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## Homework Statement

"Prove that ##\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!}\sum_{m=0}^\infty s^m e^{-\mu}\frac{\mu^m}{m!}=\sum_{m+n=0}^\infty s^{n+m} e^{-(\lambda+\mu)} \frac{(\lambda + \mu)^{m+n}}{(m+n)}!##

## Homework Equations

Binomial theorem: ##(x+y)^n=\sum_{k=0}^n x^ky^{n-k}##

Vandermonde's identity: ##\binom {n+m} m =\sum_{k=0}^m \binom n k \binom m {m-k}##

## The Attempt at a Solution

##\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!}\sum_{m=0}^\infty s^m e^{-\mu}\frac{\mu^m}{m!}=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \frac{\lambda^n \mu^m }{m!n!}=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \binom {m+n}{n} \frac{\lambda^n \mu^m }{(m+n)!}##

##=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \sum_{k=0}^n \binom m k \binom n k \frac{\lambda^n \mu^m }{(m+n)!}##

I'm afraid to go any further, because it won't get me my ##(\lambda + \mu)^{n+m}## term. If anyone has any pointers on what I should do next with this expression (or giving me another expression), or an alternate way to prove using probability-generating functions, that the sum of two independent Poisson r.v.'s have mean equal to the sum of each individual mean, then that would be much appreciated.

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