Exam MLC Sample Solutions - Society of Actuaries

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Changed on August 23, 2012: Solution 47, initial formula corrected; Solution 72, ..... 0. Prob Y=0. 1. Prob Y=1. 7.14 0.70. 5.56 0.30. 6.67. T. T. T. E E a Y. E E a ...... 235. 1.01925 0.015 1000 255 255. P. +. -. -. = 255 11.175. 235. 1.01925. P. +.
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.

Question #1 Answer: E 2

q30:34  2 p30:34  3 p30:34 2

p30   0.9  0.8   0.72

2

p34   0.5  0.4   0.20

2

p30:34   0.72  0.20   0.144 2

p30:34  0.72  0.20  0.144  0.776

3

p30   0.72  0.7   0.504

3

p34   0.20  0.3  0.06

3

p30:34   0.504  0.06   0.03024

3

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

b

g

 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) = 0.24224

Alternatively, 2

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

MLC‐09‐11

2

p30 ,

2

p34 ,

3

p30 ,

3

p34 )

1

Question #2 Answer: E 1000 Ax  1000  Ax1:10  10 Ax     10  1000   e 0.04t e 0.06t (0.06)dt  e 0.4 e 0.6  e0.05t e 0.07t (0.07)dt  0  0  10   1000 0.06  e0.1t dt  e 1 (0.07)  e 0.12t dt  0 0   0.10 t 10 0.12 t     1000 0.06   e0.10   e1 (0.07)   e0.12   0   0  

1  1  e 1   0.07  1000  0.06  0.12 e   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.

For example Ax1:10  10 E x

Ax 

  

e

1  10 Ex 

10    

  

With those relationships, the solution becomes 1000 Ax  1000  Ax1:10  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

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Question #3 Answer: D 



1 dt 20 1  100  5 0.07 t        e  0 20  7  7 2  1 1  bt vt t px  x t dt   e0.12t e 0.16t e 0.05t dt   e0.09t dt 0 20 20 0 1  100  0.09t  5      e 0 9 20  9  

E  Z    bt v t t p x  x t dt   e0.06 t e 0.08 t e 0.05 t 0

0

E  Z 2   



0

 

2

5 5 Var  Z       0.04535 9 7

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

qxb  kg ns

pxb  kg

qxb gk s

px k

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 ns A x:2  

ns v qx  

1  0.05 1.02

A x:2  1 s

v 1 1.02

ns px 

v2

1 1.022 s qx 

ns

ns

0.95  0.10  0.1403 s px 

 v2

 0.10  

qx 1

1

1.02 

2

qx 1 s

0.90  0.20  0.2710

A1x:2  weighted average = (0.75)(0.1403) + (0.25)(0.2710) = 0.1730

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3

s

Question #5 Answer: B

 x    x1   x 2    x 3  0.0001045 t

 px   e0.0001045t

  1 APV Benefits   e  t 1,000,000 t px    x  dt 0

  t

 e 0

 2 500,000 t px    x  dt

 3    e  200,000 t px    x  dt 0



1,000,000  0.0601045t 500,000  0.0601045t 250,000  0.0601045t e dt  e dt  e dt   2,000,000 0 250,000 0 10,000 0

 27.5 16.6377   457.54 Question #6 Answer: B 1 EPV Benefits  1000 A40:20 

EPV Premiums   a40:20 



 k E401000vq40k

k  20



 k E401000vq40k

k  20

Benefit premiums  Equivalence principle  1 1000 A40:20 





 k E401000vq40k   a40:20   k E401000vq40k

k  20

20



1  1000 A40: 20



/ a40:20

161.32   0.27414  369.13

14.8166   0.27414 11.1454 

 5.11 1 and was structured to take While this solution above recognized that   1000P40:20

advantage of that, it wasn’t necessary, nor would it save much time. Instead, you could do:

MLC‐09‐11

4

EPV Benefits  1000 A40  161.32 EPV Premiums = a40:20    a40:20 



20 E40  k E60 1000vq60  k k 0

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

Question #7 Answer: C

ln 1.06   0.53  0.5147 0.06 1  A70 1  0.5147 a70    8.5736 d 0.06 /1.06  0.97  a69  1  vp69 a70  1     8.5736   8.8457  1.06 

A70   A70  i

 2 a69    2  a69    2   1.00021 8.8457   0.25739

 8.5902

 m  1 works well (is closest to the exact m Note that the approximation ax   ax  2m answer, only off by less than 0.01). Since m = 2, this estimate becomes 1 8.8457   8.5957 4 Question #8 - Removed

Question #9 - Removed

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5

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



49







A40   v k 1 k q40  0.02 v1  ...  v50  0.02v 1  v50 / d  0.35076 k 0

and a40  1  A40  / d  1  0.35076  / 0.05  12.9848 so P40 

1000 A40 350.76   27.013 a40 12.9848

E  10 L K 40  10   1000 A50

Revised

 P40 a50

 549.18   27.013 9.0164   305.62





Revised

where Revised

A50

and

24

  v k 1 k q50

Revised

k 0 Revised a50 

1  A

Revised 50





 0.04 v1  ...  v 25  0.04v 1  v 25 / d  0.54918

 / d  1  0.54918 / 0.05  9.0164

Question #11 Answer: 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 1  AxS E aT Y  1  axS 







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

MLC‐09‐11

6

 

E  E aT Y 





2







2 2   7.14  0.70   5.56  0.30   44.96

 

   44.96  6.672  0.47 E  Var  aT Y     8.503 0.70    8.818  0.30 

Var E aT Y

 8.60

 

Var aT  8.60  0.47  9.07

Alternatively, here is a solution based on

 

Var(Y )  E Y 2   E Y   , a formula for the variance of any random variable. This can be 2

 

transformed into E Y 2  Var Y    E  Y   which we will use in its conditional form

 

E aT

2



2









2

NS  Var aT NS   E aT NS   

 

Var  aT   E  aT 

2



 E  aT  



2

E  aT   E  aT S  Prob S  E  aT NS  Prob  NS  0.30axS  0.70axNS 



0.30 1  AxS

  0.70 1  A  NS x

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  6.67

 

E  aT 

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

MLC‐09‐11

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Alternatively, here is a solution based on aT 

1  vT



1 v  Var aT  Var         vT   Var   since Var  X  constant   Var  X     T

 



  since Var  constant  X   constant

Var vT

 2



2

Ax   Ax 

2

2

 Var  X 

2

which is Bowers formula 5.2.9

 

This could be transformed into 2Ax   2 Var aT  Ax2 , which we will use to get 2

Ax NS and 2AxS . Ax  E  v 2T 

2

 E  v 2T NS  Prob  NS  E  v 2T S  Prob  S 2   2Var aT NS  Ax NS   Prob  NS   2   2Var aT S  AxS   Prob  S     0.01 8.503  0.2862   0.70   0.01 8.818   0.4442   0.30   0.16683 0.70    0.28532  0.30 



 



  



 0.20238 Ax  E  vT   E  vT NS  Prob  NS   E  vT S  Prob  S   0.286  0.70    0.444  0.30   0.3334

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Ax   Ax 

2

 

Var aT 

2

2

0.20238  0.33342   9.12 0.01 Question #12 - Removed

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

Prob T  t   Prob T  t NS  Prob  NS  P rob T  t S  P rob  S









 1  e0.1t  0.7  1  e0.2t  0.3  1  0.7e0.1t  0.3 e0.2t

S0 (t )  0.3e0.2 t  0.7e0.1t Want tˆ such that 0.75  1  S0  tˆ  or 0.25  S0  tˆ 



ˆ

ˆ



ˆ 2

0.25  0.3e2t  0.7e0.1t  0.3 e0.1t

 0.7e0.1t

ˆ

Substitute: let x  e0.1t 0.3x 2  0.7 x  0.25  0

ˆ

This is quadratic, so x 

0.7  0.49   0.3 0.25  4 2  0.3

x  0.3147 e0.1t  0.3147 ˆ

MLC‐09‐11

so tˆ  11.56

9

Question #14 Answer: A

At a constant force of mortality, the benefit premium equals the force of mortality and so   0.03 .  0.03 2 Ax  0.20   2   2  0.03    0.06 Var  0 L   where A 

2

Ax   Ax 

 a    



2

2



0.20   13 

2

 

0.03 1  0.09 3

0.06 2 0.09

a

 0.20 1 1     0.09

Question #15 - Removed

Question #16 Answer: A

1000 P40 

A40 161.32   10.89 a40 14.8166

 a   11.1454  1000 20V40  1000  1  60   1000  1    247.78  14.8166   a40   20V  5000 P40  (1  i)  5000q60 21V  P60 

 247.78  (5)(10.89)   1.06  5000  0.01376   255 1  0.01376

[Note: For this insurance, 20V  1000 20V40 because retrospectively, this is identical to whole life] Though it would have taken much longer, you can do this as a prospective reserve. The prospective solution is included for educational purposes, not to suggest it would be suitable under exam time constraints.

1000 P40  10.89 as above 1 1000 A40  4000 20 E40 A60:5  1000 P40  5000 P40  20 E40 a60:5  

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20 E40  5 E60

a65

1 where A60:5  A60  5 E60 A65  0.06674

a40:20  a40  20 E40 a60  11.7612 a60:5  a60  5 E60 a65  4.3407

1000  0.16132    4000  0.27414  0.06674    10.89 11.7612    510.89  0.27414  4.3407     0.27414  0.68756  9.8969  161.32  73.18  128.08  64.79 1.86544  22.32



Having struggled to solve for  , you could calculate above) calculate 21V recursively. 20V

20 V

prospectively then (as

1  4000 A60:5  1000 A60  5000 P40 a60:5   5 E60 a65

  4000  0.06674   369.13   5000  0.01089  4.3407    22.32  0.68756  9.8969   247.86 (minor rounding difference from 1000 20V40 ) Or we can continue to 21V

21V

prospectively

1  5000 A61:4  1000 4 E61 A65  5000 P40 a61:4   4 E61 a65

where

4 E61



l65 4  7,533,964  v    0.79209   0.73898 l61  8,075, 403 

1 A61:4  A61  4 E61 A65  0.38279  0.73898  0.43980

a61:4

 0.05779  a61  4 E61 a65  10.9041  0.73898  9.8969

 3.5905 21V

  5000  0.05779   1000  0.73898  0.43980    5 10.89  3.5905   22.32  0.73898  9.8969   255

Finally. A moral victory. Under exam conditions since prospective benefit reserves must equal retrospective benefit reserves, calculate whichever is simpler.

MLC‐09‐11

11

Question #17 Answer: C

Var  Z   2 A41   A41 

2

A41  A40  0.00822  A41  (vq40  vp40 A41 )

 A41  (0.0028 /1.05   0.9972 /1.05  A41 )

 A41  0.21650 2



A41  2 A40  0.00433  2 A41  v 2 q40  v 2 p40 2 A41





 2 A41  (0.0028 /1.052  0.9972 /1.052 2

A41  0.07193

Var  Z   0.07193  0.216502  0.02544 Question #18 - Removed

Question #19 - Removed

MLC‐09‐11

12



2

A41 )

Question #20 Answer: D

 x    x1t   x 2t  0.2 xt   x 2t   x 2t  0.8 xt '1 qx  k

3

 1

'1 px 

1

 1 e

  0.2 k t 2 dt 0

 1 e

0.2

k 3

 0.04

 ln 1  0.04  /  0.2   0.2041

k  0.6123  2 2 qx



2  2 px   x  dt  0.8

2

0 t

0



   0.8 2 qx   0.8 1  2 px 

t

  px   x   t  dt



 x t  d t  px   e 0 2

2

2

e 

 k t2 d t 0

8 k e 3  8   0.6123 3 e

  0.19538  2 2 qx

 0.8 1  0.19538   0.644

MLC‐09‐11

13

Question #21 Answer: A k

min( k , 3)

f(k)

f  k    min( k , 3) 

f  k    min  k ,3 

0 1 2 3+

0 1 2 3

0.1 (0.9)(0.2) = 0.18 (0.72)(0.3) = 0.216 1-0.1-0.18-0.216 = 0.504

0 0.18 0.432 1.512

0 0.18 0.864 4.536

2.124

5.580

E  min( K ,3)  2.124



E  min  K ,3 

2

  5.580

Var  min( K ,3)  5.580  2.1242  1.07 Note that E  min( K ,3)  is the temporary curtate life expectancy, ex:3 if the life is age x.

Question #22 Answer: B  0.1 60  0.08 60 e    e    S0 (60)  2  0.005354  0.1 61  0.08 61 e     e    S0 (60)  2  0.00492 0.00492 q60  1   0.081 0.005354

MLC‐09‐11

14

2

Question #23 Answer: D Let q64 for Michel equal the standard q64 plus c. We need to solve for c. Recursion formula for a standard insurance: 20V45

  19V45  P45  1.03  q64 1  20V45 

Recursion formula for Michel’s insurance 20V45

  19V45  P45  0.01 1.03   q64  c  (1  20V45 )

The values of

19 V45

and

20V45

are the same in the two equations because we are

told Michel’s benefit reserves are the same as for a standard insurance. Subtract the second equation from the first to get:

0   1.03 (0.01)  c(1  20V45 ) c

(1.03)  0.01 1  20V45 

0.0103 1  0.427  0.018 

MLC‐09‐11

15

Question #24 Answer: B K is the curtate future lifetime for one insured. L is the loss random variable for one insurance. LAGG is the aggregate loss random variables for the individual insurances.  AGG is the standard deviation of LAGG . M is the number of policies.   L  v K 1   aK 1   1   v K 1   d  d 1  Ax  E  L    Ax   ax   Ax   d  0.75095   0.24905  0.025    0.082618  0.056604    Var  L    1    d

2



2

Ax 

Ax2



2



E  LAGG   M E  L   0.082618M Var  LAGG   M Var  L   M (0.068034)   AGG  0.260833 M

L  E  LAGG   E  LAGG    Pr  LAGG  0   AGG   AGG  AGG    0.082618M   Pr  N (0,1)    M  0.260833   0.082618 M 0.260833  M  26.97  1.645 

 minimum number needed = 27

MLC‐09‐11



0.025  2   1   0.09476   0.24905   0.068034  0.056604 

16

Question #25 Answer: D

1  v K 1 for K  0,1, 2,... d Z 2  Bv K 1 for K  0,1, 2,...

Z1  12,000

Annuity benefit: Death benefit:

1  v K 1  Bv K 1 d 12,000  12,000  K 1  B v d d  

Z  Z1  Z 2  12,000

New benefit:

2





12,000   K 1 Var( Z )   B   Var v d   12,000 Var  Z   0 if B   150,000 . 0.08

In the first formula for Var  Z  , we used the formula, valid for any constants a and b and random variable X,

Var  a  bX   b 2 Var  X 

Question #26 Answer: A

 x t: y t   x t   y t  0.08  0.04  0.12 Ax   xt /   xt     0.5714





Ay   y t /  y t    0.4





Axy   x t: y t /  x t: y t    0.6667





a xy  1/  x t: y t    5.556

Axy  Ax  Ay  Axy  0.5714  0.4  0.6667  0.3047

Premium = 0.304762/5.556 = 0.0549

MLC‐09‐11

17

Question #27 Answer: B

P40  A40 / a40  0.16132 /14.8166  0.0108878 P42  A42 / a42  0.17636 /14.5510  0.0121201 a45  a45  1  13.1121 E  3 L K 42  3  1000 A45  1000 P40  1000 P42 a45  201.20  10.89  12.12 13.1121  31.39

Many similar formulas would work equally well. One possibility would be 1000 3V42  1000 P42  1000 P40  , because prospectively after duration 3, this differs from the normal benefit reserve in that in the next year you collect 1000P40 instead of 1000P42 .

Question #28 Answer: E

E  min T ,40    40  0.005  40   32 2

32   t f  t  dt   40 f  t  dt w

40

0

40

  t f  t  dt   t f  t  dt  40 .6  w

w

0

40

 86   tf  t  dt w

40

 tf  t dt  54 w

40

  t  40  f  t  dt  54  40 .6   50 w

e 40 

40

MLC‐09‐11

s  40 

.6

18

Question #29 Answer: B

d  0.05  v  0.95 Step 1 Determine px from Kevin’s work:

608  350vpx  1000vqx  1000v 2 px  px 1  qx 1 

608  350  0.95  px  1000  0.95 1  px   1000  0.9025  px 1 608  332.5 px  950 1  px   902.5 px px  342 / 380  0.9 Step 2 Calculate 1000 Px:2 , as Kira did:

608  350  0.95  0.9   1000 Px:2 1   0.95  0.9    299.25  608  489.08 1000 Px:2  1.855

The first line of Kira’s solution is that the expected present value of Kevin’s benefit premiums is equal to the expected present value of Kira’s, since each must equal the expected present value of benefits. The expected present value of benefits would also have been easy to calculate as 1000  0.95 0.1  1000  0.952  0.9   907.25





Question #30 Answer: E Because no premiums are paid after year 10 for (x), 11Vx  Ax 11 One of the recursive reserve formulas is 10V



h 1V



 hV   h  1  i   bh1qx h

 32,535  2,078  1.05  100,000  0.011  35,635.642

px  h

0.989 35,635.642  0    1.05  100,000  0.012  36,657.31  A 11V  x 11 0.988

MLC‐09‐11

19

Question #31 Answer: B t   The survival function is S0 (t )   1   :   Then, x t    ex  and t px   1   2   x 105  45   30 e45  2 105  65  20 e65  2 

e45:65   

40 0

t

p45:65dt  

40 60  t 0

FG H

60



40  t dt 40

1 60  40 2 1 3 t  t 60  40  t  60  40 2 3

IJ K

40 0

 1556 . 







e 45:65  e 45  e 65  e 45:65  30  20  1556 .  34 

In the integral for e45:65 , the upper limit is 40 since 65 (and thus the joint status also) can survive a maximum of 40 years. Question #32 Answer: E

 4   S0 (4) / S0 (4) 

d

  e4 / 100

i

1  e4 / 100



e4 / 100 1  e4 / 100



e4  1.202553 100  e4

MLC‐09‐11

20

Question # 33 Answer: A

L ln px big OP  qb g LM ln e  OP q xbi g  q xb g M M ln p b g P x M ln e-b g P bi g

N

 q xb g 

Q

x

N



Q

 bi g  b g

 x b g   x b1g   x b 2 g   x b 3g  15 . q xb g  1  e   b g  1  e 1.5 =0.7769

q xb 2 g 

b0.7769gb g  b0.5gb0.7769g 2

 b g

. 15

 0.2590 Question # 34 Answer: D 22

A 60  v 3 

B

B

 q 60  2



B

pay at end of year 3

live then die 2 years in year 3

v4

3p 60



pay at end of year 4



2 p 60

1

1.03

3

live 3 years



q60 3 then die in year 4

1  0.09  1  0.11  0.13 

1

1.03

4

1  0.09  1  0.11 1  0.13  0.15

 0.19

MLC‐09‐11

21

Question # 35 Answer: B a x  a x:5 5 E x a x 5

a x:5 5 Ex

1  e 0.07b5g   4.219 , where 0.07     for t  5 0.07

 e 0.07b5g  0.705

a x 5 

1  12.5 , where 0.08     for t  5 0.08

b

gb g

 a x  4.219  0.705 12.5  13.03 Question #36 Answer: D  1 2 p x   p ' x  p ' x   0.8  0.7   0.56

1

qx

   

 ln p ' 1  x  q   since UDD in double decrement table   ln p    x x    ln  0.8     0.44 ln 0.56      0.1693

 0.3 q x  0.1  1

0.3qx  1

 0.053  1  0.1qx 

To elaborate on the last step:

 Number dying from cause    1 1 between x  0.1 and x  0.4   q 0.3 x  0.1 Number alive at x  0.1 Since UDD in double decrement, 

 1 l x   0.3 qx 



  l x  1  0.1qx 

MLC‐09‐11

 22

Question #37 Answer: E The benefit premium is oL

1 1     0.04  0.04333 ax 12

 v T  (0.04333  0.0066)aT  0.02  0.003aT  1 vT  T v 0.04693      0.02     0.04693  0.04693  v T 1   0.02      2

 

 0.04693  Var  o L   Var v T 1    0.1(4.7230)  0.4723   

Question #38 Answer: D

 0.7 0.1 0.2    T   0.3 0.6 0.1   0 0 1  

 0.52 0.13 0.35    T   0.39 0.39 0.22   0 0 1   2

Actuarial present value (APV) prem = 800(1 + (0.7 + 0.1) + (0.52 + 0.13)) = 1,960 APV claim = 500(1 + 0.7 + 0.52) + 3000(0 + 0.1 + 0.13) = 1800 Difference = 160 Question # 39 - Removed

MLC‐09‐11

23

Question # 40 Answer: D Use Mod to designate values unique to this insured.

b

g

b

g b gb g  1000b0.36933 / 111418 . . g  3315

a60  1  A60 / d  1  0.36933 / 0.06 / 106 .  111418 . 1000 P60  1000 A60 / a60

d

i

Mod Mod  p60 A60Mod  v q60 A61 

d

b

b

i

gb

g

1 . 01376  0.8624 0.383  0.44141 . 106

g

a Mod  1  A60Mod / d  1  0.44141 / 0.06 / 106 .  9.8684

d

Mod E 0 LMod  1000 A60Mod  P60a60

i

b

 1000 0.44141  0.03315 9.8684

g

 114.27 Question # 41 Answer: D The prospective reserve at age 60 per 1 of insurance is A60 , since there will be no future premiums. Equating that to the retrospective reserve per 1 of coverage, we have: s40:10 A60  P40  P50Mod s50:10  20 k40 10 E50 1 a40:10 a A40 A40 :20 Mod 50:10 A60    P50  a40 10 E40 10 E50 10 E50 20 E40

0.36913 

016132 . 7.70 7.57 0.06   P50Mod  14.8166 0.53667 0.51081 0.51081 0.27414

b

gb

g

0.36913  0.30582  14.8196 P50Mod  0.21887 1000 P50Mod  19.04

MLC‐09‐11

24

Alternatively, you could equate the retrospective and prospective reserves at age 50. Your equation would be:

A50 

P50Mod

a50:10

1 A40 a40:10 A40:10    a40 10 E40 10 E40

1 where A40  A40 10 E40 A50 :10

b

gb

g

 016132 .  0.53667 0.24905  0.02766

d

7.70 0.02766   ib g 140.16132 .8166 0.53667 0.53667 . b1000gb014437 g  19.07 

0.24905  P50Mod 7.57 

1000 P50Mod

7.57

Alternatively, you could set the expected present value of benefits at age 40 to the expected present value of benefit premiums. The change at age 50 did not change the benefits, only the pattern of paying for them. A40  P40 a40:10  P50Mod 0.16132 

10 E40

a50:10

. FG 016132 I H 14.8166 JK b7.70g  d P ib0.53667gb7.57g b1000gb0.07748g  19.07  Mod 50

1000 P50Mod

4.0626

Question # 42 Answer: A

d xb 2g  qxb2 g  lxb g  400

b g

d xb1g  0.45 400  180 q x b 2 g 

d xb 2 g

400

  0.488 l b g  d b1g 1000  180 x

x

pxb2 g  1  0.488  0.512

MLC‐09‐11

25

Note: The UDD assumption was not critical except to have all deaths during the year so that 1000 - 180 lives are subject to decrement 2.

Question #43 Answer: D Use “age” subscripts for years completed in program. E.g., p0 applies to a person newly hired (“age” 0). Let decrement 1 = fail, 2 = resign, 3 = other. 1 1 1 Then q0b g  1 4 , q1b g  15 , q2b g  1 3

q0b2 g 

1

q b3g  1

5,

q1b2 g 

3,

q b 3g  1

10 ,

0

1

1

b

9,

gb

q2b2 g 

1

8

q b 3g  1 2

gb

4

g

This gives p0b g  1  1 / 4 1  1 / 5 1  1 / 10  0.54

p1b g  b1  1 / 5gb1  1 / 3gb1  1 / 9g  0.474 p2b g  b1  1 / 3gb1  1 / 8gb1  1 / 4g  0.438 So 1b0 g  200, 11b g  200 b0.54g  108 , and 1b2 g  108 b0.474g  512 . q b1g  log pb1g / log pb g q b g 2

2

2

2

c h / logb0.438g 1  0.438  b0.405 / 0.826gb0.562g

q2b1g  log

2 3

 0.276

d2b1g  l2b g q2b1g  512 . 0.276  14

b gb

g

Question #44 - Removed

MLC‐09‐11

26

Question #45 Answer: E 

ex 

For the given life table function:

x 2

1 k qx  x   x 1



Ax 

k b

1   x 1 k 1 v   x k b

a  x

Ax  ax 

v k 1 k qx 

x 1  Ax d



e50  25    100 for typical annuitants 

e y  15  y  Assumed age = 70 A70 

a30

 0.45883 30 a70  9.5607 500000  b a70  b  52, 297

Question #46 Answer: B 10 E30:40  10 p30 10

d p v id p v ib1  i g  b E gb E gb1  i g .  b 0.54733gb0.53667 gb179085 g

p40 v10 

10

10

30

10

10

10

40

10

10

30

10

40

 0.52604 The above is only one of many possible ways to evaluate which should give 0.52604

a30:40:10  a30:40 10 E30:40 a3010:4010

b g b gb g  b13.2068g  b0.52604gb114784 . g  a30:40  1  0.52604 a40:50  1  7.1687

MLC‐09‐11

27

10

p30

10

p40 v10 , all of

Question #47 Answer: A Equivalence Principle, where  is annual benefit premium, gives

1000 A35  ( IA)35    a35



d

1000 A35 1000  0.42898   616761 (1199143 . . ) a35  IA 35

b gi

428.98 582382 .  73.66 

We obtained a35 from a35 

1  A35 1  0.42898   11.99143 0.047619 d

Question #48 - Removed

Question #49 Answer: C

 xy   x   y  014 .  0.07 Ax  Ay    0.5833    0.07  0.05  xy 0.14 0.14 1 1 Axy     0.7368 and a xy    5.2632 .  0.05 0.19  xy   014  xy   0.14  0.05 P

Axy a xy



Ax  Ay  Axy

MLC‐09‐11

a xy



b

g

2 0.5833  0.7368  0.0817 5.2632

28

Question #50 Answer: E

 20V  P20 1  i   q40 1  21V  

V

21

b0.49  0.01gb1  ig  0.022b1  0.545g  0.545 b1  ig  b0.545gb1  0.022g  0.022 0.50

 111 .

 21V  P20 1  i   q41 1  22V   22V . g  q b1  0.605g  0.605 b0.545.01gb111 41

q41 

0.61605  0.605 0.395

 0.028 Question #51 Answer: E 1000 P60  1000 A60 / a60

b

gb g  1000bq  p A g / b1.06  p a g  c15  b0.985gb382.79gh / c106 .  b0.985gb10.9041gh  33.22

 1000 v q60  p60 A61 / 1  p60 v a61 60

60

61

60

61

Question #52 - Removed

MLC‐09‐11

29

Question #53 Answer: E

g   ln(0.96)  0.04082

 x02t: y t  0.04082  0.01  0.03082 h   ln(0.97)  0.03046

 x01t: y t  0.03046  0.01  0.02046





5

00 01 02 03 5(0.06128)  0.736 5 p xy  5 p xy  exp    x  t : y  t   x  t : y  t   x  t : y  t dt  e 0

Question #54 Answer: B Transform these scenarios into a four-state Markov chain, as shown below.

0

from year t – 3 to year t – 2 Decrease (D)

from year t – 2 to year t – 1 Decrease (I)

Probability that year t will decrease from year t - 1 0.8

1

Increase

Decrease

0.6

2

Decrease

Increase

0.75

3

Increase

Increase

0.9

State

 0.80 0.00 0.20 0.00   0.60 0.00 0.40 0.00   The transition probability matrix is   0.00 0.75 0.00 0.25    0.00 0.90 0.00 0.10  Note for example that the value of 0.2 in the first row is a transition from DD to DI, that is, the probability of an increase in year three after two successive years of decrease. The requested probability is that of starting in state 0 and then being in either state 0 or 1 after two years. That requires the first row of the square of the transition probability matrix. It can be obtained by multiplying the first row of the matrix by each column. The result is [0.64 0.15 0.16 0.05]. The two required probabilities are 0.64 and 0.15 for a total of 0.79. The last two values in the vector do not need to be calculated, but doing so provides a check (in that the four values must sum to one). Alternatively, the required probabilities are for the successive values DDID and DDDD. The first case is transitions from state 0 to 2 and then 2 to 1 for a probability of 0.2(0.75) = 0.15. The second case is transitions from 0 to 0 and then 0 to 0 for a probability of 0.8(0.8) = 0.64 for a total of 0.79. MLC‐09‐11

30

Question #55 Answer: B lx    x  105  x  t p45  l45t / l45  (60  t ) / 60

Let K be the curtate future lifetime of (45). Then the sum of the payments is 0 if K  19 and is K – 19 if K  20 .

F 60  K IJ  1 60 K b40  39...1g  b40gb41g  13.66  60 2b60g

 20 a45 

 1  GH 60

K  20

Hence,

c

h

c

Prob K  19  13.66  Prob K  32.66

h

b g  ProbbT  33g

 Prob K  33 since K is an integer

 33 p45 

l78 27  l45 60

 0.450

MLC‐09‐11

31

Question #56 Answer: C

Ax 

2

Ax 

d IAi

 E 0s x

zd 

0

 0.4

 

x

 0.25    0.04

  2 

z







z



Ax ds

0 s

Ax ds

ib g

e 0.1s 0.4 ds

F e I  b0.4 gG H 01. JK 0.1s



 0

0.4 4 01 .

Alternatively, using a more fundamental formula but requiring more difficult integration.

c IA h

x

 

z z



0  0

bg b0.04g e

t t px  x t e  t dt t e0.04 t

 0.04

z



0

0.06t

dt

t e 0.1t dt

(integration by parts, not shown)  t 1  0.04  e 0.1 t 0 01 . 0.01 0.04  4 0.01

FG H

MLC‐09‐11

IJ K

32

Question #57 Answer: E Subscripts A and B here distinguish between the tools and do not represent ages. 

We have to find e AB 

eA 



eB  

FG 1  t IJ dt  t  t H 10K 20

z z FGH z FGH 10

0

IJ K

t t2 1  dt  t  7 14

7

0

7

e AB 

2 10



1

t 7

7

0

 49 

0

49  35 . 14

IJ FG 1  t IJ dt  z FG 1  t  t  t IJ dt K H 10K H 10 7 70K 2

7

0

t2 t2 t3 t   20 14 210

7

0

 7

49 49 343    2.683 20 14 210







 10  5  5



e AB  e A  e B  e AB

 5  35 .  2.683  5817 . Question #58 Answer: A

 xt  0.100  0.004  0.104 t

pxb g  e 0.104 t

Expected present value (EPV) = EPV for cause 1 + EPV for cause 2.

z

5

b

z

g

b

5

g

2000 e 0.04 t e 0.104 t 0100 . dt  500,000 e 0.04 t e 0.104 t 0.400 dt 0

0

  2000  0.10   500, 000  0.004    e 0.144t dt  5

0

MLC‐09‐11

33





2200 1  e 0.144 5  7841 0.144

Question #59 Answer: A R  1  px  q x

S  1  px  e

k 

1

1

1

1

e 0

  xt dt   k dt  xt  k  dt 0 since e 0  e 0 

 e 0

1

  xt dt  k dt

So S  0.75R  1  px  e  k  0.75q x 1  0.75q x px 1  qx px ek   1  0.75q x 1  0.75q x e k 

k  ln

LM 1  q OP N1  0.75q Q x

x

MLC‐09‐11

34

Question #60 Answer: C

A60  0.36913 2

d  0.05660

A60  0.17741

and

2

2 A60  A60  0.202862

   Expected Loss on one policy is E  L     100,000   A60  d d    2 Variance on one policy is Var  L      100,000   2 A60  A60 d  On the 10000 lives, E  S   10,000 E  L    and Var  S   10,000 Var  L    2





The  is such that 0  E  S  / Var  S   2.326 since   2.326   0.99

    10,000   100,000   A60  d d    2.326    2 100 100,000   2 A60  A60 d      100   100,000     0.36913 d  d   2.326   100,000    0.202862  d  0.63087

 d

 36913

100,000  0.63087





 0.004719

d

 36913  471.9  0.004719

d  36913  471.9  d 0.63087  0.004719  59706   59706  d  3379

MLC‐09‐11

 d

35

Question #61 Answer: C

  0V    1  i   1000  1V  1V   q75

1V

 1.05  1000q75 Similarly, 2 V   1V     1.05  1000q76 3V

  2V     1.05  1000q77





1000 3V  1.053  1.052    1.05  1000  q75 1.052  1000 1.05  q76  1000  q77 *

  



1000  1000 1.052 q75  1.05q76  q77





1.053  1.052  1.05

1000 x 1  1.052  0.05169  1.05  0.05647  0.06168



3.310125 1000  1.17796  355.87 3.310125

* This equation is an algebraic manipulation of the three equations in three unknowns  1V , 2V ,   . One method – usually effective in problems where benefit = stated amount plus reserve, is to multiply the 1V equation by 1.052 , the 2V equation by 1.05, and add those two to the 3V equation: in the result, you can cancel out the 1V , and 2V terms. Or you can substitute the 1V equation into the 2 V equation, giving 2 V in terms of  , and then substitute that into the 3V equation. Question #62 Answer: D

A281:2 

3V

d i FG IJ H K

b g

1 .  0.05827 1  e 2  0.02622 since   ln 106 72 71  1 v  19303 . 72 

a28:2

z

2  t e 1 72 dt 0

 500,000 A281:2  6643 a28:2 = 287

MLC‐09‐11

36

Question #63 Answer: D Let Ax and a x be calculated with  x t and   0.06 Let Ax * and a x * be the corresponding values with  x t increased by 0.03 and  decreased by 0.03 ax 

1  Ax





0.4  6.667 0.06

ax  ax *

      x s  0.03 ds 0.03t * 0 e dt  Proof: ax  0 e  t





0

e 

t

  x s ds 0.03t 0.03t 0

e

e

t



 e

0

  xs ds

0

e 0.06 t dt

 ax 

Ax*  1  0.03 a x*  1  0.03 a x

b gb

 1  0.03 6.667

g

 0.8

MLC‐09‐11

37

dt

Question #64 Answer: A bulb ages Year

0

1

0 1 2 3

10000 0 1000 9000 100+2700 900 280+270+3150

2

3

0 0 6300

0 0 0

The diagonals represent bulbs that don’t burn out. E.g., of the initial 10,000, (10,000) (1-0.1) = 9000 reach year 1. (9000) (1-0.3) = 6300 of those reach year 2. Replacement bulbs are new, so they start at age 0. At the end of year 1, that’s (10,000) (0.1) = 1000 At the end of 2, it’s (9000) (0.3) + (1000) (0.1) = 2700 + 100 At the end of 3, it’s (2800) (0.1) + (900) (0.3) + (6300) (0.5) = 3700 1000 2800 3700   . 105 . 2 105 . 3 105  6688

Expected present value 

Question #65 Answer: E 

e25:25  

z z

15

0 t

p25 dt 15 p25

15 .04 t

0

e

d

z

10

0t

FG z IJ e Kz H i  e LMN.051 d1  e

dt  e



15

0

.04 ds

1 .60 1  e .60 .04 .  112797  4.3187  15.60 

MLC‐09‐11

p40 dt

10 .05t

0

dt

.50

iOPQ

38

# replaced 1000 2800 3700

Question #66 Answer: C 5

p 60 1 

e1  q je1  q jb1  q gb1  q gb1  q g  b0.89gb0.87gb0.85gb0.84gb0.83g 60 1

60  2

63

64

65

 0.4589 Question # 67 Answer: E

1      0.08      0.04  

12.50  a x 

Ax 



 0.5

  1  2  Ax    2 3

e j

Var aT 

2

Ax  Ax2

2

1



1

 3 4  52.083 0.0016 S.D.  52.083  7.217

MLC‐09‐11

39

Question # 68 Answer: D

v  0.90  d  010 . Ax  1  dax  1  010 . 5  0.5

b gb g

Benefit premium  



5000 Ax  5000vqx ax

b5000gb0.5g  5000b0.90gb0.05g  455 5

10th benefit reserve for fully discrete whole life  1  0.2  1 

ax 10 ax

ax 10  ax 10  4 5

b gb g   a  b5000gb0.6g  b455gb4g  1180

Ax 10  1  dax 10  1  010 . 4  0.6 10V

 5000 Ax 10

x 10

Question #69 Answer: D v is the lowest premium to ensure a zero % chance of loss in year 1 (The present value of the payment upon death is v, so you must collect at least v to avoid a loss should death occur). Thus v = 0.95. 2 E Z  vqx  v 2 px qx 1  0.95  0.25  0.95  0.75  0.2

bg

d i

b g

 0.3729

b g

b g

2

4

E Z2  v 2qx  v 4 px qx 1  0.95  0.25  0.95  0.75  0.2  0.3478

b g d i c b gh

Var Z  E Z2  E Z

MLC‐09‐11

2

b

g

 0.3478  0.3729

2

40

 0.21

Question #70 Answer: D Expected present value (EPV) of future benefits =   0.005  2000  0.04  1000  /1.06  1  0.005  0.04  0.008  2000  0.06 1000  /1.062

 47.17  64.60  111.77 EPV of future premiums  1  1  0.005  0.04  /1.06  50  1.9009  50  E  1 L K 55

 95.05  1  111.77  95.05  16.72

Question #71 - Removed

Question #72 Answer: A Let Z be the present value random variable for one life. Let S be the present value random variable for the 100 lives. 

E ( Z )  10  e  t e  t  dt 5

 10

  

e  (   )5

 2.426

FG  IJ e b g H 2   K F 0.04 IJ de i  11233  10 G . H 0.16 K Var b Z g  E d Z i  c E b Z gh d i

 2   5

E Z 2  102

0.8

2

2

2

.  11233  2.4262  5.348

bg bg VarbSg  100 Varb Zg  534.8 E S  100 E Z  242.6

F  242.6 .  1645  F  281 534.8

MLC‐09‐11

41

Question #73 Answer: D Prob{only 1 survives} = 1-Prob{both survive}-Prob{neither survives}

b

ge

 1 3p50  3p 50  13p50 1 3p 50

b

gb

gb

gb

j

gb

gb

g b

gb

 1  0.9713 0.9698 0.9682 0.9849 0.9819 0.9682  1  0.912012 1  0.93632    0.912012

g

0.936320

 0.140461

Question # 74 - Removed

Question #75 - Removed Question # 76 Answer: C This solution applies the equivalence principle to each life. Applying the equivalence principle to the 100 life group just multiplies both sides of the first equation by 100, producing the same result for P.

EPV  Prems   P  EPV  Benefits   10q70v  10 p70 q71v 2  Pp70 p71v 2 P

b10gb0.03318g  b10gb1  0.03318gb0.03626g  Pb1  0.03318gb1  0.03626g

108 . 108 . 2  0.3072  0.3006  0.7988 P 0.6078 P  3.02 0.2012

(EPV above means Expected Present Value).

MLC‐09‐11

42

108 . 2

Question #77 Answer: E Level benefit premiums can be split into two pieces: one piece to provide term insurance for n years; one to fund the reserve for those who survive. Then,

Px  Px1:n  Px:n1 nV And plug in to get

b

gb

g

0.090  Px1:n  0.00864 0.563 Px1:n  0.0851

Another approach is to think in terms of retrospective reserves. Here is one such solution:





V  Px  Px1:n  sx:n

n



 Px  Px1:n

 aE

x:n

n

 P P  P   P   Px  Px1:n

x

ax:n 1

x:n

ax:n

1

x

x:n

1

x:n

e

j

0.563  0.090  Px1:n / 0.00864

b

gb

g

Px1:n  0.090  0.00864 0.563  0.0851

MLC‐09‐11

43

Question #78 Answer: A

b g

  ln 1.05  0.04879

Ax  

x t

0



x

0



px  x t e  t dt

1  t e dt for the given mortality function x

1 a   x x

From here, many formulas for the reserve could be used. One approach is: Since A50 

a50 50

 1  A50  18.71  0.3742 so a50     12.83 50   



 1  A40  19.40  0.3233 so a40     13.87 60 60    0.3233 P  A40    0.02331 13.87 reserve   A50  P  A40  a50    0.3742   0.0233112.83   0.0751. A40 

a60



Question #79 Answer: D Ax  E vTx   E  vTx NS   Prob  NS  E vTx S   Prob  S  0.03  0.6       0.70     0.30  0.03  0.08   0.06  0.08   0.3195

Similarly, 2 Ax 

F H

Var aT

I b gK  x

MLC‐09‐11

2

FG 0.03 IJ  0.70  FG 0.06 IJ  0.30  01923 . . H 0.03  016 H 0.06  016 . K . K

Ax  Ax2

2



01923 .  0.31952 ..  141 0.082

44

Question #80 Answer: B 2 q80:84

 2 q80  2 q84  2 q80:84

 0.5  0.4  1  0.6   0.2  0.15  1  0.1 = 0.10136 Using new p82 value of 0.3

0.5  0.4  1  0.3  0.2  0.15  1  0.1 = 0.16118 Change = 0.16118 – 0.10136 = 0.06 Alternatively, 2 p80  0.5  0.4  0.20 3 p80  2 p80  0.6  0.12 2 p84  0.20  0.15  0.03 3 p84  2 p84  0.10  0.003 2 p80:84  2 p80  2 p84  2 p80 2 p84 since independent 3

p80:84

 0.20  0.03   0.20  0.03  0.224  3 p80  3 p84  3 p80 3 p84

 0.12  0.003   0.12  0.003  0.12264 2 q 80:84  2 p80:84  3 p80:84  0.224  0.12264  0.10136 Revised 3 p80  0.20  0.30  0.06 3

p 80:84  0.06  0.003   0.06  0.003  0.06282 2 q 80:84  0.224  0.06282  0.16118

change  0.16118  0.10136  0.06

Question #81 - Removed

MLC‐09‐11

45

Question #82 Answer: A 5

b g  pb1g pb2g p50 5 50 5 50

FG 100  55IJ e b gb g H 100  50 K  b0.9gb0.7788g  0.7009  0.05 5



Similarly 10

FG 100  60IJ e b g b g H 100  50 K  b0.8gb0.6065g  0.4852

b g  p50

 0.05 10

b g  pb g  pb g  0.7009  0.4852

5 5 q50

5 50

10 50

 0.2157 Question #83 Answer: C Only decrement 1 operates before t = 0.7

b g  b0.7g qb1g  b0.7gb010 . g  0.07

1 0.7 q40

40

Probability of reaching t = 0.7

since UDD is

1-0.07 = 0.93

Decrement 2 operates only at t = 0.7, eliminating 0.125 of those who reached 0.7

b gb

g

b2g  0.93 0125 q40 .  011625 .

MLC‐09‐11

46

Question #84 Answer: C

e

j

 1 2 p80v 2  1000 A80 

FG H

 1

0.83910 . 2 106

vq80 v 3 2 p80q82  2 2

IJ  665.75  FG 0.08030  0.83910  0.09561IJ K . g . g 2b106 H 2b106 K 3

b g b b167524 . g  665.75

g

 174680 .  665.75   0.07156   397.41

Where 2 p80 

3,284 ,542  0.83910 3,914 ,365

b

gb

g

Or 2 p80  1  0.08030 1  0.08764  0.83910

Question #85 Answer: E

At issue, expected present value (EPV) of benefits 

  bt vt t p65  0

65  t

dt

  1000  e0.04t  e0.04t  t p65 65  t  dt 

0

 1000



0

t

p65 65  t  dt  1000  q65  1000

EPV of premiums   a65  

FG 1 IJ  16.667 H 0.04  0.02 K

Benefit premium   1000 / 16.667  60 2V

 

z z



0 

0

b g

b2u v u u p67  65 2  u du   a67

b g b gb

1000 e0.04b 2 uge 0.04u u p67  65 2  u du  60 16.667

 1000e0.08

z



0 u

b g

p67  65 2  u du  1000

 1083.29  q67  1000  1083.29  1000  83.29

MLC‐09‐11

47

g

Question #86 Answer: B (1)

ax:20  ax:20  1 20 Ex

(2)

ax:20 

(3)

Ax:20 

(4)

Ax  A1x:20  20 Ex Ax 20

1  Ax:20 d

A1x:20

 Ax:201

b gb g

0.28  A1x:20  0.25 0.40 A1x:20  018 . Now plug into (3):

Ax:20  018 .  0.25  0.43

ax:20 

Now plug into (2):

Now plug into (1):

b

1  0.43 .  1197 0.05 / 105 .

g

a x:20  1197 .  1  0.25  1122 .

Question #87 - Removed

Question #88 Answer: B

ex  px  px ex 1  px 

ex 8.83   0.95048 1  ex 1 9.29

ax  1  vpx  v 2 2 px  .... a

x:2

 1  v  v 2 2 px  ...

ax:2  ax  vq x  5.6459  5.60  0.0459

v 1  0.95048  0.0459

v  0.9269 1 i   1  0.0789 v MLC‐09‐11

48

Question #89 Answer: E One approach is to enumerate the possible paths ending in F and add the probabilities: FFFF – 0.23 = 0.008 FFGF – 0.2(0.8)(0.5) = 0.080 FGFF – 0.8(0.5)(0.2) = 0.080 FGHF – 0.8(0.5)(0.75) = 0.300 The total is 0.468. An alternative is to use matrix multiplication. The desired probability is the value in the upper left corner of the transition probability matrix raised to the third power. Only the first row needs to be evaluated. The first row of the matrix squared is [0.44 0.16 0.40 0.00], obtained by multiplying the first row by each column, in turn. The first row of the matrix cubed is obtained by multiplying the first row of the squared matrix by each column. The result is [0.468 0.352 0.080 0.100]. Note that only the first of the four calculations in necessary, though doing the other three and observing that the sum is 1 provides a check. Either way, the required probability is 0.468 and the actuarial present value is 500v3 (0.468)  500(0.9)3 (0.468)  171 . Question #90 – Removed

MLC‐09‐11

49

Question #91 Answer: E 1 1  75  60 15 1 1 3 1 60F        85   60 15 5 25

60M 

t 10 t  1 25

t

M p65  1

t

F p60

Let x denote the male and y denote the female. e x  5  mean for uniform distribution over  0,10  

e y  12.5  mean for uniform distribution over  0,25   10  t  t  e xy   1   1   dt 0  10   25  10  7 t2    1  t  dt 0  50 250 



 t 



7 2 t3  t   100 750 

10 0

 10 

7 1000  100  100 750

 10  7 









exy  ex  e y  exy  5 

MLC‐09‐11

4 13  3 3

25 13 30  75  26    1317 . 2 3 6

50

Question #92 Answer: B  1   3  1 2  Ax    2 5 Ax 

c h

P Ax    0.04

F P c A hI A  A Var b Lg  G 1  j H  JK e F 0.04 IJ FG 1  FG 1IJ IJ  G1  H 0.08 K H 5 H 3K K F 3I F 4 I G J G J H 2 K H 45K 2

x

2

2 x

x

2

2

2



1 5

Question #93 Answer: A Let  be the benefit premium Let kV denote the benefit reserve at the end of year k.

For any n,  nV    1  i    q25 n  n1V  p25 n  n1V 

 Thus 1V   0 V    1  i  2V 3V

n 1V

  1V   1  i    1  i     1  i     s2





  2V   1  i     s2   1  i     s3

By induction (proof omitted) sn n V    For n  35, n V  a60 (expected present value of future benefits; there are no future premiums) a60    s35



a60  s35

For n  20,

MLC‐09‐11

20 V

 a s20   60       s35

 s    20  51

Alternatively, as above  nV    1  i   n1V Write those equations, for n  0 to n  34 0 :  0V    1  i   1V

1:  1V   1  i   2V

2 :  2V   1  i   3V 

34 :  34V    1  i   35V

Multiply equation k by 1  i 

34 k

and sum the results:

 0V    1  i 35   1V   1  i 34   2V   1  i 33     34V    1  i   34 33 32 1V 1  i   2 V 1  i   3V 1  i     34 V 1  i   35V For k  1, 2, , 34, the k V 1  i  0V

35 k

terms in both sides cancel, leaving

1  i 35   1  i 35  1  i 34    1  i   35V

Since 0V  0  s35  35V

 a60 (see above for remainder of solution)

MLC‐09‐11

52

Question #94 Answer: B

 x  t: y  t 

t qy t t qx

px  x t  t qx t p y  y t

 t p y  t px  t q y  t px  t p y

For (x) = (y) = (50)

50:50 10.5  

where 10.5

p50 

10.5 q50 10

2

 l60  l61   12 8,188,074  8,075, 403  0.90848 l50

8,950,901

 1  10.5 p50  0.09152

p50 

10.5

1

 10.5 q50  10 p50  q60  2  0.09152  0.91478 0.01376  2   0.0023   10.5 q50  10.5 p50   2   10.5 p50 2  0.09152  0.90848 2    0.908482

8,188,074  0.91478 8,950,901

p50   50  10.5   10 p50  q60

since UDD

Alternatively, 10t  p50  10 p50 t p60 10t  p50:50   10 p50 

10t  p 50:50  2 10 p50

2

t

 t p60 

2

p60   10 p50 

2

 t p60 

2

 2 10 p50 1  tq60    10 p50  1  tq60  since UDD 2

2

Derivative  2 10 p50 q60  2  10 p50  1  tq60  q60 2

Derivative at 10  t  10.5 is

2  0.91478 0.01376    0.91478 1   0.5  0.01376    0.01376   0.0023 2

10.5

p 50:50  2 10.5 p50   10.5 p50 

2

 2  0.90848    0.90848 

2

 0.99162

 (for any sort of lifetime) 

MLC‐09‐11



dp dt    0.0023  0.0023 p 0.99162

53

Question #95 Answer: D

 xt   x1t   x 2t  0.01  2.29  2.30 

P  P  vt t px   x 2t dt  50, 000 vt t px   x1t dt  50, 000 vt t px   xt dt 2

2

0 2 0.1t 2.3t

0

P  P e

e

0

2

2 0.1t 2.3t

 2.29dt  50, 000 e 0

2 2.4 

 1 e P 1  2.29  2.4  P = 11,194

e



 0.01dt  50, 000 e0.1t e2.3t  2.3dt 2

2 2.4 

  1 e   50000  0.01 2.4  

 2.3 

e

2 2.4 

Question #96 Answer: B e x  p x  2 p x  3 p x  ...  11.05

b g

Annuity  v 3 3 p x 1000  v 4 4 px  1000  104 .  ... 



. g 1000b104

k 3 k

v

k px

k 3

 1000v 3



 k px k 3

F 1 IJ  1000v be  0.99  0.98g  1000G H 104 . K

3

3

x

Let  = benefit premium.

e

j

 1  0.99v  0.98v 2  8072 2.8580  8072   2824

MLC‐09‐11

54

 9.08  8072

  2.4 

Question #97 Answer: B

b g

b ge

1  a30:10  1000 A30  P IA 30  10 :10



10

A30

j

1000 A30

b g

1  1010 A30 a30:10  IA 3010 :



b

g b

1000 0102 . 7.747  0.078  10 0.088

g

102 6.789  15.024 

Question #98 Answer: E

For the general survival function S0 (t )  1  

FG1  t IJ dt H   30K L t OP  Mt  N 2b  30g Q

e 30 

z

t



,0t  ,

  30

0

2

  30

0



  30 2

100  30  35 2

Prior to medical breakthrough

  100  e 30 

After medical breakthrough

e  30  e 30  4  39

so



e30   39 

   30 2





    108

Question #99 Answer: A

L  100, 000v 2.5  4000a3

@5%

= 77,079 MLC‐09‐11

55

Question #100 Answer: D

  accid   0.001  total   0.01   other   0.01  0.001  0.009 

Expected present value   500,000 e0.05t e0.01t  0.009  dt 0



10 50,000 e0.04t e0.05t e0.01t  0.001 dt 0

 0.009 0.001   500, 000    100, 000 0.02   0.06

Question #101 Removed

Question #102 Answer: D

1000 20V  1000 Ax  20   ax  20 

1000  19V  20 Px 1.06   qx 19 1000  px 19

 342.03  13.72 1.06   0.01254 1000   369.18 0.98746

1  0.36918  11.1445  0.06 /1.06 

so 1000 Px  20  1000

MLC‐09‐11

Ax  20 369.18   33.1 ax  20 11.1445

56

Question #103 Answer: B k

 

1

k

 xt dt  2  xt dt  e 0  e 0 

 

k px

   x1t dt    e 0    k

2

  k px  where k px is from Illustrative Life Table, since  1 follows I.L.T. 2

10

11 p60

10

6,616,155  0.80802 8,188,074 6,396,609   0.78121 8,188,074

p60 

b g  p b  g  p b  g q60 10 60 11 60 

b

10

p60

g b 2

11 p60

g

2

from I.L.T.

 0.80802 2  0.781212  0.0426

Question #104 Answer: C

Ps 

1  d , where s can stand for any of the statuses under consideration. a s

as 

1 Ps  d

1  6.25 .  0.06 01 1 axy   8.333 0.06  0.06 ax  ay 

axy  axy  ax  ay axy  6.25  6.25  8.333  4.167 Pxy 

1  0.06  018 . 4.167

MLC‐09‐11

57

Question #105 Answer: A

z

b g g j  48

d 0b g  1000 e  b   0.04 gt   0.04 dt 1

0

 1000 1  e  b   0.04

e

e  b   0.04 g  0.952

b

  0.04   ln 0.952

g

 0.049   0.009

z

b

g

d 3b1g  1000 e 0.049 t 0.009 dt 4

3

 1000

0.009  b 0.049 gb 3g  b 0.049 gb 4 g e e  7.6 0.049

e

j

Question #106 Answer: B This is a graph of lx  x .

b

g

 x would be increasing in the interval 80, 100 . The graphs of lx px , l x and lx2 would be decreasing everywhere.

Question #107 Answer: B Variance  v30 15 px 15 qx

Expected value  v15 15 px

v30 15 px 15 qx  0.065 v15 15 px v15 15qx  0.065  15 qx  0.3157

Since  is constant 15 q x

 px 

15



 1   px 

15



 0.6843

px  0.975 qx  0.025

MLC‐09‐11

58

Question #108 Answer: E

1  2

11V

A

11V

B

1   2 







10V



11V

A

10V

A

B

0

 p

1 i

 B

qx 10 1000 px 10



x 10

 p

1 i

 11V B 





x 10

10V

qx 10 1000 px 10

 10V B   B

A

 101.35  8.36 

 p

1 i x 10

1.06  1  0.004

 98.97 Question #109 Answer: A 2

EPV (x’s benefits)   v k 1bk 1 k px q x  k k 0

b g

b gb g

b gb gb g

 1000 300v 0.02  350v 2 0.98 0.04  400v 3 0.98 0.96 0.06  36,829

Question #110 Answer: E

 denotes benefit premium V  EPV future benefits - EPV future premiums 1 0.6       0.326 108 . .  q65 10 10V   108 11V  p65

19

b



gb g b gb g

. g  b010 . gb10g b5.0  0.326gb108 1  010 .

=5.28 MLC‐09‐11

59

Question #111 Answer: A Expected present value Benefits 

 0.8 0.110,000    0.8 0.9  0.097  9,000  1.062

 1, 239.75

1.063

  0.8   0.8  0.9   1, 239.75  P 1    1.062   1.06  P  2.3955  P  517.53  518

Question #112 Answer: A

1180  70a30  50a40  20a30:40

b gb g b gb g

1180  70 12  50 10  20a30:40 a30:40  8 a30:40  a30  a40  a30:40  12  10  8  14 100a30:40  1400

Question #113 Answer: B

z



bg

a  a f t dt   t 

z

 1  e 0.05t



1 0.05

0.05

zb 



te

t

1 te  t dt 2

bg

i

 te 1.05t dt

LM b g FG IJ H K N 1 L F 1 I O  . M1  G . JK PP  185941 0.05 MN H 105 Q

t 1 1   t  1 et   e 1.05t 2 . 0.05 105 . 105 2

20,000  185941 .  37,188

MLC‐09‐11

60

OP Q



0

Question #114 Answer: C

Event x0

Prob

 0.05  0.95 0.10   0.095  0.95 0.90   0.855

x 1 x2

Present Value 15

15  20 /1.06  33.87 15  20 /1.06  25 /1.062  56.12

E  X    0.0515   0.095 33.87    0.855  56.12   51.95 E  X 2    0.05 15    0.095  33.87    0.855  56.12   2813.01 2

 

2

2

Var  X   E X 2  E  X   2813.01   51.95   114.2 2

2

Question #115 Answer: B Let K be the curtate future lifetime of (x + k) k

L  1000v K 1  1000 Px:3  aK 1

When (as given in the problem), (x) dies in the second year from issue, the curtate future lifetime of  x  1 is 0, so 1L

 1000v  1000 Px:3 a1

1000  279.21 1.1  629.88  630 

The premium came from A Px:3  x:3 ax:3 Ax:3  1  d ax:3

Px:3  279.21 

MLC‐09‐11

1  d ax:3 ax:3



1 d ax:3

61

Question #116 Answer: D Let M = the force of mortality of an individual drawn at random; and T  future lifetime of the individual.

Pr T  1

n

 E Pr T  1 M

  

z zz zd 

0 0

0

bg

Pr T  1 M   f M  d

0 2 1

2

s

1 2

e   t dt d

1  e

i 21 du  21 d2  e

2

 0.56767 Question #117 Answer: E For this model: 1/ 60 1 (1) ;  40  40(1)t    20  1/ 40  0.025 1  t / 60 60  t 1/ 40 1 (1) ; 40   40(2)t   20  1/ 20  0.05 1  t / 40 40  t   40  20  0.025  0.05  0.075

MLC‐09‐11

i 21 d1  e i

1 

62

2

Question #118 Answer: D Let   benefit premium Expected present value of benefits =  0.03 200,000 v  0.97 0.06 150,000 v 2  0.97 0.94 0.09 100,000 v 3

b gb

g b gb gb

g b gb gb gb

 5660.38  7769.67  6890.08  20,32013 . Expected present value of benefit premiums  ax:3 

b gb g

 1  0.97v  0.97 0.94 v 2   2.7266  20,32013 .  7452.55  2.7266 1V



. g  b200,000gb0.03g b7452.55gb106 1  0.03

 1958.46 Initial reserve, year 2 = 1V   = 1958.56 + 7452.55 = 9411.01

Question #119 Answer: A Let  denote the premium.

b g

L  bT v T   aT  1 i

T

 v T   aT

 1   aT E L  1   ax  0

 

 L  1   aT  1 

b

g

aT ax



1

ax

d

 ax  1  v T

i

 ax

v T  1   ax v T  Ax    ax 1  Ax MLC‐09‐11

63

g

Question #120 Answer: D

(0, 1)

(1, 0.9) (1.5, 0.8775) (2, 0.885)

tp1

1

2

t

1

p1  (1  01 . )  0.9

2

p1  0.9 1  0.05  0.855

b gb

g

b

g

since uniform, 1.5 p1  0.9  0.855 / 2  0.8775 

e1:1.5  Area between t  0 and t  15 . 

FG 1  0.9 IJ b1g  FG 0.9  0.8775IJ b0.5g H 2 K H 2 K

 0.95  0.444  1394 .

MLC‐09‐11

64

Alternatively, 

e11: .5   

z z zb 1.5 t 0 1

p1dt

z

0.5

p dt 1p1 x 0 0 t 1 1

0

g

p2 dx

1  01 . t dt  0.9

 t  0.12t

2

1 0

z

0.5

0

b1  0.05xgdx

 0.9 x  0.052 x

0.5

2

0

 0.95  0.444  1394 . Question #121 Answer: A

b g

10,000 A63 112 .  5233 A63  0.4672 Ax 1  A64 

b g

Ax 1  i  qx px

. g  0.01788 b0.4672gb105 1  0.01788

 0.4813 A65 

. g  0.01952 b0.4813gb105 1  0.01952

 0.4955 Single gross premium at 65 = (1.12) (10,000) (0.4955) = 5550

b1  ig

2



5550 5233

i

5550  1  0.02984 5233

Question #122A Answer: C Because your original survival function for (x) was correct, you must have  x t  0.06   x02t: y t   x03t: y t   x02t: y t  0.02

 x02t: y t  0.04 Similarly, for (y)  y t  0.06   x01t: y t   x03t: y t   x01t: y t  0.02

 x01t: y t  0.04 MLC‐09‐11

65

The first-to-die insurance pays as soon as State 0 is left, regardless of which state is next. The force of transition from State 0 is  x01t: y t   x02t: y t   x03t: y t  0.04  0.04  0.02  0.10 . With a constant force of transition, the expected present value is   0.10 00 01 02 03  t 0.05 t 0.10 t ( ) e p      dt  t xy x t : y t x t : y t x t : y t       0 0 e e (0.10)dt  0.15

Question #122B Answer: E Because (x) is to have a constant force of 0.06 regardless of (y)’s status (and vice23 versa) it must be that  13 x  t: y  t   x  t: y  t  0.06 . There are three mutually exclusive ways in which both will be dead by the end of year 3: 1: Transition from State 0 directly to State 3 within 3 years. The probability of this is 3

0.02 0.10t 0.3 0 t p  dt  0 e 0.02dt   0.10 e 0  0.2(1  e )  0.0518 2: Transition from State 0 to State 1 and then to State 3 all within 3 years. The probability of this is 3



00 xy

3

0

3

03 x  t: y t

0.10 t

3

t

0.10 t pxy00  x01t: y t 3t p13 0.04(1  e 0.06(3t ) )dt x  t : y  t dt   e

0

3

  0.04 e 0  0.4(1  e

0.10 t

0.3

e

)e

0.18 0.04 t

0.18

e

(1  e

0.04 0.10t 0.04e0.18 0.04t     e e 0.10 0.04

0.12

3

0

)  0.00922

3: Transition from State 0 to State 2 and then to State 3 all within 3 years. By symmetry, this probability is 0.00922. The answer is then 0.0518 + 2(0.00922) = 0.0702.

MLC‐09‐11

66

Question #122C Answer: D Because the original survival function continues to hold for the individual lives, with a constant force of mortality of 0.06 and a constant force of interest of 0.05, the expected present values of the individual insurances are Ax  Ay 

0.06  0.54545 , 0.06  0.05

Then, Axy  Ax  Ay  Axy  0.54545  0.54545  0.66667  0.42423 Alternatively, the answer can be obtained be using the three mutually exclusive outcomes used in the solution to Question 122B. 1:





0



e 0.05t t pxy00  x03 t: y t dt   e 0.05t e 0.10t 0.02dt 

2 and 3:

0





0

0.02  0.13333 0.15



13 e 0.05t t pxy00  x01t: y t  e 0.05 r r p11 x  t : y  t  x  t  r : y  t  r drdt 0





0

0

  e 0.05t e 0.10t 0.04  e 0.05 r e 0.06 r 0.06drdt 

0.04 0.06  0.14545 0.15 0.11

The solution is 0.13333 + 2(0.14545) = 0.42423. The fact that the double integral factors into two components is due to the memoryless property of the exponential transition distributions.

MLC‐09‐11

67

Question #123 Answer: B 5

q35:45  5 q35  5 q45  5 q35:45  5 p35q40  5 p45q50  5 p35:45q40:50

b g  p q  p q  p  p b1  p p g  b0.9gb.03g  b0.8gb0.05g  b0.9gb0.8g 1  b0.97gb0.95g  5 p35q40  5 p45q50  5 p35  5 p45 1  p40:50 5 35 40

5 45 50

5 35

5 45

40

50

 0.01048 Alternatively, 6

p35  5 p35  p40   0.90 1  0.03  0.873

6

p45  5 p45  p50   0.80 1  0.05   0.76

5 q35:45

 5 p35:45  6 p35:45   5 p35  5 p45  5 p35:45    6 p35  6 p45  6 p35:45 

  5 p35  5 p45  5 p35  5 p45    6 p35  6 p45  6 p35  6 p45    0.90  0.80  0.90  0.80    0.873  0.76  0.873  0.76   0.98  0.96952  0.01048

Question #124 – Removed

Question #125 - Removed

MLC‐09‐11

68

Question #126 Answer: E Let

Y = present value random variable for payments on one life S   Y  present value random variable for all payments

E Y  10a40  148166 . Var Y  10

2

d

2

2 A40  A40

d

d

i

2

ib

g

2  100 0.04863  016132 106 . . / 0.06

2

 70555 . E S  100 E Y  14,816.6

Var S  100 Var Y  70,555 Standard deviation S  70,555  265.62 By normal approximation, need E [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (265.62) = 15,254

Question #127 Answer: B

e

5 A30  4 A301 :20

Initial Benefit Prem 

Where

e

j

5a30:35  4a30:20

b

g b g b

g



 4 0.02933 5 010248 . 5 14.835  4 11959 .



0.5124  011732 0.39508 .   0.015 74.175  47.836 26.339

b

g

j

1 A30  A30:20  A30:201  0.32307  0.29374  0.02933 :20

and a30:20 

1  A30:20 d



1  0.32307 .  11959 0.06 106 .

FG IJ H K

Comment: the numerator could equally well have been calculated as A30  4 20 E30 A50 = 0.10248 + (4) (0.29374) (0.24905) = 0.39510 MLC‐09‐11

69

Question #128 Answer: B 0.75

b gb g

px  1  0.75 0.05  0.9625

0.75

b gb g

p y  1  0.75 .10

 0.925 0.75 qxy  1 0.75 pxy

b p gd p i since independent = 1- b0.9625gb0.925g  1

0.75

x

0.75

y

.  01097

Question #129 Answer: D Let G be the expense-loaded premium. Expected present value (EPV) of benefits = 100,000 A35 EPV of premiums = Ga35 EPV of expenses =  0.1G  25   2.50 100   a35 Equivalence principle: Ga35  100,000 A35   0.1G  25  250  a35

A35  0.1G  275 a35 0.9G  100,000 P35  275 G  100,000

G

100  8.36   275 0.9

 1234

MLC‐09‐11

70

Question #130 Answer: A The person receives K per year guaranteed for 10 years  Ka10  8.4353K The person receives K per years alive starting 10 years from now 10 a40 K

b

g

*Hence we have 10000  8.435310 E40a50 K Derive

10 E40 :

1 A40  A4010  :

10 E40

Derive a50 



b

10 E40

1 A40  A40 :10

A50



gA

50

0.30  0.09  0.60 0.35

1  A50 1  0.35   16.90 .04 d 104 .

Plug in values: 10,000  8.4353  0.60 16.90 K

c

b gb

gh

 18.5753K K  538.35

MLC‐09‐11

71

Question #131 Answer: D 

e25:11 

STANDARD:

z

11

0

FG1  t IJ dt  t  t H 75K 2  75 2

e z

1

MODIFIED:

p25 

 0.1ds 0

e2511 : 

11

 101933 .

0

 e.1  0.90484

z

1

p dt 0 t 25

 p25

z FGH z FGH 10

0

1

IJ K

t dt 74

IJ K F t I 1 e   e Gt  01 . H 2  74JK  0.95163  0.90484b9.32432g  9.3886 

z

1 0.1t

0

e

dt  e 0.1

0.1

0.1

10

0

1

t dt 74

2

10 0

Difference

=0.8047

Question #132 Answer: B Comparing B & D: Prospectively at time 2, they have the same future benefits. At issue, B has the lower benefit premium. Thus, by formula 7.2.2, B has the higher reserve. Comparing A to B: use formula 7.3.5. At issue, B has the higher benefit premium. Until time 2, they have had the same benefits, so B has the higher reserve. Comparing B to C: Visualize a graph C* that matches graph B on one side of t=2 and matches graph C on the other side. By using the logic of the two preceding paragraphs, C’s reserve is lower than C*’s which is lower than B’s. Comparing B to E: Reserves on E are constant at 0.

MLC‐09‐11

72

Question #133 Answer: C Since only decrements (1) and (2) occur during the year, probability of reaching the end of the year is p60 b1g  p60 b 2g  1  0.01 1  0.05  0.9405

b

gb

g

Probability of remaining through the year is p60 .  0.84645 b1g  p60 b 2g  p60 b3g  1  0.01 1  0.05 1  010

b

gb

gb

g

Probability of exiting at the end of the year is b3g  0.9405  0.84645  0.09405 q60

Question #134 - Removed

Question #135 Answer: D EPV of regular death benefit 



zb 

0

gd

ib

gd

zb 

0

gd ib

gd i

100000 e- t 0.008 e-  t dt

i

100000 e-0.06t 0.008 e-0.008t dt

b

g

 100000 0.008 / 0.06  0.008  11,764.71 EPV of accidental death benefit 



zb 30

0

gd

ib

gd

i

zb 30

0

gd ib

100000 e-0.06t 0.001 e-0.008t dt

 100 1  e-2.04 / 0.068  1,279.37 Total EPV  11765  1279  13044

MLC‐09‐11

gd i

100000 e- t 0.001 e-  t dt

73

Question #136 Answer: B

b gb

g b gb

g

g b gb

g

l 60 .6  .6 79,954  .4 80,625

 80,222.4

b gb

l 60 1.5  .5 79,954  .5 78,839

 79,396.5 0.9 q 60 .6

80222.4  79,396.5 80,222.4  0.0103 

Question #137 - Removed

Question #138 Answer: A

b g  qb1g  qb2g  0.34 q40 40 40 1 2  1  p40 b g p40 b g

0.34  1  0.75 p40  b 2g

p40  b 2g  0.88

q40 . y  b 2g  012

q41  b 2g  2 y  0.24

b gb g l b g  2000b1  0.34gb1  0.392g  803 b g  1  0.8 1  0.24  0.392 q41 

42

MLC‐09‐11

74

Question #139 Answer: C

bg

Pr L  '  0  0.5 Pr 10,000 v K 1   ' aK 1  0  0.5

From Illustrative Life Table,

47

p30  0.50816 and

48

p30 .47681

Since L is a decreasing function of K, to have Pr L    0  0.5 means we must have L    0 for K  47.

b g

b g

b g

Highest value of L   for K  47 is at K = 47.

b g

L   at K  47  10,000 v 471    a471

b g

 609.98  16.589 

b

g

L    0  609.98  16.589    0

  

609.98  36.77 16.589

Question #140 Answer: B

b g Prb K  1g  p  p  0.9  0.81  0.09 Prb K  1g  p  0.81 E bY g = .1  1.09  187 . .81  2.72  2.4715 E dY i = .1  1 .09  187 . .81  2.72  6.407 VARbY g  6.407  2.4715  0.299 Pr K  0  1  px  01 . 1 x

2 x

2 x

2

2

2

2

2

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Question #141 Answer: E

E  Z   b Ax since constant force Ax   /(    ) E(Z) 

b  0.02  b   b/3    0.06 

Var  Z   Var b v T   b 2 Var v    b 2



2

Ax  Ax2



2        b      2          2 1  4   b2     b2   10 9   45  2

Var  Z   E  Z  4 b b2     45  3 4 1 b     b  3.75  45  3

Question #142 Answer: B

b g b g c AA h 2 2

In general Var L  1  P

c h

Here P Ax 

x

2 x

1 1    .08 .12 5 ax

F .12 I So Var b Lg  G 1  J c A  A h .5625 H .08K F b.12gIJ c A  A h and Var b L *g  G 1  H .08 K b1  g b0.5625g .744 So Var b L *g  b1  g E L *  A .15a  1  a b .15g  1  5b.23g  .15 E L *  Var b L *g .7125 2

2

2 x

x

2

5 4

2

x

2 x

15 2 8 12 2 8

x

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x

76

Question #143 - Removed

Question #144 Answer: B Let l0b g  number of students entering year 1 superscript (f) denote academic failure superscript (w) denote withdrawal subscript is “age” at start of year; equals year - 1

p0b g  1  0.40  0.20  0.40 l2b g  10 l2b g q2b f g  q2b f g  01 .

b

g

q2b w g  q2b g  q2b f g  10 .  0.6  01 .  0.3 l1b gq1b f g  0.4 l1b g 1  q1b f g  q1b w g

e

j

q1b f g  0.4 1  q1b f g  0.3

e

q1b f g 

j

0.28  0.2 14 .

p1b g  1  q1b f g  q1b wg  1  0.2  0.3  0.5

b g  q b w g  p b  g q b w g  p b g p b g q b w g

w 3 q0

0

0

1

0

1

2

b gb g b gb gb g

 0.2  0.4 0.3  0.4 0.5 0.3  0.38

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Question #145 Answer: D e25  p25 (1  e26 ) M e26N  e26 due to having the same  1 M 0.05 p25N  exp     25M t  0.1(1  t )dt   p25 e  0  M M e25N  p25N (1  e26 )  e 0.05 p25 (1  e26 )  0.951e25  9.51

Question #146 Answer: D

b g F c1  A hI  10,000,000  100b10,000gG H  JK

E YAGG  100E Y  100 10,000 a x

x

b10,000g 1 c A  A h b10,000g b0.25g  b016  . g  50,000 

 Y  Var Y 

2

2

2

b

x

2 x

g

 AGG  100 Y  10 50,000  500,000 0.90  Pr

LM F  E Y N 

AGG

0

AGG

.  1282 

OP Q

F  E YAGG

 AGG

.  AGG  E YAGG F  1282

b

g

F  1282 . 500,000  10,000,000  10,641,000

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Question #147 Answer: A 1 A30:3  1000vq30  500v 2 1 q30  250v3 2 q30 2

3

 1  1.53   1   1.61   1   1.70   1000     500    0.99847     250    0.99847  0.99839     1.06  1000   1.06   1000   1.06   1000 

 1.4434  0.71535  0.35572  2.51447 1

1 1  1 2  0.00153  1 a30:1  1 2  1 2   1  2 q30     0.97129   1   2 2 2  1.06    1 1    0.97129  0.999235  2 2  0.985273 2.51447 Annualized premium  0.985273  2.552  2 

2.552 2  1.28

Each semiannual premium 

Question: #148 Answer: E

b DAg

1 80:20

q80 .2

q80 .1

eb g j

 20vq80  vp80 DA

bg

1 8119 :

b g b g b g b g b gb 2.9b12.225g   12.267

20 .2 .8 1  DA 8119 : 106 . 106 . 13 106 . 4 1  DA 8119   12.225 : .8 DA801 :20  20v .1  v .9 12.225 13 

106 .

Question #149 - Removed

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g

Question #150 Answer: A t

LM N

px  exp  



z

OP Q

b

ds  exp ln 100  x  s 0 100  x  s t



g

t 0



100  x  t 100  x



e50:60  e50  e60  e50:60 

e50 

e60

LM OP  25 z N Q t O 40  t 1 L dt  z 40t  P  20 M 40 40 N 2Q F 50  t IJ FG 40  t IJ dt  z 1 d2000  90t  t idt z G H 50 K H 40 K 2000 I  14.67 t 1 F  2000t  45t  G 2000 H 3 JK 50

50 50  t

0

t2 1 dt  50t  50 50 2

0 2 40

40

0

0



e50:60

40

40

0

0

2

3

2

40 0



e50:60  25  20  14.67  30.33

Question #151 Answer: C Ways to go 0  2 in 2 years 0  0  2; p   0.7  0.1  0.07

0  1  2; p   0.2  0.25   0.05 0  2  2; p   0.11  0.1 Total = 0.22 Binomial m = 100 q = 0.22 Var = (100) (0.22) (0.78) = 17

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Question #152 Answer: A For death occurring in year 2 0.3  1000 EPV   285.71 1.05 For death occurring in year 3, two cases: (1) State 2  State 1  State 4: (0.2  0.1) = 0.02 (2) State 2  State 2  State 4: (0.5  0.3) = 0.15 0.17 Total

EPV =

0.17  1000  154.20 1.052

Total. EPV = 285.71 + 154.20 = 439.91

Question #153 - Removed

Question #154 Answer: C Let  denote the single benefit premium.   30 a35   A351 :30



30

a35

1 1  A35 :30

A e 

35:30

j

1  A35 a65 :30

1 1  A35 :30





b.21.07g9.9 b1.07g

1.386 .93  1.49 

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Question #155 Answer: E 0.4 p0

.5  e z 

0 .4

0

e F e2 x j dx

LM e22x OP.4 N Q0 e 0.8 .4 F F e 2 1 I H K e .4 F 

.5  e .4 F .6128

bg

 ln .5  .4 F .6128

 .6931  .4F .6128  F  0.20 Question #156 Answer: C

 9V  P  1.03  qx 9b  1  qx9  10V  qx 9  b  10V   10V  3431.03  0.02904 872   10V  10V  327.97

b   b  10V   10V  872  327.97  1199.97  1  1 0.03   P  b   d   1200     14.65976 1.03   ax  = 46.92 9V  benefit reserve at the start of year ten – P = 343 – 46.92 = 296.08

Question #157 Answer: B d = 0.06  V = 0.94 Step 1 Determine px

668  258vpx  1000vqx  1000v 2 px  p x 1  qx 1  668  258  0.94  px  1000  0.94  1  px   1000  0.8836  px 1 668  242.52 px  940 1  px   883.6 px

px  272 / 298.92  0.91 MLC‐09‐11

82

Step 2 Determine 1000 Px:2

668  258  0.94  0.91  1000 Px:2 1   0.94  0.91   220.69  668  479 1000 Px:2  1.8554

Question #158 Answer: D 1 1 1 100,000  IA 40:10  100,000 v p40  IA 41:10  10 v10 9 p41 q50   A40:10 100,000       8,950,901  10     0.99722  9, 287, 264  0.16736   0.00592  100,000    1.06  1.0610        0.02766  100,000 

see comment 

=15,513 1 Where A40:10  A40  10 E40 A50

 0.16132   0.53667  0.24905   0.02766

Comment: the first line comes from comparing the benefits of the two insurances. At each of age 40, 41, 42,…,49  IA40:10 provides a death benefit 1 greater than 1

 IA141:10 .

1 Hence the A40:10 term. But  IA41:10 provides a death benefit at 50 of 10, 1

while  IA40:10 provides 0. Hence a term involving 9 q41  9 p41 q50 . The various v’s 1

and p’s just get all expected present values at age 40.

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Question #159 Answer: A

1000 1V   1  i   qx 1000  1000 1V  40  80 1.1  q x 1000  40  qx 

88  40  0.05 960

1 AS

  

G 

expenses 1  i   1000qx px

100   0.4 100   1.1  1000  0.05 1  0.05

60 1.1  50  16.8 0.95

Question #160 Answer: C 1 At any age, px    e 0.02  0.9802 1 1 qx    1  0.9802  0.0198 , which is also qx  , since decrement 2 occurs only at the end of the year.

Expected present value (EPV) at the start of each year for that year’s death benefits = 10,000*0.0198 v = 188.1  px   0.9802*0.96  0.9410  Ex  px v  0.941 v  0.941*0.95  0.8940

EPV of death benefit for 3 years 188.1  E40 *188.1  E40 * E41 *188.1  506.60

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Question #161 Answer: B 40



e30:40 

 t p30dt 0

  30  t dt   30 0

40





t

t2 2   30 

40 0

800   30  27.692  40 

  95 Or, with a linear survival function, it may be simpler to draw a picture:

0

p30  1

40

30

70



e30:40  area =27.692  40 40

p30

p30  0.3846

1  40 p30  2

  70  0.3846   30   95 t

p30 

65  t 65

Var  E T    E T   2

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85

Using a common formula for the second moment: 

Var T    2 t t p x dt  ex2 

0

 65  t  t    2  t  1  dt    1   dt 2  65  0   0  65  65

 2*  2112.5  1408.333   65  65 / 2 

2

 1408.333  1056.25  352.08

Another way, easy to calculate for a linear survival function is 

Var T    t 2 t px  x  t  dt  0

65

  t2  0

t3  3  65

0

t t px  x  t  dt

1 1   65 dt    t  dt  0 65 65  

65 0





 t2   2  65

65  0



2

2

2

 

 1408.33   32.5  352.08 2

With a linear survival function and a maximum future lifetime of 65 years, you probably didn’t need to integrate to get E T30   e30  32.5 

Likewise, if you realize (after getting   95 ) that T30 is uniformly distributed on (0, 65), its variance is just the variance of a continuous uniform random variable: 2 65  0   Var   352.08

12

Question #162 Answer: E

218.15 1.06   10,000  0.02   31.88 1  0.02  31.88  218.151.06    9,000  0.021  77.66 2V  1  0.021

1V



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Question #163 Answer: D 

ex  e y   t px  0.95  0.952  ... k 1

0.95  19 1  0.95 exy  p xy  2 p xy  ... 

 1.02  0.95  0.95   1.02  0.95  0.95  ... 2

1.02  0.95 

 1.02 0.95  0.95  ...  1  0.952 exy  ex  e y  exy  28.56 2

4

2

2

 9.44152

Question #164 - Removed

Question #165 - Removed

Question #166 Answer: E 

ax   e0.08t dt  12.5 0



3  0.375 8  3 2 Ax   e 0.13t  0.03 dt   0.23077 0 13 2 1 2 2  aT  Var  aT   Ax   Ax    400 0.23077   0.375    6.0048 2        Ax   e 0.08t  0.03 dt  0

 

 

Pr  aT  ax   aT   Pr  aT  12.5  6.0048 1  vT   Pr   6.4952   Pr 0.67524  e0.05T   0.05   ln 0.67524    Pr T    Pr T  7.85374 0.05   e0.037.85374  0.79

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Question #167 Answer: A 5

 0.05 5   p50  e     e 0.25  0.7788

1 5 q55

5 1  0.03 0.02 t   55 dt    0.02 / 0.05  e0.05t t  e 0



 0.4 1  e

0.25



= 0.0885     5 q55 Probability of retiring before 60  5 p50 = 0.7788*0.0885 = 0.0689 

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88

5 0

Question #168 Answer: C Complete the table: l81  l80  d80  910

l82  l81  d81  830 (not really needed) 1  1  e x  ex   since UDD  2 2  

e x  e x  1 2  l  l  l  1   e x   81 82 83 l80   2   e  1  l  l  l   Call this equation (A)  80 2  80 81 82      e  1  l  l   Formula like (A), one age later. Call this (B)  81 2  81 82   Subtract equation (B) from equation (A) to get

    1 1   l81  e80   l80  e81   l81   2 2       910  8.5  0.51000  e81  0.5 920   

e81 

8000  460  910  8.21 920

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Alternatively, and more straightforward,

910  0.91 1000 830 p81   0.902 920 830 p81   0.912 910 p80 

 e80  1 q 80  p 80 1  e 81    2    

1 where q80 contributes since UDD 2 1    8.5  1  0.91   0.91 1  e81  2   

e81  8.291 1     e81  q81  p81 1  e82  2   1    8.291  1  0.912   0.912 1  e82  2   

e82  8.043 1     e81  q81  p81  1  e82  2   1  1  0.902    0.902 1  8.043 2  8.206 

Or, do all the recursions in terms of e, not e , starting with e80  8.5  0.5  8.0 , then 

final step e81  e81  0.5

MLC‐09‐11

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Question #169 Answer: A

T

px t

t

px

vt

vt t px

0 1 2

0.7 0.7

1 0.7

1 0.95238

1 0.6667

 

0.49

0.90703 –

0.4444

3



2

From above ax:3   vt t px  2.1111 t 0

 a 1000 2Vx:3  1000 1  x  2:1  ax:3 

 1     1000 1    526   2.1111  

Alternatively,

Px:3 

1  d  0.4261 ax:3



1000 2Vx:3  1000 v  Px:3



 1000  0.95238  0.4261  526 You could also calculate Ax:3 and use it to calculate Px:3 .

MLC‐09‐11

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Question #170 Answer: E Let G be the gross premium. Expected present value (EPV) of benefits = 1000A50 . EPV of expenses, except claim expense = 15  1 a50 EPV of claim expense = 50A50 (50 is paid when the claim is paid) EPV of premiums = G a50 Equivalence principle: Ga50  1000 A50  15  1 a50  50 A50 1050 A50  15  a50 G a50 For the given survival function, a50 1  1.0550 1 50 k 1 50 k A50  v l l v (  )  (2)    0.36512  50k 1 50k 100  l50 k 1 50 0.05(50) k 1 1  A50  13.33248 a50  d Solving for G,

G = 30.88

Question #171 Answer: A 4

 0.05 4 p50  e     0.8187

10 8

 0.05 10 p50  e     0.6065

 0.04 8 p60  e     0.7261

18

p50   10 p50  8 p60   0.6065  0.7261

 0.4404 414 q50  4 p50  18 p50  0.8187  0.4404  0.3783

MLC‐09‐11

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Question #172 Answer: D a40:5  a40  5 E40 a45

 14.8166   0.73529 14.1121  4.4401

 a40:5  100,000 A45 v5 5 p40    IA 40:5 1



1   100,000 A45  5 E40 / a40:5   IA 40:5



 100,000  0.20120  0.73529  /  4.4401  0.04042   3363 Question #173 Answer: B Calculate the probability that both are alive or both are dead. P(both alive) = k p xy  k p x  k p y P(both dead) = k qxy  k qx k q y P(exactly one alive) = 1  k pxy  k qxy Only have to do two year’s worth so have table Pr(both alive)

Pr(both dead)

Pr(only one alive)

1

0

0

(0.91)(0.91) = 0.8281

(0.09)(0.09) = 0.0081

0.1638

(0.82)(0.82) = 0.6724

(0.18)(0.18) = 0.0324

0.2952

0.8281 0.6724  0.1638 0.2952   1  0 EPV Annuity  30, 000     20, 000      80, 431 0 1 2  0 1.05 1.05  1.051 1.052   1.05  1.05

Alternatively,

0.8281 0.6724   2.3986 1.05 1.052 0.91 0.82 ax:3  ay:3  1    2.6104 1.05 1.052 EPV  20, 000 ax:3  20, 000 ay:3  10, 000 axy:3 axy:3  1 

(it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if both are alive)   20,000  2.6104    20,000  2.6104   10,000  2.3986  80, 430 MLC‐09‐11

93

Question #174 Answer: C Let P denote the gross premium. 



0 axIMP

0

P  ax   e t e   t dt   e 0.05t dt  20

E  L 

10

axIMP

 e

P

0.03t 0.02t

e

dt  e

0.0310  0.0210 

e

0



e

0.03t 0.01t

e

dt

0

l  e 0.5 e 0.5   23 0.05 0.04 E  L   23  20  3 

E  L 3   15% P 20

Question #175 Answer: C 1 A30:2  1000vq30  500v 2 1 q 30 2

 1   1   1000    0.00153  500    0.99847  0.00161  1.06   1.06   2.15875 Initial fund  2.15875 1000 participants = 2158.75

Let Fn denote the size of Fund 1 at the end of year n.

F1  2158.75 1.07   1000  1309.86 F2  1309.86 1.065   500  895.00 Expected size of Fund 2 at end of year 2 = 0 (since the amount paid was the single benefit premium). Difference is 895.

MLC‐09‐11

94

Question #176 Answer: C 2 Var  Z   E  Z 2   E  Z 



E Z   



t

t t

0



v b 



0

E  Z 2   

px  x t dt   e0.08t e0.03t e0.02t  0.02  dt 0

0.02 2   0.02  e0.07t dt  7 0.07 

0

 vt bt 

2 t

px  x t dt  



0

e



  0.02 e0.12t  x t dt  2 0

Var  Z  

 

1 2  7 6

2



12



0.05t 2

1

e0.02t  0.02  dt

6

1 4   0.08503 6 49

Question #177 Answer: C 0.1  8   311 1.1 0.1  1  6   511 1.1

 we have Ax  1  From Ax  1  d ax Ax 10

Ax  Ax  i Ax 

3 0.1   0.2861 11 ln 1.1

Ax 10 

10V



5 0.1   0.4769 11 ln 1.1

 Ax 10  P  Ax   ax 10  0.2861   0.4769   6  8  = 0.2623

There are many other equivalent formulas that could be used.

MLC‐09‐11

95

Question #178 Answer: C 

  100,000  e0.06t  e0.001t 0.001dt

Regular death benefit

0

0.001    100,000    0.06  0.001 

 1639.34 Accidental death

  100,000 e0.06t e0.001t  0.0002  dt 10

0

10

 20 e0.061t dt 0

1  e0.61   20    149.72 0.061   Expected Present Value  1639.34  149.72  1789.06

Question #179 Answer: C 00  560 / 800  0.70 p61 01 p61  160 / 800  0.20 10 p61  0, once dead, stays dead 11 p61  1, once dead by cause 1, stays dead by cause 1

00 01 10  p61  p61  p11 p61 61  0.70  0.20  0  1  1.90

MLC‐09‐11

96

Question #180 Answer: C The solution requires determining the element in row 2 column 3 of the matrix obtained by multiplying the first three transition probability matrices. Only the second row needs to be calculated. After one period, the row is [0.0 0.7 0.3]. For the second period, multiply this row by the columns of Q1 . The result is [0.07 0.49 0.44]. For the third period, multiply this row by the columns of Q2 . The result is [0.1645 0.3465 0.4890] with the final entry providing the answer. An alternative is to enumerate the transitions that result in extinction. Label the states S (sustainable), E (endangered) and X (extinct). The possible paths are: EX – 0.3 EEX – 0.7(0.2) = 0.14 EEEX – 0.7(0.7)(0.1) = 0.049 Total = 0.489 Note that if the species is not extinct by time 3, it will never become extinct. Question #181 Answer: B Pr(dies in year 1)  p 02  0.1 Pr(dies in year 2)  p 00 p 02  p 01 p12  0.8(0.1)  0.1(0.2)  0.10 Pr(dies in year 3)  p 00 p 00 p 02  p 00 p 01 p12  p 01 p11 p12  p 01 p10 p 02  0.095 EPV (benefits)  100, 000[0.9(0.1)  0.92 (0.10)  0.93 (0.095)]  24, 025.5

Pr(in State 0 at time 0)  1 Pr(in State 0 at time 1)  p 00  0.8 Pr(in State 0 at time 2)  p 00 p 00  p 01 p10  0.8(0.8)  0.1(0.1)  0.65 EPV ($1 of premium)  1  0.9(0.8)  0.92 (0.65)  2.2465 Benefit premium = 24,025.5/2.2465 = 10,695.

Question #182 - Removed Answer: A

Question #183 - Removed MLC‐09‐11

97

Question #184 Answer: B

1000 P45a45:15   a60:15  15 E45  1000 A45 1000

A45  a45  15 E45 a60     a60  15 E60 a75  15 E45   1000 A45 a45

201.20 14.1121   0.72988 0.5108111.1454  14.1121  11.1454   0.68756  0.39994  7.2170     0.72988  0.51081  201.20

where

15 E x

was evaluated as 5 Ex  10 Ex 5

14.2573  9.9568     3.4154   201.20

  17.346

Question #185 Answer: A 1V

  0 V    1  i   1000  1V  1V  qx

2V

  1V   1  i    2000  2V  2V  qx 1  2000

  1  i   1000q   1  i   2000q  2000   1.08  1000  0.1    1.08  2000  0.1  2000 x

x 1

  1027.42

MLC‐09‐11

98

Question #186 Answer: A Let Y be the present value of payments to 1 person. Let S be the present value of the aggregate payments.

E Y   500 ax  500

 Y  Var Y  

1  Ax   5572.68 d

 500 2

1 d2



2



Ax  Ax2  1791.96

S  Y1  Y2  ...  Y250

E  S   250 E Y   1,393,170

 S  250   Y  15.811388 Y  28,333  S  1,393,170 F  1,393,170  0.90  Pr  S  F   Pr   28,333   28,333 F  1,393,170    Pr  N  0,1  28,333  

0.90  Pr  N  0,1  1.28

F  1,393,170  1.28  28,333 =1.43 million Question #187 Answer: A q  b1g  1  p b1g  1  e p b g j 41

41

bg

q41 1

41

bg

q41 

  1 2 l41   l40   d 40   d 40   1000  60  55  885 1  2  d 41   l41   d 41   l42   885  70  750  65

q41  1

65   q41  135

750  p41   885 750   1  1   q41   885 

MLC‐09‐11

65

135

 0.0766

99

Question #188 Answer: D 

t   S0 (t )  1     d  t  log  S0 (t )   dt  t

ex   



t  x 1   dt   1   x

 x 

0

1     new   new  2 old  1 e0new   old 2  1  1 old old 2  1 9      old  4  0new    4 

Question #189 Answer: C Constant force implies exponential lifetime

Var T   E T    E T  2

2

  0.1

2

1 1  2     2  100    2



E  min T ,10     t  0.1e .1t dt   10  0.1e.1t dt 10

0

 te

10

.1t

 10e

1

.1t 10 0 1

 10e.1t

 10e  10e  10  10e

10 1

 10 1  e 1   6.3

MLC‐09‐11



100

Question #190 Answer: A

Gax:15

% premium amount for 15 years   100,000 Ax  0.08G  0.02Gax:15    x  5   5ax   





Per policy for life

4669.95 11.35  51, 481.97   0.08  4669.95   0.02 11.35  4669.95     x  5   5ax  ax 

1  Ax 1  0.5148197   16.66 d 0.02913

53,003.93  51, 481.97  1433.67   x  5  83.30 4.99   x  5  x  9.99

The % of premium expenses could equally well have been expressed as 0.10G  0.02G a x:14 . The per policy expenses could also be expressed in terms of an annuityimmediate.

MLC‐09‐11

101

Question #191 Answer: D For the density where Tx  Ty ,





Pr Tx  Ty  

40



y

y 0 x 0

0.0005dxdy  



40 y 0

50

y

0.0005 x dy   0

50 y  40

40

50

0

y  40

  0.0005 ydy   

40

x0 0.0005dxdy

y  40

0.0005 y 2 2

40 0

 0.02 y

0.0005 x

40

dy

0

0.02 dy 50 40

 0.40 + 0.20 = 0.60 For the overall density, Pr Tx  Ty  0.4  0  0.6  0.6  0.36





where the first 0.4 is the probability that Tx  Ty and the first 0.6 is the probability that Tx  Ty .

Question #192 Answer: B The conditional expected value of the annuity, given  , is

1 . 0.01  

The unconditional expected value is 0.02 1  0.01  0.02  ax  100  d   100 ln    40.5 0.01 0.01    0.01  0.01  100 is the constant density of  on the interval  0.01,0.02 . If the density were not constant, it would have to go inside the integral.

MLC‐09‐11

102

Question #193 Answer: E 

Recall ex  

x 2 





ex:x  ex  ex  ex:x ex:x   

t  t   dt 1   1     x    y 

 x 

0

Performing the integration we obtain x  ex:x  3 2   x   ex:x  3

2   2 a  2   3a   3  2  7 a 3 3 2   3a  2   a   k  3 3 3.5a  a  k  3.5a  3a 

(i) (ii)

k 5 The solution assumes that all lifetimes are independent.

Question #194 Answer: B Although simultaneous death is impossible, the times of death are dependent as the force of mortality increases after the first death. There are two ways for the benefit to be paid. The first is to have (x) die prior to (y). That is, the transitions are State 0 to State 2 to State 3. The EPV can be written with a double integral





0



e 0.04t t pxy00  x02t: y t  e 0.04 r r px22t: y t  x23t  r: y t  r drdt 0





0

0

  e 0.04t e 0.12t 0.06  e 0.04 r e 0.10 r 0.10drdt 

0.06 0.10  0.26786 0.16 0.14

By symmetry, the second case (State 0 to State 1 to State 3) has the same EPV. Thus the total EPV is 10,000(0.26786+0.26786) = 5,357.

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103

Question #195 Answer: E Let

k

p0  Probability someone answers the first k problems correctly.

2

p0   0.8   0.64

4

p0   0.8  0.41

2

p0:0   2 p0   0.642  0.41

4

p0:0   0.41  0.168

2

p0:0  2 p0  2 p0  2 p0:0  0.87

4

p0:0  0.41  0.41  0.168  0.652

2

4

2

2

Prob(second child loses in round 3 or 4) = 2 p0:0  4 p0:0 = 0.87-0.652 = 0.218 Prob(second loses in round 3 or 4 second loses after round 2) =

2

p0:0  4 p0:0 2

=

p0:0

0.218  0.25 0.87

Question #196 Answer: E If (40) dies before 70, he receives one payment of 10, and Y = 10. The probability of this is (70 – 40)/(110 – 40) = 3/7 If (40) reaches 70 but dies before 100, he receives 2 payments. Y  10  20v30  16.16637 The probability of this is also 3/7. If (40) survives to 100, he receives 3 payments. Y  10  20v30  30v60  19.01819 The probability of this is 1 – 3/7 – 3/7 = 1/7 E Y    3/ 7  10   3/ 7  16.16637  1/ 7  19.01819  13.93104

 

E Y 2   3/ 7  102   3/ 7  16.16637 2  1/ 7  19.018192  206.53515

 

Var Y   E Y 2   E Y    12.46 Since everyone receives the first payment of 10, you could have ignored it in the calculation.

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104

Question #197 Answer: C 2



E  Z    v k 1bk 1 k 0



k

px qx  k

 v  300   0.02  v 2  350  0.98  0.04   v3 400  0.98  0.96  0.06    36.8

 

2



E Z 2   v k 1bk 1 k 0



2 k

px qx  k

 v 2  300   0.02  v 4  350   0.98  0.04   v 6 4002  0.98  0.96  0.06 2

2

 11,773

 

Var  Z   E Z 2  E  Z 

2

 11,773  36.82  10, 419

Question #198 Answer: A Benefits + 3 0 Le  1000v 

Expenses  0.20G  8   0.06G  2  v   0.06G  2  v 2

at G  41.20 and i  0.05, 0L

 for K  2   770.59

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– Premiums  G a3

Question #199 Answer: D

P  1000 P40

 235  P 1  i   0.015 1000  255  255

[A]

 255  P 1  i   0.020 1000  272   272

[B]

Subtract [A] from [B]

20 1  i   3.385  17 1 i 

Plug into [A]

20.385  1.01925 20

 235  P 1.01925  0.015 1000  255  255 235  P 

255  11.175 1.01925

P  261.15  235  26.15 1000 25V 

 272  26.151.01925   0.0251000  1  0.025

 286

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Question #200 Answer: A 1.2 1

1

0.8 0.6 0.5 0.4

0.4

0.2 0 0

10

x

Give n 0

s  x

1

55 q35

20 55 q15

20 55 q15 55 q35

 1

20

30

15 0.70 Linear Interpolation

40

50

Give n 25

60

35

0.50

0.48 Linear Interpolation

s  90  0.16 32  1   0.6667 s  35  0.48 48



s  35   s  90  0.48  0.16 32    0.4571 s 15  0.70 70



0.4571  0.6856 0.6667

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70

Give n 75 0.4

80

90

0 100

Given 90

100

0.16 Linear Interpolation

0

Alternatively, 20 55 q15 55 q35



20 p15  55 q35 55 q35



20 p15



s  35  s 15 

0.48 0.70  0.6856 

Question #201 Answer: A

S0 (80)  1 *  e ^  0.16*50   e ^  0.08*50    0.00932555 2 S0 (81)  1 *  e ^  0.16*51  e ^  0.08*51   0.008596664 2

p80  s  81 / s  80   0.008596664 / 0.00932555  0.9218 q80  1  0.9218  0.078 Alternatively (and equivalent to the above) For non-smokers, px  e0.08  0.923116 50

px  0.018316

For smokers, px  e0.16  0.852144 50 p x  0.000335 So the probability of dying at 80, weighted by the probability of surviving to 80, is

0.018316  1  0.923116   0.000335  1  0.852144  = 0.078 0.018316  0.000335

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Question #202 Answer: B x 40 41 42

lx  2000 1920 1840 

because 2000  20  60  1920 ;

d x  20 30 40

d x  60 50

1

2

1920  30  50  1840

Let premium = P 1840 2   2000 1920 EPV premiums =   v v  P  2.749 P 2000   2000 2000 30 2 40 3   20 EPV benefits = 1000  v v  v   40.41 2000 2000   2000 40.41 P  14.7 2.749

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Question #203 Answer: A 

10

a30   e0.08t e0.05dt  10 Ex  e0.08t e0.08t dt 0 10 0.13t

 e 0

dt  e

0  1.3 0.16

0 e

e0.13t 10 e0.16t  e1.3 0.13 0 0.16 1.3 1.3 e 1 e    0.13 0.13 0.16 = 7.2992





dt  0



A30   e0.08t e0.05t  0.05  dt  e1.3  e0.16t  0.08  dt 10

0

0

 1 e  e  0.05      0.08  0.16  0.13 0.13   0.41606 1.3

1.3

= P  A30  

A30 0.41606   0.057 a30 7.29923 1 1 a40   0.08  0.08 0.16 A40  1   a40

 1   0.08 / 0.16   0.5

10V

 A30   A40  P  A30  a40

 0.5 

 0.057   0.14375 0.16

Question #204 Answer: C Let T be the future lifetime of Pat, and [T] denote the greatest integer in T. ([T] is the same as K, the curtate future lifetime). L  100,000 vT  1600 a T

 1

 100,000vT  1600 a10 1600 a10

Minimum is 1600 a10  12,973 MLC‐09‐11

0  T 10

10  t  20 201

1L

v  Px1:2

Prob qx 1

0  Px1:2

1  qx 1

So Var  1 L   v 2 qx 1 1  qx 1   0.1296 MLC‐09‐11

121

That really short formula takes advantage of Var  a X  b   a 2Var ( X ) , if a and b are constants. Here a = v; b  Px1:2 ; X is binomial with p  X  1  qx 1 . Alternatively, evaluate Px1:2  0.1303

 0.9  0.1303  0.7697 if K x  1 1 L  0  0.1303  0.1303 if K x  1 1L

E  1 L    0.2  0.7697    0.8  0.1303  0.0497

 

E 1 L2   0.2  0.7697    0.8  0.1303  0.1320 2

2

Var  1 L   0.1320   0.0497   0.1295 2

Question #228 Answer: C

P  Ax  

Ax Ax  Ax  0.1  13      0.05 ax  1  Ax  1  Ax 1  13      

 P  Ax    Var  L   1       1  0.05   1   5  0.10 



2

2



2

2



2

Ax  Ax2

Ax  Ax2







Ax  Ax2  0.08888

  Var  L   1    

2



2

Ax  Ax2



16     1    0.08888  45  0.1  2

   1   4  0.1    0.1 2

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Question #229 Answer: E





Seek g such that Pr aT  40   g  0.25 aT is a strictly increasing function of T.

Pr T  40   60  0.25 since



 Pr aT

 40 

60

p40 

100  40  0.25 120  40



 a60  0.25

g  a60  19.00

Question 230 Answer: B  0.05  A51:9  1  da51:9  1     7.1  0.6619  1.05  11V

  2000  0.6619   100  7.1  613.80

 10V  P  1.05  11V  q50  2000  11V   10V  100  1.05  613.80   0.011 2000  613.80  10V

 499.09

where q50   0.00110    0.001  0.011

Alternatively, you could have used recursion to calculate A50:10 from A51:9 , then a50:10

from A50:10 , and used the prospective reserve formula for 10V .

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Question #231 Answer: C

1000 A81  1000 A80  1  i   q80 1000  A81  689.52   679.80 1.06   q80 1000  689.52  q80 

720.59  689.52  0.10 310.48

q80  0.5q80  0.05 1000 A80  1000vq80  vp80 1000 A81  1000 

0.05 0.95  689.52   665.14 1.06 1.06

Question #232 Answer: D

lx  776 752

d x  8 8



42 43

d x  16 16

1

2

  1 2     l42 and l43 came from lx 1  lx   d x   d x 







2000 8v  8v 2  1000 16v  16v 2 EPV Benefits =

776

  76.40

 776  752v  EPV Premiums  34     34 1.92  = 65.28 776   2V

 76.40  65.28  11.12

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Question #233 Answer: B

pxx  1  qxx  0.96

px  0.96  0.9798 px 1:x 1  1  qx 1:x 1  0.99

px 1  0.99  0.995 ax  1  vpx  v 2  2 px  1 

0.9798  0.9798  0.995   1.05 1.052

 2.8174

0.96  0.96  0.99    2.7763 1.05 1.052 EPV = 2000 ax  2000 ax  6000 axx axx  1  vpxx  v 2  2 pxx  1 

  4000  2.8174    6000  2.7763  27,927

Notes: The solution assumes that the future lifetimes are identically distributed. The precise description of the benefit would be a special 3-year temporary life annuity-due.

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Question #234 Answer: B t

1 1 1 px   x   t   qx   0.20

t

2 2 px   1  tqx   1  0.08t

t

3 3 px   1  tqx   1  0.125t

qx    1

1 0t

1  1 2 3 1 1 px   x   t  dt   t px  t px  t px   xt dt 0

  1  0.08t 1  0.125t  0.20  dt 1

0

1





 0.2 1  0.205t  0.01t 2 dt 0

1

 0.205t 2 0.01t 3   0.2 t    2 3 0  0.01     0.2  1  0.1025   0.1802 3  

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Question #235 Answer: B

1

d   w G  0.1G  1.50 1  1.06   1000q40  2.93  q40  AS      1  q40  q40 d



w

 0.9G  1.50 1.06   1000  0.00278   2.93 0.2  1  0.00278  0.2



0.954G  1.59  2.78  0.59 0.79722

 1.197G  6.22

2

d   w  1 AS  G  0.1G  1.50 1  1.06   1000q41  2 CV  q41 AS      1  q41  q41 d





w

1.197G  6.22  G  0.1G  1.50 1.06   1000  0.00298  2 CV  0 1  0.00298  0

 2.097G  7.72 1.06   2.98 0.99702

 2.229G  11.20 2.229G  11.20  24 G  15.8

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Question #236 Answer: A

5

1  2  4 AS  G 1  c4   e4  1  i   1000q x  4  5 CV  q x  4 AS 

1  qx4  qx 4 1





2

 396.63  281.77 1  0.05  7  1  i   90  572.12  0.26 1  0.09  0.26

 657.311  i   90  148.75 0.65

 694.50

 657.311  i   90  148.75   0.65  694.50  690.18  1.05 657.31 i  0.05

1 i 

Question #237 - Removed

Question #238 – Removed

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Question #239 Answer: B

Let P denote the annual premium The EPV of benefits is 25, 000 Ax:20  25, 000(0.4058)  10,145 . The EPV of premiums is Pax:20  12.522 P The EPV of expenses is 25, 000  (15  3)  3ax:20 1, 000  0.20 P  0.6261P  194.025  12  37.566  0.8261P  243.591

 0.25  0.05 P  0.05 P ax:20

  2.00  0.50   0.50 ax:20 

Equivalence principle: 12.522 P  10,145  0.8261 P  243.591 10,389.591 P 12.522  0.8261  888.31

Question #240 Answer: D Let G denote the premium. Expected present value (EPV) of benefits = 1000A40:20 EPV of premiums = G a40:10 EPV of expenses   0.04  0.25  G  10   0.04  0.05  G a40:9  5a40:19  0.29G  10  0.09G a40:9  5a40:19  0.2G  10  0.09G a40:10  5a40:19

(The above step is getting an a40:10 term since all the answer choices have one. It could equally well have been done later on).

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Equivalence principle:

G a40:10  1000 A40:20  0.2G  10  0.09G a40:10  5 a40:19





G a40:10  0.2  0.09 a40:10  1000 A40:20  10  5 a40:19 G

1000 A40:20  10  5 a40:19 0.91a40:10  0.2

Question #241 - Removed

Question #242 Answer: C

11

d   w  10 AS  G  c10 G  e10  1  i   10,000q x 10  11 CV q x 10 AS 

1  qx10  qx10 d



w

1600  200   0.04  200   70  1.05  10,000  0.02   1700  0.18



1  0.02  0.18

1302.1 0.8

 1627.63

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Question #243 Answer: E The benefit reserve at the end of year 9 is the certain payment of the benefit one year from now, less the premium paid at time 9. Thus, it is 10,000v – 76.87. The gross premium reserve adds expenses paid at the beginning of the tenth year and subtracts the gross premium rather than the benefit premium. Thus it is 10,000v + 5 + 0.1G – G where G is the gross premium. Then,

10, 000v  76.87  (10, 000v  5  0.9G )  1.67 0.9G  81.87  1.67 0.9G  83.54 G  92.82

Question #244 Answer: C 4 AS 

 3 AS  G  c4G  e4  1  i   1000qxd3  4 CVqxw3 1  qx 3  qx 3 d

w

Plugging in the given values: 4 AS

 

 25.22  30   0.02  30   5 1.05  1000  0.013  75  0.05 1  0.013  0.05 35.351 0.937

= 37.73

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With higher expenses and withdrawals: 4 AS

revised



25.22  30  1.2    0.02  30   5  1.05   1000  0.013  75 1.2  0.05  1  0.013  1.2  0.05 

 

 48.51.05  13  4.5 0.927 33.425 0.927

= 36.06 4 AS

 4 AS revised  37.73  36.06 = 1.67

Question #245 Answer: E

Let G denote the gross premium. EPV (expected present value) of benefits  100010 20 A30 . EPV of premiums  Ga30:5 .

EPV of expenses   0.05  0.25  G  20 first year

  0.05  0.10  G  10  a 30:4 years 2-5 10 5 a35:4 years 6-10 (there is no premium)  0.30G  0.15G a30:4  20  10 a30:4  10 5 a 30:5  0.15G  0.15Ga30:5  20  10 a30:9

(The step above is motivated by the form of the answer. You could equally well put it that form later). Equivalence principle: Ga30:5  100010 20 A30  0.15G  0.15G a30:5  20  10 a30:9 G



MLC‐09‐11

1000

10 20 A30

 20  10 a 30:9



 20  10 a30:9



1  0.15 a30:5  0.15

1000

10 20 A30

0.85 a30:5  0.15 132

Question #246 Answer: E

Let G denote the gross premium EPV (expected present value) of benefits   0.1 3000  v   0.9  0.2  2000  v 2   0.9  0.8 1000v 2 

300 360 720    1286.98 2 1.04 1.04 1.042

EPV of premium = G EPV of expenses  0.02G  0.03G  15   0.9  2  v

 0.05G  16.73 Equivalence principle: G  1286.98  0.05G  16.73 1303.71 G  1372.33 1  0.05 Question #247 Answer: C

EPV (expected present value) of benefits = 3499 (given) EPV of premiums  G   0.9  G  v G

0.9G  1.8571G 1.05

EPV of expenses, except settlement expenses,   25   4.5 10   0.2 G    0.9  10  1.5 10   0.1G  v   0.9  0.85  10  1.5 10   v 2

0.9  25  0.1G  0.765  25   1.05 1.052 =108.78+0.2857G  70  0.2G 

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Settlement expenses are 20  110   30 , payable at the same time the death benefit is paid.  30  So EPV of settlement expenses    EPV of benefits  10,000    0.003 3499   10.50

Equivalence principle:

1.8571G  3499  108.78  0.2857G  10.50 3618.28 G  2302.59 1.8571  0.2857 Question #248 Answer: D a50:20  a50  20 E50 a70

 13.2668   0.23047  8.5693  11.2918

 0.06  A50:20  1  d a50:20  1    11.2918   1.06   0.36084

Expected present value (EPV) of benefits  10,000A50:20

 3608.40 EPV of premiums  495 a50:20  5589.44 EPV of expenses   0.35  495   20  15 10    0.05  495   5  1.50 10   a50:19

 343.25   44.75 11.2918  1  803.81

EPV of amounts available for profit and contingencies = EPV premium – EPV benefits – APV expenses = 5589.44 – 3608.40 – 803.81 = 1177.23

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Question #249 Answer: B 1

1

0

0

q1xy   t pxy  x t dt   t px t p y  x t dt 1

  qx e 0.25t dt (under UDD, t px  x t  qx ) 0

1

0.125  qx (4e0.25t )  qx (4)(1  e 0.25 )  0.8848qx 0

qx  0.1413

Question #250 Answer: C 2

p[11x ]1  p[11x ]1 p[11x ] 2  p[12x ]1 p[21x ] 2 0.1   0.1   0.1   0.2     0.7    0.7     0.3    0.4   2  3   2  3    0.75(0.7333)  0.25(0.3333)  0.6333

Note that Anne might have changed states many times during each year, but the annual transition probabilities incorporate those possibilities.

Questions #251-260 – Removed

Question #261 Answer: A The insurance is payable on the death of (y) provided (x) is already dead. 

E ( Z )  Axy2   e t t qx t p y  y t dt 0



 e

0.06 t

0

(1  e0.07 t )e 0.09t 0.09dt



 0.09 e 0.15t  e0.22t dt 0

1   1  0.09     0.191  0.15 0.22 

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Question #262 Answer: C

t

px 

95  x  t 1 ,  x t  , 95  x 95  x  t

t

p y  e  t

Pr(x dies within n years and before y)  

n

0 t



n

0

px t p y  x t dt

n 95  x  t  t 1 1 1  e  n  t  e dt  e dt 95  x 95  x  t 95  x 0  (95  x)

Question #263 Answer: A 0.25

2 q30.5:40.5 

0.25

0

t



0.25

0.4 0.6t dt 1  0.5(0.4) 1  0.5(0.6)

0

p30.5 30.5t t q40.5 dt

0.4(0.6) t 2  0.8(0.7) 2

0.25

 0.0134 0

Question #264 – Removed

Question #265 Answer: D t 2 px  exp    5rdr   e 2.5t  0   t  0.5t 2 t p y  exp   0 rdr   e   t

qx1: y   t p y t px  x t dt   e0.5t e2.5t 5tdt  5 e3t tdt 1

1

0

0

2

2

1

0

1

5 2 5  e 3t  (1  e 3 )  0.7918 6 6 0

MLC‐09‐11

136

2

Question #266 Answer: B 5

G

5

0

t2 1 t 1 dt   30 25 30(25)(2) 0 60

H  5 p80:85  10 p80:85  GH 

25 20 20 15 200 4    30 25 30 25 750 15

1 16 17    0.2833 60 60 60

Question #267 Answer: D t t S0 (t )  exp    (80  x) 0.5 dx   exp  2(80  x)0.5   exp  2((80  t )0.5  800.5 )  0  0  

F  S0 (10.5)  exp  2(69.50.5  800.5 )   0.29665 S0 (10)  0.31495 S0 (11)  0.27935 G  S0 (10.5)exp  [0.31495(0.27935)]0.5  0.29662 F  G  0.00003

Question #268 Answer: A 4

4

E ( Z )  500  0.2(1  0.25t )dt  1000  0.25(0.2t )dt 0

0

 500(0.2)  t  0.125t 2   1000(0.25) (0.1t 2 ) 4

4

0

0

 100(4  2)  250(1.6)  600 Question #269 Answer: A 10

q30:50  10 q30 10 q50  (1  e0.5 )(1  e0.5 )  0.1548

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137

Question #270 Answer: C

e30:50  e30  e50  e30:50 









e30  e50   e0.05t dt  20 



0



e30:50   e 0.10t dt  10 

0

e30:50  20  20  10  30 

Question #271 Answer: B 



0

0

1 A30:50   e  t t p30 t p50 30 t dt   e 0.03t e 0.05t e 0.05t 0.05dt 

0.05  0.3846 0.13

Question #272 Answer: B

T30:50 has the exponential distribution with parameter 0.05 + 0.05 = 0.10 and so its mean is 10 and its variance is 100.

Question #273 Answer: D

    Cov[T30:50 , T30:50 ]   e30  e30:50  e50  e30:50   (20  10)(20  10)  100    See solution to Question #270 for the individual values.

Question #274 Answer: E

V  ( 2V   3 )(1  i3 )  qx  2 (b3  3V )

3

96  (84  18)(1.07)  qx  2 (240  96) qx  2  13.14 /144  0.09125

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138

Question #275 Answer: A ( 3V   4 )(1  i4 )  qx 3b4 (96  24)(1.06)  0.101(360)   101.05 px 3 0.899

V

4

Question #276 Answer: D Under UDD: 0.5qx 3 0.5(0.101)   0.0532 0.5 q x  3.5  1  0.5qx 3 1  0.5(0.101)

Question #277 Answer: E

V  v 0.5 0.5 p3.5 4V  v 0.5 0.5 q3.5b4

3.5

 1.060.5 (0.9468)(101.05)  1.060.5 (0.0532)(360)  111.53

Question #278 Answer: D 1  10 p30:40  1  10 p30 10 p40  1 

60 50 2  70 60 7

Question #279 Answer: A

10

2  q30:40

10

0

t

p30 30t (1  t p40 )dt  

10

0

1 t 50  0.0119 dt  70 60 70(60)

Question #280 Answer: A



20

10 t



p30 30t t q40 dt  

20

10

20

10 t

p40 40t t q30 dt

20 1 t 1 t 1 400  100 1 400  100 dt   dt    0.0714 10 70 60 60 70 70 2(60) 60 2(70)

MLC‐09‐11

139

Question #281 Answer: C 140, 000 

30 t

0

p30 30t t p40 dt  180, 000 

30

0

t

p40  40t t p30 dt

30 1 70  t 1 60  t dt  180, 000  dt 0 70 60 0 60 70 1 602  302 1 702  402  140, 000  180, 000  115, 714 70 2(60) 60 2(70)

 140, 000 

30

Question #282 Answer: B P

20

0



t

p30 t p40 dt  P 

20

0

70  t 60  t P 20 dt  4200  130t  t 2 dt  0 70 60 4200

P [4200(20)  130(200)  8000 / 3]  14.444 P 4200

Question #283 Answer: A Note that this is the same as Question 33, but using multi-state notation rather than multiple-decrement notation. The only way to be in State 2 one year from now is to stay in State 0 and then make a single transition to State 2 during the year. 1

1

1

p   t p  dt   e 02 x

0

00 x

MLC‐09‐11

02 x t

0

 (0.3 0.5  0.7) t

e 1.5t 1  (1  e 1.5 )  0.259 0.5dt  0.5 1.5 0 3

140

Question #284 Answer: C

m  1 m2  1  (  80 ) . Then, 2m 12m 2 2  1 22  1 4  1 42  1 (4)  a80(2)  a80 a    a80   (  )    (  80 ) 80 80 2(2) 12(2) 2 2(4) 12(4) 2 8.29340  8.16715  (1/ 4)  (3 / 8)  [(3 / 48)  (15 /192)](  80 )

(m) Woolhouse’s formula to three terms is a80  a80 

0.00125  0.015625(  80 )

  80  0.08 (2) a80  a80  1/ 4  (3 / 48)(0.08)  8.5484.

The answer is: (12) a80  a80 

12  1 122  1 11 143 (  80 )  8.54840   (0.08)  8.08345.  2 2(12) 12(12) 24 1728

Question #285 Answer: B 1

| a70:2  v 2 2 p70  v3 3 p70  B  x t  c ( c 1)  At  ln c 

t

p70  e

2

p70  e 0.0002(2) (0.9754483)1.211  0.9943956

3

p70  e0.0002(3) (0.9754483)1.3311  0.9912108

1

e

e

 0.000003  70 t  1.1 (1.1 1) 0.0002 t  ln(1.1) 

e

 e0.0002t (0.9754483)1.1 1 t

| a70:2  0.9943956 /1.052  0.9912108 /1.053  1.7582.

Question #286 Answer: E 1 A50:2  vq50  v 2 p50 q51

px  e

 B  x   c ( c 1)  ln c 

e

 0.000005  x  1.2 (0.2)  ln(1.2) 

 e 0.0000054848(1.2)

50

p50  0.951311 p51  0.941861 1 A50:2  0.048689 /1.03  0.951311(0.058139) /1.032  0.09940.

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Question #287 Answer: E

The reserve at the end of year 2 is 10, 000vq52   3 where  3 is the benefit premium in year 3. From the equivalence principle at time zero: 10, 000(vq50  v 2 p50 q51  v 3 p50 p51q52 )  0.5 3  0.5 3vp50   3v 2 p50 p51 0.95 0.95(0.94)   0.05 0.95(0.06) 0.95(0.94)(0.07)      10, 000     0.5  0.5 3 2 3 1.04 1.04 1.04 1.042   1.04   1563.4779  1.78236 3

 3  877.1953 V  10, 000(0.7) /1.04  877.1953  204.12

2

Question #288 Answer: C

The benefit premium in year one is vq50 . By the equivalence principle: 10, 000(vq50  v 2 p50 q51  v 3 p50 p51q52 )  10, 000vq50  (vp50  v 2 p50 p51 )  0.95(0.06) 0.95(0.94)(0.07)   0.95 0.95(0.94)  10, 000      2 1.043 1.042   1.04   1.04 1082.708665  1.739090   622.5719 2V  10, 000(0.07) /1.04  622.5719  50.51.

Question #289 Answer: E t-th year 0* 1 2 3

V

P

E

I

EDB

E tV

0

0

1000

0

0

0

-1000 1.000

0 700 700

14500 14500 14500

100 100 100

864 906 906

14000 15000 16000

690.2 689.5 0

573.8 1.000 316.5 0.986 6 0.971

t 1

Pr

t 1

px

P 1000.0 573.8 312.1 5.8

NPV  1000  573.8v  312.1v 2  5.8v 3  216.08 using a 10% discount rate.

*The 1000 at time 0 is neither accumulated nor discounted. The value is treated as occurring at the end of time 0 and not as occurring at the beginning of year 1.

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Question #290 Answer: C NPV  140v  0.95(135) p67 v 2  (0.95) 2 (130) 2 p67 v 3  314.09 at a 10% discount rate.

Question #291 Answer: B Solution Pr  (6, 000  700  10)(1.06)1/12  0.002(200, 000)  0.0015(50, 000  15, 000)  0.9965(6, 200)  46.7.

Question #292 Answer: C

245  p40 274 and 300  2 p40 395 Present value of premium: 1000[1  (245 / 274)(1/1.12)  (300 / 395)(1/1.122 )]  2403.821. Present value of profit: 400  150 / 1.12  245 / 1.12 2  300 / 1.123  142.775. PV Profit / PV premium = 5.94%

Question #293 Answer: B

P[1  0.97(0.98624)  0.92(0.98624)(0.98499)]  100, 000[0.97(0.01376)  0.92(0.98264)(0.01501)  0.87(0.98264)(0.98499)(0.01638)] P  1431.74.

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Question #294 Answer: A Let IF equal 1 if the index drops below its current level in the next year and equals 0 otherwise. E[ X N | I F  1]  N (1000)vqx  48.54 N E[ X N | I F  0]  0 E[ X N ]  0.1(48.54 N )  0  4.854 N Var[ X N | I F  1]  (1000v) 2 Nqx (1  qx )  44, 773.31N Var[ X N | I F  0]  0 E[Var ( X N | I F )]  0.1(44, 773.31N )  0  4, 477.33 N Var[ E ( X N | I F )]  0.1(48.54 N ) 2  0  (4.854 N ) 2  212.05 N 2 Var[ X N ]  4, 477.33 N  212.05 N 2 Var ( X 10 ) 10 lim

 25.69

Var ( X N )

N 

 lim

N  N 25.69  14.56  11.13.

4, 477.33 N  212.05 N 2  212.05  14.56 N

Question #295 Answer: E The actuarial present value of the death benefit is

S62 (4)d 62(2) v 0.5  S63 (4)d 63(2) v1.5  S64 (4)d 64(2) v 2.5 100, 000 S62l62  100, 000

3.589(4)(213)v 0.5  3.643(4)(214)v1.5  3.698(4)(215)v 2.5 3.589(52,860)

 4,585.

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Question #296 Answer: A Policy 1: AV371  [ AV36  G  E  (100, 000  AV371 ) 1/12 q63 /1.004](1.004)

 ( AV36  G  E )1.004  100, 000 1/12 q63  AV371 1/12 q63  [( AV36  G  E )1.004  100, 000 1/12 q63 ] / (1  1/12 q63 ) Policy 2 AV372  [ AV36  G  E  100, 000 1/12 q63 /1.004](1.004)

 ( AV36  G  E )1.004  100, 000 1/12 q63 Because the starting account value, G and E are identical for both policies: AV361 / AV362  1/ (1  1/12 q63 )  1 / [1  (1 / 12)(0.01788)]  1.0015.

Question #297 Answer: E The recursive formula for the account values is: AVk 1  ( AVk  0.95G  50)1.06  (100, 000  AVk 1 )q50 k . This is identical to the recursive formula for benefit reserves for a 20-year term insurance where the benefit premium is 0.95G – 50. Because the benefit reserve is zero after 20 years, using this premium will ensure that the account value is zero after 20 years. Therefore,

0.95G  50  100, 000

1 A50:20 A  E A 0.24905  0.23047(0.51495)  50 20 50 70  100, 000  1154.55 13.2668  0.23047(8.5693) a50:20 a50  20 E50 a70

G = (1154.55+50)/0.95 = 1,267.95.

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Question #298 Answer: A 2

 5, 000  E ( Z )  (5, 000 /1.030) (0.005)    (1  0.005)(0.006) 1.030(1.032)  2

2

2

  5, 000   (1  0.005)(1  0.006)(0.007) 1.030(1.032)(1.035)   392,917

Question #299 Answer: E 44.75  0.00004(1.144.75 )  0.002847

44.5  0.00004(1.144.5 )  0.002780 1000  E ( 4.75 L)  0.25{0.04 E ( 4.75 L)  150  150(0.05)  0.002847[10, 000  100  E ( 4.75 L)]} E ( 4.75 L)  {1000  0.25[150  150(0.05)  0.002847(10, 000  100)]} / [1  0.25(0.04  0.002847)]  961.27 961.27  E ( 4.5 L)  0.25{0.04 E ( 4.5 L)  150  150(0.05)  0.002780[10, 000  100  E ( 4.5 L)]} E ( 4.5 L)  {961.27  0.25[150  150(0.05)  0.002780(10, 000  100)]} / [1  0.25(0.04  0.002780)]  922.795

Question #300 - Removed Question #301 Answer: E The death benefit is max(100, 000,1.3  85, 000  110, 500)  110,500 . The withdrawal benefit is 85, 000  1, 000  84, 000 . (death) (withdrawal) ( 9 AS  P  E )(1  i )  110,500q79  84, 000q79 AS  10 (death) (withdrawal) 1  q79  q79 (75, 000  9, 000  900)(1.08)  110,500(0.01)  84, 000(0.03) 1  0.01  0.03  89, 711. Note – companies find asset share calculations less useful for universal life policies than for traditional products. This is because policyholders who choose different premium payment patterns have different asset shares. However, for any pattern of premiums, an asset share can be calculated. 

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Question #302 Answer: B 10V 



( 9V  G  E )(1  i )  1000qx(death) )qx(withdrawal)  10 CV (1  qx(death) 9 9 9 (death) ( withdrawal)   1  qx(death) (1 q ) q   9 x 9 x 9 (115  16  3)(1.06)  1000(0.01)  110(1  0.01)(0.10)  128.83. 1  0.01  (1  0.01)(0.10)

Expected deaths = 1000(0.01) = 10; actual deaths = 15. Expected withdrawals = 1000(1 – 0.01)(0.1) = 99; actual withdrawals = 100. Gain from mortality and withdrawals is equal to (Expected deaths – Actual deaths)(Death benefit – End of Year Reserve) + (Expected withdrawals – Actual withdrawals)(Withdrawal benefit – End of Year Reserve) = (10 – 15)(1000 – 128.83) + (99 – 100)(110 – 128.83) = -4337.

Question #303 Answer: C Because there are no cash flows at the beginning of the year, the only item earning interest is the reserve from the end of the previous year. The gain is 1000(10,994.49)(0.05 – 0.06) = –109,944.90

Question #304 Answer: B Expenses are incurred for all who do not die. Because gain from mortality has not yet been calculated, the anticipated experience should be used. Thus, expenses are assumed to be incurred for 1000(1 – 0.01) = 990 survivors. The gain is 990(50 – 60) = -9,900.

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Question #305 Answer: E The tenth reserve is (d ) 9V (1  i )  (1  q74 )(1000  50)  V 10 (d ) 1  q74 10,994.49(1.06)  (1  0.01)(1050) 1  0.01  V 10, 721.88. 10 V

10

Note that reserves are prospective calculations using anticipated experience. There were 2 more deaths than expected. For each extra death, an annuity benefit is not paid, an expense is not paid and a reserve does not have to be maintained. Thus, the saving is 1000 + 60 + 10,721.88 = 11,781.88. The total gain is 2(11,781.88) = 23,563.76. Because the gain from expenses has already been calculated, the actual value is used.

As an aside, the total gain from questions 303-305 is -96,281. The total gain can be determined by first calculating the assets at the end of the year. Begin with 1000(10,994.49) = 10,994,490. They earn 5% interest, to accumulate to 11,544,215. At the end of the year, expenses are 60 for each of 988 who did not die, for 59,280. Annuity benefits of 1000 are paid to the same 988 people, for 988,000. Assets at the end of the year are 11,544,215 – 59,280 – 988,000 = 10,496,935. Reserves must be held for the 988 continuing policyholders. That is, 988(10,721.88) = 10,593,217. The difference, 10,496,935 – 10,593,217 = -96,282.

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Question #306 Answer: E d tV  G   ( tV )  t (bt  tV ) dt At t = 4.5,  4.5V  25  0.05( 4.5V )  0.02(4.5)(100  4.5V )  16  0.14( 4.5V ) Euler’s formula in this case is 5.0V  4.5V  (5.0  4.5) 4.5V  . V '

t

Because the endowment benefit is 100, 5.0V  100 and thus, 100  4.5V  0.5( 4.5V )  4.5V  0.5[16  0.14( 4.5V )]  8  1.07( 4.5V )

V  85.981. Similarly,  4.5V  4V  (4.5  4.0) 4V and  4.0V  25  0.05( 4.0V )  0.02(4.0)(100  4.0V )  17  0.13( 4.0V ) 85.981  4.0V  0.5( 4.0V )  4.0V  0.5[17  0.13( 4.0V )]  8.5  1.065( 4.0V ) 4.5

V  72.752.

4.0

Note that if smaller step sizes were used (which would be inappropriate for an exam question, where the step size must be specified), the estimate of the time 4 reserve converges to its true value of 71.96. Question #307 Answer: A d tV  G   ( tV )  t (bt  tV ) dt At t = 5.0,  5.0V  25  0.05( 5.0V )  0.02(5.0)(100  5.0V )  15  0.15( 5.0V ) V '

t

Because the endowment benefit is 100, 5.0V  100 and thus,  5.0V  15  0.15(100)  30 . Euler’s formula in this case is 4.5V  5.0V  (4.5  5.0) 5.0V   100  0.5(30)  85. Similarly,  4.0V  4.5V  (4.0  4.5) 4.5V and  4.5V  25  0.05( 4.5V )  0.02(4.5)(100  4.5V )  16  0.14( 4.5V )  16  0.14(85)  27.9 . 4.0V  85  (4.0  4.5)(27.9)  71.05. Note that if smaller step sizes were used (which would be inappropriate for an exam question, where the step size must be specified), the estimate of the time 4 reserve converges to its true value of 71.96.

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Question #308 Answer: A t

t

0 s ds  e  0 (0.05 0.02 s ) ds  e (0.05t  0.01t 2 ) t p0  e 

1

p0  e 0.06  0.94176 4

APV   80(1  0.25t )e t t p0 t dt 0

f (1)  80(1  0.25)e 0.05 (0.94176)(0.05  0.02)  3.7625. Question #309 Answer: D Let P be the benefit premium. There are three ways to approach this problem. The first two are intuitive: The actuarial present value of the death benefit of 1000 is 1000 10 | A40 . The return of premium benefit can be thought of as two benefits. First, provide a ten-year temporary annuity-due to everyone, with actuarial present value Pa40:10 . However, those who live ten years must then return the accumulated value of the premiums. This forms a pure endowment with actuarial present value Ps10 10 E40 . The total actuarial present value of all benefits is 1000 10 | A40  Pa40:10  Ps10 10 E40 . Setting this equal to the actuarial present value of benefit premiums ( Pa40:10 ) and solving gives 1000 10 | A40  Pa40:10  Ps10 10 E40  Pa40:10 1000 10 | A40  Ps10 P

1000 10 | A40 1000 10 E40 A50 1000 A50   .    s10 10 E40 s10 10 E40 s10

The second intuitive approach examines the reserve at time 10. Retrospectively, premiums were returned to those who died, so per survivor, the accumulated premiums are only the ones paid by the survivors, that is Ps10 . There are no other past benefits so this is the reserve (it is easy to show this using a recursive formula) Prospectively, the reserve is the actuarial present value of benefits (there are no future premiums), or 1000A50 . Setting the two reserves equal to each other produces the premium: 1000A50 . P  s10

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The final approach is to work from basic principles: APV (benefit premiums) = Pa40:10 . 9 9k

APV benefits = 1000 10 | A40  P  v k  j 1 k p40 (1  i) j 1 j | q40 k k 0 j 0

In that double sum, k is the time the premium is paid, with probability k p40 that it is paid. j is the curtate time, from when the premium was paid, until death. The amount of premium refunded, with interest is P(1  i ) j 1 . The probability, given that it was paid, that it will be refunded at time j+1 is j | q40  k . The interest discount factor, from the date of refund to age 40 is v j  k 1 vk+j+1. Then, 9 9k

9 9k

P  v k  j 1 k p40 (1  i ) j 1 j | q40 k  P  v k k p40 j | q40 k k 0 j 0

k 0 j 0

 P  v k  k p40  10 p40   P  a40:10  10 p40 a10  9

k 0

Setting APV benefit premiums = APV benefits: Pa40:10  1000 10 | A40  P (a40:10  10 p40 a10 ) P 10 p40 a10  1000 10 | A40 P

10

E40 (1000 A50 ) v10 10 p40 (1000 A50 ) 1000 A50 .      s10 10 p40 a10 10 p40 a10

The numerical answer is 249.05 P  17.83. 13.97164

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