# Conditional mean and variance related to Poisson distribution

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.

• Which reference on Poisson distributions do you use? – Did Nov 5 '12 at 11:12
• @did I'm not sure if I understand your question correctly, but the amount of tickets he buys has a pmf $\frac{\lambda^x}{x!}e^{-\lambda}$. – woaini Nov 5 '12 at 11:20
• Which textbooks do you follow, if you follow some? In other words: what can you do with Poisson distributions? – Did Nov 5 '12 at 11:25
• @did There isn't a specific textbook that I follow. – woaini Nov 5 '12 at 11:26
• So... all you know is the definition of the Poisson distribution you reproduced in your comment? – Did Nov 5 '12 at 11:28

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 $E(X) = \sigma^2(X) = \lambda p$.