119883 (0) = 1199090

(12)

A strategy 120572 is said to be admissible if forall119905 isin

[0 119879] (119886(119905) 120587(119905)) is F(119905) progressively measurable and119864[int

infin

0120587

2(119905)119883

2(119905)119878

2120573(119905)] lt infin 119886(119905) isin [0 119863] and 120587(119905) isin

[0 +infin) The set of all admissible strategies denoted byΛ and (12) has a unique (strong) solution Suppose theinsurer has a utility function 119880(119909) which is strictly concaveand continuously differentiable on (minusinfin +infin) and aims tomaximize the expected utility of hisher terminal wealth thatis

max120572isinΛ

119864 [119880 (119883 (119879))] (13)

3 Main Results

In this section we solve the excess-of-loss reinsurance andinvestment problem with independence of two Brownianmotions119882(119905) and119882

1(119905) By maximizing the expected utility

of terminal wealth the optimal strategy and value functionare given At the beginning let us give the following expo-nential utility function of an risk aversion insurer

119880 (119909) = minus1

119902119890minus119902119909

119902 gt 0 (14)

This utility function has a constant absolute risk aversionparameter 119902 and is the only utility function under the princi-ple of ldquozero utilityrdquo giving a fair premium that is independentof the level of reserves of insurers For an admissible strategy120572 = (119886(119905) 120587(119905)) we define the value function as

119867120572(119905 119909 119904) = 119864 [119880 (119883 (119879)) | 119883 (119905) = 119909 119878 (119905) = 119904] (15)

and the optimal value function is

119867(119905 119909 119904) = sup120572isinΛ

119867120572(119905 119909 119904) (16)

with the boundary condition

119867(119879 119909 119904) = minus1

119902119890minus119902119909

(17)

The objective of the insurer is to find an optimal strategy120572

lowast= (119886

lowast(119905) 120587

lowast(119905)) such that 119867(119905 119909 119904) = 119867

120572lowast

(119905 119909 119904) where119886

lowast(119905) is called the optimal reinsurance strategy and 120587

lowast(119905) is

called the optimal investment strategyFor any119867120572

isin 119862122

([0 119879]times119877+times119877

+) define the generator

A119867120572(119905 119909 119904)

= 119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909

+ 120583119904119867120572

119904+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+ (120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909+ 119896

2120587119909119904

2120573+1119867

120572

119909119904

+ 119867120572

119909(1 + 120579) 120582int

119886

0

119865(119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

(18)

where 119867120572

119905 119867120572

119909 119867120572

119904 119867120572

119909119909 and 119867

120572

119904119904denote the corresponding

first- and second-order partial derivatives of 119867120572(119905 119909 119904) with

respect to (wrt) the corresponding variables respectivelyApplying the classical tools of stochastic optimal control

we can derive the following Hamilton-Jacobi-Bellman (HJB)equation for problem (16)

sup120572isinΛ

A119867120572(119905 119909 119904) = 0 (19)

with the boundary condition (17)

4 Abstract and Applied Analysis

Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem

Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast

= (119886lowast 120587

lowast) in feedback form where

one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the

whole interval [0 119879] always is

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (20)

and the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(21)

The optimal value function is119881 (119905 119909 119904)

= minus1

119902exp minus 119902

times [119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909) d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

times 119865 (119909) d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

]

(22)

(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(23)

And the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(24)

The optimal value function is given by

119881 (119905 119909 119904)

= minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

=

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]

119892 (119905 119904)

= [(minus(120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573) 119890

2120573119903119905+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573

+ ((120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573)1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ (2120573 + 1) (120583 minus 119903)2

times (4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573) 119905 isin [0 119905

0)

119892 (119905 119904)

=

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

119905 isin [1199050 119879]

119891 (119905)

=(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [119905

0 119879]

Abstract and Applied Analysis 5

119891 (119905)

=(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

minus(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905

0)

+ (1 + 120579) 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [0 119905

0)

(25)Proof See Appendix A

Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= (120583 minus 119903) [1

119902minus

(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)

)

2119902119903]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587

lowast(119905) = ((120583 minus 119903)

1199091199021198962)119890

minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]

A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy

Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem

Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions

(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt

(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)

2minus 1205832120583)

then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast

= (119886lowast(119905) 120587

lowast(119905)) given in Theorem 1

Proof See Appendix B

Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump

4 Numerical Results

In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge

ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar

Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890

minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583

infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1

120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the

optimal strategy when 120573 ge 0

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= [

[

120583 minus 119903

119902minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

2119902119903

]

]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(27)

where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price

6 Abstract and Applied Analysis

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of x on the optimal investment strategy 120587

x = 5

x = 3

x = 1

Figure 1 The optimal investment strategy 120587 decreases with respectto 119909

Deriving 120587lowast with respect to 119909 we have

120597120587lowast

120597119909= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(28)

because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Remark 5 Obviously the optimal investment strategy 120587lowast

decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1

However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset

Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890

2119903119905120573 and 120583 gt (120583minus119903)we have

120597120587lowast

120597120583=

119890119903(119905minus119879(1+2120573))

119904minus2120573

[1198902119903119879120573

120583 minus 1198902119903119905120573

(120583 minus 119903)]

1198962119902119903119909gt 0 (29)

Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2

010 011 012 013 014 015 016

06

08

10

12

14

120583

120587

The effect of 120583 on the optimal investment strategy 120587

t = 8

t = 5

t = 1

Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset

Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt

119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)

Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions

120597120587lowast

120597120573= minus (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [1198902119903119905120573

(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

= (119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

(30)

Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573

gt 1198902119903119905120573

gt 0 and ln[119904] gt 0If 1199042 gt 119890

(120583minus119903)(119879minus119905) take logarithm on both sides

ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)

We have

1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905)

gt 1198902119903119905120573

119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0

(32)

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 4: Optimal Control of Investment-Reinsurance Problem for an

4 Abstract and Applied Analysis

Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem

Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast

= (119886lowast 120587

lowast) in feedback form where

one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the

whole interval [0 119879] always is

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (20)

and the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(21)

The optimal value function is119881 (119905 119909 119904)

= minus1

119902exp minus 119902

times [119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909) d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

times 119865 (119909) d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

]

(22)

(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(23)

And the optimal investment strategy is given by

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

(24)

The optimal value function is given by

119881 (119905 119909 119904)

= minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

=

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)

minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]

119892 (119905 119904)

= [(minus(120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573) 119890

2120573119903119905+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573

+ ((120583 minus 119903)

2

119890minus2119903119879120573

41199021199031198962120573)1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ (2120573 + 1) (120583 minus 119903)2

times (4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573) 119905 isin [0 119905

0)

119892 (119905 119904)

=

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

119905 isin [1199050 119879]

119891 (119905)

=(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [119905

0 119879]

Abstract and Applied Analysis 5

119891 (119905)

=(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

minus(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905

0)

+ (1 + 120579) 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [0 119905

0)

(25)Proof See Appendix A

Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= (120583 minus 119903) [1

119902minus

(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)

)

2119902119903]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587

lowast(119905) = ((120583 minus 119903)

1199091199021198962)119890

minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]

A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy

Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem

Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions

(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt

(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)

2minus 1205832120583)

then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast

= (119886lowast(119905) 120587

lowast(119905)) given in Theorem 1

Proof See Appendix B

Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump

4 Numerical Results

In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge

ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar

Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890

minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583

infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1

120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the

optimal strategy when 120573 ge 0

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= [

[

120583 minus 119903

119902minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

2119902119903

]

]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(27)

where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price

6 Abstract and Applied Analysis

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of x on the optimal investment strategy 120587

x = 5

x = 3

x = 1

Figure 1 The optimal investment strategy 120587 decreases with respectto 119909

Deriving 120587lowast with respect to 119909 we have

120597120587lowast

120597119909= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(28)

because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Remark 5 Obviously the optimal investment strategy 120587lowast

decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1

However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset

Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890

2119903119905120573 and 120583 gt (120583minus119903)we have

120597120587lowast

120597120583=

119890119903(119905minus119879(1+2120573))

119904minus2120573

[1198902119903119879120573

120583 minus 1198902119903119905120573

(120583 minus 119903)]

1198962119902119903119909gt 0 (29)

Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2

010 011 012 013 014 015 016

06

08

10

12

14

120583

120587

The effect of 120583 on the optimal investment strategy 120587

t = 8

t = 5

t = 1

Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset

Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt

119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)

Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions

120597120587lowast

120597120573= minus (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [1198902119903119905120573

(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

= (119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

(30)

Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573

gt 1198902119903119905120573

gt 0 and ln[119904] gt 0If 1199042 gt 119890

(120583minus119903)(119879minus119905) take logarithm on both sides

ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)

We have

1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905)

gt 1198902119903119905120573

119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0

(32)

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 5: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 5

119891 (119905)

=(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

minus(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905

0)

+ (1 + 120579) 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909) d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909) d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903 119905 isin [0 119905

0)

(25)Proof See Appendix A

Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= (120583 minus 119903) [1

119902minus

(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)

)

2119902119903]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587

lowast(119905) = ((120583 minus 119903)

1199091199021198962)119890

minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]

A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy

Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem

Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions

(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt

(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)

2minus 1205832120583)

then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast

= (119886lowast(119905) 120587

lowast(119905)) given in Theorem 1

Proof See Appendix B

Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump

4 Numerical Results

In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge

ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar

Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890

minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583

infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1

120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the

optimal strategy when 120573 ge 0

120587lowast(119905) = [

[

120583 minus 119903

1199091199021198962minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

21199021199091199031198962

]

]

119890minus119903(119879minus119905)

119904minus2120573

= [

[

120583 minus 119903

119902minus

(120583 minus 119903)2

(1198902119903120573(119905minus119879)

minus 1)

2119902119903

]

]

sdot 119890minus119903(119879minus119905)

sdot1

119909sdot

1

(119896119904120573)2

(27)

where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price

6 Abstract and Applied Analysis

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of x on the optimal investment strategy 120587

x = 5

x = 3

x = 1

Figure 1 The optimal investment strategy 120587 decreases with respectto 119909

Deriving 120587lowast with respect to 119909 we have

120597120587lowast

120597119909= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(28)

because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Remark 5 Obviously the optimal investment strategy 120587lowast

decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1

However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset

Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890

2119903119905120573 and 120583 gt (120583minus119903)we have

120597120587lowast

120597120583=

119890119903(119905minus119879(1+2120573))

119904minus2120573

[1198902119903119879120573

120583 minus 1198902119903119905120573

(120583 minus 119903)]

1198962119902119903119909gt 0 (29)

Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2

010 011 012 013 014 015 016

06

08

10

12

14

120583

120587

The effect of 120583 on the optimal investment strategy 120587

t = 8

t = 5

t = 1

Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset

Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt

119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)

Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions

120597120587lowast

120597120573= minus (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [1198902119903119905120573

(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

= (119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

(30)

Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573

gt 1198902119903119905120573

gt 0 and ln[119904] gt 0If 1199042 gt 119890

(120583minus119903)(119879minus119905) take logarithm on both sides

ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)

We have

1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905)

gt 1198902119903119905120573

119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0

(32)

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 6: Optimal Control of Investment-Reinsurance Problem for an

6 Abstract and Applied Analysis

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of x on the optimal investment strategy 120587

x = 5

x = 3

x = 1

Figure 1 The optimal investment strategy 120587 decreases with respectto 119909

Deriving 120587lowast with respect to 119909 we have

120597120587lowast

120597119909= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(28)

because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Remark 5 Obviously the optimal investment strategy 120587lowast

decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1

However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset

Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890

2119903119905120573 and 120583 gt (120583minus119903)we have

120597120587lowast

120597120583=

119890119903(119905minus119879(1+2120573))

119904minus2120573

[1198902119903119879120573

120583 minus 1198902119903119905120573

(120583 minus 119903)]

1198962119902119903119909gt 0 (29)

Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2

010 011 012 013 014 015 016

06

08

10

12

14

120583

120587

The effect of 120583 on the optimal investment strategy 120587

t = 8

t = 5

t = 1

Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset

Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt

119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)

Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions

120597120587lowast

120597120573= minus (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [1198902119903119905120573

(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

= (119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])

minus 1198902119903119879120573

(119903 + 120583) ln [119904]]) times (1198962119902119903119909)

minus1

(30)

Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573

gt 1198902119903119905120573

gt 0 and ln[119904] gt 0If 1199042 gt 119890

(120583minus119903)(119879minus119905) take logarithm on both sides

ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)

We have

1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905)

gt 1198902119903119905120573

119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0

(32)

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 7: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 7

so that120597120587

lowast

120597120573= (119896

2119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [1198902119903119905120573

(120583 minus 119903) ln [119904]

+ 1198902119903119905120573

(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573

times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]

= (1198962119902119903119909)

minus1

119890119903(119905minus119879(1+2120573))

119904minus2120573

(120583 minus 119903)

times [ minus (1198902119903119879120573

minus 1198902119903119905120573

) (120583 minus 119903) ln [119904]

minus (1198902119903119879120573

2119903 ln [119904] minus 1198902119903119905120573

119903 (120583 minus 119903) (119879 minus 119905))] lt 0

(33)

which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))

120597120587lowast

120597119904= (119890

119903(119905minus119879(1+2120573))119904minus2120573

(119903 minus 120583)

times [(1198902119903119905120573

(119903 minus 120583) + 1198902119903119879120573

(119903 + 120583))]) times (21198962119902119903119909

2)minus1

= minus ((120583 minus 119903) 119904minus2120573

119890119903(119905minus119879(1+2120573))

times [119903 (1198902119903119879120573

+ 1198902119903119905120573

) + 120583 (1198902119903119879120573

minus 1198902119903119905120573

)] )

times (21198962119902119903119909

2)minus1

lt 0

(34)

because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))

gt 0 and120583(119890

2119903119879120573minus 119890

2119903119905120573) gt 0

Consider

120597120587lowast

120597119896= minus

119890minus119903(119879minus119905)

119904minus2120573

(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573

(120583 minus 119903))

1198963119902119903119909lt 0

(35)

Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)

All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price

Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3

(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which

makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence

119881 (119905 119909 119904)

= minus1

119902exp

times minus119902[119909119890119903(119879minus119905)

minus120590

2119902119890

2119903(119879minus119905)

4119903

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2

)

times (81199021205731199032)minus1

+120590

2119902

4119903]

(36)

Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 119896 is the volatility coefficient

of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively

Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then

(21198963119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

gt 0 (37)

so that

120597119881

120597119896= minus(2119896

3119902119903120573)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573) )] ]

lt 0

(38)

which means the value function decreases with respect to 119896See Figure 4(b)

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 8: Optimal Control of Investment-Reinsurance Problem for an

8 Abstract and Applied Analysis

05 10 15 20 25 30 35 4000

01

02

03

k

120587

t = 7

The optimal investment strategy 120587 with k

t = 4

t = 1

(a)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

s = 15

The effect of s on the optimal investment strategy 120587

s = 10

s = 5

(b)

0 2 4 6 8 1000

01

02

03

04

05

06

07

t

120587

The effect of 120573 on the optimal investment strategy 120587

120573 = 15

120573 = 1

120573 = 05

(c)

Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 9: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 9

0 2 4 6 8 10

minus25

minus20

minus15

minus10

minus05

00

V

t = 9

t = 4

t = 1

s

The optimal value function V varies with s

(a)

01 02 03 04 05

minus20

minus15

minus10

minus05

00

k

V

t = 7

t = 3

The optimal value function V decreases with k

t = 1

(b)

Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896

Completely similar to the analysis in (1) we also have

120597119881

120597119904= minus(2119896

2119902119903)

minus1

(1198902119903119879120573

minus 1198902119903119905120573

) 119904minus1minus2120573

(119903 minus 120583)2

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

lt 0

(39)

which shows that the value function decreases with respect to119904 See Figure 4(a)

Let

119863120573(119905) ≜ 119896

21199042120573

times (1198902119903119879120573

(1 + 4119903 (119879 minus 119905) 1205732)

minus 1198902119903119905120573

(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])120597119881

120597120573= (8119896

2119902119903

2120573

2)minus1

119904minus2120573

(119903 minus 120583)2

sdot 11989621199042120573(119890

2119903119879120573(1 + 4119903 (119879 minus 119905) 120573

2) minus 119890

2119903119905120573

times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))

+ 21198902119903119905120573

119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])

minus 21198902119903119879120573

119903 (1 + 2120573 ln [119904])

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 10: Optimal Control of Investment-Reinsurance Problem for an

10 Abstract and Applied Analysis

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

times [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]

(40)

Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with

respect to 119905 we have

120597119863120573

120597119905= 4119903120573

2minus119890

211990311987912057311989621199042120573+ 2119890

2119903119905120573119903 ln [119904]

+ 1198902119903119905120573

(21199032(minus119905 + 119879) + 119896

21199042120573

times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )

(41)

Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0

which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to

119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as

119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879

which means the value function decreases with respect to 120573

when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573

In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found

Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast

120587lowast) and 120583 is the expected return of risk

asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583

Proof Letting

119863120583(119905) ≜ [119896

21199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

(42)

obviously119863120583(119879) = 0 and

120597119863120583

120597119905= [119896

21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890

2119903119905120573)

+ 211990312057311989621199042120573(1 + 2120573) 119890

2119903119879120573

= minus21199031205731198902119903119905120573

11989621199042120573(1 + 2120573)

+ 41199032120573119890

2119903119905120573+ 2119903120573119896

21199042120573(1 + 2120573) 119890

2119903119879120573

= 211990312057311989621199042120573(1 + 2120573) (119890

2119903119879120573minus 119890

2119903119905120573)

+ 41199032120573119890

2119903119905120573gt 0

(43)

which shows that 119863120583(119905) increases to 119863120583

(119879) = 0 with respectto time 119905 Thus

[11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573le 0

(44)

So

120597119881

120597120583= minus(4119896

2119902119903

2120573)

minus1

119904minus2120573

(120583 minus 119903)

sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890

2119903119879120573minus 119890

2119903119905120573)

minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890

2119903119879120573

sdot exp minus 2119903119879120573 minus1

8119902

times [8119890119903(minus119905+119879)

119909 minus

2 (minus1 + 1198902119903(minus119905+119879)

) 1199021205902

119903

minus (1198962119902119903

2120573)

minus1

119890minus2119903119879120573

119904minus2120573

(119903 minus 120583)2

sdot [2 (1198902119903119905120573

minus 1198902119903119879120573

) 119903 + 11989621199042120573(1 + 2120573)

times (minus1198902119903119905120573

+ 1198902119903119879120573

(1 + 2119903 (119905 minus 119879) 120573))] ]

ge 0

(45)

The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate

5 Conclusion

In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 11: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 11

010 015 020 025 030

minus090

minus085

minus080

minus075

minus070

minus065

minus060

120583

V

The value function V varies with 120583

Figure 5 The value function 119881 varies with 120583

stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper

Appendices

A Proof of Theorem 1

This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix

Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy

120587lowast= minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A1)

Putting (A1) into (19) after simplification we have

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A2)

That is

119867120572

119905+ [(120578 minus 120579) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904+1

2120590

2119867

120572

119909119909

+1

211989621199042120573+2

119867120572

119904119904minus(120583 minus 119903)

2

(119867120572

119909)2

211989621199042120573119867120572

119909119909

minus11989621199042120573+2

(119867120572

119909119904)2

2119867120572

119909119909

minus(120583 minus 119903) 119904119867

120572

119909119867

120572

119909119904

119867120572

119909119909

+max119886

119867120572

119909(1 + 120579) 120582int

119886

0

119865 (119909)d119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 0

(A3)

We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form

119881 (119905 119909 119904) = minus1

119902119890minus119902[119909119890

119903(119879minus119905)+119891(119905)+119892(119905119904)]

119905 isin [0 119879]

119891 (119879) = 0

119892 (119879 119904) = 0

119881 (119879 119909 119904) = minus1

119902119890minus119902119909

(A4)

Differentiating the solution (A4) with respect to the corre-sponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 119891

1015840(119905) + 119892

119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

1198922

119904+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119886) 119904) minus 119867

120572(119905 119909 119904)]

= 119881119902119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

(A5)

Plugging the above derivatives (A5) into (A1) gives

120587lowast=

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A6)

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 12: Optimal Control of Investment-Reinsurance Problem for an

12 Abstract and Applied Analysis

and substituting the above derivatives into (A2) yields

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

+ 119903119904119892119904

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+max119886

(1 + 120579) 120582119890119903(119879minus119905)

int

119886

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

119886

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A7)

Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860

lowast(119905) satisfies

[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)

] 119865 (1198860(119905)) = 0 (A8)

If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus

1198901199021198860(119905)119890119903(119879minus119905)

= 0 which gives

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

(A9)

To ensure that 0 le 1198860

lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)

119902)119890minus119903(119879minus119905)

lt 119863 it must be ensured that

119905 lt 119879 +1

119903ln [

119902119863

ln (1 + 120579)] ≜ 119905

0 (A10)

which leads to two different cases

Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and

1198860

lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905) (A11)

always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy

is given by

120587lowast(119905) =

120583 minus 119903

11990911990211989621199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905)(A12)

and 119886lowast= 119886

0(119905) = (ln(1 + 120579)119902)119890

minus119903(119879minus119905) The correspondingHJB equation (A7) becomes

1198911015840(119905) + 119892

119905+ 119903119904119892

119904+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)+1

211989621199042120573+2

119892119904119904

+(120583 minus 119903)

2

211990211989621199042120573+ 119890

119903(119879minus119905)(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909 minus 120582119890119903(119879minus119905)

times int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A13)

To solve (A13) we decompose (A13) into two equations

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119890119903(119879minus119905)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909

= 0

(A14)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

211990211989621199042120573= 0 (A15)

By simple transposition and integration for (A14) with119891(119879) = 0 we have

119891 (119905) =(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

(A16)

To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0

119892 (119905 119904) = 119860 (119905) 119904minus2120573

+ 119861 (119905) (A17)

Differentiating (A17) with respect to the correspondingvariables respectively

119892119905= 119860

1015840(119905) 119904

minus2120573+ 119861

1015840(119905) 119892

119904= minus2120573119904

minus2120573minus1119860 (119905)

119892119904119904= 2120573 (2120573 + 1) 119904

minus2120573minus2119860 (119905)

(A18)

Substituting the above derivatives into (A20) yields

1198601015840(119905) 119904

minus2120573+ 119861

1015840(119905) minus 2119903120573119904

minus2120573119860 (119905)

+ 1198962120573 (2120573 + 1)119860 (119905) +

(120583 minus 119903)2

21199021198962119904minus2120573

= 0

(A19)

To solve (A20) we decompose (A20) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0 (A20)

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0 (A21)

Obviously 119904minus2120573gt 0 (A20) is equivalent to

1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962= 0 (A22)

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 13: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 13

Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is

119860 (119905) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962 (A23)

Inserting (A23) into (A21) with 119861(119879) = 0 we have

119861 (119905) = (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A24)

Substituting (A23) and (A24) back into (A17) we obtain

119892 (119905 119904) =

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

(A25)

Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by

119881 (119905 119909 119904)

= minus1

119902exp

times

minus 119902

times [

[

119909119890119903(119879minus119905)

+(120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

+ int

119879

119905

(119890119903(119879minus119910)

(1 + 120579) 120582

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119865 (119909)d119909) d119910

minus int

119879

119905

(120582119890119903(119879minus119910)

times int

(ln(1+120579)119902)119890minus119903(119879minus119910)

0

119890119902119909119890119903(119879minus119910)

119865 (119909)d119909) d119910

minus(120578 minus 120579) 120582120583

infin

119903+120590

2119902

4119903

+ (119890minus2119903120573119879

(1 + 2120573)

times (1198902119903120573119905

+ 1198902119903120573119879

(minus1 + 2119903 (119879 minus 119905) 120573))

times (120583 minus 119903)2

) times (81199021205731199032)minus1

+

119890minus2119903120573119879

(minus1198902119903120573119905

+ 1198902119903120573119879

) (120583 minus 119903)2

41199021199031205731198962119904minus2120573]

]

119905 isin [0 119879]

(A26)

Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval

[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860

(119905) =

(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on

[0 1199050) however 1198860

(119905) ge 119863 on [1199050 119879] we can only take 119863 as

the retention level naturally Therefore the optimal retentionlevel is denoted by

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050)

119863 119905 isin [1199050 119879]

(A27)

The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively

(1) For

119886lowast(119905) = 119863 119905 isin [119905

0 119879] (A28)

the corresponding HJB equation is

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

+max120587

(120583 minus 119903) 120587119909119867120572

119909+1

2120587

2119909

211989621199042120573119867

120572

119909119909

+1198962120587119909119904

2120573+1119867

120572

119909119904

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A29)

The first-order maximizing condition for the optimal invest-ment strategy is

120587lowast(119905) = minus

(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904

11990911989621199042120573119867120572

119909119909

(A30)

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 14: Optimal Control of Investment-Reinsurance Problem for an

14 Abstract and Applied Analysis

Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes

119867120572

119905+ [(1 + 120578) 120582120583

infin+ 119903119909]119867

120572

119909+ 120583119904119867

120572

119904

+1

2120590

2119867

120572

119909119909+1

211989621199042120573+2

119867120572

119904119904

minus

[(120583 minus 119903)119867120572

119909+ 119896

21199042120573+1

119867120572

119909119904]2

211989621199042120573119867120572

119909119909

+ 120582119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)] = 0

(A31)

We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [1199050 119879]

119881 (119879 119909 119904) = 119880 (119909)

119891 (119879) = 0

119892 (119879 119904) = 0

(A32)

Differentiating the conjecture (A32) with respect to thecorresponding variables

119881119905= 119881 (minus119902) [minus119903119909119890

119903(119879minus119905)+ 1198911015840

(119905) + 119892119905]

119881119909= 119881 (minus119902) 119890

119903(119879minus119905) 119881

119909119909= 119881(minus119902)

2

1198902119903(119879minus119905)

119881119904= 119881 (minus119902) 119892

119904 119881

119904119904= 119881 [(minus119902)

2

119892119904

2+ (minus119902) 119892

119904119904]

119881119909119904= 119881(minus119902)

2

119890119903(119879minus119905)

119892119904

119864 [119867120572(119905 119909 minus (119885

119894and 119863) 119904) minus 119867

120572(119905 119909 119904)]

= 119881int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) minus 119881

(A33)

Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives

120587lowast(119905) =

(120583 minus 119903)

11990211989621199091199042120573119890119903(119879minus119905)minus

119904119892119904

119909119890119903(119879minus119905) (A34)

1198911015840(119905) + 119892

119905+ (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ 119903119904119892119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A35)

Decompose (A35) into two equations

1198911015840(119905) + (1 + 120578) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

minus120582

119902int

119863

0

119890119902119909119890119903(119879minus119905)

d119865 (119909) + 120582

119902= 0

(A36)

119892119905+ 119903119904119892

119904+1

211989621199042120573+2

119892119904119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A37)

By transposition and integration for (A36) with 119891(119879) = 0

119891 (119905) =(1 + 120578) 120582120583

infin119890119903(119879minus119905)

119903minus120590

2119902119890

2119903(119879minus119905)

4119903

minus int

119879

119905

[120582

119902int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus120582 (119905 minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A38)

For (A37) we try to find a solution as follows

119892 (119905 119904) = 119860 (119905)119904minus2120573

+ 119861 (119905) (A39)

satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into

1198601015840(119905)119904

minus2120573+ 1198611015840

(119905) minus 2119903120573119860 (119905)119904minus2120573

+ 1198962120573 (2120573 + 1)119860 (119905)

+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0

(A40)

Decompose (A40) into two equations

[1198601015840(119905) minus 2119903120573119860 (119905) +

(120583 minus 119903)2

21199021198962] 119904

minus2120573= 0

1198611015840(119905) + 119896

2120573 (2120573 + 1)119860 (119905) = 0

(A41)

Solving the above simple ordinary differential equationswe get

119860 (119905) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573

119861 (119905) =

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

(A42)

So

119892 (119905 119904) =

(1 minus 1198902119903120573(119905minus119879)

) (120583 minus 119903)2

41199021198962119903120573119904minus2120573

+

(1 + 2120573) [1198902119903120573(119905minus119879)

minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2

81199021199032120573

120587lowast(119905) =

2119903 (120583 minus 119903) + (120583 minus 119903)2

(1 minus 1198902119903120573(119905minus119879)

)

211990211989621199031199091199042120573119890119903(119879minus119905)

(A43)

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 15: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 15

(2) For

119886lowast(119905) =

ln (1 + 120579)

119902119890minus119903(119879minus119905)

119905 isin [0 1199050) (A44)

the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891 (119905) + 119892 (119905 119904)]

119905 isin [0 1199050]

119891(1199050) =

119891(1199050)

119892(1199050 119904) =

119892(1199050 119904)

(A45)

After the same calculation as (1) we obtain the followingresults

1198911015840(119905) + 119892

119905+ (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A46)

Splitting (A46) into two equations

119892119905+1

211989621199042120573+2

119892119904119904+ 119903119904119892

119904+(120583 minus 119903)

2

21199021198962119904minus2120573

= 0 (A47)

The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]

1198911015840(119905) + (120578 minus 120579) 120582120583

infin119890119903(119879minus119905)

minus1

2120590

2119902119890

2119903(119879minus119905)

+ (1 + 120579) 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119865 (119909)d119909

minus 120582119890119903(119879minus119905)

int

(ln(1+120579)119902)119890minus119903(119879minus119905)

0

119890119902119909119890119903(119879minus119905)

119865 (119909)d119909 = 0

(A48)

By transposition and integration to (A48) the result isgiven by

119891 (119905) =(120578 minus 120579) 120582120583

infin

119903119890119903(119879minus119905)

minus120590

2119902

41199031198902119903(119879minus119905)

minus int

119905

0

[(1 + 120579) 120582119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

+ 120582int

119905

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+ 1198623

(A49)

where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider

1198623= minus

(120578 minus 120579) 120582120583infin

119903119890119903(119879minus119905

0)+ (1 + 120579) 120582

times int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119865 (119909)d119909] d120591

minus 120582int

1199050

0

[119890119903(119879minus120591)

int

(ln(1+120579)119902)119890minus119903(119879minus120591)

0

119890119902119909119890119903(119879minus120591)

119865 (119909)d119909] d120591

+(1 + 120578) 120582120583

infin

119903119890119903(119879minus119905

0)

minus120582

119902int

119879

1199050

[int

119863

0

119890119902119909119890119903(119879minus120591)

d119865 (119909)] d120591

minus

120582 (1199050minus 119879)

119902minus(1 + 120578) 120582120583

infin

119903+120590

2119902

4119903

(A50)

So

119891lowast(119905) =

119891 (119905) 119905 isin [0 1199050)

119891 (119905) 119905 isin [1199050 119879]

(A51)

With the same calculations as before we obtain

119892 (119905 119904) = [11986211198902120573119903119905

+(120583 minus 119903)

2

41199021198962120573119903] 119904

minus2120573minus 119862

1

1198962(2120573 + 1)

21199031198902120573119903119905

minus(2120573 + 1)

119902119903(120583 minus 119903)

2

+ 1198622

(A52)

here the undetermined constants 1198621and 119862

2will be deter-

mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905

0 we have

119892 (1199050 119904) =119892(1199050 119904) (A53)

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 16: Optimal Control of Investment-Reinsurance Problem for an

16 Abstract and Applied Analysis

Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862

1and 119862

2

1198621= minus

(120583 minus 119903)2

119890minus2119903119879120573

41199021199031198962120573

1198622= (2120573 + 1) (120583 minus 119903)

2

(4 + 119879 minus 119905

0

4119902119903minus

1

81199021199032120573)

(A54)

So

119892lowast(119905 119904) =

119892 (119905 119904) 119905 isin [0 1199050)

119892 (119905 119904) 119905 isin [1199050 119879]

(A55)

And the optimal value function is

119881 (119905 119909 119904) = minus1

119902exp minus119902 [119909119890119903(119879minus119905)

+ 119891lowast(119905) + 119892

lowast(119905 119904)]

(A56)

B Proof of Theorem 3

A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)

Lemma B1 Take a sequence of bounded open sets 1198761 119876

2

1198763 with 119876

119894sub 119876

119894+1sub 119876 119894 = 1 2 and 119876 = cup

119894119876

119894 Let 120591

119894

denote the exit time of (119883120572lowast

(119905) 119878(119905)) from 119876119894 If the conditions

inTheorem 3 hold then one has119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and

119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2

Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum

119873(119905)

119894=1119885

119894in terms of a

Poisson random measure Suppose 120574 is the Poisson randommeasure then

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

119911120574 (d119911 d119906)

119862119886(119905) =

119873(119905)

sum

119894=1

119885119894= int

119905

0

int119877+

(119911 and 119886) 120574 (d119911 d119906)

(B1)

The compensator ] of the random measure 120574 is

] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)

so the compensated Poisson random measure of 119862(119905) is

120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)

Set (119905) = 119881(119905 119883120572lowast

(119905) 119878(119905)) Applying Itorsquos formula to

2(119905)

d2(119905)

= 2 (119905) 119881119909120590d119882(119905) + 119896119878

120573(119905)

times [119881119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905) +A (119905)

+ [1198812

119909120590

2+ 119881

2

119909120587

2119883(119905)

211989621198782120573(119905) + 119881

2

11990411989621198782120573+2

(119905)

+ 2119881119909119881

119904120587 (119905)119883 (119905) 119896

21198782120573+1

(119905)] d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))

sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905

+ int119877+

[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B4)

Since 120572lowast is the optimal strategy for (19) thenA120572lowast

(119905) =

0 Plugging the expression of119881119909119881

119904 and 119886 lowast (119905) into (B4) we

obtain the following equation by simple calculations

d2(119905)

2(119905)

=

minus 2119902120590119890119903(119879minus119905)d119882(119905) minus

2 (120583 minus 119903) 119904minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903] d119882

1(119905)

+

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119905)

+

1199022120590

21198902119903(119879minus119905)

+(120583 minus 119903)

2

119904minus2120573

1198962[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

2

+

(1 minus 119890minus2119903120573(119879minus119905)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962

minus

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times[1 +

(1 minus 119890minus2119903120573(119879minus119905)

) (120583 minus 119903)

2119903]

d119905

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 17: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 17

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119905)minus 1] 120582d119865 (119911) d119905

+ int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119905)

minus 1] 120574 (d119911 d119905)

(B5)

The solution of (B5) is

2(119905)

2(0)

= exp

int

119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906)

minus1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

+ int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1] 120582d119865 (119911) d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906)

(B6)

where (1199090 119904

0) is the state at time 0 Applying Itorsquos formula to

(11) we can derive

d119878minus2120573(119905) = (120573 (2120573 + 1) 119896

2minus 2120573119903119878

minus2120573(119905)) d119905

minus 2120573119896radic119878minus2120573(119905)d119882

1(119905)

(B7)

According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that

expint119905

0

minus2119902120590119890119903(119879minus119906)d119882(119906) minus

1

2int

119905

0

41199022120590

21198902119903(119879minus119906)d119906

exp

int

119905

0

minus2 (120583 minus 119903) 119904

minus120573

119896

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119882

1(119906)

minus1

2int

119905

0

4(120583 minus 119903)2

119904minus2120573

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

exp

int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)2

119904minus120573

119903119896d119882

1(119906)

minus1

2int

119905

0

(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

11990321198962d119906

(B8)

are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt

+infin and

119864[int

119879

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1]

2

120582d119865 (119911) d119906] lt infin (B9)

then the process

int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

minus 1] 120574 (d119911 d119906) (B10)

is also a martingaleAccording to Taksar and Zeng [16]

119864[expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906] lt infin (B11)

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 18: Optimal Control of Investment-Reinsurance Problem for an

18 Abstract and Applied Analysis

when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields

119864 [2(119905)]

= 2(0) 119864

times [

[

exp

int

119905

0

31199022120590

21198902119903(119879minus119906)d119906

+ int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

+ int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

+ int

119905

0

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903] d119906

+ int

119905

0

int119877+

[119890minus2119902(119911and119886

lowast)119890119903(119879minus119906)

+ 2119902 (119911 and 119886lowast) 119890

119903(119879minus119906)minus 1]

times 120582d119865 (119911) d119906

]

]

(B12)

If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)

) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have

exp

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

2

d119906

lt expint119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962[1 +

(120583 minus 119903)

2119903]

2

d119906

= exp[1 +(120583 minus 119903)

2119903]

2

int

119905

0

3(120583 minus 119903)2

119904minus2120573

(119906)

1198962d119906

exp

int

119905

0

3(1 minus 119890minus2119903120573(119879minus119906)

)2

(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

lt expint119905

0

3(120583 minus 119903)4

119904minus2120573

(119906)

411990321198962d119906

exp

int

119905

0

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)3

119904minus2120573

(119906)

1199031198962

times [1 +

(1 minus 119890minus2119903120573(119879minus119906)

) (120583 minus 119903)

2119903]

10038161003816100381610038161003816100381610038161003816100381610038161003816

d119906

lt expint119905

0

(120583 minus 119903)3

119904minus2120573

(119906)

1199031198962[1 +

(120583 minus 119903)

2119903] d119906

(B13)

For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)

| lt 119890minus2119903120573119879 which is

similar After simple calculation and analysis we obtain

119864 [1198812(119905 119883

120572lowast

(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)

So119864[1198812(120591

119894and119879119883

120572lowast

(120591119894and119879) 119878(120591

119894and119879))] lt infin for 119894 = 1 2

We use Lemma B1 to prove the verificationTheorem 3 asfollows

Proof By Itorsquos formula for Ito-Levy process we have

d119881 (119905 119883120572(119905) 119878 (119905))

= A120572119881 (119905 119883

120572(119905) 119878 (119905)) + 119881

119909120590d119882(119905)

+ 119896119878120573(119905) [119881

119909120587 (119905)119883 (119905) + 119881

119904119878 (119905)] d119882

1(119905)

+ int119877+

[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))

minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)

(B15)

Take a sequence of bounded open sets1198761119876

2119876

3 with

119876119894sub 119876

119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876

1 let 120591

119894denote the

exit time of (119909 119904) from 119876119894 Then 120591

119894and 119879 rarr 119879 when 119894 rarr infin

Integrating both sides of the above equation

119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

= 119881 (119905 119909 119904) + int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

+ int

120591119894and119879

119905

119881119909120590d119882(119906)

+ int

120591119894and119879

119905

119896119878120573(119906) [119881

119909120587119909 + 119881

119904119878 (119906)] d119882

1(119906)

+ int

120591119894and119879

119905

int119877+

[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))

minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)

(B16)

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975

Page 19: Optimal Control of Investment-Reinsurance Problem for an

Abstract and Applied Analysis 19

Since sup120572isinΛ

A120572119881(119905 119909 119904) = 0 then A120572

119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))

| 119883120572(119905) = 119909 119878 (119905) = 119904]

= 119881 (119905 119909 119904)

+ 119864 [int

120591119894and119879

119905

A120572119881 (119906119883

120572(119906) 119878 (119906)) d119906

| 119883120572(119905) = 119909 119878 (119905) = 119904]

le 119881 (119905 119909 119904)

(B17)

From Lemma B1 119881(120591119894and 119879119883

120572(120591

119894and 119879) 119878(120591

119894and 119879)) 119894 =

1 2 are uniformly integrable Thus we have

119867(119905 119909 119904)

= sup120572isinΛ

119864 [119880 (119883 (119879)) | 119883120572lowast

(119905) = 119909 119878 (119905) = 119904]

= lim119905rarrinfin

119864 [119881 (120591119894and 119879119883

120572(120591

119894and 119879) 119878 (120591

119894and 119879))]

le 119881 (119905 119909 119904)

(B18)

When 120572 = 120572lowast the inequality in the above formula

becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished

Conflict of Interests

The authors declare that they have no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)

References

[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960

[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982

[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997

[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010

[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012

[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013

[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005

[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009

[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013

[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976

[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008

[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009

[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006

[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010

[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011

[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009

[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009

[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993

[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013

[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975