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Exam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A 27 E 8 D 28 D 9 B 29 A 10 A 30 D 11 A 31 A 12 A 32 A 13 D 33 B 14 C 34 C 15 A 35 A 16 D 36 A 17 D 37 C 18 D 38 C 19 B 39 E 20 B 40 B
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Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

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Page 1: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M Fall 2005

PRELIMINARY ANSWER KEY

Question # Answer Question # Answer

1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A 27 E 8 D 28 D 9 B 29 A

10 A 30 D 11 A 31 A 12 A 32 A 13 D 33 B 14 C 34 C 15 A 35 A 16 D 36 A 17 D 37 C 18 D 38 C 19 B 39 E 20 B 40 B

Page 2: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -1- GO ON TO NEXT PAGE

**BEGINNING OF EXAMINATION**

1. For a special whole life insurance on (x), you are given:

(i) Z is the present value random variable for this insurance.

(ii) Death benefits are paid at the moment of death.

(iii) ( ) 0.02, 0x t tµ = ≥

(iv) 0.08δ =

(v) 0.03 , 0t

tb e t= ≥ Calculate ( )Var Z . (A) 0.075

(B) 0.080

(C) 0.085

(D) 0.090

(E) 0.095

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2. For a whole life insurance of 1 on (x), you are given: (i) Benefits are payable at the moment of death.

(ii) Level premiums are payable at the beginning of each year.

(iii) Deaths are uniformly distributed over each year of age.

(iv) 0.10i =

(v) 8xa =

(vi) 10 6xa + = Calculate the 10th year terminal benefit reserve for this insurance. (A) 0.18

(B) 0.25

(C) 0.26

(D) 0.27

(E) 0.30

Page 4: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

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3. A special whole life insurance of 100,000 payable at the moment of death of (x) includes a double indemnity provision. This provision pays during the first ten years an additional benefit of 100,000 at the moment of death for death by accidental means. You are given: (i) µ τ

x t tb gb g = ≥0 001 0. ,

(ii) µ x t t1 0 0002 0b gb g = ≥. , , where µ x1b g is the force of decrement due to death by

accidental means.

(iii) δ = 0 06.

Calculate the single benefit premium for this insurance. (A) 1640

(B) 1710

(C) 1790

(D) 1870

(E) 1970

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4. Kevin and Kira are modeling the future lifetime of (60). (i) Kevin uses a double decrement model:

x ( )xlτ ( )1

xd ( )2xd

60 1000 120 80 61 800 160 80 62 560 − −

(ii) Kira uses a non-homogeneous Markov model:

(a) The states are 0 (alive), 1 (death due to cause 1), 2 (death due to cause 2). (b) 60Q is the transition matrix from age 60 to 61; 61Q is the transition matrix

from age 61 to 62. (iii) The two models produce equal probabilities of decrement. Calculate 61Q .

(A) 1.00 0.12 0.08

0 1.00 00 0 1.00

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(B) 0.80 0.12 0.080.56 0.16 0.08

0 0 1.00

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(C) 0.76 0.16 0.08

0 1.00 00 0 1.00

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(D) 0.70 0.20 0.10

0 1.00 00 0 1.00

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(E) 0.60 0.28 0.12

0 1.00 00 0 1.00

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Page 6: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

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5. A certain species of flower has three states: sustainable, endangered and extinct. Transitions between states are modeled as a non-homogeneous Markov chain with transition matrices iQ as follows:

1

EndangeredSustainable ExtinctSustainable 0.85 0.15 0

0 0.7 0.3EndangeredExtinct 0 0 1

Q

⎛ ⎞⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

2

0.9 0.1 00.1 0.7 0.20 0 1

Q⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

3

0.95 0.05 00.2 0.7 0.10 0 1

Q⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

0.95 0.05 00.5 0.5 0 , 4,5,...0 0 1

iQ i⎛ ⎞⎜ ⎟= =⎜ ⎟⎜ ⎟⎝ ⎠

Calculate the probability that a species endangered at the start of year 1 will ever become extinct. (A) 0.45

(B) 0.47

(C) 0.49

(D) 0.51

(E) 0.53

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6. For a special 3-year term insurance: (i) Insureds may be in one of three states at the beginning of each year: active, disabled,

or dead. All insureds are initially active. The annual transition probabilities are as follows:

Active Disabled Dead Active 0.8 0.1 0.1 Disabled 0.1 0.7 0.2 Dead 0.0 0.0 1.0

(ii) A 100,000 benefit is payable at the end of the year of death whether the insured was

active or disabled.

(iii) Premiums are paid at the beginning of each year when active. Insureds do not pay any annual premiums when they are disabled.

(iv) d = 0.10 Calculate the level annual benefit premium for this insurance. (A) 9,000

(B) 10,700

(C) 11,800

(D) 13,200

(E) 20,800

Page 8: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

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7. Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20% of them make only a deposit, 30% make only a withdrawal and the remaining 50% are there only to complain. Deposit amounts are distributed with mean 8000 and standard deviation 1000. Withdrawal amounts have mean 5000 and standard deviation 2000. The number of customers and their activities are mutually independent. Using the normal approximation, calculate the probability that for an 8-hour day the total withdrawals of the bank will exceed the total deposits. (A) 0.27 (B) 0.30 (C) 0.33 (D) 0.36 (E) 0.39

Page 9: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

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8. A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponential with mean 1 year. The first battery is activated when the probe lands on Mars. The second battery is activated when the first fails. Battery lifetimes after activation are independent. The probe transmits data until both batteries have failed. Calculate the probability that the probe is transmitting data three years after landing. (A) 0.05

(B) 0.10

(C) 0.15

(D) 0.20

(E) 0.25

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9. For a special fully discrete 30-payment whole life insurance on (45), you are given: (i) The death benefit of 1000 is payable at the end of the year of death.

(ii) The benefit premium for this insurance is equal to 451000P for the first 15 years

followed by an increased level annual premium of π for the remaining 15 years.

(iii) Mortality follows the Illustrative Life Table.

(iv) 0.06i = Calculate π . (A) 16.8

(B) 17.3

(C) 17.8

(D) 18.3

(E) 18.8

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10. For a special fully discrete 2-year endowment insurance on (x): (i) The pure endowment is 2000.

(ii) The death benefit for year k is ( )1000k plus the benefit reserve at the end of year k,

1,2k = .

(iii) π is the level annual benefit premium.

(iv) i = 0.08

(v) 1 0.9, 1, 2x kp k+ − = = Calculate π . (A) 1027

(B) 1047

(C) 1067

(D) 1087

(E) 1107

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11. For a group of 250 individuals age x, you are given: (i) The future lifetimes are independent.

(ii) Each individual is paid 500 at the beginning of each year, if living.

(iii) 0.369131xA =

(iv) 2 0.1774113xA =

(v) 0.06i = Using the normal approximation, calculate the size of the fund needed at inception in order to be 90% certain of having enough money to pay the life annuities. (A) 1.43 million

(B) 1.53 million

(C) 1.63 million

(D) 1.73 million

(E) 1.83 million

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12. For a double decrement table, you are given:

Age ( )xlτ ( )1

xd ( )2xd

40 1000 60 55 41 − − 70 42 750 − −

Each decrement is uniformly distributed over each year of age in the double decrement table. Calculate ( )1

41q′ . (A) 0.077

(B) 0.078

(C) 0.079

(D) 0.080

(E) 0.081

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13. The actuarial department for the SharpPoint Corporation models the lifetime of pencil

sharpeners from purchase using a generalized DeMoivre model with ( ) ( )1 /s x x αω= − , for 0α > and 0 x ω≤ ≤ .

A senior actuary examining mortality tables for pencil sharpeners has determined that the original value of α must change. You are given: (i) The new complete expectation of life at purchase is half what it was previously.

(ii) The new force of mortality for pencil sharpeners is 2.25 times the previous force of

mortality for all durations.

(iii) ω remains the same. Calculate the original value of α . (A) 1

(B) 2

(C) 3

(D) 4

(E) 5

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14. You are given: (i) T is the future lifetime random variable.

(ii) ( )tµ µ= , 0t ≥

(iii) [ ]Var 100T = . Calculate [ ]E 10T ∧ . (A) 2.6

(B) 5.4

(C) 6.3

(D) 9.5

(E) 10.0

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15. For a fully discrete 15-payment whole life insurance of 100,000 on (x), you are given: (i) The expense-loaded level annual premium using the equivalence principle is 4669.95.

(ii) 100,000 51,481.97xA =

(iii) :15 11.35xa =

(iv) 0.02913d =

(v) Expenses are incurred at the beginning of the year.

(vi) Percent of premium expenses are 10% in the first year and 2% thereafter.

(vii) Per policy expenses are K in the first year and 5 in each year thereafter until death. Calculate K. (A) 10.0

(B) 16.5

(C) 23.0

(D) 29.5

(E) 36.5

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16. For the future lifetimes of (x) and (y): (i) With probability 0.4, ( ) ( )T x T y= (i.e., deaths occur simultaneously).

(ii) With probability 0.6, the joint density function is

( ) ( ), ( , ) 0.0005T x T yf t s = , 0 40t< < , 0 50s< <

Calculate ( ) ( )Prob T x T y<⎡ ⎤⎣ ⎦ . (A) 0.30

(B) 0.32

(C) 0.34

(D) 0.36

(E) 0.38

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17. The length of time, in years, that a person will remember an actuarial statistic is modeled by an exponential distribution with mean 1

Y . In a certain population, Y has a gamma distribution with 2α θ= = . Calculate the probability that a person drawn at random from this population will remember an actuarial statistic less than 1

2 year. (A) 0.125

(B) 0.250

(C) 0.500

(D) 0.750

(E) 0.875

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18. In a CCRC, residents start each month in one of the following three states: Independent Living (State #1), Temporarily in a Health Center (State #2) or Permanently in a Health Center (State #3). Transitions between states occur at the end of the month. If a resident receives physical therapy, the number of sessions that the resident receives in a month has a geometric distribution with a mean which depends on the state in which the resident begins the month. The numbers of sessions received are independent. The number in each state at the beginning of a given month, the probability of needing physical therapy in the month, and the mean number of sessions received for residents receiving therapy are displayed in the following table:

State # Number in state

Probability of needing therapy

Mean number of visits

1 400 0.2 2 2 300 0.5 15 3 200 0.3 9

Using the normal approximation for the aggregate distribution, calculate the probability that more than 3000 physical therapy sessions will be required for the given month. (A) 0.21

(B) 0.27

(C) 0.34

(D) 0.42

(E) 0.50

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19. In a given week, the number of projects that require you to work overtime has a geometric distribution with 2β = . For each project, the distribution of the number of overtime hours in the week is the following:

x ( )f x 5 0.2 10 0.3 20 0.5

The number of projects and number of overtime hours are independent. You will get paid for overtime hours in excess of 15 hours in the week. Calculate the expected number of overtime hours for which you will get paid in the week. (A) 18.5

(B) 18.8

(C) 22.1

(D) 26.2

(E) 28.0

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20. For a group of lives age x, you are given: (i) Each member of the group has a constant force of mortality that is drawn from the

uniform distribution on [ ]0.01, 0.02 .

(ii) 0.01δ = For a member selected at random from this group, calculate the actuarial present value of a continuous lifetime annuity of 1 per year. (A) 40.0

(B) 40.5

(C) 41.1

(D) 41.7

(E) 42.3

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21. For a population whose mortality follows DeMoivre’s law, you are given:

(i) 40:40 60:603e e=

(ii) 20:20 60:60e k e=

Calculate k. (A) 3.0

(B) 3.5

(C) 4.0

(D) 4.5

(E) 5.0

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22. For an insurance on (x) and (y): (i) Upon the first death, the survivor receives the single benefit premium for a whole life

insurance of 10,000 payable at the moment of death of the survivor.

(ii) ( ) ( ) 0.06x yt tµ µ= = while both are alive.

(iii) ( ) 0.12x y tµ =

(iv) After the first death, ( ) 0.10tµ = for the survivor.

(v) 0.04δ = Calculate the actuarial present value of this insurance on (x) and (y). (A) 4500

(B) 5400

(C) 6000

(D) 7100

(E) 7500

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23. Kevin and Kira are in a history competition: (i) In each round, every child still in the contest faces one question. A child is out as

soon as he or she misses one question. The contest will last at least 5 rounds.

(ii) For each question, Kevin’s probability and Kira’s probability of answering that question correctly are each 0.8; their answers are independent.

Calculate the conditional probability that both Kevin and Kira are out by the start of round five, given that at least one of them participates in round 3. (A) 0.13

(B) 0.16

(C) 0.19

(D) 0.22

(E) 0.25

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24. For a special increasing whole life annuity-due on (40), you are given: (i) Y is the present-value random variable.

(ii) Payments are made once every 30 years, beginning immediately.

(iii) The payment in year 1 is 10, and payments increase by 10 every 30 years.

(iv) Mortality follows DeMoivre’s law, with 110ω = .

(v) 0.04i = Calculate ( )Var Y . (A) 10.5

(B) 11.0

(C) 11.5

(D) 12.0

(E) 12.5

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25. For a special 3-year term insurance on ( )x , you are given: (i) Z is the present-value random variable for this insurance.

(ii) q kx k+ = +0 02 1. ( ) , k = 0, 1, 2

(iii) The following benefits are payable at the end of the year of death:

k bk+1 0 300 1 350 2 400

(iv) i = 0 06.

Calculate Var Zb g . (A) 9,600

(B) 10,000

(C) 10,400

(D) 10,800

(E) 11,200

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26. For an insurance: (i) Losses have density function

( )0.02 0 100 elsewhereX

x xf x

< <⎧= ⎨⎩

(ii) The insurance has an ordinary deductible of 4 per loss.

(iii) PY is the claim payment per payment random variable. Calculate E PY⎡ ⎤⎣ ⎦ .

(A) 2.9

(B) 3.0

(C) 3.2

(D) 3.3

(E) 3.4

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27. An actuary has created a compound claims frequency model with the following properties: (i) The primary distribution is the negative binomial with probability generating function

( ) ( ) 21 3 1P z z −

= − −⎡ ⎤⎣ ⎦ .

(ii) The secondary distribution is the Poisson with probability generating function

( ) ( )1zP z eλ −= .

(iii) The probability of no claims equals 0.067. Calculate λ . (A) 0.1

(B) 0.4

(C) 1.6

(D) 2.7

(E) 3.1

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28. In 2005 a risk has a two-parameter Pareto distribution with 2α = and 3000θ = . In 2006 losses inflate by 20%. An insurance on the risk has a deductible of 600 in each year. iP , the premium in year i, equals 1.2 times the expected claims. The risk is reinsured with a deductible that stays the same in each year. iR , the reinsurance premium in year i, equals 1.1 times the expected reinsured claims.

20052005

0.55RP =

Calculate 2006

2006

RP .

(A) 0.46

(B) 0.52

(C) 0.55

(D) 0.58

(E) 0.66

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29. For a fully discrete whole life insurance of 1000 on (60), you are given: (i) The expenses, payable at the beginning of the year, are:

Expense Type First Year Renewal Years % of Premium 20% 6% Per Policy 8 2

(ii) The level expense-loaded premium is 41.20.

(iii) i = 0.05 Calculate the value of the expense augmented loss variable, 0 eL , if the insured dies in the third policy year. (A) 770

(B) 790

(C) 810

(D) 830

(E) 850

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30. For a fully discrete whole life insurance of 1000 on (45), you are given:

t 451000tV 45 tq + 22 235 0.015 23 255 0.020 24 272 0.025

Calculate 25 451000 V . (A) 279

(B) 282

(C) 284

(D) 286

(E) 288

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31. The graph of a piecewise linear survival function, ( )s x , consists of 3 line segments with endpoints (0, 1), (25, 0.50), (75, 0.40), (100, 0).

Calculate 1520 55

55 35

q

q.

(A) 0.69

(B) 0.71

(C) 0.73

(D) 0.75

(E) 0.77

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32. For a group of lives aged 30, containing an equal number of smokers and non-smokers, you are given: (i) For non-smokers, ( ) 0.08n xµ = , 30x ≥

(ii) For smokers, ( ) 0.16,s xµ = 30x ≥ Calculate 80q for a life randomly selected from those surviving to age 80. (A) 0.078

(B) 0.086

(C) 0.095

(D) 0.104

(E) 0.112

Page 34: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -33- GO ON TO NEXT PAGE

33. For a 3-year fully discrete term insurance of 1000 on (40), subject to a double decrement model: (i)

x ( )xlτ ( )1

xd ( )2xd

40 2000 20 60 41 − 30 50 42 − 40 −

(ii) Decrement 1 is death. Decrement 2 is withdrawal.

(iii) There are no withdrawal benefits.

(iv) 0.05i = Calculate the level annual benefit premium for this insurance. (A) 14.3

(B) 14.7

(C) 15.1

(D) 15.5

(E) 15.7

Page 35: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -34- GO ON TO NEXT PAGE

34. Each life within a group medical expense policy has loss amounts which follow a compound Poisson process with 0.16λ = . Given a loss, the probability that it is for Disease 1 is 1

16 . Loss amount distributions have the following parameters:

Mean per loss

Standard Deviation per loss

Disease 1 5 50 Other diseases 10 20

Premiums for a group of 100 independent lives are set at a level such that the probability (using the normal approximation to the distribution for aggregate losses) that aggregate losses for the group will exceed aggregate premiums for the group is 0.24. A vaccine which will eliminate Disease 1 and costs 0.15 per person has been discovered. Define: A = the aggregate premium assuming that no one obtains the vaccine, and B = the aggregate premium assuming that everyone obtains the vaccine and the cost of the

vaccine is a covered loss. Calculate A/B. (A) 0.94

(B) 0.97

(C) 1.00

(D) 1.03

(E) 1.06

Page 36: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -35- GO ON TO NEXT PAGE

35. An actuary for a medical device manufacturer initially models the failure time for a particular device with an exponential distribution with mean 4 years. This distribution is replaced with a spliced model whose density function: (i) is uniform over [0, 3]

(ii) is proportional to the initial modeled density function after 3 years

(iii) is continuous Calculate the probability of failure in the first 3 years under the revised distribution. (A) 0.43

(B) 0.45

(C) 0.47

(D) 0.49

(E) 0.51

Page 37: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -36- GO ON TO NEXT PAGE

36. For a fully continuous whole life insurance of 1 on (30), you are given: (i) The force of mortality is 0.05 in the first 10 years and 0.08 thereafter.

(ii) 0.08δ =

Calculate the benefit reserve at time 10 for this insurance. (A) 0.144

(B) 0.155

(C) 0.166

(D) 0.177

(E) 0.188

Page 38: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -37- GO ON TO NEXT PAGE

37. For a 10-payment, 20-year term insurance of 100,000 on Pat: (i) Death benefits are payable at the moment of death.

(ii) Contract premiums of 1600 are payable annually at the beginning of each year for 10

years.

(iii) i = 0.05

(iv) L is the loss random variable at the time of issue. Calculate the minimum value of L as a function of the time of death of Pat. (A) −21,000

(B) −17,000

(C) −13,000

(D) −12,400

(E) −12,000

Page 39: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -38- GO ON TO NEXT PAGE

38. For an insurance: (i) The number of losses per year has a Poisson distribution with 10λ = .

(ii) Loss amounts are uniformly distributed on (0, 10).

(iii) Loss amounts and the number of losses are mutually independent.

(iv) There is an ordinary deductible of 4 per loss. Calculate the variance of aggregate payments in a year. (A) 36

(B) 48

(C) 72

(D) 96

(E) 120

Page 40: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -39- GO ON TO NEXT PAGE

39. For an insurance portfolio: (i) The number of claims has the probability distribution

n np 0 0.1 1 0.4 2 0.3 3 0.2

(ii) Each claim amount has a Poisson distribution with mean 3; and

(iii) The number of claims and claim amounts are mutually independent. Calculate the variance of aggregate claims. (A) 4.8

(B) 6.4

(C) 8.0

(D) 10.2

(E) 12.4

Page 41: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Exam M: Fall 2005 -40- STOP

40. Lucky Tom deposits the coins he finds on the way to work according to a Poisson process with a mean of 22 deposits per month. 5% of the time, Tom deposits coins worth a total of 10. 15% of the time, Tom deposits coins worth a total of 5. 80% of the time, Tom deposits coins worth a total of 1. The amounts deposited are independent, and are independent of the number of deposits. Calculate the variance in the total of the monthly deposits. (A) 180

(B) 210

(C) 240

(D) 270

(E) 300

**END OF EXAMINATION**

Page 42: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Fall 2005 Exam M Solutions

Question #1 Key: C

[ ] [ ]22Var Z E Z E Z⎡ ⎤= −⎣ ⎦

[ ] ( ) ( ) ( )0.08 0.03 0.020 0

0.02t t t tt t x xE Z v b p t dt e e e dtµ

∞ ∞ − −= =∫ ∫

( ) 0.070

0.02 20.02 70.07te dt

∞ −= = =∫

( ) ( ) ( ) ( )2 22 0.05 0.02

0 00.02t t

t t t x xE Z v b p t dt e e dtµ∞ ∞ − −⎡ ⎤ = =⎣ ⎦ ∫ ∫

( )0.120

2 10.02 12 6t

xe t dtµ∞ −= = =∫

[ ] ( )21 1 42 0.0850376 6 49Var Z = − = − =

Question #2 Key: C

From 1xA d ax= − we have ( )0.1 31 8 111.1xA = − =

( )100.1 51 6 111.1xA + = − =

x xiA A δ= ×

( )

( )10

3 0.1 0.286111 ln 1.1

5 0.1 0.476911 ln 1.1

x

x

A

A +

= × =

= × =

( )10 10 10x x x xV A P A a+ += − ×

0.28610.4769 68

⎛ ⎞= − ⎜ ⎟⎝ ⎠

= 0.2623 There are many other equivalent formulas that could be used.

Page 43: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #3 Key: C

Regular death benefit 0.06 0.0010

100,000 0.001t te e dt∞ − −= × ×∫

0.001100,0000.06 0.001

⎛ ⎞= ⎜ ⎟+⎝ ⎠

= 1639 34.

Accidental death ( )10 0.06 0.0010

100,000 0.0002t te e dt− −= ∫

10 0.0610

20 te dt−= ∫

0.61120 149.72

0.061e−⎡ ⎤−

= =⎢ ⎥⎣ ⎦

Actuarial Present Value = + =1639 34 149 72 1789 06. . . Question #4 Key: D Once you are dead, you are dead. Thus, you never leave state 2 or 3, and rows 2 and 3 of the matrix must be (0 1 0) and (0 0 1). Probability of dying from cause 1 within the year, given alive at age 61, is 160/800 = 0.20. Probability of dying from cause 2 within the year, given alive at age 61, is 80/800 = 0.10 Probability of surviving to 62, given alive at 61, is 560/800 = 0.70 (alternatively, 1 – 0.20 – 0.10), so correct answer is D.

Page 44: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #5 Key: C This first solution uses the method on the top of page 9 of the study note. Note that if the species is it is not extinct after 3Q it will never be extinct. This solution parallels the example at the top of page 9 of the Daniel study note. We want the second entry of the product ( )1 2 3 3Q Q Q e× × which is equal to ( )( )1 2 3 3Q Q Q e× × × .

3

0 00 0.11 1

Q =

2

0 0.010.1 0.271 1

Q =

1

0.01 0.0490.27 0.4891 1

Q =

The second entry is 0.489; that’s our answer. Alternatively, start with the row matrix (0 1 0) and project it forward 3 years. (0 1 0 ) 1Q = (0.00 0.70 0.30)

(0 0.70 0.30) 2Q = (0.07 0.49 0.44)

(0.07 0.49 0.44) 3Q = (0.16 0.35 0.49) Thus, the probability that it is in state 3 after three transitions is 0.49. Yet another approach would be to multiply 1 2 3Q Q Q× × , and take the entry in row 2, column 3. That would work but it requires more effort.

Page 45: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #6 Key: B Probabilities of being in each state at time t:

t Active Disabled Dead Deaths 0 1.0 0.0 0.0 1 0.8 0.1 0.1 0.1 2 0.65 0.15 0.2 0.1 3 not needed not needed 0.295 0.095

We built the Active Disabled Dead columns of that table by multiplying each row times the transition matrix. E.g., to move from t = 1 to t = 2, (0.8 0.1 0.1) Q = (0.65 0.15 0.2) The deaths column is just the increase in Dead. E.g., for t = 2, 0.2 – 0.1 = 0.1. v = 0.9 APV of death benefits = ( )2 3100,000* 0.1 0.1 0.095 24,025.5v v v+ + =

APV of $1 of premium = 21 0.8 0.65 2.2465v v+ + =

24,025.5Benefit premium = 10,6952.2465

=

Page 46: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #7 Key: A Split into three independent processes: Deposits, with ( )( )( )* 0.2 100 8 160λ = = per day

Withdrawals, with ( )( )( )* 0.3 100 8 240λ = = per day Complaints. Ignore, no cash impact. For aggregate deposits, ( ) ( )( )160 8000 1,280,000E D = =

( ) ( )( ) ( )( )2 2160 1000 160 8000Var D = +

101.04 10= × For aggregate withdrawals ( ) ( )( )240 5000 1,200,000E W = =

( ) ( )( ) ( )( )2 2240 2000 240 5000Var W = +

100.696 10= × ( ) 1,200,000 1,280,000 80,000E W D− = − = −

( ) 10 10 100.696 10 1.04 10 1.736 10Var W D− = × + × = ×

( ) 131,757SD W D− =

( ) ( ) 80,000 80,000Pr Pr 0 Pr131,757 131,757

W DW D W D − +⎛ ⎞> = − > = >⎜ ⎟⎝ ⎠

( )1 0.6070.27

= −Φ

=

Page 47: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #8 Key: D Exponential inter-event times and independent implies Poisson process (imagine additional batteries being activated as necessary; we don’t care what happens after two have failed). Poisson rate of 1 per year implies failures in 3 years is Poisson with 3λ = .

x f(x) F(x) 0 0.050 0.050 1 0.149 0.199

Probe works provided that there have been fewer than two failures, so we want F(1) = 0.199. Alternatively, the sum of two independent exponential 1θ = random variables is Gamma with

2, 1α θ= =

( ) ( ) ( )( )

3

0

3

03

13 2;32

1

1 40.80 is probability 2 have occurred

t

t

F t e dt

t e

e

= Γ =Γ

= − −

= −=

1 – 0.80 = 0.20

Page 48: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #9 Key: B

45 15 45 4545:15 60:151000 1000P a a E Aπ+ × =

( ) ( )( )4545 15 45 60 60 15 60 75 15 45 45

451000 1000A a E a a E a E A

aπ− + − =

( )( )(( )201.20 14.1121 0.72988 0.51081 11.145414.1121

( )( )( )( ) ( )( )11.1454 0.68756 0.39994 7.2170 0.72988 0.51081 201.20π+ − × = where 15 xE was evaluated as 5 10 5x xE E +×

( ) ( )( )14.2573 9.9568 3.4154 201.20π+ =

17.346π = Question #10 Key: A

( ) ( ) ( )1 0 1 11 1000 xV V i V V qπ= + + − + −

( )( ) ( )2 1 2 2 11 2000 2000xV V i V V qπ += + + − + − =

( )(( ) )( )

( )( )( )( )11 1000 1 2000 2000

1.08 1000 0.1 1.08 2000 0.1 2000

x xi q i qπ π

π π

++ − + + − =

− × + − × =

1027.42π =

Page 49: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #11 Key: A Let Y be the present value of payments to 1 person. Let S be the present value of the aggregate payments.

[ ] ( )1500 500 5572.68x

xA

E Y ad−

= = =

[ ] ( ) ( )2 2 22

1Var 500 1791.96Y x xY A Ad

σ = = − =

( ) [ ]1 2 250...

250 1,393,170S Y Y YE S E Y= + + +

= =

250 15.811388 28,333S Y Yσ σ σ= × = =

( ) 1,393,170 1,393,1700.90 Pr Pr28,333 28,333

S FS F − −⎡ ⎤= ≤ = ≤⎢ ⎥⎣ ⎦

( ) 1,393,170Pr 0,128,333

FN −⎡ ⎤≈ ≤⎢ ⎥⎣ ⎦

( )( )0.90 Pr 0,1 1.28N= ≤

( )1,393,170 1.28 28,333F = + =1.43 million

Page 50: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #12 Key: A

′ = − ′ = −q p pq

q41

141

1411 1

41 1

41b g b g b ge jb gb gτ τ

( ) ( ) ( ) ( )1 2

41 40 40 40 1000 60 55 885l l d dτ τ= − − = − − =

( ) ( ) ( ) ( )1 241 41 41 42 885 70 750 65d l d lτ τ= − − = − − =

( )41

750885

p τ = ( )

( )

141

41

65135

qq τ =

( )65

1351

417501 0.0766885

q ⎛ ⎞′ = − =⎜ ⎟⎝ ⎠

Question #13 Key: D

( ) 1 xs xα

ω⎛ ⎞= −⎜ ⎟⎝ ⎠

( ) ( )( )logdx s xd x x

αµω

= =−

01

1x

xt xe dt

x

αω ω

ω α− −⎛ ⎞= − =⎜ ⎟− +⎝ ⎠∫

new new old0 old new

1 2 12 1 1

e ω ω α αα α

= × = ⇒ = ++ +

( )old old

new old0

2 1 9 44

α αµ αω ω+

= = × ⇒ =

Page 51: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #14 Key: C Constant force implies exponential lifetime where 1/θ µ= using the Loss Models parameterization

[ ] [ ]( )2

222 2

2 1 1 100Var T E T E Tµµ µ

⎛ ⎞⎡ ⎤= − = − = =⎜ ⎟⎣ ⎦ ⎝ ⎠

0.1µ =

[ ]10 10

0

110 6.3t eE T e dtµ

µ

µ

−− −

∧ = = =∫

Alternatively, the formula for ( )E X x∧ is given in the tables handout.

Note that since T is the future lifetime random variable, ( )10E T ∧ can also be written as

:10xe , which for the exponential distribution (constant force of mortality) is independent of x. Question #15 Key: A % premium amount for 15 years

( ) ( )( ):15 :15100,000 0.08 0.02 5 5x xx xGa A G Ga x a= + + + − +

Per policy for life

( ) )(( ) ( )( )( ) ( )( )4669.95 11.35 51,481.97 0.08 4669.95 0.02 11.35 4669.95 5 5 xx a= + + + − +

1 1 0.5148197 16.660.02913x

Axad− −

= = =

( )53,003.93 51,481.97 1433.67 5 83.30x= + + − +

( )4.99 59.99

xx

= −

=

The % of premium expenses could equally well have been expressed as :140.10 0.02 xG G a+ . The per policy expenses could also be expressed in terms of an annuity-immediate.

Page 52: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #16 Key: D For the density where ( ) ( ) ,T x T y≠

( ) ( )( ) 40

0 0Pr 0.0005

y

y xT x T y dxdy

= =< = ∫ ∫

40

0 00.0005

y

yx dy

== ∫

40

00.0005ydy= ∫

2 40

0

0.00052

y=

0.40= For the overall density,

( ) ( )( )Pr 0.4 0 0.6 0.4 0.24T x T y< = × + × =

where the first 0.4 is the probability that ( ) ( )T x T y= and 0.6 is the probability that

( ) ( )T x T y≠ .

Page 53: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #17 Key: D The following derives the general formula for the statistic to be forgotten by time x. It would

work fine, and the equations would look simpler, if you immediately plugged in 12

x = , the only

value you want. Then the 12

x + becomes 1.

Let X be the random variable for when the statistic is forgotten. Then ( ) 1 xyXF x y e−= −

For the unconditional distribution of X, integrate with respect to y

( ) ( ) ( )

( )

( )

12

22

0

0

212

112 2

114

114

yxy

X

y x

yF x e e dyy

y e d y

x

Γ

∞−−

∞− +

⎛ ⎞= − ⎜ ⎟⎝ ⎠

= −

= −+

( )( )

12 21 1

2 2

11 0.754

F = − =+

There are various ways to evaluate the integral in the second line: 1. Calculus, integration by parts

2. Recognize that 12

0

12

y xy x e dy

⎛ ⎞− +⎜ ⎟∞ ⎝ ⎠⎛ ⎞+⎜ ⎟⎝ ⎠∫

is the expected value of an exponential random variable with 112

xθ =

+

3. Recognize that ( )12212

2

y xx y e

⎛ ⎞− +⎜ ⎟⎝ ⎠⎛ ⎞Γ +⎜ ⎟

⎝ ⎠ is the density function for a Gamma random

variable with α = 2 and 112

xθ =

+, so it would integrate to 1.

(Approaches 2 and 3 would also work if you had plugged in 12

x = at the start. The resulting θ

becomes 1).

Page 54: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #18 Key: D State# Number Probability

of needing Therapy

Mean Number of visits

E(X)

E(N) Var(N) Var(X) E(S) Var(S)

1 400 0.2 2 80 64 6 160 736 2 300 0.5 15 150 75 240 2,250 52,875 3 200 0.3 9 60 42 90 540 8,802 2,950 62,413

( )Std Dev 62413 250S = =

( ) ( )2950 50Pr 3000 Pr 1 0.2 0.42250 250

SS −⎛ ⎞> = > = −Φ =⎜ ⎟⎝ ⎠

The ( )Var X column came from the formulas for mean and variance of a geometric distribution. Using the continuity correction, solving for ( )Pr 3000.5S > , is theoretically better but does not affect the rounded answer.

Page 55: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #19 Key: B Frequency is geometric with 2β = , so

0 1/ 3,p = 1 2 / 9,p = 2 4 / 27p = Convolutions of ( )Xf x needed are

x f 2*f 5 0.2 0 10 0.3 0.04

so ( )0 1/ 3,Sf = ( ) ( )5 2 / 9 0.2 0.044Sf = = , ( ) ( ) ( )10 2 / 9 0.3 4 / 27 0.04 0.073Sf = + =

( ) ( )( ) ( )( ) ( )( )0.2 5 0.3 10 0.5 20 14E X = + + =

[ ] ( )2 28E S E X= =

[ ] [ ] ( )( ) ( )( ) ( )( )15 5 1 0 5 1 5 5 1 10E S E S F F F+

− = − − − − − −

( ) ( ) ( )28 5 1 1/ 3 5 1 1/ 3 0.044 5 1 1/ 3 0.044 0.073= − − − − − − − − − 28 3.33 3.11 2.75 18.81= − − − = Alternatively, [ ] [ ] ( ) ( ) ( )

( ) ( ) ( )

15 15 15 0 10 5 5 10

128 15 15 10 0.044 5 0.0733

18.81

S S SE S E S f f f+

− = − + + +

⎛ ⎞= − + + +⎜ ⎟⎝ ⎠

=

Question #20 Key: B

The conditional expected value of the annuity, given µ , is 10.01 µ+

.

The unconditional expected value is 0.02

0.01

1 0.01 0.02100 100 ln 40.50.01 0.01 0.01xa dµ

µ+⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠∫

100 is the constant density of µ on the internal [ ]0.01,0.02 . If the density were not constant, it would have to go inside the integral.

Page 56: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #21 Key: E

Recall 2x

xe ω −=

:: x x x xx xe e e e= + −

: 01 1

xx x

t te dtx y

ω

ω ω− ⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

Performing the integration we obtain

: 3x xxe ω −

=

( ):

23x x

xe

ω −=

(i) ( ) ( )2 2 2 33 2 7

3 3a a

aω ω

ω− −

= × ⇒ =

(ii) ( ) ( )2 323 3

aa k

ωω

−− = ×

( )3.5 3.5 3a a k a a− = − 5k =

The solution assumes that all lifetimes are independent. Question #22 Key: B

Upon the first death, the survivor receives 0.1010,000 10,000 71430.10 0.04

µµ δ

⎛ ⎞= =⎜ ⎟+ +⎝ ⎠

The actuarial present value of the insurance of 7143 is

( ) 0.127,143 7,143 53570.12 0.04

xy

xy

µµ δ

⎛ ⎞= =⎜ ⎟+ +⎝ ⎠

If the force of mortality were not constant during each insurance period, integrals would be required to express the actuarial present value.

Page 57: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #23 Key: E Let 0k p = Probability someone answers the first k problems correctly.

( )22 0 0.8 0.64p = = ( )4

4 0 0.8 0.41p = =

( )2 22 0:0 2 0 0.64 0.41p p= = = ( )2

4 0:0 0.41 0.168p = =

2 2 0 2 0 2 0:00:0 0.87p p p p= + − = 4 0:0 0.41 0.41 0.168 0.652p = + − = Prob(second child loses in round 3 or 4) = 2 40:0 0:0p p− = 0.87-0.652 = 0.218

Prob(second loses in round 3 or 4 second loses after round 2) = 2 40:0 0:0

2 0:0

p pp−

= 0.218 0.250.87

=

Question #24 Key: E If (40) dies before 70, he receives one payment of 10, and Y = 10. Under DeMoivre, the probability of this is (70 – 40)/(110 – 40) = 3/7 If (40) reaches 70 but dies before 100, he receives 2 payments.

3010 20 16.16637Y v= + = The probability of this is also 3/7. (Under DeMoivre, all intervals of the same length, here 30 years, have the same probability). If (40) survives to 100, he receives 3 payments.

30 6010 20 30 19.01819Y v v= + + = The probability of this is 1 – 3/7 – 3/7 = 1/7 ( ) ( ) ( ) ( )3/ 7 10 3/ 7 16.16637 1/ 7 19.01819 13.93104E Y = × + × + × =

( ) ( ) ( ) ( )2 2 2 23/ 7 10 3/ 7 16.16637 1/ 7 19.01819 206.53515E Y = × + × + × =

( ) ( ) ( ) 22 12.46Var Y E Y E Y= − =⎡ ⎤⎣ ⎦

Since everyone receives the first payment of 10, you could have ignored it in the calculation.

Page 58: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #25 Key: C

( ) ( )( ) ( )( )( ) ( )( )( )

21

10

2 3

E

300 0.02 350 0.98 0.04 400 0.98 0.96 0.06

36.8

kk k x x k

kZ v b p q

v v v

++ +

=

=

⎡ ⎤= × + +⎣ ⎦=

( ) ( )( ) ( ) ( )( ) ( )( )

2 22 11

02 22 4 6 2300 0.02 350 0.98 0.04 400 0.98 0.96 0.06

11,773

kk k x x k

kE Z v b p q

v v v

++ +

=

=

= × + +

=

[ ] ( ) ( )22

211,773 36.810,419

Var Z E Z E Z= −

= −=

Question #26 Key: E

( ) ( )4 4

0 0420

4 1 1 0.02

1 0.01

0.84

X XS f x dx x dx

x

= − = −

= −

=

∫ ∫

( ) ( )( )

( ) ( )24 0.02 40.0238 4

4 0.84pX

YX

f y yf y y

S+ +

= = = +

( ) ( )( )3 2 66

0 0

40.0238 4 0.02383 2

3.4272

p y yE Y y y dy⎛ ⎞

= + = +⎜ ⎟⎝ ⎠

=

Page 59: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #27 Key: E By Theorem 4.51 (on page 93 of the second edition of Loss Models), probability of zero claims = pgf of negative binomial applied to the probability that Poisson equals 0. For the Poisson, ( )0f e λ−=

So ( ) ( ) 20.067 1 1 1 3 1

re eλ λβ

− −− −⎡ ⎤ ⎡ ⎤= − − = − −⎣ ⎦ ⎣ ⎦

Solving gives 3λ =

Page 60: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #28 Key: D For any deductible d and the given severity distribution ( ) ( )

( )

( )

30003000 3000 13000

300030003000

E X d E X E X d

d

d

+− = − ∧

⎛ ⎞= − −⎜ ⎟+⎝ ⎠⎛ ⎞= ⎜ ⎟+⎝ ⎠

So ( )( )200530001.2 3000 30003600

P ⎛ ⎞= =⎜ ⎟⎝ ⎠

The following paragraph just clarifies the notation in the rest of the solution: Let r denote the reinsurer’s deductible relative to losses (not relative to reinsured claims). Thus if 1000r = (we are about to solve for r), then on a loss of 4000, the insured collects 4000 – 600 = 3400, the reinsurer pays 4000 – 1000 = 3000, leaving the primary insurer paying 400. Another way, exactly equivalent, to express that reinsurance is that the primary company pays the insured 3400. The reinsurer reimburses the primary company for its claims less a deductible of 400 applied to claims. So the reinsurer pays 3400 – 400 = 3000, the same as before. Expected reinsured claims in 2005

( ) 3000 9,000,00030003000 3000r r

⎛ ⎞= =⎜ ⎟+ +⎝ ⎠

( ) ( )2005 20059,000,0001.1 0.553000

R Pr

⎛ ⎞= =⎜ ⎟+⎝ ⎠

( )( )9,900,000 0.55 3000 16503000 r

= =+

3000r = In 2006, after 20% inflation, losses will have a two-parameter Pareto distribution with 2α = and

( )( )1.2 3000 3600θ = = . The general formula for claims will be

( ) ( ) 3600 12,960,00036003600 3600

E X dd d+

⎛ ⎞− = =⎜ ⎟+ +⎝ ⎠

200612,960,0001.2 37033000 600

P ⎛ ⎞= =⎜ ⎟+⎝ ⎠

Page 61: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

200612,960,0001.1 21603600 3000

R ⎛ ⎞= =⎜ ⎟+⎝ ⎠

2006 2006/ 0.5833R P =

[If you applied the reinsurer’s deductible to the primary insurer’s claims, you would solve that the deductible is 2400, and the answer to the problem is the same]. Question #29 Key: A

Benefits + Expenses – Premiums

0 eL = 31000v + ( ) ( ) ( ) 20.20 8 0.06 2 0.06 2G G v G v+ + + + + 3G a−

( )0

at 41.20 and 0.05, for 2 770.59e

G iL K

= =

= =

Page 62: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #30 Key: D

401000P P= ( )( ) ( )235 1 0.015 1000 255 255P i+ + − − = [A] ( )( ) ( )255 1 0.020 1000 272 272P i+ + − − = [B] Subtract [A] from [B] ( )20 1 3.385 17i+ − =

20.3851 1.0192520

i+ = =

Plug into [A] ( )( ) ( )235 1.01925 0.015 1000 255 255P+ − − =

255 11.1752351.01925

P ++ =

261.15 235 26.15P = − =

( )( ) ( )( )25 40

272 26.15 1.01925 0.025 10001000

1 0.025V

+ −=

286=

Page 63: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #31 Key: A

1

0

0.50.4

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40 50 60 70 80 90 100

Given Given Given Given x 0 15 25 35 75 90 100 ( )s x 1 0.70 0.50 0.48 0.4 0.16 0

Linear Interpolation Linear

Interpolation Linear Interpolation

( )( )55 3590 0.16 321 1 0.666735 0.48 48

sq

s= − = − = =

( ) ( )

( )1520 5535 90 0.48 0.16 32 0.4571

15 0.70 70s s

qs− −

= = = =

1520 55

55 35

0.4571 0.68560.6667

q

q= =

Alternatively,

( )( )

1520 55 20 15 55 3520 15

55 35 55 35

3515

0.480.700.6856

q sp q pq q s

×= = =

=

=

Page 64: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #32 Key: A ( ) ( ) ( )( )180 * ^ 0.16*50 ^ 0.08*50 0.009325552s e e= − + − =

( ) ( ) ( )( )181 * ^ 0.16*51 ^ 0.08*51 0.0085966642s e e= − + − =

( ) ( )80 81 / 80 0.008596664 / 0.00932555 0.9218p s s= = =

80 1 0.9218 0.078q = − = Alternatively (and equivalent to the above) For non-smokers, 0.08 0.923116xp e−= = 50 0.018316xp =

For smokers, 0.16 0.852144xp e−= = 50 0.000335xp = So the probability of dying at 80, weighted by the probability of surviving to 80, is

( ) ( )0.018316 1 0.923116 0.000335 1 0.8521440.018316 0.000335

× − + × −+

= 0.078

Page 65: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #33 Key: B

x ( )xlτ ( )1

xd ( )2xd

40 2000 20 60 41 1920 30 50 42 1840 40

because 2000 20 60 1920− − = ; 1920 30 50 1840− − = Let premium = P

APV premiums = 22000 1920 1840 2.7492000 2000 2000

v v P P⎛ ⎞+ + =⎜ ⎟⎝ ⎠

APV benefits = 2 320 30 401000 40.412000 2000 2000

v v v⎛ ⎞+ + =⎜ ⎟⎝ ⎠

40.41 14.72.749

P = =

Page 66: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #34 Key: C Consider Disease 1 and other Diseases as independent Poisson processes with respective

( ) 1' 0.16 0.0116

sλ ⎛ ⎞= =⎜ ⎟⎝ ⎠

and ( ) 150.16 0.1516⎛ ⎞ =⎜ ⎟⎝ ⎠

respectively. Let 1S = aggregate losses from

Disease 1; 2S = aggregate losses from other diseases. ( )1 100 0.01 5 5E S = × × =

( ) ( )( )

2 21

2

100 0.01 50 5 2525

100 0.15 10 150

Var S

E S

= × × + =

= × × =

( ) ( )2 22 100 0.15 20 10 7500Var S = × × + =

If no one gets the vaccine: ( )( )

5 150 155

2525 7500 10,025

E S

Var S

= + =

= + =

( )0.7 1 0.24

155 0.7 10,025 225.08A

Φ = −

= + =

If all get the vaccine, vaccine cost = ( )( )100 0.15 15= No cost or variance from Disease 1

15 150 0.7 7500 225.62/ 0.998

BA B= + + =

=

Page 67: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #35 Key: A

For current model f x exb g = −1

44

Let g(x) be the new density function, which has (i) g(x) = c, 0 3x≤ ≤ (ii) ( ) / 4 , 3*xg x ke x−= >

(iii) 3/ 4c ke−= , since continuous at x = 3 Since g is density function, it must integrate to 1.

3 34 4 4

311 3 3 4 3 4 7

xc ke d x ke ke c c c

− − −∞= + = + = + ⇒ =∫

( )3 3

3700

13 0.437

F cd x d x= = = =∫ ∫

*This could equally well have been written ( ) / 414

xg x d e−⎛ ⎞= ×⎜ ⎟⎝ ⎠

, then let k = d/4, or even carry

the d/4 throughout.

Page 68: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #36 Key: A

10 0.08 0.05 0.08 0.0830 100 0

t t txa e e dt E e e dt

∞− − − −= +∫ ∫

( )

10 0.13 1.3 0.160 00.13 0.1610 1.3

000.13 0.16

t

t t

e dt e e dt

e ee

∞− − −

− − ∞−

= +

− −+

∫ ∫

1.3 1.31

0.13 0.13 0.16e e−−

= + +

= 7.2992

( ) ( )

( )

10 0.08 0.05 1.3 0.1630 0 0

1.3 1.3

0.05 0.08

10.05 0.080.13 0.13 0.16

0.41606

t t tA e e dt e e dt

e e

∞− − − −

− −

= +

⎛ ⎞= − +⎜ ⎟

⎝ ⎠=

∫ ∫

= ( ) 3030

30

0.41606 0.0577.29923

AP Aa

= = =

401 1

0.08 0.08 0.16a = =

+

40 401A aδ= − ( )1 0.08/ 0.16 0.5= − =

( ) ( )10 40 40 40 40V A A P A a= −

( )0.0570.5 0.14375

0.16= − =

Question #37 Key: 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).

[ ] 1100,000 1600T

TL v a

+= − 0 10T< ≤

10100,000 1600Tv a= − 10 20t< ≤ 101600a− 20<t Minimum is 101600a− when evaluated at i = 0.05 12,973= −

Page 69: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #38 Key: C Since loss amounts are uniform on (0, 10), 40% of losses are below the deductible (4), and 60% are above. Thus, claims occur at a Poisson rate ( )( )* 0.6 10 6λ = = . Since loss amounts were uniform on (0, 10), claims are uniform on (0, 6). Let N = number of claims; X = claim amount; S = aggregate claims. ( ) ( ) * 6E N Var N λ= = =

( ) ( )( ) ( )2

6 0 / 2 3

6 0 /12 3

E X

Var X

= − =

= − =

( ) ( ) ( ) ( ) ( ) 2

26*3 6*3

Var S E N Var X Var N E X= + ⎡ ⎤⎣ ⎦

= +

72= Question #39 Key: E

n np nn p× 2nn p×

0 0.1 0 0 1 0.4 0.4 0.4 2 0.3 0.6 1.2 3 0.2 0.6 1.8

[ ]E N =1.6 2E N⎡ ⎤ =⎣ ⎦ 3.4

( ) 23.4 1.6Var N = − = 0.84

[ ]( )

3

3

E X

Var X

λ

λ

= =

= =

( )Var S =

[ ] ( ) ( ) ( )2E N Var X E X Var N= + ×

( ) ( ) ( )21.6 3 3 0.84= + 12.36=

Page 70: Exam M Fall 2005 PRELIMINARY ANSWER KEY - MATHExam M Fall 2005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 21 E 2 C 22 B 3 C 23 E 4 D 24 E 5 C 25 C 6 B 26 E 7 A

Question #40 Key: B Method 1: as three independent processes, based on the amount deposited. Within each process, since the amount deposited is always the same, ( ) 0Var X = . Rate of depositing 10 = 0.05 * 22 = 1.1 Rate of depositing 5 = 0.15 * 22 = 3.3 Rate of depositing 1 = 0.80 * 22 = 17.6 Variance of depositing 10 = 1.1 * 10 * 10 = 110 Variance of depositing 5 = 3.3 * 5 * 5 = 82.5 Variance of depositing 1 = 17.6 * 1 *1 = 17.6 Total Variance = 110 + 82.5 + 17.6 = 210.1 Method 2: as a single compound Poisson process ( ) 0.8 1 0.15 5 0.05 10 2.05E X = × + × + × =

( )2 2 2 20.8 1 0.15 5 0.05 10 9.55E X = × + × + × =

( ) ( ) ( ) ( ) ( )( )( )( ) ( )( )

2

222 5.3475 22 2.05

210.1

Var S E N Var X Var N E X= +

= +

=