Contents
1
Probability Review
1.1 Functions and moments . . . . . . . . .
1.2 Probability distributions . . . . . . . . .
1.2.1 Bernoulli distribution . . . . . .
1.2.2 Uniform distribution . . . . . . .
1.2.3 Exponential distribution . . . . .
1.3 Variance . . . . . . . . . . . . . . . . . .
1.4 Normal approximation . . . . . . . . . .
1.5 Conditional probability and expectation
1.6 Conditional variance . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . .
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1
1
2
3
3
4
4
5
7
9
10
14
2
Survival Distributions: Probability Functions
2.1 Probability notation . . . . . . . . . . . . .
2.2 Actuarial notation . . . . . . . . . . . . . .
2.3 Life tables . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . .
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19
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25
31
3
Survival Distributions: Force of Mortality
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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51
4
Survival Distributions: Mortality Laws
4.1 Mortality laws that may be used for human mortality . . . . .
4.1.1 Gompertz’s law . . . . . . . . . . . . . . . . . . . . . . .
4.1.2 Makeham’s law . . . . . . . . . . . . . . . . . . . . . . .
4.1.3 Weibull Distribution . . . . . . . . . . . . . . . . . . . .
4.2 Mortality laws for exam questions . . . . . . . . . . . . . . . .
4.2.1 Exponential distribution, or constant force of mortality
4.2.2 Uniform distribution . . . . . . . . . . . . . . . . . . . .
4.2.3 Beta distribution . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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61
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73
5
Survival Distributions: Moments
5.1 Complete . . . . . . . . . . . .
5.1.1 General . . . . . . . . .
5.1.2 Special mortality laws
5.2 Curtate . . . . . . . . . . . . .
Exercises . . . . . . . . . . . .
Solutions . . . . . . . . . . . .
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79
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6
Survival Distributions: Percentiles and Recursions
6.1 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
111
111
LC Study Manual—1st edition 44th printing
Copyright ©2015 ASM
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iii
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iv
CONTENTS
6.2
Recursive formulas for life expectancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
112
113
118
7
Survival Distributions: Fractional Ages
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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130
134
8
Insurance: Annual—Moments
8.1 Review of Financial Mathematics . . .
8.2 Moments of annual insurances . . . .
8.3 Standard insurances and notation . . .
8.4 Illustrative Life Table . . . . . . . . . .
8.5 Constant force and uniform mortality
8.6 Normal approximation . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . .
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143
143
144
145
147
149
151
152
165
9
Insurance: Continuous—Moments—Part 1
9.1 Definitions and general formulas . . .
9.2 Constant force of mortality . . . . . . .
Exercises . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . .
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177
177
178
184
192
10 Insurance: Continuous—Moments—Part 2
10.1 Uniform survival function . . . . . . . . . . . . .
10.2 Other mortality functions . . . . . . . . . . . . .
10.2.1 Integrating at n e −ct (Gamma Integrands)
10.3 Variance of endowment insurance . . . . . . . .
10.4 Normal approximation . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . .
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201
201
203
203
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206
207
213
11 Insurance: Probabilities and Percentiles
11.1 Introduction . . . . . . . . . . . . . . . . . . . . .
11.2 Probabilities for Continuous Insurance Variables
11.3 Probabilities for Discrete Variables . . . . . . . .
11.4 Percentiles . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . .
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223
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224
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234
12 Insurance: Recursive Formulas
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
241
243
246
13 Annuities: Continuous, Expectation
13.1 Whole life annuity . . . . . . . . . . .
13.2 Temporary and deferred life annuities
13.3 n-year certain-and-life annuity . . . .
Exercises . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . .
251
252
254
257
259
264
LC Study Manual—1st edition 44th printing
Copyright ©2015 ASM
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v
CONTENTS
14 Annuities: Discrete, Expectation
14.1 Annuities-due . . . . . . . . .
14.2 Annuities-immediate . . . . .
14.3 Actuarial Accumulated Value
Exercises . . . . . . . . . . . .
Solutions . . . . . . . . . . . .
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271
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290
15 Annuities: Variance
15.1 Whole Life and Temporary Life Annuities . . . . . . . . . . .
15.2 Other Annuities . . . . . . . . . . . . . . . . . . . . . . . . . .
15.3 Typical Exam Questions . . . . . . . . . . . . . . . . . . . . .
15.4 Combinations of Annuities and Insurances with No Variance
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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301
301
303
304
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307
317
16 Annuities: Probabilities and Percentiles
16.1 Probabilities for continuous annuities . . . . . .
16.2 Distribution functions of annuity present values
16.3 Probabilities for discrete annuities . . . . . . . .
16.4 Percentiles . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . .
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333
333
336
337
338
340
345
17 Annuities: Recursive Formulas
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
353
355
357
18 Premiums: Net Premiums for Fully Continuous Insurances
18.1 Future loss . . . . . . . . . . . . . . . . . . . . . . . . . .
18.2 Net premium . . . . . . . . . . . . . . . . . . . . . . . .
18.3 Expected value of future loss . . . . . . . . . . . . . . .
18.4 International Actuarial Premium Notation . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
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361
361
362
364
366
366
373
19 Premiums: Net Premiums from Life Tables
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
381
382
386
20 Premiums: Net Premiums from Formulas
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
391
393
401
21 Premiums: Variance of Future Loss, Continuous
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
411
414
419
22 Premiums: Variance of Future Loss, Discrete
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
427
429
434
LC Study Manual—1st edition 44th printing
Copyright ©2015 ASM
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vi
CONTENTS
23 Premiums: Probabilities and Percentiles of Future Loss
23.1 Probabilities . . . . . . . . . . . . . . . . . . . . . . .
23.1.1 Fully continuous insurances . . . . . . . . . .
23.1.2 Discrete insurances . . . . . . . . . . . . . . .
23.1.3 Annuities . . . . . . . . . . . . . . . . . . . .
23.2 Percentiles . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . .
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441
441
441
445
445
448
449
452
24 Multiple Lives: Joint Life Probabilities
24.1 Introduction . . . . . . . . . . . . .
24.2 Independence . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . .
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459
459
460
462
467
25 Multiple Lives: Last Survivor Probabilities
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
471
475
480
26 Multiple Lives: Moments
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
485
490
494
27 Multiple Decrement Models: Probabilities
27.1 Introduction . . . . . . . . . . . . . . . . . . . .
27.2 Life Tables . . . . . . . . . . . . . . . . . . . . .
27.3 Examples of Multiple Decrement Probabilities
27.4 Discrete Insurances . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . .
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501
501
503
505
506
509
519
28 Multiple Decrement Models: Forces of Decrement
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
527
529
536
29 Multi-State Models (Markov Chains): Probabilities
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
545
553
561
30 Multi-State Models (Markov Chains): Premiums and Reserves
30.1 Prototype Example . . . . . . . . . . . . . . . . . . . . . . .
30.2 Insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30.3 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30.4 Premiums and Reserves . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
569
569
569
572
574
576
585
Practice Exams
1
Practice Exam 1
LC Study Manual—1st edition 44th printing
Copyright ©2015 ASM
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593
595
CONTENTS
vii
2
Practice Exam 2
601
3
Practice Exam 3
607
4
Practice Exam 4
613
5
Practice Exam 5
619
6
Practice Exam 6
625
7
Practice Exam 7
631
8
Practice Exam 8
637
9
Practice Exam 9
643
10 Practice Exam 10
649
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 . .
655
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657
657
662
667
672
677
683
689
694
700
706
B Solutions to Old Exams
B.1 Solutions to CAS Exam 3, Spring 2005 . .
B.2 Solutions to CAS Exam 3, Fall 2005 . . . .
B.3 Solutions to CAS Exam 3, Spring 2006 . .
B.4 Solutions to CAS Exam 3, Fall 2006 . . . .
B.5 Solutions to CAS Exam 3, Spring 2007 . .
B.6 Solutions to CAS Exam 3, Fall 2007 . . . .
B.7 Solutions to CAS Exam 3L, Spring 2008 .
B.8 Solutions to CAS Exam 3L, Fall 2008 . . .
B.9 Solutions to CAS Exam 3L, Spring 2009 .
B.10 Solutions to CAS Exam 3L, Fall 2009 . . .
B.11 Solutions to CAS Exam 3L, Spring 2010 .
B.12 Solutions to CAS Exam 3L, Fall 2010 . . .
B.13 Solutions to CAS Exam 3L, Spring 2011 .
B.14 Solutions to CAS Exam 3L, Fall 2011 . . .
B.15 Solutions to CAS Exam 3L, Spring 2012 .
B.16 Solutions to SOA Exam MLC, Spring 2012
B.17 Solutions to CAS Exam 3L, Fall 2012 . . .
B.18 Solutions to SOA Exam MLC, Fall 2012 . .
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713
713
717
721
724
727
731
734
737
739
742
745
748
751
753
756
759
761
764
LC Study Manual—1st edition 44th printing
Copyright ©2015 ASM
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viii
B.19
B.20
B.21
B.22
B.23
B.24
B.25
B.26
B.27
B.28
CONTENTS
Solutions to CAS Exam 3L, Spring 2013 .
Solutions to SOA Exam MLC, Spring 2013
Solutions to CAS Exam 3L, Fall 2013 . . .
Solutions to SOA Exam MLC, Fall 2013 . .
Solutions to CAS Exam LC, Spring 2014 .
Solutions to SOA Exam MLC, Spring 2014
Solutions to CAS Exam LC, Fall 2014 . . .
Solutions to SOA Exam MLC, Fall 2014 . .
Solutions to CAS Exam LC, Spring 2015 .
Solutions to SOA Exam MLC, Spring 2015
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C Lessons Corresponding to Questions on Released and Practice Exams
LC Study Manual—1st edition 44th printing
Copyright ©2015 ASM
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766
770
775
779
782
788
790
794
796
800
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 (q x ) or lives (l x ) for integral ages only. Often, it is necessary to determine
lives at fractional ages (like l x+0.5 for x an integer) or mortality rates for fractions of a year. We need some
way to interpolate between ages.
The easiest interpolation method is linear interpolation, or uniform distribution of deaths between
integral ages (UDD). This means that the number of lives at age x + s, 0 ≤ s ≤ 1, is a weighted average of
the number of lives at age x and the number of lives at age x + 1:
l x+s (1 − s ) l x + sl x+1 l x − sd x
The graph of l x+s is a straight line between s 0 and s 1 with
slope −d x . The graph at the right portrays this for a mortality rate
q 100 0.45 and l 100 1000.
Contrast UDD with an assumption of a uniform survival function.
If age at death is uniformly distributed, then l x as a function of x is a
straight line. If UDD is assumed, l x is a straight line between integral
ages, but the slope may vary for different ages. Thus if age at death is
uniformly distributed, UDD holds at all ages, but not conversely.
Using l x+s , we can compute s q x :
s qx
1 − s px
l x+s
1−
1 − (1 − sq x ) sq x
lx
(7.1)
l 100+s
1000
550
0
0
1
s
(7.2)
That is one of the most important formulas, so let’s state it again:
s qx
More generally, for 0 ≤ s + t ≤ 1,
s q x+t
sq x
l x+s+t
l x+t
sq x
l x − ( s + t ) dx
sd x
1−
l x − td x
l x − td x 1 − tq x
(7.2)
1 − s p x+t 1 −
(7.3)
where the last equation was obtained by dividing numerator and denominator by l x . The important point
to pick up is that while s q x is the proportion of the year s times q x , the corresponding concept at age x + t,
s q x+t , is not sq x , but is in fact higher than sq x . The number of lives dying in any amount of time is constant,
and since there are fewer and fewer lives as the year progresses, the rate of death is in fact increasing
over the year. The numerator of s q x+t is the proportion of the year being measured s times the death
rate, but then this must be divided by 1 minus the proportion of the year that elapsed before the start of
measurement.
For most problems involving death probabilities, it will suffice if you remember that l x+s is linearly
interpolated. It often helps to create a life table with an arbitrary radix. Try working out the following
example before looking at the answer.
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Copyright ©2015 ASM
125
126
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
Example 7A You are given:
• q x 0.1
• Uniform distribution of deaths between integral ages is assumed.
Calculate 1/2 q x+1/4 .
Answer: Let l x 1. Then l x+1 l x (1 − q x ) 0.9 and d x 0.1. Linearly interpolating,
l x+1/4 l x − 14 d x 1 − 41 (0.1) 0.975
l x+3/4 l x − 43 d x 1 − 43 (0.1) 0.925
l x+1/4 − l x+3/4 0.975 − 0.925
0.051282
1/2 q x+1/4 l x+1/4
0.975
You could also use equation (7.3) to work this example.
Example 7B For two lives age ( x ) with independent future lifetimes,
Deaths are uniformly distributed between integral ages.
Calculate the probability that both lives will survive 2.25 years.
k| q x
0.1 ( k + 1) for k 0, 1, 2.
Answer: Since the two lives are independent, the probability of both surviving 2.25 years is the square
of 2.25 p x , the probability of one surviving 2.25 years. If we let l x 1 and use d x+k l x k| q x , we get
q x 0.1 (1) 0.1
l x+1 1 − d x 1 − 0.1 0.9
1| q x 0.1 (2) 0.2
2| q x
l x+2 0.9 − d x+1 0.9 − 0.2 0.7
0.1 (3) 0.3
l x+3 0.7 − d x+2 0.7 − 0.3 0.4
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
l x+2.25
0.625
2.25p x lx
Squaring, the answer is 0.6252 0.390625 .
The probability density function of Tx , s p x µ x+s , is the constant q x ,
the derivative of the conditional cumulative distribution function s q x sq x with respect to s. That is another important formula, since the density is needed to compute expected values, so let’s repeat it:
s px
µ x+s q x
(7.4)
µ100+s
1
0.45
0.55
0.45
It follows that the force of mortality is q x divided by 1 − sq x :
µ x+s qx
qx
p
1
−
sq x
s x
(7.5)
0
0
1
The force of mortality increases over the year, as illustrated in the graph for q 100 0.45 to the right.
?
Quiz 7-1 You are given:
• µ50.4 0.01
• Deaths are uniformly distributed between integral ages.
Calculate 0.6q 50.4 .
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s
127
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
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
1
. It
if uniform distribution of deaths is not assumed. Since S is uniform on [0, 1) , E[S] 21 and Var ( S ) 12
1
follows from E[S] 2 that
e̊ x e x + 12
(7.6)
More common on exams are questions asking you to evaluate the temporary complete expectancy of
life under UDD. You can always evaluate the temporary complete expectancy, whether or not UDD is
assumed, by integrating t p x , as indicated by formula (5.6) on page 80. For UDD, t p x is linear between
integral ages. Therefore, a rule we learned in Lesson 5 applies for all integral x:
e̊ x:1 p x + 0.5q x
(5.13)
This equation will be useful. In addition, the method for generating this equation can be used to work
out questions involving temporary complete life expectancies for short periods. The following example
illustrates this. This example will be reminiscent of calculating temporary complete life expectancy for
uniform mortality.
Example 7C You are given
• q x 0.1.
• Deaths are uniformly distributed between integral ages.
Calculate e̊ x:0.4 .
Answer: We will discuss two ways to solve this: an algebraic method and a geometric method.
The algebraic method is based on the double expectation theorem, equation (1.11). It uses the fact that
for a uniform distribution, the mean is the midpoint. If deaths occur uniformly between integral ages, then
those who die within a period contained within a year survive half the period on the average.
In this example, those who die within 0.4 survive an average of 0.2. Those who survive 0.4 survive an
average of 0.4 of course. The temporary life expectancy is the weighted average of these two groups, or
0.4 q x (0.2) + 0.4 p x (0.4) . This is:
0.4 q x
(0.4)(0.1) 0.04
0.4 p x
1 − 0.04 0.96
e̊ 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)
A
0
(1.0, 0.9)
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
128
?
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
Quiz 7-2 As in Example 7C, you are given
• q x 0.1.
• Deaths are uniformly distributed between integral ages.
Calculate e̊ x+0.4:0.6 .
Let’s now work out an example in which the duration crosses an integral boundary.
Example 7D You are given:
• q x 0.1
• q x+1 0.2
• Deaths are uniformly distributed between integral ages.
Calculate e̊ 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 q x+0.5
0.5 p x+0.5
0.5|0.5 q x+0.5
1 p x+0.5
0.5 (0.1)
5
1 − 0.5 (0.1) 95
5
90
1−
95
95
!
9
90
0.5 (0.2) 95
95
5
9
81
1−
−
95 95 95
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
90
5
0.5 q x+0.5 1 −
95 95
90 − 81
9
0.5|0.5 q x+0.5 95
95
l x+1.5 81
1 p x+0.5 l x+0.5 95
Either way, we’re now ready to calculate e̊ x+0.5:1 .
e̊ x+0.5:1 5 (0.25) + 9 (0.75) + 81 (1)
89
95
95
For the geometric method we draw the following graph:
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129
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
t px +0.5
0.5, 90
95
1
1.0, 81
95
A
B
0
x + 0.5
0.5
x +1
t
1.0
x + 1.5
The heights at x + 1 and
Then
x + 1.5 are as we computed above.
we compute each area separately. The
185
1 90
81
171
(
0.5
)
.
The
area
of
B
is
+
area of A is 21 1 + 90
95
95 (4)
2 95
95 (0.5) 95 (4) . Adding them up, we get
185+171
95 (4)
?
89
95
.
Quiz 7-3 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 e̊ 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 e̊ x:n since they die on the
average in the middle of the year. Thus the difference is 21 n q x :
e̊ x:n e x:n + 0.5n q x
(7.7)
Example 7E You are given:
• q x 0.01 for x 50, 51, . . . , 59.
• Deaths are uniformly distributed between integral ages.
Calculate e̊50:10 .
Answer: As we just said, e̊50:10 e50:10 + 0.510 q50 . The first summand, e50:10 , is the sum of k p50 0.99k
for k 1, . . . , 10. This sum is a geometric series:
e50:10 10
X
0.99k k1
0.99 − 0.9911
9.46617
1 − 0.99
The second summand, the probability of dying within 10 years is 10 q 50 1 − 0.9910 0.095618. Therefore
e̊50:10 9.46617 + 0.5 (0.095618) 9.51398
The formulas are summarized in Table 7.1.
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130
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
Table 7.1: Summary of formulas for fractional ages
l x+s l x − sd x
sqx
spx
sq x
s p x 1 − sq x
sq x
s q x+t 1 − tq x
qx
µ x+s 1 − sq x
µ x+s q x
e̊ x e x + 0.5
e̊ x:n e x:n + 0.5 n q x
e̊ x:1 p x + 0.5q x
Exercises
7.1. [CAS4-S85:16] (1 point) Deaths are uniformly distributed between integral ages.
Which of the following represents 3/4p x +
A.
3/4p x
7.2.
B.
3/4q x
C.
1
2 1/2p x
µ x+1/2 ?
1/2p x
D.
1/2q x
E.
1/4p x
[Based on 150-S88:25] You are given:
• 0.25 q x+0.75 3/31.
• Mortality is uniformly distributed within age x.
Calculate q x .
Use the following information for questions 7.3 and 7.4:
You are given:
• Deaths are uniformly distributed between integral ages.
• q x 0.10.
• q x+1 0.15.
7.3. Calculate 1/2q x+3/4 .
7.4. Calculate 0.3|0.5 q x+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).
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Exercises continue on the next page . . .
131
EXERCISES FOR LESSON 7
7.6. [160-F90:5] You are given:
• A survival distribution is defined by
l x 1000 1 −
•
•
x
100
2!
, 0 ≤ x ≤ 100.
µ x denotes the actual force of mortality for the survival distribution.
µ Lx 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
7.7. A survival distribution is defined by
• S0 ( k ) 1/ (1 + 0.01k ) 4 for k a non-negative integer.
• Deaths are uniformly distributed between integral ages.
Calculate 0.4 q20.2 .
7.8. [Based on150-S89:15] You are given:
• Deaths are uniformly distributed over each year of age.
•
x
lx
35
36
37
38
39
100
99
96
92
87
Which of the following are true?
I.
II.
III.
1|2q 36
0.091
µ37.5 0.043
0.33q 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.
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D. I, II and III
Exercises continue on the next page . . .
132
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
7.9. [150-82-94:5] You are given:
• Deaths are uniformly distributed over each year of age.
• 0.75 p x 0.25.
Which of the following are true?
I.
0.25 q x+0.5
II.
0.5 q x
III.
0.5
0.5
µ x+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.
7.10.
D. I, II and III
[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
7.11.
B. 0.0785
C. 0.0790
D. 0.0796
E. 0.0800
You are given:
• Deaths are uniformly distributed between integral ages.
• q x 0.1.
• q x+1 0.3.
Calculate e̊ x+0.7:1 .
7.12.
You are given:
• Deaths are uniformly distributed between integral ages.
• q45 0.01.
• q46 0.011.
Calculate Var min T45 , 2 .
7.13.
You are given:
• Deaths are uniformly distributed between integral ages.
• 10 p x 0.2.
Calculate e̊ x:10 − e x:10 .
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Exercises continue on the next page . . .
133
EXERCISES FOR LESSON 7
7.14.
[4-F86:21] You are given:
• q60 0.020
• q61 0.022
• Deaths are uniformly distributed over each year of age.
Calculate e̊60:1.5 .
A. 1.447
7.15.
B. 1.457
C. 1.467
D. 1.477
E. 1.487
[150-F89:21] You are given:
• q70 0.040
• q71 0.044
• Deaths are uniformly distributed over each year of age.
Calculate e̊70:1.5 .
A. 1.435
7.16.
B. 1.445
C. 1.455
D. 1.465
E. 1.475
[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, e̊1:1.5 .
A. 1.25
7.17.
B. 1.30
C. 1.35
D. 1.40
E. 1.45
You are given:
• Deaths are uniformly distributed between integral ages.
• e̊ x+0.5:0.5 5/12.
Calculate q x .
7.18.
You are given:
• Deaths are uniformly distributed over each year of age.
• e̊55.2:0.4 0.396.
Calculate µ55.2 .
7.19.
[150-S87:21] You are given:
• d x k for x 0, 1, 2, . . . , ω − 1
• e̊20:20 18
• Deaths are uniformly distributed over each year of age.
Calculate 30|10 q30 .
A. 0.111
B. 0.125
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C. 0.143
D. 0.167
E. 0.200
Exercises continue on the next page . . .
134
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
[150-S89:24] You are given:
7.20.
• Deaths are uniformly distributed over each year of age.
• µ45.5 0.5
Calculate e̊45:1 .
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 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.
B.
C.
D.
E.
Less than 515
At least 515, but less than 525
At least 525, but less than 535
At least 535, but less than 545
At least 545
Additional old CAS Exam 3/3L questions: S05:31, F05:13, S06:13, F06:13, S07:24, S08:16, S09:3, F09:3,
S10:4, F10:3, S11:3, S12:3, F12:3, S13:3, F13:3
Additional old CAS Exam LC questions: S14:4, F14:4, S15:3
Solutions
7.1. In the second summand, 1/2 p x µ x+1/2 is the density function, which is the constant q x under UDD.
The first summand 3/4 p x 1 − 34 q x . So the sum is 1 − 14 q x , or 1/4 p x . (E)
7.2. Using equation (7.3),
3
0.25 q x+0.75
31
3
2.25
−
qx
31
31
3
31
qx
0.25q x
1 − 0.75q x
0.25q x
10
qx
31
0.3
7.3. 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 (7.3),
1 − 0.10
0.9
1/4p x+3/4 1 − 0.75 (0.10) 0.925
The probability of ( x + 1) surviving to x + 1.25 is
1/4p x+1
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Copyright ©2015 ASM
1 − 0.25 (0.15) 0.9625
135
EXERCISE SOLUTIONS FOR LESSON 7
The answer to the question is then the complement of the product of these two numbers:
1/2q x+3/4 1 − 1/2p x+3/4 1 − 1/4p x+3/4 1/4p x+1 1 −
!
0.9
(0.9625) 0.06351
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
l x+1.25 0.86625
0.93649
1/2 p 3/4 l x+0.75
0.925
1/2 q 3/4
7.4.
0.3|0.5q x+0.4
1 − 1/2 p3/4 1 − 0.93649 0.06351
is 0.3p x+0.4 − 0.8p x+0.4 . The first summand is
0.3 p x+0.4
1 − 0.7q x 1 − 0.07 93
1 − 0.4q x 1 − 0.04 96
The probability that ( x + 0.4) survives to x + 1 is, by equation (7.3),
0.6p x+0.4
and the probability ( x + 1) survives to x + 1.2 is
0.2p x+1
So
1 − 0.10 90
1 − 0.04 96
1 − 0.2q x+1 1 − 0.2 (0.15) 0.97
0.3|0.5q x+0.4 late:
!
93
90
−
(0.97) 0.059375
96
96
Alternatively, you could use the life table from the solution to the last question, and linearly interpol 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
0.93 − 0.873
0.059375
0.3|0.5 q x+0.4 0.96
7.5. 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, l45.5 12 (9,164,051 + 9,127,426) 9,145,738.5. Median future lifetime occurs when l x 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 ( 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
4,828,182 − 4,572,869 255,313
s
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 .
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136
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
7.6.
x p0
lx
l0
1−
x 2
100 .
The force of mortality is calculated as the negative derivative of ln x p0 :
x
1
d ln x p0 2 100
2x
100
µx −
2
x
dx
1002 − x 2
1 − 100
100.5
µ50.25 0.0134449
1002 − 50.252
For UDD, we need to calculate q 50 .
l51 1 − 0.512
0.986533
l50 1 − 0.502
1 − 0.986533 0.013467
p 50 q 50
so under UDD,
L
µ50.25
q50
0.013467
0.013512.
1 − 0.25q 50 1 − 0.25 (0.013467)
L
is 0.013445 − 0.013512 −0.000067 . (B)
The difference between µ50.25 and µ50.25
7.7.
S0 (20) 1/1.24 and S0 (21) 1/1.214 , so q 20 1 − (1.2/1.21) 4 0.03265. Then
0.4 q 20.2
0.4q20
0.4 (0.03265)
0.01315
1 − 0.2q 20 1 − 0.2 (0.03265)
7.8.
I.
Calculate 1|2 q36 .
1|2q 36
2 d37
l 36
96 − 87
0.09091
99
!
This statement does not require uniform distribution of deaths.
II.
By equation (7.5),
µ37.5 III.
Calculate 0.33 q38.5 .
q37
4/96
4
0.042553
1 − 0.5q37 1 − 2/96 94
0.33q 38.5
0.33 d38.5
l 38.5
(0.33)(5)
89.5
0.018436
I can’t figure out what mistake you’d have to make to get 0.021. (A)
7.9. First calculate q x .
1 − 0.75q x 0.25
qx 1
Then by equation (7.3), 0.25 q x+0.5 0.25/ (1 − 0.5) 0.5, making I true.
By equation (7.2), 0.5 q x 0.5q x 0.5, making II true.
By equation (7.5), µ x+0.5 1/ (1 − 0.5) 2, making III false. (A)
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!
#
137
EXERCISE SOLUTIONS FOR LESSON 7
7.10.
We use equation (7.5) to back out q x for each age.
µ x+0.5 µ x+0.5
qx
⇒ qx 1 − 0.5q x
1 + 0.5µ x+0.5
0.0202
q 80 0.02
1.0101
0.0408
0.04
q 81 1.0204
0.0619
q 82 0.06
1.03095
Then by equation (7.3), 0.5 p80.5 0.98/0.99. p81 0.96, and 0.5 p82 1 − 0.5 (0.06) 0.97. Therefore
2 q 80.5 1 −
!
0.98
(0.96)(0.97) 0.0782
0.99
(A)
7.11. 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
Survival
time
Probability
of group
Average
survival time
(0, 0.3]
(0.3, 1]
(1, ∞)
1 − 0.3 p x+0.7
p
0.3 x+0.7 − 1 p x+0.7
1 p x+0.7
0.15
0.65
1
I
II
III
We calculate the required probabilities.
0.3 p x+0.7
1 p x+0.7
1 − 0.3 p x+0.7
0.9
0.967742
0.93
0.9
1 − 0.7 (0.3) 0.764516
0.93
1 − 0.967742 0.032258
0.3 p x+0.7 − 1 p x+0.7 0.967742 − 0.764516 0.203226
e̊ 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
e̊ 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
7.12.
For the expected value, we’ll use the recursive formula. (The trapezoidal rule could also be used.)
e̊45:2 e̊45:1 + p45 e̊46:1
(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
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Z
2
0
t t p x dt
138
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
Z
2
1
0
t (1 − 0.01t ) dt +
!
Z
2
1
t (0.99) 1 − 0.011 ( t − 1) dt
!
!
1
(1.011)(22 − 12 )
23 − 13 ++
1
*
*
//
+ 0.99 .
− 0.011
2 . − 0.01
2
3
2
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 difference is
1
1
10 q x (1 − 0.2) 0.4
2
2
7.14. 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
p60 0.98
1.5 p 60
0.98 1 − 0.5 (0.022) 0.96922
e̊60:1.5 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 p60 0.98 and
width 1. The second trapezoid has heights p60 0.98 and 1.5 p 60 0.96922 and width 1/2.
e̊60:1.5
1
1
(1 + 0.98) +
2
2
1.477305
!
(D)
!
1
(0.98 + 0.96922)
2
7.15. p70 1 − 0.040 0.96, 2p 70 (0.96)(0.956) 0.91776, and by linear interpolation, 1.5 p70 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
e̊70:1.5 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
e̊70:1.5 0.5 (1 + 0.96) + 0.5 (0.5)(0.96 + 0.93888) 1.45472
7.16.
(C)
First we calculate t p1 for t 1, 2.
p 1 1 − q 1 0.90
2 p1
(1 − q1 )(1 − q2 ) (0.90)(0.95) 0.855
By linear interpolation, 1.5 p1 (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
e̊1:1.5 0.10 (0.5) + (0.90 − 0.8775)(1.25) + 0.8775 (1.5) 1.394375
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(D)
139
EXERCISE SOLUTIONS FOR LESSON 7
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)
e̊1:1.5 (1)(0.5)(1 + 0.9) + (0.5)(0.5)(0.90 + 0.8775)
0.95 + 0.444375 1.394375
7.17.
(D)
0
1
1.5
2
Those who die survive 0.25 years on the average and survivors survive 0.5 years, so we have
0.25 0.5 q x+0.5 + 0.5 0.5 p x+0.5 !
!
1 − qx
0.5q x
+ 0.5
1 − 0.5q x
1 − 0.5q x
0.25
t
5
12
5
12
5
5
− 24
qx
0.125q x + 0.5 − 0.5q x 12
1
5
5
1 1
qx
−
− + −
2 12
24 2 8
qx
1
12
6
qx 1
2
Alternatively, complete life expectancy is the area of the trapezoid shown on the right, so
t px +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 32 , from which it follows
0
1 − qx
2
3 1 − 12 q x
qx 7.18.
1
2
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 + 0.20.4 q 55.2
Using the formula 0.4 q55.2 0.4q55 / (1 − 0.2q55 ) (equation (7.3)), we have
0.4
!
!
0.4q 55
1 − 0.6q55
+ 0.2
0.396
1 − 0.2q55
1 − 0.2q 55
0.4 − 0.24q 55 + 0.08q 55 0.396 − 0.0792q 55
0.0808q 55 0.004
0.004
q 55 0.0495
0.0808
q 55
0.0495
µ55.2 0.05
1 − 0.2q 55 1 − 0.2 (0.0495)
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0.5 px +0.5
0.5
t
140
7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
7.19. Since d x is constant for all x and deaths are uniformly distributed within each year of age, mortality
is uniform globally. We back out ω using equation (5.12), e̊ x:n n p x ( n ) + n q x ( n/2) :
10 20 q 20 + 20 20 p 20 18
10
!
!
20
ω − 40
+ 20
18
ω − 20
ω − 20
200 + 20ω − 800 18ω − 360
2ω 240
ω 120
x −20 p20
Alternatively, we can back out ω using the trapezoidal rule. Complete life expectancy is the area of the trapezoid shown to the right.
e̊20:20 18 (20)(0.5) 1 +
ω − 40
ω − 20
0.8ω − 16 ω − 40
0.8 ω − 40
ω − 20
1
ω − 40
ω − 20
18
20
40
x
0.2ω 24
ω 120
Once we have ω, we compute
30|10 q 30
7.20.
We use equation (7.5) to obtain
10
10
0.1111
ω − 30 90
0.5 (A)
qx
1 − 0.5q x
q x 0.4
Then e̊45:1 0.5 1 + (1 − 0.4) 0.8 . (E)
7.21. According to the Illustrative Life Table, l30 9,501,381, so we are looking for the age x such that l x 0.75 (9,501,381) 7,126,036. This is between 67 and 68. Using linear interpolation, since l 67 7,201,635
and l 68 7,018,432, we have
x 67 +
This is 37.4127 years into the future.
3
4
7,201,635 − 7,126,036
67.4127
7,201,635 − 7,018,432
of the people collect 50,000. We need 50,000
540.32 per person. (D)
Quiz Solutions
7-1. Notice that µ50.4 q 50
1−0.4q50
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while 0.6q 50.4 0.6q50
1−0.4q50 ,
so 0.6q 50.4 0.6 (0.01) 0.006
3
4
!
1
1.1237.4127
!
141
QUIZ SOLUTIONS FOR LESSON 7
7-2. The algebraic method goes: those who die will survive 0.3 on the average, and those who survive
will survive 0.6.
0.6 q x+0.4
0.6 p x+0.4
e̊ x+0.4:0.6
0.6 (0.1)
6
1 − 0.4 (0.1) 96
90
6
1−
96 96
6
90
55.8
(0.3) + (0.6) 0.58125
96
96
96
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 .
7-3. 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, 2.25 q 0 0.25 (0.6) + 0.75 (0.2) 0.3. So we have
e̊0:2.25 q 0 (0.5) + 1| q 0 (1.5) + 2|0.25 q 0 (2.125) + 2.25 p0 (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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7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES
Practice Exam 1
1. You are given:
• The following life table.
2 q 52
•
x
lx
dx
50
51
52
53
1000
20
35
37
0.07508.
Determine d51 .
A.
B.
C.
D.
E.
Less than 20
At least 20, but less than 22
At least 22, but less than 24
At least 24, but less than 26
At least 26
Mortality follows Gompertz’s law with B 0.002 and c 1.03.
2.
Calculate the 40th percentile of future lifetime for (45).
A.
B.
C.
D.
E.
Less than 35.0
At least 35.0, but less than 40.0
At least 40.0, but less than 45.0
At least 45.0, but less than 50.0
At least 50.0
You are given:
3.
•
•
•
l x 100 (120 − x ) for x ≤ 90
µ x µ for x > 90
e̊80 has the same value that it would have if l x 100 (120 − x ) for x ≤ 120.
Determine µ.
A. 0.04
B. 0.05
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C. 0.067
595
D. 0.075
E. 0.083
Exercises continue on the next page . . .
596
PRACTICE EXAMS
4. Three independent probes are sent to Mars. The probability distribution for the amount of time
each one will function is given in the following table.
Years
(n)
Probability of functioning
n years
1
2
3
0.9
0.8
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 31 years.
A.
B.
C.
D.
E.
Less than 0.40
At least 0.40, but less than 0.55
At least 0.55, but less than 0.70
At least 0.70, but less than 0.85
At least 0.85
5. For three independent lives ( x ) , ( y ) , and ( z ) , you are given
•
•
•
µ x+t 0.01, t > 0
µ y+t 0.02, t > 0
µ z+t 0.03, t > 0
Calculate the probability that at least one life survives 30 years.
A. 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 e̊30:40 .
A.
B.
C.
D.
E.
Less than 24
At least 24, but less than 26
At least 26, but less than 28
At least 28, but less than 30
At least 30
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597
PRACTICE EXAM 1
7. A discrete Markov chain model for insurance on two lives ( x ) and ( y ) has the following states:
0 Both alive
1 Only ( x ) alive
2 Only ( y ) alive
3 Neither alive
Let p i 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.
p 01 p 23
II.
p 12 p 21
III.
p 03 0
A. I and II only
B. I and III only
C. II and III only
E. The correct answer is not given by A. , B. , C. , or D.
D. I, II, and III
8. For an insurance policy on ( x ) , decrements occur due to death (1), withdrawal (2), and conversion (3). Decrement rates are as follows:
t
1
2
3
(1)
q x+t−1
0.01
0.01
0.01
(2)
q x+t−1
0.09
0.06
0.05
(3)
q x+t−1
0.01
0.01
0.04
Determine the probability that conversion will occur between 1 and 3 years from now.
A.
B.
C.
D.
E.
Less than 0.042
At least 0.042, but less than 0.047
At least 0.047, but less than 0.052
At least 0.052, but less than 0.057
At least 0.057
9. Persistency of an insurance contract is affected 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 policyholders who eventually withdraw.
A.
B.
C.
D.
E.
Less than 15
At least 15, but less than 17
At least 17, but less than 19
At least 19, but less than 21
At least 21
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598
PRACTICE EXAMS
You are given:
10.
Z1 is the present value random variable for a 10-year term insurance paying 1 at the moment of
death of (45).
• Z2 is the present value random variable for a 20-year deferred whole life insurance paying 1 at
the moment of death of (45).
• µ 0.02
• δ 0.04
•
Calculate Cov ( Z1 , Z2 ) .
A.
B.
C.
D.
E.
Less than −0.04
At least −0.04, but less than −0.03
At least −0.03, but less than −0.02
At least −0.02, but less than −0.01
At least −0.01
For a 30-year deferred whole life annuity on (35):
11.
•
•
•
•
•
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.
B.
C.
D.
E.
Less than 204
At least 204, but less than 205
At least 205, but less than 206
At least 206, but less than 207
At least 207
For a special fully discrete life insurance on (45),
12.
•
•
•
•
•
•
The benefit is 1000 if death occurs before age 65, 500 otherwise.
Premiums are payable at the beginning of the first 20 years only.
i 0.05
A45 0.2
20 E45 0.3
ä 45:20 12.6
Determine the annual net premium.
A.
B.
C.
D.
E.
Less than 11.50
At least 11.50, but less than 12.00
At least 12.00, but less than 12.50
At least 12.50, but less than 13.00
At least 13.00
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Exercises continue on the next page . . .
599
PRACTICE EXAM 1
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 fixed during the day and
working at the beginning of the next day is 0.6.
You are given that the probability that the copier is working at the beginning of Tuesday is 0.77.
Determine the probability that the copier was working at the beginning of the previous day, Monday.
A.
B.
C.
D.
E.
Less than 0.75
At least 0.75, but less than 0.78
At least 0.78, but less than 0.81
At least 0.81, but less than 0.84
At least 0.84
14. You are given the following transition matrices for a Markov chain with three states numbered 1,
2, and 3.
1 − 0.2n 0.1n 0.1n



*. 0.2


0.6
0.2 +/ 0 ≤ n < 5




0
1 , 0
Qn 


0 0 1


*.0 0 1+/



n≥5


 ,0 0 1-
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.
B.
C.
D.
E.
Less than 2.6
At least 2.6, but less than 2.8
At least 2.8, but less than 3.0
At least 3.0, but less than 3.2
At least 3.2
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Exercises continue on the next page . . .
600
PRACTICE EXAMS
15. Life insurance policies include a disability waiver rider that waives premiums while the insured is
disabled.
Initially, an insured ( x ) is active. At the beginning of each subsequent year, an insured may either be
in one of three states:
State 1 = active
State 2 = disabled
State 3 = dead
The transition probability matrix is
0.80
*
Q .0.50
, 0
0.15
0.40
0
0.05
0.10+/
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 using
i 0.05.
Determine π.
A.
B.
C.
D.
E.
Less than 59
At least 59, but less than 61
At least 61, but less than 63
At least 63, but less than 65
At least 65
Solutions to the above questions begin on page 657.
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Appendix A. Solutions to the Practice Exams
Answer Key for Practice Exam 1
1
2
3
4
5
B
B
C
D
D
6
7
8
9
10
B
E
A
B
D
11
12
13
14
15
C
B
E
B
C
Practice Exam 1
1. [Lesson 2] 0.07508 2 q 52 ( d52 + d53 ) /l52 72/l 52 , 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 µ x Bc x 0.002 (1.03x ) . We have
t p 45 exp −
Z
45+t
45
0.002 (1.03u ) du
!
0.002 (1.0345+t − 1.0345 )
exp −
ln 1.03
!
and we want to set this equal to 0.6, since the 40th percentile of future lifetime corresponds to a 40% chance
of dying before that time, or a 60% probability of survival to that time. We log the expression and set it
equal to ln 0.6.
−
0.002 1.0345+t − 1.0345
ln 1.03
ln 0.6
1.0345+t −
(ln 0.6)(ln 1.03)
+ 1.0345 11.33129
0.002
ln 11.33129
45 + t 82.12674
ln 1.03
t 37.12674
(B)
3.
[Lessons 5 and 6] By the recursive formula (6.1)
e̊80 e̊80:10 + 10p 80 e̊90
Since mortality for the first ten years is unchanged, e̊80:10 and 10p 80 are unchanged. We only need to equate
e̊90 between the two mortality assumptions. Under uniform mortality through age 120, e̊90 15, whereas
1
under constant force mortality, e̊90 1/µ. We conclude that 1/µ 15, or µ 15
0.066667 . (C)
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PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 4–8
4. [Lesson 7] The probability of one probe surviving 2 31 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
5.
(D)
[Lesson 25] The probability that all three fail within 10 years is
30q x 30q y 30q z
(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]
e̊30:40 Z
Z
∞
0
0
80
tpx y
dt
!
80 − t −0.02t
e
dt
80
Z
80
1 80 − t −0.02t 80
1
e
−
e −0.02t dt
80 −0.02
0.02 0
0
80
50
1
−
(1 − e −0.02(80) )
80 0.02 0.02
25.0592
(B)
7.
!
[Lesson 25 and 29] The transition matrix with two independent lives looks like this:
p00
*. 0
..
0
,0
p01
p11
0
0
p02
0
p22
0
0
p02 +/
/
p01 /
1 -
The probability of death for each life must be the same before and after the death of the other, so p01 , the
probability of the death of ( y ) , must also be p23 , and p02 , the probability of the death of ( x ) , must also be
p13 . Thus I is true. While p12 p21 0, this is true even if the lives are not independent, so II is false. III is
false, 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 conditional
probabilities of decrement (3) given survival to the beginning of each year, so we need probabilities of
survival to the beginning of years 2 and 3, t p 00
x , t 1, 2.
p x( τ ) 1 − q x(1) − q x(2) − q x(3) 1 − 0.01 − 0.09 − 0.01 0.89
(τ)
2p x
(1)
(2)
(3)
− q x+1
− q x+1
0.89 (1 − 0.01 − 0.06 − 0.01) 0.8188
0.89 1 − q x+1
Now we calculate the probabilities of decrement (3) in years 2 and 3.
q (3)
1| x
q (3)
2| x
q (3)
1|2 x
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(0.89)(0.01) 0.0089
(0.8188)(0.04) 0.032752
0.0089 + 0.032752 0.041652
(A)
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PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 9–13
9.
[Lesson 28] The expected amount of time to withdrawal is the integral
R
Z
∞
0
(τ)
(2)
t t p x µ x+t dt Z
∞
0
0.05te −0.06t dt
∞
and this is 5/6 of 0 0.06te −0.06t dt, 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 withdrawal
is the ratio of the µ’s, or 0.05/ (0.05 + 0.01) 5/6. So the expected amount of time to the next decrement
given that it is a withdrawal is
5
6 (1/0.06)
16 32
(B)
5/6
10.
[Lesson 10] Since E[Z1 Z2 ] 0, the covariance is − E[Z1 ] E[Z2 ].
E[Z1 ] E[Z2 ] Z
Z
10
0
∞
20
e −0.06t 0.02 dt 1 − e −0.6
0.150396
3
e −0.06t 0.02 dt e −1.2
0.100398
3
Cov ( Z1 , Z2 ) − (0.150396)(0.100398) −0.015099
(D)
11. [Lesson 14] Let Y1 be the value of the benefits if death occurs after age 65 and Y2 the value of the
benefits if death occurs before age 65.
30 E 35
10 E35 20 E45 (0.54318)(0.25634) 0.139239
E[Y1 ] 10030 E35 ä65 100 (0.139239)(9.8969) 137.803
E[Y2 ] 1000 ( A35 − 30 E35 A65 )
1000 0.12872 − 0.139239 (0.43980) 67.483
E[Y] 137.803 + 67.483 205.286
12.
(C)
[Lesson 20] Let P be the annual net premium.
A45:20 1 − d ä45:20
1
A45
:20
1−
1
21 (12.6)
0.4
A45:20 − 20 E45 0.4 − 0.3 0.1
1
500A45
:20 + 500A45
P
ä 45:20
500 (0.1) + 500 (0.2)
11.9048
12.6
(B)
13. [Lesson 29] Let p be the probability that the copier was working at the beginning of the previous
day. Then
0.8p + 0.6 (1 − p ) 0.77
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PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 14–15
0.6 + 0.2p 0.77
0.17
p
0.85
0.2
14.
(E)
[Lesson 29] The state vectors vn at time n before the transition are
v1 1
0
v2 1
0
0.8
0 *.0.2
,0
0
v3 0.8
0.1
v4 0.5
0.22
v5 0.244
vn 0
0
0.6
0.1 *.0.2
,0
0.2
0.6
0
0.2
0.2+/ 0.5
1-
0.1
0.2
0.474 *.0.2
,0
0.4
0.6
0
n>5
0.22
0.4 0.3 0.3
0.28 *.0.2 0.6 0.2+/ 0.244
0
1,0
0.282
1
0.1 0.1
0.6 0.2+/ 0.8 0.1
0
1-
0.28
0.282 0.474
0.4
0.2+/ 0.1052
1-
0.2668 0.628
Expected amount of time in State 1 is the sum of the first 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
Start of year 3
0.80
0.15
0.80
0.05
0.15
Start of year 4
0.715
0.180
0.80 0.15 0.05
0.05 *.0.50 0.40 0.10+/ 0.715
0
1 , 0
0.80 0.15 0.05
0.105 *.0.50 0.40 0.10+/ 0.662
0
1 , 0
0.180 0.105
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 first three years are the differences in the probabilities of being in the
third 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. In
year 4, the probability of death is 0.662 (0.05) + 0.17925 (0.1) 0.051025. The present value of the benefits
is
0.05 0.055 0.05375 0.051025
+
+
1000
+
185.9153
1.05 1.052
1.053
1.054
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PRACTICE EXAM 1, SOLUTION TO QUESTION 15
The net premium is π 185.9153/2.9823 62.34 . (C)
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