# How can I find the expected value of a random variable with terms that increase until infinity?

Here is the question

A company buys a policy to insure its revenue in the event of major snow storms 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 snow storms 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'm pretty sure that I know how to solve most of the problem. The payment about will look like this 0 x <=1 10,000(x-1) x > 1 The Poisson distribution will look like this P(X=x)= e^1.5 (1.5^x/x!) I know that the expected value should look like this E[X] = P(X=0)(0) + P(X=1)(0) + P(X=2)(10,000) + P(X=3)(20,000) + ... I know how to calculate the probability that X is greater than 1 P(X>1) = 1 - P(X=0) - P(X=1) I'm just not sure how to incorporate the increasing payment sizes into the calculation. Can somebody offer some guidance please? Does this have something to do with summing infinite sequences? • Consider reformulating the pay-out terms: The insurance pays \$10,000 for each snowstorm, minus a fixed extra premium of \$10,000, plus a bonus of \$10,000 in years that have no snowstorms at all. It's easy to find the expectation of each of these terms, now just add. – Henning Makholm Sep 25 '14 at 23:00

First of all, you should have $P(X=x) = e^{-1.5} \frac{1.5^x}{x!}$, and the sum you have written down is correct, if $Y$ is the payout random variable: \begin{align*} \mathbb{E}Y &= \sum_{n=1}^\infty 10000n P(X=n+1) \\ &= 10000e^{-1.5} \sum_{n=1}^\infty n\frac{1.5^{n+1}}{(n+1)!} \\ &= 10000e^{-1.5} \left[\sum_{n=1}^\infty (n+1)\frac{1.5^{n+1}}{(n+1)!} - \sum_{n=1}^\infty \frac{1.5^{n+1}}{(n+1)!}\right] \\ &= 10000e^{-1.5} \left[1.5\sum_{n=1}^\infty \frac{1.5^{n}}{n!} - \sum_{n=1}^\infty \frac{1.5^{n+1}}{(n+1)!}\right] \end{align*} I'm not sure if what I've written is legitimate (the third line), but the two series can be evaluated as something close to $e^{1.5}$ with few of the first terms of the Taylor expansion missing.
• For example, if you were to evaluate the first sequence in the brackets, we know that $\sum_{n=0}^\infty x^n/n! = e^x$ from calculus (Taylor series). Hence, $\sum_{n=1}^\infty 1.5^n/n! = \sum_{n=0}^\infty 1.5^n/n! - 1 = e^{1.5} - 1$ (the 1 is the value of the summand when $n=0$.) – symmetricuser Sep 25 '14 at 23:13