contents · 2015. 6. 1. · b.24 solutions to soa exam mlc, spring 2014 . . . . . . . . . . . . . ....

Contents

1 Probability Review 11.1 Functions and moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Probability distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Bernoulli distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.2 Uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.3 Exponential distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Normal approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Conditional probability and expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Conditional variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Survival Distributions: Probability Functions 192.1 Probability notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Actuarial notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Life tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23


3 Survival Distributions: Force of Mortality 37Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 Survival Distributions: Mortality Laws 614.1 Mortality laws that may be used for human mortality . . . . . . . . . . . . . . . . . . . . . 61

4.1.1 Gompertz’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.1.2 Makeham’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.1.3 Weibull Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.2 Mortality laws for exam questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.2.1 Exponential distribution, or constant force of mortality . . . . . . . . . . . . . . . . 664.2.2 Uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.2.3 Beta distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5 Survival Distributions: Moments 795.1 Complete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.1.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.1.2 Special mortality laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.2 Curtate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

6 Survival Distributions: Percentiles and Recursions 1116.1 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

LC Study Manual—1st edition 44th printingCopyright ©2015 ASM

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6.2 Recursive formulas for life expectancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

7 Survival Distributions: Fractional Ages 125Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

8 Insurance: Annual—Moments 1438.1 Review of Financial Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1438.2 Moments of annual insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1448.3 Standard insurances and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1458.4 Illustrative Life Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1478.5 Constant force and uniform mortality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1498.6 Normal approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151


9 Insurance: Continuous—Moments—Part 1 1779.1 De�nitions and general formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1779.2 Constant force of mortality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178


10 Insurance: Continuous—Moments—Part 2 20110.1 Uniform survival function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20110.2 Other mortality functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

10.2.1 Integrating atn e−ct (Gamma Integrands) . . . . . . . . . . . . . . . . . . . . . . . . 20310.3 Variance of endowment insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20510.4 Normal approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206


11 Insurance: Probabilities and Percentiles 22311.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22311.2 Probabilities for Continuous Insurance Variables . . . . . . . . . . . . . . . . . . . . . . . . 22411.3 Probabilities for Discrete Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22611.4 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227


12 Insurance: Recursive Formulas 241Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

13 Annuities: Continuous, Expectation 25113.1 Whole life annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25213.2 Temporary and deferred life annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25413.3 n-year certain-and-life annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257



CONTENTS v

14 Annuities: Discrete, Expectation 27114.1 Annuities-due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27114.2 Annuities-immediate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27514.3 Actuarial Accumulated Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277


15 Annuities: Variance 30115.1 Whole Life and Temporary Life Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30115.2 Other Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30315.3 Typical Exam Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30415.4 Combinations of Annuities and Insurances with No Variance . . . . . . . . . . . . . . . . . 306


16 Annuities: Probabilities and Percentiles 33316.1 Probabilities for continuous annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33316.2 Distribution functions of annuity present values . . . . . . . . . . . . . . . . . . . . . . . . 33616.3 Probabilities for discrete annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33716.4 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338


17 Annuities: Recursive Formulas 353Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

18 Premiums: Net Premiums for Fully Continuous Insurances 36118.1 Future loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36118.2 Net premium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36218.3 Expected value of future loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36418.4 International Actuarial Premium Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366


19 Premiums: Net Premiums from Life Tables 381Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386

20 Premiums: Net Premiums from Formulas 391Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

21 Premiums: Variance of Future Loss, Continuous 411Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

22 Premiums: Variance of Future Loss, Discrete 427Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434


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23 Premiums: Probabilities and Percentiles of Future Loss 44123.1 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

23.1.1 Fully continuous insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44123.1.2 Discrete insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44523.1.3 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

23.2 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452

24 Multiple Lives: Joint Life Probabilities 45924.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45924.2 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460


25 Multiple Lives: Last Survivor Probabilities 471Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480

26 Multiple Lives: Moments 485Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494

27 Multiple Decrement Models: Probabilities 50127.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50127.2 Life Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50327.3 Examples of Multiple Decrement Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . 50527.4 Discrete Insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506


28 Multiple Decrement Models: Forces of Decrement 527Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536

29 Multi-State Models (Markov Chains): Probabilities 545Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561

30 Multi-State Models (Markov Chains): Premiums and Reserves 56930.1 Prototype Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56930.2 Insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56930.3 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57230.4 Premiums and Reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574


Practice Exams 593

1 Practice Exam 1 595


CONTENTS vii










Appendices 655

A Solutions to the Practice Exams 657Solutions for Practice Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 657Solutions for Practice Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662Solutions for Practice Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667Solutions for Practice Exam 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672Solutions for Practice Exam 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677Solutions for Practice Exam 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683Solutions for Practice Exam 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689Solutions for Practice Exam 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694Solutions for Practice Exam 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 700Solutions for Practice Exam 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706

B Solutions to Old Exams 713B.1 Solutions to CAS Exam 3, Spring 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713B.2 Solutions to CAS Exam 3, Fall 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717B.3 Solutions to CAS Exam 3, Spring 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721B.4 Solutions to CAS Exam 3, Fall 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724B.5 Solutions to CAS Exam 3, Spring 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 727B.6 Solutions to CAS Exam 3, Fall 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 731B.7 Solutions to CAS Exam 3L, Spring 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734B.8 Solutions to CAS Exam 3L, Fall 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 737B.9 Solutions to CAS Exam 3L, Spring 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739B.10 Solutions to CAS Exam 3L, Fall 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 742B.11 Solutions to CAS Exam 3L, Spring 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745B.12 Solutions to CAS Exam 3L, Fall 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 748B.13 Solutions to CAS Exam 3L, Spring 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751B.14 Solutions to CAS Exam 3L, Fall 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753B.15 Solutions to CAS Exam 3L, Spring 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 756B.16 Solutions to SOA Exam MLC, Spring 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 759B.17 Solutions to CAS Exam 3L, Fall 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761B.18 Solutions to SOA Exam MLC, Fall 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764


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B.19 Solutions to CAS Exam 3L, Spring 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766B.20 Solutions to SOA Exam MLC, Spring 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 770B.21 Solutions to CAS Exam 3L, Fall 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775B.22 Solutions to SOA Exam MLC, Fall 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779B.23 Solutions to CAS Exam LC, Spring 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782B.24 Solutions to SOA Exam MLC, Spring 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 788B.25 Solutions to CAS Exam LC, Fall 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 790B.26 Solutions to SOA Exam MLC, Fall 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794B.27 Solutions to CAS Exam LC, Spring 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796B.28 Solutions to SOA Exam MLC, Spring 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 800

C Lessons Corresponding to Questions on Released and Practice Exams 803


Lesson 7

Survival Distributions: Fractional Ages

Reading: Models for Quantifying Risk (4th or 5th edition) 6.6.1, 6.6.2

Life tables list mortality rates (qx) or lives (lx) for integral ages only. Often, it is necessary to determinelives at fractional ages (like lx+0.5 for x an integer) or mortality rates for fractions of a year. We need someway to interpolate between ages.

The easiest interpolation method is linear interpolation, or uniform distribution of deaths betweenintegral ages (UDD). This means that the number of lives at age x + s, 0 ≤ s ≤ 1, is a weighted average ofthe number of lives at age x and the number of lives at age x + 1:

lx+s � (1 − s)lx + slx+1 � lx − sdx (7.1)

l100+s

1000

00 1

s

550

The graph of lx+s is a straight line between s � 0 and s � 1 withslope −dx . The graph at the right portrays this for a mortality rateq100 � 0.45 and l100 � 1000.

Contrast UDDwith an assumption of a uniform survival function.If age at death is uniformly distributed, then lx as a function of x is astraight line. If UDD is assumed, lx is a straight line between integralages, but the slope may vary for di�erent ages. Thus if age at death isuniformly distributed, UDD holds at all ages, but not conversely.

Using lx+s , we can compute sqx :

s qx � 1 − s px

� 1 − lx+s

lx� 1 − (1 − sqx ) � sqx (7.2)

That is one of the most important formulas, so let’s state it again:

s qx � sqx (7.2)

More generally, for 0 ≤ s + t ≤ 1,

s qx+t � 1 − s px+t � 1 − lx+s+t

lx+t

� 1 − lx − (s + t)dx

lx − tdx�

sdx

lx − tdx�

sqx

1 − tqx(7.3)

where the last equation was obtained by dividing numerator and denominator by lx . The important pointto 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 , is not sqx , but is in fact higher than sqx . 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 increasingover the year. The numerator of s qx+t is the proportion of the year being measured s times the deathrate, but then this must be divided by 1 minus the proportion of the year that elapsed before the start ofmeasurement.

For most problems involving death probabilities, it will su�ce if you remember that lx+s is linearlyinterpolated. It often helps to create a life table with an arbitrary radix. Try working out the followingexample before looking at the answer.


125

126 7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

Example 7A You are given:• qx � 0.1• Uniform distribution of deaths between integral ages is assumed.

Calculate 1/2qx+1/4.

Answer: Let lx � 1. Then lx+1 � lx (1 − qx ) � 0.9 and dx � 0.1. Linearly interpolating,

lx+1/4 � lx − 14 dx � 1 − 1

4 (0.1) � 0.975lx+3/4 � lx − 3

4 dx � 1 − 34 (0.1) � 0.925

1/2qx+1/4 �lx+1/4 − lx+3/4

lx+1/4�

0.975 − 0.9250.975 � 0.051282

You could also use equation (7.3) to work this example. �

Example 7B 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 squareof 2.25px , the probability of one surviving 2.25 years. If we let lx � 1 and use dx+k � lx k |qx , we get

qx � 0.1(1) � 0.1 lx+1 � 1 − dx � 1 − 0.1 � 0.91|qx � 0.1(2) � 0.2 lx+2 � 0.9 − dx+1 � 0.9 − 0.2 � 0.72|qx � 0.1(3) � 0.3 lx+3 � 0.7 − dx+2 � 0.7 − 0.3 � 0.4

Then linearly interpolating between lx+2 and lx+3, we get

lx+2.25 � 0.7 − 0.25(0.3) � 0.625

2.25px �lx+2.25

lx� 0.625

Squaring, the answer is 0.6252 � 0.390625 . �

µ100+s

s

1

0

0.45

0.450.55

0 1

The probability density function of Tx , spx µx+s , is the constant qx ,the derivative of the conditional cumulative distribution function s qx �

sqx with respect to s. That is another important formula, since the den-sity is needed to compute expected values, so let’s repeat it:

s px µx+s � qx (7.4)

It follows that the force of mortality is qx divided by 1 − sqx :

µx+s �qx

s px�

qx

1 − sqx(7.5)

The force of mortality increases over the year, as illustrated in the graph for q100 � 0.45 to the right.

?Quiz 7-1 You are given:

• µ50.4 � 0.01• Deaths are uniformly distributed between integral ages.

Calculate 0.6q50.4.


7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES 127

Complete Expectation of Life Under UDDIf the complete future lifetime random variable T is written as T � K + S, where K is the curtate futurelifetime and S is the fraction of the last year lived, then K and S are independent. This is usually not trueif uniform distribution of deaths is not assumed. Since S is uniform on [0, 1), E[S] � 1

2 and Var(S) � 112 . It

follows from E[S] � 12 that

ex � ex + 12 (7.6)

More common on exams are questions asking you to evaluate the temporary complete expectancy oflife under UDD. You can always evaluate the temporary complete expectancy, whether or not UDD isassumed, by integrating t px , as indicated by formula (5.6) on page 80. For UDD, t px is linear betweenintegral ages. Therefore, a rule we learned in Lesson 5 applies for all integral x:

ex:1 � px + 0.5qx (5.13)

This equation will be useful. In addition, the method for generating this equation can be used to workout questions involving temporary complete life expectancies for short periods. The following exampleillustrates this. This example will be reminiscent of calculating temporary complete life expectancy foruniform mortality.Example 7C You are given

• qx � 0.1.• Deaths are uniformly distributed between integral ages.

Calculate ex:0.4 .

Answer: We will discuss two ways to solve this: an algebraic method and a geometric method.The algebraic method is based on the double expectation theorem, equation (1.11). It uses the fact that

for a uniform distribution, the mean is the midpoint. If deaths occur uniformly between integral ages, thenthose 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 anaverage of 0.4 of course. The temporary life expectancy is the weighted average of these two groups, or0.4qx (0.2) + 0.4px (0.4). This is:

0.4qx � (0.4)(0.1) � 0.040.4px � 1 − 0.04 � 0.96

ex:0.4 � 0.04(0.2) + 0.96(0.4) � 0.392

An equivalent geometric method, the trapezoidal rule, is to draw the t px function from 0 to 0.4. Theintegral of t px is the area under the line, which is the area of a trapezoid: the average of the heights timesthe width. The following is the graph (not drawn to scale):

A B

(0.4, 0.96)(1.0, 0.9)

0 0.4 1.0

1

t px

t

Trapezoid A is the area we are interested in. Its area is 12 (1 + 0.96)(0.4) � 0.392 . �



?Quiz 7-2 As in Example 7C, you are given

• qx � 0.1.• Deaths are uniformly distributed between integral ages.

Calculate ex+0.4:0.6 .

Let’s now work out an example in which the duration crosses an integral boundary.Example 7D You are given:

• qx � 0.1• qx+1 � 0.2• Deaths are uniformly distributed between integral ages.

Calculate ex+0.5:1 .

Answer: Let’s start with the algebraic method. Since the mortality rate changes at x+1, we must split thegroup into those who die before x + 1, those who die afterwards, and those who survive. Those who diebefore x + 1 live 0.25 on the average since the period to x + 1 is length 0.5. Those who die after x + 1 livebetween 0.5 and 1 years; the midpoint of 0.5 and 1 is 0.75, so they live 0.75 years on the average. Thosewho survive live 1 year.

Now let’s calculate the probabilities.

0.5qx+0.5 �0.5(0.1)

1 − 0.5(0.1)�

595

0.5px+0.5 � 1 − 595 �

9095

0.5|0.5qx+0.5 �

(9095

) (0.5(0.2)

)�

995

1px+0.5 � 1 − 595 −

995 �

8195

These probabilities could also be calculated by setting up an lx table with radix 100 at age x and interpo-lating within it to get lx+0.5 and lx+1.5. Then

lx+1 � 0.9lx � 90lx+2 � 0.8lx+1 � 72

lx+0.5 � 0.5(90 + 100) � 95lx+1.5 � 0.5(72 + 90) � 81

0.5qx+0.5 � 1 − 9095 �

595

0.5|0.5qx+0.5 �90 − 81

95 �995

1px+0.5 �lx+1.5lx+0.5

�8195

Either way, we’re now ready to calculate ex+0.5:1 .

ex+0.5:1 �5(0.25) + 9(0.75) + 81(1)

95 �8995

For the geometric method we draw the following graph:


7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES 129

A B

�0.5, 90

95

��1.0, 81

95

�

0x +0.5

0.5x +1

1.0x +1.5

1

t px+0.5

t

The heights at x + 1 and x + 1.5 are as we computed above. Then we compute each area separately. Thearea of A is 1

2

(1 + 90

95

)(0.5) �

18595(4) . The area of B is 1

2

(9095 +

8195

)(0.5) �

17195(4) . Adding them up, we get

185+17195(4) �

8995 . �

?Quiz 7-3 The probability that a battery fails by the end of the kth month is given in the following table:

kProbability of battery failure by

the end of month k1 0.052 0.203 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 ex:n in terms of ex:n when x and n are both integers, note that those who survive n yearscontribute the same to both. Those who die contribute an average of 1

2 more to ex:n since they die on theaverage in the middle of the year. Thus the di�erence is 1

2 n qx :

ex:n � ex:n + 0.5n qx (7.7)

Example 7E You are given:• qx � 0.01 for x � 50, 51, . . . , 59.• Deaths are uniformly distributed between integral ages.

Calculate e50:10 .

Answer: As we just said, e50:10 � e50:10 + 0.510q50. The �rst summand, e50:10 , is the sum of k p50 � 0.99k

for k � 1, . . . , 10. This sum is a geometric series:

e50:10 �

10∑

k�10.99k

�0.99 − 0.9911

1 − 0.99 � 9.46617

The second summand, the probability of dying within 10 years is 10q50 � 1− 0.9910 � 0.095618. Therefore

e50:10 � 9.46617 + 0.5(0.095618) � 9.51398 �

The formulas are summarized in Table 7.1.



Table 7.1: Summary of formulas for fractional ages

lx+s � lx − sdx

sqx � sqx

spx � 1 − sqx

sqx+t �sqx

1 − tqx

µx+s �qx

1 − sqx

spx µx+s � qx

ex � ex + 0.5ex:n � ex:n + 0.5 n qx

ex:1 � px + 0.5qx

Exercises

7.1. [CAS4-S85:16] (1 point) Deaths are uniformly distributed between integral ages.Which of the following represents 3/4px + 1

2 1/2px µx+1/2?

A. 3/4px B. 3/4qx C. 1/2px D. 1/2qx E. 1/4px

7.2. [Based on 150-S88:25] You are given:

• 0.25qx+0.75 � 3/31.• Mortality is uniformly distributed within age x.

Calculate qx .Use the following information for questions 7.3 and 7.4:

You are given:• Deaths are uniformly distributed between integral ages.• qx � 0.10.• qx+1 � 0.15.

7.3. Calculate 1/2qx+3/4.

7.4. Calculate 0.3|0.5qx+0.4.

7.5. You are given:

• Deaths are uniformly distributed between integral ages.• Mortality follows the Illustrative Life Table.

Calculate the median future lifetime for (45.5).


Exercises continue on the next page . . .

EXERCISES FOR LESSON 7 131

7.6. [160-F90:5] You are given:

• A survival distribution is de�ned by

lx � 1000(1 −

( x100

)2), 0 ≤ x ≤ 100.

• µx denotes the actual force of mortality for the survival distribution.• µL

x denotes the approximation of the force ofmortality based on the uniformdistribution of deathsassumption for lx , 50 ≤ x < 51.

Calculate µ50.25 − µL50.25.

A. −0.00016 B. −0.00007 C. 0 D. 0.00007 E. 0.00016

7.7. A survival distribution is de�ned by

• S0(k) � 1/(1 + 0.01k)4 for k a non-negative integer.• Deaths are uniformly distributed between integral ages.

Calculate 0.4q20.2.

7.8. [Based on150-S89:15] You are given:

• Deaths are uniformly distributed over each year of age.• x lx

35 10036 9937 9638 9239 87

Which of the following are true?

I. 1|2q36 � 0.091II. µ37.5 � 0.043III. 0.33q38.5 � 0.021

A. I and II only B. I and III only C. II and III only D. I, II and IIIE. The correct answer is not given by A. , B. , C. , or D.




7.9. [150-82-94:5] You are given:

• Deaths are uniformly distributed over each year of age.• 0.75px � 0.25.

Which of the following are true?

I. 0.25qx+0.5 � 0.5II. 0.5qx � 0.5III. µx+0.5 � 0.5

A. I and II only B. I and III only C. II and III only D. I, II and IIIE. The correct answer is not given by A. , B. , C. , or D.

7.10. [3-S00:12] For a certain mortality table, you are given:

• µ80.5 � 0.0202• µ81.5 � 0.0408• µ82.5 � 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

7.11. You are given:

• Deaths are uniformly distributed between integral ages.• qx � 0.1.• qx+1 � 0.3.

Calculate ex+0.7:1 .


• Deaths are uniformly distributed between integral ages.• q45 � 0.01.• q46 � 0.011.

Calculate Var(min

(T45 , 2

)).


• Deaths are uniformly distributed between integral ages.• 10px � 0.2.

Calculate ex:10 − ex:10 .



EXERCISES FOR LESSON 7 133

7.14. [4-F86:21] You are given:

• q60 � 0.020• q61 � 0.022• Deaths are uniformly distributed over each year of age.

Calculate e60:1.5 .

A. 1.447 B. 1.457 C. 1.467 D. 1.477 E. 1.487

7.15. [150-F89:21] You are given:

• q70 � 0.040• q71 � 0.044• Deaths are uniformly distributed over each year of age.

Calculate e70:1.5 .

A. 1.435 B. 1.445 C. 1.455 D. 1.465 E. 1.475

7.16. [3-S01:33] For a 4-year college, you are given the following probabilities for dropout from all causes:

q0 � 0.15q1 � 0.10q2 � 0.05q3 � 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, e1:1.5 .

A. 1.25 B. 1.30 C. 1.35 D. 1.40 E. 1.45


• Deaths are uniformly distributed between integral ages.• ex+0.5:0.5 � 5/12.

Calculate qx .


• Deaths are uniformly distributed over each year of age.• e55.2:0.4 � 0.396.

Calculate µ55.2.

7.19. [150-S87:21] You are given:

• dx � k for x � 0, 1, 2, . . . , ω − 1• e20:20 � 18• Deaths are uniformly distributed over each year of age.

Calculate 30|10q30.

A. 0.111 B. 0.125 C. 0.143 D. 0.167 E. 0.200




7.20. [150-S89:24] You are given:

• Deaths are uniformly distributed over each year of age.• µ45.5 � 0.5

Calculate e45:1 .

A. 0.4 B. 0.5 C. 0.6 D. 0.7 E. 0.8

7.21. [CAS3-S04:10] 4,000 people age (30) each pay an amount, P, into a fund. Immediately after the1,000th 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 515B. At least 515, but less than 525C. At least 525, but less than 535D. At least 535, but less than 545E. At least 545

Additional old CAS Exam 3/3L questions: S05:31, F05:13, S06:13, F06:13, S07:24, S08:16, S09:3, F09:3,S10:4, F10:3, S11:3, S12:3, F12:3, S13:3, F13:3Additional old CAS Exam LC questions: S14:4, F14:4, S15:3

Solutions

7.1. In the second summand, 1/2px µx+1/2 is the density function, which is the constant qx under UDD.The �rst summand 3/4px � 1 − 3

4 qx . So the sum is 1 − 14 qx , or 1/4px . (E)

7.2. Using equation (7.3),

331 � 0.25qx+0.75 �

0.25qx

1 − 0.75qx

331 −

2.2531 qx � 0.25qx

331 �

1031 qx

qx � 0.3

7.3. Wecalculate 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 for0.25 years is, by equation (7.3),

1/4px+3/4 �1 − 0.10

1 − 0.75(0.10)�

0.90.925

The probability of (x + 1) surviving to x + 1.25 is

1/4px+1 � 1 − 0.25(0.15) � 0.9625


EXERCISE SOLUTIONS FOR LESSON 7 135

The answer to the question is then the complement of the product of these two numbers:

1/2qx+3/4 � 1 − 1/2px+3/4 � 1 − 1/4px+3/4 1/4px+1 � 1 −(0.90.925

)(0.9625) � 0.06351

Alternatively, you could build a life table starting at age x, with lx � 1. Then lx+1 � (1− 0.1) � 0.9 andlx+2 � 0.9(1 − 0.15) � 0.765. Under UDD, lx at fractional ages is obtained by linear interpolation, so

lx+0.75 � 0.75(0.9) + 0.25(1) � 0.925lx+1.25 � 0.25(0.765) + 0.75(0.9) � 0.86625

1/2p3/4 �lx+1.25lx+0.75

�0.866250.925 � 0.93649

1/2q3/4 � 1 − 1/2p3/4 � 1 − 0.93649 � 0.06351

7.4. 0.3|0.5qx+0.4 is 0.3px+0.4 − 0.8px+0.4. The �rst summand is

0.3px+0.4 �1 − 0.7qx

1 − 0.4qx�

1 − 0.071 − 0.04 �

9396

The probability that (x + 0.4) survives to x + 1 is, by equation (7.3),

0.6px+0.4 �1 − 0.101 − 0.04 �

9096

and the probability (x + 1) survives to x + 1.2 is

0.2px+1 � 1 − 0.2qx+1 � 1 − 0.2(0.15) � 0.97

So

0.3|0.5qx+0.4 �9396 −

(9096

)(0.97) � 0.059375

Alternatively, you could use the life table from the solution to the last question, and linearly interpo-late:

lx+0.4 � 0.4(0.9) + 0.6(1) � 0.96lx+0.7 � 0.7(0.9) + 0.3(1) � 0.93lx+1.2 � 0.2(0.765) + 0.8(0.9) � 0.873

0.3|0.5qx+0.4 �0.93 − 0.873

0.96 � 0.059375

7.5. Under uniform distribution of deaths between integral ages, lx+0.5 �12 (lx + lx+1), since the sur-

vival function is a straight line between two integral ages. Therefore, l45.5 �12 (9,164,051 + 9,127,426) �

9,145,738.5. Median future lifetime occurs when lx �12 (9,145,738.5) � 4,572,869. This happens between

ages 77 and 78. We interpolate between the ages to get the exact median:

l77 − s(l77 − l78) � 4,572,8694,828,182 − 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

297,822 �255,313297,822 � 0.8573

So the median age at death is 77.8573, and median future lifetime is 77.8573 − 45.5 � 32.3573 .



7.6. x p0 �lxl0� 1 − ( x

100)2. The force of mortality is calculated as the negative derivative of ln x p0:

µx � −d ln x p0

dx�

2( x100

) ( 1100

)

1 − ( x100

)2 �2x

1002 − x2

µ50.25 �100.5

1002 − 50.252� 0.0134449

For UDD, we need to calculate q50.

p50 �l51l50

�1 − 0.512

1 − 0.502� 0.986533

q50 � 1 − 0.986533 � 0.013467

so under UDD,µL50.25 �

q501 − 0.25q50

�0.013467

1 − 0.25(0.013467)� 0.013512.

The di�erence between µ50.25 and µL50.25 is 0.013445 − 0.013512 � −0.000067 . (B)

7.7. S0(20) � 1/1.24 and S0(21) � 1/1.214, so q20 � 1 − (1.2/1.21)4 � 0.03265. Then

0.4q20.2 �0.4q20

1 − 0.2q20�

0.4(0.03265)1 − 0.2(0.03265)

� 0.01315

7.8.I. Calculate 1|2q36.

1|2q36 �2d37l36

�96 − 87

99 � 0.09091 !

This statement does not require uniform distribution of deaths.II. By equation (7.5),

µ37.5 �q37

1 − 0.5q37�

4/961 − 2/96 �

494 � 0.042553 !

III. Calculate 0.33q38.5.

0.33q38.5 �0.33d38.5

l38.5�

(0.33)(5)89.5 � 0.018436 #

I can’t �gure out what mistake you’d have to make to get 0.021. (A)

7.9. First calculate qx .

1 − 0.75qx � 0.25qx � 1

Then by equation (7.3), 0.25qx+0.5 � 0.25/(1 − 0.5) � 0.5, making I true.By equation (7.2), 0.5qx � 0.5qx � 0.5, making II true.By equation (7.5), µx+0.5 � 1/(1 − 0.5) � 2, making III false. (A)



7.10. We use equation (7.5) to back out qx for each age.

µx+0.5 �qx

1 − 0.5qx⇒ qx �

µx+0.5

1 + 0.5µx+0.5

q80 �0.02021.0101 � 0.02

q81 �0.04081.0204 � 0.04

q82 �0.06191.03095 � 0.06

Then by equation (7.3), 0.5p80.5 � 0.98/0.99. p81 � 0.96, and 0.5p82 � 1 − 0.5(0.06) � 0.97. Therefore

2q80.5 � 1 −(0.980.99

)(0.96)(0.97) � 0.0782 (A)

7.11. To do this algebraically, we split the group into those who die within 0.3 years, those who diebetween 0.3 and 1 years, and those who survive one year. Under UDD, those who die will die at themidpoint of the interval (assuming the interval doesn’t cross an integral age), so we have

Survival Probability AverageGroup time of group survival time

I (0, 0.3] 1 − 0.3px+0.7 0.15II (0.3, 1] 0.3px+0.7 − 1px+0.7 0.65III (1,∞) 1px+0.7 1

We calculate the required probabilities.

0.3px+0.7 �0.90.93 � 0.967742

1px+0.7 �0.90.93

(1 − 0.7(0.3)

)� 0.764516

1 − 0.3px+0.7 � 1 − 0.967742 � 0.0322580.3px+0.7 − 1px+0.7 � 0.967742 − 0.764516 � 0.203226

ex+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 ofthe �rst 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

ex+0.7:1 � (0.3)(0.5)(1 + 0.967742) + (0.7)(0.5)(0.967742 + 0.764516)

� 0.295161 + 0.606290 � 0.901451

7.12. For the expected value, we’ll use the recursive formula. (The trapezoidal rule could also be used.)

e45:2 � e45:1 + p45 e46:1� (1 − 0.005) + 0.99(1 − 0.0055)� 1.979555

We’ll use equation (5.7)to calculate the second moment.

E[min(T45 , 2)2] � 2∫ 2

0t t px dt



� 2(∫ 1

0t(1 − 0.01t) dt +

∫ 2

1t(0.99)

(1 − 0.011(t − 1)

)dt

)

� 2 *.,12 − 0.01

(13

)+ 0.99 *.,

(1.011)(22 − 12)2 − 0.011

(23 − 13

3

)+/-+/-

� 2(0.496667 + 1.475925) � 3.94518

So the variance is 3.94518 − 1.9795552 � 0.02654 .7.13. As discussed on page 129, by equation (7.7), the di�erence is

12 10qx �

12 (1 − 0.2) � 0.4

7.14. Those who die in the �rst year survive 12 year on the average and those who die in the �rst half of

the second year survive 1.25 years on the average, so we have

p60 � 0.98

1.5p60 � 0.98(1 − 0.5(0.022)

)� 0.96922

e60:1.5 � 0.5(0.02) + 1.25(0.98 − 0.96922) + 1.5(0.96922) � 1.477305 (D)

Alternatively, we use the trapezoidal method. The �rst trapezoid has heights 1 and p60 � 0.98 andwidth 1. The second trapezoid has heights p60 � 0.98 and 1.5p60 � 0.96922 and width 1/2.

e60:1.5 �12 (1 + 0.98) +

(12

) (12

)(0.98 + 0.96922)

� 1.477305 (D)

7.15. p70 � 1−0.040 � 0.96, 2p70 � (0.96)(0.956) � 0.91776, and by linear interpolation, 1.5p70 � 0.5(0.96+0.91776) � 0.93888. Those who die in the �rst year survive 0.5 years on the average and those who die inthe �rst half of the second year survive 1.25 years on the average. So

e70:1.5 � 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 �rst year’s trapezoid has heights 1 and 0.96and width 1 and the second year’s trapezoid has heights 0.96 and 0.93888 and width 1/2, so

e70:1.5 � 0.5(1 + 0.96) + 0.5(0.5)(0.96 + 0.93888) � 1.45472 (C)

7.16. First we calculate t p1 for t � 1, 2.

p1 � 1 − q1 � 0.902p1 � (1 − q1)(1 − q2) � (0.90)(0.95) � 0.855

By linear interpolation, 1.5p1 � (0.5)(0.9 + 0.855) � 0.8775.The algebraic method splits the students into three groups: �rst 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 �rst group, 1.25 years for the second group) and survivors survive 1.5 years. Therefore

e1:1.5 � 0.10(0.5) + (0.90 − 0.8775)(1.25) + 0.8775(1.5) � 1.394375 (D)



0 1 1.5 2

1

t

t p1

(1, 0.9) (1.5,0.8775)

Alternatively, we could sum the two trapezoids making up theshaded area at the right.

e1:1.5 � (1)(0.5)(1 + 0.9) + (0.5)(0.5)(0.90 + 0.8775)

� 0.95 + 0.444375 � 1.394375 (D)

7.17. 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 0.5px+0.5 �512

0.25(

0.5qx

1 − 0.5qx

)+ 0.5

(1 − qx

1 − 0.5qx

)�

512

0.125qx + 0.5 − 0.5qx �512 − 5

24 qx

12 −

512 �

(− 524 + 1

2 −18

)qx

112 �

qx

6qx �

12

5

12

0 0.5t

t px+0.5

1

0.5px+0.5

Alternatively, complete life expectancy is the area of the trape-zoid shown on the right, so

512 � 0.5(0.5)(1 + 0.5px+0.5)

Then 0.5px+0.5 �23 , from which it follows

23 �

1 − qx

1 − 12 qx

qx �12

7.18. Survivors live 0.4 years and those who die live 0.2 years on the average, so

0.396 � 0.40.4p55.2 + 0.20.4q55.2

Using the formula 0.4q55.2 � 0.4q55/(1 − 0.2q55) (equation (7.3)), we have

0.4(1 − 0.6q551 − 0.2q55

)+ 0.2

(0.4q55

1 − 0.2q55

)� 0.396

0.4 − 0.24q55 + 0.08q55 � 0.396 − 0.0792q550.0808q55 � 0.004

q55 �0.0040.0808 � 0.0495

µ55.2 �q55

1 − 0.2q55�

0.04951 − 0.2(0.0495)

� 0.05



7.19. Since dx is constant for all x and deaths are uniformly distributedwithin each year of age, mortalityis uniform globally. We back out ω using equation (5.12), ex:n � n px (n) + n qx (n/2):

10 20q20 + 20 20p20 � 18

10(

20ω − 20

)+ 20

(ω − 40ω − 20

)� 18

200 + 20ω − 800 � 18ω − 3602ω � 240ω � 120

18

20 40x

x−20p20

1ω−40

ω−20

Alternatively, we can back out ω using the trapezoidal rule. Com-plete life expectancy is the area of the trapezoid shown to the right.

e20:20 � 18 � (20)(0.5)(1 + ω − 40

ω − 20

)

0.8 �ω − 40ω − 20

0.8ω − 16 � ω − 400.2ω � 24ω � 120

Once we have ω, we compute

30|10q30 �10

ω − 30 �1090 � 0.1111 (A)

7.20. We use equation (7.5) to obtain

0.5 �qx

1 − 0.5qx

qx � 0.4

Then e45:1 � 0.5(1 + (1 − 0.4)

)� 0.8 . (E)

7.21. According to the Illustrative Life Table, l30 � 9,501,381, sowe are looking for the age x such that lx �

0.75(9,501,381) � 7,126,036. This is between 67 and 68. Using linear interpolation, since l67 � 7,201,635and l68 � 7,018,432, we have

x � 67 + 7,201,635 − 7,126,0367,201,635 − 7,018,432 � 67.4127

This is 37.4127 years into the future. 34 of the people collect 50,000. We need 50,000

(34

) (1

1.1237.4127

)�

540.32 per person. (D)

Quiz Solutions

7-1. Notice that µ50.4 �q50

1−0.4q50 while 0.6q50.4 �0.6q50

1−0.4q50 , so 0.6q50.4 � 0.6(0.01) � 0.006


QUIZ SOLUTIONS FOR LESSON 7 141

7-2. The algebraic method goes: those who die will survive 0.3 on the average, and those who survivewill survive 0.6.

0.6qx+0.4 �0.6(0.1)

1 − 0.4(0.1)�

696

0.6px+0.4 � 1 − 696 �

9096

ex+0.4:0.6 �696 (0.3) + 90

96 (0.6) �55.896 � 0.58125

The geometric method goes: we need the area of a trapezoid having height 1 at x + 0.4 and height90/96 at x + 1, where 90/96 is 0.6px+0.4, as calculated above. The width of the trapezoid is 0.6. The answeris therefore 0.5 (1 + 90/96) (0.6) � 0.58125 .7-3. Batteries failing in month 1 survive an average of 0.5 month, those failing in month 2 survive anaverage of 1.5 months, and those failing in month 3 survive an average of 2.125 months (the average of 2and 2.25). By linear interpolation, 2.25q0 � 0.25(0.6) + 0.75(0.2) � 0.3. So we have

e0:2.25 � q0(0.5) + 1|q0(1.5) + 2|0.25q0(2.125) + 2.25p0(2.25)

� (0.05)(0.5) + (0.20 − 0.05)(1.5) + (0.3 − 0.2)(2.125) + 0.70(2.25) � 2.0375


Practice Exam 1

1. You are given:

• The following life table.

x lx dx

50 1000 205152 3553 37

• 2q52 � 0.07508.

Determine d51.A. Less than 20B. At least 20, but less than 22C. At least 22, but less than 24D. At least 24, but less than 26E. At least 26

2. Mortality follows Gompertz’s law with B � 0.002 and c � 1.03.Calculate the 40th percentile of future lifetime for (45).A. Less than 35.0B. At least 35.0, but less than 40.0C. At least 40.0, but less than 45.0D. At least 45.0, but less than 50.0E. At least 50.0

3. You are given:

• lx � 100(120 − x) for x ≤ 90• µx � µ for x > 90• e80 has the same value that it would have if lx � 100(120 − x) for x ≤ 120.

Determine µ.

A. 0.04 B. 0.05 C. 0.067 D. 0.075 E. 0.083


595 Exercises continue on the next page . . .

596 PRACTICE EXAMS

4. Three independent probes are sent to Mars. The probability distribution for the amount of timeeach one will function is given in the following table.

Years Probability of functioning(n) n years1 0.92 0.83 0.5

The time to failure is assumed to be uniformly distributed between integral numbers of years.Calculate the probability that at least two of the probes function for at least 2 1

3 years.A. Less than 0.40B. At least 0.40, but less than 0.55C. At least 0.55, but less than 0.70D. At least 0.70, but less than 0.85E. At least 0.85

5. For three independent lives (x), (y), and (z), you are given

• µx+t � 0.01, t > 0• µy+t � 0.02, t > 0• µz+t � 0.03, t > 0

Calculate the probability that at least one life survives 30 years.

A. 0.78 B. 0.83 C. 0.88 D. 0.93 E. 0.98

6. For two independent lives ages 30 and 40:

• µ30(t) � 0.02• µ40(t) � 1

80−t , t < 80

Calculate e30:40.A. Less than 24B. At least 24, but less than 26C. At least 26, but less than 28D. At least 28, but less than 30E. At least 30



PRACTICE EXAM 1 597

7. A discrete Markov chain model for insurance on two lives (x) and (y) has the following states:0 Both alive1 Only (x) alive2 Only (y) alive3 Neither alive

Let pi j be the (i , j) entry in the transition matrix.Which of the following is true if the two lives are independent but may not be true if the lives are

dependent?

I. p01 � p23

II. p12 � p21

III. p03 � 0

A. I and II only B. I and III only C. II and III only D. I, II, and IIIE. The correct answer is not given by A. , B. , C. , or D.

8. For an insurance policy on (x), decrements occur due to death (1), withdrawal (2), and conver-sion (3). Decrement rates are as follows:

t q (1)x+t−1 q (2)

x+t−1 q (3)x+t−1

1 0.01 0.09 0.012 0.01 0.06 0.013 0.01 0.05 0.04

Determine the probability that conversion will occur between 1 and 3 years from now.A. Less than 0.042B. At least 0.042, but less than 0.047C. At least 0.047, but less than 0.052D. At least 0.052, but less than 0.057E. At least 0.057

9. Persistency of an insurance contract is a�ected by two decrements: death (1) and withdrawal (2).The force of mortality is µ(1) � 0.01. The force of withdrawal is µ(2) � 0.05.Determine the expected amount of time until withdrawal for those policyholderswho eventuallywith-

draw.A. Less than 15B. At least 15, but less than 17C. At least 17, but less than 19D. At least 19, but less than 21E. At least 21



598 PRACTICE EXAMS

10. You are given:

• Z1 is the present value random variable for a 10-year term insurance paying 1 at the moment ofdeath of (45).

• Z2 is the present value random variable for a 20-year deferred whole life insurance paying 1 atthe moment of death of (45).

• µ � 0.02• δ � 0.04

Calculate Cov(Z1 , Z2).A. Less than −0.04B. At least −0.04, but less than −0.03C. At least −0.03, but less than −0.02D. At least −0.02, but less than −0.01E. At least −0.0111. For a 30-year deferred whole life annuity on (35):

• The annuity will pay 100 per year at the beginning of each year, starting at age 65.• If death occurs during the deferral period, the contract will pay 1000 at the end of the year of

death.• Mortality follows the Illustrative Life Table.• i � 0.06.• Y is the present value random variable for the contract.

Calculate E[Y].A. Less than 204B. At least 204, but less than 205C. At least 205, but less than 206D. At least 206, but less than 207E. At least 207

12. For a special fully discrete life insurance on (45),

• The bene�t is 1000 if death occurs before age 65, 500 otherwise.• Premiums are payable at the beginning of the �rst 20 years only.• i � 0.05• A45 � 0.2• 20E45 � 0.3• a45:20 � 12.6

Determine the annual net premium.A. Less than 11.50B. At least 11.50, but less than 12.00C. At least 12.00, but less than 12.50D. At least 12.50, but less than 13.00E. At least 13.00



PRACTICE EXAM 1 599

13. A copier is either working or out-of-order.If the copier is working at the beginning of a day, the probability that it is working at the beginning of

the next day is 0.8.If the copier is out-of-order at the beginning of a day, the probability that it is �xed during the day and

working at the beginning of the next day is 0.6.You are given that the probability that the copier is working at the beginning of Tuesday is 0.77.Determine the probability that the copier was working at the beginning of the previous day, Monday.A. Less than 0.75B. At least 0.75, but less than 0.78C. At least 0.78, but less than 0.81D. At least 0.81, but less than 0.84E. At least 0.84

14. You are given the following transition matrices for a Markov chain with three states numbered 1,2, and 3.

Qn �

*.,1 − 0.2n 0.1n 0.1n

0.2 0.6 0.20 0 1

+/- 0 ≤ n < 5

*.,0 0 10 0 10 0 1

+/- n ≥ 5

n is time measured in years. Each transition is made at the beginning of the year.Calculate the expected amount of time in State 1 (not necessarily continuously) for someone starting

out in State 1 at time n � 0.A. Less than 2.6B. At least 2.6, but less than 2.8C. At least 2.8, but less than 3.0D. At least 3.0, but less than 3.2E. At least 3.2



600 PRACTICE EXAMS

15. Life insurance policies include a disability waiver rider that waives premiums while the insured isdisabled.

Initially, an insured (x) is active. At the beginning of each subsequent year, an insured may either bein one of three states:

State 1 = activeState 2 = disabledState 3 = dead

The transition probability matrix is

Q �*.,0.80 0.15 0.050.50 0.40 0.100 0 1

+/-A 4-year term insurance policy pays 1000 at the end of the year of death. Net premiums of π are paid

at the beginning of each year if the insured is in the active state. Net premiums are determined usingi � 0.05.

Determine π.A. Less than 59B. At least 59, but less than 61C. At least 61, but less than 63D. At least 63, but less than 65E. At least 65

Solutions to the above questions begin on page 657.


Appendix A. Solutions to the Practice Exams

Answer Key for Practice Exam 11 B 6 B 11 C2 B 7 E 12 B3 C 8 A 13 E4 D 9 B 14 B5 D 10 D 15 C

Practice Exam 1

1. [Lesson 2] 0.07508 � 2q52 � (d52 + d53)/l52 � 72/l52, so l52 � 72/0.07508 � 959. But l52 � l50 − d50 −d51 � 1000 − 20 − d51, so d51 � 21 . (B)

2. [Lesson 4] Gompertz’s law is µx � Bcx � 0.002(1.03x ). We have

t p45 � exp(−

∫ 45+t

450.002(1.03u )du

)

� exp(−0.002(1.0345+t − 1.0345)

ln 1.03

)

andwewant to set this equal to 0.6, since the 40th percentile of future lifetime corresponds to a 40% chanceof dying before that time, or a 60% probability of survival to that time. We log the expression and set itequal to ln 0.6.

−0.002

(1.0345+t − 1.0345

)

ln 1.03 � ln 0.6

1.0345+t� − (ln 0.6)(ln 1.03)

0.002 + 1.0345 � 11.33129

45 + t �ln 11.33129

ln 1.03 � 82.12674

t � 37.12674 (B)

3. [Lessons 5 and 6] By the recursive formula (6.1)

e80 � e80:10 + 10p80 e90

Sincemortality for the �rst ten years is unchanged, e80:10 and 10p80 are unchanged. We only need to equatee90 between the two mortality assumptions. Under uniform mortality through age 120, e90 � 15, whereasunder constant force mortality, e90 � 1/µ. We conclude that 1/µ � 15, or µ �

115 � 0.066667 . (C)


657

658 PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 4–8

4. [Lesson 7] The probability of one probe surviving 2 13 years is linearly interpolated between 0.8

and 0.5, 0.8− (13)(0.8−0.5) � 0.7. The probability of at least two surviving is the probability of exactly two

surviving (and there are three ways to choose 2 out of 3) or three surviving, or

3(0.72)(0.3) + 0.73 � 0.784 (D)

5. [Lesson 25] The probability that all three fail within 10 years is

30qx 30q y 30qz � (1 − e−0.3)(1 − e−0.6)(1 − e−0.9) � 0.06940

so the probability that at least one survives is 1 − 0.06940 � 0.93060 . (D)

6. [Lesson 26]

e30:40 �∫ ∞

0tpx y dt

�

∫ 80

0

(80 − t80

)e−0.02t dt

�180

(80 − t−0.02 e−0.02t ��

80

0− 1

0.02

∫ 80

0e−0.02tdt

)

�180

( 800.02 −

500.02 (1 − e−0.02(80))

)

� 25.0592 (B)

7. [Lesson 25 and 29] The transition matrix with two independent lives looks like this:

*...,

p00 p01 p02 00 p11 0 p020 0 p22 p010 0 0 1

+///-The probability of death for each life must be the same before and after the death of the other, so p01, theprobability of the death of (y), must also be p23, and p02, the probability of the death of (x), must also bep13. Thus I is true. While p12 � p21 � 0, this is true even if the lives are not independent, so II is false. III isfalse, since p03 is not necessarily 0 regardless of whether the two lives are dependent or independent. (E)

8. [Lesson 27] We need the probability of decrement (3) in years 2 and 3. The table gives conditionalprobabilities of decrement (3) given survival to the beginning of each year, so we need probabilities ofsurvival to the beginning of years 2 and 3, tp

00x , t � 1, 2.

p (τ)x � 1 − q (1)

x − q (2)x − q (3)

x � 1 − 0.01 − 0.09 − 0.01 � 0.89

2p(τ)x � 0.89

(1 − q (1)

x+1 − q (2)x+1 − q (3)

x+1

)� 0.89(1 − 0.01 − 0.06 − 0.01) � 0.8188

Now we calculate the probabilities of decrement (3) in years 2 and 3.

1|q(3)x � (0.89)(0.01) � 0.0089

2|q(3)x � (0.8188)(0.04) � 0.032752

1|2q(3)x � 0.0089 + 0.032752 � 0.041652 (A)


PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 9–13 659

9. [Lesson 28] The expected amount of time to withdrawal is the integral∫ ∞

0t tp

(τ)x µ(2)

x+tdt �∫ ∞

00.05te−0.06tdt

and this is 5/6 of∫ ∞0 0.06te−0.06tdt, the expected amount of time to the next decrement, which we know

is 1/0.06 since it is exponential with mean 1/0.06. The probability that a decrement will be a withdrawalis the ratio of the µ’s, or 0.05/(0.05 + 0.01) � 5/6. So the expected amount of time to the next decrementgiven that it is a withdrawal is

56 (1/0.06)

5/6 � 16 23 (B)

10. [Lesson 10] Since E[Z1Z2] � 0, the covariance is −E[Z1]E[Z2].

E[Z1] �∫ 10

0e−0.06t0.02 dt �

1 − e−0.6

3 � 0.150396

E[Z2] �∫ ∞

20e−0.06t0.02 dt �

e−1.2

3 � 0.100398

Cov(Z1 , Z2) � −(0.150396)(0.100398) � −0.015099 (D)

11. [Lesson 14] Let Y1 be the value of the bene�ts if death occurs after age 65 and Y2 the value of thebene�ts if death occurs before age 65.

30E35 � 10E35 20E45 � (0.54318)(0.25634) � 0.139239E[Y1] � 10030E35 a65 � 100(0.139239)(9.8969) � 137.803E[Y2] � 1000 (A35 − 30E35 A65)

� 1000(0.12872 − 0.139239(0.43980)

)� 67.483

E[Y] � 137.803 + 67.483 � 205.286 (C)

12. [Lesson 20] Let P be the annual net premium.

A45:20 � 1 − da45:20� 1 − ( 1

21)(12.6) � 0.4

A451:20 � A45:20 − 20E45 � 0.4 − 0.3 � 0.1

P �500A45

1:20 + 500A45

a45:20

�500(0.1) + 500(0.2)

12.6 � 11.9048 (B)

13. [Lesson 29] Let p be the probability that the copier was working at the beginning of the previousday. Then

0.8p + 0.6(1 − p) � 0.77


660 PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 14–15

0.6 + 0.2p � 0.77

p �0.170.2 � 0.85 (E)

14. [Lesson 29] The state vectors vn at time n before the transition are

v1 �(1 0 0

)

v2 �(1 0 0

) *.,0.8 0.1 0.10.2 0.6 0.20 0 1

+/- �

(0.8 0.1 0.1

)

v3 �(0.8 0.1 0.1

) *.,0.6 0.2 0.20.2 0.6 0.20 0 1

+/- �

(0.5 0.22 0.28

)

v4 �(0.5 0.22 0.28

) *.,0.4 0.3 0.30.2 0.6 0.20 0 1

+/- �

(0.244 0.282 0.474

)

v5 �(0.244 0.282 0.474

) *.,0.2 0.4 0.40.2 0.6 0.20 0 1

+/- �

(0.1052 0.2668 0.628

)

vn �

(0 0 1

)n > 5

Expected amount of time in State 1 is the sumof the �rst entries of the vectors or 1+0.8+0.5+0.244+0.1052 �

2.6492 . (B)

15. [Lesson 30] The probabilities of the states are:

Start of year 2(0.80 0.15 0.05

)

Start of year 3

(0.80 0.15 0.05

) *.,0.80 0.15 0.050.50 0.40 0.100 0 1

+/- �

(0.715 0.180 0.105

)

Start of year 4

(0.715 0.180 0.105

) *.,0.80 0.15 0.050.50 0.40 0.100 0 1

+/- �

(0.662 0.17925 0.15875

)

The expected present value of 1 per year in the active state is 1+0.80/1.05+0.715/1.052 +0.662/1.053 �2.9823.

The probabilities of death in the �rst three years are the di�erences in the probabilities of being in thethird state, or 0.05 in year 1, 0.105 − 0.05 � 0.055 in year 2, and 0.15875 − 0.105 � 0.05375 in year 3. Inyear 4, the probability of death is 0.662(0.05) + 0.17925(0.1) � 0.051025. The present value of the bene�tsis

1000(0.051.05 + 0.055

1.052+ 0.05375

1.053+ 0.051025

1.054)� 185.9153


PRACTICE EXAM 1, SOLUTION TO QUESTION 15 661

The net premium is π � 185.9153/2.9823 � 62.34 . (C)


contents · 2015. 6. 1. · b.24 solutions to soa exam mlc, spring 2014 . . . . . . . . . . . . . ....

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