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• SOCIETY OF ACTUARIES

EXAM MLC Models for Life Contingencies

EXAM MLC SAMPLE SOLUTIONS

The following questions or solutions have been modified since this document was prepared to use with the syllabus effective spring 2012,

Prior to March 1, 2012: Questions: 151, 181, 289, 300 Solutions: 2, 284, 289, 290, 295, 300 Changed on March 19, 2012: Questions: 20, 158, 199 (all are minor edits) Changed on April 24, 2012: Solution: 292 Changed on August 20, 2012: Questions and Solutions 38, 54, 89, 180, 217 and 218 were restored from MLC-09-08 and reworded to conform to the current presentation. Question 288 was reworded to resolve an ambiguity. A typo in Question 122 was corrected. Questions and Solutions 301-309 are new questions Changed on August 23, 2012: Solution 47, initial formula corrected; Solution 72, minus signs added in the first integral Changed on December 11, 2012: Question 300 deleted Copyright 2011 by the Society of Actuaries MLC-09-11 PRINTED IN U.S.A.

• MLC0911 1

2 32 30:34 30:34 30:34

2 30

2 34

2 30:34

0.9 0.8 0.72

0.5 0.4 0.20

0.72 0.20 0.144

q p p

p

p

p

2 30:34

3 30

3 34

3 30:34

3 30:34

0.72 0.20 0.144 0.776

0.72 0.7 0.504

0.20 0.3 0.06

0.504 0.06 0.030240.504 0.06 0.03024

0.53376

p

p

p

pp

2 30:34 0.776 0.53376

0.24224

q

Alternatively, 2 30 34 2 30 2 34 2 30 34

2 30 32 2 34 36 2 30 34 32 361

q q q q

p q p q p p: :

: :

b g = (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) = 0.24224 Alternatively,

3 30 3 34 2 30 2 342 30:34

3 30 3 34 2 30 2 341 1 1 1

1 0.504 1 0.06 1 0.72 1 0.200.24224

q q q q q

p p p p

(see first solution for 2 30 2 34 3 30 3 34, , , p p p p )

• MLC0911 2

0.10 0.12

110:10

10 0.04 0.06 0.4 0.6 0.05 0.070 0

10 0.1 1 0.120 0

10 10.10 0.120

1000 1000

1000 (0.06) (0.07)

1000 0.06 (0.07)

1000 0.06 (0.07)t t

x xx

t t t t

t t

e e

A A A

e e dt e e e e dt

e dt e e dt

e

0

1 10.06 0.070.10 0.121000 1

1000 0.37927 0.21460 593.87

e e

Because this is a timed exam, many candidates will know common results for constant force and constant interest without integration.

110:10

1010

For example 1 xx

x

x

A E

E e

A

With those relationships, the solution becomes

110 10:10

0.06 0.04 10 0.06 0.04 10

1 1

1000 1000

0.06 0.071000 10.06 0.04 0.07 0.05

1000 0.60 1 0.5833

593.86

x x xxA A E A

e e

e e

• MLC0911 3

0.06 0.08 0.050 0

0.070

120

1 100 520 7 7

t t t tt t x x t

t

E Z b v p dt e e e dt

e

22 0.12 0.16 0.05 0.090 0 00.09

0

1 120 20

1 100 520 9 9

t t t t tt t x x t

t

E Z b v p dt e e e dt e dt

e

25 5 0.04535

9 7Var Z

Question #4 Answer: C Let ns = nonsmoker and s = smoker

k = qx knsb g px k

nsb g qx k

sb g sx kp

0 .05 0.95 0.10 0.90 1 .10 0.90 0.20 0.80 2 .15 0.85 0.30 0.70

1 2

1:2ns ns ns ns

x x xxA v q v p q

21 10.05 0.95 0.10

1.02 1.02 0.1403

1 2

1:2s s s s

x x xxA v q v p q

2

1 10.10 0.90 0.201.02 1.02

0.2710

1

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

• MLC0911 4

1 2 3

0.0001045

0.0001045x x x xt

t xp e

10

20

30

0.0601045 0.0601045 0.06010450 0 0

APV Benefits 1,000,000

500,000

200,000

1,000,000 500,000 250,0002,000,000 250,000 10,00027.5 16.6377 45

tt x x

tt x x

t x x

t t t

e p dt

e p dt

e p dt

e dt e dt e dt

7.54

140 4040:20

20

40 4040:2020

140 40 40 4040:20 40:20

20 20

40:

Benefits 1000 1000

1000 1000 1000

1000

k kk

k kk

k k k kk

EPV A E vq

EPV a E vq

A E vq a E vq

A

120 40:20/

161.32 0.27414 369.1314.8166 0.27414 11.14545.11

a

While this solution above recognized that 1

40:201000P and was structured to take

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

• MLC0911 5

40 Benefits 1000 161.32EPV A

20 40 60 6040:200

20 40 6040:20

1000

14.8166 0.27414 11.1454 0.27414 369.13

11.7612 101.19

k kk

EPV a E E vq

a E A

11.7612 101.19 161.32

161.32 101.19 5.1111.7612

70 70

7070

69 69 70

269 69

ln 1.060.53 0.5147

0.061 1 0.5147 8.5736

0.06 /1.060.971 1 8.5736 8.84571.06

2 2 1.00021 8.8457 0.257398.5902

A AiAa

d

a vp a

a a

Note that the approximation 12

mx x

ma a

m

works well (is closest to the exact

answer, only off by less than 0.01). Since m = 2, this estimate becomes 18.8457 8.59574

Question #8 - Removed Question #9 - Removed

• MLC0911 6

Question #10 Answer: E d = 0.05 v = 0.95 At issue

491 1 50 50

40 400

40 40

4040

40

0.02 ... 0.02 1 / 0.35076

and 1 / 1 0.35076 / 0.05 12.98481000 350.76so 27.013

12.9848

kk

kA v q v v v v d

a A dAP

a

RevisedRevised10 40 50 40 5010 1000 549.18 27.013 9.0164 305.62E L K A P a where

24Revised Revised1 1 25 2550 50

0RevisedRevised

50 50

0.04 ... 0.04 1 / 0.54918

and 1 / 1 0.54918 / 0.05 9.0164

kk

kA v q v v v v d

a A d

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

Var Var VarX E X Y E X Y Let Y = 1 if smoker; Y = 0 if non-smoker

111 0.444 5.56

0.1

SS xxT

AE a Y a

Similarly 1 0.2860 7.140.1TE a Y

0 Prob Y=0 1 Prob Y=1

7.14 0.70 5.56 0.306.67

T T TE E a Y E E a E E a

• MLC0911 7

2 2 27.14 0.70 5.56 0.3044.96

TE E a Y

2Var 44.96 6.67 0.47TE a Y Var 8.503 0.70 8.818 0.30

8.60TE a Y

Var 8.60 0.47 9.07Ta Alternatively, here is a solution based on

22Var( )Y E Y E Y , a formula for the variance of any random variable. This can be

transformed into 22 VarE Y Y E Y which we will use in its conditional form

22 NS Var NS NST T TE a a E a

22Var

S Prob S NS Prob NS

T T T

T T T

a E a E a

E a E a E a

S NS

S NS

0.30 0.70

0.30 1 0.70 1

0.1 0.10.30 1 0.444 0.70 1 0.286

0.30 5.56 0.70 7.140.1

1.67 5.00 6.67

x x

x x

a a

A A

2 2 2

2

2

2 2

S Prob S NS Prob NS

0.30 Var S S

0.70 Var NS NS

0.30 8.818 5.56 0.70 8.503 7.14

T T T

T T

T T

E a E a E a

a E a

a E a

11.919 + 41.638 = 53.557

2Var 53.557 6.67 9.1Ta

• MLC0911 8

Alternatively, here is a solution based on 1T

Tva

22

22

2

1Var Var

Var since Var constant Var

Var since Var constant constant Var

which is Bowers formula 5.2.9

T

T

T

T

x x

va

v X X

vX X

A A

This could be transformed into 2 2 2Varx xTA a A , which we will use to get 2 NS 2 Sand x xA A .

2 2

2 2

22 NS

22 S

2

2

NS Prob NS S Prob S

Var NS Prob NS

Var S Prob S

0.01 8.503 0.286 0.70

0.01 8.818 0.444 0.30

0.16683 0.70 0.28532 0.300.20238

Tx

T T

xT

xT

A E v

E v E v

a A

a A

NS Prob NS S Prob S

0.286 0.70 0.444 0.300.3334

Tx

T T

A E v

E v E v

• MLC0911 9

22

2

2

Var

0.20238 0.3334 9.120.01

x xT

A Aa

Question #12 - Removed Question #13 Answer: D Let NS denote non-smokers, S denote smokers.

0.1 0.2

0.1 0.2

Prob Prob NS Prob NS P rob S P rob S

1 0.7 1 0.3

1 0.7 0.3

t t

t t

T t T t T t

e e

e e

0.2 0.1

0 ( ) 0.3 0.7t tS t e e

Want t such that 0 0 0.75 1 or 0.25S t S t

2 2 0.1 0.1 0.10.25 0.3 0.7 0.3 0.7t t t te e e e Substitute: let 0.1tx e

20.3 0.7 0.25 0 x x

0.7 0.49 0.3 0.25 4

2 0.3x

0.3147x

0.1 0.3147 so 11.56te t

• MLC0911 10

At a constant force of mortality, the benefit premium equals the force of mortality and so 0.03 . 2 0.030.20

2 2 0.030.06

xA

2 22 13

0 2 20.060.09

0.20Var 0.20

0.03 1 1 1where 0.09 3 0.09

x xA ALa

A a

Question #15 - Removed Question #16 Answer: A

4040

40

6020 40

40

161.321000 10.8914.8166

11.14541000 1000 1 1000 1