0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Which reference on Poisson distributions do you use? $\endgroup$ – Did Nov 5 '12 at 11:12
  • $\begingroup$ @did I'm not sure if I understand your question correctly, but the amount of tickets he buys has a pmf $\frac{\lambda^x}{x!}e^{-\lambda}$. $\endgroup$ – woaini Nov 5 '12 at 11:20
  • $\begingroup$ Which textbooks do you follow, if you follow some? In other words: what can you do with Poisson distributions? $\endgroup$ – Did Nov 5 '12 at 11:25
  • $\begingroup$ @did There isn't a specific textbook that I follow. $\endgroup$ – woaini Nov 5 '12 at 11:26
  • $\begingroup$ So... all you know is the definition of the Poisson distribution you reproduced in your comment? $\endgroup$ – Did Nov 5 '12 at 11:28
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.