Computing Expected Values for Discrete Random Variables: Practice Problems and Solutions

This section of sample problems and solutions is a part of The Actuary's Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 12 of the Study Guide. See an index of all sections by following the link in this paragraph.

The following theorem, as given by Bowers, Gerber, et. al., is useful for computing the expected values of continuous random variables.

Theorem 12.1: "If K is a discrete random variable with probability only on the non-negative integers, with cumulative distribution function (c. d. f.) G(k) and probability density function (p. d. f.) g(k) = ΔG(k-1), and z(k) is a nonnegative, monotonic function such that E[z(K)] exists, then

E[z(K)] = _k=0^∞Σz(k)*g(k) = z(0) + _k=0^∞Σ[1 - G(k)]Δz(k).

When K is the curate-future-lifetime of a life (x), then we have E[K] = e_x = _k=0^∞Σ_k+1p_x.

Note that E[K] can also be referred to as e_x.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 64-65.

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

Problem S3L12-1. Prxqaks have a 0.36 probability of senselessly perishing (i.e., dying) on their 3^rd birthday, 0.41 probability of dying on their 7^th birthday, and 0.23 probability of dying on their 13^th birthday. They cannot die at any other time. What is the expected value for the age of death of a prxqak?

Solution S3L12-1. We are trying to find E[k]. We use the formula E[z(K)] = _k=0^∞Σz(k)*g(k) = _k=0^∞Σk*g(k) in this case. g(k) is a discrete p.d.f. with

g(k) = 0.36 for k = 3;

g(k) = 0.41 for k = 7;

g(k) = 0.23 for k = 13.

Thus, E[k] = _k=0^∞Σk*g(k) = 3*0.36 + 7*0.41 + 13*0.23 = 6.94 years.

Problem S3L12-2. Flying goblins can only die on their 2^nd, 3^rd, and 6^th birthdays. For each of the times t when it can die, the probability that a particular flying goblin will senselessly perish is 1/t. Find E[t²].

Solution S3L12-2. We use the formula E[z(K)] = _k=0^∞Σz(k)*g(k), where K = T and z(t) = t² and g(t) = 1/t for t = 2, 3, and 6. Thus, E[t²] = 2²(1/2) + 3²(1/3) + 6²(1/6) = 2 + 3 + 6 = E[t²] = 11