A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and \( \$ 9300 \) for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business has a Poisson distribution with mean 1.9.
Find the expected amount paid to the company under this policy during a one-year period.
Answer
Calculate the expected amount paid under the policy during a one-year period: \(E[Paid] = \$ 9300 \cdot E[Y] = \$ 9300 \cdot (e^{-1.9} \cdot (e^{1.9} - 1.9 - 1)) = \$ 12127.5764\)
Step 1 ：Let X be the number of major snowstorms per year. X follows a Poisson distribution with \(\lambda = 1.9\). Then the expected number of snowstorms is equal to \(E[X] = 1.9\).
Step 2 ：Define Y as the number of major snowstorms with payments, so \(Y = max(0, X - 1)\). Calculate the expected value of Y: \(E[Y] = \sum_{y=0}^{\infty} y \cdot P(Y = y) = \sum_{y=1}^{\infty} (y-1) \cdot P(X = y)\)
Step 3 ：Calculate E[Y]: \(E[Y] = \sum_{y=2}^{\infty} (y - 1) \cdot \frac{e^{-1.9} \cdot 1.9^y}{y!} = e^{-1.9} \cdot \sum_{y=2}^{\infty} \frac{1.9^y}{(y-1)!}\)
Step 4 ：Recognize that the remaining sum is the sum of a Poisson distribution with mean \(\lambda' = 1.9\) shifted by one unit: \(E[Y] = e^{-1.9} \cdot (\sum_{y=1}^{\infty} \frac{1.9^y}{y!} - \frac{1.9}{1!} - \frac{1}{0!}) = e^{-1.9} \cdot (e^{1.9} - 1.9 - 1)\)
Step 5 ：Calculate the expected amount paid under the policy during a one-year period: \(E[Paid] = \$ 9300 \cdot E[Y] = \$ 9300 \cdot (e^{-1.9} \cdot (e^{1.9} - 1.9 - 1)) = \$ 12127.5764\)
