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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0Which reference on Poisson distributions do you use? – 2012-11-05
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0@did I'm not sure if I understand your question correctly, but the amount of tickets he buys has a pmf $\frac{\lambda^x}{x!}e^{-\lambda}$. – 2012-11-05
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0Which textbooks do you follow, if you follow some? In other words: what can you do with Poisson distributions? – 2012-11-05
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0@did There isn't a specific textbook that I follow. – 2012-11-05
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0So... all you know is the definition of the Poisson distribution you reproduced in your comment? – 2012-11-05
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0@did I guess I'm not very familiar with Poisson random variables. – 2012-11-05
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0Nevertheless, surely you can compute the expected value of the number $W$ of winning tickets conditionally on $N=n$? – 2012-11-05
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0@did $E(W) = np$. Is that correct? – 2012-11-05
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0More precisely, E(W|N=n)=np. Hence E(W|N)=Np. Hence E(W)=... – 2012-11-05
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0let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6335/discussion-between-woaini-and-did) – 2012-11-05
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0@did Sorry, I can't seem to log in to use the chat. So would $E(W)=p(\frac{\lambda^x}{x!}e^{-\lambda})$? – 2012-11-05
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0Aaaargh, no! (But you seem to have a full bolts-on solution at your disposal, now.) – 2012-11-05
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0@did I looked through the other solution and I understand how martini arrived at it, but I still want to figure it out this way. If you don't mind, can you guide me through it a bit more? – 2012-11-05
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0Re the expectation, this can be quick: an important property of conditional expectations is the **tower property**, saying that E(E(U|V))=E(U) for every U and V. Here, E(W)=E(E(W|N))=E(Np)=pE(N)=pλ. Re the variance, one should start from an expression of E(W^2|N=n) and proceed similarly. – 2012-11-05
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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 $E(X) = \sigma^2(X) = \lambda p$.