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TABLE OF CONTENTS

1. Survival 1

2. Multiple Lives 95

3. Multiple Decrements 135

4. Daniel Markov 1 169

5. Daniel Poisson 185

6. Insurance 279

7. Annuities 367

8. Premiums 449

9. Reserves 517

10. Insurances/Annuities 599

11. Daniel Markov 2 651

12. Statistics 669

ii

© 2010 ACTEX Publications, Inc. CAS Exam 3L – Peter J. Murdza

NOTES

Questions and parts of some solutions have been taken from material copyrighted by the Casualty Actuarial Society and the Society of Actuaries. They are reproduced in this study manual with the permission of the CAS and SoA solely to aid students studying for the actuarial exams. Some editing of questions has been done. Students may also request past exams directly from both societies. I am very grateful to these organizations for their cooperation and permission to use this material. They are, of course, in no way responsible for the structure or accuracy of the manual. Exam questions are identified by numbers in parentheses at the end of each question. CAS questions have four numbers separated by hyphens: the year of the exam, the number of the exam, the number of the question, and the points assigned. SoA or joint exam questions usually lack the number for points assigned. W indicates a written answer question; for questions of this type, the number of points assigned are also given. A indicates a question from the afternoon part of an exam. MC indicates that a multiple choice question has been converted into a true/false question. Page references refer to Bowers et al., Actuarial Mathematics (1997), Cunningham et al. Models for Quantifying Risk (2008); Daniel, “Multi-state Transition Models with Actuarial Applications (2007); Daniel “Poisson Processes (and mixture distributions)” (2008); Hoel, Introduction to Mathematical Statistics (1971), Hogg et al., Introduction to Mathematical Statistics, (2004); Hogg/Tanis, Probability and Statistical Inference (2006): Larsen/Marx, An Introduction to Mathematical Statistics and Its Applications (2006); and Mood, Introduction to the Theory of Statistics (1974). Although I have made a conscientious effort to eliminate mistakes and incorrect answers, I am certain some remain. I am very grateful to students who discovered errors in the past and encourage those of you who find others to bring them to my attention. Please check our web site for corrections subsequent to publication. I would also like to thank Chip Cole, Graham Lord, and Katy Murdza for their help in preparation of the manual. Hanover, NH 11/15/10 PJM

Daniel Poisson 185


DANIEL POISSON : PAST CAS AND SoA EXAMINATION QUESTIONS A. Definition of a Poisson Process A1. For a Poisson distribution, the mean and variance are equal. (86–4–50–MC) A2. With respect to the Poisson process for the number of claims, the distribution of the increase in the

number of claims in a time interval is Poisson, with parameter proportional to the length of the interval. (90F–151–16–MC)

A3. The mean of the Poisson distribution is equal to the variance. (92S–4B–3–MC) A4. You are given the following data: Year Number of Claims Number of Claims Squared

1994 1,076 1,157,776 1995 947 896,809 1996 926 857,476 1997 1,184 1,401,856 1998 1,057 1,117,249 Average 1,038 1,086,233

Based on the data above, a Poisson distribution would be suitable to model the number of claims. (99S–5A–11–.5)

A5. Daniel in "Poisson Processes" identifies four requirements that a counting process N(t) must satisfy. Which of the following is not one of them?

A. N(t) must be greater than or equal to zero. B. N(t) must be an integer. C. If s < t, then N(s) must be less than or equal to N(t). D. For s < t , N(t) − N(s) must equal the number of events that have occurred in the interval

(s, t]. E. All of these are requirements. (03F–3C–32–2)

A6. A counting process is said to possess independent increments if the number of events that occur between time s and t is independent of the number of events that occur between time s and (t + u) for all u > 0. (06F–3–26–2–MC)

186 Daniel Poisson


A1. T, p. 4. A2. T, p. 13.

A3. T, p. 4. A4. F, p. 4 – The mean (1,038) is less than the variance, which equals 1,086,233 − (1,038)2 = 8,789. A5. A. T, p. 4

B. T, p. 4 C. T, p. 4 D. T, p. 4 Answer: E

A6. F, p. 3 – Intervals must be disjoint.

Daniel Poisson 187


B. Calculating Poisson Probabilities and Moments B1. A random variable X has a Poisson distribution with a mean of 3. What is the probability that X is

bounded by 1 and 3, that is, P(1.00 ≤ x ≤ 3.00)? A. 15e-3 B. 12e-3 C. (13/2)e-3 D. (3/2)e-3 E. (1/2)e-3 (791–2–6) B2. If the number of typographical errors per page follows a Poisson distribution with parameter λ, what is

the probability that the total number of errors in 10 randomly selected pages is 10? A. (λe-λ)10 B. (1 − e-λ)10 C. 10(λe-λ) D. 10λ10e-10λ/10! E. (10λ)10e-10λ/10! (791–2–20) B3. Suppose X has a Poisson distribution with a standard deviation of 4. What is the conditional probability

that X is exactly 1 given that X ≥ 1? A. 4(e4 − 1)-1 B. 2(e2 − 1)-1 C. 16e-16 D. 4e-4 E. 16(e16 − 1)-1 (792–2–2) B4. If X has a Poisson distribution and P(X ≥ 1) = 1/2, what is the variance of X? A. e-1/2 B. 1/2 C. 2 D. 1/(log 2) E. log 2 (792–2–4) B5. The number of potholes on a certain highway has a Poisson distribution with an average of three

potholes per mile. What is the probability that a given two-mile portion of this highway has no more than three potholes?

A. 25e-6 B. 61e-6 C. (9/2)e-3 D. 13e-3 E. 1 − 25e-6 (82S–2–20) B6. The number of traffic accidents per week in a small city has a Poisson distribution with mean equal to 3.

What is the probability of exactly 2 accidents in 2 weeks? A. 9e-6 B. 18e-6 C. 25e-6 D. 4.5e-3 E. 9.5e-3 (83S–2–28) B7. Let X have a Poisson distribution with parameter λ = 1. What is the probability that X ≥ 2, given that

X ≤ 4 ? A. 5/65 B. 5/41 C. 17/65 D. 17/41 E. 3/5 (83S–2–45) B8. Let X be a Poisson random variable with mean λ. If P(X = 1 | X ≤ 1) = .8, what is the value of λ? A. 4 B. −log .2 C. .8 D. .25 E. −log .8 (86F–110–12) B9. The number of power surges in an electric grid has a Poisson distribution with a mean of one power

surge every twelve hours. What is the probability that there will be no more than one power surge in a 24-hour period?

A. 2e-2 B. 3e-2 C. e-1/2 D. (3/2)e-1/2 E. 3e-1 (88S–110–49)

188 Daniel Poisson


Solutions are based on pp. 4–6, 14. B1. P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) P(1 ≤ X ≤ 3) = (3)1e-3/1! + (3)2e-3/2! + (3)3e-3/3! = 12e-3 Answer: B B2. λ' = 10λ P(X = 10) = (10λ)10e-10λ/10! Answer: E

B3. λ = (4)2 = 16 P(X = 1 | X ≥ 1) = P(X = 1)1 − P(0) =

16e-16

1 − e-16 = 16(e16 − 1)-1

Answer: E B4. P(0) = e-λ = 1 − P(X ≥ 1) = 1 − 1/2 = 1/2 λ = log 2 Answer: E B5. λ = (2)(3) = 6 P(X ≤ 3) = (6)0e-6/0! + (6)1e-6/1! + (6)2e-6/2! + (6)3e-6/3! = 61e-6 Answer: B B6. λ = (2)(3) = 6 P(X = 2) = (6)2e-6/2! = 18e-6 Answer: B B7. P(X = 0) + P(X = 1) = e-1 + e-1 = 2e-1 P(X = 2) + P(X = 3) + P(X = 4) = e-1/2! + e-1/3! + e-1/4! = (17/24)e-1

P(X ≥ 2 | X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X ≥ 2 | X ≤ 4) = (17/24)e-1

2e-1 + (17/24)e-1 = 17/65

Answer: C

B8. .8 = P(X = 1 | X ≤ 1) = P(X = 1)

P(0) + P(1) = λe-λ

e-λ + λe-λ = λ

1 + λ λ = 4

Answer: A B9. λ = 24/12 = 2 P(X ≤ 1) = P(X = 0) + P(X = 1) = e-2 + 2e-2 = 3e-2 Answer: B

Daniel Poisson 189


B10. The density function for a certain parameter, α, is:

f(α) = (4.6)α e-4.6

α! , where α = 0, 1, 2, . . .

Which of the following statements are true concerning the distribution function for α? 1. The mode is less than the mean.

2. The variance is greater than the mean. 3. The median is less than the mean.

A. 1 B. 2 C. 3 D. 1,2 E. 1,3 (94F–4B–19–3) B11. Given the following, determine the expected number of risks in the group that will have four claims.

i) The number of claims for each risk in a group of identical risks follows a Poisson distribution. ii) The expected number of risks in the group that will have no claims is 96. iii) The expected number of risks in the group that will have two claims is 3.

A. < .01 B. ≥ .01 but < .05 C. ≥ .05 but < .1 D. ≥ .1 but < .2 E. ≥ .2 (95S–4B–9–2) B12. The random variable X has a Poisson distribution with mean λ. Determine the expected value of:

X(X − 1) . . . (X − 9)

A. 1 B. λ C. λ(λ − 1) . . . (λ − 9) D. λ10 E. λ(λ + 1) . . . (λ + 9) (97F–4B–21–2)

B13. The random variable X has a Poisson distribution with mean (n − 1/2), where n is a positive integer greater than 1. Determine the mode of X.

A. n − 2 B. n − l C. n D. n + l E. n + 2 (98F–4B–2–2) B14. You are given the following:

i) The number of annual claims for a risk follows a distribution with probability function

p(n) = λn e-λ

n! , n = 0, 1, . . . , where λ > 0.

ii) Two claims were observed for this risk during year 1, and one claim was observed for this risk

during year 2. Determine the smallest value of λ for which the probability of observing three or more claims during

two given years combined is greater than .1. A. < .7 B. ≥ .7 but < 1.0 C. ≥ 1.0 but < 1.3 D. ≥ 1.3 but < 1.6 E. ≥ 1.6 (98F–4B–18–2) B15. You are given the following:

i) The number of claims per year for risk A follows a Poisson distribution with mean m. ii) The number of claims per year for risk B follows a Poisson distribution with mean m + 1. iii) The probability of selecting risk A is equal to the probability of selecting risk B.

One of the risks is randomly selected, and zero claims are observed for this risk during one year. Determine the probability that the selected risk is risk A.

A. < .3 B. ≥ .3 but < .5 C. ≥ .5 but < .7 D. ≥ .7 but < .9 E. ≥ .9 (99S–4B–16–2)

190 Daniel Poisson


B10. f(0) = .010 f(1) = .046 f(2) = .106 f(3) = .163 f(4) = .187 f(5) = .173

1. T – For a Poisson, the mode is the largest integer less than or equal to the mean, i.e., 4. 2. F – For a Poisson, the variance equals the mean 3. T – The median equals 4 since f(0) + f(1) + f(2) + f(3) + f(4) = .010 + .046 + .106 + .163 + .187

= .512; this is less than the mean.

Answer: E B11. P(X = 0) = Ne-λ/N = 96/N P(X = 2) = Nλ2e-λ/2N = 96λ2/2N = 3/N λ2 = 1/16 N[P(X = 4)] = Nλ4e-λ/24 = (96)(1/16)2/24 = .0156 Answer: B B12. Let the expression be represented by Y and let W = X − 10. For the integers from 0 to 9, the value of Y

equals zero. Thus, we get:

E[Y] = E[X!/(X − 10)!] = ∑x=10

∞

x! e-λ λx

(x − 10)! x! = ∑x=10

∞

e-λ λx

(x − 10)! = ∑w=0

∞ e-λ λw+10

w!

E[Y] = λ10 ∑w=0

∞

e-λ λw

w! = (λ10)(1)

Answer: D

B13. For a Poisson, the mode is the largest integer less than or equal to the mean, i.e., (n − 1). Answer: B B14. P(N > 2) = 1 − P(0) − P(1) − P(2) = 1 − e-2λ − (2λ)e-2λ − [(2λ)2/2]e-2λ P(N > 2) = 1 − (e-2λ)(1 + 2λ + 2λ2) f(.7) = 1 − [e-(2)(.7)][1 + (2)(.7) + (2)(.7)2] = 1 − (.24660)(3.38) = .16649 Answer: A B15. 1) Calculate the unconditional probabilities P(A and 0 Claims) = (1/2)(e-m) P(B and 0 Claims) = (1/2)(e-(m+1)) P(0 Claims) = (1/2)(e-m + e-(m+1)) 2) Calculate the conditional probability:

P(A | 0 Claims) = (1/2)(e-m)

(1/2)(e-m + e-(m+1)) = 1

1 + e-1 = .731

Answer: D

Daniel Poisson 191


B16. You are given the following:

i) The annual number of claims follows a Poisson distribution with mean λ. ii) The null hypothesis, H0: λ = m, is to be tested against the alternative hypothesis,

H1: λ < m, based on one year of data. iii) The significance level must not be greater than .05.

Determine the smallest value of m for which the critical region could be nonempty.

A. < .5 B. ≥ .5 but < 1.5 C. ≥ 1.5 but < 2.5 D. ≥ 2.5 but < 3.5 E. ≥ 3.5 (99F–4B–15–2) B17. You are given the following:

i) The number of claims per year for risk A follows a Poisson distribution with mean m. ii) The number of claims per year for risk B follows a Poisson distribution with mean 2m. iii) The probability of selecting risk A is equal to the probability of selecting risk B.

iv) A risk is randomly selected, and no claims are observed for this risk during one year.

Determine the probability that the selected risk will have at least one claim during the next year.

A. 1 − e-m

1 + e-m B. 1 − e-3m

1 + e-m C. 1 − e-m D. 1 − e-2m E. 1 − e-2m − e-4m (99F–4B–28–2)

B18. An insurance company designates 10% of its customers as high risk and 90% as low risk. The number

of claims made by a customer in a calendar year is Poisson distributed with mean λ and is independent of the number of claims made by a customer in the previous calendar year. For high- risk customers, λ = .6, while for low-risk customers, λ = .1. Calculate the expected number of claims made in calendar year 1998 by a customer who made one claim in calendar year 1997.

A. .15 B. .18 C. .24 D. .30 E. .40 (Sample2–1–29) B19. An actuary has discovered that policyholders are three times as likely to file two claims as to file four

claims. If the number of claims filed has a Poisson distribution, what is the variance of the number of claims filed?

A. 1/ 3 B. 1 C. 2 D. 2 E. 4 (00S–1–24) B20. A company buys a policy to insure its revenue in the event of major snowstorms that shut down

business. The policy pays nothing for the first such snowstorm of the year and 10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. What is the expected amount paid to the company under this policy during a one-year period?

A. 2,769 B. 5,000 C. 7,231 D. 8,347 E. 10,578 (00F–1–23) B21. You are given the following for a dental insurer:

i) Claim counts for individual insureds follow a Poisson distribution. ii) Half of the insureds are expected to have 2.0 claims per year. iii) The other half of the insureds are expected to have 4.0 claims per year.

A randomly selected insured has made four claims in each of the first two policy years. Determine the Bayesian estimate of this insured's claim count in the next (third) policy year.

A. 3.2 B. 3.4 C. 3.6 D. 3.8 E. 4.0 (00F–4–3)

192 Daniel Poisson


B16. For the critical region to be nonempty, f(0) must be less than .05: f(0) = e-λ = .05 λ = − log .05 = 2.9957

Answer: D B17. 1) Calculate the unconditional probabilities: P(Risk A and 0 Claims) = (1/2)(e-m) P(Risk B and 0 Claims) = (1/2)(e-2m) P(0 Claims) = (1/2)(e-m + e-2m) 2) Calculate the conditional probabilities:

P(Risk A | 0 Claims) = (1/2)(e-m)

(1/2)(e-m + e-2m) = 1

1 + e-m

P(Risk B | 0 Claims) = (1/2)(e-2m)

(1/2)(e-m + e-2m) ) = e-m

1 + e-m

3) Calculate the probabilities of at least one claim: P(≥ 1 Claim | Risk A) = 1 – e-m P(≥ 1 Claim | Risk B) = 1 − e-2m 4) Calculate the overall probability of one claim:

P(≥ 1 Claim | 0 Claims) = 1 − e-m

1 + e-m + e-m − e-3m

1 + e-m = 1 − e-3m

1 + e-m

Answer: B B18. 1) Calculate the unconditional probabilities:

P(λ = .1 and 1 Claim) = (.9)(.1)e-.1 = .08144 P(λ = .6 and 1 Claim) = (.1)(.6)e-.6 = .03293 P(1 Claim) = .08144 + .03293 = .11437

2) Calculate the conditional probabilities: P(λ = .1 | 1 Claim) = .08144/.11437 = .71207 P(λ = .6 | 1 Claim) = .03293/.11437 = .28793

3) Calculate the expected number of claims: E[N | 1 Claim] = (.1)(.71207) + (.6)(.28793) = .24

Answer: C B19. P(2) = 3 P(4) λ2e-λ/2! = 3λ4e-λ/4! λ = 2

Answer: D

B20. P(X ≥ 1) = 1 − P(0) = 1 − e-1.5 = .77687 E[X | X ≥ 1] = λ

P(X ≥ 1) = 1.5

.77687 = 1.93083

E[Payment] = Pr (X ≥ 1) 10,000 E[(X − 1) | X ≥ 1)] E[Payment] = (.77687)(10,000)(1.93083 − 1) = 7,231

Answer: C B21. 1) Calculate the unconditional probabilities:

P(λ = 2 and Outcome) = [1/2][(2)4e-2/24]2 = 2e-4/9 = .0040701 P(λ = 4 and Outcome) = [1/2][(4)4e-4/24]2 = 512e-8/9 = .019084 P(Outcome) = .0040701 + .019084 = .023154

2) Calculate the conditional probabilities: P(Class A | Outcome) = .0040701/.023154 = .176

P(Class B | Outcome) = .019084/.023154 = .824 3) Calculate the overall expected value: E[X | Outcome] = (.176)(2) + (.824)(4) = 3.6

Answer: C

Daniel Poisson 193


B22. A driver is selected at random. If the driver is a good driver, he is from a Poisson population with a mean of one claim per year. If the driver is a bad driver, he is from a Poisson population with a mean of five claims per year. There is equal probability that the driver is either a good driver or a bad driver. If the driver had three claims last year, calculate the probability that the driver is a good driver.

A. < .325 B. ≥ .325 but < .375 C. ≥ .375 but < .425 D. ≥ .425 but < .475 E. ≥ .475 (03F–3C–12–2) B23. The Allerton Insurance Company insures three indistinguishable populations. The claims frequency of

each insured follows a Poisson process. You are given: Population Expected Time Probability of Claim (class) Between Claims Being in the Class Cost I 12 months 1/3 1,000 II 15 months 1/3 1,000 III 18 months 1/3 1,000 Calculate the expected loss in year 2 for an insured that had no claims in year 1. A. < 810 B. ≥ 810 but < 910 C. ≥ 910 but < 1,010 D. ≥ 1,010 but < 1,110 E. ≥ 1,110 (03F–3C–13–2) B24. A member of a high school math team is practicing for a contest. Her advisor has given her three

practice problems: #l, #2, and #3. She randomly chooses one of the problems, and works on it until she solves it. Then she randomly chooses one of the remaining unsolved problems, and works on it until solved. Then she works on the last unsolved problem. She solves problems at a Poisson rate of one problem per five minutes. Calculate the probability that she has solved problem #3 within ten minutes of starting the problems.

A. .18 B. .34 C. .45 D. .51 E. .59 (03F–3S–26) (Sample–M–19) B25. The number of major hurricanes that hit the island nation of Justcoast is given by a Poisson process with

.100 storms expected per year.

i) Justcoast establishes a fund that will pay 100/storm. ii) The fund charges an annual premium, payable at the start of each year, of 10. iii) At the start of this year (before the premium is paid) the fund has 65. iv) Claims are paid immediately when there is a storm. v) If the fund ever runs out of money, it immediately ceases to exist. vi) Assume no investment income and no expenses.

What is the probability that the fund is still functioning in ten years?

A. < 60% B. ≥ 60% but < 61% C. ≥ 61% but < 62% D. ≥ 62% but < 63% E. ≥ 63% (04S–3C–16–2) B26. XYZ Insurance introduces a new policy and starts a sales contest for 1,000 of its agents. Each agent

makes a sale of the new product at a Poisson rate of 1 per week. Once an agent has made four sales, he gets paid a bonus of $1,000. The contest ends after three weeks. Assuming 0% interest, what is the expected cost of the contest?

A. $18,988 B. $57,681 C. $168,031 D. $184,737 E. $352,768 (04F–3C–19–2)

194 Daniel Poisson


B22. 1) Calculate the unconditional probabilities: P(N = 3 and λ = 1) = (.5)(1)3e-1/3! = .030657 P(N = 3 and λ = 5) = (.5)(5)3e-5/3! = .070187 P(N = 3) = .030657 + .070187 = .100844

2) Calculate the conditional probability:

P(Mean of 1 | N = 3) = .030657/.100844 = .30400 Answer: A B23. 1) Calculate the expected frequencies: λI = 12/12 = 1 λII = 12/15 = .8 λIII = 12/18 = 2/3 2) Calculate the unconditional probabilities. P(N = 0 and λ = 1) = (1/3)e-1 = .12263 P(N = 0 and λ = .8) = (1/3)e-.8 = .14978 P(N = 0 and λ = 2/3) = (1/3)e-2/3 = .17114 P(N = 0) = .12263 + .14978 + .17114 = .44355

3) Calculate the conditional probabilities:

P(λ = 1 | N = 0) = .12263/.44355 = .27647 P(λ = .8 | N = 0) = .14978/.44355 = .33768 P(λ = 2/3 | N = 0) = .17114/.44355 = .38584 4) Calculate the posterior mean: E[X | N = 0] = [1,000][(1)(.27647) + (.8)(.33768) + (2/3)(.38584)] = 803.84 Answer: A B24. λ = (10/5)(1) = 2 P(N = 0) = e-2 P(N = 1) = 2e-2 P(N = 2) = (2)2e-2/2! = 2e-2 P(Solving #3) = [P(N = 1)][1/3] + [P(N = 2)][2/3] + P(N ≥ 3) P(Solving #3) = [P(N = 1)][1/3] + [P(N = 2)][2/3] + [1 − P(N = 0) − P(N = 1) − P(N = 2)] P(Solving #3) = 1 − P(N = 0) − P(N = 1)][2/3] − [P(N = 2)][1/3] P(Solving #3) = 1 − e-2 − (4/3)e-2 – (2/3)e-2 = 1 − 3e-2 = .59399 Answer: E B25. P(Functioning) = (No Storm in 10 Years) + (No Storm in First 3 Years)(Only One Storm in Next 7 Years) P(Functioning) = e-(10)(.1) + [e-(3)(.1)][(.7)e-(7)(.1)] = 1.7e-1 = .62540 Answer: D B26. λ = (3)(1) = 3 (1,000)2 P(N > 3) = [1,000]2[1 − P(0) − P(1) − P(2) − P(3)] (1,000)2 P(N > 3) = [1,000]2[1 − e-3 − 3e-3 − (3)2e-3/2 − (3)3e-3/6] (1,000)2 P(N > 3) = (1,000)2(1 − 13e-3) = 352,768 Answer: E

Daniel Poisson 195


B27. Dental Insurance Company sells a policy that covers two types of dental procedures: root canals and fillings. There is a limit of one root canal per year and a separate limit of two fillings per year. The number of root canals a person needs in a year follows a Poisson distribution with λ = 1, and the number of fillings a person needs in a year is Poisson with λ = 2. The company is considering replacing the single limits with a combined limit of three claims per year, regardless of the type of claim. Determine the change in the expected number of claims per year if the combined limit is adopted. A. There is no change. B. > 0 but < .20 claims C. ≥ .20 but < .25 claims D. ≥ .25 but < .30 claims E. ≥ .30 claims (04F–3C–23–2)

B28. A baseball team has scheduled its opening game for April 1. If it rains on April 1, the game is postponed

and will be played on the next day that it does not rain. The team purchases insurance against rain. The policy will pay 1,000 for each day, up to two days, that the opening game is postponed. The insurance company determines that the number of consecutive days of rain beginning on April 1 is a Poisson random variable with mean .6. What is the standard deviation of the amount the insurance company will have to pay?

A. 668 B. 699 C. 775 D. 817 E. 904 (Sample–P–67) B29. An insurer selects risks from a population that consists of three independent groups.

i) The claims generation process for each group is Poisson. ii) The first group consists of 50% of the population. These individuals are expected to generate

one claim per year. iii) The second group consists of 35% of the population. These individuals are expected to generate

two claims per year. iv) Individuals in the third group are expected to generate three claims per year.

A certain insured has two claims in year 1. What is the probability that this insured has more than two claims in year 2?

A. < 21% B. ≥ 21% but < 25% C. ≥ 25% but < 29% D. ≥ 29% but < 33% E. ≥ 33% (05S–3–17–2)

B30. Long-term Insurance Company insures 100,000 drivers who have each been driving for at least five

years. Each driver gets violations at a Poisson rate of .5/year. Currently, drivers with 1 or more violations in the past three years pay a premium of 1,000. Drivers with 0 violations in the past three years pay 850. Your marketing department wants to change the pricing so that drivers with 2 or more accidents in the past five years pay 1,000 and drivers with zero or one violation in the past five years pay X. Find X so that the total premium revenue for your firm remains constant when this change is made.

A. < 900 B. ≥ 900 but < 925 C. ≥ 925 but < 950 D. ≥ 950 but < 975 E. ≥ 975 (05S–3–39–2) B31. In a certain town the number of common colds an individual will get in a year follows a Poisson

distribution that depends on the individual's age and smoking status. The distribution of the population and the mean number of colds are as follows:

Proportion of Population Mean Number of Colds Children .30 3 Adult nonsmokers .60 1 Adult smokers .10 4

Calculate the conditional probability that a person with exactly three common colds in a year is an adult smoker.

A. .12 B. .16 C. .20 D. .24 E. .28 (05S–M–39)

196 Daniel Poisson


B27. 1) Calculate the expected number of claims under the current system: P(1RC) = 1 − P(0RC) = 1 − e-1 P(2F) = 1 − P(0F + 1F) = 1 − 3e-2 ENC = P(1RC + 0F) + P(0RC + 1F) + 2 P(1RC +1F) + 2 P(0RC + 2F) + 3 P(1RC + 2F)

ENC = (1 − e-1)(e-2) + (e-1)(2e-2) + (2)(1 − e-1)(2e-2) + (2)(e-1)(1 − 3e-2) + (3)(1 − e-1)(1 − 3e-2) ENC = .08555 + .09957 + (2)(.17110) + (2)(.21852) + (3)(.37547) = 2.0908 2) Calculate the expected number of claims under the proposed system:

ENC' = P(1) + (2)P(2) + [3][1 − P(0) − P(1) − P(2)] ENC' = 3e-3 + (2)(3)2e-3/2 + [3][1 − e-3 − 3e-3 − (3)2e-3/2] = 3 − 13.5e-3 = 2.32787

3) Calculate the increase in the expected number of claims: ENC' − ENC = 2.32787 − 2.0908 = .23707

Answer: C B28. P(N = 0) = e-.6 = .54881 P(N = 1) = .6e-.6 = .32929 P(N ≥ 2) = 1 − P(N = 0) − P(N = 1) = 1 − .54881 − .32929 = .12190 E[X] = [1,000][(1)(.32929) + (2)(.12190)] = 573 E[X2] = [1,000]2[(1)(.32929) + (4)(.12190)] = 816,890 Var(X) = E[X2] − (E[X])2 = 816.890 − (573)2 = 488,561 SD(X) = 699

Answer: B B29. 1) Calculate the unconditional probabilities. P(N = 2 and λ = 1) = (.5)(1)2e-1/2! = .09197 P(N = 2 and λ = 2) = (.35)(2)2e-2/2! = .09473 P(N = 2 and λ = 3) = (.15)(3)2e-3/2! = .03361 P(N = 2) = .09197 + .09473 + .03361 = .22031

2) Calculate the conditional probabilities: P(λ = 1 | N = 2) = .09197/.22031 = .41746 P(λ = 2 | N = 2) = .09473/.22031 = .42998 P(λ = 3 | N = 2) = .03361/.22031 = .15256 3) Calculate the probabilities of more than two claims for each group: P(N > 2 | λ = 1) = 1 − [e-1 + 1e-1 + (1)2e-1/2!] = 1 − 2.5e-1 = .08030 P(N > 2 | λ = 2) = 1 − [e-2 + 2e-2 + (2)2e-2/2!] = 1 − 5e-2 = .32332 P(N > 2 | λ = 3) = 1 − [e-3 + 3e-3 + (3)2e-3/2!] = 1 − 8.5e-3 = .57681

4) Calculate the posterior probability P(N > 2) = (.08030)(.417456) + (.32332)(.42998) + (.57681)(.15256) = .26054

Answer: C B30. Current Drivers with Discount = 100,000e-1.5 = 22,313 Proposed Drivers with Discount = (100,000)(e-2.5 + 2.5e-2.5) = 28,730 Current Premium = (22,313)(850) + (77,687)(1,000) = 96,653,050

Proposed Discount = Current Premium − (Proposed Drivers without Discount)(1,000)

Proposed Drivers with Discount

PD = [96,653,050 − (100,000 − 28,730)(1,000)]/28,730 = 884

Answer: A B31. 1) Calculate the unconditional probabilities. P(N=3 and λ=3) = (.3)(3)3e-3/3! = .06721 P(N = 3 and λ = 1) = (.6)(1)3e-1/3! = .03679 P(N=3 and λ=4) = (.1)(4)3e-4/3! = .01954 P(N=3) = .06721 + .03679 + .01954 = .12354

2) Calculate the conditional probability: P(λ = 4 | N = 3) = .01954/.12354 = .15817 Answer: B

Daniel Poisson 197


B32. ABC Insurance Company plans to sell insurance policies that provide coverage for damage due to hailstorms. The annual premium of $125 per policy is collected at the moment the policy is sold. ABC expects policies to be sold according to a Poisson process, with a rate of 200 per year. Hailstorm events also follow a Poisson process, with a rate of ten per year. Calculate the probability that ABC collects at least $10,000 in premiums by the time the first storm occurs.

A. .01 B. .02 C. .03 D. .04 D. .05 (05F–3–25–2)

B33. An insurance company sells three types of policies with the following characteristics: Type of Policy Proportion of Total Policies Annual Claim Frequency I 5% Poisson with λ = .25 II 20% Poisson with λ = .50 III 75% Poisson with λ = 1.00

A randomly selected policyholder is observed to have a total of one claim for year 1 through year 4. For the same policyholder, determine the Bayesian estimate of the expected number of claims in year 5.

A. < .4 B. ≥ .4 but < .5 C. ≥ .5 but < .6 D. ≥ .6 but < .7 E. ≥ .7 (06F–C–2) B34. You are given the following: i) Hurricanes occur at a Poisson rate of 1/4 per week during the hurricane season. ii) The hurricane season lasts for exactly fifteen weeks. Prior to the next hurricane season, a weather forecaster makes the statement, "There will be at least

three and no more than five hurricanes in the upcoming hurricane season." Calculate the probability that this statement will be correct.

A. < 54% B. ≥ 54%, but < 56% C. ≥ 56%, but < 58% D. ≥ 58%, but < 60% E. ≥ 60%

(08F–3L–2–2) B35. You are given the following information:

i) An insurance company processes policy endorsements at a Poisson rate of λ = 200 per day. ii) 45% of these endorsements result in a premium increase. iii) Policy endorsements are observed for one day. Using the normal approximation with no continuity correction, calculate the probability that there are

more endorsements resulting in a premium increase than endorsements not resulting in a premium increase.

A. < 5.0% B. ≥ 5.0% but < 7.5% C. ≥ 7.5% but < 10.0% D. ≥ 10.0% but < 12.5% E. ≥ 12.5% (09F–3L–10–2)

198 Daniel Poisson


B32. 10,000/125 = 80 p{S1n < S2

1} = ( λ1λ1 + λ2

)n = ( 200

200 + 10)80 = .02018

Answer: B B33. The four-year claim frequencies for the three types of policies are 1, 2, and 4. 1) Calculate the unconditional probabilities. P(N = 1 and λ = 1) = (.05)(1)e-1 = .01839 P(N = 1 and λ = 2) = (.20)(2)e-2 = .05413 P(N = 1 and λ = 4) = (.75)(4)e-4 = .05495 P(N = 1) = .01839 + .05413 + .05495 = .12747

2) Calculate the conditional probabilities: P(λ = 1 | N = 1) = .01839/.12747 = .14427 P(λ = 2 | N = 1) = .05413/.12747 = .42465 P(λ = 4 | N = 1) = .05495/.12747 = .43108

3) Calculate the expected number of claims in year 5: E[λ | N = 1) = (.25)(.14427) + (.50)(.42465) + (1)(.43108) = .67947

Answer: D B34. λ = (15)(1/4) = 3.75 P(N = 3) = (3.75)3e-3.75/3! = 8.78906e-3.75

P(N = 4) = (3.75)4e-3.75/4! = 8.23975e-3.75

P(N = 5) = (3.75)5e-3.75/5! = 6.17981e-3.75

P(3 ≤ N ≤ 5) = P(N = 3) + P(N = 4) + P(N = 5) = (8.78906 + 8.23975 + 6.17981e-3.75) P(3 ≤ N ≤ 5) = (23.20862)(.02352) = 54.6%

Answer: B B35. Let N represent a policy with no premium increase and I a policy with a premium increase. 200 E[N – I] = (200)(.55 – .45) = 20 Var(N – I) = E[X] = 200 SD(N – I) = 200

P[(N – I) < 0) ≈ P(Z < 0 − E[N – I]SD(N – I) ) = P(Z <

0 − 20200

) = Φ(.–1.414)

P[(N – I) < 0) = 1 – .92131 = .07869 Answer: C

Daniel Poisson 199


B36. You are given the following information regarding bank collapses: i) Bank collapses occur Monday through Thursday and are classified into two categories: severe

and mild. ii) Severe bank collapses occur only on Mondays and Tuesdays, and follow a Poisson distribution

with λ = 1 on each of those days. iii) Mild bank collapses can occur only on Wednesdays and Thursdays. iv) If there was no more than one severe bank collapse earlier in the week, then mild bank

collapses follow a Poisson distribution with λ = 1 for Wednesdays and Thursdays. v) If there was more than one severe bank collapse earlier in the week, then mild bank

collapses follow a Poisson distribution with λ = 2 for Wednesdays and Thursdays. Calculate the probability that no mild bank collapses occur during one week. A. < 6.5% B. ≥ 6.5% but < 7.0% C. ≥ 7.0% but < 7.5% D. ≥ 7.5% but < 8.0% E. ≥ 8.0%

(10S–3L–13–2) B37. You are given the following information:

i) A Poisson process N has a rate function λ(t) = 3t2. ii) You have observed 50 events by time t = 2.1.

Calculate Var[N(3) | N(2.1) = 50]. A. < 10 B. ≥ 10 but < 20 C. ≥ 20 but < 30 D. ≥ 30 but < 40 E. ≥ 40 (10F–3L–11–2)

200 Daniel Poisson


B36. The mean for the two-day period of Monday and Tuesday equals 2. The means for the two situations for the two-day period of Wednesday and Thursday equal 2 and 4 respectively.

P(No Mild Bank Collapse) = P(0 or 1 Severe Bank Collapses) P(No Mild Bank Collapse | 0 or 1 Severe Bank Collapses) + P( > 1 Severe Bank Collapses) P(No Mild Bank Collapse | > 1 Severe Bank Collapses) P(Outcome) = (e-2 + 2e-2)(e-2) + (1 – e-2 – 2e-2)(e-4) = 4e-4 – 3e-6

P(Outcome) = .07326 – .00744 = .06582 Answer: B B37. Since the Poisson is memoryless, calculate the value of λ(t) from 2.1 to 3.0.

⌡⌠2.1

3 3t2 dt = t3|3

2.1 = (3)3 – (2.1)3 = 17.739

Answer: B

Daniel Poisson 201


C. Waiting Times C1. The number of cars crossing a certain intersection during any time interval of length t minutes between

3:00 p.m. and 4:00 p.m. has a Poisson distribution with mean t. Let W be time elapsed after 3:00 p.m. before the first car crosses the intersection. What is the probability that W is less than two minutes?

A. 1 − 2e-1 − e-2 B. e-2 C. 2e-1 D. 1 − e-2 E. 2e-1 + e-2 (85S–2−9) C2. Customers arrive randomly and independently at a service window, and the time between arrivals has an

exponential distribution with a mean of 12 minutes. Let X equal the number of arrivals per hour. What is P(N = 10)?

A. 10e-12/10! B. (10)12e-10/10! C. (12)10e-10/10! D. (12)10e-12/10! E. (5)10e-5/10! (90S–110−48)

C3. If the expectation of the increase in the number of claims in an interval of length t is λt, then the

distribution of the waiting time between claims is exponential with parameter λ. (90F–151–16–MC) C4. An aggregate claims process is compound Poisson. The probability that the waiting time until the next

claim will be at least two years is .60. Determine the probability that exactly four claims will occur within five years. (Use the approximation log .6 = −.51083.) A. .03 B. .11 C. .19 D. .24 E. .28 (91S–151–14)

C5. For a claims process (N(t), t > 0), you are given that the elapsed times between successive claims are

independent and identically distributed with distribution function F(t) = 1 − e-2t. Determine the probability that exactly three claims will occur in an interval of length 1.5. A. .20 B. .22 C. .24 D. .26 E. .28 (93F–151–12–2)

C6. For a claim number process {N(t), t ≥ 0} you are given that the elapsed time between successive claims

are mutually independent and identically distributed with distribution function F(t) = 1 − e-3t, t ≥ 0 Determine the probability of exactly four claims in an interval of length 2. A. .11 B. .13 C. .15 D. .17 E. .19 (97F–151–7–1) C7. Subway trains arrive at a station at a Poisson rate of 20 per hour. 25% of the trains are express and 75%

are local. The types of each train are independent. An express gets you to work in 16 minutes and a local gets you there in 28 minutes. You always take the first train to arrive. Your coworker always takes the first express. You both are waiting at the same station. Which of the following is true?

A. Your expected arrival time is 6 minutes earlier than your coworker's. B. Your expected arrival time is 4.5 minutes earlier than your coworker's. C. Your expected arrival times are the same. D. Your expected arrival time is 4.5 minutes later than your coworker's. E. Your expected arrival time is 6 minutes later than your coworker's. (02F–3–20) (Sample–M–48)

C8. Justin takes the train to work each day. It takes ten minutes for Justin to walk from home to the train

station. In order to get to work on time, Justin must board the train by 7:50 a.m. Trains arrive at the station at a Poisson rate of 1 every eight minutes. What is the latest time he must leave home each morning so that he is on time for work at least 90% of the time?

A. 7:21 a.m. B. 7:22 a.m. C. 7:31 a.m. D. 7:32 a.m. E. 7:41 a.m. (04F–3C–18–2)

202 Daniel Poisson


Solutions are based on pp. 13–14. C1. Since the number of automobiles has a Poisson distribution, the waiting time follows an exponential

distribution with λ = 1. P(W < 2) = 1 − e-W = 1 − e-2

Answer: D C2. Since the time between arrivals has an exponential distribution, the number of arrivals has a Poisson

distribution with λ = 60/θ = 60/12 = 5. P(N = 10) = (5)10e-5/10! Answer: E C3. T. C4. Since the number of claims has a Poisson distribution, waiting time has an exponential distribution. e-2λ = .60 −2λ = −.51083 λ = .25542 λ' = 5λ = 1.2771 P(N = 4) = (1.2771)4e-1.2771/4! = .03 Answer: A C5. Since elapsed time has an exponential distribution, the number of claims has a Poisson distribution with

λ' = λt = (2)(1.5) = 3. P(N = 3) = (3)3e-3/3! = .22 Answer: B C6. Since elapsed time has an exponential distribution with λ = 3, the number of claims has a Poisson with

parameter λ' = λt = (3)(2) = 6. P(N = 4) = (6)4e-6/4! = .13 Answer: B C7. Express trains arrive at the rate of (.25)(20) = 5 per hour. E[Waiting Time]Any = 60/λ = 60/20 = 3 E[Waiting Time]Express = 60/5 = 12 E[Travel Time]You = (.75)(Travel Time)Local + (.25)(Travel Time)Express = (.75)(28) + (.25)(16) = 25 E[Arrival Time]You = E[Waiting Time] + E[Travel Time] = 3 + 25 = 28 E[Arrival Time]Co-worker = 12 + 16 = 28 Answer: C C8. Calculate the 90th percentile for interarrival times, which follow an exponential distribution since a

Poisson process describes the number of arrivals. .90 = F(x) = 1 − e-x/θ = 1 − e-x/8 .10 = e-x/8 x = − 8 ln .10 = (−8)(−2.30259) x = 18.42 or 18 minutes and 25 seconds Adding 10 minutes for walking, we get 28 minutes, 25 seconds and subtracting this from 7:50, we get

7:21:35 Answer: A

Daniel Poisson 203


C9. Subway trains arrive at your station at a Poisson rate of 20 per hour. 25% of the trains are express and 75% are local. The types and number of trains arriving are independent. An express gets you to work in 16 minutes and a local gets you there in 28 minutes. You always take the first train to arrive. Your coworker always takes the first express. You are both waiting at the same station. Calculate the conditional probability that you arrive at work before your coworker, given that a local arrives first.

A. 37% B. 40% C. 43% D. 46% E. 49% (05S–M–24) C10. A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponential with mean

one 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. .05 B. .10 C. .15 D. .20 E. .25 (05F–M–8) C11. The time elapsed between claims processed is modeled such that Vk represents the time elapsed

between processing the (k − 1)th and kth claim. (V1 is the time until the first claim is processed.) You are given:

i) V1, V2, . . ., are mutually independent. ii) The pdf of each Vk is f(t) = .2e-.2t, t > 0, where t is measured in minutes. Calculate the probability of at least two claims being processed in a ten-minute period.

A. .2 B. .3 C. .4 D. .5 E. .6 (06F–M–8) C12. You arrive at a subway station at 6:15. Until 7:00, trains arrive at a Poisson rate of one train per 30

minutes. Starting at 7:00, they arrive at a Poisson rate of two trains per 30 minutes. Calculate your expected waiting time (in minutes) until a train arrives.

A. 24 B. 25 C. 26 D. 27 E. 28 (06F–M–10) C13. You are given the following information:

i) The number of wild fires per day in a state follows a Poisson distribution. ii) The expected number of wild fires in a thirty-day time period is 15. Calculate the probability that the time between the eighth and ninth fire will be greater than three days. A. < 20% B. ≥ 20% but < 25% C. ≥ 25% but < 30% D. ≥ 30% but < 35% E. ≥ 35% (07S–3–1–2)

C14. Heart/lung transplant claims in 2007 have interarrival times that are independent with a common distribution, which is exponential with mean one month. At of the end of January 2007, no transplant claims have arrived. Calculate the probability that at least three heart/lung transplant claims will have arrived by the end of March 2007. A. .18 B. .25 C. .32 D. .39 E. .45 (07S–MLC–5)

C15. You are given the following information about the interarrival times for tornadoes in county XYZ. The waiting time in days between tornadoes follows an exponential distribution and remains constant throughout the year. The probability that more than 30 days elapses between tornadoes is .60. Calculate the expected number of tornadoes in the next 90 days.

A. < 1.0 B. ≥ 1.0 but < l.5 C. ≥ 1.5 but < 2.0 D. ≥ 2.0 but < 2.5 E. ≥ 2.5 (07F–3–1–2)

204 Daniel Poisson


C9. The coworker will arrive first if an express arrives within 28 − 16 = 12 minutes. Interarrival times follow an exponential distribution since a Poisson process describes the number of arrivals.

λ = (.25)(20) = 5 θ = 60/5 = 12 P(You Arrive First) = 1 − F(12) = e-12/θ = e-12/12 = 36.8% Answer: A C10. Since the future lifetime is exponential with mean 1, the probability of failure is a Poisson with λ = 1. P(Transmission) = P(No Failures in 3 Years) + P(One Failure in 3 Years)

P(Transmission) = e-(1)(3) + ⌡⌠0

3 e-te-(3-t) dt = e-3 + te-3|3

0 = 4e-3 = .19915

Answer: D C11. Since the waiting time is an exponential with mean 5, the number of claims is a Poisson with mean

1/5 = .2. λ = (10)(.2) = 2 P(N ≥ 2) = 1 − P(N = 0) − P(N = 1) P(N ≥ 2) = 1 − e-2 − 2e-2 = 1 − 3e-2 = .59399 Answer: E C12. Since the number of claims is Poisson with mean (per minute) 1/30, the waiting time is exponential with

mean 30 before 7:00. Since the number of claims is Poisson with mean (per minute) 1/15, the waiting time is exponential with mean 15 after 7:00.

E[Waiting Time 6:15 to 7:00] = ⌡⌠0

45 e-x/30 dx = (30)(1 − e-45/30) = 23.306

E[Waiting Time After 7:00] = P(No Trains Before 7) E[Waiting Time | θ = 15] = (e-45/30)(15) = 3.35 Total Waiting Time = 23.306 + 3.35 = 26.656 Answer: D C13. λ = 15/30 = .5 P[T11 > 3] = e-3λ = e-(3)(.5) = .22313 Answer: B C14. λ = (2)(1) = 2 P(N ≥ 3) = 1 − P(0) − P(1) − P(2) = 1 − e-2 − 2e-2 − (2)2e-2/2 = 1 − 5e-2 = .32332, Answer: C

C15. .60 = e-30/θ θ = −30

ln .60 = −30

−.51083 = 58.72795 λ = 90/θ = 90/58.72795 = 1.53249

Answer: C

Daniel Poisson 205


C16. Car crashes occur according to a Poisson process at a rate of two per hour. There have been twelve crashes between 9:00 a.m. and l0:00 a.m. Given that it is now 10:00 a.m., at what time do you expect the 13th crash?

A. Before 10:05 a.m. B. On or after 10:05 a.m. but before 10:15 a.m. C. On or after 10:15 a.m. but before 10:25 a.m. D. On or after 10:25 a.m. but before 10:35 a.m. E. On or after 10:35 a.m. (07F–3–2–2) C17. Hurricanes make landfall according to a homogeneous Poisson process, with a rate of 1.5 per month. Calculate the variance (in months squared) of the waiting time until the third hurricane makes landfall. A. < 1.35 B. ≥ 1.35 but < 1.45 C. ≥ 1.45 but < 1.55 D. ≥ 1.55 but < 1.65 E. ≥ 1.65 (10F–3L–10–2)

206 Daniel Poisson


C16. Since the number of crashes is Poisson with mean 2 per hour , the waiting time is exponential with mean 1/2. Ignoring the number of crashes before 10:00, the next crash is expected 1/2 hour after 10:00.

Answer: D

C17. Event times are gamma random variables with α = n and θ = 1/λ and variance equal to n/λ2. Variance (T3) = n/λ2 = 3/(1.5)2 = 1.33333 Answer: A

Daniel Poisson 207


D. Probabilities of Compound Poisson Distributions D1. The distribution of aggregate claims arising from a portfolio is compound Poisson with the first two

values of its distribution function given by: x FS(x) 0 1/e 1 2/e Individual claims may only have integral values. Determine FS(3). A. 6/3e B. 7/3e C. 8/3e D. 9/3e E. 10/3e (84F–5–18) D2. Aggregate claims (S) can be modeled by a compound Poisson distribution with λ = 10. Calculate fS(5) if

the distribution of the claim amount (X) is as follows:

Claim amount 1 2 3 5 Probability .2 .4 .2 .2

A. (178/5)e-10 B. (181/5)e-10 C. (183/5)e-10 D. (188/5)e-10 E. (193/5)e-10 (85F–5–7) D3. A compound Poisson process has expected number of claims of 1. Claim amounts are 1 and 2 with

probabilities 2/3 and 1/3 respectively. Calculate the expected aggregate claims given that aggregate claims are less than or equal to 2.

A. 16/9e B. 4/5 C. 3/e D. 4/3 E. 14/9 (85F–5–10) D4. S1 has a compound Poisson distribution with λ = 3 and f1

X(1) = f1X(2) = f1

X(3) = 1/3. S2 has a

compound distribution with λ = 2 and f2X(1) = f2

X(2) = 1/2. S1 and S2 are independent. Determine

f1+2X (2) for the distribution S1 + S2.

A. 1/6 B. 1/5 C. 2/5 D. 5/12 E. 13/20 (86S–5–8)

D5. A compound Poisson distribution has expected amount of claims of 3. All claim amounts are positive

integers.

fS(1) = e-3 fS(2) = (3/2)e-3

Determine fX(2). A. 1/6 B. 1/3 C. 1/2 D. 2/3 E. 3/4 (86F–151–7)

D6. S has a compound Poisson distribution with λ = 2. The distribution of claim amounts is as follows:

fX(200) = .2, fX(300) = .3, and fX(500) = .6. Calculate the probability that S ≤ 500. (Use the approximation e-2 = .135.)

A. .43 B. .45 C. .47 D. .49 E. .51 (87S–151–9)

D7. The distribution of aggregate claims, S, has a compound Poisson distribution with λ = 2 and possible

claim amounts 1, 2 and 3. Determine fX(3), given that fS(1) = e-2 and fS(2) = 1.1e-2.

A. .2 B. .3 C. .4 D. .5 E. .6 (88S–151–8)

208 Daniel Poisson


Solutions are based on pp. 6–8, 14. D1. fS(0) = e-1 λ = 1 fS(1) = P(N = 1) P(X = 1) = e-1 fX(1)

fX(1) = 1 FS(3) = P(N = 0) + P(N = 1) + P(N = 2) + P(N = 3) FS(3) = e-1 + e-1 + e-1/2 + e-1/6 = 8e-1/3

Answer: C

D2. fS(5) = P(N = 1 and X = 5) + P(N = 5 and X = 1) + P(N = 2, X = 2, and X = 3) + P(N = 3, X = 1, X = 2, and X = 2) + P(N = 3, X = 1, X = 1, and X = 3) + P(N = 4, X = 1, X = 1, X = 1, and X = 2)

fS(5) = (10)(e-10)(.2) + (10)5(e-10/5!)(.2)5 + (2)(10)2(e-10/2)(.4)(.2) + (3)(10)3(e-10/3!)(.2)(.4)(.4) + (3)(10)3(e-10/3!)(.2)(.2)(.2) + (4)(10)4(e-10/4!)(.2)(.2)(.2)(.4) = (2 + 4/15 + 8 + 16 + 4 + 16/3) e-10 = e-10(178/5)

Answer: A

D3. fS(0) = e-1 fS(1) = P(N = 1 and X = 1) = (e-1)(2/3) fS(2) = P(N = 1 and X = 2) + P(N = 2 and X = 1) = (e-1)(1/3) + (e-1/2)(2/3)2 = (e-1)(5/9)

FS(2) = fS(0) + fS(1) + fS(2) = e-1 + (e-1)(2/3) + (e-1)(5/9) = (e-1)(20/9) E[S | S ≤ 2] = [(0)(e-1) + (1)(e-1)(2/3) + (2)(e-1)(5/9)/FS(2)] = 16/9e-1/(20/9)e-1 = 4/5

Answer: B D4. Take a weighted average of the probabilities for each distribution: fX(2) = (3/5)(1/3) + (2/5)(1/2) = 2/5

Answer: C D5. fX(1) = fS(1)/P(N = 1) = e-3/3e-3 = 1/3

fX(2) = fS(2) − [P(N = 2)][fX(1)]2

P(N = 1) = (3/2)e-3 − [(9/2)e-3][1/3]2

3e-3 = 1/3

Answer: B D6. fS(0) = e-2 fS(200) = P(N = 1 and X = 200) = (2)(e-2)(.2) = .4e-2 fS(300) = P(N = 1 and X = 300) = (2)(e-2)(.3) = .6e-2 fS(400) = P(N = 2 and X = 200) = (2)2(e-2/2)(.2)2 = .08e-2 fS(500) = P(N = 1 and X = 500) + P(N = 2, X = 200, and X = 300) fS(500) = (2)(e-2)(.5) + (2)(2)2(e-2/2)(.2)(.3) = 1.24e-2 FS(500) = fS(0) + fS(200) + fS(300) + fS(400) + fS(500) FS(500) = e-2 + .4e-2 + .6e-2 + .08e-2 + 1.24e-2 = 3.32e-2 = .45

Answer: B D7. fX(1) = fS(1)/P(N = 1) = e-2/2e-2 = .5

fX(2) = fS(2) − [P(N = 2)][fX(1)]2

P(N = 1) = 1.1e-2 − [2e-2][(.5)2]

2e-2 = .3

fX(3) = 1 − fX(1) − fX(2) = 1 − .5 − .3 = .2

Answer: A

Daniel Poisson 209


D8. Aggregate claims, S, has a compound Poisson distribution with λ = 1. Individual claim amounts are 1, 2, or 3. Given fS(1) = .5e-1 and fS(2) = 3/8e-1, determine fX(2).

A. 1/8 B. 1/4 C. 3/8 D. 1/2 E. 5/8 (88F–151–10) D9. The distribution of aggregate claims is compound Poisson. The first three values of the distribution

function are as follows;

FS(0) = 1/e FS(1) = 1/e FS(2) = 2/e Individual claim amounts can have only positive integral values. Determine FS(4). A. 5/2e B. 18/7e C. 13/5e D. 8/3e E. 19/7e (89S–151–7) D10. S has a compound Poisson distribution with λ = 1.0 and claim amounts that are 1, 2, or 3 with

probabilities .50, .25, and .25, respectively. P(N1 = 0) = .61. Determine fS(3).

A. .13 B. .15 C. .17 D. .19 E. .21 (89F–151–10) D11. S has a compound Poisson distribution with λ = .6 and individual claim amounts that are 1, 2 or 3 with

probability .2, .3 and .5, respectively. Determine Pr (S ≥ 3). (Use the approximation e-.6 = .55.) A. .269 B. .271 C. .275 D. .279 E. .281 (90S–151–7) D12. Let N = number of claims and S = Xl + X2 + . . . + XN. Suppose S has a compound Poisson distribution

with Poison parameter λ = .6. The only possible individual claim amounts are $2,000, $5,000, and $10,000 with probabilities .6, .3, and .1, respectively. Calculate P(S ≤ $7,0 00 | N ≤ 2). Note: X is Poisson distributed with parameter λ and the density function, f(x) = λxe-λ/x!. Show all work. (94F–5A–37–2)

D13. Suppose that the aggregate loss S has a compound Poisson distribution with expected number of claims

equal to 3 and the following claim amount distribution: individual claim amounts can be 1, 2 or 3 with probabilities of .6, .3, and .1, respectively. Calculate the probability that S = 2. Show all work. Note: X is Poisson distributed with parameter (λ) and the density function, f(x) = (λxe-λ)/x! (95F–5A–36–2)

D14. Claims arising from a particular insurance policy have a compound Poisson distribution. The expected

number of claims is five. The claim amount density function is given by: P(X = 1,000) = .8 P(X = 5,000) = .2 Compute the probability that losses from this policy will total 6,000. Show all work. (96F–5A–37–2) D15. Given the following, determine fS(4). i) S has a compound Poisson distribution with λ = 2.

ii) Individual claim amounts (x) are distributed as follows: p(1) = .4 and p(2) = .6 A. .05 B. .07 C. .10 D. .15 E. .21 (98S–151–12–2)

210 Daniel Poisson


D8. fX(1) = fS(1)/P(N = 1) = .5e-1/e-1 = .5

fX(2) = fS(2) − [P(N = 2)][fX(1)]2

P(N = 1) = (3/8)e-1 − [e-1/2][(.5)2]

e-1 = 1/4

Answer: B D9. fS(0) = e-1 fS(1) = FS(1) − FS(0) = e-1 − e-1 = 0 fS(2) = FS(2) − FS(1) = 2e-1 − e-1 = e-1 = P(N = 1) Thus, all claims must be of size two and fS(4) = P(N = 2) = e-1/2 Since fS(3) must equal zero, FS(4) = FS(2) + fS(4) = 2/e + 1/2e = 5/2e Answer: A D10. fS(3) = P(N = 1 and X = 3) + P(N = 2, X = 1, and X = 2) + P(N = 3, and X = 1) fS(3) = e-1(.25) + (2)(e-1/2)(.50)(.25) + (e-1/6)(.50)3 = .3958e-1 = .15 Answer: B D11. fS(0) = e-.6 fS(1) = P(N = 1 and X = 1) = (.6)(e-.6)(.2) fS(2) = P(N = 1 and X = 2) + P(N = 2 and X = 1) = (.6)(e-.6)(.3) + (.6)2(e-.6/2)(.2)2 = .1872e-1

FS(2) = fS(0) + fS(1) + fS(2) = e-.6 + .12e-.6 + .1872e-.6 = 1.3072e-.6 = (1.3072)(.55) FS(2) = .719 P(S ≥ 3) = 1 − FS(2) = 1 − .719 = .281

Answer: E D12. P(N ≤ 2) = P(N = 0) + P(N = 1) + P(N = 2) = e-.6 + .6e-.6 + (.6)2(e-.6/2) = 1.78e-.6 P(S ≤ 7,000 and N ≤ 2) = P(N = 0) + P(N = 1 and X = 2,000) + P(N = 1 and X = 5,000)

+ P(N = 2 and X = 2,000) + P(N = 2, X = 2,000, and X = 5,000) P(S ≤ 7,000 and N ≤ 2) = e-.6 + (.6)(e-.6)(.6) + (.6)(e-.6)(.3) + (.6)2(e-.6/2)(.6)2 + (2)(.6)2(e-.6/2)(.6)(.3) = 1.6696e.6 P(S ≤ 7,000 | N ≤ 2) = P(S ≤ 7,000 and N ≤ 2)/P(N ≤ 2) = 1.6696e.6/1.78e.6 = .938 D13. fS(2) = P(N = 1 and X = 2) + P(N = 2 and X = 1) = (3)(e-3)(.3) + (3)2(e-3/2)(.6)2 fS(2) = 2.52e-3 = .1255 D14. P(S = 6,000) = P(N = 6 and X = 1,000) + P(N = 2, X = 1,000, and X = 5,000) P(S = 6,000) = (5)6(e-5/6!)(.8)6 + (2)(5)2(e-5/2)(.8)(.2) = (436/45)e-5 = .0653 D15. fS(4) = P(N = 2 and X = 2) + P(N = 3, X = 1, X = 1, and X = 2) + P(N = 4, and X = 1) fS(4) = (2)2(e-2/2)(.6)2 + (2)3(e-2/6)(3)(.4)2(.6) + (2)4(e-2/24)(.4)4 = 1.12107e-2 = .15172 Answer: D

Daniel Poisson 211


D16. You are given:

i) An aggregate loss distribution has a compound Poisson distribution with expected number of claims equal to 1.25.

ii) Individual claim amounts can take only the values 1, 2 or 3, with equal probability. Determine the probability that aggregate losses exceed 3. (Sample–3–14) D17. A compound Poisson distribution has λ = 5 and claim amount distribution as follows: p(100) = .80 p(500) = .16 p(1,000) = .04 Calculate the probability that aggregate claims will be exactly 600. A. .022 B. .038 C. .049 D. .060 E. .070 (06F–M–40) D18. Annual aggregate losses for a dental policy follow the compound Poisson distribution with λ = 3. The

distribution of individual losses is:

Loss 1 2 3 4 Probability .4 .3 .2 .1

Calculate the probability that aggregate losses in one year do not exceed 3. A. < .20 B. ≥ .20 but < .40 C. ≥ .40 but < .60 D. ≥ .60 but < .80 E. ≥ .80 (07S–C–8) D19. You are given the following information:

i) An insurance policy covers a dwelling for damage only due to hailstorms and floods. ii) Aggregate hailstorm damage follows a compound Poisson process. iii) The average time between hailstorms is three months. iv) The damage caused by an individual hailstorm follows an exponential distribution with

θ = 2,000. v) Aggregate flood damage follows a compound Poisson process. vi) The average time between floods is twelve months. vii) The damage caused by an individual flood follows an exponential distribution with θ = 10,000.

Calculate the probability that the damage caused by an insured event exceeds 7,500.

A. < 5% B. ≥ 5% but < 7.5% C. ≥ 7.5% but < 10% D. ≥ 10% but < 12.5% E. ≥ 12.5% (10F–3L–12–2)

212 Daniel Poisson


D16. P(S ≤ 3) = P(N = 0) + P(N = 1) + P(N = 2, X = 1, and X = 1) + P(N = 2, X = 1, and X = 2) + P(N = 3, X = 1, X = 1, and X = 1) P(S ≤ 3) = e-1.25 + 1.25e-1.25 + (1.25)2(e-1.25/2)(1/3)2 + (2)(1.25)2(e-1.25/2)(1/3)2 + (1.25)3(e-1.25/3)(1/3)3 = .28650 + .35813 + .02487 + .04974 + .00691 P(S ≤ 3) = .72615 P(S > 3) = 1 − P(S ≤ 3) = 1 − .72615 = .274 D17. fS(600) = P(N = 6 and X = 100) + P(N = 2 and X = 100 and X = 500) fS(600) = (5)6(e-5/6!)(.8)6 + (5)2(e-5/2)(2)(.8)(.16) = .03833 + .02156 = .05989 Answer: D D18. P(S ≤ 3) = P(N = 0) + P(N = 1, X = 1, 2, or 3) + P(N = 2, X = 1, and X = 1) + P(N = 2, X = 1, and X = 2) + P(N = 3, X = 1, X = 1, and X = 1) P(S ≤ 3) = e-3 + 3e-3(.9) + (3)2(e-3/2)(.4)2 + (2)(3)2(e-3/2)(.4)(.3) + (3)3(e-3/6)(.4)3 = .04979 + .13443 + .03585 + .05377 + .01434 = .28818 Answer: B D19. Calculate a weighted average of the probabilities for an excess hail loss and an excess flood loss. Based

on the interevent times, weight hailstorms 80% and floods 20%: Pr(Hailstorm > 7,500) = 1 – FH(7,500) = e-7,500/2,000 = .02352 Pr(Flood > 7,500) = 1 – FF(7,500) = e-7,500/10,000 = .47237 Pr(Event > 7,500) = .8 Pr(Hailstorm > 7,500) + .2 Pr(Flood > 7,500) Pr(Event > 7,500) = (.8)(.02352) + (.2)(.47237) = .11329 Answer: D

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