A baseball team has scheduled its opening game for April 1. If it rains on April 1, the game is postponed and will be played on the next day that it does not rain. The team purchases insurance against rain. The policy will pay 1000 for each day, up to 2 days, that the opening game is postponed. The insurance company determines that the number of consecutive days of rain beginning on April 1 is a Poisson random variable with mean 0.6. Calculate the standard deviation of the amount the insurance company will have to pay.

• 668
• 699
• 775
• 817
• 904

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

Solution: B

The amount of money the insurance company will have to pay is defined by the random variable

[[math]] Y = \begin{cases} 1000x, \quad x \lt 2 \\ 2000, \quad x \geq 2 \end{cases} [[/math]]

where x is a Poisson random variable with mean 0.6 . The probability function for X is

[[math]] p(x) = \frac{e^{-0.6}(0.6)^k}{k!} \quad k = 0,1,2,3,\ldots [[/math]]

and

[[math]] \begin{align*} \operatorname{E}[Y] &= 0 + 1000 (0.6)e^{-0.6} + 2000 e^{-0.6} \sum_{k=2}^{\infty} \frac{0.6^k}{k!} \\ &= 1000 (0.6) e^{-0.6} + 2000 \left( e^{-0.6} \sum_{k=0}^{\infty} \frac{0.6^k}{k!} - e^{-0.6} - (0.6)e^{-0.6}\right) \\ &= 2000 e^{-0.6} \sum_{k=0}^{\infty} \frac{0.6}{k!} - 2000e^{-0.6} - 1000(0.6)e^{-0.6}\\ &= 2000 - 2000e^{-0.6} - 600e^{-0.6} \\ &= 573 \end{align*} [[/math]]

[[math]] \begin{align*} \operatorname{E}[Y^2] &= 1000^2 (0.6)e^{-0.6} + (2000)^2 e^{-0.6} \sum_{k=2}^{\infty} \frac{0.6^k}{k!} \\ &= (2000)^2 e^{-0.6} \sum_{k=0}^{\infty} \frac{0.6^k}{k!} - (2000)^2e^{-0.6} - \left[ (2000)^2 - (1000)^2\right](0.6)e^{-0.6} \\ &= (2000)^2 - (2000)^2e^{-0.6}-[(2000)^2 - (1000)^2](0.6)e^{-0.6} \\ &= 816,893 \end{align*} [[/math]]

[[math]] \operatorname{Var}[Y] = \operatorname{E}[Y^2 ] − { \operatorname{E}[Y ]} = 816,893 − ( 573) = 488,564, \, \sqrt{\operatorname{Var}[Y]} = 699. [[/math]]

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