# actuary paper 1

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Actuarial Science Paper 1TRANSCRIPT

SOCIETY OF ACTUARIES

EXAM MLC ACTUARIAL MODELS

EXAM MLC SAMPLE SOLUTIONS

Copyright 2008 by the Society of Actuaries

Some of the questions in this study note are taken from past SOA examinations.

MLC-09-08

PRINTED IN U.S.A.

Question #1 Key: E2

q30:34 = 2 p30:34 3 p30:342 2 2

p30 = ( 0.9 )( 0.8 ) = 0.72 p34 = ( 0.5 )( 0.4 ) = 0.20 p30:34 = ( 0.72 )( 0.20 ) = 0.1442 3 3 3 3

p30 = ( 0.72 )( 0.7 ) = 0.504 p34 = ( 0.20 )( 0.3) = 0.06

p30:34 = 0.72 + 0.20 0.144 = 0.776

p30:34 = ( 0.504 )( 0.06 ) = 0.03024 p30:34 = 0.504 + 0.06 0.03024 = 0.53376

2

q30:34 = 0.776 0.53376 = 0.24224

Alternatively,2 q30:34 = 2 q30 + 2 q34 2 q30:34

= 2 p30q32 + 2 p34q36 2 p30:34 1 p32:36 = (0.9)(0.8)(0.3) + (0.5)(0.4)(0.7) (0.9)(0.8)(0.5)(0.4) [1-(0.7)(0.3)] = 0.216 + 0.140 0.144(0.79) = 0.24224Alternatively,2

b

g

q30:34 = 3 q30 3 q34 2 q30 2 q34 = (1 3 p30 )(1 3 p34 ) (1 2 p30 )(1 2 p34 ) = (1 0.504 )(1 0.06 ) (1 0.72 )(1 0.20 ) = 0.24224

(see first solution for

2

p30 ,

2

p34 ,

3

p30 ,

3

p34 )

MLC-09-08

-1-

Question #2

Key: E1 1000 Ax = 1000 Ax:10 + 10 Ax 10 0.04t 0.06t e (0.06)dt + e0.4e0.6 e0.05t e0.07t (0.07)dt = 1000 e 0 0 0.12t 10 0.1t dt = 1000 0.06 e dt + e 1 (0.07) e 0 0 0.10 t 10 + e 1 (0.07) e0.12t = 1000 0.06 e 0.10 0 0.12 0 = 1000 0.06 1 e 1 + 0.07 e 1 1 e1.2 0.12 0.10

= 1000 ( 0.37927 + 0.21460 ) = 593.87

Because this is a timed exam, many candidates will know common results for constant force and constant interest without integration.1 For example Ax:10 = 10 E x

+

(1 10 Ex )

=e

10 ( + )

Ax =

+

With those relationships, the solution becomes 1 1000 Ax = 1000 Ax:10 + 10 Ex Ax +10 0.06 0.07 ( 0.06+ 0.04 )10 0.06+ 0.04 )10 = 1000 +e ( 1 e 0.07 + 0.05 0.06 + 0.04 = 1000 ( 0.60 ) 1 e 1 + 0.5833 e 1 = 593.86

(

)

(

)

Question #3 Key: D

E [ Z ] = bt v t t p x ( x + t ) dt = e0.06 t e 0.08 t e0.05 t0 0

1 dt 20 1 100 5 e 0.07 t = = 0 7 20 7

MLC-09-08

-2-

E Z 2 =

0

(b v )t t

2t

px ( x + t ) dt = e0.12t e0.16t e0.05t0

1 1 dt = e0.09t dt 20 20 0 1 100 0.09t 5 = e 0 = 9 20 9

5 5 Var [ Z ] = = 0.04535 9 7

2

Question #4 Key: C Let ns = nonsmoker and s = smoker k= 0 1 21 ( ns A x:2 ) =

b q x + kgns

b px + kgns

b q x +gks

( px +)ks

.05 .10 .15( ns v qx ) +

0.95 0.90 0.85( ns px ) ( qx +1)ns

0.10 0.20 0.30

0.90 0.80 0.70

v2

1 ( 0.05) 1.02( A x:2 )1 s

1 1.022(s qx )

0.95 0.10 = 0.1403(s px ) ( q x +)1s

v

+ v2

1 1.02

( 0.10 ) +

1

(1.02 )2

0.90 0.20 = 0.2710

A1 :2 = weighted average = (0.75)(0.1403) + (0.25)(0.2710) x= 0.1730

Question #5 Key: B( ( ( ( x ) = x1) + x2 ) + x3) = 0.0001045t

( px ) = e 0.0001045t

MLC-09-08

-3-

1 APV Benefits = e t 1,000,000 t px ( ) x( ) dt 0

+ e0 0

t

2 500,000 t px ( ) x( ) dt

3 + e 200,000 t px ( ) x( ) dt

=

1,000,000 0.0601045t 500,000 0.0601045t 250,000 0.0601045t dt + dt + e dt 0 e 0 e 2,000,000 250,000 10,000 0

= 27.5 (16.6377 ) = 457.54

Question #6

Key: B1 APV Benefits = 1000 A40:20 +

k = 20

k E401000vq40+k

&& APV Premiums = a40:20 +

k = 20

k E401000vq40+k

Benefit premiums Equivalence principle 1 1000 A40:20 + k = 20

&& k E401000vq40+k = a40:20 + k E401000vq40+k20

1 = 1000 A40:20

=

161.32 ( 0.27414 )( 369.13) 14.8166 ( 0.27414 )(11.1454 )

&& / a40:20

= 5.111 While this solution above recognized that = 1000 P40:20 and was structured to take

advantage of that, it wasnt necessary, nor would it save much time. Instead, you could do:

APV Benefits = 1000 A40 = 161.32&& APV Premiums = a40:20 + 20 E40 k E60 1000vq60+ kk =0

&& = a40:20 + 20 E40 1000 A60

= 14.8166 ( 0.27414 )(11.1454 ) + ( 0.27414 )( 369.13) = 11.7612 + 101.19 11.7612 + 101.19 = 161.32 161.32 101.19 = = 5.11 11.7612

MLC-09-08

-4-

Question #7

Key: CA70 = A70 = i ln (1.06 ) ( 0.53) = 0.5147 0.06 1 A70 1 0.5147 && = = 8.5736 a70 = 0.06 /1.06 d 0.97 && && a69 = 1 + vp69a70 = 1 + ( 8.5736 ) = 8.8457 1.06 = 8.5902

&&( 2 && a69) = ( 2 ) a69 ( 2 ) = (1.00021)( 8.8457 ) 0.25739

( m 1) works well (is closest to the exact answer, &&( m && Note that the approximation ax ) ax 2m 1 only off by less than 0.01). Since m = 2, this estimate becomes 8.8457 = 8.5957 4Question #8 Key: C The following steps would do in this multiple-choice context: 1. From the answer choices, this is a recursion for an insurance or pure endowment. 2. Only C and E would satisfy u(70) = 1.0. 1+ i u ( k 1) 3. It is not E. The recursion for a pure endowment is simpler: u ( k ) = pk 1 4. Thus, it must be C. More rigorously, transform the recursion to its backward equivalent, u ( k 1) in terms of u ( k ) : q 1+ i u ( k ) = k 1 + u ( k 1) pk 1 pk 1 pk 1u ( k ) = qk 1 + (1 + i ) u ( k 1) u ( k 1) = vqk 1 + vpk 1 u ( k )

This is the form of (a), (b) and (c) on page 119 of Bowers with x = k 1 . Thus, the recursion could be:

MLC-09-08

-5-

Ax = vqx + vpx Ax +1or orA1: y x = vq x + vp x Ax1 1: y x 1 x +

Ax: y x = vqx + vpx Ax +1: y x 1

Condition (iii) forces it to be answer choice Cu ( k 1) = Ax fails at x = 69 since it is not true that

A69 = vq69 + vp69 (1)u ( k 1) = A1: y x fails at x = 69 since it is not true that x1 A69:1 = vq69 + vp69 (1)

(

)

( (

) )

u ( k 1) = Ax: y x is OK at x = 69 since A69:1 = vq69 + vp69 (1)Note: While writing recursion in backward form gave us something exactly like page 119 of Bowers, in its original forward form it is comparable to problem 8.7 on page 251. Reasoning from that formula, with h = 0 and bh +1 = 1 , should also lead to the correct answer.

MLC-09-08

-6-

Question #9 Key: A You arrive first if both (A) the first train to arrive is a local and (B) no express arrives in the 12 minutes after the local arrives. P ( A) = 0.75 Expresses arrive at Poisson rate of ( 0.25 )( 20 ) = 5 per hour, hence 1 per 12 minutes.

e 110 = 0.368 0! A and B are independent, so P ( A and B ) = ( 0.75 )( 0.368 ) = 0.276 f ( 0) =

Question #10

Key: Ed = 0.05 v = 0.095 At issueA40 = v k +1 k q40 = 0.02 v1 + ... + v50 = 0.02v 1 v50 / d = 0.35076 && and a40 = (1 A40 ) / d = (1 0.35076 ) / 0.05 = 12.9848 so P40 = 1000 A40 350.76 = = 27.013 && a40 12.9848k =0 49

(

)

(

)

E

(

10 L

K ( 40 ) 10 = 1000 A50

)

Revised

&& P40a50

Revised

= 549.18 ( 27.013)( 9.0164 ) = 305.62

whereA50Revised k =0 Revised && a50 =

= v k +1 k q50

24

Revised

= 0.04 v1 + ... + v 25 = 0.04v 1 v 25 / d = 0.54918

(

)

(

)

and

(1 A

Revised 50

) / d = (1 0.54918) / 0.05 = 9.0164

MLC-09-08

-7-

Question #11 Key: E Let NS denote non-smokers and S denote smokers. The shortest solution is based on the conditional variance formula

Var ( X ) = E Var ( X Y ) + Var E ( X Y )

(

)

(

)

Let Y = 1 if smoker; Y = 0 if non-smoker S 1 Ax S E aT Y = 1 = ax =

(

)

1 0.444 = 5.56 0.1 1 0.286 Similarly E aT Y = 0 = = 7.14 0.1 =

(

)

E E aT Y

( (

) ) = E ( E ( aT 0 ) ) Prob ( Y=0 ) + E ( E ( aT 1) ) Prob ( Y=1)= ( 7.14 )( 0.70 ) + ( 5.56 )( 0.30 ) = 6.67

E E aT Y

( (

))

2

Var E aT Y

) ) = 44.96 6.672 = 0.47 E ( Var ( aT Y ) ) = ( 8.503)( 0.70 ) + ( 8.818 )( 0.30 )

( (

2 2 = 7.14 ( 0.70 ) + 5.56 ( 0.30 ) = 44.96

(

)

(

)

Var aT = 8.60 + 0.47 = 9.07Alternatively, here is a solution based onVar(Y ) = E Y 2 E (Y ) , a formula for the variance of any random variable. This can be2

( )

= 8.60

( )

transformed into E Y 2 = Var (Y ) + E (Y ) which we will use in its conditional form 2

E aT

(( )

2

NS = Var aT NS + E aT NS

)

( )

(

)

(

)

2

Var aT = E aT

( )

2

E aT

(

)

2

E aT = E aT S Prob [S] + E aT NS Prob [ NS]

MLC-09-08

-8-

= 0.30axS + 0.70axNS =S 0.30 1 Ax

(

0.1 0.1 0.30 (1 0.444 ) + 0.70 (1 0.286 ) = = ( 0.30 )( 5.56 ) + ( 0.70 )( 7.14 ) 0.1 = 1.67 + 5.00 = 6.67E aT

) + 0.70 (1 A )NS x

( )

2

= E aT 2 S Prob [S] + E aT 2 NS Prob [ NS] 2 = 0.30 Var aT S + E aT S

( ((

) ()

)

+0.70 Var aT NS + E aT NS

(

(

)

2

)

2 2 = 0.30 8.818 + ( 5.56 ) + 0.70 8.503 + ( 7.14 ) 11.919 + 41.638 = 53.557

Var aT = 53.557 ( 6.67 ) = 9.1 2

Alternatively, here is a solution based on aT =1 v Var aT = Var

1 vT

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