For an insurance:

1. The number of losses per year has a Poisson distribution with [math]\lambda = 10 [/math] .
2. Loss amounts are uniformly distributed on (0, 10).
3. Loss amounts and the number of losses are mutually independent.
4. There is an ordinary deductible of 4 per loss.

Calculate the variance of aggregate payments in a year.

• 36
• 48
• 72
• 96
• 120

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

Key: C

Since loss amounts are uniform on (0, 10), 40% of losses are below the deductible (4), and 60% are above. Thus, claims occur at a Poisson rate [math]\lambda^* = 0.6(10) = 6 . [/math]

Since loss amounts were uniform on (0, 10), claims are uniform on (0, 6).

Let N = number of claims; X = claim amount; S = aggregate claims.

[[math]] \begin{aligned} & \operatorname{E}[ N ] = \operatorname{E}[ N ] = \lambda^* = 6\\ & \operatorname{E}[ X ] = (6 − 0) / 2 = 3\\ &\operatorname{E}[ X ] = (6 − 0)2 /12 = 3 \\ &\operatorname{E}[ S ] = \operatorname{E}[ N ]\operatorname{E}[ X ] + \operatorname{E}[ X ] 2\operatorname{E}[ N ] = 6(3) + 32 (6) = 72 \end{aligned} [[/math]]

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