A random variable [math]X[/math] has the cumulative distribution function

[[math]] F(x) = \begin{cases} 0, \, x \lt 1 \\ \frac{x^2-2x +2}{2}, \, 1 \leq x \lt 2 \\ 1, \, x \geq 2. \end{cases} [[/math]]

Calculate the variance of [math]X[/math].

• 7/72
• 1/8
• 5/36
• 4/3
• 23/12

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

First note that the distribution function jumps ½ at x = 1, so there is discrete probability at that point. From 1 to 2, the density function is the derivative of the distribution function, x – 1. Then,

[[math]] \begin{align*} \operatorname{E}(X) = \frac{1}{2} + \int_{1}^2 x(x-1) dx = \frac{1}{2} + \int_{1}^2 (x^2-x) dx &= \frac{1}{2} + (\frac{1}{3}x^3 - \frac{1}{2} x^2) \Big |_1^2 \\ &= 1/2 + 8/3 - 4/2 - 1/3 + 1/2 \\ &= 7/3 -1 = 4/3 \end{align*} [[/math]]

[[math]] \begin{align*} \operatorname{E}(X^2) = \frac{1}{2} + \int_{1}^2 x^2(x-1) dx = \frac{1}{2} + \int_{1}^2 (x^3-x^2) dx &= \frac{1}{2} + (\frac{1}{4}x^4 - \frac{1}{3} x^3) \Big |_1^2 \\ &= 1/2 +16/4 - 8/3 - 1/4 + 1/3 \\ &= 17/4 -7/3 \\ &= 23/12 \end{align*} [[/math]]

[[math]] \operatorname{Var}(X) = \operatorname{E}[X^2] - (\operatorname{E}[X])^2 = \frac{23}{12} - \left( \frac{4}{3} \right)^2 = \frac{23}{12} - \frac{16}{9} = \frac{5}{36} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.