Exam M Fall 2005 PRELIMINARY ANSWER KEY

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1 Exam M Fall 005 PRELIMINARY ANSWER KEY Question # Answer Question # Answer 1 C 1 E C B 3 C 3 E 4 D 4 E 5 C 5 C 6 B 6 E 7 A 7 E 8 D 8 D 9 B 9 A 10 A 30 D 11 A 31 A 1 A 3 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 0 B 40 B

2 **BEGINNING OF EXAMINATION** 1. For a special whole life insurance on (x), you are given: (i) (ii) Z is the present value random variable for this insurance. Death benefits are paid at the moment of death. (iii) µ x () t = 0.0, t 0 (iv) δ = 0.08 (v) 0.03 t b = e, t 0 t Calculate Var ( Z ). (A) (B) (C) (D) (E) Exam M: Fall GO ON TO NEXT PAGE

3 . For a whole life insurance of 1 on (x), you are given: (i) (ii) (iii) Benefits are payable at the moment of death. Level premiums are payable at the beginning of each year. Deaths are uniformly distributed over each year of age. (iv) i = 0.10 (v) a = 8 (vi) a + 10 = 6 x x Calculate the 10 th year terminal benefit reserve for this insurance. (A) 0.18 (B) 0.5 (C) 0.6 (D) 0.7 (E) 0.30 Exam M: Fall GO ON TO NEXT PAGE

4 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 bg b g = x t , t 0 bg b g = 1 (ii) µ x t , t 0, where µ bg 1 x is the force of decrement due to death by accidental means. (iii) δ = 006. Calculate the single benefit premium for this insurance. (A) 1640 (B) 1710 (C) 1790 (D) 1870 (E) 1970 Exam M: Fall GO ON TO NEXT PAGE

5 4. Kevin and Kira are modeling the future lifetime of (60). (i) Kevin uses a double decrement model: x ( ) l τ ( 1) x d x ( ) d x (ii) (iii) Kira uses a non-homogeneous Markov model: (a) The states are 0 (alive), 1 (death due to cause 1), (death due to cause ). (b) Q 60 is the transition matrix from age 60 to 61; Q 61 is the transition matrix from age 61 to 6. The two models produce equal probabilities of decrement. Calculate Q 61. (A) (B) (C) (D) (E) Exam M: Fall GO ON TO NEXT PAGE

6 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 Q i as follows: Q 1 = Sustainable Endangered Extinct Sustainable Endangered Extinct Q Q 3 Qi = = = , i = 4,5, 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 Exam M: Fall GO ON TO NEXT PAGE

7 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 Disabled Dead (ii) (iii) A 100,000 benefit is payable at the end of the year of death whether the insured was active or disabled. 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) 13,00 (E) 0,800 Exam M: Fall GO ON TO NEXT PAGE

8 7. Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 0% 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 Withdrawal amounts have mean 5000 and standard deviation 000. 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) 0.7 (B) 0.30 (C) 0.33 (D) 0.36 (E) 0.39 Exam M: Fall GO ON TO NEXT PAGE

9 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) 0.0 (E) 0.5 Exam M: Fall GO ON TO NEXT PAGE

10 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 1000P 45 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) i = 0.06 Calculate π. (A) 16.8 (B) 17.3 (C) 17.8 (D) 18.3 (E) 18.8 Exam M: Fall GO ON TO NEXT PAGE

11 10. For a special fully discrete -year endowment insurance on (x): (i) The pure endowment is 000. (ii) The death benefit for year k is ( 1000k ) plus the benefit reserve at the end of year k, k = 1,. (iii) π is the level annual benefit premium. (iv) i = 0.08 (v) px+ k 1 0.9, k 1, Calculate π. (A) 107 (B) 1047 (C) 1067 (D) 1087 (E) 1107 Exam M: Fall GO ON TO NEXT PAGE

12 11. For a group of 50 individuals age x, you are given: (i) (ii) The future lifetimes are independent. Each individual is paid 500 at the beginning of each year, if living. (iii) A x = (iv) A = x (v) i = 0.06 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) (B) (C) (D) (E) 1.43 million 1.53 million 1.63 million 1.73 million 1.83 million Exam M: Fall GO ON TO NEXT PAGE

13 1. For a double decrement table, you are given: Age ( ) l τ ( 1) ( ) x d x d x Each decrement is uniformly distributed over each year of age in the double decrement table. Calculate () 1 q. 41 (A) (B) (C) (D) (E) Exam M: Fall GO ON TO NEXT PAGE

14 13. The actuarial department for the SharpPoint Corporation models the lifetime of pencil α s x = 1 x/ ω, for sharpeners from purchase using a generalized DeMoivre model with ( ) ( ) α > 0 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) (ii) (iii) The new complete expectation of life at purchase is half what it was previously. The new force of mortality for pencil sharpeners is.5 times the previous force of mortality for all durations. ω remains the same. Calculate the original value of α. (A) 1 (B) (C) 3 (D) 4 (E) 5 Exam M: Fall GO ON TO NEXT PAGE

15 14. You are given: (i) T is the future lifetime random variable. (ii) µ () t = µ, t 0 (iii) Var[ T ] = 100. Calculate E[ T 10]. (A).6 (B) 5.4 (C) 6.3 (D) 9.5 (E) 10.0 Exam M: Fall GO ON TO NEXT PAGE

16 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 (ii) 100,000A x = 51, (iii) a = x:15 (iv) d = (v) (vi) (vii) Expenses are incurred at the beginning of the year. Percent of premium expenses are 10% in the first year and % thereafter. 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) 3.0 (D) 9.5 (E) 36.5 Exam M: Fall GO ON TO NEXT PAGE

17 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 ft( x), T( y) ( t, s ) = , 0< t < 40, 0< s < 50 Calculate Prob T( x) < T( y). (A) 0.30 (B) 0.3 (C) 0.34 (D) 0.36 (E) 0.38 Exam M: Fall GO ON TO NEXT PAGE

18 17. The length of time, in years, that a person will remember an actuarial statistic is modeled by an exponential distribution with mean 1. In a certain population, Y has a gamma Y distribution with α = θ =. Calculate the probability that a person drawn at random from this population will remember an actuarial statistic less than 1 year. (A) 0.15 (B) 0.50 (C) (D) (E) Exam M: Fall GO ON TO NEXT PAGE

19 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 #) 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 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) 0.1 (B) 0.7 (C) 0.34 (D) 0.4 (E) 0.50 Exam M: Fall GO ON TO NEXT PAGE

20 19. In a given week, the number of projects that require you to work overtime has a geometric distribution with β =. For each project, the distribution of the number of overtime hours in the week is the following: x f ( x ) 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).1 (D) 6. (E) 8.0 Exam M: Fall GO ON TO NEXT PAGE

21 0. 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, 0.0 ]. (ii) δ = 0.01 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) 4.3 Exam M: Fall GO ON TO NEXT PAGE

22 1. For a population whose mortality follows DeMoivre s law, you are given: (i) e = 3e 40:40 60:60 (ii) e = ke 0:0 60:60 Calculate k. (A) 3.0 (B) 3.5 (C) 4.0 (D) 4.5 (E) 5.0 Exam M: Fall GO ON TO NEXT PAGE

23 . 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) µ () t µ () t 0.06 x (iii) µ () t = 0.1 xy = = while both are alive. y (iv) After the first death, µ () t = 0.10 for the survivor. (v) δ = 0.04 Calculate the actuarial present value of this insurance on (x) and (y). (A) 4500 (B) 5400 (C) 6000 (D) 7100 (E) 7500 Exam M: Fall GO ON TO NEXT PAGE

24 3. Kevin and Kira are in a history competition: (i) (ii) 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. 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. (E) 0.5 Exam M: Fall GO ON TO NEXT PAGE

25 4. For a special increasing whole life annuity-due on (40), you are given: (i) (ii) (iii) Y is the present-value random variable. Payments are made once every 30 years, beginning immediately. The payment in year 1 is 10, and payments increase by 10 every 30 years. (iv) Mortality follows DeMoivre s law, with ω = 110. (v) i = 0.04 Calculate Var( Y ). (A) 10.5 (B) 11.0 (C) 11.5 (D) 1.0 (E) 1.5 Exam M: Fall GO ON TO NEXT PAGE

26 5. For a special 3-year term insurance on ( x ), you are given: (i) Z is the present-value random variable for this insurance. (ii) qx+ k= 00. ( k + 1), k = 0, 1, (iii) The following benefits are payable at the end of the year of death: k b k (iv) i = 006. Calculate VarbZg. (A) 9,600 (B) 10,000 (C) 10,400 (D) 10,800 (E) 11,00 Exam M: Fall GO ON TO NEXT PAGE

27 6. For an insurance: (i) Losses have density function f X ( x) 0.0x 0 < x< 10 = 0 elsewhere (ii) (iii) The insurance has an ordinary deductible of 4 per loss. P Y is the claim payment per payment random variable. P Calculate E Y. (A).9 (B) 3.0 (C) 3. (D) 3.3 (E) 3.4 Exam M: Fall GO ON TO NEXT PAGE

28 7. 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 ( ) = 1 3( z 1) P z. (ii) The secondary distribution is the Poisson with probability generating function ( z 1) P z = e λ. ( ) (iii) The probability of no claims equals Calculate λ. (A) 0.1 (B) 0.4 (C) 1.6 (D).7 (E) 3.1 Exam M: Fall GO ON TO NEXT PAGE

29 8. In 005 a risk has a two-parameter Pareto distribution with losses inflate by 0%. α = and θ = In 006 An insurance on the risk has a deductible of 600 in each year. equals 1. times the expected claims. P i, the premium in year i, The risk is reinsured with a deductible that stays the same in each year. premium in year i, equals 1.1 times the expected reinsured claims. R i, the reinsurance R005 P = R Calculate 006 P. 006 (A) 0.46 (B) 0.5 (C) 0.55 (D) 0.58 (E) 0.66 Exam M: Fall GO ON TO NEXT PAGE

30 9. 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 0% 6% Per Policy 8 (ii) The level expense-loaded premium is (iii) i = 0.05 Calculate the value of the expense augmented loss variable, 0 L e, if the insured dies in the third policy year. (A) 770 (B) 790 (C) 810 (D) 830 (E) 850 Exam M: Fall GO ON TO NEXT PAGE

31 30. For a fully discrete whole life insurance of 1000 on (45), you are given: t 1000 t V 45 q 45 + t Calculate 10005V 45. (A) 79 (B) 8 (C) 84 (D) 86 (E) 88 Exam M: Fall GO ON TO NEXT PAGE

32 31. The graph of a piecewise linear survival function, s( x ), consists of 3 line segments with endpoints (0, 1), (5, 0.50), (75, 0.40), (100, 0). Calculate 0 55 q q (A) 0.69 (B) 0.71 (C) 0.73 (D) 0.75 (E) 0.77 Exam M: Fall GO ON TO NEXT PAGE

33 3. For a group of lives aged 30, containing an equal number of smokers and non-smokers, you are given: (i) For non-smokers, ( x) 0.08 (ii) For smokers, ( x) 0.16, s n µ =, x 30 µ = x 30 Calculate q 80 for a life randomly selected from those surviving to age 80. (A) (B) (C) (D) (E) 0.11 Exam M: Fall GO ON TO NEXT PAGE

34 33. For a 3-year fully discrete term insurance of 1000 on (40), subject to a double decrement model: (i) (ii) (iii) x ( ) l τ ( 1) x d x ( ) d x Decrement 1 is death. Decrement is withdrawal. There are no withdrawal benefits. (iv) i = 0.05 Calculate the level annual benefit premium for this insurance. (A) 14.3 (B) 14.7 (C) 15.1 (D) 15.5 (E) 15.7 Exam M: Fall GO ON TO NEXT PAGE

35 34. Each life within a group medical expense policy has loss amounts which follow a compound Poisson process with λ = Given a loss, the probability that it is for Disease 1 is Loss amount distributions have the following parameters: Standard Mean per loss Deviation per loss Disease Other diseases 10 0 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 0.4. 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 Exam M: Fall GO ON TO NEXT PAGE

36 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) (iii) is proportional to the initial modeled density function after 3 years 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 Exam M: Fall GO ON TO NEXT PAGE

37 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) δ = 0.08 Calculate the benefit reserve at time 10 for this insurance. (A) (B) (C) (D) (E) Exam M: Fall GO ON TO NEXT PAGE

38 37. For a 10-payment, 0-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) 1,000 (B) 17,000 (C) 13,000 (D) 1,400 (E) 1,000 Exam M: Fall GO ON TO NEXT PAGE

39 38. For an insurance: (i) The number of losses per year has a Poisson distribution with λ = 10. (ii) Loss amounts are uniformly distributed on (0, 10). (iii) (iv) Loss amounts and the number of losses are mutually independent. There is an ordinary deductible of 4 per loss. Calculate the variance of aggregate payments in a year. (A) 36 (B) 48 (C) 7 (D) 96 (E) 10 Exam M: Fall GO ON TO NEXT PAGE

40 39. For an insurance portfolio: (i) The number of claims has the probability distribution p n n (ii) (iii) Each claim amount has a Poisson distribution with mean 3; and 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. (E) 1.4 Exam M: Fall GO ON TO NEXT PAGE

41 40. Lucky Tom deposits the coins he finds on the way to work according to a Poisson process with a mean of deposits per month. 5% of the time, Tom deposits coins worth a total of % 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) 10 (C) 40 (D) 70 (E) 300 **END OF EXAMINATION** Exam M: Fall STOP

42 Fall 005 Exam M Solutions Question #1 Key: C [ ] = [ ] Var Z E Z E Z t 0.08t 0.03t 0.0t [ ] = ( ) µ () = ( 0.0) E Z v b p t dt e e e dt 0 t t x x t 0.0 = ( 0.0) e dt = = t t t x x 0 0.1t = 0.0e () 1 µ 0 x t dt = = t t ( ) µ () ( ) ( 0.0) E Z = v b p t dt = e e dt 1 6 Var [ Z ] = 1 ( ) = 6 49 = Question # Key: C 0.1 From Ax = 1 dax we have A 1 ( 8) 3 x = = A 10 1 ( 6) 5 x + = = A i x = Ax δ A x A x = = ln ( 1.1) = = ln 1.1 ( ) ( ) V = A P A a 10 x x+ 10 x x = = 0.63 There are many other equivalent formulas that could be used.

43 Question #3 Key: C Regular death benefit t 0.001t = 100,000 e e 0.001dt = 100, = t 0.001t Accidental death 100,000 ( 0.000) = 0 e e dt = e 0.061t dt e = 0 = Actuarial Present Value = = Question #4 Key: D Once you are dead, you are dead. Thus, you never leave state or 3, and rows 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 = 0.0. Probability of dying from cause within the year, given alive at age 61, is 80/800 = 0.10 Probability of surviving to 6, given alive at 61, is 560/800 = 0.70 (alternatively, ), so correct answer is D.

44 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 Q 3 it will never be extinct. This solution parallels the example at the top of page 9 of the Daniel study note. We want the Q Q Q e Q Q Q e. second entry of the product ( ) which is equal to ( ) ( ) Q 0 = Q 0.1 = Q 0.7 = 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 ) Q 1 = ( ) ( ) Q = ( ) ( ) Q 3 = ( ) Thus, the probability that it is in state 3 after three transitions is Yet another approach would be to multiply 1 3 Q Q Q, and take the entry in row, column 3. That would work but it requires more effort.

45 Question #6 Key: B Probabilities of being in each state at time t: t Active Disabled Dead Deaths not needed not needed 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 =, ( ) Q = ( ) The deaths column is just the increase in Dead. E.g., for t =, = 0.1. v = 0.9 APV of death benefits = ( v v v 3 ) 100,000* = 4,05.5 APV of $1 of premium = v+ 0.65v =.465 4,05.5 Benefit premium = 10, =

46 Question #7 Key: A Split into three independent processes: * λ = = 160 per day Deposits, with ( )( )( ) * Withdrawals, with ( )( )( ) λ = = 40 per day Complaints. Ignore, no cash impact. For aggregate deposits, E D = = 1,80,000 ( ) ( )( ) ( ) ( 160)( 1000) ( 160)( 8000) Var D = + 10 = For aggregate withdrawals EW = = 1,00,000 ( ) ( )( ) ( ) ( 40)( 000) ( 40)( 5000) Var W = + 10 = ( D) = 1,00,000 1,80,000 = 80, ( D) = = ( D) = 131,757 EW Var W SD W W D+ 80,000 80,000 Pr( W > D) = Pr( W D> 0) = Pr > 131, ,757 = 1 Φ( 0.607) = 0.7

47 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 λ = 3. x f(x) F(x) Probe works provided that there have been fewer than two failures, so we want F(1) = Alternatively, the sum of two independent exponential θ = 1 random variables is Gamma with α =, θ = t F( 3) =Γ ( ;3) = te dt Γ ( ) 0 t 3 = t 1 e ( ) 3 0 = 1 4e = 0.80 is probability have occurred = 0.0

48 Question #9 Key: B 1000P a + π a E = 1000A 45 45:15 60: A ( ) π ( )( ) a45 15E45 a60 + a60 15 E60 a75 15 E45 = 1000A45 a ( ( )( )( ) π ( ( )( )( 7.170) ) ( )( ) = 01.0 where 15 E x was evaluated as 5 E x 10 E x+ 5 ( ) ( π )( ) = 01.0 π = Question #10 Key: A 1 = ( 0 + π )( + ) ( ) ( )( ) ( ) (( π( ) x) π)( ) (( π( ) ) π)( ) V V 1 i 1000 V V qx V = 1V + π 1+ i V V q x + 1 = i 1000q + 1+ i 000q = 000 x = 000 π = 107.4

49 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. [ ] ( 1 A ) EY = 500a x = 500 = d 1 σ Y = Var[ Y] = ( 500) ( Ax Ax) = d S = Y1 + Y Y50 E S = 50E Y = 1,393,170 ( ) [ ] x σs = 50 σy = σY = 8,333 S 1,393,170 F 1,393, = Pr( S F) = Pr 8,333 8,333 F 1,393,170 Pr N ( 0,1) 8, = Pr N 0,1 1.8 ( ( ) ) F = 1,393, ( 8,333) =1.43 million

50 Question #1 Key: A 1 41 bg bg bg 1 41 q = 1 p = 1 p e τ 41 j bg bg q41 1 q τ 41 ( τ) ( τ) ( 1) ( ) l41 = l40 d40 d40 = = 885 ( 1) ( τ) ( ) ( τ) d41 = l41 d41 l4 = = 65 ( ) 750 p τ 41 = 885 () 1 q41 65 = ( ) q τ () q41 = 1 = Question #13 Key: D α x s( x) = 1 ω d α µ ( x) = log ( s( x) ) = dx ω x ex ω x t ω x = 1 dt 0 = ω x α + 1 α new 1 new old e0 = ω ω α α 1 old new α + 1 = α + 1 = + old old ( new) α α old µ 0 = = α = 4 ω 4 ω

51 Question #14 Key: C Constant force implies exponential lifetime where θ = 1/ µ using the Loss Models parameterization 1 1 Var[ T ] = E T ( E[ T ]) = 100 = = µ µ µ µ = 0.1 [ ] 10 10µ µ t 1 e ET 10 = e dt= = 6.3 µ 0 Alternatively, the formula for E( X x) is given in the tables handout. Note that since T is the future lifetime random variable, ET ( 10) can also be written as e, which for the exponential distribution (constant force of mortality) is independent of x. x:10 Question #15 Key: A % premium amount for 15 years Ga = 100,000 A ( ) (( 5) 5 x:15 x + G + Ga + x + a x:15 x ) Per policy for life ( ) = + ( )( ) + ( )( )( ) + (( x ) + a ) , x a x 1 Ax = = = d ( x ) 53, = 51, ( x ) 4.99 = 5 x = 9.99 The % of premium expenses could equally well have been expressed as 0.10G+ 0.0Ga. The per policy expenses could also be expressed in terms of an annuity-immediate. x:14

52 Question #16 Key: D For the density where T( x) T( y), ( ( ) ( )) 40 y y= 0 x= 0 Pr T x T y dxdy < = = x y dy y= = ydy 0 = = 0.40 y 40 For the overall density, Pr ( T( x) < T( y) ) = = 0.4 T x where the first 0.4 is the probability that ( ) T( y) T( x) T( y). 0 = and 0.6 is the probability that

53 Question #17 Key: D The following derives the general formula for the statistic to be forgotten by time x. It would 1 work fine, and the equations would look simpler, if you immediately plugged in x =, the only 1 value you want. Then the x + becomes 1. xy Let X be the random variable for when the statistic is forgotten. Then FX ( x y) = 1 e For the unconditional distribution of X, integrate with respect to y y 1 y FX x = e e dy y xy ( ) ( 1 ) Γ( ) ( ) 0 1 = 1 4 = ye 1 ( x + ) 1 ( 1 ) dy y x+ 1 F = 1 = ( ) There are various ways to evaluate the integral in the second line: 1. Calculus, integration by parts 1 1 y x+. Recognize that y x+ e dy 0 1 is the expected value of an exponential random variable with θ = 1 x + 3. Recognize that ( ) 1 y x+ 1 Γ x+ ye is the density function for a Gamma random 1 variable with α = and θ =, so it would integrate to 1. 1 x + (Approaches and 3 would also work if you had plugged in becomes 1). 1 x = at the start. The resulting θ

54 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) ,50 5, ,80,950 6,413 Std Dev ( S ) = 6413 = 50 S Pr( S > 3000) = Pr > = 1 Φ ( 0.) = The Var( X ) column came from the formulas for mean and variance of a geometric distribution. Using the continuity correction, solving for Pr( S > ), is theoretically better but does not affect the rounded answer.

55 Question #19 Key: B Frequency is geometric with β =, so p 0 = 1/3, p 1 = /9, p = 4/7 Convolutions of f ( x ) needed are X x f f * so f S ( 0) = 1/3, f S ( 5) = / 9( 0.) = 0.044, S ( ) ( ) ( ) E( X ) = ( 0.)( 5) + ( 0.3)( 10) + ( 0.5)( 0) = 14 E[ S] = E( X) = 8 E[ S 15] + = E[ S] 51 ( F( 0) ) 51 ( F( 5) ) 51 ( F( 10) ) = 8 5( 1 1/ 3) 5( 1 1/ ) 5( 1 1/ ) = = f 10 = / / = Alternatively, [ 15] = [ ] ( 0) + 10 ( 5) + 5 ( 10) E S E S f f f + S S S 1 = = ( ) ( ) ( ) Question #0 Key: B The conditional expected value of the annuity, given µ, is The unconditional expected value is ax = 100 dµ = 100 ln = µ µ. 100 is the constant density of µ on the internal [ 0.01,0.0 ]. If the density were not constant, it would have to go inside the integral.

56 Question #1 Key: E Recall x e ω x = e = e xx : x + ex e x: x xx : e ω x t t = 1 1 dt 0 ω x ω y Performing the integration we obtain ω x exx : = 3 ( ω x) e = xx : 3 (i) ( ω a) ( ω a) 3 = 3 ω = 7a 3 3 ( ω 3a) ω a = k a a= k 3.5a 3a (ii) ( ) k = 5 ( ) The solution assumes that all lifetimes are independent. Question # Key: B µ 0.10 Upon the first death, the survivor receives 10,000 = 10,000 = 7143 µ + δ The actuarial present value of the insurance of 7143 is µ xy 0.1 7,143 = ( 7,143) = 5357 µ + δ xy If the force of mortality were not constant during each insurance period, integrals would be required to express the actuarial present value.

57 Question #3 Key: E Let k p 0 = Probability someone answers the first k problems correctly. ( ) p 0 = 0.8 = 0.64 p = ( ) 4 = ( ) p = p = = ( ) : p 0:0 = 0.41 = p = p0 + 0:0 p0 p0:0 = p = = :0 Prob(second child loses in round 3 or 4) = p 4 p = = :0 0:0 Prob(second loses in round 3 or 4 second loses after round ) = p 4 p p = = 0:0 0:0 0:0 Question #4 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 payments. 30 Y = 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 Y = v + 30v = The probability of this is 1 3/7 3/7 = 1/7 EY = 3/ / / = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) EY = 3/ / / = ( ) ( ) ( ) VarY = EY EY = 1.46 Since everyone receives the first payment of 10, you could have ignored it in the calculation.

58 Question #5 Key: C E k+ 1 ( ) = ( 1) Z v b p q k= 0 k+ k x x+ k 3 ( 300) 0.0 ( 350)( 0.98)( 0.04) 400( 0.98)( 0.96)( 0.06) = v + v + v = 36.8 k+ 1 ( ) = ( 1) k+ k x x+ k k= 0 E Z v b p q ( ) ( ) ( )( ) ( )( ) 4 6 = v v v = 11,773 [ ] = ( ) ( ) Var Z E Z E Z = 11, = 10, 419 Question #6 Key: E ( ) ( ) X 0 S 4 = 1 f x dx= xdx X = x = 0.84 fx ( y+ 4) 0.0( y+ 4) f p ( y) = = = y+ 4 Y S X ( ) 4 0 ( ) p 6 y 4y EY ( ) = y( 0.038( y+ 4) ) dy= =

59 Question #7 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, f ( 0) = e λ r λ λ So = 1 β ( e 1) = 1 3( e 1) Solving gives λ = 3

60 Question #8 Key: D For any deductible d and the given severity distribution E X d = E( X) E X d ( ) ( ) + So P ( )( ) = d 3000 = ( 3000) d 3000 = = 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 r = 1000 (we are about to solve for r), then on a loss of 4000, the insured collects = 3400, the reinsurer pays = 3000, leaving the primary insurer paying 400. Another way, exactly equivalent, to express that reinsurance is that the primary company pays the insured The reinsurer reimburses the primary company for its claims less a deductible of 400 applied to claims. So the reinsurer pays = 3000, the same as before. Expected reinsured claims in ,000,000 = ( 3000) = r r ( ) 9,000,000 R = 1.1 = ( 0.55) P r 9,900,000 ( 0.55)( 3000) r = = r = In 006, after 0% inflation, losses will have a two-parameter Pareto distribution with α = and θ = ( 1.)( 3000) = The general formula for claims will be ,960,000 E( X d) = ( 3600) + = d d 1,960,000 P006 = 1. =

61 R 006 1,960,000 = 1.1 = R006 / P 006 = [If you applied the reinsurer s deductible to the primary insurer s claims, you would solve that the deductible is 400, and the answer to the problem is the same]. Question #9 Key: A Benefits + Expenses Premiums 3 L = 1000v + ( 0.0G+ 8) + ( 0.06G+ ) v+ ( 0.06G+ ) v Ga 3 0 e at G = 41.0 and i = 0.05, 0 L e ( K ) for = =

62 Question #30 Key: D P= 1000P 40 ( P)( i) ( ) = 55 [A] ( P)( i) ( ) = 7 [B] Subtract [A] from [B] ( i) = i = = Plug into [A] ( P)( ) ( ) 1000 V 5 40 = = P = P = = 6.15 ( )( ) ( 0.05)( 1000) =

63 Question #31 Key: A Given Given Given Given x s( x ) q q 15 ( ) ( ) Linear Interpolation s = 1 = 1 = = s ( ) s( ) s( ) Linear Interpolation s = = = = Linear Interpolation 0 55 q q = = Alternatively, q p q s = = 0 p15 = q q s = 0.70 = ( 35) ( 15)

64 Question #3 Key: A ( ) ( ) ( ) 1 ( ) ( ) s 80 = * e^ 0.16*50 + e^ 0.08*50 = s( 81 ) = 1 * e^ ( 0.16*51) + e^ ( 0.08*51) = ( ) ( ) p80 = s 81 / s 80 = / = q 80 = = Alternatively (and equivalent to the above) 0.08 For non-smokers, px = e = p x = For smokers, px = e = p x = So the probability of dying at 80, weighted by the probability of surviving to 80, is ( ) ( ) =

65 Question #33 Key: B x ( ) l τ ( ( ) x d x d x because = 190 ; = 1840 Let premium = P APV premiums = + v+ v P=.749P APV benefits = 1000 v+ v + v = P = =

66 Question #34 Key: C Consider Disease 1 and other Diseases as independent Poisson processes with respective 1 λ ' s = ( 0.16) = 0.01 and ( ) = 0.15 respectively. Let S 1 = aggregate losses from Disease 1; S = aggregate losses from other diseases. E( S 1) = = 5 Var S ( 1) ( ) ( ) E S = = 55 = = 150 ( ) ( ) Var S = = 7500 If no one gets the vaccine: E( S) = = 155 Var S = = 10,05 ( ) ( ) Φ 0.7 = A = ,05 = 5.08 If all get the vaccine, vaccine cost = ( 100)( 0.15) = 15 No cost or variance from Disease 1 B = = 5.6 A/ B= 0.998

67 Question #35 Key: A For current model f x b g = e x Let g(x) be the new density function, which has (i) g(x) = c, 0 x 3 (ii) x g( x) = ke /4, x> 3* (iii) c= ke 3/4, since continuous at x = 3 Since g is density function, it must integrate to 1. 3 x = 3c+ ke dx= 3ke + 4ke = 3c+ 4c c= F( 3) = 3 cdx= dx= 7 = *This could equally well have been written ( ) /4 the d/4 throughout. g x d 1 e x =, then let k = d/4, or even carry 4

68 Question #36 Key: A t t 0.08t 30 = x t = e dt+ e 0 e dt t t e 1.3 e + ( e ) 0 0 a e e dt E e e dt e 1 e = = t 0.05t t A = e e 0.05 dt+ e e 0.08 dt 30 ( ) ( ) e e = ( 0.08) = A = P( A30 ) = = = a a 40 = = A40 = 1 δ a40 = / 0.16 = 0.5 ( ) ( ) = ( ) V A A P A a ( 0.057) = 0.5 = 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). T L 100,000v 1600a [ T ] < T 10 T = 100,000v 1600a < t a 0<t Minimum is a when evaluated at i = = 1,973

69 Question #38 Key: C Since loss amounts are uniform on (0, 10), 40% of losses are below the deductible (4), and 60% λ * = = 6. are above. Thus, claims occur at a Poisson rate ( )( ) 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. E( N) = Var( N) = λ* = 6 E( X) = ( 6 0 )/= 3 Var ( X ) = ( 6 0 ) /1= 3 ( ) = ( ) ( ) + ( ) ( ) Var S E N Var X Var N E X = 6*3+ 6*3 = 7 Question #39 Key: E n p n n pn n p n Var ( N ) = = 0.84 E[ X] = λ = 3 Var ( X ) = λ = 3 Var( S ) = [ ] ( ) ( ) ( ) ( ) ( ) ( ) = E N Var X + E X Var N = = 1.36 E[ N ] = 1.6 = E N 3.4

70 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, ( ) 0 Var X =. Rate of depositing 10 = 0.05 * = 1.1 Rate of depositing 5 = 0.15 * = 3.3 Rate of depositing 1 = 0.80 * = 17.6 Variance of depositing 10 = 1.1 * 10 * 10 = 110 Variance of depositing 5 = 3.3 * 5 * 5 = 8.5 Variance of depositing 1 = 17.6 * 1 *1 = 17.6 Total Variance = = 10.1 Method : as a single compound Poisson process E( X ) = =.05 ( ) E X = = 9.55 ( ) ( ) = ( ) ( ) + ( ) ( ) Var S E N Var X Var N E X ( )( ) ( )(.05 ) = + = 10.1