Contents
I
Probability Review
1
1 Probability Basics
1.1 Functions and Moments . . . . . . . . . . . .
1.2 Percentiles . . . . . . . . . . . . . . . . . . . . . .
1.3 Conditional Probability and Expectation
1.4 Probability Distributions . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . .
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3
3
4
6
8
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12
2 Variance
2.1 Additivity . . . . . . . . . .
2.2 Normal Approximation
2.3 Mixtures . . . . . . . . . . .
2.4 Bernoulli Shortcut . . .
2.5 Conditional Variance .
Exercises . . . . . . . . . .
Solutions . . . . . . . . . .
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17
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19
21
22
23
25
II
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Life Contingencies
29
3 Survival Distributions: Probability Functions, Life Tables
3.1 Probability Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Life Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Relation of Actuarial Functions to Mathematical Probability Functions
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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31
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34
35
40
4 Survival Distributions: Force of Mortality
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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49
58
5 Survival Distributions: Mortality Laws
5.1 Mortality Laws for Exam Questions . . . . . . . . . . . . . . . . . . . . .
5.1.1 Exponential Distribution, or Constant Force of Mortality
5.1.2 Uniform Distribution, or De Moivre’s Law . . . . . . . . . . .
5.1.3 Beta Distribution, or Generalized De Moivre’s Law . . . . .
5.2 Mortality Laws that May be Used for Human Mortality . . . . . . .
5.2.1 Gompertz’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.2 Makeham’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.3 Weibull Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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67
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69
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71
71
74
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
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iv
6 Survival Distributions: Moments
6.1 Complete . . . . . . . . . . . . . .
6.1.1 General . . . . . . . . . .
6.1.2 Special Distributions
6.2 Curtate . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . .
CONTENTS
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79
79
79
81
84
89
94
7 Survival Distributions: Percentiles, Recursions, and Life Table Concepts
7.1 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Recursive Formulas for Life Expectancy . . . . . . . . . . . . . . . . . . . . . .
7.3 Central Death Rate, L x , Tx , a (x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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105
105
106
107
109
117
8 Survival Distributions: Fractional Ages
8.1 Uniform Distribution of Deaths . . .
8.2 Constant Force of Mortality . . . . . .
8.3 Hyperbolic Assumption . . . . . . . .
8.4 Summary . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . .
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127
127
132
133
136
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147
9 Survival Distributions: Select Mortality
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161
165
171
10 Insurance: Payable at Moment of Death—Moments—Part 1
10.1 Definitions and General Formulas . . . . . . . . . . . . . . . . .
10.2 Constant Force of Mortality . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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181
181
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191
199
11 Insurance: Payable at Moment of Death—Moments—Part 2
11.1 De Moivre’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 Other Mortality Functions . . . . . . . . . . . . . . . . . . . . . . .
11.2.1 Integrating c t n e −δt (Gamma Integrands) . . . . . .
11.3 Variance of Endowment Insurance . . . . . . . . . . . . . . . .
11.4 Normal Approximation . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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209
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213
214
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221
12 Insurance Payable at End of Year: Moments
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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234
246
13 Insurance: Probabilities and Percentiles
13.1 Probabilities for Continuous Variables
13.2 Probabilities for Discrete Variables . .
13.3 Percentiles . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . .
259
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263
266
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
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v
CONTENTS
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 Insurance: Recursive Formulas, Varying Insurances
14.1 Recursive Formulas . . . . . . . . . . . . . . . . . . . . . .
14.2 Increasing and Decreasing Insurance . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
271
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281
281
284
289
299
15 Insurance: Discrete to Continuous
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
309
312
315
16 Annuities: Continuous, Expectation
16.1 Whole Life Annuity . . . . . . . . . . . . . . . . . .
16.2 Temporary Life and Deferred Life Annuities
16.3 n -year certain and life annuity . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . .
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321
322
324
327
329
333
17 Annuities: Discrete, Expectation
17.1 Annuities-Due . . . . . . . . . . . . . . . .
17.2 Annuities-Immediate . . . . . . . . . . .
17.3 Increasing and Decreasing Annuities
17.4 Actuarial Accumulated Value . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . .
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339
339
343
346
347
352
366
18 Annuities: Variance
18.1 Whole Life and Temporary Annuities . . . . . . . . . . . . . . . . . .
18.2 Deferred Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.3 Other Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.4 Typical Exam Questions . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.5 Combinations of Annuities and Insurances with No Variance
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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379
379
381
382
383
385
387
394
19 Annuities: Probabilities, Percentiles, Recursive Calculations
19.1 Probabilities for Continuous Annuities . . . . . . . . . . . . .
19.2 Probabilities for Discrete Annuities . . . . . . . . . . . . . . . .
19.3 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19.4 Recursive calculations . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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407
407
411
412
414
415
424
20 Annuities: m -thly Payments
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
435
436
440
21 Premiums: Fully Continuous Expectation
21.1 Benefit Premium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21.2 Loss at Issue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
445
445
448
450
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
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vi
CONTENTS
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 Premiums: Discrete Expectation
22.1 Benefit Premium . . . . . . . .
22.2 Three Premium Principle . .
Exercises . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . .
455
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461
461
466
468
484
23 Premiums: Variance of Loss at Issue, Continuous
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
503
506
511
24 Premiums: Variance of Loss at Issue, Discrete
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
517
519
524
25 Premiums: Probabilities and Percentiles of Loss at Issue
25.1 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25.2 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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531
531
535
538
541
26 Premiums: True Fractional Premiums
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
549
550
553
27 Reserves: Prospective Formula
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
557
561
568
28 Reserves: Other Formulas
28.1 Premium Difference and Paid Up Insurance Formulas
28.2 Retrospective Formula . . . . . . . . . . . . . . . . . . . . . . .
28.3 Three Premium Principle . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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577
577
580
583
585
591
29 Reserves: Special Formulas for Whole Life and Endowment Insurance
29.1 Annuity-ratio formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29.2 Insurance-ratio formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29.3 Premium-ratio formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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601
601
602
603
604
614
30 Reserves: Variance of Loss
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
625
627
631
31 Reserves: Recursive Formulas
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
639
642
652
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
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vii
CONTENTS
32 Reserves: Advanced Topics
32.1 General Insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32.2 Refund of Reserve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32.3 Insurance Payable at Moment of Death and Fractional Premiums
32.4 Fractional Durations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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663
663
664
667
671
674
681
33 Reserves: Hattendorf’s Theorem
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
691
694
698
34 Multiple Lives: Joint Life Probabilities
34.1 Introduction . . . . . . . . . . . . . . . . . .
34.2 The Joint Life Distribution Function
34.3 Independence . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . .
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703
703
704
706
708
714
35 Multiple Lives: Last Survivor Probabilities
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
719
724
729
36 Multiple Lives: Moments
36.1 Expected Value . . . . . . . .
36.2 Variance and Covariance .
Exercises . . . . . . . . . . . .
Solutions . . . . . . . . . . . .
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735
735
739
741
744
37 Multiple Lives: Insurances
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
751
754
760
38 Multiple Lives: Annuities
38.1 Introduction . . . . . . . . . . . . . . . . . . . . . .
38.2 Three Techniques for Handling Annuities
38.3 Other Techniques . . . . . . . . . . . . . . . . . .
38.4 Variance . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . .
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767
767
768
771
772
773
782
39 Multiple Lives: Contingent Survival Functions
39.1 Contingent Probabilities . . . . . . . . . . . . .
39.2 Contingent Insurances . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . .
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791
791
796
799
805
40 Multiple Lives: Common Shock
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
815
819
821
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SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
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viii
CONTENTS
41 Multiple Decrement Models: Probabilities
41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . .
41.2 Life Tables . . . . . . . . . . . . . . . . . . . . . . . . . .
41.3 Examples of Multiple Decrement Probabilities
41.4 Discrete Insurances . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
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827
827
828
830
831
833
842
42 Multiple Decrement Models: Forces of Decrement
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
847
850
857
43 Multiple Decrement Models: Associated Single Decrement Tables
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
865
868
873
44 Multiple Decrement Models: Going Between Probabilities and Rates
44.1 Uniform Distribution Assumed in the Double Decrement Table . . . . . . . .
44.2 Uniform Distribution Assumed in the Associated Single-Decrement Tables
44.3 Discrete Decrements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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881
881
883
886
890
897
45 Multiple Decrement Models: Continuous Insurances
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
907
909
918
46 Expenses: Expense-Loaded Premium and Reserve
46.1 Expense-Loaded Premium . . . . . . . . . . . . . .
46.2 Expense-Loaded Reserve . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
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929
929
931
933
939
47 Expenses: Other Topics
47.1 Expense Premium and Expense Reserve .
47.2 Expense-Loaded Loss Random Variable .
47.3 Policy Fee . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . .
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945
945
945
948
950
958
48 Expenses: Asset Shares
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
967
972
977
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III Stochastic Processes
49 Multi-State Models (Markov Chains): Probabilities
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50 Multi-State Models (Markov Chains): Premiums and Reserves
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
983
985
993
998
1005
ix
CONTENTS
50.1
50.2
50.3
50.4
Prototype Example . . . .
Insurances . . . . . . . . . .
Annuities . . . . . . . . . . .
Premiums and Reserves
Exercises . . . . . . . . . . .
Solutions . . . . . . . . . . .
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1005
1005
1008
1010
1012
1018
51 The Poisson Process: Probabilities of Events
51.1 Introduction . . . . . . . . . . . . . . . . . . . . . . .
51.2 Probabilities—Homogeneous Process . . . .
51.3 Probabilities—Non-Homogeneous Process
Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . .
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1025
1025
1026
1027
1030
1034
52 The Poisson Process: Time To Next Event
1039
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1042
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043
53 The Poisson Process: Thinning
53.1 Constant Probabilities . . . . .
53.2 Non-Constant Probabilities .
Exercises . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . .
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1045
1045
1047
1049
1053
54 The Poisson Process: Sums and Mixtures
54.1 Sums of Poisson Processes . . . . . . .
54.2 Mixtures of Poisson Processes . . . . .
Exercises . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . .
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1059
1059
1060
1065
1068
55 Compound Poisson Processes
55.1 Definition and Moments . . . . . . .
55.2 Sums of Compound Distributions
Exercises . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . .
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1073
1073
1075
1077
1082
IV
Practice Exams
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1089
1 Practice Exam 1
1091
2 Practice Exam 2
1099
3 Practice Exam 3
1109
4 Practice Exam 4
1115
5 Practice Exam 5
1123
6 Practice Exam 6
1129
7 Practice Exam 7
1137
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
x
CONTENTS
8 Practice Exam 8
1145
9 Practice Exam 9
1153
10 Practice Exam 10
1161
Appendices
A Solutions to the Practice Exams
Solutions for Practice Exam 1 . .
Solutions for Practice Exam 2 . .
Solutions for Practice Exam 3 . .
Solutions for Practice Exam 4 . .
Solutions for Practice Exam 5 . .
Solutions for Practice Exam 6 . .
Solutions for Practice Exam 7 . .
Solutions for Practice Exam 8 . .
Solutions for Practice Exam 9 . .
Solutions for Practice Exam 10 .
1169
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1171
1171
1180
1190
1200
1210
1220
1231
1243
1254
1263
B Solutions to Old Exams
B.1 Solutions to CAS Exam 3, Spring 2005 . . .
B.2 Solutions to SOA Exam M, Spring 2005 . . .
B.3 Solutions to CAS Exam 3, Fall 2005 . . . . . .
B.4 Solutions to SOA Exam M, Fall 2005 . . . . .
B.5 Solutions to CAS Exam 3, Spring 2006 . . .
B.6 Solutions to CAS Exam 3, Fall 2006 . . . . . .
B.7 Solutions to SOA Exam M, Fall 2006 . . . . .
B.8 Solutions to CAS Exam 3, Spring 2007 . . .
B.9 Solutions to SOA Exam MLC, Spring 2007 .
B.10 Solutions to CAS Exam 3, Fall 2007 . . . . . .
B.11 Solutions to CAS Exam 3L, Spring 2008 . .
B.12 Solutions to CAS Exam 3L, Fall 2008 . . . . .
B.13 Solutions to CAS Exam 3L, Spring 2009 . .
B.14 Solutions to CAS Exam 3L, Fall 2009 . . . . .
B.15 Solutions to CAS Exam 3L, Spring 2010 . .
B.16 Solutions to CAS Exam 3L, Fall 2010 . . . . .
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1277
1277
1281
1289
1293
1301
1306
1311
1319
1324
1332
1336
1339
1343
1347
1351
1355
C Exam Question Index
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
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1359
Lesson 8
Survival Distributions: Fractional Ages
Reading: Actuarial Mathematics 3.6 or Models for Quantifying Risk (3rd edition) 4.5
Life tables, such as the Illustrative Life Table, list mortality rates (qx ) 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.
There are three commonly used interpolation methods: linear, geometric, and harmonic. We will consider all
three of them. Linear interpolation is usually referred to as uniform distribution of deaths between integral ages
or UDD. Geometric interpolation is usually referred to as constant force of mortality between integral ages, and
may also be called exponential interpolation. Harmonic interpolation is referred to as hyperbolic interpolation
or Balducci’s hypothesis.
8.1
Uniform Distribution of Deaths
The easiest interpolation method is linear interpolation, or uniform distribution of deaths. 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 + s l x +1 = l x − s d x
(8.1)
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 q100 = 0.45 and
l 100 = 1000.
Contrast UDD with de Moivre. If mortality follows de Moivre’s law, 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
mortality follows de Moivre’s law, UDD holds at all ages, but not conversely.
Using l x +s , we can compute s qx :
s qx
l 100+s
1000
= 1 − s px
=1−
550
0
l x +s
= 1 − (1 − s qx ) = s qx
lx
(8.2)
0
1
s
That is one of the most important formulas, so let’s state it again:
s qx
More generally, for 0 ≤ s + t ≤ 1,
s qx +t
= s qx
l x +s +t
l x +t
l x − (s + t )d x
sdx
s qx
=1−
=
=
lx − t dx
lx − t dx
1 − t qx
(8.2)
= 1 − s p x +t = 1 −
(8.3)
where the last equation was obtained by dividing numerator and denominator by l x . The important point to
pick up is that while s qx is the proportion of the year s times qx , the corresponding concept at age x + t , s qx +t ,
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127
128
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
is not s qx , but is in fact higher than s qx . 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 qx +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.
EXAMPLE 8A You are given:
(i) qx = 0.1
(ii) Uniform distribution of deaths between integral ages is assumed.
Calculate 1/2qx +1/4 .
ANSWER: Let l x = 1. Then l x +1 = l x (1 − qx ) = 0.9 and d x = 0.1. Linearly interpolating,
l x +1/4 = l x − 41 d x = 1 − 14 (0.1) = 0.975
l x +3/4 = l x − 34 d x = 1 − 34 (0.1) = 0.925
1/2 qx +1/4
=
l x +1/4 − l x +3/4 0.975 − 0.925
=
= 0.051282
l x +1/4
0.975
ƒ
You could also use equation (8.3) to work this example.
EXAMPLE 8B For two lives age (x ) with independent future lifetimes, k |qx = 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 |qx , we get
l x +1 = 1 − d x = 1 − 0.1 = 0.9
qx = 0.1(1) = 0.1
l x +2 = 0.9 − d x +1 = 0.9 − 0.2 = 0.7
1| qx = 0.1(2) = 0.2
2| qx
l x +3 = 0.7 − d x +2 = 0.7 − 0.3 = 0.4
= 0.1(3) = 0.3
Then linearly interpolating between l x +2 and l x +3 , we get l x +2.25 = 0.7−0.25(0.3) = 0.625, and 2.25 p x = l x +2.25 /l x =
ƒ
0.625. Squaring, the answer is 0.6252 = 0.390625 .
The probability density function of T (x ), s p x µx +s , is the constant qx ,
the derivative of the conditional cumulative distribution function s qx = s qx
with respect to s . That is another important formula, since the density is
needed to compute expected values, so let’s repeat it:
s px
(8.4)
µx +s = qx
µ100+s
1
0.45
0.55
0.45
It follows that the force of mortality is qx divided by 1 − s qx :
µx +s =
qx
qx
=
1 − s qx
s px
(8.5)
s
0
0
1
The force of mortality increases over the year, as illustrated in the graph for q100 = 0.45 to the right.
Under the uniform distribution of deaths hypothesis, the fraction of the year lived by those dying during the
year is a (x ) = 21 . Therefore, L x = 21 (l x + l x +1 ) and
mx =
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2d x
qx
=
l x + l x +1 1 − 0.5qx
(8.6)
129
8.1. UNIFORM DISTRIBUTION OF DEATHS
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
1
distribution of deaths is not assumed. Since S is uniform on [0, 1), E[S] = 21 and Var(S) = 12
. It follows from
1
E[S] = 2 that
e x = e x + 12
(8.7)
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 (6.6) on page 80. For UDD, t p x is linear between integral ages. Therefore,
a rule we learned in Lesson 6 applies for all integral x :
e x :1 = p x + 0.5qx
(6.10)
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 de Moivre.
EXAMPLE 8C You are given
(i) qx = 0.1.
(ii) 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 (1.3). 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 0.4qx (0.2) +
0.4 p x (0.4). This is:
0.4 qx
= (0.4)(0.1) = 0.04
0.4 p x
= 1 − 0.04 = 0.96
e x :0.4 = 0.04(0.2) + 0.96(0.4) = 0.392
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 px
1
(0.4, 0.96)
(1.0, 0.9)
A
0
B
0.4
Trapezoid A is the area we are interested in. Its area is 12 (1 + 0.96)(0.4) = 0.392 .
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1.0
t
ƒ
130
?
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
Quiz 8-1 As in Example 8C, you are given
(i) qx = 0.1.
(ii) 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 8D You are given:
(i) qx = 0.1
(ii) qx +1 = 0.2
(iii) 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.
0.5(0.1)
5
=
1 − 0.5(0.1) 95
5
90
=
0.5 p x +0.5 = 1 −
95 95
Š
9
90 €
0.5(0.2) =
0.5|0.5 qx +0.5 =
95
95
5
9
81
−
=
1 p x +0.5 = 1 −
95 95 95
0.5 qx +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 +0.5 = 0.5(90 + 100) = 95
l x +1.5 = 0.5(72 + 90) = 81
5
90
=
95 95
90 − 81
9
=
0.5|0.5 qx +0.5 =
95
95
l x +1.5 81
=
1 p x +0.5 =
l x +0.5 95
0.5 qx +0.5
=1−
Either way, we’re now ready to calculate e x +0.5:1 .
e x +0.5:1 =
89
5(0.25) + 9(0.75) + 81(1)
=
95
95
For the geometric method we draw the following graph:
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131
8.1. UNIFORM DISTRIBUTION OF DEATHS
t p x +0.5
€
1
90
0.5, 95
Š
A
€
1.0, 81
95
Š
B
0
x + 0.5
0.5
x +1
t
1.0
x + 1.5
The heights at x + 1 and x + 1.5 are as we computed above. Then we compute each area separately. The area of
€
Š
€
Š
90
185
81
171
A is 21 1 + 95
(0.5) = 95(4)
. The area of B is 21 90
+ 95
(0.5) = 95(4)
. Adding them up, we get 185+171
= 89
.
ƒ
95
95(4)
95
?
Quiz 8-2 The probability that a battery fails by the end of the k th month is given in the following table:
k
Probability of battery failure by the
end of month k
1
2
3
0.05
0.20
0.60
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 12 more to e x :n since they die on the
1
average in the middle of the year. Thus the difference is 2 n qx :
e x :n = e x :n + 0.5n qx
(8.8)
EXAMPLE 8E You are given:
(i) qx = 0.01 for x = 50, 51, . . . , 59.
(ii) Deaths are uniformly distributed between integral ages.
Calculate e 50:10 .
ANSWER: As we just said, e 50:10 = e 50:10 + 0.510q50 . The first summand, e 50:10 , is the sum of k p 50 = 0.99k for
k = 1, . . . , 10. This sum is a geometric series:
e 50:10 =
10
X
k =1
0.99k =
0.99 − 0.9911
= 9.46617
1 − 0.99
The second summand, the probability of dying within 10 years is 10q50 = 1 − 0.9910 = 0.095618. Therefore
e 50:10 = 9.46617 + 0.5(0.095618) = 9.51398
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ƒ
132
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.2
Constant Force of Mortality
The constant force of mortality
method sets µx +s equal to a constant for x an integral age and
R 1 interpolation
0 < s ≤ 1. Since p x = exp − 0 µx +s ds and µx +s = µ is constant,
p x = e −µ
(8.9)
µ = − ln p x
(8.10)
Moreover, s p x = e −µs = p xs . In fact, s p x +t is independent of t :
s p x +t
= p xs
(8.11)
for any 0 ≤ t ≤ 1−s . Figure 8.1 shows l 100+s and µ100+s for l 100 = 1000 and q100 = 0.45 if constant force of mortality
is assumed.
l 100+s
1000
µ100+s
1
550
0
0
1
s
− ln 0.55
0
− ln 0.55
0
(a) l 100+s
1
s
(b) µ100+s
Figure 8.1: Example of constant force of mortality
As discussed in Section 7.3 in the answer to the third part of Example 7E on page 109, m x = µx if constant
force of mortality is assumed.
Contrast constant force of mortality between integral ages to global constant force of mortality, which was
introduced in Subsection 5.1.1. The method discussed here allows µx to vary for different integers x .
We will now repeat some of the earlier examples but using constant force of mortality.
EXAMPLE 8F You are given:
(i) qx = 0.1
(ii) The force of mortality is constant between integral ages.
Calculate 1/2qx +1/4 .
ANSWER:
1/2 qx +1/4
= 1 − 1/2 p x +1/4 = 1 − p x1/2 = 1 − 0.91/2 = 1 − 0.948683 = 0.051317
EXAMPLE 8G You are given:
(i) qx = 0.1
(ii) qx +1 = 0.2
(iii) The force of mortality is constant between integral ages.
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ƒ
133
8.3. HYPERBOLIC ASSUMPTION
Calculate e x +0.5:1 .
R1
ANSWER: We calculate 0 t p x +0.5 dt . We split this up into two integrals, one from 0 to 0.5 for age x and one from
0.5 to 1 for age x + 1. The first integral is
Z
Z
0.5
Z
0.5
0.5
p xt dt =
t p x +0.5 dt =
0
0
0.9t dt = −
0
1 − 0.90.5
= 0.487058
ln 0.9
For t > 0.5,
t p x +0.5
= 0.5 p x +0.5 t −0.5 p x +1 = 0.90.5 t −0.5 p x +1
so the second integral is
Z
Z
1
0.5
0.9
0.5
t −0.5 p x +1 dt
0.5
0.5
Š 1 − 0.80.5
0.8 dt = − 0.9
= (0.948683)(0.473116) = 0.448837
ln 0.8
€
t
= 0.9
0
0.5
The answer is e x +0.5:1 = 0.487058 + 0.448837 = 0.935895 .
ƒ
Although constant force of mortality is not used as often as UDD, it can be useful for simplifying formulas
under certain circumstances. Calculating the expected present value of an insurance where the death benefit
within a year follows an exponential pattern (this can happen when the death benefit is the discounted present
value of something) may be easier with constant force of mortality than with UDD.
8.3
Hyperbolic Assumption
Under the hyperbolic assumption for distribution of deaths between integral ages,
1−s
s
1
=
+
l x +s
lx
l x +1
or
l x +s =
l x +1
l x l x +1
=
l x +1 + s d x
p x + s qx
(8.12)
where the last equality is obtained by dividing numerator and denominator by l x . Dividing both sides by l x , we
have
px
px
=
p x + s qx
1 − (1 − s )qx
s qx
s qx =
1 − (1 − s )qx
s px
More generally,
=
s qx +t
=
€
s qx
Š
1 − 1 − (s + t ) qx
(8.13)
(8.14)
Notice that the denominator increases as t increases, so that s qx +t is a decreasing function of age x + t . Not very
plausible.
d
One way to derive µx +s is to use µx +s = − ds
ln s p x , or
µx +s

Š
qx p x (1 − (1 − s )qx )2
d
px
qx

Š =
= − ln
=
ds
1 − (1 − s )qx
1
−
(1
− s )qx
p x 1 − (1 − s )qx
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(8.15)
134
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
l 100+s
1000
µ100+s
1
0.45
0.55
550
0
0
1
0.45
s
0
0
1
(a) l 100+s
s
(b) µ100+s
Figure 8.2: Example of hyperbolic assumption
and we see that the force of mortality decreases over the year.
Figure 8.2 shows l 100+s and µ100+s for l 100 = 1000 and q100 = 0.45 if a hyperbolic assumption for fractional
ages is assumed.
The density function s p x µx +s is the product of s p x and µx +s :
s px
µx +s = €
p x qx
1 − (1 − s )qx
Š2
and can also be derived as the negative derivative of s p x with respect to s . If you use that alternative, you can
divide the result by s p x as an alternative method for deriving µx +s .
Calculating m x is messy. First we calculate L x .
Z1
Lx =
l x +t dt
Z
0
1
=
0
l x +1 dt
p x + t qx
= l x +1
=−
by equation (8.12)
1
ln(p x + t qx ) qx
0
l x +1 ln p x
qx
Then
mx =
dx
Lx
=−
=−
d x qx
l x p x ln p x
qx2
p x ln p x
We will now repeat some of the earlier examples but using the hyperbolic assumption.
EXAMPLE 8H You are given:
(i) qx = 0.1
(ii) Mortality follows the Balducci hypothesis between integral ages.
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8.3. HYPERBOLIC ASSUMPTION
Calculate 1/2qx +1/4 .
ANSWER:
1/2 qx +1/4
=
(1/2)qx
(0.5)(0.1)
0.05
€
Š =
=
= 0.051282
1 − 0.25(0.1) 0.975
1 − 1 − (1/2 + 1/4) qx
In this case, the answer is the same as with uniform distribution of deaths.
ƒ
EXAMPLE 8I You are given:
(i) qx = 0.1
(ii) qx +1 = 0.2
(iii) Deaths between integral ages follow the hyperbolic assumption.
Calculate e x +0.5:1 .
R1
ANSWER: We calculate 0 t p x +0.5 dt . We split this up into two integrals, one from 0 to 0.5 for age x and one from
0.5 to 1 for age x + 1. By equation (8.14),
s p x +t
=1−
s qx
1 − (1 − t )qx
€
Š =
€
Š
1 − 1 − (s + t ) qx
1 − 1 − (s + t ) qx
The first integral is
Z
Z
1 − (1 − 0.5)(0.1)
€
Š dt
1 − 1 − (0.5 + t )(0.1)
0
0.5
€
Š
0.95
=
ln 1 − 1 − (0.5 + t ) (0.1) 0.1
0
0.95
(− ln 0.95) = 0.487286
=
0.1
0.5
0.5 t p x +0.5 dt =
0
Under the hyperbolic assumption, 0.5qx +0.5 = 0.5qx , so 0.5 p x +0.5 in this example is 1 − 0.5(0.1) = 0.95. For t > 0.5,
t p x +0.5
= 0.5 p x +0.5 t −0.5 p x +1 = 0.95 t −0.5 p x +1
so the second integral is (changing the variable, reducing t by 0.5)
Z
Z
0.5
0.95
t p x +1 dt
0
0.5
p x +1 dt
1
−
(1 − t )qx +1
0
€
Š0.5
(0.95)(0.8)
=
ln 1 − (1 − t )(0.2) 0.2
0
= 0.95
(0.95)(0.8)(ln 0.9 − ln 0.8)
0.2
= 0.447576
=
The answer is 0.487286 + 0.447576 = 0.934862
ƒ
One convenient property of the hyperbolic hypothesis is that 1−s qx +s = (1 − s )qx . This can be convenient if
the exact starting age at which a population is observed varies within an age.
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8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
EXAMPLE 8J A population of 60-year olds with life insurance policies has mortality rate q60 = 0.01. Their birthdays are uniformly distributed throughout the year.
50 members of this population surrender their contracts on Jan. 1. They are assumed to have randomly
distributed birthdays.
Calculate as of Jan. 1, for the 50 lives who surrendered their contracts, the expected number of deaths before
their next birthdays if:
1. Deaths between integral ages follow the hyperbolic assumption.
2. Deaths between integral ages are uniformly distributed.
ANSWER:
1. The expected number of deaths for each life is (1 − s )qx = 0.01(1 − s ), where 1 − s is the amount
of time to the next birthday. We integrate this over the uniform distribution of birthdays (density 1) from
0 to 1, and multiply by 50:
Z1
50
0
0.01(1 − s )ds = 0.25
2. The expected number of deaths for each life is
1−s qx +s
=
(1 − s )qx
0.01(1 − s )
=
1 − s qx
1 − 0.01s
We temporarily drop the multiplicative constant 0.01. We then integrate the expression as follows:
99
1−s
= 100 −
1 − 0.01s
1 − 0.01s
Z1
Z1
(1 − s )ds
ds
= 100 − 99
1
−
0.01s
1
−
0.01s
0
0
1
= 100 + 9900 ln(1 − 0.01s )
0
= 100 + 9900 ln 0.99 = 0.501675
Multiplying this result by 0.01 and by 50 lives, we get 0.5(0.501675) = 0.25084 .
ƒ
The implausible hyperbolic hypothesis is mainly of historical interest. It used to be thought that this hypothesis could justify a linear method for estimating mortality rates. Because of the awkwardness of the formulas,
exam questions have generally been limited to calculating survival probabilities, and on rare occasions force of
mortality.
8.4
Summary
The formulas are summarized in Table 8.1. Figure 8.3 compares l x +s and µx +s for the three assumptions, using
x = 100 and qx = 0.45.
What do you need to memorize for the exam? For UDD, if you know the linear formula for l x , you will be
able to answer all probability questions. You will be able to rederive the formula for µx +s , or you may memorize
the µx +s formula.
The constant force of mortality appears least often on the released exams, but it is easy. If you know that
−s µx , you have the whole story.
p
s x =e
The hyperbolic assumption is the hardest one to memorize, but it is unlikely they will ask anything beyond a
probability question on it. If you memorize the formula for s qx +t , you should be able to get by. To play safe, you
may also memorize the formula for µx +s .
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8.4. SUMMARY
Table 8.1: Summary of formulas for fractional ages
Function
Uniform distribution of
deaths
Constant force of
mortality
Hyperbolic assumption
lx − s dx
l x p xs
s px
1 − s qx
p xs
s qx +t
s qx /(1 − t qx )
1 − p xs
µx +s
qx /(1 − s qx )
− ln p x
qx
qx /(1 − 0.5qx )
− ln p x
−qx2 /(p x ln p x )
l x +s
s px
1 − p xs
s qx
s qx
µx +s
mx
qx
l x − 0.5d x
Lx
ex
e x + 0.5
e x :n
e x :n + 0.5 n qx
e x :1
p x + 0.5qx
l x +1 /(p x + s qx )

Š
s qx 1 − (1 − s )qx

Š
p x 1 − (1 − s )qx
.
€
Š s qx
1 − 1 − (s + t ) qx
l 100+s
1000
−p xs (ln p x )

Š
1 − (1 − s )qx

Š2
p x qx 1 − (1 − s )qx
−d x / ln p x
−l x +1 ln p x /qx
µ100+s
1
550
400
0
1
s
(a) l x +s
0.4
0
1
s
LEGEND
Uniform distribution of deaths
Constant force of
mortality
Hyperbolic assumption
(b) µx +s
Figure 8.3: Comparison of the three assumptions for distribution of deaths between integral ages
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138
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
You should also understand the picture, Figure 8.3. Namely, UDD kills off lives the slowest, and hyperbolic
the fastest. UDD has an increasing force of mortality, and hyperbolic a decreasing force. The constant force of
mortality assumption has a constant force of course, and is in the middle of the other two.
Exercises
8.1. [CAS4-S85:16] (1 point) Deaths are uniformly distributed between integral ages.
Which of the following represents 3/4 p x + 12 1/2 p x µx +1/2 ?
(A)
3/4 p x
8.2.
(B)
3/4 qx
(C)
1/2 p x
(D)
1/2 qx
(E)
1/4 p x
[Based on 150-S88:25] You are given:
(i) 0.25qx +0.75 = 3/31.
(ii) Mortality is uniformly distributed within age x .
Calculate qx .
Use the following information for questions 8.3 and 8.4:
You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii) qx = 0.10.
(iii) qx +1 = 0.15.
8.3. Calculate 1/2qx +3/4 .
8.4. Calculate 0.3|0.5qx +0.4 .
8.5. You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii) Mortality follows the Illustrative Life Table.
Calculate the median future lifetime for (45.5).
8.6. Deaths are uniformly distributed between integral ages. You are given that µx +0.75 = 0.04.
Calculate m x .
8.7. You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii)
x
qx
52
53
54
0.10
0.11
0.13
Calculate 3 m 52 .
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139
EXERCISES FOR LESSON 8
8.8. [160-F90:5] You are given:
(i) A survival distribution is defined by

x
l x = 1000 1 −
100
‹2 , 0 ≤ x ≤ 100.
(ii) µx denotes the actual force of mortality for the survival distribution.
(iii) µxL denotes the approximation of the force of mortality based on the uniform distribution of deaths
assumption for l x , 50 ≤ x < 51.
L
Calculate µ50.25 − µ50.25
.
(A) −0.00016
(B) −0.00007
(C) 0
(D) 0.00007
(E) 0.00016
8.9. A survival distribution is defined by
(i)
(ii)
s (k ) = 1/(1 + 0.01k )4 for k a non-negative integer.
Deaths are uniformly distributed between integral ages.
Calculate 0.4q20.2 .
8.10. [150-S89:15] You are given:
(i) Deaths are uniformly distributed over each year of age.
(ii)
x
lx
35
36
37
38
39
100
99
96
92
87
Which of the following are true?
I.
1|2 q 36
= 0.091
II. m 37 = 0.043
III.
0.33 q 38.5
= 0.021
(A) I and II only
(B) I and III only
(C) II and III only
(E) The correct answer is not given by (A) , (B) , (C) , or (D) .
(D) I, II and III
8.11. [150-82-94:5] You are given:
(i) Deaths are uniformly distributed over each year of age.
(ii) 0.75 p x = 0.25.
Which of the following are true?
I.
II.
III.
0.25 qx +0.5
0.5 qx
= 0.5
= 0.5
µx +0.5 = 0.5
(A) I and II only
(B) I and III only
(C) II and III only
(E) The correct answer is not given by (A) , (B) , (C) , or (D) .
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(D) I, II and III
Exercises continue on the next page . . .
140
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.12. [3-S00:12] For a certain mortality table, you are given:
(i)
(ii)
(iii)
(iv)
µ(80.5) = 0.0202
µ(81.5) = 0.0408
µ(82.5) = 0.0619
Deaths are uniformly distributed between integral ages.
Calculate the probability that a person age 80.5 will die within two years.
(A) 0.0782
(B) 0.0785
(C) 0.0790
(D) 0.0796
(E) 0.0800
8.13. You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii) qx = 0.1.
(iii) qx +1 = 0.3.
Calculate e x +0.7:1 .
8.14. You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii) q45 = 0.01.
(iii) q46 = 0.011.
Calculate the variance of the 2-year temporary complete life expectancy on (45).
8.15. You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii) 10 p x = 0.2.
Calculate e x :10 − e x :10 .
8.16. [4-F86:21] You are given:
(i) q60 = 0.020
(ii) q61 = 0.022
(iii) Deaths are uniformly distributed over each year of age.
Calculate e 60:1.5 .
(A) 1.447
(B) 1.457
(C) 1.467
(D) 1.477
(E) 1.487
(D) 1.465
(E) 1.475
8.17. [150-F89:21] You are given:
(i) q70 = 0.040
(ii) q71 = 0.044
(iii) Deaths are uniformly distributed over each year of age.
Calculate e 70:1.5 .
(A) 1.435
(B) 1.445
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(C) 1.455
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141
EXERCISES FOR LESSON 8
8.18. [3-S01:33] For a 4-year college, you are given the following probabilities for dropout from all causes:
q0 = 0.15
q1 = 0.10
q2 = 0.05
q3 = 0.01
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) 1.25
(B) 1.30
(C) 1.35
(D) 1.40
(E) 1.45
(D) 0.167
(E) 0.200
(D) 0.7
(E) 0.8
8.19. You are given:
(i) Deaths are uniformly distributed between integral ages.
(ii) e x +0.5:0.5 = 5/12.
Calculate qx .
8.20. You are given:
(i) Deaths are uniformly distributed over each year of age.
(ii) e 55.2:0.4 = 0.396.
Calculate µ55.2 .
8.21. [150-S87:21] You are given:
(i) d x = k for x = 0, 1, 2, . . . , ω − 1
(ii) e 20:20 = 18
(iii) Deaths are uniformly distributed over each year of age.
Calculate 30|10q30 .
(A) 0.111
(B) 0.125
(C) 0.143
8.22. [150-S89:24] You are given:
(i) Deaths are uniformly distributed over each year of age.
(ii) µ45.5 = 0.5
Calculate e 45:1 .
(A) 0.4
(B) 0.5
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(C) 0.6
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142
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.23. [CAS3-S04:10] 4,000 people age (30) each pay an amount, P, into a fund. Immediately after the 1,000th
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
8.24. [160-F87:5] Based on given values of l x and l x +1 ,
force of mortality.
1/4 p x +1/4
= 49/50 under the assumption of constant
Calculate 1/4 p x +1/4 under the uniform distribution of deaths hypothesis.
(A) 0.9799
(B) 0.9800
(C) 0.9801
(D) 0.9802
(E) 0.9803
8.25. [160-S89:5] A mortality study is conducted for the age interval (x , x + 1].
If a constant force of mortality applies over the interval, 0.25qx +0.1 = 0.05.
Calculate 0.25qx +0.1 assuming a uniform distribution of deaths applies over the interval.
(A) 0.044
(B) 0.047
(C) 0.050
(D) 0.053
(E) 0.056
8.26. [150-F89:29] You are given that qx = 0.25.
Based on the constant force of mortality assumption, the force of mortality is µxA+s , 0 < s < 1.
Based on the uniform distribution of deaths assumption, the force of mortality is µxB+s , 0 < s < 1.
Calculate the smallest s such that µxB+s ≥ µxA+s .
(A) 0.4523
(B) 0.4758
(C) 0.5001
(D) 0.5242
(E) 0.5477
8.27. [160-S91:4] From a population mortality study, you are given:
(i) Within each age interval, (x + k , x + k + 1], the force of mortality, µx +k , is constant.
1 − e −µx +k
(ii)
k
e −µx +k
µx +k
0
1
0.98
0.96
0.99
0.98
Calculate e x :2 , the expected lifetime in years over (x , x + 2].
(A) 1.92
(B) 1.94
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(C) 1.95
(D) 1.96
(E) 1.97
Exercises continue on the next page . . .
143
EXERCISES FOR LESSON 8
8.28. [150-81-94:50] You are given:
s (x ) =
(10 − x )2
100
0 ≤ x ≤ 10
Instead of using s (x ) within each year of age, use the constant force of mortality assumption to calculate the
average number of years lived between ages 1 and 2 by those of the survivorship group who die between those
ages.
(A) 0.461
(B) 0.480
(C) 0.490
(D) 0.500
(E) 0.508
8.29. [3-S01:27] An actuary is modeling the mortality of a group of 1000 people, each age 95, for the next three
years.
The actuary starts by calculating the expected number of survivors at each integral age by
l 95+k = 1000 k p 95 ,
k = 1, 2, 3
The actuary subsequently calculates the expected number of survivors at the middle of each year using the
assumption that deaths are uniformly distributed over each year of age.
This is the result of the actuary’s model:
Age
95
95.5
96
96.5
97
97.5
98
Survivors
1000
800
600
480
—
288
—
The actuary decides to change his assumption for mortality at fractional ages to the constant force assumption. He retains his original assumption for each k p 95 .
Calculate the revised expected number of survivors at age 97.5
(A) 270
(B) 273
(C) 276
(D) 279
(E) 282
8.30. [160-S87:3] You are analyzing death rates within the age interval [x , x + 1]. You are given that d x > 0.
Death rates are calculated according to the following assumptions:
Mortality Rate
Assumption
U
1q
2 x
B
1q
2 x
C
1q
2 x
Uniform distribution of deaths
Balducci hypothesis
Constant force of mortality
Which of the following is correct?
(A)
(B)
(C)
(D)
(E)
1
2
qxU < 1 qxB < 1 qxC
1
2
qxU < 1 qxC < 1 qxB
1
2
qxB < 1 qxC < 1 qxU
1
2
qxC < 1 qxU < 1 qxB
1
2
qxC < 1 qxB < 1 qxU
2
2
2
2
2
2
2
2
2
2
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144
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.31. [150-F86:17] You are given 1000 1/4qx = 2.
Rank 1/4qx +3/4 under the Balducci hypothesis, the uniform distribution of deaths (UDD) hypothesis, and the
constant force of mortality hypothesis.
(A)
Balducci < constant force < UDD
(B)
Constant force < Balducci < UDD
(C)
UDD < constant force < Balducci
(D)
UDD < Balducci < constant force
(E)
Balducci < UDD < constant force
8.32. [160-F86:13] You are given that l x = 284 and l x +1 = 236.
For some t , 0 < t < 1, the hyperbolic assumption for deaths within a year produces l x +t = 248.
For some u , 0 < u < 1, the assumption of uniform deaths within a year produces l x +u = 248.
Determine t − u .
(A) −0.079
(B) −0.036
(C) 0.000
(D) 0.036
(E) 0.079
8.33. [160-S88:6] There is an age (x + s ) in the interval (x , x + 12 ) such that 13/16−s qx +s has the same value under
both the hyperbolic and uniform assumptions over (x , x + 1].
Determine s .
(A) 3/16
(B) 4/16
(C) 5/16
(D) 6/16
(E) 7/16
(D) 0.40
(E) 0.45
8.34. [150-S88:25] You are given:
(i) 0.5 p x µx +0.5 = 12/49
(ii) Mortality follows the Balducci assumption.
(iii) qx < p x
Calculate qx .
(A) 0.25
(B) 0.30
(C) 0.35
8.35. [160-S89:4] Under the hyperbolic survival assumption for l x +s in the interval (x , x + 1], which of the
following are true?
I.
1/2 qx
≥ 1/2qx +1/2
II.
s px
= p x /(p x + s qx )
III.
s qx
= s µx +s
IV. l x + 1 = (2l x l x +1 )/(l x + l x +1 )
2
(A) All but I
(B) All but II
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(C) All but III
(D) All but IV
(E) All
Exercises continue on the next page . . .
145
EXERCISES FOR LESSON 8
8.36. [150-F88:29] For integral values of x and 0 < s < 1, you are given:
µxA+s is based on the uniform distribution of deaths assumption.
µxB+s is based on the Balducci distribution of deaths assumption.
(i)
(ii)
Which of the following are true for 0.5 ≤ t < 1?
I. µxA+t = t d x /(l x − t d x )
II. µxB+t = d x /(l x − (1 − t )d x )
µxA+t ≤ µxB+t
III.
(A) I and II only
(B) I and III only
(C) II and III only
(E) The correct answer is not given by (A) , (B) , (C) , or (D) .
8.37.
(D) I, II and III
[160-S91:3] You are given:
l x = 100, 000/x 2 , for positive integer values of x .
Under the hyperbolic assumption over (y , y + 1] for some integer y , l y +s = 911, where 0 < s < 1.
(i)
(ii)
Calculate 1−s qy +s under the uniform distribution of deaths assumption.
(A) 0.087
8.38.
(B) 0.091
(C) 0.093
(D) 0.095
(E) 0.101
[150-S91:20] You are given:
(i) e x :1 = F under the uniform distribution of deaths assumption.
(ii) e x :1 = G under the Balducci assumption.
(iii) qx = 0.1.
Calculate 1000(F − G ).
(A) 0.00
(B) 0.24
(C) 1.00
(D) 1.76
(E) 2.52
8.39. [150-S91:28] You are given that qx = 0.1200.
Which of the following are true?
(i)
(ii)
(iii)
1/3 qx +1/2
= 0.0426 under the uniform distribution of deaths assumption.
q
=
0.0435
under the Balducci assumption.
1/3 x
1/2 qx = 0.0619 under the constant force of mortality assumption.
(A) I and II only
(B) I and III only
(C) II and III only
(E) The correct answer is not given by (A) , (B) , (C) , or (D) .
(D) I, II and III
8.40. [150-81-94:23] You are given:
(i) q60 = 0.3
(ii) q61 = 0.4
(iii) f is the probability that (60) will die between ages 60.5 and 61.5 under the uniform distribution of
deaths assumption.
(iv) g is the probability that (60) will die between ages 60.5 and 61.5 under the Balducci assumption.
Calculate 10,000(g − f ).
(A) 0
(B) 85
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(C) 94
(D) 178
(E) 213
Exercises continue on the next page . . .
146
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.41. [3-S01:13] Mr. Ucci has only 3 hairs left on his head and he won’t be growing any more.
(i) The future mortality of each hair follows
k | qx
= 0.1(k + 1),
k = 0, 1, 2, 3 and x is Mr. Ucci’s age
(ii) Hair loss follows the hyperbolic assumption at fractional ages.
(iii) The future lifetimes of the 3 hairs are independent.
Calculate the probability that Mr. Ucci is bald (has no hair left) at age x + 2.5.
(A) 0.090
(B) 0.097
(C) 0.104
(D) 0.111
(E) 0.118
8.42. [SOA3-F03:16] For a space probe to Mars:
(i) The probe has three radios, whose future lifetimes are independent, each with mortality following
k |q0
= 0.1(k + 1),
k = 0, 1, 2, 3
where time 0 is the moment the probe lands on Mars.
(ii) The failure time of each radio follows the hyperbolic assumption within each year.
(iii) The probe will transmit until all three radios have failed.
Calculate the probability that the probe is no longer transmitting 2.25 years after landing on Mars.
(A) 0.053
(B) 0.059
(C) 0.063
(D) 0.067
(E) 0.069
8.43. [SOA3-F03:36] Consider the 1-year temporary complete life expectancy of (90), i.e. e 90:1 , as based on the
Illustrative Life Table. Let C , H , and U be its values under the constant force, the hyperbolic, and the uniform
distribution assumptions respectively.
Which of the following is true?
(A)
C > H >U
(B)
C >U > H
(C)
U >C >H
(D)
U >H >C
(E)
H =C =U
8.44.
[160-S91:5] With respect to the interval (x , x + 1], you are given:
(i) p x = 0.5
(ii) m xH denotes the central death rate under the hyperbolic assumption.
(iii) m xE denotes the central death rate under the exponential assumption.
Calculate m xH − m xE .
(A) −0.028
(B) −0.026
8.45. You are given that
assumption.
30 p 50
= 0.52 and
(C) 0.000
31 p 50
(D) 0.026
(E) 0.028
= 0.48. Deaths between integral ages follow the hyperbolic
Calculate median future lifetime for (50).
Additional released exam questions: CAS3-S05:31, CAS3-F05:13, CAS3-S06:13, CAS3-F06:13, SOA M-F06:16,
CAS3-S07:24, CAS3L-S08:16, CAS3L-S09:3, CAS3L-F09:3, CAS3L-S10:4, CAS3L-F10:3
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147
EXERCISE SOLUTIONS FOR LESSON 8
Solutions
8.1. In the second summand, 1/2 p x µx +1/2 is the density function, which is the constant qx under UDD. The
first summand 3/4 p x = 1 − 34 qx . So the sum is 1 − 14 qx , or 1/4 p x . (E)
8.2. Using equation (8.3),
3
0.25qx
= 0.25qx +0.75 =
31
1 − 0.75qx
3
2.25
−
qx = 0.25qx
31
31
3
10
= qx
31 31
qx = 0.3
8.3. We calculate the probability that (x + 34 ) 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 (8.3),
1 − 0.10
0.9
=
1/4 p x +3/4 =
1 − 0.75(0.10) 0.925
The probability of (x + 1) surviving to x + 1.25 is
1/4 p x +1
= 1 − 0.25(0.15) = 0.9625
The answer to the question is then the complement of the product of these two numbers:
0.9
(0.9625) = 0.06351
1/2 qx +3/4 = 1 − 1/2 p x +3/4 = 1 − 1/4 p x +3/4 1/4 p x +1 = 1 −
0.925
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) = 0.765. Under UDD, l x at fractional ages is obtained by linear interpolation, so
l x +0.75 = 0.75(0.9) + 0.25(1) = 0.925
l x +1.25 = 0.25(0.765) + 0.75(0.9) = 0.86625
8.4.
0.3|0.5 qx +0.4
l x +1.25 0.86625
=
= 0.93649
l x +0.75
0.925
1/2 p 3/4
=
1/2 q 3/4
= 1 − 1/2 p 3/4 = 1 − 0.93649 = 0.06351
is 0.3 p x +0.4 − 0.8 p x +0.4 . The first summand is
0.3 p x +0.4
=
1 − 0.7qx
1 − 0.07 93
=
=
1 − 0.4qx
1 − 0.04 96
The probability that (x + 0.4) survives to x + 1 is, by equation (8.3),
0.6 p x +0.4
and the probability (x + 1) survives to x + 1.2 is
0.2 p x +1
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=
1 − 0.10 90
=
1 − 0.04 96
= 1 − 0.2qx +1 = 1 − 0.2(0.15) = 0.97
148
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
So
93
90
−
(0.97) = 0.059375
96
96
0.3|0.5 qx +0.4 =
Alternatively, you could use the life table from the solution to the last question, and linearly interpolate:
l x +0.4 = 0.4(0.9) + 0.6(1) = 0.96
l x +0.7 = 0.7(0.9) + 0.3(1) = 0.93
l x +1.2 = 0.2(0.765) + 0.8(0.9) = 0.873
0.3|0.5 qx +0.4
=
0.93 − 0.873
= 0.059375
0.96
8.5. Under uniform distribution of deaths between integral ages, l x +0.5 = 12 (l x + l x +1 ), since the survival function is a straight line between two integral ages. Therefore, l 45.5 = 21 (9,164,051+9,127,426) = 9,145,738.5. Median
future lifetime occurs when l x = 21 (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
4,828,182 − 297,822s = 4,572,869
s=
4,828,182 − 4,572,869 255,313
=
= 0.8573
297,822
297,822
So the median age at death is 77.8573, and median future lifetime is 77.8573 − 45.5 = 32.3573 .
8.6. First we use equation (8.5) to get qx .
0.04 = µx +0.75 =
0.04 − 0.03qx = qx
4
qx =
103
qx
1 − 0.75qx
Then from equation (8.6),
mx =
4
4/103
=
1 − 2/103
101
8.7. We use the fact that for each age, L x = 12 (l x + l x +1 ), so 3 L 52 = 12 (l 52 + 2l 53 + 2l 54 + l 55 ), and we have
3 m 52
=
3 d 52
1
2
(l 52 + 2l 53 + 2l 54 + l 55 )
Dividing numerator and denominator by l 52 ,
3 m 52
=
=
23q52
1 + 2p 52 + 22 p 52 + 3 p 52
€
Š
2 1 − (0.90)(0.89)(0.87)
1 + 2(0.90) + 2(0.90)(0.89) + (0.90)(0.89)(0.87)
0.60626
=
= 0.11890
5.09887
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EXERCISE SOLUTIONS FOR LESSON 8
8.8.
x p0
=
lx
l0
=1−
x 2
.
100
The force of mortality is calculated as the negative derivative of ln x p 0 :
x 1 d ln x p 0 2 100 100
2x
µx = −
=
=
2 −x2
x 2
dx
100
1 − 100
µ50.25 =
For UDD, we need to calculate q50 .
100.5
1002 − 50.252
p 50 =
= 0.0134449
l 51 1 − 0.512
=
= 0.986533
l 50 1 − 0.502
q50 = 1 − 0.986533 = 0.013467
so under UDD,
L
µ50.25
=
q50
0.013467
=
= 0.013512.
1 − 0.25q50 1 − 0.25(0.013467)
L
is 0.013445 − 0.013512 = −0.000067 . (B)
The difference between µ50.25 and µ50.25
8.9. s (20) = 1/1.24 and s (21) = 1/1.214 , so q20 = 1 − (1.2/1.21)4 = 0.03265. Then
0.4 q 20.2
=
0.4q20
0.4(0.03265)
=
= 0.01315
1 − 0.2q20 1 − 0.2(0.03265)
8.10.
I. Calculate 1|2q36 .
1|2 q 36
=
2 d 37
l 36
=
96 − 87
= 0.09091
99
!
This statement does not require uniform distribution of deaths.
II. By equation (8.6),
m 37 =
2d 37
2(4)
=
= 0.042553
l 37 + l 38 96 + 92
III. Calculate 0.33q38.5 .
0.33 q 38.5
=
0.33 d 38.5
l 38.5
=
(0.33)(5)
= 0.018436
89.5
I can’t figure out what mistake you’d have to make to get 0.021. (A)
8.11. First calculate qx .
1 − 0.75qx = 0.25
qx = 1
Then by equation (8.3), 0.25qx +0.5 = 0.25/(1 − 0.5) = 0.5, making I true.
By equation (8.2), 0.5qx = 0.5qx = 0.5, making II true.
By equation (8.5), µx +0.5 = 1/(1 − 0.5) = 2, making III false. (A)
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!
#
150
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.12. We use equation (8.5) to back out qx for each age.
µx +0.5 =
qx
µx +0.5
⇒ qx =
1 − 0.5qx
1 + 0.5µx +0.5
0.0202
q80 =
= 0.02
1.0101
0.0408
q81 =
= 0.04
1.0204
0.0619
q82 =
= 0.06
1.03095
Then by equation (8.3), 0.5 p 80.5 = 0.98/0.99. p 81 = 0.96, and 0.5 p 82 = 1 − 0.5(0.06) = 0.97. Therefore
0.98
(A)
q
=
1
−
(0.96)(0.97) = 0.0782
2 80.5
0.99
8.13. 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
Group
I
II
III
Survival
time
Probability
of group
Average
survival time
(0, 0.3]
(0.3, 1]
(1, ∞)
0.3 p x +0.7 − 1 p x +0.7
1 − 0.3 p x +0.7
0.15
0.65
1
1 p x +0.7
We calculate the required probabilities.
0.9
= 0.967742
0.93
Š
0.9 €
1 − 0.7(0.3) = 0.764516
1 p x +0.7 =
0.93
1 − 0.3 p x +0.7 = 1 − 0.967742 = 0.032258
0.3 p x +0.7
0.3 p x +0.7 − 1 p x +0.7
=
= 0.967742 − 0.764516 = 0.203226
e x +0.7:1 = 0.032258(0.15) + 0.203226(0.65) + 0.764516(1) = 0.901452
Alternatively, we can use trapezoids. We already know from the above solution that the heights of the first
trapezoid are 1 and 0.967742, and the heights of the second trapezoid are 0.967742 and 0.764516. So the sum of
the area of the two trapezoids is
e x +0.7:1 = (0.3)(0.5)(1 + 0.967742) + (0.7)(0.5)(0.967742 + 0.764516)
= 0.295161 + 0.606290 = 0.901451
8.14. 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
= (1 − 0.005) + 0.99(1 − 0.0055)
= 1.979555
We’ll use equation (6.7)to calculate the second moment.
Z2
E[(T ∧ 2)2 ] = 2
t t p x dt
0
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151
EXERCISE SOLUTIONS FOR LESSON 8
Z
=2

Z
1
0
t (1 − 0.01t ) dt +
2
1
€
Š
t (0.99) 1 − 0.011(t − 1) dt
!


3
2
2
3
1
2 − 1 
1
 (1.011)(2 − 1 )
= 2  − 0.01
+ 0.99 
− 0.011

2
3
2
3
= 2(0.496667 + 1.475925) = 3.94518
So the variance is 3.94518 − 1.9795552 = 0.02654 .
8.15. As discussed on page 131, by equation (8.8), the difference is
1
1
10 qx = (1 − 0.2) = 0.4
2
2
8.16. Those who die in the first year survive 21 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 60 = 0.98
1.5 p 60
€
Š
= 0.98 1 − 0.5(0.022) = 0.96922
e 60:1.5 = 0.5(0.02) + 1.25(0.98 − 0.96922) + 1.5(0.96922) = 1.477305
(D)
Alternatively, we use the trapezoidal method. The first trapezoid has heights 1 and p 60 = 0.98 and width 1.
The second trapezoid has heights p 60 = 0.98 and 1.5 p 60 = 0.96922 and width 1/2.
1
1 1
e 60:1.5 = (1 + 0.98) +
(0.98 + 0.96922)
2
2 2
= 1.477305
(D)
8.17. p 70 = 1 − 0.040 = 0.96, 2 p 70 = (0.96)(0.956) = 0.91776, and by linear interpolation, 1.5 p 70 = 0.5(0.96 +
0.91776) = 0.93888. 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
e 70:1.5 = 0.5(0.04) + 1.25(0.96 − 0.93888) + 1.5(0.93888) = 1.45472
(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 0.93888 and width 1/2, so
e 70:1.5 = 0.5(1 + 0.96) + 0.5(0.5)(0.96 + 0.93888) = 1.45472
(C)
8.18. First we calculate t p 1 for t = 1, 2.
p 1 = 1 − q1 = 0.90
2p 1
= (1 − q1 )(1 − q2 ) = (0.90)(0.95) = 0.855
By linear interpolation, 1.5 p 1 = (0.5)(0.9 + 0.855) = 0.8775.
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 1:1.5 = 0.10(0.5) + (0.90 − 0.8775)(1.25) + 0.8775(1.5) = 1.394375
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(D)
152
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
t p1
Alternatively, we could sum the two trapezoids making up the shaded
area at the right.
1
(1, 0.9)
(1.5,0.8775)
e 1:1.5 = (1)(0.5)(1 + 0.9) + (0.5)(0.5)(0.90 + 0.8775)
= 0.95 + 0.444375 = 1.394375
(D)
0
1
1.5
8.19. Those who die survive 0.25 years on the average and survivors survive 0.5 years, so we have
0.25 0.5qx +0.5 + 0.5 0.5 p x +0.5 =
0.5qx
1 − qx
0.25
+ 0.5
=
1 − 0.5qx
1 − 0.5qx
2
t
5
12
5
12
0.125qx + 0.5 − 0.5qx =
5
12
qx =
1
2
5
− 24
qx
5
5
1 1
1
−
= − + −
qx
2 12
24 2 8
qx
1
=
12
6
Alternatively, complete life expectancy is the area of the trapezoid
shown on the right, so
t p x +0.5
1
5
12
5
= 0.5(0.5)(1 + 0.5 p x +0.5 )
12
Then 0.5 p x +0.5 = 23 , from which it follows
0
1 − qx
2
=
3 1 − 1 qx
2
qx =
1
2
8.20. Survivors live 0.4 years and those who die live 0.2 years on the average, so
0.396 = 0.40.4 p 55.2 + 0.20.4q55.2
Using the formula 0.4q55.2 = 0.4q55 /(1 − 0.2q55 ) (equation (8.3)), we have
1 − 0.6q55
0.4q55
0.4
+ 0.2
= 0.396
1 − 0.2q55
1 − 0.2q55
0.4 − 0.24q55 + 0.08q55 = 0.396 − 0.0792q55
0.0808q55 = 0.004
0.004
= 0.0495
0.0808
q55
0.0495
µ55.2 =
=
= 0.05
1 − 0.2q55 1 − 0.2(0.0495)
q55 =
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0.5 p x +0.5
0.5
t
153
EXERCISE SOLUTIONS FOR LESSON 8
8.21. Since d x is constant for all x and deaths are uniformly distributed within each year of age, mortality
follows de Moivre’s law. We back out ω using equation (6.9), e x :n = n p x (n ) + n qx (n /2):
10 20q20 + 20 20 p 20 = 18
20
ω − 40
10
+ 20
= 18
ω − 20
ω − 20
200 + 20ω − 800 = 18ω − 360
2ω = 240
ω = 120
x −20 p 20
Alternatively, we can back out ω using the trapezoidal rule. Complete
life expectancy is the area of the trapezoid shown to the right.
ω − 40
e 20:20 = 18 = (20)(0.5) 1 +
ω − 20
ω − 40
0.8 =
ω − 20
0.8ω − 16 = ω − 40
1
18
20
ω − 40
ω − 20
40
x
0.2ω = 24
ω = 120
Once we have ω, we compute
30|10 q 30
8.22. We use equation (8.5) to obtain
=
10
10
=
= 0.1111
ω − 30 90
0.5 =
Then e 45:1
€
Š
= 0.5 1 + (1 − 0.4) = 0.8 . (E)
(A)
qx
1 − 0.5qx
qx = 0.4
8.23. 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
7,201,635 − 7,126,036
x = 67 +
= 67.4127
7,201,635 − 7,018,432
3
1
This is 37.4127 years into the future. 34 of the people collect 50,000. We need 50,000
= 540.32
4 1.1237.4127
per person. (D)
8.24. Under constant force, s p x +t = p xs , so p x = 1/4 p x4 +1/4 = 0.984 = 0.922368 and qx = 1 − 0.922368 = 0.077632.
Under uniform distribution of deaths,
1/4 p x +1/4
(1/4)qx
1 − (1/4)qx
(1/4)(0.077632)
=1−
1 − (1/4)(0.077632)
=1−
= 1 − 0.019792 = 0.980208
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(D)
154
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
8.25. Under constant force, s p x +t = p xs , so p x = 0.954 = 0.814506, qx = 1 − 0.814506 = 0.185494. Then under a
uniform assumption,
0.25qx
(0.25)(0.185494)
=
= 0.047250
(B)
0.25 qx +0.1 =
1 − 0.1qx
1 − 0.1(0.185494)
8.26. Using constant force, µA is a constant equal to − ln p x = − ln 0.75 = 0.2877. Then
µxB+s =
qx
= 0.2877
1 − s qx
0.25
= 0.2877
1 − 0.25s
0.2877 − 0.25(0.2877)s = 0.25
s=
0.2877 − 0.25
= 0.5242
(0.25)(0.2877)
(D)
8.27. We integrate t p x from 0 to 2. Between 0 and 1, t p x = e −t µx .
Z
1
e −t µx dt =
0
Between 1 and 2, t p x = p x t −1 p x +1
1 − e µx
= 0.99
µx
= 0.98e −(t −1)µx +1 .
Z
2
e −(t −1)µx +1 dt =
1
1 − e µx +1
= 0.98
µx +1
So the answer is 0.99 + 0.98(0.98) = 1.9504 . (C)
8.28. We calculate the temporary expectation of life for one-year, e 1:1 , subtract p 1 for the amount lived by
those who survive, then divide by q1 . We don’t use the constant force of mortality assumption to evaluate p 1
and q1 .
s (2) (10 − 2)2 64
=
=
s (1) (10 − 1)2 81
17
q1 = 1 − p 1 =
81
p1 =
For constant force, t p 1 = e −t µ for t ≤ 1. In particular, p 1 = e −µ , so µ = − ln(64/81) = 0.235566. Then
Z
1
e 1:1 =
t p 1 dt
Z
0
1
e −t µ dt
=
0
1 − e −µ
=
µ
q1
17/81
=
=
= 0.890946
µ
0.235566
Then
a (1) =
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0.890946 − (64/81)
= 0.48039
17/81
(B)
155
EXERCISE SOLUTIONS FOR LESSON 8
8.29. Under uniform distribution, the numbers of deaths in each half of the year are equal, so if 120 deaths
occurred in the first half of x = 96, then 120 occurred in the second half, and l 97 = 480 − 120 = 360. Then
p if
0.5
0.6.
q
=
(360
−
288)/360
=
0.2,
then
q
=
2
q
=
0.4,
so
p
=
0.6.
Under
constant
force,
p
=
p
=
0.5 97
97
0.5 97
97
1/2 97
97
p
The answer is 360 0.6 = 278.8548 . (D)
8.30. We are being asked to order the hypotheses based on which one puts the most deaths in the first half of
the year. The Balducci hypothesis speeds deaths the most, since it has a decreasing hazard rate, and the uniform
hypothesis speeds them up the least, since it has an increasing hazard rate, so the answer is (B).
8.31. In general, Balducci speeds up the deaths the most and UDD the least. When looking at deaths in the last
quarter of the year, this means Balducci will have the least and UDD the most, or (A).
They gave you 1/4qx to help make it more concrete. Under constant force of mortality, this equals 1/4qx +3/4 ,
while under Balducci 1/4qx +s is a decreasing function of s and under UDD 1/4qx +s is an increasing function of s .
8.32. For the hyperbolic assumption,
l x l x +1
l x +1 + t (l x − l x +1 )
(236)(284)
248 =
236 + 48t
(236)(284)
236 + 48t =
= 270.2581
248
270.2581 − 236
t=
= 0.71371
48
l x +t =
For the uniform assumption, 248 = 284 + u (236 − 284) = 284 − 48u which yields u = 0.75. t − u = 0.71371 −
0.75 = −0.03629 . (B)
8.33. For the uniform hypothesis,
€ 13
13/16−s qx +s
=
For the hyperbolic hypothesis,
13/16−s p x +s
=
=
=
13/16−s qx +s
=
16
Š
− s qx
1 − s qx
.
13/16 p x
s px
1 − (1 − s )qx
€
Š
1 − 1 − (13/16) qx
1 − (1 − s )qx
1 − (3/16)qx
€ 13
Š
− s qx
16
1 − (3/16)qx
The numerators are equal. Equating the denominators, we get s =
8.34. Use the formula in Table 8.1 for t p x µx +t .
p x qx
(p x + t qx )2
12
(1 − qx )(qx )
=€
Š2
49
1 − (1/2)qx
€
Š2
49qx (1 − qx ) = 12 1 − 12 qx
t px
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µx +t =
3
16
. (A)
156
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
49qx − 49qx2 = 12 − 12qx + 3qx2
52qx2 − 61qx + 12 = 0
p
61 ± 612 − 4(52)(12)
qx =
104
Since qx < p x , we choose the lower solution to the quadratic, (61 − 35)/104 = 0.25 . (A)
8.35. I is true because mortality drops over the year under the hyperbolic assumption. The other formulas are
correct. (E)
8.36.
I. Dividing numerator and denominator by l x , we have t qx /(1 − t qx ). There should not be a t in the numerator. #

Š
II. Dividing the numerator and denominator by l x we have the formula qx 1 − (1 − t )qx . !
III. This is the only statement depending on t ≥ 0.5. It is false since µx +t is an increasing function for t under
UDD and decreasing under Balducci. Alternatively, compare the corrected version of I to II; the denominator of I is less than or (at t = 0.5, or if qx = 0) equal to the denominator of II when t ≥ 0.5. #
(E)
8.37. First we determine y .
100,000
= 911
x2
Ç
100,000
= 10.477
x=
911
so y must be 10 so that l 10 > 911 and l 11 ≤ 911. Then p 10 =
solve for s .
102
112
= 0.826446 and q10 = 1 − 0.826446 = 0.173554. We
p 10
0.826446
=
p 10 + s q10 0.826446 + 0.173554s
0.826446
− 0.826446 = 0.080740
0.173554s =
0.911
0.080740
s=
= 0.465214
0.173554
0.911 =
Under uniform distribution of deaths,
1−0.465214 q 10.465214
=
(1 − 0.465214)(0.173554)
= 0.100966
1 − (0.465214)(0.173554)
(E)
R1
8.38. The best approach is to use e x :1 = 0 t p x dt . Under UDD, the integral is the area of a trapezoid with
heights 1 and 1 − qx = 0.9 and width 1; the area is F = 0.95. Alternatively, e x :1 = p x + 0.5qx (see Table 8.1) which
is F = 0.9 + 0.5(0.1) = 0.95. Under Balducci,
px
0.9
9
=
=
p x + t qx
0.9 + 0.1t
9+t
Z1
Z1
9dt
G=
= 9(ln 10 − ln 9) = 0.948245
t p x dt =
9+t
0
0
t px
=
1000(F − G ) = 1000(0.95 − 0.948245) = 1.755
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(D)
157
EXERCISE SOLUTIONS FOR LESSON 8
8.39.
I.
II.
III.
(D)
1/3 qx +1/2
1/3 qx =
1/2 qx
=
1
q
3 x
1 − 21 qx
1
q
3 x
1 − 32 qx
=
=
0.04
= 0.042553. !
0.94
0.04
= 0.043478. !
0.92
= 1 − p x1/2 = 1 − 0.881/2 = 0.061917. !
8.40. We’ll calculate 0.5 p 60 and 1.5 p 60 . The former minus the latter is the probability of death between ages 60.5
and 61.5.
Under UDD, half the deaths occur in the first half of the year and half in the second half, so the probability
of surviving half a year is 0.5 p 60 = 1 − 0.5(0.3) = 0.85 and the probability of surviving 1.5 years is
€
Š
1.5 p 60 = p 60 0.5 p 61 = 0.7 1 − 0.5(0.4) = 0.56
Therefore f = 0.85 − 0.56 = 0.29.
Under Balducci, we have
1/2 qx
=
1
q
2 x
1 − 12 qx
so 1/2q60 = 0.15/0.85 and 1/2q61 = 0.2/0.8. Then 1/2 p 60 = 0.7/0.85 = 0.823529 and 3/2 p 60 = (0.7)(0.6/0.8) = 0.525.
The difference is g = 0.823529 − 0.525 = 0.298529. So the answer is
10,000(g − f ) = 10,000(0.298529 − 0.29) = 85.29
(B)
8.41. Since the future lifetimes are independent, we calculate the probability of one hair not surviving and
cube it. The probability of dying in year 1 is 0.1 and in year 2 it is 0.2, which adds up to 0.3. In the third year,
qx +2 =
2| qx
2px
=
Applying Balducci,
3
0.3
=
1 − 0.3 7
0.5qx +2
(0.5)(3/7)
3
=
=
1 − 0.5qx +2 1 − (0.5)(3/7) 11
3
2|0.5 qx = 2 p x 0.5 qx +2 = (0.7)
11
3
= 0.490909
2.5 qx = 0.1 + 0.2 + 0.7
11
0.5 qx +2
=
0.4909093 = 0.118305
(E)
8.42. We calculate the probability of one probe failing and cube it. Probability of failure in the first 2 years is
0.1 + 0.2 = 0.3. In the third year,
0.3 3
2| q 0
q2 =
=
=
p
0.7 7
2 0
By hyperbolic assumption,
0.25 q 2
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=
0.25q2
0.25(3/7)
= 0.157895
=
1 − 0.75q2 1 − 0.75(3/7)
158
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
2|0.25 q 0
2.25 q 0
= 2 p 0 0.25q2 = (0.7)(0.157895) = 0.110526
= 0.1 + 0.2 + 0.110526 = 0.410526
0.4105263 = 0.069187
(E)
8.43. You don’t really need to look up the Illustrative Life Table, except to check that q90 6= 0. (If it equals 0,
(E) would be correct.) Since the hyperbolic assumption kills off lives the earliest (with a decreasing force) and
uniform the latest (with an increasing force), the answer must be (C).
8.44. For the exponential assumption, m x = µx = − ln p x = 0.69315. For hyperbolic,
Z
1
l x l x +1 dt
l
x +1 + t d x
0
Š1
l x l x +1 €
=
ln(l x +1 + t d x ) 0
dx
l x +1
=
(ln l x − ln l x +1 )
qx
l x p x ln p x
=−
qx
qx2
dx
mx =
=−
Lx
p x ln p x

Š
Substituting p = 0.5, we have m x = 0.25 0.5(0.69315) = 0.72135. The difference between exponential m x and
Lx =
hyperbolic m x is 0.72135 − 0.69315 = 0.02820 . (E)
8.45. In order to get 30+s p 50 = 0.5, we need to solve s p 80 = 0.5/0.52. From the two given survival probabilities,
we have p 80 = 0.48/0.52. So
s p 80
p 80
1 − (1 − s )q80
48/52
=
1 − (1 − s )(4/52)
48
=
48 + 4s
=
Setting this equal to 0.5/0.52,
48
50
=
48 + 4s
52
(48)(52)
48 + 4s =
= 49.92
50
1.92
s=
= 0.48
4
Median future lifetime for (50) is 30.48 .
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159
QUIZ SOLUTIONS FOR LESSON 8
Quiz Solutions
8-1. The algebraic method goes: those who die will survive 0.3 on the average, and those who survive will
survive 0.6.
0.6(0.1)
6
=
1 − 0.4(0.1) 96
90
6
=
0.6 p x +0.4 = 1 −
96 96
6
90
55.8
e x +0.4:0.6 =
(0.3) + (0.6) =
= 0.58125
96
96
96
0.6 qx +0.4
=
The geometric method goes: we need the area of a trapezoid having height 1 at x + 0.4 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) = 0.58125 .
8-2. 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 2.125 months (the average of 2 and 2.25). By
linear interpolation, 2.25q0 = 0.25(0.6) + 0.75(0.2) = 0.3. So we have
e 0:2.25 = q0 (0.5) + 1|q0 (1.5) + 2|0.25q0 (2.125) + 2.25 p 0 (2.25)
= (0.05)(0.5) + (0.20 − 0.05)(1.5) + (0.3 − 0.2)(2.125) + 0.70(2.25) = 2.0375
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160
SOA MLC Study Manual—10th edition 2nd printing
Copyright ©2010 ASM
8. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES