An insurance company sells a one-year automobile policy with a deductible of 2. The probability that the insured will incur a loss is 0.05. If there is a loss, the probability of a loss of amount [math]N[/math] is [math]KN[/math], for [math]N = 1, . . . , 5 [/math] and [math]K[/math] a constant. These are the only possible loss amounts and no more than one loss can occur.

Calculate the expected payment for this policy

• 0.031
• 0.066
• 0.072
• 0.110
• 0.150

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

Solution: A

Let us first determine [math]K[/math]. Observe that

[[math]] 1 = K(1 + 1/2 + 1/3 + 1/4 + 1/5) = K\frac{60 + 30 + 20 + 15 + 12}{60} = K \frac{137}{60}. [[/math]]

Hence [math]K = \frac{60}{137}[/math]. It then follows that

[[math]] \begin{align*} \operatorname{P}[ N = n ] &= \operatorname{P}[ N = n | \textrm{Insured Suffers a Loss}] \operatorname{P}[ \textrm{Insured Suffers a Loss} ] \\ &= \frac{60}{137N}(0.05) = \frac{3}{137N}, \, N = 1,\ldots,5 \end{align*} [[/math]]

Now because of the deductible of 2, the net annual premium [math]P = \operatorname{E}[X] [/math] where

[[math]] X = \begin{cases} 0, \quad N \leq 2 \\ N-2, \quad N \gt 2 \end{cases} [[/math]]

Then,

[[math]] P = \operatorname{E}[X] = \sum_{N=3}^5 (N-2) \frac{3}{137N} = 0.0314. [[/math]]

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