first hitting time law for some jump-diffusion processes ...docs.isfa.fr/labo/2009.5(wp2105).pdf ·...

First hitting time law for some jump-diffusionprocesses : existence of a density

- Laure COUTIN (Université de Toulouse, IMT)- Diana Dorobantu (Université Lyon 1, Laboratoire SAF)

2009.5 (WP 2105)

Laboratoire SAF – 50 Avenue Tony Garnier - 69366 Lyon cedex 07 http://www.isfa.fr/la_recherche

First hitting time law for some jump-diffusion

processes : existence of a density∗

Laure Coutin

IMT, University of Toulouse

118 route de Narbonne, 31062, Toulouse

[email protected]

Diana Dorobantu

Université de Lyon, Université Lyon 1, Laboratoire de Science Actuarielle et Financière,

ISFA, 50 avenue Tony Garnier, F-69007, Lyon

[email protected]

Abstract. Let (Xt, t ≥ 0) be a diffusion process with jumps, sum of a Brownian motionwith drift and a compound Poisson process. We consider τx the first hitting time of a fixed levelx > 0 by (Xt, t ≥ 0). We prove that the law of τx has a density (defective when E(X1) < 0)with respect to the Lebesgue measure.

Keywords : Lévy process, jump-diffusion process, hitting time law.

∗The authors thank M. Pontier and Ph. Carmona for their careful reading and G. Letac for his helpfull

comments

1

1 Introduction

The main purpose of this paper is to show that the first hitting time distribution associated witha jump-diffusion process (sum of a Brownian motion with drift and an independent compoundPoisson process) has a density with respect to the Lebesgue measure.

Let (Xt, t ≥ 0) be a càd-làg process started at 0 and τx the first hitting time of level x > 0by X.

Lévy, in [16], computes the law of τx when X is a Brownian motion with drift. This resultis extended by Alili, Patie and Petersen [1] or Leblanc [13] to the case where X is an Ornstein-Uhlenbeck process. The case where X is a Bessel process was studied by Borodin and Salminenin [4].

Some results are also available when the process X has jumps. The first results are obtainedby Zolotarev [23] and Borokov [5] when X is a spectrally negative Lévy process. They give thelaw of τx . Moreover, if Xt has the probability density with respect to the Lebesgue measurep(t, x), then the law of τx has the density with repect to the Lebesgue measure f(t, x), wherexf(t, x) = tp(t, x). In this case where X has only negative jumps, Xτx = x almost surely.

If X is a spectrally positive Lévy process, Doney [7] gives an explicit formula for the jointLaplace transform of τx and the overshoot Xτx − x. When X is a stable Lévy process, Peskir[17], and Bernyk, Dalang and Peskir [2] obtain an explicit formula for the hitting time density.

The case where X has signed jumps is more recently studied. In [9], the authors give the lawof τx when X is the sum of a decreasing Lévy process and an independent compound processwith exponential jump sizes. This result is extended by Kou et Wang in [11] to the case ofa diffusion process with jumps where the jump sizes follow a double exponential law. Theycompute the Laplace transform of τx and derive an expression for the density of τx. For a moregeneral jump-diffusion process, Roynette, Vallois and Volpi [20] show that the Laplace transformof (τx, x − Xτx−, Xτx − x) is solution of some kind of random integral.

Doney and Kyprianou [8] studied the problem for general Lévy processes. They give thequintuple law of (Gτx−, τx − Gτx−, Xτx − x, x − Xτx−, x − Xτx−) where Xt = sups≤t Xs andGt = sup{s < t, Xs = Xs}.

Results are also available for some Lévy processes without Gaussian component, see Lefèvre,Loisel and Picard [14, 15, 18, 19]. Blanchet [3] considers a process satisfying the followingstochastic equation : dSt = St−(µdt + σ1φ(t)=0dWt + φ1φ(t)=φdNt), t ≤ T where T is a finite

horizon, µ ∈ R, σ > 0, φ(.) is a function taking two values 0 or φ, W is a Brownian motion, Nis a Poisson process with intensity 1

φ2 1φ(t)=φ and N is the compensated Poisson process.

The aim of our paper is to add to these studies the law of a first hitting time by a Lévyprocess which is the sum of a Brownian motion with drift and a compound Poisson process. Wedo not limit our study to a particular distribution of the jumps sizes.

This paper is organized as follows : Section 2 contains the main result (Theorem 2.1) whichgives the first hitting time law by a jump Lévy process. The following two sections (Section2.1 and Section 2.2) are dedicated to the proof of Theorem 2.1. In these sections we computethe derivative at t = 0 (Section 2.1) and at t > 0 (Section 2.2) of the hitting time distributionfunction. Section 3 contains the proofs of some useful results.

2

2 Hitting time law

Let m ∈ R, (Wt, t ≥ 0) be a standard Brownian motion, (Nt, t ≥ 0) be a Poisson processwith constant positive intensity a and (Yi, i ∈ N

∗) be a sequence of independent identicallydistributed random variables with distribution function FY . We suppose that the followingσ−fields σ(Yi, i ∈ N

∗), σ(Nt, t ≥ 0) and σ(Wt, t ≥ 0) are independent. Let (Tn, n ∈ N∗) be

the sequence of the jump times of the process N .

Let X be the process defined by

Xt = mt + Wt +

Nt∑

i=1

Yi, t ≥ 0. (1)

Let τx be the first hitting time of level x > 0 by the stochastic process (Xt, t ≥ 0) :τx = inf{u > 0 : Xu ≥ x}.

The main result of this paper is the following theorem. It gives the law of τx.

Theorem 2.1 If there exists β > 0 such that E(

eβ|Y1|)

< ∞, then the distribution function of τx

has a right derivative at 0 and is differentiable at every point of ]0, ∞[. The derivative, denotedf(., x), is equal to

f(0, x) =a

2

(

2 − FY (x) − FY (x−))

+a

4

(

FY (x) − FY (x−))

and for every t > 0 f(t, x) = aE (1τx>t(1 − FY ) (x − Xt)) + E

(

1τx>TNtf

(

t − TNt , x − XTNt

))

where

f(u, z) =| z |√2πu3

exp

[

−(z − mu)2

2u

]

1]0,∞[(u), u ∈ R, z ∈ R.

Furthermore, P(τx = ∞) = 0 if and only if m + aE(Y1) ≥ 0.

The proof of Theorem 2.1 is given in Sections 2.1 and 2.2.

Let X be the m drifted Brownian motion defined by

Xt = mt + Wt, for all t ≥ 0,

and τz be the first hitting time of z > 0 by X. Then, according to [10], f(., z) is the derivativeof the distribution function of τz (see Section 3 for more details).

Let (Ft)t≥0 be the completed natural filtration generated by the processes (Wt, t ≥ 0),(Nt, t ≥ 0) and the random variables (Yi, i ∈ N

∗).

Remark 2.2 This result is already known in the following cases :When X has no positive jumps, the law of τx is given in Theorem 46.4 page 348 of [21].When X is a stable Lévy process, with no negative jumps, the law of τx is given in [2].When X is a jump-diffusion where the jump sizes follow a double exponential law, the result isgiven in [11].

3

2.1 Existence of the right derivative at t = 0

In this section we prove the first part of Theorem 2.1. Our goal is to show that the distributionfunction of τx (i.e. t 7→ P(τx ≤ t)) has a right derivative at 0 and we compute this derivative.

For this purpose we split the probability P(τx ≤ h) according to the values of Nh :

P(τx ≤ h) = P(τx ≤ h, Nh = 0) + P(τx ≤ h, Nh = 1) + P(τx ≤ h, Nh ≥ 2).

Note that P(τx ≤ h, Nh ≥ 2) ≤ P(Nh ≥ 2) = 1 − e−ah − ahe−ah, thus

limh→0P(τx ≤ h, Nh ≥ 2)

h= 0.

The proof of the first part of Theorem 2.1 will be complete when the following lemma andproposition are shown :

Lemma 2.3 The term P(τx≤h, Nh=0)h converges to 0 when h goes to 0.

Proposition 2.4 If there exists β > 0 such that E(

eβ|Y1|)

< ∞, then for every x > 0 the termP(τx≤h, Nh=1)

h converges to a2 (2 − FY (x) − FY (x−)) + a

4 (FY (x) − FY (x−)) when h goes to 0.

Proof of Lemma 2.3.The essential observation is that on the set {ω : Nh(ω) = 0}, the processes (Xt, 0 ≤ t ≤ h) and(Xt, 0 ≤ t ≤ h) are equal , and P-almost surely τx ∧ h = τx ∧ h. Since τx is independent of N ,then

P(τx ≤ h, Nh = 0)

h=

e−ahP(τx ≤ h)

h.

The law of τx has a C∞ density (possibly defective) with respect to the Lebesgue measure, null

on ] −∞, 0]. Thus the limit of P(τx≤h, Nh=0)h exists and is equal to 0 when h goes to 0. 2

To prove Proposition 2.4, we use the same type of arguments as in [20] (for the proof ofTheorem 2.4). In [20], the authors compute the joint Laplace transform of (a passage time of aLévy process, overshoot) as solution of an integral equation.Proof of Proposition 2.4.We split the probability P(τx ≤ h, Nh = 1) into three parts according to the relative position ofτx and T1, the first jump time of the Poisson process N :

P(τx ≤ h, Nh = 1) = P(τx < T1, Nh = 1) + P(τx = T1, Nh = 1) + P(T1 < τx ≤ h, Nh = 1).(2)

Step 1 : We prove that the contribution to the limit limh→0+ h−1P(τx ≤ h, Nh = 1) of the firstterm on the right hand side of (2) is null.

Since the processes (Xt, 0 ≤ t < T1) and (Xt, 0 ≤ t < T1) are equal, then on the set{ω, τx(ω) < T1(ω)}, the stopping time τx and τx are equal. Thus

P(τx < T1, Nh = 1) = P(τx < T1 ≤ h, Nh = 1) ≤ P(τx ≤ h).

4

The law of τx has a C∞ density (possibly defective) with respect to the Lebesgue measure,

null on ] −∞, 0]. Thus P(τx<T1, Nh=1)h converges to 0 when h goes to 0.

Step 2 : We prove that the contribution to the limit limh→0+ h−1P(τx ≤ h, Nh = 1) of thesecond term on the right hand side of (2) is a

2 (2 − FY (x) − FY (x−)).

Note that

P(τx = T1, Nh = 1) = P(τx > T1, XT1 + Y1 ≥ x, T1 ≤ h < T2).

Here, for every n ∈ N∗, Tn = S1 + ... + Sn where (Si, i ≥ 1) is a sequence of independent

identically distributed random variables with exponential distribution with parameter a. Usingthe independence between (Si, i ≥ 1) and (Y1, X, τx) we get :

P(τx = T1, Nh = 1) =

∫ h

0ae−as1

∫ ∞

h−s1

ae−as2E

(

1τx>s11Y1≥x−Xs1

)

ds2ds1

=ae−ah

∫ h

0E

(

1τx>s1Y1≥x−Xs

)

ds.

Integrating with respect to Y1, we obtain :

P(τx = T1, Nh = 1) = ae−ah

∫ h

0E

(

(1 − FY )((x − Xs)−)

)

ds−ae−ah

∫ h

0E

(

1τx≤s(1 − FY )((x − Xs)−)

)

ds.

On the one hand, since FY is a càdlàg bounded function and X is a Brownian motion with drift,we get

lims→0E

(

FY ((x − Xs)−)

)

=FY (x) + FY (x−)

2.

On the other hand, since the distribution function of τx is differentiable, then

lims→0E

(

1τx≤s(1 − FY )((x − Xs)−)

)

= 0.

We deduce that

limh→0P(τx = T1, Nh = 1)

h=

a

2

(

2 − FY (x) − FY (x−))

.

Step 3 : We prove that the contribution to the limit limh→0+ h−1P(τx ≤ h, Nh = 1) of thethird term on the right hand side of (2) is a

4 (FY (x) − FY (x−)).

To this end, we state the following lemma :

Lemma 2.5 The term P(T1<τx≤h, Nh=1)h converges to a

4 (FY (x) − FY (x−)) when h goes to 0.

Proof Note that

P(T1 < τx ≤ h, Nh = 1) = P(T1 < τx ≤ h, T1 ≤ h < T2)

and T2 = T1 + S2 ◦ θT1 where θ is the translation operator. Here S2 is a random variable withexponential distribution with parameter a, independent of (T1, (Wt, t ≥ 0), (Yi, i ∈ N

∗)).

5

Moreover, on {ω : T1(ω) < τx(ω) ≤ h < T2(ω)}, Xs = XT1 + Xs−T1 ◦ θT1 when T1 < s ≤ hand τx = T1 + τx−XT1

◦ θT1 .

Strong Markov Property at the (Ft, t ≥ 0)-stopping time T1 gives :

P(T1 < τx ≤ h, Nh = 1) = E

(

1τx>T11h≥T1ET1

(

1τx−XT1≤h−T11h−T1<S2

))

where ET1(.) is E(. | FT1).

Integrating with respect to S2, we obtain :

P(T1 < τx ≤ h, Nh = 1) = E

(

1τx>T11h≥T1e−a(h−T1)

ET1

(

1τx−XT1≤h−T1

))

.

Remark that {ω : τx(ω) > T1(ω)} = {ω : τx(ω) > T1(ω)} ∩ {ω : XT1(ω) < x}. Consequently

P(T1 < τx ≤ h, Nh = 1) = − E

(

1τx≤T1≤h1XT1<xe−a(h−T1)

ET1

(

1τx−XT1≤h−T1

))

+ E

(

1h≥T11XT1<xe−a(h−T1)

ET1

(

1τx−XT1≤h−T1

))

.

Since the distribution function of τx has a null derivative at 0, then

limh→01

hE

(

1τx≤T1≤h1XT1<xe−a(h−T1)

ET1

(

1τx−XT1≤h−T1

))

= 0.

It remains to show that

limh→0+

G(h)

h=

a

4[F (x) − F (x−)] (3)

whereG(h) = E

(

1h≥T11XT1<xe−a(h−T1)

ET1

(

1τx−XT1≤h−T1

))

.

Integrating with respect to T1 and then using the fact that f(., z) is the derivative of the distri-bution function of τz, we get :

G(h) =

∫ h

0ae−as

E

[

1Xs+Y1<xe−a(h−s)E

s(1τx−Xs−Y1

≤h−s)]

ds

=ae−ah

∫ h

0

∫ h−s

0E

[

1Xs+Y1<xf(u, x − Xs − Y1)]

duds.

Since Xs = ms + Ws, we may apply Lemma 3.1 to µ = x − ms − Y1 and σ =√

s, then

G(h) =ae−ah

√2π

∫ h

0

∫ h−s

0E

[

e− (x−m(u+s)−Y1)2

2(u+s)

(

x − Y1

(u + s)3/2+

G√

s√u(u + s)

)

+

]

duds.

We make the following change of variable r = u + s.

G(h) =ae−ah

√2π

∫ h

0

∫ h

sE

[

e−(x−mr−Y1)2

2r

(

x − Y1

r3/2+

G√

s

r√

r − s

)

+

]

drds.

6

Firstly, we apply Fubini’s Theorem and secondly, we make the following change of variable v = sr .

G(h) =ae−ah

√2π

∫ h

0

∫ r

0E

[

e−(x−mr−Y1)2

2r

(

x − Y1

r3/2+

G√

s

r√

r − s

)

+

]

dsdr

=ae−ah

√2π

∫ h

0g(r)dr,

where

g(r) =

∫ 1

0E

[

e−(x−mr−Y1)2

2r

(

x − Y1√r

+G√

v√1 − v

)

+

]

dv.

But,

limr→0+

e−(x−mr−Y1)2

2r

(

x − Y1√r

+G√

v√1 − v

)

+

=

√v√

1 − vG+1x=Y1 ,

and

sup0≤r≤1

e−(x−mr−Y1)2

2r

(

x − Y1√r

+G√

v√1 − v

)

+

≤ supz≥0

ze−z2

2 + |m| +√

v√1 − v

|G|.

Then from Lebesgue’s Dominated Convergence Theorem we obtain :

limr→0g(r) = P(Y1 = x)E(G+)

∫ 1

0

√v√

1 − vdv =

√2π

4P (Y1 = x) .

We deduce the identity (3), i.e.

limh→0G(h)

h=

a

4P (Y1 = x) ,

which achieves the proof of Lemma 2.5 2

Propostion 2.4 is a consequence of the Steps 1 to 3. 2

2.2 Existence of the derivative at t > 0

Our task is now to show that the distribution function of τx is differentiable on R∗+ and to

compute its derivative. For this purpose we split the probability P(t < τx ≤ t + h) according tothe values of Nt+h − Nt :

P(t < τx ≤ t + h) =P(t < τx ≤ t + h, Nt+h − Nt = 0) + P(t < τx ≤ t + h, Nt+h − Nt = 1)

+P(t < τx ≤ t + h, Nt+h − Nt ≥ 2). (4)

The third term on the right hand side of (4) is upper bounded by

P(t < τx ≤ t + h, Nt+h − Nt ≥ 2) ≤ P(Nt+h − Nt ≥ 2) = 1 − e−ah − ahe−ah.

7

Therefore limh→0P(t<τx≤t+h, Nt+h−Nt≥2)

h = 0.

Let us study the second term on the right hand side of (4). Markov Property at t gives :

P(t < τx ≤ t + h, Nt+h − Nt = 1) = E(1τx>tPt(τx−Xt ≤ h, Nh = 1)),

where Pt(.) = P(.|Ft).

In virtue of Lemma 2.4,P

t(τx−Xt

≤h, Nh=1)

h converges to

a

2

[

2 − FY (x − Xt) − FY

(

(x − Xt)−)]

+a

4

[

FY (x − Xt) − FY

(

(x − Xt)−)]

and is upper bounded by P(Nh=1)h = ae−ah ≤ a. Dominated Convergence Theorem gives :

limh→0P(t < τx ≤ t + h, Nt+h − Nt = 1)

h= aE (1τx>t(1 − FY )(x − Xt))+

3a

4E (1τx>t∆FY (x − Xt))

where ∆FY (z) = FY (z) − FY (z−).

However the jumps set of FY (the distribution function of Y ) is countable and X has a density(cf. Proposition 3.12 page 90 of [6]). Thus E (1τx>t∆FY (x − Xt)) = 0 : indeed0 ≤ E (1τx>t∆FY (x − Xt)) ≤ E(1Y1=x−Xt) = E(1Xt=x−Y1) = 0. Therefore


h= aE (1τx>t(1 − FY ) (x − Xt)) .

The proof of the second part of Theorem 2.1 will be complete when the following propositionis shown :

Proposition 2.6 If there exists β > 0 such that E(

eβ|Y1|)

< ∞, then


h= E

(

1τx>TNtf

(

t − TNt , x − XTNt

))

,

where f is the function introduced in Theorem 2.1.

Proof Since TNt is not a stopping time, we can not apply Strong Markov Property. We split

P(t < τ ≤ t + h, Nt+h − Nt = 0) =∞

∑

k=0

P(t < τx ≤ t + h, Nt+h = Nt = k)

=P(t < τx ≤ t + h < T1) +∞

∑

k=1

P (t < τx ≤ t + h, Tk < t < t + h < Tk+1) .

On the set {(ω, t) : Tk(ω) < t}, we have Xt(ω) = XTk(ω) + Xt−Tk

◦ θTk(ω), hence on the set

{ω, τx(ω) > Tk(ω)}, τx = Tk + τx−XTk◦ θTk

. Strong Markov Property at the stopping time Tk

gives

P(t < τ ≤ t + h, Nt+h − Nt = 0) =e−a(t+h)P(t < τx ≤ t + h)

+∞

∑

k=1

E

(

1Tk<t1τx>TkE

Tk

(

1t−Tk<τx−XTk≤t+h−Tk

1t+h−Tk<Sk+1

))

.

8

On the set {(ω, t) : τz(ω) ≤ t < Sk+1(ω)}, we have τz = τz for every z < 0. Therefore

P(t < τx ≤ t + h, Nt+h − Nt = 0) =e−a(t+h)P(t < τx ≤ t + h)

+∞

∑

k=1

E

(

1Tk<t1τx>Tke−a(t+h−Tk)

ETk

(

1t−Tk<τx−XTk≤t+h−Tk

))

.

The FTk-conditional law of τx−XTk

has the density (possibly defective) f(., x − XTk), thus

P(t < τx ≤ t + h, Nt+h − Nt = 0) =e−a(t+h)P(t < τx ≤ t + h)

+∞

∑

k=1

E

(

1Tk<t1τx>Tke−a(t+h−Tk)

∫ t+h−Tk

t−Tk

f(u, x − XTk)du

)

.

Let us point out that e−a(t−Tk) = ETk

(

1Tk+1>t

)

then

P(t < τx ≤ t + h, Nt+h − Nt = 0) =e−ah

∫ t+h

tE (10≤t<T1) f(u, x)du

+ e−ah∞

∑

k=1

∫ t+h

tE

(

1Tk≤t<Tk+11τx>Tk

f(u − Tk, x − XTk))

du,

or shortly

P(t < τx ≤ t + h, Nt+h − Nt = 0) = e−ah

∫ t+h

tE

(

1TNt<τx f(u − TNt , x − XTNt

))

du. (5)

Since f is continuous with respect to u, then

limu→t+

1TNt<τx f(u − TNt , x − XTNt

) = 1TNt<τx f(t − TNt , x − XTNt

),

From Lemma 3.2, Propositions 3.5 and 3.6, f(u − TN−t, x − XTNt) is dominated uniformly in u

by a integrable random variable

f(u − TN−t, x − XTNt) ≤ cε,M (t − TNt)

1−ε

[

1

(x − XTNt)4ε

+ exp

(

2m

M(x − XTNt

)

)

]

.

Here 0 < ε < 1/4 and M > max(1, 2|m|β−1), (where E(eβ|Y |) < +∞, ) and cε,M , is a con-stant defined in Lemma 3.2 depending only on ε and M . Then, using Lebesgue’s DominatedConvergence Theorem in equation (5) we obtain


h= E

(

1τx>TNtf(t − TNt , x − XTNt

))

.

2

Using Proposition 2.6, the limit limh→0P(t<τx≤t+h)

h exists and is equal to

aE (1τx>t(1 − FY )(x − Xt)) + E

(

1τx>TNtf(t − TNt , x − XTNt

))

.

The following lemma allows to conclude the proof of Theorem 2.1. The results are known,for example in [15] and [22] but for sake of completeness we give a proof.

9

Lemma 2.7 For all x > 0, the stopping time τx is finite almost surely if and only if m+aE(Y1) ≥0.

Proof Remark thatP(τx = ∞) = P(supt≥0Xt < x).

Thanks to Theorem 7.2 page 183 of [12] which is a consequence of Strong Law of Large Numbers,

if m + aE(Y1) > 0, then limt→∞Xt = +∞ and

if m + aE(Y1) = 0, then limsupt→∞Xt = −liminft→∞Xt = ∞.

Therefore (see Exercise 39.11 page 271 of [21]), if m + aE(Y1) ≥ 0, then supt≥0Xt = +∞. Thisproves the first part of the lemma.

Conversely, let us suppose that m + aE(Y1) < 0. Then limt→∞ Xt = −∞, and accordingto Theorem 48.1 page 363 of [21], supt≥0 Xt < ∞. Assume that there exists x0 > 0, such thatP(τx0 < +∞) = 1. Then from all x, such that x ≤ x0 we have P(τx < +∞) = 1.

Now we use recurrence reasoning. Assume that for n ≥ 1 and for all x ≤ nx0, we haveP(τx < +∞) = 1. Let x such that nx0 < x ≤ (n + 1)x0, then using Strong Markov Property,

P(τx < +∞) = E(1τnx0<+∞1Xτnx0≥x) + E(1τnx0<+∞1Xτnx0

<xPτnx0 (τx−Xτnx0

< ∞)).

Using the recurrence hypothesis, since x − Xτnx0< x0, almost surely P

τnx0 (τx−Xτnx0< ∞) = 1

and then

P(τx < +∞) = 1.

We have proved that if there exists x0 > 0 such that P(τx0 < ∞) = 1, then for all x > 0,P(τx < ∞) = P(supt≥0 Xt ≥ x) = 1. This contradicts the fact that supt≥0 Xt < ∞. Then, ifm + aE(Y1) < 0, for all x > 0, P(τx < ∞) < 1. 2

3 Appendix

A Brownian motion with drift is a process

Xt = mt + Wt, t ≥ 0

with m ∈ R.

Let z > 0 and τz be the passage time defined by τz = inf{t ≥ 0 : Xt ≥ z}. By (5.12) page197 of [10], τz has the following law on R+ :

f(u, z)du + P(τz = ∞)δ∞(du)

where

f(u, z) =| z |√2πu3

exp

[

−(z − mu)2

2u

]

1]0,∞[(u), u ∈ R, and P(τz = ∞) = 1 − emz−|mz|. (6)

For a fixed z, the function f(., z) and all its derivatives admit 0 as limit at 0+. The functionf admits an extension (denoted f) of class C∞ on R, defined by f(u, z) = 0 for u ≤ 0. Moreoverit checks the two following lemmas :

10

Lemma 3.1 Let G be a Gaussian random variable N (0, 1) and let µ ∈ R, σ ∈ R+. Then for

every u ∈ R

E[f(u, µ + σG)1µ+σG>0] =1√2π

E

[

e− (µ−mu)2

2(σ2+u)

(

µ + σ2m

(σ2 + u)3/2+

σG√u(σ2 + u)

)

+

]

where (x)+ = max(x, 0).

Proof Using the probability density function of G and the definition of f (see (6)), we get :

E[f(u, µ + σG)1µ+σG>0] =1√

2πu3

∫

R

1√2π

|µ + σg|e−(µ+σg−mu)2

2u e−g2

2 1µ+σg>0dg.

Since |µ + σg|1µ+σg>0 = (µ + σg)1µ+σg>0 = (µ + σg)+, then

E[f(u, µ + σG)1µ+σG>0] =1√

2πu3

∫

R

1√2π

(µ + σg)+e−(µ+σg−mu)2

2u e−g2

2 dg.

A simple computation shows that (µ+σg−mu)2

u + g2 = σ2+uu

(

g + σ(µ−mu)σ2+u

)2+ (µ−mu)2

σ2+u. Therefore

E[f(u, µ + σG)1µ+σG>0] =e− (µ−mu)2

2(σ2+u)

√2πu3

∫

R

1√2π

(µ + σg)+e−σ2+u

2u

(

g+σ(µ−mu)

σ2+u

)2

dg.

Making the following change of variable x =√

σ2+uu

(

g + σ(µ−mu)σ2+u

)

, we conclude the proof

E[f(u, µ + σG)1µ+σG>0] =e− (µ−mu)2

2(σ2+u)

√2π

∫

R

1√2π

(

µ2 + σ2m

(σ2 + u)3/2+

σx√u(σ2 + u)

)

+

e−x2

2 dx.

2

Lemma 3.2 Let f : R × R → R+ be the function defined by

f(u, z) =

{

|z|√2πu3

exp[− (z−mu)2

2u ] si u > 0

0 si u ≤ 0,

where m ∈ R. Then for every ε > 0 and M ≥ 1, there exists a constant cε,M > 0 such that

f(u, z) ≤ cε,Mu−1+ε

[

1

| z |4ε+ exp

(

2mz

M

)]

.

Proof Let ε > 0 and M ≥ 1 be fixed. Remark that it is enough to prove that there exists aconstant cε,M > 0 such that

f(u, z) ≤ cε,Mu−1+ε 1

| z |2εexp

(mz

M

)

.

11

Then we conclude the proof using the inequality x1x2 ≤ x21+x2

22 for x1 = 1

|z|2ε and x2 = exp(

mzM

)

.

We now seek to upper bound the quotient f(u,z)

u−1+ε 1|z|2ε exp[mz

M], z < 0, u ∈ R+.

Since exp[− (z−mu)2

2u ] ≤ exp[− (z−mu)2

2uM ], then

f(u, z)


M ]≤

|z|√2πu3

exp[− (z−mu)2

2uM ]


M ]

=1√2π

(

z2

u

)

12+ε

exp

[

− z2

2uM

]

exp

[

−m2u

2M

]

≤M12+ε2ε

√π

(

z2

2uM

)

12+ε

exp

[

− z2

2uM

]

.

Since the function x 7→ x12+εe−x is continuous, null at 0 and at +∞, then it is bounded on R+.

Hence there exists cε > 0 such that x12+εe−x ≤ cε for any x ∈ R+. We apply this inequality to

x = z2

2uM , so ( z2

2uM )12+ε exp[− z2

2uM ] ≤ cε and the proof is complete. 2

Lemma 3.3 Let G be a Gaussian random variable N (0, 1), µ ∈ R, σ > 0 and 0 < α < 1. Thenthere exists two constants k1,α > 0, k2,α > 0, depending only on α, such that

E(|µ + σG|−α) ≤ k1,α

σ|µ|1−α + k2,ασ−α.

Proof Using the probability density function of G, we get :

E(|µ + σG|−α) =

∫ ∞

−µ

σ

1√2π

(µ + σx)−αe−x2

2 µx +

∫ −µ

σ

−∞

1√2π

(−µ − σx)−αe−x2

2 dx.

Integration by parts gives :

E(|µ + σG|−α) =

[

1√2π

e−x2

2(µ + σx)1−α

σ(1 − α)

]∞

−µ

σ

+1

σ(1 − α)

∫ ∞

−µ

σ

1√2π

(µ + σx)1−αxe−x2

2 dx

+

[

1√2π

e−x2

2(−µ − σx)1−α

−σ(1 − α)

]−µ

σ

−∞− 1

σ(1 − α)

∫ −µ

σ

−∞

1√2π

(−µ − σx)1−αxe−x2

2 dx

=1

σ(1 − α)

∫

R

1√2π

|µ + σx|1−αe−x2

2

(

x1x>−µ

σ− x1x≤−µ

σ

)

dx

=1

σ(1 − α)E

[

|µ + σG|1−αG(1G>−µ

σ− 1G≤−µ

σ)]

.

However for every µ ∈ R and σ > 0, we have

|G(1G>−µ

σ− 1G≤−µ

σ)| ≤ G − 2G1G≤0 = |G|.

Consequently E(|µ + σG|−α) ≤ 1σ(1−α)E(|G||µ + σG|1−α).

The inequality |µ + σG|1−α ≤ (|µ| + σ|G|)1−α ≤ |µ|1−α + σ1−α|G|1−α concludes the proof. 2

12

Lemma 3.4 For every t > 0 and (α, γ) ∈] − 1,∞[2, the following two series

∞∑

i≥1

E (1t>Ti(t − Ti)

αT γi ) zi

∞∑

i≥1

E (1t>Ti(t − Ti)

αT γi ) izi,

have an infinite radius of convergence.

Proof It is enough to prove that the first series has an infinite radius of convergence.

Note that for i ≥ 1, Ti admits as density the function u 7→ ai

(i−1)!ui−1e−au, thus

E (1t>Ti(t − Ti)

αT γi ) =

ai

(i − 1)!

∫ t

0e−au(t − u)αuγ+i−1du

≤ ai

(i − 1)!

∫ t

0(t − u)αuγ+i−1du

=ai

(i − 1)!tγ+i+α Γ(γ + i)Γ(α + 1)

Γ(γ + i + α + 1)

and the conclusion holds. 2

As a consequence of Lemma 3.4, we have the following two propositions which are used toprove Theorem 2.1.

Proposition 3.5 Assume that there exists β > 0 such that E(

eβ|Y1|)

< ∞. For every t > 0,ε > 0 and M ≥ 1 such that |2m

M | ≤ β, the random variable (t − TNt)−1+ε exp 2m

M (x − XTNt)

P-integrable.

Proof Note that

(t − TNt)−1+ε exp

2m

M(x − XTNt

) = t−1+ε exp2mx

M1t<T1 +

∞∑

i=1

1Ti≤t<Ti+1(t − Ti)−1+ε exp

2m

M(x − XTi

),

≤ t−1+ε exp2mx

M+

∞∑

i=1

1t>Ti(t − Ti)

−1+ε exp

[

2m

M(x − XTi

)

]

.

Let σ(Ti) be the σ-field generated by Ti, i > 0.Conditioning by Ti, we obtain that

E

[

exp

[

2m

M(x − XTi

)

]

|σ(Ti)

]

= e2mM

xe−2m2

MTi+

2m2

M2 TiE

(

e−2mM

Y1

)i.

Since M ≥ 1, then e−2m2

MTi+

2m2

M2 Ti ≤ 1, therefore

E

[

exp

[

2m

M(x − XTi

)

]

|σ(Ti)

]

≤ e2mM

xE

(

e−2mM

Y1

)i.

13

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