Aggregate losses are modeled as follows:

1. The number of losses has a Poisson distribution with [math]\lambda = 3 [/math]
2. The amount of each loss has a Burr distribution with [math]\alpha = 3, \theta = 2, \gamma = 1 [/math] .
3. The number of losses and the amounts of the losses are mutually independent.

Calculate the variance of aggregate losses.

• 12
• 14
• 16
• 18
• 20

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

Key: A

Let S = aggregate losses, X = severity

Since the frequency is Poisson,

[[math]] \begin{aligned} \operatorname{E}(S) - \lambda \operatorname{E}(X^2) \\ \operatorname{E}(X^2) = \frac{2^2 \Gamma(3) \Gamma(1) }{\Gamma(3)} = 4 \, \textrm{(table lookup)} \\ \operatorname{E}(S) = 3(4) = 12 \end{aligned} [[/math]]

You would get the same result if you used

[[math]] \operatorname{E}[ S ] = \operatorname{E}[ N ]\operatorname{E}[ X ] + \operatorname{E}[ N ] \operatorname{E}[ X ]^2 [[/math]]

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