Life Table Probability Functions: Practice Problems and Solutions: Part 3

If we have a particular group of lives, to which no new members are added, we can let l₀be the number of lives in the group at time 0. We can also let ℓ(x) be the number of survivors in this group at age x. Then the expected number of survivors E(ℓ(x)) can be expressed as follows:

E(ℓ(x)) = l_x = l₀*s(x)

If we assume that the indicators for the survival of each life are independent, then ℓ(x) has a binomial distribution with parameters n = l₀ and p = s(x). Recall that a binomial probability mass function is of the following form: p(r) = C(n, r)p^r(1-p)^n-r

We can also let _nD_x be the number of deaths between ages x and x + n and we can define the expected number of such deaths as E(_nD_x) =_nd_x. When n = 1, _nd_x is simply written as d_x.

The following formulas hold with respect to l_x and _nd_x:

_nd_x = l₀[s(x) - s(x + n)]

_nd_x = l_x - l_x+n

(-1/l_x)(dl_x/dx) = (-1/[s(x)])ds(x)/dx = μ_x

-dl_x = l_x*μ_x dx

Meaning of Terms:

X refers to the age of death of a particular life.

s(x) is the survival function of X.

μ_xis the force of mortality for life (x).

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. p. 52-53.

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

Problem S3L6-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Out of a group of 9035 newborn triceratopses, how many would you expect to remain alive at age 8? Round down to the nearest whole triceratops.

Solution S3L6-1. We want to find l₈, given l₀ = 9035 and s(x) = e^-0.34x. We use the formula l_x = l₀*s(x). Thus, l₈ = l₀*s(8) = 9035* e^-0.34*8 = 595.1784062 = about 595 triceratopses.

Problem S3L6-2. The lives of Unicorn-Pegasi can be modeled by the following cumulative distribution function: F(x) = x/36 - e^-0.45x. If any Unicorn-Pegasus reaches the age of 35, it will get to live forever. For a group of newborn 3612 Unicorn-Pegasi, find ₂d₇. Round down to the nearest whole Unicorn-Pegasus.

Solution S3L6-2. We first find s(x) = 1 - F(x) = 1 - x/36 + e^-0.45x. We are also given l₀ = 3612. So we use the formula _nd_x = l₀[s(x) - s(x + n)]. ₂d₇ = 3612[s(7) - s(9)] =

3612[(1 - 7/36 + e^-0.45*7) - (1 - 9/36 + e^-0.45*9)] = 292.5189317 = about 292 Unicorn-Pegasi.

Problem S3L6-3. 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. There is a cohort of 25 newborn black swans. Assuming that the survival of every black swan is independent of the survival of every other black swan, what is the probability that exactly 9 black swans of this group are alive at age 23? (Hint: The answer is extremely small.)

Solution S3L6-3. We want to find Pr(ℓ(23) = 9). We know that ℓ(23) follows a binomial distribution with n = l₀= 25 and p = s(23) = 1 - F(23) = 4*23^-1/2 = 0.8340576562.

We use the binomial probability mass function p(r) = C(n, r)p^r(1-p)^n-r, for r = 9.

p(9) = C(25, 9)*0.8340576562⁹*(1 - 0.8340576562)¹⁶

p(9) = about 1.319298986*10^-7

Problem S3L6-4. For a particular group of polka-dotted zebras, you are given l_x = 95/x⁴, for all x > 1.

Find μ_x for all x > 1.

Solution S3L6-4. We use the formula (-1/l_x)(dl_x/dx) = μ_x.

(dl_x/dx) = -4*95x^-5 = -380x^-5

Thus, μ_x = (-x⁴/95)(-380x^-5) = μ_x = 4/x, for x > 1.

Problem S3L6-5. The lives of Unicorn-Pegasi can be modeled by the following cumulative distribution function: F(x) = x/36 - e^-0.45x. If any Unicorn-Pegasus reaches the age of 35, it will get to live forever. For a group of 69 newborn Unicorn-Pegasi, find the probability that at least one of them will get to live forever.

Solution S3L6-5. We want to find 1 - Pr(ℓ(35) = 0). We know that ℓ(35) follows a binomial distribution with n = l₀= 69 and p = s(35) = 1 - 35/36 + e^-0.45*35 = p = 0.0277779223.

We use the binomial probability mass function p(r) = C(n, r)p^r(1-p)^n-r, for r = 0.

p(0) = C(69, 0)*p⁰(1-p)⁶⁹

p(0) = (1-p)⁶⁹

p(0) = (1-0.0277779223)⁶⁹

p(0) = 0.1431588013

Our desired answer is 1 - p(0) = 0.8568411987

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