Let [math]X[/math] be a continuous random variable with density function

[[math]] f(x) = \begin{cases} \frac{p-1}{x^p}, \, x \gt 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate the value of [math]p[/math] such that [math]\operatorname{E}(X) = 2 [/math].

• 1
• 2.5
• 3
• 5
• There is no such [math]p[/math]

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

Solution: C

[[math]] \begin{align*} \operatorname{E}[X] = \int_{1}^{\infty} x \frac{p-1}{x^p} dx &= (p-1) \int_{1}^{\infty} x^{1-p} dx \\ &= (p-1) \frac{x^{2-p}}{2-p} \Big |_1^{\infty} \\ &= \frac{p-1}{p-2} = 2. \end{align*} [[/math]]

Hence [math]p = 3[/math].

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