Exercise


May 13'23

Answer

Key: A

Under the initial assumptions, losses, [math]X[/math], have [math]\operatorname{E}(X) = \exp(6 + \frac{1.5^2}{2}) = 1242.65 [/math]

After 5 years: exponential distribution with mean 1242.65(1.045 ) = 1511.87 and LER = 0.33, so

[[math]] 0.33 = \frac{\operatorname{E}[Y \wedge d )}{\operatorname{E}(Y)} = \frac{\theta(1-e^{-d/\theta}}{\theta} = 1- e^{-d/1511.87} \Rightarrow d = −1511.87 \ln(1 − 0.33) = 605.47 [[/math]]

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

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