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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

    Question #1 Answer: E

    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

    Question #2 Answer: E

    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

    Question #3 Answer: D

    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

    Question #5 Answer: B

    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

    Question #6 Answer: B

    140 4040:20

    20

    40 4040:2020

    140 40 40 4040:20 40:20

    20 20

    40:

    Benefits 1000 1000

    Premiums 1000

    Benefit premiums Equivalence principle

    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

    Premiums = 1000

    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

    Question #7 Answer: C

    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

    This is quadratic, so

    0.7 0.49 0.3 0.25 4

    2 0.3x

    0.3147x

    0.1 0.3147 so 11.56te t

  • MLC0911 10

    Question #14 Answer: A

    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