A lottery player decides to use a Poisson random variable generator to help him decide how many ticket to buy. He generates a Poisson random variable $N$ with mean $\lambda$, and then purchases $N$ lottery tickets. If each of the tickets he buys has (independently of the other tickets) the probability $p$ of winning, calculate the mean and the variance of the number of winning tickets the player buys.
Conditional mean and variance related to Poisson distribution
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0@did I'm trying to figure out what $E(W^2|N=n)$ equals, but unlike for $E(W|N=n)$ I can't seem to reason it out. – 2012-11-05
1 Answers
Let $X$ be the number of winning tickets, we have for $k \in \mathbb N$ \begin{align*} P(X=k) &= \sum_{\ell=k}^\infty P(N=\ell, X=k)\\ &= \sum_{\ell=k}^\infty P(X = k\mid N = \ell)P(N = \ell)\\ &= \sum_{\ell=k}^\infty \binom{\ell}k p^k(1-p)^{\ell-k}\frac{\lambda^\ell}{\ell!}\exp(-\lambda)\\ &= \frac {(\lambda p)^k}{k!}\sum_{\ell=k}^\infty \frac 1{(\ell-k)!}(\lambda - \lambda p)^{\ell-k}\exp(-\lambda)\\ &= \frac {(\lambda p)^k}{k!}\sum_{\ell=0}^\infty \frac 1{\ell!}(\lambda -\lambda p)^{\ell} \exp(-\lambda)\\ &= \frac{(\lambda p)^k}{k!}\exp(-\lambda) \sum_{\ell=0}^\infty \frac 1{\ell!} \left({\lambda - \lambda p}\right)^\ell\\ &= \frac{(\lambda p)^k}{k!}\exp(-\lambda p) \end{align*} So we have $X \sim \text{Poisson}(\lambda p)$, from here one can easily give $E(X) = \sigma^2(X) = \lambda p$.