An insurer's annual weather-related loss, [math]X[/math], is a random variable with density function

[[math]] f(x) = \begin{cases} \frac{2.5(200)^{2.5}}{x^{3.5}}, \, x\gt200 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the difference between the 30^th and 70^th percentiles of [math]X[/math].

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

Solution: B

The distribution function of X is given by

[[math]] F(x) = \int_{200}^x \frac{2.5 (200)^{2.5}}{t^{3.5}} dt = \frac{-(200)^{2.5}}{t^{2.5}} \Big |_{200}^{x} = 1- \frac{(200)^{2.5}}{x^{2.5}}, \, x \gt 200. [[/math]]

Therefore, the [math]p^{th}[/math] percentile [math]x_p[/math] of [math]X[/math] is given by

[[math]] \begin{align*} \frac{p}{100} = F(x_p) = 1- \frac{(200)^{2.5}}{x_p^{2.5}} \\ 1-0.01p = \frac{(200)^{2.5}}{x_p^{2.5}} \\ (1-0.01p)^{2/5} = \frac{200}{x_p} \\ x_p = \frac{200}{(1-0.01p)^{2/5}}. \end{align*} [[/math]]

It follows that

[[math]] x_{70} - x_{30} = \frac{200}{(0.30)^{2/5}} - \frac{200}{(0.70)^{2/5}} = 93.06. [[/math]]

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