A group insurance policy covers the medical claims of the employees of a small company. The value, [math]V[/math], of the claims made in one year is described by [math]V = 100,000Y[/math] where [math]Y[/math] is a random variable with density function

[[math]] f(y) = \begin{cases} k(1-y)^4, \, 0\lt y \lt 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

where [math]k[/math] is a constant. Calculate the conditional probability that [math]V[/math] exceeds 40,000, given that [math]V[/math] exceeds 10,000.

• 0.08
• 0.13
• 0.17
• 0.20
• 0.51

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

Solution: B

To determine [math]k[/math], note that

[[math]] 1 = \int_0^1 k(1-y)^4 dy = -\frac{k}{5}(1-y)^{5} \Big | _0 ^1 = \frac{k}{5}. [[/math]]

Hence [math] k = 5 [/math]. We need to find

[[math]] \begin{align*} \operatorname{P}[V \gt 10,000] &= \operatorname{P}[100,000 Y \gt 10,000] = \operatorname{P}[Y \gt 0.1] \\ &= \int_{0.1}^1 5(1-y)^4 dy = -(1-y)^5 \Big |_{0.1}^1 \\ &= (0.9)^5 \\ &= 0.59 \end{align*} [[/math]]

and

[[math]] \begin{align*} \operatorname{P}[V \gt 40,000] = \operatorname{P}[100,000 Y \gt 40,000] = \operatorname{P}[Y \gt 0.4] &= \int_{0.4}^1 5(1-y)^4 dy \\ &= -(1-y)^5 \Big |_{0.4}^1 \\ &= (0.6)^5 \\ &= 0.078. \end{align*} [[/math]]

It now follows that

[[math]] \begin{align*} \operatorname{P}[V \gt 40,000 | V \gt 10,000] &= \frac{\operatorname{P}[V \gt 40, 000 ∩ V \gt 10, 000]}{\operatorname{P}[V \gt 10, 000]} \\ &= \frac{\operatorname{P}[V \gt 40, 000]}{\operatorname{P}[V \gt 10, 000]} \\ &= \frac{0.078}{0.59}\\ &= 0.132. \end{align*} [[/math]]

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