The loss [math]L[/math] due to a boat accident is exponentially distributed. Boat insurance policy A covers up to 1 unit for each loss. Boat insurance policy B covers up to 2 units for each loss. The probability that a loss is fully covered under policy B is 1.9 times the probability that it is fully covered under policy A.

Calculate the variance of [math]L[/math].

• 0.1
• 0.4
• 2.4
• 9.5
• 90.1

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

Solution: E

The distribution function of [math]L[/math] is [math]F(x) = 1-e^{-\lambda x} [/math] and its variance is [math]1/\lambda^2 [/math]. We are given

[[math]] \begin{align*} 1-e^{-2\lambda} = 1.9(1-e^{-\lambda}) \\ (1-e^{-\lambda})(1+e^{-\lambda}) = 1.9(1-e^{-\lambda}) \\ e^{-\lambda} = 0.9 \\ \lambda = -\ln(0.9) = 0.10536 \\ \operatorname{Var}(L) = 1/0.10536^2 = 90.1 \end{align*} [[/math]]

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