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1 Contents 1 Probability Review Functions and moments Probability distributions Bernoulli distribution Uniform distribution Exponential distribution Variance Normal approximation Conditional probability and expectation Conditional variance Exercises Solutions Survival Distributions: Probability Functions Probability notation Actuarial notation Life tables Exercises Solutions Survival Distributions: Force of Mortality 37 Exercises Solutions Survival Distributions: Mortality Laws Mortality laws that may be used for human mortality Gompertz s law Makeham s law Weibull Distribution Mortality laws for exam questions Exponential distribution, or constant force of mortality Uniform distribution Beta distribution Exercises Solutions Survival Distributions: Moments Complete General Special mortality laws Curtate Exercises Solutions Survival Distributions: Percentiles and Recursions Percentiles iii

2 iv CONTENTS 6.2 Recursive formulas for life expectancy Exercises Solutions Survival Distributions: Fractional Ages 125 Exercises Solutions Insurance: Annual Moments Review of Financial Mathematics Moments of annual insurances Standard insurances and notation Illustrative Life Table Constant force and uniform mortality Normal approximation Exercises Solutions Insurance: Continuous Moments Part Definitions and general formulas Constant force of mortality Exercises Solutions Insurance: Continuous Moments Part Uniform survival function Other mortality functions Integrating at n e ct (Gamma Integrands) Variance of endowment insurance Normal approximation Exercises Solutions Insurance: Probabilities and Percentiles Introduction Probabilities for Continuous Insurance Variables Probabilities for Discrete Variables Percentiles Exercises Solutions Insurance: Recursive Formulas 241 Exercises Solutions Annuities: Continuous, Expectation Whole life annuity Temporary and deferred life annuities n-year certain-and-life annuity Exercises Solutions

3 CONTENTS v 14 Annuities: Discrete, Expectation Annuities-due Annuities-immediate Actuarial Accumulated Value Exercises Solutions Annuities: Variance Whole Life and Temporary Life Annuities Other Annuities Typical Exam Questions Combinations of Annuities and Insurances with No Variance Exercises Solutions Annuities: Probabilities and Percentiles Probabilities for continuous annuities Distribution functions of annuity present values Probabilities for discrete annuities Percentiles Exercises Solutions Annuities: Recursive Formulas 353 Exercises Solutions Premiums: Net Premiums for Fully Continuous Insurances Future loss Net premium Expected value of future loss International Actuarial Premium Notation Exercises Solutions Premiums: Net Premiums from Life Tables 381 Exercises Solutions Premiums: Net Premiums from Formulas 391 Exercises Solutions Premiums: Variance of Future Loss, Continuous 411 Exercises Solutions Premiums: Variance of Future Loss, Discrete 427 Exercises Solutions

4 vi CONTENTS 23 Premiums: Probabilities and Percentiles of Future Loss Probabilities Fully continuous insurances Discrete insurances Annuities Percentiles Exercises Solutions Multiple Lives: Joint Life Probabilities Introduction Independence Exercises Solutions Multiple Lives: Last Survivor Probabilities 471 Exercises Solutions Multiple Lives: Moments 485 Exercises Solutions Multiple Decrement Models: Probabilities Introduction Life Tables Examples of Multiple Decrement Probabilities Discrete Insurances Exercises Solutions Multiple Decrement Models: Forces of Decrement 527 Exercises Solutions Multi-State Models (Markov Chains): Probabilities 545 Exercises Solutions Multi-State Models (Markov Chains): Premiums and Reserves Prototype Example Insurances Annuities Premiums and Reserves Exercises Solutions Practice Exams Practice Exam 1 595

5 CONTENTS vii 2 Practice Exam Practice Exam Practice Exam Practice Exam Practice Exam Practice Exam Practice Exam Practice Exam Practice Exam Appendices 655 A Solutions to the Practice Exams 657 Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam Solutions for Practice Exam B Solutions to Old Exams 713 B.1 Solutions to CAS Exam 3, Spring B.2 Solutions to CAS Exam 3, Fall B.3 Solutions to CAS Exam 3, Spring B.4 Solutions to CAS Exam 3, Fall B.5 Solutions to CAS Exam 3, Spring B.6 Solutions to CAS Exam 3, Fall B.7 Solutions to CAS Exam 3L, Spring B.8 Solutions to CAS Exam 3L, Fall B.9 Solutions to CAS Exam 3L, Spring B.10 Solutions to CAS Exam 3L, Fall B.11 Solutions to CAS Exam 3L, Spring B.12 Solutions to CAS Exam 3L, Fall B.13 Solutions to CAS Exam 3L, Spring B.14 Solutions to CAS Exam 3L, Fall B.15 Solutions to CAS Exam 3L, Spring B.16 Solutions to SOA Exam MLC, Spring B.17 Solutions to CAS Exam 3L, Fall B.18 Solutions to SOA Exam MLC, Fall

6 viii CONTENTS B.19 Solutions to CAS Exam 3L, Spring B.20 Solutions to SOA Exam MLC, Spring B.21 Solutions to CAS Exam 3L, Fall B.22 Solutions to SOA Exam MLC, Fall B.23 Solutions to CAS Exam LC, Spring B.24 Solutions to SOA Exam MLC, Spring B.25 Solutions to CAS Exam LC, Fall B.26 Solutions to SOA Exam MLC, Fall B.27 Solutions to CAS Exam LC, Spring B.28 Solutions to SOA Exam MLC, Spring C Lessons Corresponding to Questions on Released and Practice Exams 803

7 Lesson 7 Survival Distributions: Fractional Ages Reading: Models for Quantifying Risk (4 th or 5 th edition) 6.6.1, Life tables list mortality rates (q x ) or lives (l x ) for integral ages only. Often, it is necessary to determine lives at fractional ages (like l x+0.5 for x an integer) or mortality rates for fractions of a year. We need some way to interpolate between ages. The easiest interpolation method is linear interpolation, or uniform distribution of deaths between integral ages (UDD). This means that the number of lives at age x + s, 0 s 1, is a weighted average of the number of lives at age x and the number of lives at age x + 1: l x+s (1 s)l x + sl x+1 l x sd x (7.1) The graph of l x+s is a straight line between s 0 and s 1 with slope d x. The graph at the right portrays this for a mortality rate q and l Contrast UDD with an assumption of a uniform survival function. If age at death is uniformly distributed, then l x as a function of x is a straight line. If UDD is assumed, l x is a straight line between integral ages, but the slope may vary for different ages. Thus if age at death is uniformly distributed, UDD holds at all ages, but not conversely. Using l x+s, we can compute s q x : sq x 1 s p x 1 l x+s l x 1 (1 sq x ) sq x (7.2) That is one of the most important formulas, so let s state it again: l 100+s s 550 sq x sq x (7.2) More generally, for 0 s + t 1, sq x+t 1 s p x+t 1 l x+s+t l x+t 1 l x (s + t)d x l x td x sd x l x td x sq x 1 tq x (7.3) where the last equation was obtained by dividing numerator and denominator by l x. The important point to pick up is that while s q x is the proportion of the year s times q x, the corresponding concept at age x + t, sq x+t, is not sq x, but is in fact higher than sq x. The number of lives dying in any amount of time is constant, and since there are fewer and fewer lives as the year progresses, the rate of death is in fact increasing over the year. The numerator of s q x+t is the proportion of the year being measured s times the death rate, but then this must be divided by 1 minus the proportion of the year that elapsed before the start of measurement. For most problems involving death probabilities, it will suffice if you remember that l x+s is linearly interpolated. It often helps to create a life table with an arbitrary radix. Try working out the following example before looking at the answer. 125

8 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES Example 7A You are given: q x 0.1 Uniform distribution of deaths between integral ages is assumed. Calculate 1/2 q x+1/4. Answer: Let l x 1. Then l x+1 l x (1 q x ) 0.9 and d x 0.1. Linearly interpolating, l x+1/4 l x 1 4 d x (0.1) l x+3/4 l x 3 4 d x (0.1) /2q x+1/4 l x+1/4 l x+3/ l x+1/ You could also use equation (7.3) to work this example. Example 7B For two lives age (x) with independent future lifetimes, k q x 0.1(k + 1) for k 0, 1, 2. Deaths are uniformly distributed between integral ages. Calculate the probability that both lives will survive 2.25 years. Answer: Since the two lives are independent, the probability of both surviving 2.25 years is the square of 2.25 p x, the probability of one surviving 2.25 years. If we let l x 1 and use d x+k l x k q x, we get q x 0.1(1) 0.1 l x+1 1 d x q x 0.1(2) 0.2 l x d x q x 0.1(3) 0.3 l x d x Then linearly interpolating between l x+2 and l x+3, we get l x (0.3) p x l x+2.25 l x Squaring, the answer is The probability density function of T x, s p x µ x+s, is the constant q x, the derivative of the conditional cumulative distribution function s q x sq x with respect to s. That is another important formula, since the density is needed to compute expected values, so let s repeat it: µ 100+s sp x µ x+s q x (7.4) It follows that the force of mortality is q x divided by 1 sq x : 0.45 µ x+s q x q x (7.5) sp x 1 sq x 0 s 0 1 The force of mortality increases over the year, as illustrated in the graph for q to the right.? Quiz 7-1 You are given: µ Deaths are uniformly distributed between integral ages. Calculate 0.6 q 50.4.

9 7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES 127 Complete Expectation of Life Under UDD If the complete future lifetime random variable T is written as T K + S, where K is the curtate future lifetime and S is the fraction of the last year lived, then K and S are independent. This is usually not true if uniform distribution of deaths is not assumed. Since S is uniform on [0, 1), E[S] 1 2 and Var(S) It follows from E[S] 1 2 that e x e x (7.6) More common on exams are questions asking you to evaluate the temporary complete expectancy of life under UDD. You can always evaluate the temporary complete expectancy, whether or not UDD is assumed, by integrating t p x, as indicated by formula (5.6) on page 80. For UDD, t p x is linear between integral ages. Therefore, a rule we learned in Lesson 5 applies for all integral x: e x:1 p x + 0.5q x (5.13) This equation will be useful. In addition, the method for generating this equation can be used to work out questions involving temporary complete life expectancies for short periods. The following example illustrates this. This example will be reminiscent of calculating temporary complete life expectancy for uniform mortality. Example 7C You are given q x 0.1. Deaths are uniformly distributed between integral ages. Calculate e x:0.4. Answer: We will discuss two ways to solve this: an algebraic method and a geometric method. The algebraic method is based on the double expectation theorem, equation (1.11). It uses the fact that for a uniform distribution, the mean is the midpoint. If deaths occur uniformly between integral ages, then those who die within a period contained within a year survive half the period on the average. In this example, those who die within 0.4 survive an average of 0.2. Those who survive 0.4 survive an average of 0.4 of course. The temporary life expectancy is the weighted average of these two groups, or 0.4q x (0.2) p x (0.4). This is: 0.4q x (0.4)(0.1) p x e x: (0.2) (0.4) An equivalent geometric method, the trapezoidal rule, is to draw the t p x function from 0 to 0.4. The integral of t p x is the area under the line, which is the area of a trapezoid: the average of the heights times the width. The following is the graph (not drawn to scale): t p x 1 (0.4, 0.96) (1.0, 0.9) A B t Trapezoid A is the area we are interested in. Its area is 1 2 ( )(0.4)

10 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES? Quiz 7-2 As in Example 7C, you are given q x 0.1. Deaths are uniformly distributed between integral ages. Calculate e x+0.4:0.6. Let s now work out an example in which the duration crosses an integral boundary. Example 7D You are given: q x 0.1 q x Deaths are uniformly distributed between integral ages. Calculate e x+0.5:1. Answer: Let s start with the algebraic method. Since the mortality rate changes at x + 1, we must split the group into those who die before x + 1, those who die afterwards, and those who survive. Those who die before x + 1 live 0.25 on the average since the period to x + 1 is length 0.5. Those who die after x + 1 live between 0.5 and 1 years; the midpoint of 0.5 and 1 is 0.75, so they live 0.75 years on the average. Those who survive live 1 year. Now let s calculate the probabilities. 0.5q x (0.1) 1 0.5(0.1) p x ( ) (0.5(0.2) ) p x q x+0.5 These probabilities could also be calculated by setting up an l x table with radix 100 at age x and interpolating within it to get l x+0.5 and l x+1.5. Then l x+1 0.9l x 90 l x+2 0.8l x+1 72 l x ( ) 95 l x ( ) q x p x+0.5 l x l x q x+0.5 Either way, we re now ready to calculate e x+0.5:1. e x+0.5:1 5(0.25) + 9(0.75) + 81(1) For the geometric method we draw the following graph:

11 7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES 129 t p x , , A B 0 x x + 1 t 1.0 x The heights at ( x + 1 and ) x are as we computed above. ( Then ) we compute each area separately. The area of A is (0.5) (4). The area of B is (0.5) (4). Adding them up, we get (4) ? Quiz 7-3 The probability that a battery fails by the end of the kth month is given in the following table: Probability of battery failure by k the end of month k Between integral months, time of failure for the battery is uniformly distributed. Calculate the expected amount of time the battery survives within 2.25 months. To calculate e x:n in terms of e x:n when x and n are both integers, note that those who survive n years contribute the same to both. Those who die contribute an average of 1 2 more to e x:n since they die on the average in the middle of the year. Thus the difference is 1 2 n q x : e x:n e x:n n q x (7.7) Example 7E You are given: q x 0.01 for x 50, 51,..., 59. Deaths are uniformly distributed between integral ages. Calculate e 50:10. Answer: As we just said, e 50:10 e 50: q 50. The first summand, e 50:10, is the sum of k p k for k 1,..., 10. This sum is a geometric series: e 50:10 10 k k The second summand, the probability of dying within 10 years is 10 q Therefore The formulas are summarized in Table 7.1. e 50: ( )

12 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES Table 7.1: Summary of formulas for fractional ages l x+s l x sd x sq x sq x sp x 1 sq x sq x+t sq x 1 tq x q x µ x+s 1 sq x sp x µ x+s q x e x e x e x:n e x:n n q x e x:1 p x + 0.5q x Exercises 7.1. [CAS4-S85:16] (1 point) Deaths are uniformly distributed between integral ages. Which of the following represents 3/4 p x /2 p x µ x+1/2? A. 3/4p x B. 3/4 q x C. 1/2p x D. 1/2 q x E. 1/4 p x 7.2. [Based on 150-S88:25] You are given: 0.25 q x /31. Mortality is uniformly distributed within age x. Calculate q x. Use the following information for questions 7.3 and 7.4: You are given: Deaths are uniformly distributed between integral ages. q x q x Calculate 1/2 q x+3/ Calculate q x You are given: Deaths are uniformly distributed between integral ages. Mortality follows the Illustrative Life Table. Calculate the median future lifetime for (45.5). Exercises continue on the next page...

13 EXERCISES FOR LESSON [160-F90:5] You are given: A survival distribution is defined by l x 1000 ( 1 ( ) x 2 ), 0 x µ x denotes the actual force of mortality for the survival distribution. µ L x denotes the approximation of the force of mortality based on the uniform distribution of deaths assumption for l x, 50 x < 51. Calculate µ µ L A B C. 0 D E A survival distribution is defined by S 0 (k) 1/( k) 4 for k a non-negative integer. Deaths are uniformly distributed between integral ages. Calculate 0.4 q [Based on150-s89:15] You are given: Deaths are uniformly distributed over each year of age. x l x Which of the following are true? I. 1 2q II. µ III. 0.33q A. I and II only B. I and III only C. II and III only D. I, II and III E. The correct answer is not given by A., B., C., or D. Exercises continue on the next page...

14 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES 7.9. [ :5] You are given: Deaths are uniformly distributed over each year of age p x Which of the following are true? I. 0.25q x II. 0.5q x 0.5 III. µ x A. I and II only B. I and III only C. II and III only D. I, II and III E. The correct answer is not given by A., B., C., or D [3-S00:12] For a certain mortality table, you are given: µ µ µ Deaths are uniformly distributed between integral ages. Calculate the probability that a person age 80.5 will die within two years. A B C D E You are given: Deaths are uniformly distributed between integral ages. q x 0.1. q x Calculate e x+0.7: You are given: Deaths are uniformly distributed between integral ages. q q Calculate Var ( min ( T 45, 2 )) You are given: Deaths are uniformly distributed between integral ages. 10 p x 0.2. Calculate e x:10 e x:10. Exercises continue on the next page...

15 EXERCISES FOR LESSON [4-F86:21] You are given: q q Deaths are uniformly distributed over each year of age. Calculate e 60:1.5. A B C D E [150-F89:21] You are given: q q Deaths are uniformly distributed over each year of age. Calculate e 70:1.5. A B C D E [3-S01:33] For a 4-year college, you are given the following probabilities for dropout from all causes: q q q q Dropouts are uniformly distributed over each year. Compute the temporary 1.5-year complete expected college lifetime of a student entering the second year, e 1:1.5. A B C D E You are given: Deaths are uniformly distributed between integral ages. e x+0.5:0.5 5/12. Calculate q x You are given: Deaths are uniformly distributed over each year of age. e 55.2: Calculate µ [150-S87:21] You are given: d x k for x 0, 1, 2,..., ω 1 e 20:20 18 Deaths are uniformly distributed over each year of age. Calculate q 30. A B C D E Exercises continue on the next page...

16 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES [150-S89:24] You are given: Deaths are uniformly distributed over each year of age. µ Calculate e 45:1. A. 0.4 B. 0.5 C. 0.6 D. 0.7 E [CAS3-S04:10] 4,000 people age (30) each pay an amount, P, into a fund. Immediately after the 1,000 th death, the fund will be dissolved and each of the survivors will be paid \$50,000. Mortality follows the Illustrative Life Table, using linear interpolation at fractional ages. i 12% Calculate P. A. Less than 515 B. At least 515, but less than 525 C. At least 525, but less than 535 D. At least 535, but less than 545 E. At least 545 Additional old CAS Exam 3/3L questions: S05:31, F05:13, S06:13, F06:13, S07:24, S08:16, S09:3, F09:3, S10:4, F10:3, S11:3, S12:3, F12:3, S13:3, F13:3 Additional old CAS Exam LC questions: S14:4, F14:4, S15:3 Solutions 7.1. In the second summand, 1/2 p x µ x+1/2 is the density function, which is the constant q x under UDD. The first summand 3/4 p x q x. So the sum is q x, or 1/4 p x. (E) 7.2. Using equation (7.3), q x q x q x q x 0.25q x q x q x We calculate the probability that (x+ 3 4 ) survives for half a year. Since the duration crosses an integer boundary, we break the period up into two quarters of a year. The probability of (x + 3/4) surviving for 0.25 years is, by equation (7.3), /4p x+3/ (0.10) The probability of (x + 1) surviving to x is 1/4p x (0.15)

17 EXERCISE SOLUTIONS FOR LESSON The answer to the question is then the complement of the product of these two numbers: ( ) 0.9 1/2q x+3/4 1 1/2 p x+3/4 1 1/4 p x+3/4 1/4 p x+1 1 (0.9625) Alternatively, you could build a life table starting at age x, with l x 1. Then l x+1 (1 0.1) 0.9 and l x+2 0.9(1 0.15) Under UDD, l x at fractional ages is obtained by linear interpolation, so l x (0.9) (1) l x (0.765) (0.9) /2p 3/4 l x l x /2q 3/4 1 1/2 p 3/ q x+0.4 is 0.3 p x p x+0.4. The first summand is 0.3p x q x q x The probability that (x + 0.4) survives to x + 1 is, by equation (7.3), and the probability (x + 1) survives to x is So 0.6p x p x q x (0.15) q x ( ) 90 (0.97) Alternatively, you could use the life table from the solution to the last question, and linearly interpolate: l x (0.9) + 0.6(1) 0.96 l x (0.9) + 0.3(1) 0.93 l x (0.765) + 0.8(0.9) q x Under uniform distribution of deaths between integral ages, l x (l x + l x+1 ), since the survival function is a straight line between two integral ages. Therefore, l (9,164, ,127,426) 9,145, Median future lifetime occurs when l x 1 2 (9,145,738.5) 4,572,869. This happens between ages 77 and 78. We interpolate between the ages to get the exact median: l 77 s(l 77 l 78 ) 4,572,869 4,828,182 s(4,828,182 4,530,360) 4,572,869 s 4,828, ,822s 4,572,869 4,828,182 4,572, , , ,822 So the median age at death is , and median future lifetime is

18 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES 7.6. xp 0 l x l0 1 ( x 100) 2. The force of mortality is calculated as the negative derivative of ln x p 0 : µ x d ln xp 0 dx µ For UDD, we need to calculate q 50. 2( x ) ( ( x 100 ) ) p 50 l l q x x 2 so under UDD, µ L q q ( ) The difference between µ and µ L is (B) 7.7. S 0 (20) 1/1.2 4 and S 0 (21) 1/1.21 4, so q 20 1 (1.2/1.21) Then 0.4q q ( ) q ( ) 7.8. I. Calculate 1 2 q q 36 2 d l This statement does not require uniform distribution of deaths. II. By equation (7.5), III. Calculate 0.33 q µ 37.5 q 37 4/ q / q d 38.5 (0.33)(5) l I can t figure out what mistake you d have to make to get (A) 7.9. First calculate q x q x 0.25 q x 1 Then by equation (7.3), 0.25 q x /(1 0.5) 0.5, making I true. By equation (7.2), 0.5 q x 0.5q x 0.5, making II true. By equation (7.5), µ x+0.5 1/(1 0.5) 2, making III false. (A)

19 EXERCISE SOLUTIONS FOR LESSON We use equation (7.5) to back out q x for each age. µ x+0.5 q x 1 0.5q x q x q q q µ x µ x+0.5 Then by equation (7.3), 0.5 p /0.99. p , and 0.5 p (0.06) Therefore ( ) q (0.96)(0.97) (A) To do this algebraically, we split the group into those who die within 0.3 years, those who die between 0.3 and 1 years, and those who survive one year. Under UDD, those who die will die at the midpoint of the interval (assuming the interval doesn t cross an integral age), so we have We calculate the required probabilities. Survival Probability Average Group time of group survival time I (0, 0.3] p x II (0.3, 1] 0.3p x p x III (1, ) 1p x p x p x ( ) 1 0.7(0.3) p x p x p x e x+0.7: (0.15) (0.65) (1) Alternatively, we can use trapezoids. We already know from the above solution that the heights of the first trapezoid are 1 and , and the heights of the second trapezoid are and So the sum of the area of the two trapezoids is e x+0.7:1 (0.3)(0.5)( ) + (0.7)(0.5)( ) For the expected value, we ll use the recursive formula. (The trapezoidal rule could also be used.) e 45:2 e 45:1 + p 45 e 46:1 ( ) ( ) We ll use equation (5.7)to calculate the second moment. E[min(T 45, 2) 2 ] t t p x dt

20 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES ( 1 2 t(1 0.01t) dt + So the variance is ( ) (1.011)( ) 3 2 2( ) As discussed on page 129, by equation (7.7), the difference is q x 1 (1 0.2) t(0.99) ( (t 1) ) dt ( ) Those who die in the first year survive 1 2 year on the average and those who die in the first half of the second year survive 1.25 years on the average, so we have p p ( 1 0.5(0.022) ) e 60: (0.02) ( ) + 1.5( ) (D) Alternatively, we use the trapezoidal method. The first trapezoid has heights 1 and p and width 1. The second trapezoid has heights p and 1.5 p and width 1/2. e 60:1.5 1 ( ) ( ) ( ) + ( ) p , 2 p 70 (0.96)(0.956) , and by linear interpolation, 1.5 p ( ) Those who die in the first year survive 0.5 years on the average and those who die in the first half of the second year survive 1.25 years on the average. So (D) e 70: (0.04) ( ) + 1.5( ) (C) Alternatively, we can use the trapezoidal method. The first year s trapezoid has heights 1 and 0.96 and width 1 and the second year s trapezoid has heights 0.96 and and width 1/2, so First we calculate t p 1 for t 1, 2. e 70: ( ) + 0.5(0.5)( ) (C) p 1 1 q p 1 (1 q 1 )(1 q 2 ) (0.90)(0.95) By linear interpolation, 1.5 p 1 (0.5)( ) The algebraic method splits the students into three groups: first year dropouts, second year (up to time 1.5) dropouts, and survivors. In each dropout group survival on the average is to the midpoint (0.5 years for the first group, 1.25 years for the second group) and survivors survive 1.5 years. Therefore e 1: (0.5) + ( )(1.25) (1.5) (D) )

21 EXERCISE SOLUTIONS FOR LESSON Alternatively, we could sum the two trapezoids making up the shaded area at the right. 1 (1,0.9) (1.5,0.8775) t p 1 e 1:1.5 (1)(0.5)( ) + (0.5)(0.5)( ) (D) t Those who die survive 0.25 years on the average and survivors survive 0.5 years, so we have q x p x ( ) ( ) 0.5q x 1 qx q x 1 0.5q 12 x 0.125q x q x q x ( q x 6 q x 1 2 ) q x Alternatively, complete life expectancy is the area of the trapezoid shown on the right, so (0.5)( p x+0.5 ) Then 0.5 p x , from which it follows t p x t 0.5p x q x q x q x Survivors live 0.4 years and those who die live 0.2 years on the average, so p q 55.2 Using the formula 0.4 q q 55 /(1 0.2q 55 ) (equation (7.3)), we have ( ) ( ) 1 0.6q55 0.4q q q q q q q q q µ q (0.0495)

22 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES Since d x is constant for all x and deaths are uniformly distributed within each year of age, mortality is uniform globally. We back out ω using equation (5.12), e x:n n p x (n) + n q x (n/2): q p ( ) ( ) 20 ω ω 20 ω ω ω 360 2ω 240 ω 120 Alternatively, we can back out ω using the trapezoidal rule. Complete life expectancy is the area of the trapezoid shown to the right. ( e 20:20 18 (20)(0.5) 1 + ω 40 ) ω ω 40 ω ω 16 ω ω 24 ω 120 x 20p x ω 40 ω 20 Once we have ω, we compute We use equation (7.5) to obtain 30 10q (A) ω q x q x q x 0.4 Then e ( 45: (1 0.4) ) 0.8. (E) According to the Illustrative Life Table, l 30 9,501,381, so we are looking for the age x such that l x 0.75(9,501,381) 7,126,036. This is between 67 and 68. Using linear interpolation, since l 67 7,201,635 and l 68 7,018,432, we have x ,201,635 7,126, ,201,635 7,018,432 This is years into the future. 3 4 of the people collect 50,000. We need 50,000 ( per person. (D) ) ( ) Quiz Solutions 7-1. Notice that µ 50.4 q q 50 while 0.6 q q q 50, so 0.6 q (0.01) 0.006

23 QUIZ SOLUTIONS FOR LESSON The algebraic method goes: those who die will survive 0.3 on the average, and those who survive will survive q x (0.1) 1 0.4(0.1) p x e x+0.4: (0.3) + (0.6) The geometric method goes: we need the area of a trapezoid having height 1 at x and height 90/96 at x + 1, where 90/96 is 0.6 p x+0.4, as calculated above. The width of the trapezoid is 0.6. The answer is therefore 0.5 (1 + 90/96) (0.6) Batteries failing in month 1 survive an average of 0.5 month, those failing in month 2 survive an average of 1.5 months, and those failing in month 3 survive an average of months (the average of 2 and 2.25). By linear interpolation, 2.25 q (0.6) (0.2) 0.3. So we have e 0:2.25 q 0 (0.5) + 1 q 0 (1.5) q 0 (2.125) p 0 (2.25) (0.05)(0.5) + ( )(1.5) + ( )(2.125) (2.25)

24 SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

25 Practice Exam 1 1. You are given: The following life table. x l x d x q Determine d 51. A. Less than 20 B. At least 20, but less than 22 C. At least 22, but less than 24 D. At least 24, but less than 26 E. At least Mortality follows Gompertz s law with B and c Calculate the 40 th percentile of future lifetime for (45). A. Less than 35.0 B. At least 35.0, but less than 40.0 C. At least 40.0, but less than 45.0 D. At least 45.0, but less than 50.0 E. At least You are given: l x 100(120 x) for x 90 µ x µ for x > 90 e 80 has the same value that it would have if l x 100(120 x) for x 120. Determine µ. A B C D E Exercises continue on the next page...

26 596 PRACTICE EXAMS 4. Three independent probes are sent to Mars. The probability distribution for the amount of time each one will function is given in the following table. Years (n) Probability of functioning n years The time to failure is assumed to be uniformly distributed between integral numbers of years. Calculate the probability that at least two of the probes function for at least years. A. Less than 0.40 B. At least 0.40, but less than 0.55 C. At least 0.55, but less than 0.70 D. At least 0.70, but less than 0.85 E. At least For three independent lives (x), (y), and (z), you are given µ x+t 0.01, t > 0 µ y+t 0.02, t > 0 µ z+t 0.03, t > 0 Calculate the probability that at least one life survives 30 years. A B C D E For two independent lives ages 30 and 40: µ 30 (t) 0.02 µ 40 (t) 1 80 t, t < 80 Calculate e 30:40. A. Less than 24 B. At least 24, but less than 26 C. At least 26, but less than 28 D. At least 28, but less than 30 E. At least 30 Exercises continue on the next page...

27 PRACTICE EXAM A discrete Markov chain model for insurance on two lives (x) and (y) has the following states: 0 Both alive 1 Only (x) alive 2 Only (y) alive 3 Neither alive Let p ij be the (i, j) entry in the transition matrix. Which of the following is true if the two lives are independent but may not be true if the lives are dependent? I. p 01 p 23 II. p 12 p 21 III. p 03 0 A. I and II only B. I and III only C. II and III only D. I, II, and III E. The correct answer is not given by A., B., C., or D. 8. For an insurance policy on (x), decrements occur due to death (1), withdrawal (2), and conversion (3). Decrement rates are as follows: t q (1) q (2) q (3) x+t 1 x+t 1 x+t Determine the probability that conversion will occur between 1 and 3 years from now. A. Less than B. At least 0.042, but less than C. At least 0.047, but less than D. At least 0.052, but less than E. At least Persistency of an insurance contract is affected by two decrements: death (1) and withdrawal (2). The force of mortality is µ (1) The force of withdrawal is µ (2) Determine the expected amount of time until withdrawal for those policyholders who eventually withdraw. A. Less than 15 B. At least 15, but less than 17 C. At least 17, but less than 19 D. At least 19, but less than 21 E. At least 21 Exercises continue on the next page...

28 598 PRACTICE EXAMS 10. You are given: Z 1 is the present value random variable for a 10-year term insurance paying 1 at the moment of death of (45). Z 2 is the present value random variable for a 20-year deferred whole life insurance paying 1 at the moment of death of (45). µ 0.02 δ 0.04 Calculate Cov(Z 1, Z 2 ). A. Less than 0.04 B. At least 0.04, but less than 0.03 C. At least 0.03, but less than 0.02 D. At least 0.02, but less than 0.01 E. At least For a 30-year deferred whole life annuity on (35): The annuity will pay 100 per year at the beginning of each year, starting at age 65. If death occurs during the deferral period, the contract will pay 1000 at the end of the year of death. Mortality follows the Illustrative Life Table. i Y is the present value random variable for the contract. Calculate E[Y]. A. Less than 204 B. At least 204, but less than 205 C. At least 205, but less than 206 D. At least 206, but less than 207 E. At least For a special fully discrete life insurance on (45), The benefit is 1000 if death occurs before age 65, 500 otherwise. Premiums are payable at the beginning of the first 20 years only. i 0.05 A E ä 45: Determine the annual net premium. A. Less than B. At least 11.50, but less than C. At least 12.00, but less than D. At least 12.50, but less than E. At least Exercises continue on the next page...

29 PRACTICE EXAM A copier is either working or out-of-order. If the copier is working at the beginning of a day, the probability that it is working at the beginning of the next day is 0.8. If the copier is out-of-order at the beginning of a day, the probability that it is fixed during the day and working at the beginning of the next day is 0.6. You are given that the probability that the copier is working at the beginning of Tuesday is Determine the probability that the copier was working at the beginning of the previous day, Monday. A. Less than 0.75 B. At least 0.75, but less than 0.78 C. At least 0.78, but less than 0.81 D. At least 0.81, but less than 0.84 E. At least You are given the following transition matrices for a Markov chain with three states numbered 1, 2, and n 0.1n 0.1n Q n n < 5 n 5 n is time measured in years. Each transition is made at the beginning of the year. Calculate the expected amount of time in State 1 (not necessarily continuously) for someone starting out in State 1 at time n 0. A. Less than 2.6 B. At least 2.6, but less than 2.8 C. At least 2.8, but less than 3.0 D. At least 3.0, but less than 3.2 E. At least 3.2 Exercises continue on the next page...

30 600 PRACTICE EXAMS 15. Life insurance policies include a disability waiver rider that waives premiums while the insured is disabled. Initially, an insured (x) is active. At the beginning of each subsequent year, an insured may either be in one of three states: State 1 = active State 2 = disabled State 3 = dead The transition probability matrix is Q A 4-year term insurance policy pays 1000 at the end of the year of death. Net premiums of π are paid at the beginning of each year if the insured is in the active state. Net premiums are determined using i Determine π. A. Less than 59 B. At least 59, but less than 61 C. At least 61, but less than 63 D. At least 63, but less than 65 E. At least 65 Solutions to the above questions begin on page 657.

31 Appendix A. Solutions to the Practice Exams Answer Key for Practice Exam 1 1 B 6 B 11 C 2 B 7 E 12 B 3 C 8 A 13 E 4 D 9 B 14 B 5 D 10 D 15 C Practice Exam 1 1. [Lesson 2] q 52 (d 52 + d 53 )/l 52 72/l 52, so l 52 72/ But l 52 l 50 d 50 d d 51, so d (B) 2. [Lesson 4] Gompertz s law is µ x Bc x 0.002(1.03 x ). We have ( 45+t tp 45 exp 0.002(1.03 u )du 45 ( ) exp 0.002( t ) ln 1.03 and we want to set this equal to 0.6, since the 40 th percentile of future lifetime corresponds to a 40% chance of dying before that time, or a 60% probability of survival to that time. We log the expression and set it equal to ln 0.6. ( t ) ln 0.6 ln t (ln 0.6)(ln 1.03) ln t ln 1.03 t (B) ) 3. [Lessons 5 and 6] By the recursive formula (6.1) e 80 e 80: p 80 e 90 Since mortality for the first ten years is unchanged, e 80:10 and 10 p 80 are unchanged. We only need to equate e 90 between the two mortality assumptions. Under uniform mortality through age 120, e 90 15, whereas under constant force mortality, e 90 1/µ. We conclude that 1/µ 15, or µ (C) 657

32 658 PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS [Lesson 7] The probability of one probe surviving years is linearly interpolated between 0.8 and 0.5, 0.8 ( 1 3) ( ) 0.7. The probability of at least two surviving is the probability of exactly two surviving (and there are three ways to choose 2 out of 3) or three surviving, or 3(0.7 2 )(0.3) (D) 5. [Lesson 25] The probability that all three fail within 10 years is 30q x 30 q y 30 q z (1 e 0.3 )(1 e 0.6 )(1 e 0.9 ) so the probability that at least one survives is (D) 6. [Lesson 26] e 30: tp x y dt ( ) 80 t e 0.02t dt 80 ( 80 t e 0.02t 1 80 ) e 0.02t dt ( ) 0.02 (1 e 0.02(80) ) (B) 7. [Lesson 25 and 29] The transition matrix with two independent lives looks like this: p 00 p 01 p p 11 0 p p 22 p The probability of death for each life must be the same before and after the death of the other, so p 01, the probability of the death of (y), must also be p 23, and p 02, the probability of the death of (x), must also be p 13. Thus I is true. While p 12 p 21 0, this is true even if the lives are not independent, so II is false. III is false, since p 03 is not necessarily 0 regardless of whether the two lives are dependent or independent. (E) 8. [Lesson 27] We need the probability of decrement (3) in years 2 and 3. The table gives conditional probabilities of decrement (3) given survival to the beginning of each year, so we need probabilities of survival to the beginning of years 2 and 3, t p 00 x, t 1, 2. p (τ) x 1 q (1) x q (2) x q (3) x ( 2 p(τ) x q (1) x+1 q(2) x+1 x+1) q(3) 0.89( ) Now we calculate the probabilities of decrement (3) in years 2 and 3. 1 q(3) x (0.89)(0.01) q(3) x (0.8188)(0.04) q(3) x (A)

33 PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS [Lesson 28] The expected amount of time to withdrawal is the integral 0 t t p (τ) x µ (2) x+t dt te 0.06t dt and this is 5/6 of 0.06te 0.06t dt, the expected amount of time to the next decrement, which we know 0 is 1/0.06 since it is exponential with mean 1/0.06. The probability that a decrement will be a withdrawal is the ratio of the µ s, or 0.05/( ) 5/6. So the expected amount of time to the next decrement given that it is a withdrawal is 5 6 (1/0.06) 16 2 (B) 5/ [Lesson 10] Since E[Z 1 Z 2 ] 0, the covariance is E[Z 1 ] E[Z 2 ]. E[Z 1 ] E[Z 2 ] e 0.06t 0.02 dt 1 e e 0.06t 0.02 dt e Cov(Z 1, Z 2 ) ( )( ) (D) 11. [Lesson 14] Let Y 1 be the value of the benefits if death occurs after age 65 and Y 2 the value of the benefits if death occurs before age E E E 45 ( )( ) E[Y 1 ] E 35 ä ( )(9.8969) E[Y 2 ] 1000 (A E 35 A 65 ) 1000 ( ( ) ) E[Y] (C) 12. [Lesson 20] Let P be the annual net premium. A 45:20 1 dä 45:20 1 ( 1 21) (12.6) :20 A 45:20 20 E A 45 P 500A 45 1 : A 45 ä 45:20 500(0.1) + 500(0.2) (B) 13. [Lesson 29] Let p be the probability that the copier was working at the beginning of the previous day. Then 0.8p + 0.6(1 p) 0.77

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