TABLE OF CONTENTS. 4. Daniel Markov 1 173

TABLE OF CONTENTS 1. Survival A. Time of Death for a Person Aged x 1 B. Force of Mortality 7 C. Life Tables and the Deterministic Survivorship Group 19 D. Life Table Characteristics: Expectation of Life 27 E. Other Life Table Characteristics 39 F. Different Mortality Rates 47 G. Assumptions for Fractional Ages: Uniform Distribution 53 H. Assumptions for Fractional Ages: Balducci and Constant Force 61 I. De Moivre s Law of Mortality 73 J. Other Laws of Mortality: Gompertz, Makeham, and Weibull 85 2. Multiple Lives A. Joint Distributions of Future Lifetimes and the Joint Life Status 95 B. The Last Survivor Status 111 C. Expectation and Probabilities 123 3. Multiple Decrements A. Multiple Random Variables: Basic Probabilities 137 B. Multiple Random Variables: Other Problems 145 C. Random Survivorship Groups: Basic Problems Involving Double-Decrement Tables 155 4. Daniel Markov 1 173 5. Daniel Poisson A. Definition of a Poisson Process 191 B. Calculating Poisson Probabilities and Moments 193 C. Waiting Times 207 D. Probabilities of Compound Poisson Distributions 213 E. Moments of the Compound Poisson Distribution 219 F. Approximations: Compound Poisson 233 G. Thinning 243 H. Sums of Poisson Processes 253 I. Mixture Distributions: Gamma and Negative Binomials 263 J. Mixture Distributions: Other Distributions 273 K. Nonhomogeneous Poisson Processes 279

ii 6. Insurance A. Whole Life and Term Insurance at the Moment of Death: Calculating Expected Values 285 B. Whole Life and Term Insurance at the Moment of Death: Calculating Variances 291 C. Endowment Insurance Payable at the Moment of Death 301 D. Deferred Insurance Payable at the Moment of Death 313 E. Varying Benefit Insurance Payable at the Moment of Death 323 F. Nonvariable Insurance Payable at the End of the Year of Death: Expected Values 331 G. Nonvariable Insurance Payable at the End of the Year of Death: Variances 347 H. Variable Insurance Payable the End of the Year 355 I. Relationships Between Insurances at the Moment of Death and End of the Year of Death 369 7. Annuities A. Continuous Life Annuities: Expected Values 373 B. Continuous Life Annuities: Other Problems 385 C. Annuities: Calculating Probabilities 395 D. Discrete Whole Life Annuities 405 E. Discrete Temporary Life Annuities 415 F. Deferred Discrete Life Annuities 427 G. Other Discrete Life Annuities 433 H. Reinsurance and Reserves 439 8. Premiums A. Fully Continuous Level Annual Benefit Premiums 455 B. Fully Continuous Premiums: Loss-At-Issue Random Variable 463 C. Fully Discrete Level Annual Benefit Premiums 477 D. Fully Discrete Whole Life Premiums: Loss-At-Issue Random Variable 493 E. Fully Discrete Term and Endowment Premiums: Loss-at-Issue Random Variable 507 F. Semicontinuous, Variable, and Other Premiums 513 9. Reserves A. Fully Continuous Benefit Reserves: Prospective Formulas 523 B. Fully Continuous Benefit Reserves: Variance of Prospective Loss 533 C. Fully Continuous Benefit Reserves: Retro, Paid-Up, and Premium Difference Formulas 537 D. Fully Continuous Benefit Reserves: Special Formulas 543 E. Fully Discrete Benefit Reserves: Prospective Formulas 547 F. Fully Discrete Benefit Reserves: Prospective Loss 559 G. Fully Discrete Benefit Reserves: Retro, Paid-Up, and Premium Difference Formulas 569 H. Fully Discrete Benefit Reserves: Special Formulas 581 I. Recursive Relations for Fully Discrete Benefit Reserves 591 10. Insurances/Annuities A. Multilife Survival Statuses 607 B. Special Annuity Benefits 619 C. Multiple Decrement Insurances and Annuities 637 11. Daniel Markov 2 659

iii 12. Statistics A. Maximum Likelihood Estimation 679 B. Method-of-Moments Estimation 713 C. Measures of Quality 729 D. Hypothesis Testing: Normal Distributions 741 E. Hypothesis Testing: Other Questions 759 F. Order Statistics 773 G. The t, F, and Chi-Square Distributions 785 H. Least-Squares Estimation 809 I. Likelihood Ratio Tests 827

iv 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 7/15/11 PJM

Insurance 313 D. Deferred Insurance Payable at the Moment of Death D1. You are given the following for (40): i) Mortality follows de Moivre's law with 100, ii) Z is the present value random variable for a five-year deferred life insurance of 1. iii) M is the mode of Z. Which of the following are true? 1. F Z (0) > 0 2. M v 5 3. P(Z < v 30 ).5 A. 1, 2 B. 1,3 C. 2,3 D. 1,2,3 E. None of these answers are correct. (86S 4 21) D2. Z A is the present value random variable for a whole life insurance issued to (x) that pays 2 at the moment of death if death occurs within n years and 1 at the moment of death if death occurs after n years. Z B is the present value random variable for a whole life insurance issued to (x) that pays 1 at the moment of death if death occurs within n years and 2 at the moment of death if death occurs after n years. You are given that E[Z A ] E[Z B ]. a. Write expressions for Z A and Z B in terms of v T where T T(x). b. Determine E[Z A ] in terms of net single premiums. c. Determine E[(Z A ) 2 ] and E[(Z B ) 2 ] in terms of net single premiums. d. Demonstrate that Var(Z A ) Var(Z B ) 3( 2 A _ 1 45 : 20 2 n A _ x) (87F 150 A6 4) D3. A five-year deferred whole life policy pays $10,000 at the moment of death for a person aged 30. Assume a constant force of mortality.05 and a force of interest.10. What is the standard deviation of the present value of the benefit payment? You may use: e.75 2.117 e 1.00 2.718 e 1.25 3.490 e 1.50 4.482 A. < $1,820 B. $1,820 but < $1,840 C. $1,840 but < $1,860 D. $1,860 but < $1,880 E. $1,880 (88 4 18 2) D4. A twenty-year endowment policy is purchased by a man who just turned 30 years old. The policy will pay $10,000 at the moment of death, if he dies within 20 years, or $20,000 if he survives for twenty years. The force of mortality is.01 until age 40 and at age 40 jumps to.02. The force of interest is.10. What is the actuarial present value at policy issue of the benefit payment? A. < $3,100 B. $3,100 but < $3,200 C. $3,200 but < $3,300 D. $3,300 but < $3,400 E. $3,400 (89 4 14 3)

314 Insurance Solutions are based on Cunningham, pp. 138 45, and the pages from Bowers listed below. D1. 1. T, p. 104 For deferred insurance, there is no payout during the deferral period so F Z (0) > 0. 2. F, p. 104 The mode is 0 since this is the probability mass for the deferral period. 3. F, pp. 78, 104 5 Z < v 30 over two intervals: when t is between 0 and 5, the deferral period, and when t is between 30 and 60. Since probability is uniform under de Moivre's law, their combined probability equals (5 + 30)/60 >.5. Answer: E D2. a. When 0 T < n, Z A 2v T and Z B v T. When T n, Z A v T and Z B 2v T. b. E[Z A ] 2A _ 1 _ x:n + n A _ x c. E[(Z A ) 2 ] 4( 2 A _ 1 _ n:x ) + 2 n A _ x E[(Z B ) 2 ] 2 A _ 1 _ n:x + 4( 2 n A _ x) d. Since (E[Z A ]) 2 (E[Z B ]) 2, we get: Var(Z A ) Var(Z B ) E[(Z A ) 2 ] E[(Z B ) 2 ] 4( 2 A _ 1 _ n:x ) + 2 n A _ x 2 A _ 1 _ n:x 4( 2 n A _ x) D3. 5 A _ 30 e-(5)( + Var(Z A ) Var(Z B ) 3( 2 A _ 1 _ n:x ) 3( 2 n A _ x), pp. 95 96, 103. 2 5 A_ 30 e-(5)( +2 + 2.05 e-(5)(.05+.10).05 +.10.05 e-(5)(.05+.20).05 +.20 e-.75 3 e-1.25 5.15746.05730 (10,000) 2 Var( 5 A _ 30) (10,000) 2 [ 2 5 A_ 30 ( 5 A _ 30) 2 ] [(10,000) 2 ][.05730 (.15746) 2 ] (10,000) 2 Var( 5 A _ 30) 3,250,635 Aggregate SD 1,803, p. 103. Answer: A D4. The present value has three components: a ten-year term component, a ten-year term component, deferred ten years, and a pure endowment: A _ 1 (1 e ) 30:10 (.01)(1 e-(10)(.01+.1) ).01 +.10 10 A_ 1 (e ' 1 e -(10)( ' ) 40:10 10 A_ 1 40:10 (.33287)(1/6)(1.30119).03877 1 2A 30: 20 1 2A 30: (1/11)(1.33287).06065 (e-(10)(.01+.1) )(.02)(1 e -(10)(.02+.1) ).02 +.10 2(e -(10)( )(e -(10)( '+ ) 2(e -(10)(.01+.1) )(e -(10)(.02+1) ) (2)(.33287)(.30119).20051 20 APV [10,000][A _ 1 30:10 + 10 A _ 1 1 40:10 + 2A30: ] APV [10,000] [.06065 +.03877 +.20051] 2,999, pp. 102 3. Answer: A 20

Insurance 315 D5. n A x n+1 A x vq x+n n E x (89F 150 28 MC) D6. You are given: i) X is the present value random variable for the 25-year term insurance of 7 on (35). ii) Y is the present value random variable for the 25-year deferred, 10-year term insurance of 4 on the same life. iii) E[X] 2.80 and E[Y].12 iv) Var(X) 5.76 and Var(Y).10 Calculate Var(X + Y). A. 4.75 B. 5.19 C. 5.51 D. 5.86 E. 6.14 (90S 150 15) D7. A life aged x is subject to a constant force of mortality, (x).06, and a constant force of interest ( ). A whole life insurance with a death benefit of $20 payable at the moment of death is purchased. The actuarial present value is $12. Determine the actuarial present value if a ten-year deferred whole life insurance with a $20 death benefit was purchase instead. A. < $5 B. $5 but < $6 C. $6 but < $7 D. $7 but < $8 E. $8 (94F 4A 19 2) D8. For a life aged 35, you are given a force of interest,.10, and a force of mortality,.06. This life purchases a ten-year deferred whole life insurance with a benefit of one payable at the moment of death. Z is the present value at policy issue of the benefit for this insurance. Determine the 90th percentile of Z. A. <.30 B..30 but <.40 C..40 but <.50 D..50 but <.60 E..60 (95F 4A 14 2) D9. An endowment insurance has the following provisions: i) ii) A benefit of $0 if the insured dies within five years of policy issue. A benefit of $1 if the insured survives five years but dies within the subsequent seven years. iii) A benefit of $2 at the end of twelve years if the insured survives at least twelve years after policy issue. The death benefit is payable at the moment of death. You also are given the following values with t equal to the number of years from policy issue: t (t) t 0 t 8.05.10 t > 8.10.10 Determine the actuarial present value for this insurance. A. <.25 B..25 but <.35 C..35 but <.45 D..45 but <.55 E..55 (95F 4A 17 2)

316 Insurance D5. T, pp. 101, 118 The expression on the left-hand side of the equation provides term coverage for age (x + n) discounted n years. The term coverage equals vq x+n and the discount factor is n E x, which together comprise the right-hand side. D6. Since X and Y are mutually exclusive, E[XY] 0. E[X 2 ] Var(X) + (E[X]) 2 5.76 + (2.8) 2 13.6 E[Y 2 ] Var(Y) + (E[Y]) 2.1 + (.12) 2.1144 Var(X + Y) E[(X + Y) 2 ] (E[X + Y]) 2 E[X 2 ] + 2 E[XY] + E[Y 2 ] (E[X] + E[Y]) 2 Var(X + Y) 13.6 + (2)(0) +.1144 (2.8 +.12) 2 5.19, p. 96. Answer: B D7. (20 20A_ x) 20A _ x (20 12)(.06) 12.04 20 10 A _ x (20)(.06)e-(10)(.06+.04).06 +.04 Answer: A 4.41, pp. 99, 103. D8. 1) Calculate the probability Z equals 0: 10 P(Z 0) P(T 10) e -.06t.06 dt 1 e -.6.45119 0 2) Calculate the probability Z is greater than zero but less than the 90th percentile: P(0 < Z.9 ).9.45119.44881 3) Calculate the 90th percentile, using the same procedure as in A11. log.9 log.44881 log.9 / log.44881.9 (.44881).10/.06.26309, pp. 96 98, 103 5. Answer: A D9. The present value has three components: a three-year term component deferred five years, a four-year term component deferred eight years, and a pure endowment: 5 3A _ x v 5 5p x A _ 1 _ x:3 e -(5)( + ) ( )(1 e-(3)( + ) ) 5 3A _ x (.47237)(1/3)(1.63763).05706 8 4A _ x v 8 8p x A _ 1 _ x:4 e -(8)( + ) '(1 e-(4)( '+ ') ) ' + ' e -(5)(.05+.10) (.05)(1 e-(3)(.05+.10) ).05 +.10 e -(8)(.05+.10).10(1 e-(4)(.10+.10) ).10 +.10 8 4A _ x (.30119)(1/2)(1.44933).08293 1 2A 2v 12 x :12 12px 2e -(8)( + ) e -(4)( '+ ') 2e -(8)(.05+.10) e -(4)(.10+.10) 1 2A (2)(.30119)(.44933).27067 x :12 APV 5 3 A _ x + 8 4 A _ x + Answer: C 1 2A.05706 +.08293 +.27067.41066, pp. 99 100, 102 3. x :12

Insurance 317 D10. For a special whole life insurance on (x), you are given: i) (x + t), t 0 ii) t, t 0 iii) The death benefit, payable at the moment of death, is 1 for the first ten years and.5 thereafter. iv) The single benefit premium is.3324. v) Z is the present value random variable at issue of the death benefit. Calculate Var (Z). A. <.07 B..07 but.08 C..08 but <.09 D..09 but <.10 E..10 (96F 150 11) D11. The benefit payable under an m-year deferred whole life policy, with benefit payable at the moment of death, is twice that of a similar nondeferred whole life insurance. The actuarial present values for these insurances are equal. You may assume constant forces of mortality and interest,.08 and.06. Determine m. A. < 2 B. 2 but < 4 C. 4 but < 6 D. 6 but < 8 E. 8 (97F 4A 10 2) D12. For a special whole life insurance on (t), you are given: i) Benefits are payable at the moment of death. ii) b t 200 for 0 t < 65 iii) b t 100 for t 65 iv) 0 (t).03 for t 0 v) t.01 for 0 t < 65 vi) t.02 for t 65 Calculate the actuarial present value at issue of this insurance. A. 140 B. 141 C. 142 D. 143 E. 144 (98S 150 18) D13. A special insurance program is designed to pay a benefit in the event a product fails. You are given: i) Benefits are payable at the moment of failure. ii) b t 300 and.02 for 0 t < 25 iii) b t 100 and.03 for t 25 iv) (t).04 for t 0 Calculate the actuarial present value of this special insurance. A. 165 B. 168 C. 171 D. 210 E. 213 (Sample 3 11) D14. For a ten-year deferred whole life insurance of 1 payable at the moment of death on a life aged 35, you are given: i) The force of interest is.10. ii) iii) The force of mortality is.06. Z is the present value random variable for this insurance. Determine the 90th percentile of Z. A..1335 B..1847 C..2631 D..4488 E..4512 (Sample 3 44)

318 Insurance D10..3324 E[Z] A _ 1 _ x:n + (.5)( n A _ x) (1 e-10( + ) ).3324 (1 e-10( + ) ) E[Z 2 ] (1 e-10( +2 ) ) + 2 + (e-10( + ) ) 2() + (e -10( + ) ) 2() 1/2 e -20 /4.02 + (e-10( +2 ) ) (2) 2 1/3 e ( + 2 ) (-10)(.06) /4.19613 Var(Z) E[Z] (E[Z]) 2.19613 (.3324) 2.085640, pp. 99, 103. Answer: C D11. 2 (e -m( + ) ) 2e -m( +.06) 1 m log.5.14.69315.14 4.95, pp. 99, 103. Answer: C D12. A _ 1 ( )(1 e -(65)( + ) ) 0:65 65 A _ 0 e-(65)( + ' APV 200A Answer: D 1 0:25 D13. A _ 1 ( )(1 e -(25)( + ) ) 0:25 25 A _ 0 e-(5)( + ' APV 300A Answer: B 1 (.03)(1 e-(65)(.03+.01) ).03 +.01.03 e-(65)(.03+.01).03 +.02 3e-2.6 5 (3)(1 e-2.6 ) 4.044564.69429 + 100 25 A _ 0 (200)(.69429) + (100)(.044564) 143, pp. 99 100, 102 3. 0:25 (.04)(1 e-(25)(.04+.02) ).04 +.02.04 e-(25)(.04+.02).04 +.03 4e-1.5 7 (2)(1 e-1.5 ) 3.12750.51791 + 100 25 A _ 0 (300)(.51791) + (100)(.12750) 168, pp. 99 100, 102 3. D14. 1) Calculate the probability Z equals 0: 10 P(Z 0) P(T 10) 0 e -.06t.06 dt 1 e -.6.45119 2) Calculate the probability Z is greater than zero but less than the 90th percentile: P(0 < Z.9 ).9.45119.44881 3) Calculate the 90th percentile, using the same procedure as in A11. log.9 log.44881 log.9 / log.44881.9 (.44881).10/.06.26309, pp. 96 98, 103 5. Answer: C

Insurance 319 D15. An investment fund is established to provide benefits on 400 independent lives age x. i) On January 1, 2001, each life is issued a 10-year deferred whole life insurance of 1,000, payable at the moment of death. ii) Each life is subject to a constant force of mortality of.05. iii) The force of interest is.07. Calculate the amount needed in the investment fund on January 1, 2001, so that the probability, as determined by the normal approximation, is.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 (00S 3 13) D16. Each of 100 independent lives purchases a single premium five-year deferred whole life insurance of 10 payable at the moment of death. You are given: i).04 ii).06 iii) F is the aggregate amount the insurer receives from the 100 lives. Using the normal approximation, calculate F such that the probability the insurer has sufficient funds to pay all claims is.95. A. 280 B. 390 C. 500 D. 610 E. 720 (01F 3 8) D17. Given the following, Calculate 20 A _ x. i) i 5% ii) The force of mortality is constant. iii) e x 16.0 A. <.050 B..050 but <.075 C..075 but <.100 D..100 but <.125 E..125 (03F 3C 7 2) D18. For a whole life insurance of 1,000 on (x) with benefits payable at the moment of death:.04, 0 < t 10 i) t.05, 10 < t.06, 0 < t 10 ii) (x + t).07, 10 < t Calculate the single benefit premium for this insurance. A. 379 B. 411 C. 444 D. 519 E. 594 (03F 3S 2) (Sample M 2) D19. For a five-year deferred whole life insurance of 1, payable at the moment of death of (x), you are given: i) Z is the present value random variable of this insurance. ii).10 iii).04 Calculate Var(Z). A. <.035 B..035 but <.045 C..045 but <.055 D..055 but <.065 E..065 (04F 3C 2 2)

320 Insurance D15. 10 A _ x e-(10)( +.05 e-(10)(.05+.07).05 +.07 5e-1.2 12.12550 E[S] (400)(1,000) 10 A _ x (400)(1,000)(.12550) 50,200 2 10 A_ x e-(10)( +2 + 2.05 e-(10)(.05+.14).05 +.14 5e-1.9 19.03936 Var(Z) (400)(1,000) 2 Var( 10 A _ x) (400)(1,000) 2 [ 2 10 A_ x ( 10 A _ x) 2 ] Var(Z) [400][(1,000) 2 ][.03936 (.12550) 2 ] (1,000) 2 (9.4439) SD(Z) 3,073 APV E[Z] + 1.645SD(Z) 50,200 + (1.645)(3,073) 55,255, pp. 99, 103. Answer: A D16. 5 A _ x e-(5)( +.04 e-(5)(.04+.06).04 +.06.4e -.5.24261 E[Z] (100)(10) 5 A _ x (100)(10)(.24261) 242.61 2 5 A_ x e-(5)( +2 + 2.04 e-(5)(.04+.12).04 +.12 e-.8 4.11233 Var(Z) (100)(10) 2 Var( 5 A _ x) (100)(10) 2 [ 2 5 A_ x ( 5 A _ x) 2 ] Var(Z) [100][(10) 2 ][.11233 (.24261) 2 ] (100) 2 (.05347) SD(Z) 23.12 F E[Z] + 1.645SD(Z) 242.61 + (1.645)(23.12) 280.64, pp. 99, 103. Answer: A D17. See exercise 3.18. log 1.05.048790 1/ e x 1/16.0625 20 A _ x e-(20)( + pp. 87, 99, 103 Answer: B.0625e-(20)(.0625+.048790).0625 +.048790.56160e -2.2258.060642, D18. A _ 1 x:10 ( ) )(1 e-10( + ) (.06)(1 e-(10)(.06+.04) ).06 +.04 10 A _ x 'e-(10)( +.07e-(10)(.06+.04) ' + '.07 +.05 7e-1 12.21460 APV (1,000)(.37927 +.21460) 593.87, pp. 99, 103. (.6)(1 e -1 ).37927 Answer: E D19. 5 A _ x e-(5)( + 2 5 A_ x e-(5)( +2 + 2.04 e-(5)(.04+.10).04 +.10.04 e-(5)(.04+.20).04 +.20 (2/7)e -.7.14188 e-1.2 6.05020 Var(Z) 2 5 A_ x ( 5 A _ x) 2.05020 (.14188) 2.03007, p. 99, 103. Answer: A

Insurance 321 D20. For a five-year deferred whole life insurance of one on (x), you are given: i).06 ii).04 iii) The benefit is paid at the moment of death. iv) Z is the present value random variable of the insurance benefit. Calculate Var(Z). A. <.05 B..05 but <.06 C..06 but <.07 D..07 but <.08 E..08 (06F 3 33 2) D21. An appliance store sells microwave ovens with a three-year warranty against failure. At the time of purchase, the consumer may buy a two-year extended warranty that would pay half of the original purchase price at the moment of failure. You are given: i) The extended warranty period begins exactly three years after the time of purchase, but only if the oven has not failed by then. ii) Any failure is considered permanent. iii) 4% iv) Failure of the ovens follows the mortality table below, with uniform distribution of failure within each year: Age (x) 0 1 2 3 4 5 q x.008.015.026.042.063.089 Calculate the actuarial present value of the extended warranty as a percent of the purchase price. A. < 3.8% B. 3.8% but < 4.1% C. 4.1% but < 4.4% D. 4.4% but < 4.7% E. 4.7% (06F 3 35 2) D22. For a whole life insurance of 1 on (x) with benefits payable at the moment of death, you are given:.02, t < 12 t {.03, t 12 x (t) {.04, t < 5.05, t 5 Calculate the actuarial present value of this insurance. A..59 B..61 C..64 D..66 E..68 (06F M 12)

322 Insurance D20. 5 A _ x e-5( +.04e-(5)(.04+.06).04 +.06.24261 2 5 A_ x e-5( +2 + 2.04e-[5][.04+(2)(.06)].04 + (2)(.06).11233 Var(Z) 2 5 A_ x ( 5 A _ x) 2.11233 (.24261) 2.05347, pp. 99, 103. Answer: B D21. Approximate the answer by assuming that failures occur at time t 3.5 and t 4.5.5 3 2 A _ x (.5)(1 q 0 )(1 q 1 )(1 q 2 )e -3.5 (q 3 + e - q 4 ).5 3 2 A _ x (.5)(1.008)(1.015)(1.026)e -(.04)(3.5) (.042 +.063e ).5 3 2 A _ x (.5)(.95171)(.86936)(.10253).04242, pp. 74, 103, 109. Answer: C D22. A _ x A _ 1 _ + 5 7 A _ x + 12 A _ x x:5 A _ x ( ) )(1 e-5( + ) + (e )( -5( + ) ' ) ' + )(1 e-7( '+ ) + (e -5( + ) )(e )( -7( '+ ) ' ) ' + ' A _ x (.04.04 +.02)(1 e ) -5(.04+.02) + (e )( -5( +.02).05.05 +.02)(1 e ) -7(.05+.02) + (e -5(.04+.02) )(e )( -7(.05+.02).05.05 +.03) A _ x (2/3)(.25918) + (.74081)(5/7)(.38737) + (.74081)(.61263)(5/8) A _ x.17279 +.20498 +.28365.66142, pp. 99, 103. Answer: D