Let [math]X[/math] represent the age of an insured automobile involved in an accident. Let [math]Y[/math] represent the length of time the owner has insured the automobile at the time of the accident. [math]X[/math] and [math]Y[/math] have joint probability density function

[[math]] f(x,y) = \begin{cases} \frac{10-xy^2}{64}, \,\, 2 \leq x \leq 10, 0 \leq y \leq 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected age of an insured automobile involved in an accident.

• 4.9
• 5.2
• 5.8
• 6.0
• 6.4

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

Solution: C

The marginal density of X is given by

[[math]] f_X(x) = \int_0^1 \frac{1}{64}(10-xy^2) dy = \frac{1}{64}(10y - \frac{xy^3}{3}) \Big |_0^1 = \frac{1}{64}(10-\frac{x}{3}). [[/math]]

Then

[[math]] \begin{align*} \operatorname{E}(X) = \int_2^{10}xf_X(x) dx &= \int_2^{10} \frac{1}{64}(10x - \frac{x^2}{3}) dx \\ &= \frac{1}{64}(5x^2 - \frac{x^3}{9})\Big |_2^{10} \\ &= \frac{1}{64} \left [ \left( 500 - \frac{1000}{9}\right ) - \left( 20 - \frac{8}{9}\right)\right ] \\ &= 5.778. \end{align*} [[/math]]

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