Median Future Lifetime: Practice Problems and Solutions

m(x) is the median future lifetime of life (x). It can be determined by solving either of the following equations:

Pr[T(x) > m(x)] = 1/2

s[x + m(x)]/s(x) = 1/2

Meaning of terms:

T(x) = future lifetime of the life that has already reached the age of x.

s(x) = the survival function for life (x).

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. p. 63.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L11-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Find m(24).

Solution S3L11-1. We use the formula s[x + m(x)]/s(x) = ½. Thus,

e^{-0.34(24+m(24))}/e^-0.34(24) = ½

e^-0.34(m(24)) = ½

-0.34*m(24) = ln(1/2)

m(24) = ln(1/2)/-0.34 = m(24) = 2.038668178 years.

Problem S3L11-2. Zigzag-striped plankton never survive past the age of 2. The cumulative distribution function for the lives of zigzag-striped plankton is as follows: G(t) = t³/8, 0 ≤ t ≤ 2, 0 otherwise. Find m(1).

Solution S3L11-2. We use the formula s[x + m(x)]/s(x) = ½. We can find s(x) = 1 - G(t) = 1 - t³/8.

Thus, s[1 + m(1)]/s(1) = ½

(1 - [1 + m(1)]³/8)/(1 - 1³/8) = ½

(1 - (1+ m(1))³/8)/(7/8) = ½

(8 - (1+ m(1))³)/7 = ½

(8 - (1+ m(1))³) = 7/2

(1+ m(1))³ = 9/2

1 + m(1) = 1.6509636244

m(1) = about 0.6509636244 years

Problem S3L11-3. Black swans always survive until age 16. After age 16, the lifetime of a black swan can be modeled by the cumulative distribution function F(x) = 1 - 4x^-1/2, x > 16. What is the median age to which newborn black swans live?

Solution S3L11-3. The median age to which newborn black swans live is the same as 16 plus the median age to which 16-year-old black swans live. We know that s(x) = 1 - F(x) = 4x^-1/2, x > 16, and we want to find [16 + m(16)]. We use the formula s[x + m(x)]/s(x) = ½.

s[16 + m(16)]/s(16) = ½

4[16 + m(16)]^-1/2/(4(16)^-1/2) = ½

[16 + m(16)]^-1/2/(16)^-1/2 = ½

[16 + m(16)]^-1/2/(1/4) = ½

[16 + m(16)]^-1/2 = 1/8

([16 + m(16)]^-1/2)^-2 = (1/8)^-2

[16 + m(16)] = 64 years.

Problem S3L11-4. Three-headed donkeys always survive until age 1. Thereafter, the survival function for the life of a three-headed donkey is s(x) = 1/x for all x > 1. What is the median age to which three-headed donkeys that have survived for three years live?

Solution S3L11-4. We want to find [3 + m(3)], since m(3) only tells us the median number of years to which three-year-old three-headed donkeys live after their third birthday.

We use the formula s[x + m(x)]/s(x) = ½.

s[3 + m(3)]/s(3) = ½

(1/[3 + m(3)])/(1/3) = ½

3/[3 + m(3)] = ½

We cross-multiply to get 3 + m(3) = 6 years.

Problem S3L11-5. You know the following about the lives of jumping slugs. For a jumping slug that has survived to age 39, the probability of surviving to age 45 is 0.835. For a jumping slug that has survived to age 45, the probability of dying within the subsequent 8 years is 0.23. A jumping slug that has survived to age 53 has the survival function s(x) = 1/x^1/2 for all x > 53. Find the median age to which jumping slugs that have survived for 39 years live.

Solution S3L11-5. We seek to find 39 + m(39), i.e., the age at which one-half of the slugs which were alive at age 39 will have died.

By age 45, 1 - 0.835 = 0.165 of the slug population alive at age 39 will have died.

By age 53, another 0.835*0.23 = 0.19205 of the slug population alive at age 39 will have died.

At age 53, only 1 - 0.165 - 0.19205 = 0.64295 of the slug population alive at age 39 is still alive.

We want to know the time at which 0.5 of the slug population alive at age 39 will be alive.

This is 0.5/0.64295 = 0.7776654483 of the slug population alive at age 53.

Thus, we modify our formula thus:

s[39 + m(39)]/s(53) = 0.7776654483

We have a tidy survival function, s(x) = 1/x^1/2, to work with for all x > 53. And we know that

39 + m(39) > 53, because half of the jumping slug population has not yet senselessly perished (i.e., died) at age 53. Thus,

[39 + m(39)]^-1/2/(53)^-1/2 = 0.7776654483

[39 + m(39)]^-1/2 = 0.1068205645

39 + m(39) = 87.63755693 years.

See other sections of The Actuary's Free Study Guide for Exam 3L.