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Let $X_1, \dots, X_n$ be independent and identically distributed Poisson(µ) random variables. Derive the distribution of $W= \sum X_i$.

I'm not sure how to answer this, should I use the moment generating function?

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  • $\begingroup$ Just do it for the sum of two variables and use the pdf. It falls right out. Extend to $n$ by induction. $\endgroup$ – Scott Burns Dec 12 '16 at 12:59
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Yes, we can use the moment generating functions to prove this. We have that $$ M_W(t)=M_{X_1}(t)M_{X_2}(t)\ldots M_{X_n}(t)=[\exp(\mu(e^t-1))]^n=\exp(n\mu(e^t-1)) $$ using the independence and identical distributions. This means that the sum is a Poisson random variable with the parameter equal to $n\mu$.

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