Under a liability insurance policy, losses are uniformly distributed on [math][0, b][/math], where [math]b[/math] is a positive constant. There is a deductible of [math]b/2[/math]. Calculate the ratio of the variance of the claim payment (greater than or equal to zero) from a given loss to the variance of the loss.

• 1:8
• 3:16
• 1:4
• 5:16
• 1:2

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

Solution: D

Let X represent the loss. The variance for a uniform distribution is the square of the interval length, divided by 12. Thus,

[[math]] \operatorname{Var}(X) = \frac{b^2}{12}. [[/math]]

Let [math]C[/math] represent the claim payment from the loss. Then [math]C = 0[/math] for [math]X \lt b/2 [/math] and [math]C = X – b/2[/math], otherwise. Then,

[[math]] \begin{align*} \operatorname{E}(C) &= \int_9^{b/2}0(1/b) dx + \int_{b/2}^b (x-b/2) (1/b) dx \\ &= 0 + (x-b/2)^2 /(2b) |_{b/2}^{b} = (b/2)^2 / (2b) = b/8. \end{align*} [[/math]]

[[math]] \begin{align*} \operatorname{E}(C^2) &= \int_9^{b/2}0^2(1/b) dx + \int_{b/2}^b (x-b/2)^2 (1/b) dx \\ &= 0 + (x-b/2)^3 /(3b) |_{b/2}^{b} = (b/2)^3 / (3b) = b^2/24. \end{align*} [[/math]]

[[math]] \operatorname{Var}(C) = b^2/24 - (b/8)^2 = 5b^2/192. [[/math]]

The ratio is

[[math]][5b^2 /192] / [b/12] = 60/192 = 5 /16.[[/math]]

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