Makeham's Law of Mortality: Practice Problems and Solutions

Makeham's law of mortality corresponds to the following survival distribution.

s(x) = exp[-Ax -m(c^x-1)]

The force of mortality under Makeham's Law is as follows.

µ_x = A + Bc^x

Under Makeham's Law, A, B, and c are given constants, and m is defined as m = B/ln(c)

The restrictions for using Makeham's Law are as follows:

B > 0, A ≥ -B, c > 1, x ≥ 0.

A special case of Makeham's Law is Gompertz's Law, where A = 0.

Thus, under Gompertz's law,s(x) = exp[-m(c^x-1)] and µ_x = Bc^x.

If c = 1 under Gompertz's and Makeham's laws, we get a simple exponential distribution.

A under Makeham's law could be interpreted as the accident hazard, while Bc^x could be interpreted as the hazard of aging.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 78.

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

Problem S3L24-1. The lives of three-horned rhinoceroses follow Makeham's law with an accident hazard of 0.31, B = 0.43, and c = 2. Find µ₁ for three-horned rhinoceroses.

Solution S3L24-1. We use the formula µ_x = A + Bc^x.

We are given that x = 1, A = 0.31, B = 0.43, and c = 2. Thus, µ₁ = 0.31 + 0.43*2¹ = µ₁ = 1.17.

Problem S3L24-2. The lives of three-horned rhinoceroses follow Makeham's law with an accident hazard of 0.31, B = 0.43, and c = 2. Find s(3) for three-horned rhinoceroses.

Solution S3L24-2. We use the formula s(x) = exp[-Ax -m(c^x-1)].

We are given that x = 3, A = 0.31, B = 0.43, and c = 2. Thus, m = B/ln(c) = 0.43/ln(2) = 0.6203588676.

Hence, s(3) = exp[-0.31*3 - 0.6203588676(2³-1)] = s(3) = about 0.005130705663.

Problem S3L24-3. The lives of orange orangutans follow Gompertz's law with s(40) = 0.65 and c = 1.03. Find µ₄₀ for orange orangutans.

Solution S3L24-3. We first find B using the formula

s(x) = exp[-m(c^x-1)] = exp[-B(c^x-1)/ln(c)]

We are given that s(40) = 0.65, x = 40, and c = 1.03. Thus,

0.65 = exp[-B(1.03⁴⁰-1)/ln(1.03)]

0.65 = exp[-76.52670678B]

ln(0.65) = -76.52670678B

B = ln(0.65)/-76.52670678

B = about 0.005629184.

µ₄₀ = Bc^x = 0.005629184*1.03⁴⁰ = about 0.0183626111.

Problem S3L24-4. The lives of pterodactyls follow Makeham's law with accident hazard of 0.23 and hazard of aging of 0.02 for 20-year-old pterodactyls and 0.06 for 60-year-old pterodactyls. Find s(20) for pterodactyls.

Solution S3L24-4. We are given that A = 0.23, Bc²⁰ = 0.02, and Bc⁶⁰ = 0.06

Thus, c⁴⁰ = (Bc⁶⁰)/(Bc²⁰)= 0.06/0.02 = 3. Thus, c = 3^1/40 = 1.027845956.

Hence, B = 0.02c^-20 = 0.02*1.027845956^-20 = B = 0.0115470054.

We now use the formula

s(x) = exp[-Ax - m(c^x-1)] = exp[-Ax - B(c^x-1)/ln(c)]

s(20) = exp[-0.23*20 - 0.0115470054(1.027845956²⁰-1)/ln(1.027845956)]

s(20) = exp[-4.907769891]

s(20) = about 0.0073889481.

Problem S3L24-5. The lives of jabberwocks follow Makeham's law with B = 0.1, c = 1.0003, and s(35) = 0.02. Find µ₅ for jabberwocks.

Solution S3L24-5. We use the formula µ_x = A + Bc^x and note that A is at present an unknown quantity. We try to find A by using the formula

s(x) = exp[-Ax - m(c^x-1)] = exp[-Ax - B(c^x-1)/ln(c)]

We are given s(35), so we work with the conditions for x = 35:

0.02 = exp[-35A - 0.1(1.0003³⁵-1)/ln(1.0003)]

0.02 = exp[-35A - 3.518436707]

ln(0.02) = -35A - 3.518436707

-35A = ln(0.02) + 3.518436707

A = [ln(0.02) + 3.518436707]/-35

A = 0.0112453228.

So µ₅ = A + Bc⁵ = 0.0112453228+ 0.1*1.0003⁵ = µ₅ = about 0.1113954128.
See other sections of The Actuary's Free Study Guide for Exam 3L.