An insurance policy is written to cover a loss [math]X[/math] where [math]X[/math] has density function

[[math]] f(x) = \begin{cases} \frac{3}{8}x^2, \, 0 \leq x \leq 2 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

The time (in hours) to process a claim of size [math]x[/math], where [math]0\lt x \lt 2[/math] , is uniformly distributed on the interval from [math]x[/math] to [math]2x[/math].

Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more.

• 0.17
• 0.25
• 0.32
• 0.58
• 0.83

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

Solution: A

We are given that [math]X[/math] denotes loss. In addition, denote the time required to process a claim by T.

Then the joint pdf of [math]X[/math] and [math]T[/math] is

[[math]] f(x,t) = \begin{cases} \frac{3}{x}x^2 \frac{1}{x} = \frac{3}{8}x, \, x \lt t \lt 2x, 0 \leq x \leq 2 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Now we can find

[[math]] \begin{align*} P[T \geq 3] = \int_2^4 \int_{t/2}^2 \frac{3}{8} x dx dt = \int_3^4 \left[ \frac{3}{16}x^2\right]_{t/2}^2 dt &= \int_3^4 \left( \frac{12}{16} - \frac{3}{64}t^2\right) dt\\ &= \left [ \frac{12}{16} - \frac{1}{64}t^3\right ]_3^4 \\ &= \frac{12}{4} - 1 - \left( \frac{36}{16} - \frac{27}{64}\right) \\ &= \frac{11}{64} = 0.17 \end{align*} [[/math]]

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