Force of Mortality: Practice Problems and Solutions

This section will explore the force of mortality, alternatively known as the failure rate or the hazard rate function. It is abbreviated as μ_x. The force of mortality describes an "instantaneous rate of death" and can be found via the following formulas:

μ_x = f(x)/(1 - F(x))

μ_x = -s'(x)/s(x)

The force of mortality is always greater than or equal to zero, since s(x) is always nonnegative and its derivative, s'(x), is always non-positive, so -s'(x) is always nonnegative.

Meaning of Terms:

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

F(x) is the cumulative distribution function of X.

s(x) is the survival function of X.

f(x) is the probability density function of X.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. p. 48-49.

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

Problem S3L4-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Find μ_x.

Solution S3L4-1. We use the formula μ_x = -s'(x)/s(x). Since s(x) = e^-0.34x, it follows that

s'(x) = -0.34e^-0.34x and so μ_x = (0.34e^-0.34x)/(e^-0.34x) = μ_x = 0.34

Problem S3L4-2. The life of a giant pin-striped cockroach has the following survival function associated with it: s(x) = 1 - x/94, for 0 ≤ x ≤ 94 and 0 otherwise. Find μ_x.

Solution S3L4-2. We use the formula μ_x = -s'(x)/s(x). Since s(x) = 1 - x/94, it follows that

s'(x) = -1/94 and so μ_x = (1/94)/(1 - x/94) = (1/94)/((94 - x)/94) = μ_x = 1/(94 - x)

Problem S3L4-3. 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 force of mortality for three-headed donkeys? Warning: Make sure you examine all possible ages!

Solution S3L4-3. Prior to age 1, no three-headed donkeys die, so the force of mortality is 0. (We can confirm this by recognizing that s(x) is 1 for x ≤ 1 and so s'(x) = 0 and thus μ_x = 0).

After age 1, s(x) is 1/x, so s'(x) = -1/x². We now apply the formula μ_x = -s'(x)/s(x) = (1/x²)/(1/x) = x/x² = μ_x = 1/x.

Thus, μ_x = 0, for 0 ≤ x ≤ 1 and μ_x = 1/x for x > 1.

Problem S3L4-4. 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 force of mortality for black swans? Warning: Make sure you examine all possible ages!

Solution S3L4-4. Prior to age 16, no black swans die, so the force of mortality for x ≤ 16 is 0.

After age 16, F(x) = 1 - 4x^-1/2, and f(x) = F'(x) = (-1/2)(-4)x^-3/2 = f(x) = 2x^-3/2. So we can use the formula μ_x = f(x)/(1 - F(x)) = (2x^-3/2)/(4x^-1/2) = 1/(2x).

Thus, μ_x = 0, for 0 ≤ x ≤ 16 and μ_x = 1/(2x) for x > 16.

Problem S3L4-5. 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. What is the force of mortality for Unicorn-Pegasi? Warning: Make sure you examine all possible ages!

Solution S3L4-5. For all x < 35, Unicorn-Pegasi are still mortal. It is given that F(x) = x/36 - e^-0.45x, and so f(x) = F'(x) = 1/36 + 0.45e^-0.45x. So we can use the formula

μ_x = f(x)/(1 - F(x)) = (1/36 + 0.45e^-0.45x)/(1 - x/36 + e^-0.45x) =

(1 + 16.2e^-0.45x)/(36 - x + 36e^-0.45x).

After reaching age 35, Unicorn-Pegasi no longer die, so their force of mortality becomes 0.

Thus, μ_x = (1 + 16.2e^-0.45x)/(36 - x + 36e^-0.45x), for 0 ≤ x < 35 and μ_x = 0 for x ≥ 35.

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