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

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

Problem S3L12-3. The cumulative distribution function for the lives of brontosauruses is as follows:

G(k) = 0 for k ≤ 3;

G(k) = 0.2 for 3 < k ≤ 19;

G(k) = 0.3 for 19 < k ≤ 34;

G(k) = 0.33 for 34 < k ≤ 36;

G(k) = 0.6 for 36 < k ≤ 55;

G(k) = 0.9 for 55 < k ≤ 77;

G(k) = 1 for k > 77;

Find E[k³ + 2].

Solution S3L12-3. We use the formula

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

Here, z(0) = 2, z(k) = k³ + 2, and G(k) is as given above.

Thus, we have

E[k³ + 2] = 2 + Δz(3) + [1 - 0.2]Δz(19) + [1 - 0.3]Δz(34) + [1 - 0.33]Δz(36) + [1 - 0.6]Δz(55) + [1 - 0.9]Δz(77)

E[k³ + 2] = 2 + Δz(3) + 0.8Δz(19) + 0.7Δz(34) + 0.67Δz(36) + 0.4Δz(55) + 0.1Δz(77)

z(3) = 29, so Δz(3) = 29 - 2 = 27

z(19) = 6861, so Δz(19) = 6861 - 29 = 6832

z(34) = 39306, so Δz(34) = 39306 - 6861 = 32445

z(36) = 46658, so Δz(36) = 46658 - 39306 = 7352

z(55) = 166377, so Δz(55) = 166377 - 46658 = 119719

z(77) = 456535, so Δz(77) = 456535 - 166377 = 290158

Thus, E[k³ + 2] = 2 + 27 + 0.8*6832 + 0.7*32445 + 0.67*7352 + 0.4*119719 + 0.1*290158

E[k³ + 2] = 110035.31

Problem S3L12-4. 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 e₄ for Prxqaks? (Note: assume that _k+1p_x takes into account lives that have survived to the (k+1)st birthday but will die on said birthday, immediately after survival statistics are compiled.)

Solution S3L12-4. We use the formula e_x = _k=0^∞Σ_k+1p_x. In this case, the relevant upper bound of the summation is 8, since no prxqaks live beyond 4+(8+1) = 13 years.

Thus, e₄ = ₁p₄ + ₂p₄ + ₃p₄ + ₄p₄ + ₅p₄ + ₆p₄ + ₇p₄ + ₈p₄ + ₉p₄

At age 4, 1-0.36 = 0.64 of the original cohort of newborn prxqaks is alive.

No prxqak dies prior to age 7, so

₁p₄ = ₂p₄ = ₃p₄ = 1

But at age 7, only 0.23 of the original population or 0.23/0.64 = 0.359375 of the population alive at age 4 remains alive. No prxqak dies thereafter, until age 13, when all remaining prxqaks die. Thus,

₄p₄ = ₅p₄ = ₆p₄ = ₇p₄ = ₈p₄ = ₉p₄ = 0.359375.

Hence, e₄ = 3(1) + 6(0.359375) = e₄ = 5.15625 years

Problem S3L12-5. 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 e₁. (Note: assume that _k+1p_x takes into account lives that have survived to the (k+1)st birthday but will die on said birthday, immediately after survival statistics are compiled.)

Solution S3L12-5. We use the formula e_x = _k=0^∞Σ_k+1p_x. In this case, the relevant upper bound of the summation is 4, since no prxqaks live beyond 1+(4+1) = 6 years.

Thus, e₁ = ₁p₁ + ₂p₁ + ₃p₁ + ₄p₁ + ₅p₁

At age 2, ½ of the goblins will die, so ₁p₁ = 1 and ₂p₁ = 1/2

At age 2, another 1/3 of the goblins will die. No further goblins die until age 6, when the remaining goblins die. So ₃p₁ = ₄p₁ = ₅p₁ = ½ - 1/3 = 1/6.

Thus, e₁ = 1 + ½ + 3(1/6) = e₁ = 2 years.

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