# Deriving the distribution of poisson random variables.

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?

• Just do it for the sum of two variables and use the pdf. It falls right out. Extend to $n$ by induction. – Scott Burns Dec 12 '16 at 12:59

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 $n\mu$.