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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 $10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. What is the expected amount paid to the company under this policy during a one-year period?

I know how to calculate the expectation and what the series is. I'm having problems with the summations. I know it should involve:

$$\sum_{k=2}^{+\infty} \frac{(1.5)^k}{k!}$$

Diego Silvera
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user60497
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4 Answers4

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Let $X$ be the number of snowstorms occurring in the given year and let $Y$ be the amount paid to the company. Call one unit of money $\$ 10{,}000$.

Then $Y$ takes the value $0$ when $X=0$ or $X=1$, the value $1$ when $X=2$, the value $2$ when $X=3$, etc..

The expected payment is $$\eqalign{ \Bbb E(Y) &=\sum_{k=2}^\infty (k-1)P[X=k]\cr &=\sum_{k=2}^\infty (k-1) e^{-1.5}{(1.5)^k\over k!}\cr &=\sum_{k=1}^\infty (k-1) e^{-1.5}{(1.5)^k\over k!}\cr &= \sum_{k=1}^\infty k e^{-1.5}{(1.5)^k\over k!} -\sum_{k=1}^\infty e^{-1.5}{(1.5)^k\over k!}\cr &=\underbrace{ \sum_{k=0}^\infty k e^{-1.5}{(1.5)^k\over k!}}_{\text{mean of } X} - \biggl(-e^{-1.5}+\underbrace{\sum_{k=0}^\infty e^{-1.5}{(1.5)^k\over k!}}_{=1}\biggr)\cr &=1.5+e^{-1.5}- 1\cr &=0.5+e^{-1.5}\cr &\approx .7231\,\text{units}. }$$

David Mitra
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The mean of the plain Poisson is $1.5$. So if it paid for all snowstorms, the mean outlay would be $15000$.

The probability of one storm is $1.5e^{-1.5}$. The company does not pay for this, so subtract $15000e^{-1.5}$ from $15000$.

André Nicolas
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  • Wouldn't this assume that $$20,000$ is paid when there are exactly two snowstorms, $$30,000$ for three, etc? The way I read the problem, $$10,000$ is paid for two storms, $$20,000$ for three, etc. Or, perhaps I'm off... – David Mitra Feb 01 '13 at 01:35
  • Yes, I believe this is correct. – user60497 Feb 01 '13 at 02:07
  • It says $10000$ for each one thereafter (my emphasis). It seems clear that this means $20$ K for $2$, $30$ K for $3$, and so on. – André Nicolas Feb 01 '13 at 06:17
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The expected payout is actually

$$\begin{align} 10,000 \sum_{k=2}^{\infty} (k-1) \frac{(1.5)^k}{k!} e^{-1.5} &= 10,000 (1.5) \sum_{k=1}^{\infty} \frac{(1.5)^k}{k!} e^{-1.5} - 10,000 \sum_{k=2}^{\infty} \frac{(1.5)^k}{k!} e^{-1.5} \\ &= 10,000 (1.5) (1-e^{-1.5}) - 10,000 (1-2.5 e^{-1.5})\\ &\approx 7231.30 \\ \end{align} $$

Ron Gordon
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If they got $\$10,000$ every time, it would be $\$10,000\cdot(1.5)$. From that subtract $\$10,000$ times the probability that there's exactly one such storm, which is $1.5e^{-1.5}$. So you get $$ \$15,000 - \$10,000\cdot1.5e^{-1.5}. $$

Michael Hardy
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