For a certain company, losses follow a Poisson frequency distribution with mean 2 per year, and the amount of a loss is 1, 2, or 3, each with probability 1/3. Loss amounts are independent of the number of losses, and of each other.

An insurance policy covers all losses in a year, subject to an annual aggregate deductible of 2.

Calculate the expected claim payments for this insurance policy.

• 2.00
• 2.36
• 2.45
• 2.81
• 2.96

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

Key: B

Let S denote aggregate loss before deductible.

[[math]] \operatorname{E}(S) = 2(2) = 4, [[/math]]

since mean severity is 2.

[[math]] f_S(0) = \frac{e^{-2}2^0}{0!} = 0.1353, [[/math]]

since must have 0 losses to get aggregate loss = 0.

[[math]] f_S(1) = \frac{}{} \frac{1}{3} = 0.0902, [[/math]]

since must have 1 loss whose size is 1 to get aggregate loss = 1.

[[math]] \begin{aligned} &\operatorname{E}[ S \wedge 2) = 0 f S (0) + 1 f S (1) + 2[1 − f S (0) − f S (1)] \\ &= 0(0.1353) + 1(0.0902) + 2(1 − 0.01353 − 0.0902) = 1.6392 \\ &\operatorname{E}[(S − 2)+ ] = \operatorname{E}[ S ) − \operatorname{E}[ S \wedge 2) = 4 − 1.6392 = 2.3608 \end{aligned} [[/math]]

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