The number of annual losses has a Poisson distribution with a mean of 5. The size of each loss has a Pareto distribution with [math]\theta = 10 [/math] and [math] \alpha = 2.5 [/math]. An insurance policy for the losses has an ordinary deductible of 5 per loss.

Calculate the expected value of the aggregate annual payments for this policy.

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

Key: C

Let [math]X[/math] be the loss random variable. Then, [math]( X − 5)_+[/math] is the claim random variable.

[[math]] \begin{aligned} &\operatorname{E}(X) = \frac{10}{2.5-1} = 6.667 \\ &\operatorname{E}(X \wedge 5) = \left( \frac{10}{2.5 -1}\right) \left [ 1-(\frac{10}{5+10})^{2.5-1}\right] = 3.038 \\ &\operatorname{E}[( X − 5)_+ ] = \operatorname{E}[ X ) − \operatorname{E}[ X \wedge 5) = 6.667 − 3.038 = 3.629 \end{aligned} [[/math]]

Expected aggregate claims =

[[math]]\operatorname{E}( N ) \operatorname{E}[( X − 5) + ] = 5(3.629) = 18.15.[[/math]]

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