An insurance company insures a large number of drivers. Let [math]X[/math] be the random variable representing the company’s losses under collision insurance, and let [math]Y[/math] represent the company’s losses under liability insurance. [math]X[/math] and [math]Y[/math] have joint density function

[[math]] f(x,y) = \begin{cases} \frac{2x-2-y}{4}, \, 0 \lt x \lt 1 \,\, \textrm{and} \,\, 0 \lt y \lt 2 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the probability that the total company loss is at least 1.

• 0.33
• 0.38
• 0.41
• 0.71
• 0.75

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

Solution: D

We want to find [math]\operatorname{P}[X + Y \gt 1][/math] . To this end, note that [math]\operatorname{P}[X + Y \gt 1][/math]

[[math]] \begin{align*} \int_0^1 \int_{1-x}^2 \left [\frac{2x+2-y}{4}\right] dy dx &= \int_0^1 \left[\frac{1}{2} xy + \frac{1}{2}y + \frac{1}{8}y^2 \right]^2_{1-x} \, dx \\ &= \int_0^1 \left [ x + 1 - \frac{1}{2} - \frac{1}{2} x(1-x) - \frac{1}{2} (1-x) + \frac{1}{8}(1-x)^2 \right] dx \\ &= \int_0^1 \left [ x + \frac{1}{2}x^2 + \frac{1}{8} - \frac{1}{4} x + \frac{1}{8} x^2\right ] \, dx \\ &= \int_0^1 \left [ \frac{5}{8}x^2 + \frac{3}{4}x + \frac{1}{8}\right ] \, dx \\ &= \left [ \frac{5}{24} x^3 + \frac{3}{8}x^2 + \frac{1}{8}x\right ]_0^1 \\ &= \frac{5}{24} + \frac{3}{8} + \frac{1}{8} \\ &= \frac{17}{24}. \end{align*} [[/math]]

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