A company has purchased a policy that will compensate for the loss of revenue due to severe weather events. The policy pays 1000 for each severe weather event in a year after the first two such events in that year. The number of severe weather events per year has a Poisson distribution with mean 1.

Calculate the expected amount paid to this company in one year.

• 80
• 104
• 368
• 512
• 632

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

Solution: B

The payment random variable is 1000([math]X[/math] – 2) if positive, where [math]X[/math] has a Poisson distribution with mean 1. The expected value is

[[math]] \begin{align*} 1000 \sum_{x=3}^{\infty} (x-2) \frac{e^{-1}}{x!} &= 1000 \left( \sum_{x=0}^{\infty} (x-2) \frac{e^{-1}}{x!} - \sum_{x=0}^2 (x-2) \frac{e^{-1}}{x!}\right ) \\ &= 1000(1-2-[-2e^{-1}-e^{-1}]) \\ &= 1000(-1 + 3e^{-1}) \\ &= 104. \end{align*} [[/math]]

Note the first sum splits into the expected value of [math]X[/math], which is 1, and 2 times the sum of the probabilities (also 1).

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