Computing Expected Values for Continuous 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 9.1: "If T is a continuous type random variable with cumulative distribution function (c.d.f.) G(t) such that G(0) = 0 and probability density function (p.d.f.) G'(t) = g(t), and z(t) is such that it is a nonnegative, monotonic, differentiable function and E[z(T)] exists, then

E[z(T)] = ₀^∞∫z(t)g(t)dt = z(0) + ₀^∞∫z'(t)[1 - G(t)]dt."

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. p. 62.

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

Problem S3L9-1. The random variable T has a probability density function of g(t) = (1/76)e^-(t/76). Find E[t].

Solution S3L9-1. Here, our z(t) = t and so we can use the formula E[z(T)] = ₀^∞∫z(t)g(t)dt =

₀^∞∫t(1/76)e^-(t/76)dt. We use the tabular method of integration by parts:

Sign-----u-----dv

+...(1/76)t....e^-(t/76)

-....(1/76)....-76e^-(t/76)

+........0......5776e^-(t/76)

Thus, E[x] = -te^-(t/76) -76e^-(t/76)│₀^∞ = - 0 - 0 + 0 + 76 = E[x] = 76.

Problem S3L9-2. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Find E[x³+5].

Solution S3L9-2. We use the formula E[z(T)] =z(0) + ₀^∞∫z'(t)[1 - G(t)]dt. Here, T = X and z(X) = x³+5, so z(0) = 5. z'(x) = 3x². G(x) is the cumulative distribution function, which is one minus the survival function. Thus, [1 - G(x)] = s(x) = e^-0.34x. Hence, E[x³+5] = 5 + ₀^∞∫3x²e^-0.34xdx.

We use the tabular method of integration by parts:

Sign-----u-----dv

+.......3x².....e^-0.34x

-........6x....-(50/17)e^-0.34x

+.......6......(2500/289)e^-0.34x

-........0......-(125000/4913)e^-0.34x

Thus, E[x³+5] = 5 + (-3x²(50/17)e^-0.34x - 6x(2500/289)e^-0.34x - 6(125000/4913)e^-0.34x│₀^∞ ) =

5 + 6(125000/4913) = about 157.6562182

Problem S3L9-3. 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 E[x⁶ + 2].

Solution S3L9-3. We use the formula E[z(T)] =z(0) + ₀^∞∫z'(t)[1 - G(t)]dt. Here, T = X and

z(X) = x⁶ + 2, so z(0) = 2. z'(x) = 6x⁵. G(x) is the cumulative distribution function, which is one minus the survival function. Thus, [1 - G(x)] = s(x) = 1 - x/94, for 0 ≤ x ≤ 94 and 0 otherwise. Hence,

E[x⁶ + 2] = 2 + ₀⁹⁴∫6x⁵(1 - x/94)dx = 2 + ₀⁹⁴∫(6x⁵ - 3x⁶/47)dx = 2 + (x⁶ - 3x⁷/329)│₀⁹⁴ =

2 + (94⁶ - 3*94⁷/329) = about 98,552,825,867.

Problem S3L9-4. 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 expected value of the lifetime of a three-headed donkey?

Solution S3L9-4. We use the formula E[z(T)] = ₀^∞∫z(t)g(t)dt. Here, X = T and s(x) = 1/x, so G(x) = 1 - 1/x, and g(x) = 1/x². Since z(x) = x and every donkey survives to age 1, it follows that E(x) = ₁^∞∫(x/x²)dx = ₁^∞∫(1/x)dx = ln(x)│₁^∞ = ln(∞) - ln(1) = ∞ - 0 = ∞. So the expected value of the lifetime of a three-headed donkey is infinite. Even though some three-headed donkeys will die, this is still the case, because some small number of donkeys will get to live forever - since the survival function asymptotically approaches zero, but never gets there.

Problem S3L9-5. Zigzag-striped plankton never survive past the age of 2. The cumulative distribution function for the lives of zigzag-striped plankton is as follows: G(t) = t³/8, 0 ≤ t ≤ 2, 0 otherwise. What is the expected value of the lifetime of a zigzag-striped plankton?

Solution S3L9-5. We use the formula E[z(T)] = ₀^∞∫z(t)g(t)dt.Here, our upper bound is 2 rather than infinity, and z(t) = t. Moreover, g(t) = G'(t) = 3t²/8. Thus, E[t] = ₀²∫3t³/8dt = 3t⁴/32│₀² = 3*16/32 =

E[t] = 3/2

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