A manufacturer’s annual losses follow a distribution with density function

[[math]] f(x) = \begin{cases} \frac{2.5(0.6)^{2.5}}{x^{3.5}}, \, x \gt 0.6 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

To cover its losses, the manufacturer purchases an insurance policy with an annual deductible of 2. Calculate the mean of the manufacturer’s annual losses not paid by the insurance policy.

• 0.84
• 0.88
• 0.93
• 0.95
• 1.00

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

Solution: C

The expected payment is

[[math]] \begin{align*} \int_{0.6}^2 x [\frac{2.5(0.6)^{2.5}}{x^{3.5}}] \, dx + \int_2^{\infty} [\frac{2.5(0.6)^{2.5}}{x^{3.5}}] \, dx &= 2.5(0.6)^{3.5} \left( \frac{-x^{1.5}}{1.5} \Big |_{0.6}^{2} + \frac{-x^{2.5}}{2.5} \Big |_2^{\infty} \right) \\ &= 2.5(0.6)^{2.5} \left ( \frac{-2^{-1.5}}{1.5} + \frac{0.6^{-1.5}}{1.5} + 2 \frac{2^{-2.5}}{2.5}\right ) \\ &= 0.9343. \end{align*} [[/math]]

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