**BEGINNING OF EXAMINATION** 1. Given: (i) e o 0 = 5 (ii) l = ω, 0 ω (iii) is the future lifetime random variable. T Calculate Var Tb10g. (A) 65 (B) 93 (C) 133 (D) 178 (E) 333 COURSE/EXAM 3: MAY 000-1 - GO ON TO NEXT PAGE
. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins/minute. The denominations are randomly distributed: (i) 60% of the coins are worth 1; (ii) 0% of the coins are worth 5; and (iii) 0% of the coins are worth 10. Calculate the conditional epected value of the coins Tom found during his one-hour walk today, given that among the coins he found eactly ten were worth 5 each. (A) 108 (B) 115 (C) 18 (D) 165 (E) 180 COURSE/EXAM 3: MAY 000 - - GO ON TO NEXT PAGE
3. For a fully discrete two-year term insurance of 400 on ( ): (i) i = 01. 1 (ii) 400 P : = 74. 33 1 (iii) 400 V : = 16. 58 1 (iv) The contract premium equals the benefit premium. Calculate the variance of the loss at issue. (A) 1,615 (B) 3,15 (C) 7,450 (D) 31,175 (E) 34,150 COURSE/EXAM 3: MAY 000-3 - GO ON TO NEXT PAGE
4. You are given: (i) The claim count N has a Poisson distribution with mean Λ. (ii) Λ has a gamma distribution with mean 1 and variance. Calculate the probability that N = 1. (A) 0.19 (B) 0.4 (C) 0.31 (D) 0.34 (E) 0.37 COURSE/EXAM 3: MAY 000-4 - GO ON TO NEXT PAGE
5. An insurance company has agreed to make payments to a worker age who was injured at work. (i) The payments are 150,000 per year, paid annually, starting immediately and continuing for the remainder of the worker s life. (ii) (iii) t p After the first 500,000 is paid by the insurance company, the remainder will be paid by a reinsurance company. (iv) i = 0. 05 R S t t = T 0. 7, 0 55. 0, 55. < t Calculate the actuarial present value of the payments to be made by the reinsurer. (A) Less than 50,000 (B) At least 50,000, but less than 100,000 (C) At least 100,000, but less than 150,000 (D) At least 150,000, but less than 00,000 (E) At least 00,000 COURSE/EXAM 3: MAY 000-5 - GO ON TO NEXT PAGE
6. A special purpose insurance company is set up to insure one single life. The risk consists of a single possible claim. (i) The claim amount distribution is: Amount 100 00 Probability 0.60 0.40 (ii) The probability that the claim does not occur by time t is 1 1+t. (iii) The insurer s surplus at time t is U t = 60 + 0 t S t, where S t is the aggregate claim amount paid by time t. (iv) The claim is payable immediately. Calculate the probability of ruin. (A) (B) (C) (D) (E) 4 7 3 5 3 3 4 7 8 COURSE/EXAM 3: MAY 000-6 - GO ON TO NEXT PAGE
7. In a triple decrement table, lives are subject to decrements of death ( d), disability big, and withdrawal (w). You are given: (i) (ii) The total decrement is uniformly distributed over each year of age. τ l = 5, 000 bτg, (iii) l + 1 = 3 000 (iv) (v) mbdg = 00. mbwg = 0. 05 Calculate qbig, the probability of decrement by disability at age. (A) 0.0104 (B) 0.011 (C) 0.010 (D) 0.018 (E) 0.0136 COURSE/EXAM 3: MAY 000-7 - GO ON TO NEXT PAGE
8. For a two-year term insurance on a randomly chosen member of a population: (i) (ii) (iii) 1/3 of the population are smokers and /3 are nonsmokers. The future lifetimes follow a Weibull distribution with: τ = and θ = 15. for smokers τ = and θ =. 0 for nonsmokers The death benefit is 100,000 payable at the end of the year of death. (iv) i = 0. 05 Calculate the actuarial present value of this insurance. (A) 64,100 (B) 64,300 (C) 64,600 (D) 64,900 (E) 65,100 COURSE/EXAM 3: MAY 000-8 - GO ON TO NEXT PAGE
9. For a 10-year deferred whole life annuity of 1 on (35) payable continuously: (i) Mortality follows De Moivre s law with ω = 85. (ii) i = 0 (iii) Level benefit premiums are payable continuously for 10 years. Calculate the benefit reserve at the end of five years. (A) 9.38 (B) 9.67 (C) 10.00 (D) 10.36 (E) 10.54 COURSE/EXAM 3: MAY 000-9 - GO ON TO NEXT PAGE
10. Taicabs leave a hotel with a group of passengers at a Poisson rate λ =10 per hour. The number of people in each group taking a cab is independent and has the following probabilities: Number of People Probability 1 0.60 0.30 3 0.10 Using the normal approimation, calculate the probability that at least 1050 people leave the hotel in a cab during a 7-hour period. (A) 0.60 (B) 0.65 (C) 0.70 (D) 0.75 (E) 0.80 COURSE/EXAM 3: MAY 000-10 - GO ON TO NEXT PAGE
11. A company provides insurance to a concert hall for losses due to power failure. You are given: (i) The number of power failures in a year has a Poisson distribution with mean 1. (ii) The distribution of ground up losses due to a single power failure is: 10 0 50 Probability of 0.3 0.3 0.4 (iii) The number of power failures and the amounts of losses are independent. (iv) There is an annual deductible of 30. Calculate the epected amount of claims paid by the insurer in one year. (A) 5 (B) 8 (C) 10 (D) 1 (E) 14 COURSE/EXAM 3: MAY 000-11 - GO ON TO NEXT PAGE
1. For a certain mortality table, you are given: = = (i) µ 80. 5 0. 00 (ii) µ 815. 0. 0408 (iii) µ b8. 5g = 0. 0619 (iv) Deaths are uniformly distributed between integral ages. Calculate the probability that a person age 80.5 will die within two years. (A) 0.078 (B) 0.0785 (C) 0.0790 (D) 0.0796 (E) 0.0800 COURSE/EXAM 3: MAY 000-1 - GO ON TO NEXT PAGE
13. An investment fund is established to provide benefits on 400 independent lives age. (i) On January 1, 001, each life is issued a 10-year deferred whole life insurance of 1000, payable at the moment of death. (ii) Each life is subject to a constant force of mortality of 0.05. (iii) The force of interest is 0.07. Calculate the amount needed in the investment fund on January 1, 001, so that the probability, as determined by the normal approimation, is 0.95 that the fund will be sufficient to provide these benefits. (A) 55,300 (B) 56,400 (C) 58,500 (D) 59,300 (E) 60,100 COURSE/EXAM 3: MAY 000-13 - GO ON TO NEXT PAGE
14. The Rejection Method is used to generate a random variable with density function f the density function g( ) and constant c as the basis, where: c 3 f ( ) = 1 +, 0 < < 1 g( ) = 1, 0 < < 1 h by using The constant c has been chosen so as to minimize n, the epected number of iterations needed to generate a random variable from f. Calculate n. (A) 1.5 (B) 1.78 (C) 1.86 (D).05 (E).11 COURSE/EXAM 3: MAY 000-14 - GO ON TO NEXT PAGE
15. In a double decrement table: (i) lbτg 30 = 1000 (ii) q b1g 30 = 0100. (iii) q bg 30 = 0. 300 1 (iv) 1 30 q =. 0 075 (v) lbτg 3 = 47 Calculate qbg 31. (A) 0.11 (B) 0.13 (C) 0.14 (D) 0.15 (E) 0.17 COURSE/EXAM 3: MAY 000-15 - GO ON TO NEXT PAGE
16. You are given: Standard Mean Deviation Number of Claims 8 3 Individual Losses 10,000 3,937 Using the normal approimation, determine the probability that the aggregate loss will eceed 150% of the epected loss. Φ b. g. Φ (A) Φ 15. (B) Φ 15. (C) 1 15 (D) 1 Φ 15. (E) 15 1 COURSE/EXAM 3: MAY 000-16 - GO ON TO NEXT PAGE
17. The future lifetimes of a certain population can be modeled as follows: (i) Each individual s future lifetime is eponentially distributed with constant hazard rate θ. (ii) Over the population, θ is uniformly distributed over (1,11). Calculate the probability of surviving to time 0.5, for an individual randomly selected at time 0. (A) 0.05 (B) 0.06 (C) 0.09 (D) 0.11 (E) 0.1 COURSE/EXAM 3: MAY 000-17 - GO ON TO NEXT PAGE
18. The pricing actuary at Company XYZ sets the premium for a fully continuous whole life insurance of 1000 on (80) using the equivalence principle and the following assumptions: (i) The force of mortality is 0.15. (ii) i = 0. 06 The pricing actuary s supervisor believes that the Illustrative Life Table with deaths uniformly distributed over each year of age is a better mortality assumption. Calculate the insurer s epected loss at issue if the premium is not changed and the supervisor is right. (A) (B) 14 6 (C) 0 (D) 37 (E) 0 COURSE/EXAM 3: MAY 000-18 - GO ON TO NEXT PAGE
19. An insurance company sold 300 fire insurance policies as follows: Number of Policies 100 00 Policy Maimum 400 300 Probability of Claim Per Policy 0.05 0.06 You are given: (i) The claim amount for each policy is uniformly distributed between 0 and the policy maimum. (ii) The probability of more than one claim per policy is 0. (iii) Claim occurrences are independent. Calculate the variance of the aggregate claims. (A) 150,000 (B) 300,000 (C) 450,000 (D) 600,000 (E) 750,000 COURSE/EXAM 3: MAY 000-19 - GO ON TO NEXT PAGE
0. For a last-survivor insurance of 10,000 on independent lives (70) and (80), you are given: (i) (ii) The benefit, payable at the end of the year of death, is paid only if the second death occurs during year 5. Mortality follows the Illustrative Life Table. (iii) i = 0. 03 Calculate the actuarial present value of this insurance. (A) 35 (B) 45 (C) 55 (D) 65 (E) 75 COURSE/EXAM 3: MAY 000-0 - GO ON TO NEXT PAGE
1. A risky investment with a constant rate of default will pay: (i) (ii) principal and accumulated interest at 16% compounded annually at the end of 0 years if it does not default; and zero if it defaults. A risk-free investment will pay principal and accumulated interest at 10% compounded annually at the end of 0 years. The principal amounts of the two investments are equal. The actuarial present values of the two investments are equal at time zero. Calculate the median time until default or maturity of the risky investment. (A) 9 (B) 10 (C) 11 (D) 1 (E) 13 COURSE/EXAM 3: MAY 000-1 - GO ON TO NEXT PAGE
. For a special annual whole life annuity-due on independent lives (30) and (50): (i) (ii) (iii) Y is the present-value random variable. The benefit is 1000 while both are alive and 500 while only one is alive. Mortality follows the Illustrative Life Table. (iv) i = 0. 06 (v) You are doing a simulation of K(30) and K(50) to study the distribution of Y, using the Inverse Transform Method (where small random numbers correspond to early deaths). (vi) In your first trial, your random numbers from the uniform distribution on [0,1] are 0.63 and 0.40 for generating K(30) and K(50) respectively. (vii) F is the simulated value of Y in this first trial. Calculate F. (A) 15,150 (B) 15,300 (C) 15,450 (D) 15,600 (E) 15,750 COURSE/EXAM 3: MAY 000 - - GO ON TO NEXT PAGE
3. For a birth and death process, you are given: (i) There are four possible states {0,1,,3}. (ii) These limiting probabilities: P 0 = 010. P 1 = 030. (iii) These instantaneous transition rates: q 1 = 0. 40 q 3 = 01. (iv) If the system is in state, the time until it leaves state is eponentially distributed with mean 0.5. Calculate the limiting probability P. (A) 0.04 (B) 0.07 (C) 0.7 (D) 0.33 (E) 0.55 COURSE/EXAM 3: MAY 000-3 - GO ON TO NEXT PAGE
4. For a fully discrete whole life insurance with non-level benefits on (70): (i) The level benefit premium for this insurance is equal to P 50. (ii) q70+ k = q50+ k + 0. 01, k = 0, 1,..., 19 (iii) q 60 = 001368. (iv) kv = kv50, k = 0, 1,..., 19 (v) 11V 50 = 0. 16637 Calculate b 11, the death benefit in year 11. (A) 0.48 (B) 0.64 (C) 0.636 (D) 0.648 (E) 0.834 COURSE/EXAM 3: MAY 000-4 - GO ON TO NEXT PAGE
5. An insurance agent will receive a bonus if his loss ratio is less than 70%. You are given: (i) (ii) His loss ratio is calculated as incurred losses divided by earned premium on his block of business. The agent will receive a percentage of earned premium equal to 1/3 of the difference between 70% and his loss ratio. (iii) The agent receives no bonus if his loss ratio is greater than 70%. (iv) His earned premium is 500,000. (v) His incurred losses are distributed according to the Pareto distribution: F( ) = 1 F 600, 000 I, HG +, K J > 0 600 000 3 Calculate the epected value of his bonus. (A) 16,700 (B) 31,500 (C) 48,300 (D) 50,000 (E) 56,600 COURSE/EXAM 3: MAY 000-5 - GO ON TO NEXT PAGE
6. For a fully discrete 3-year endowment insurance of 1000 on bg: (i) q = q+1 = 00. (ii) i = 0. 06 (iii) 1000P : = 373. 63 3 Calculate 1000 V 3 1 V 3 e j : :. (A) 30 (B) 35 (C) 330 (D) 335 (E) 340 COURSE/EXAM 3: MAY 000-6 - GO ON TO NEXT PAGE
7. For an insurer with initial surplus of : (i) The annual aggregate claim amount distribution is: Amount 0 3 8 Probability 0.6 0.3 0.1 (ii) (iii) Claims are paid at the end of the year. A total premium of is collected at the beginning of each year. (iv) i = 0. 08 Calculate the probability that the insurer is surviving at the end of year 3. (A) 0.74 (B) 0.77 (C) 0.80 (D) 0.85 (E) 0.86 COURSE/EXAM 3: MAY 000-7 - GO ON TO NEXT PAGE
8. For a mortality study on college students: (i) Students entered the study on their birthdays in 1963. (ii) You have no information about mortality before birthdays in 1963. (iii) (iv) (v) (vi) Dick, who turned 0 in 1963, died between his 3 nd and 33 rd birthdays. Jane, who turned 1 in 1963, was alive on her birthday in 1998, at which time she left the study. All lifetimes are independent. Likelihoods are based upon the Illustrative Life Table. Calculate the likelihood for these two students. (A) 0.00138 (B) 0.00146 (C) 0.00149 (D) 0.00156 (E) 0.00169 COURSE/EXAM 3: MAY 000-8 - GO ON TO NEXT PAGE
9. For a whole life annuity-due of 1 on (), payable annually: (i) q = 0. 01 (ii) q + 1 = 0. 05 (iii) i = 0. 05 (iv) a&& + 1 = 6. 951 Calculate the change in the actuarial present value of this annuity-due if p +1 is increased by 0.03. (A) 0.16 (B) 0.17 (C) 0.18 (D) 0.19 (E) 0.0 COURSE/EXAM 3: MAY 000-9 - GO ON TO NEXT PAGE
30. X is a random variable for a loss. Losses in the year 000 have a distribution such that: E X d = 0. 05d + 1475. d. 5, d = 10, 11, 1,..., 6 Losses are uniformly 10% higher in 001. An insurance policy reimburses 100% of losses subject to a deductible of 11 up to a maimum reimbursement of 11. Calculate the ratio of epected reimbursements in 001 over epected reimbursements in the year 000. (A) 110.0% (B) 110.5% (C) 111.0% (D) 111.5% (E) 11.0% COURSE/EXAM 3: MAY 000-30 - GO ON TO NEXT PAGE
31. Company ABC issued a fully discrete three-year term insurance of 1000 on Pat whose stated age at issue was 30. You are given: (i) 30 31 3 33 q 0.01 0.0 0.03 0.04 (ii) i = 0. 04 (iii) Premiums are determined using the equivalence principle. During year 3, Company ABC discovers that Pat was really age 31 when the insurance was issued. Using the equivalence principle, Company ABC adjusts the death benefit to the level death benefit it should have been at issue, given the premium charged. Calculate the adjusted death benefit. (A) 646 (B) 664 (C) 71 (D) 750 (E) 963 COURSE/EXAM 3: MAY 000-31 - GO ON TO NEXT PAGE
3. Insurance for a city s snow removal costs covers four winter months. (i) (ii) (iii) (iv) There is a deductible of 10,000 per month. The insurer assumes that the city s monthly costs are independent and normally distributed with mean 15,000 and standard deviation,000. To simulate four months of claim costs, the insurer uses the Inverse Transform Method (where small random numbers correspond to low costs). The four numbers drawn from the uniform distribution on [0,1] are: 0.5398 0.1151 0.0013 0.7881 Calculate the insurer s simulated claim cost. (A) 13,400 (B) 14,400 (C) 17,800 (D) 0,000 (E) 6,600 COURSE/EXAM 3: MAY 000-3 - GO ON TO NEXT PAGE
33. In the state of Elbonia all adults are drivers. It is illegal to drive drunk. If you are caught, your driver s license is suspended for the following year. Driver s licenses are suspended only for drunk driving. If you are caught driving with a suspended license, your license is revoked and you are imprisoned for one year. Licenses are reinstated upon release from prison. Every year, 5% of adults with an active license have their license suspended for drunk driving. Every year, 40% of drivers with suspended licenses are caught driving. Assume that all changes in driving status take place on January 1, all drivers act independently, and the adult population does not change. Calculate the limiting probability of an Elbonian adult having a suspended license. (A) 0.019 (B) 0.00 (C) 0.08 (D) 0.036 (E) 0.047 COURSE/EXAM 3: MAY 000-33 - GO ON TO NEXT PAGE
34. For a last-survivor whole life insurance of 1 on () and (y): (i) The death benefit is payable at the moment of the second death. (ii) The independent random variables T *, T * y, Z common shock model. (iii) T (iv) T y T* * has an eponential distribution with µ btg = T* y * has an eponential distribution with µ y btg = and are the components of a 003., t 0. 0. 05, t 0. (v) Z, the common shock random variable, has an eponential distribution with µ Z t 00., t 0. = (vi) δ = 006. Calculate the actuarial present value of this insurance. (A) 0.16 (B) 0.71 (C) 0.36 (D) 0.368 (E) 0.43 COURSE/EXAM 3: MAY 000-34 - GO ON TO NEXT PAGE
35. The distribution of Jack s future lifetime is a two-point miture: (i) (ii) With probability 0.60, Jack s future lifetime follows the Illustrative Life Table, with deaths uniformly distributed over each year of age. With probability 0.40, Jack s future lifetime follows a constant force of mortality µ= 00.. A fully continuous whole life insurance of 1000 is issued on Jack at age 6. Calculate the benefit premium for this insurance at i = 0. 06. (A) 31 (B) 3 (C) 33 (D) 34 (E) 35 COURSE/EXAM 3: MAY 000-35 - GO ON TO NEXT PAGE
36. A new insurance salesperson has 10 friends, each of whom is considering buying a policy. (i) (ii) (iii) (iv) (v) (vi) Each policy is a whole life insurance of 1000, payable at the end of the year of death. The friends are all age and make their purchase decisions independently. Each friend has a probability of 0.10 of buying a policy. The 10 future lifetimes are independent. S is the random variable for the present value at issue of the total payments to those who purchase the insurance. Mortality follows the Illustrative Life Table. (vii) i = 0. 06 Calculate the variance of S. (A) 9,00 (B) 10,800 (C) 1,300 (D) 13,800 (E) 15,400 COURSE/EXAM 3: MAY 000-36 - GO ON TO NEXT PAGE
37. Given: (i) p k denotes the probability that the number of claims equals k for k = 0,1,, (ii) p p n m m =! n!, m 0, n 0 Using the corresponding zero-modified claim count distribution with p M = 01., calculate p. M 0 1 (A) 0.1 (B) 0.3 (C) 0.5 (D) 0.7 (E) 0.9 COURSE/EXAM 3: MAY 000-37 - GO ON TO NEXT PAGE
38. For Shoestring Swim Club, with three possible financial states at the end of each year: (i) (ii) State 0 means cash of 1500. If in state 0, aggregate member charges for the net year are set equal to operating epenses. State 1 means cash of 500. If in state 1, aggregate member charges for the net year are set equal to operating epenses plus 1000, hoping to return the club to state 0. (iii) State means cash less than 0. If in state, the club is bankrupt and remains in state. (iv) (v) (vi) (vii) The club is subject to four risks each year. These risks are independent. Each of the four risks occurs at most once per year, but may recur in a subsequent year. Three of the four risks each have a cost of 1000 and a probability of occurrence 0.5 per year. The fourth risk has a cost of 000 and a probability of occurrence 0.10 per year. Aggregate member charges are received at the beginning of the year. (viii) i = 0 Calculate the probability that the club is in state at the end of three years, given that it is in state 0 at time 0. (A) 0.4 (B) 0.7 (C) 0.30 (D) 0.37 (E) 0.56 COURSE/EXAM 3: MAY 000-38 - GO ON TO NEXT PAGE
39. For a continuous whole life annuity of 1 on (): (i) Tbg, the future lifetime of ( ), follows a constant force of mortality 0.06. (ii) The force of interest is 0.04. e Calculate Pr a > a. T j (A) 0.40 (B) 0.44 (C) 0.46 (D) 0.48 (E) 0.50 COURSE/EXAM 3: MAY 000-39 - GO ON TO NEXT PAGE
40. Rain is modeled as a Markov process with two states: (i) If it rains today, the probability that it rains tomorrow is 0.50. (ii) If it does not rain today, the probability that it rains tomorrow is 0.30. Calculate the limiting probability that it rains on two consecutive days. (A) 0.1 (B) 0.14 (C) 0.16 (D) 0.19 (E) 0. **END OF EXAMINATION** COURSE/EXAM 3: MAY 000-40 - STOP
Solutions to Course 3 Eam May 000 Question # 1 Key: C o e0 o e10 z F ω t I = 1 dt 5 50 0 HG K J = ω ω ω = = ω = ω ω F I 40 HG K J = = bgb40g t z40 = tf I 1 dt HG K J b0g 0 40 40 L 3 t t = N M O Q P 3 40 b0g 0 z0 40 t = 1 dt 40 40 Var T = 133 0 Question # Key: C A priori, epect 30 coins: 18 worth one, 6 worth 5, 6 worth 10. Given 10 worth 5, epect: 18 @ 1; 10 @ 5; 6 @ 10. Total = 18 + 50 + 60 = 18 Course 3 May 000
Question # 3 Key: E Need to determine q and q. +1 400q + 1 Formula 7.3: 1658. = 1V = 74. 33 q+ 1 = 05. 11. Formula 8.39: 0V = 0 = 400vq π + v Vp 1 V q = π 11. 1 400 V 1 = 017. Loss (Eample 8.51): 0 400 1 11. F I HG K J = F 74. 33 89. 30 1 400 1 I 1 7433. 1 HG 11. K J + 11. F HG 1 7433. 1+ 11. Since E L = 0, I K J = F I HG K J = 14190. 18868. Var = 89. 30 0. 17 + 188. 68 0. 5 1 0. 17 + 14190. 1 0. 17 1 0. 5 = 34150 Course 3 May 000
Question # 4 Key: A The distribution is negative binomial E N = 1 Var N = E Var N Λ + Var E N Λ Λ = E Λ + Var Λ = 1+ = 3 c h Λ c h From the KPW appendi, we have: rβ = 1 b g rβ 1+ β = 3 3 b1 + βg = β = 1 1 rβ = 1 r = rβ 1 Pr N = 1 = = = 01945. r+ 1 3 1+ β 3 b g Λ Λ Question # 5 Key: B If the worker survives for three years, the reinsurance will pay 100,000 at t=3, and everything after that. So the actuarial present value of the reinsurer s portion of the claim F = H G 3 I K J F F + H G 4 I K J F + H G 5. 7. 7. 7 I HG K J I 100, 000 150, 000 105. 105. 105. KJ = 79, 01 Course 3 May 000
Question # 6 Key: D S t bruing F = < HG Surplus = U t = 60 + 0 t S t. P P t 60 0 I KJ Hence claims of 100, 00 causing ruin can only occur on the intervals L 100 60 00 60 0, = O, L 0, = 7O NM 0 QP NM 0 QP respectively. t PbClaim by tg = 1 Pbno claimg = 1 + t Therefore, PbRuing = pbsgpbclaim of size s causes ruing s = 60% + + 40% 7 1 1+ 7 = 75% Question # 7 Key: D L q m m τ i d w l = τ + l big d = 5, 000 bdg τ +1 bdg d d d = d τ L = = 0. 0 = 480 4, 000 bwg d d w = d τ L = = 0. 05 = 100 4, 000 τ bwg i d d = 000 = d + d + d d = 30 i 30 q = = 0018. 5, 000 = 4, 000 w i Course 3 May 000
Question # 8 Key: C Z q = R S T 10 v 0 5 k+ 1 k = 0, 1 k =, 3,... E Z = E Z S 1 3+ E Z N 3 s N e 1 j e 1 j { } 5 s N = 10 vq + v q 1 3+ vq + v q 3 = 1 p p e 1 q = p p p Now plug in = F H G 1I K J θ τ = e F H G τ I K J θ Smokers Nonsmokers p 0.64118 0.77880 p 0.16901 0.36788 R F = + H G L I K J O NM + + QP F H G L I K J O 5 1 10 S NM T 105 03588 1 1 0 4717 1 3 105 105 01 1 b. g b. g b. g b. g... 105. QP = b58, 338g 3 + b77, 000g1 3 = 64, 559 04109 3 U V W Course 3 May 000
Question # 9 Key: A 10 5 e j e j V a P a s 10 35 = 10 35 35: 5 e P a = 10 35 j a s 45 35: 10 z z 40 40 45 t 45 t 45 0 0 40 t a = v p dt = p dt i = 0 F HG o 1 t = e45 40 40 I KJ = = 0 0 b g s = a / E = a / v p 35: 10 35: 10 10 35 35: 10 1 10 = 50 t dt l 50 0 l = 450 40 z b g 35 45 10 10 35 e j. 0 P a = 40 = 1778 10 35 450 F = H G I K J F I = HG KJ 10 1778. 5 35 5V = Pe10 a35j z s = 50 t dt l 35: 5 50 0 l40 5 1 50 1778 50 45 50 t. t 938. 0 Course 3 May 000
Question # 10 Key: D Let X(t) be the number leaving by cab in a t hour interval and let of people in the ith group. Then: E Y i E Y E X Var X c h c h c h b7g 10 7 15 1 080 b7g =. =, = 10 7. 7 = 1944, = 1 0. 6 + 0. 3 + 3 01. = 15. = 1 06. + 0. 3 + 3 01. =. 7 i Y i denote the number, = P 1049. 5 L Xb7g 1080. = P NM 1944 = 1 Φ b 0. 691754g = Φb0. 691754g P X 7 1050 = 0. 7553 1049 5 1080 1944 The answer is D with or without using the 0.5 continuity correction. O QP Course 3 May 000
Question # 11 Key: E Calculate convolutions of fbg: f f * 10 0 0.3 0.3 0.00 0.09 n 0 1 3 4 Pr N = n f f f s s s 0.37 0.37 0.18 0.06 0.0 b0g = 037. Fs b10g = 037. 03. = 011. Fs b0g = 018. 009. + 03. 037. = 013. Fs E S = 1 03. 10 + 03. 0 + 0. 4 50 = 9 0 = 037. 10 = 037. + 011. = 0. 48 0 = 0. 48 + 013. = 0. 61 + c Sh c Sh c Sh = 9 10b3 037. 048. 061. g = 9 154. = 136. E S 30 = E S 10 1 F 0 10 1 F 10 10 1 F 0 Question # 1 Key: A µ b 81g q81 0. 0408 = 815. = q81 = 00400. 1 1 q Similarly, q = 0. 000 and q = 0. 0600 80 8 q = q + p q + p q 80. 5 1/ 80. 5 1/ 80. 5 81 81 1/ 8 001. 098. = + 004. + 096. 0. 03 = 0. 078 0. 99 099. b g Course 3 May 000
Question # 13 Key: A µ 10 EbZg = e bµ + δ 1000 g µ + δ = F H G I K J 1000 5 1 = 155. e 1. F F H G I HG K J µ 10 5 Var Z = 1000 e e µ + δ 1 bsg E S Var 095. = Pr = 3, 61016. = 400E Z = 50, 00 = 400Var Z = 9, 444, 064 F H G S E S VarbSg µ + δ. 4 I K I KJ k 50, 00 J k = 1645. 9, 444, 064 + 50, 00 = 55, 55 9, 444, 064 Course 3 May 000
Question # 14 Key: B According to the theorem on p. 67, the epected number of interations is c 178., as follows: d d d d F HG f 3 = 1 gbg d + i f I gbgkj = d + 1 1 4 3 i = 0 1b3 1gb 1g = 0 1 = 3 F HG I f = 1 4 + 6 1 = 1 4 + = 4 < 0 g KJ 3 F HG L NM 1 3 c = 1 1 1 + 3 9 7 O f = gbg 1 3 1 I K J + = 1 9 6 1 QP = 4 1 7 = 48 7 7 178. Course 3 May 000
Question # 15 Key: B 30 30 30 τ 1 p = p p = 09. 07. = 063. lbτg τ τ = pl = 31 30 30 630 db1g 1 q lbτ = g = 31 1 30 30 75 b τ g b τ g b τ g d31 = l31 l3 = 630 47 = 158 dbg 1 dbτ = g d = 158 75 = 83 q 31 31 31 31 31 31 d 83 = 01317 013 l = =.. τ 630 Question # 16 Key: C Let S = the aggregate loss. = E S Var S 8 10, 000 = 8 3937 + 3 b 10 000 g = b 3 000,, g F 80, 000 40, 000 c I > 15. h = > HG K J 3, 000 3, 000 = PbZ > 15. g = Φb 15. g = 1 Φb15. g P S E S P S Course 3 May 000
Question # 17 Key: E c h 11 ce e h θ s = Pr X > = E Pr X > θ = 01. e dθ = 01. So: sb05. g = 0105. z1 11 Question # 18 Key: A µ δ + µ µ δ µ π = 1000 e t dt 1000 0 = + δ + µ e t dt 1 0 δ + µ z z = 1000µ = 150 Epected loss = 1000A 80 π a 80. i 1000A 80 = 1000 A80 = 1. 0970867 665. 75 = 685. 53 δ 1 068553 a 80 =. = 53969. δ Epected loss = 685. 53 150 5. 3969 = 14 b g Course 3 May 000
Question # 19 Key: D is binomial = Var = b g E S = E N Var X + E X Var N N E N nq, N nq 1 q, where q is the probability of a claim. Ma Ma X is uniform. E X =, Var X = 1 Hence, Total variance R = S F + H G I K J U T V #policies Ma Ma q qb 1 qg policies 1 W L F = + H G I K J N O M Q L F P + + H G I 100 005. 400 b005. gb0. 95g 400 00 K J N 0. 06 1 M 300 b006. gb094. g 300 1 = 600, 46667. O Q P Course 3 May 000
Question # 0 Key: A c APV = v 5 p p p 4 70 p 4 80 4 70: 80 5 7080 : 70 84 = = = l80 3914365,. 4 70: 80 4 4 80 4 70 4 80 5 70 5 80 l74 56, 64051. = = = 0856094. l 66, 16155. l 75 = = = l70 66, 16155. 85 = = = l80 3914365,. 5 70: 80 5 70 5 80 5 70 5 80 h 6, 60734. 0679736. p = p + p p p = 095391. p p l l 70 53, 96081. 3, 5846. 081559. 0. 60459 p = p + p p p = 096690. APV b g 5 007. = 095391. 096690. v = = 0. 0348 5 103. Course 3 May 000
Question # 1 Key: E 0 0 0 v and 116 0 0 µ 0 b. g e v µ The epected values at the end of 0 years are 11. = 116. e. Actuarial present values at time zero are 11 0 0. So 11. = 116. e µ 116 log. 11. So µ = F H G I K J = 0. 0531 µ log 05. 06931. b 05. g = = 0. 5 = = = S e 0. 5 05. µ 00531. 1305. Question # Key: B Note: These values of l are 1/100 of the ones in the Eam booklet, which does not affect the answer. l b 1 0. 63 l = 35155, l 30 = 95, 014 g 30 < 35155, < l 81 8 So: K 30 = 51 l b 1 0. 4 l = 53, 705 l 50 = 89, 509 g 50 < 53, 705 < l 75 76 So: K 50 = 5 6 e 5 6j b. g b.. g Simulated Y = 1000a&& + 500 a&& a&& = 1000 13 783 + 500 16 813 13 783 = 15, 98 Course 3 May 000
Question # 3 Key: A 1 Mean time until leaving state is q0 + q1 + q 3 =. b q 0 + 0. 4+ q = q 3 0 = 0 for birth and death process = 16. 3 g Pi = 1 P = 0. 6 P 3 Rate entering 3 = rate leaving 3 P q = P q 3 3 3 16. P = 06. P 01. 17. P = 007. P = 0. 04 Course 3 May 000
Question # 4 Key: D (1) V = V + P 1+ i b V q 11 10 50 11 11 80 But by traditional formula: and since 10 V = 10 V 50 and 11 V = 11 V 50 V = V + P 1+ i 1 V q 11 50 10 50 50 11 50 60 So: () V = V + P 1+ i 1 V q 11 10 50 11 60 ( 1) ( ) b V q = 1 V q 11 11 80 11 60 So: b 11 b 1 11V q60 = + 11V q g 80 b1 016637. gb0. 01368g = 0. 0368 = 064796. 0648. + 016637. Course 3 May 000
Question # 5 Key: E 1 Let L be the loss, then the bonus = 3 500 000 7 L. If L 350, 000, 500, 000 0 If L > 350, 000 350, 000 1 EbBonusg = E L 350, 000 3 3 R S T 350, 000 F = H G 1 I K J 600, 000F F H G 600, 000I 1 HG K J 3 3 950, 000 = 56, 556 I KJ Question # 6 Key: B 1000 V 1 : 3 1000 V : 3 e = = L NM L NM 08. L O NM QP e1000p : 3 jb1+ ig b1000qg p 37363. 106. 1000 0. 396. 0478 00 = 08. = 4506. = = O QP O QP L p+ 1 NM L NM L O NM QP e1000p : 3 + 1V : 3jb1 + ig b1000q+ 1g 37363. + 45. 06 106. 1000 0. 08. 655. 8114 00 = 08. = 56976. j 1000 V : 3 1V : 3 = 569. 76 45. 06 = 34. 70 O QP O QP Course 3 May 000
Question # 7 Key: A Cash Before Probability Receiving Prem. T=0 1.000 4 T=1 0.6 4.3 6.3 0.3 1.3 3.3 0.1 Ruin -- T= 0.36 0.18 0.06 0.18 0.09 0.03 0.10 T=3 0.16 0.108 0.036 0.108 0.054 0.018 0.108 0.054 0.018 0.054 0.07 0.009 0.190 6.83 3.83 Ruin 3.59 0.59 Ruin Ruin (from above) 9.53 6.53 1.53 6.9 3.9 Ruin 6.03 3.03 Ruin.79 Ruin Ruin Ruin (From above) Cash After Premium 8.83 5.83 -- 5.59.59 -- -- Course 3 May 000
Question # 8 Key: C For Dick: q 1 0 l q l 3 3 = = = 0 9471591 0. 00170 961780 0. 001674 l56 8563435 For Jane: 35 p1 = = = 089191. l 9607896 1 For both: b0. 001674gb0. 89191g = 0. 00149 Question # 9 Key: C a&& = 1+ vp + v pp+ a&& + 1 Let y denote the change in p +1. = a&& bwithoutg + yv p&& a+ a&& with increase = 1+ vp + v p p + y a&& + 1 + yv p a&& + = change in actuarial present value. 1 a&& = 6951. = 1+ vp && a = 1+ 1 005. a&& 6. 577 = a&& 105. + 1 + 1 + + + F I HG K J = 0. 03 1 105. 099. 6577. 0177. b g Course 3 May 000
Question # 30 Key: D Epected claim payments in 000 = E X ^ E X ^11 d i d i = 005. + 1475.. 5 005. 11 + 1475. 11 5. b = 1810. 10. 95 = 715. g From theorem.5: Epected claim payments in 001 = 11. E X ^0 E X ^10 = 11. 0. 05 0 + 1475. 0. 5 0. 05 10 + 1. 475 10. 5 = 11. 175. 10 = 7975. d i d i b g 7975. = 1115. 715. Course 3 May 000
Question # 31 Key: B A a&& 001. 099. 00. 099. 0. 98 003. = + + = 0. 05379675 3 104. 104. 104. 1 30: 3 1 0. 99 0. 99 0. 98 = + + =. 84897515 104. 104. 30: 3 1 1000 1000 005379675. P 18 88315 30 = :. 3 84897515. = 1 A 31 : 3 && 00. 098. 0. 03 098. 097. 004. = + + = 0. 08015919 3 104. 104. 104.... : 1 098 098 097 = + + = 8119. 104. 104. a 31 3 1 1 1000P 303 : a&& 313 : = BA313 : 53. 7954 = B 0. 08015919 B = 664 b g Course 3 May 000
Question # 3 Key: B Results from the standard normal cdf are converted to N 15, 000; 000 X 15, 000 000Z using = + Random Number 0.5398 0.1151 0.0013 0.7881 Standard Normal 0.1-1. -3.0 0.8 Cost 1500 1600 9000 16600 Claim Paid 500 600 0 6600 Total claims = 500 + 600 + 0 + 6600 = 14,400 Question # 33 Key: E Given: b0. 95gπ + b060. gπ + π = π b 005. g π = π 040. π = π 0 1 0 b 0 1 1 π + π + π = 1 g 0 1 5 Solve for π 1 = ~ 0. 047. 107 Course 3 May 000
Question # 34 Key: D T µ bg T* Z = µ + µ = 0. 03+ 0. 0 = 0. 05 T y T* y Z µ = µ + µ = 005. + 00. = 007. y y T* T* y Z µ = µ + µ + µ = 0. 03+ 0. 05+ 0. 0 = 010. y y A = µ + = 0. 05 µ δ 011 = 04545.. T Tbg A y T y µ y = + = 007. µ δ 013 = 05385.. T y y A y = µ y + = 010. µ δ 016 = 0. 650. y Ay = A + Ay Ay = 04545. + 05385. 0650. = 0. 3680 Course 3 May 000
Question # 35 Key: A Distribution of moment of death is 60% ILT and 40% constant force. Therefore: AJack = 0. 6 A 6 using ILT + 0.4 A 6 constant force. A ILT = 006. 6 039670. 040849 ln 106. =. A 6 constant force = 0. 0 0. 0 + ln 106. = 0. 5553 A Jack = 06. 040849. + 0. 4 0. 5553 = 03473. a Jack 1 AJack = 1 = 03473. = 11013. ln 106. 0. 0587 b g b1000gb0. 3473g 1000P Jack = = 3101. 11013. Course 3 May 000
Question # 36 Key: E S = N Z j j= 1 bsg = EbNg bz1g + EbZ1g bng Var Var Var = 10 01. 10, 77918. + 7135. 10 01. 09. = 15361, Where Z T = 1000 v j Var Z = 10 A A = 10, 77918. E Z j 6 b 1g d i = A 10 = 7135. 1 3 Question # 37 Key: C The given relationship can be written p k F HG = pk 1 0 1 + k I K J so p k is an (a,b,1) class with a = 0 and b = 1, so p k is Poisson with λ = 1, λ so p = e = 0368. = p. 0 1 By 3.14 of Loss Models, p M 1 = 1 p0 1 p M 0 p 1 0. 9 = 0368. = 05. 0. 63 Course 3 May 000
Question # 38 Key: E P = P, and therefore it is not necessary to track states 0 and 1 separately. 0 1 P 1 = Pr 1 big or ( no big and or more smalls) 3 = 01. + 09. 3 05. 0. 75 + 0. 5 = 01. + 014. = 04. If you enter, you stay there. 3. Probability of not entering in 3 years = 1 0. 4 = 0. 44 Probability of being in after 3 years = 1 0. 44 = 056.. Question # 39 Key: C e Pr a T > a = Pr j 1 v δ > a = Pr T > 1 µ log δ δ + µ = t p t = 1 µ where log 0 0 δ δ + µ = e = F HG F HG L N M µ t 0 µ µ + δ 6 10 3 I KJ = F H G I K J = T µ δ I KJ F HG 04648. IO KJ Q P F HG I KJ Course 3 May 000
Question # 40 Key: D p = limiting probability of rain. p = 05. p + 0. 3 1 p b 0. 3 p = = 0. 375 08. g Prob (rain on two consecutive days) = Prob (rain on first) Prob (rain on second, given rain on first) = (0.375)(0.5) = 0.1875 Course 3 May 000