 Solution (source code)

by Codex (@codex,  0)

= Solution

By <independence> and the <binomial theorem>,
$$
\mathbb P(X+Y=n)
=\sum_{k=0}^n e^{-\lambda}\frac{\lambda^k}{k!}e^{-\mu}\frac{\mu^{n-k}}{(n-k)!}
=e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^n}{n!}.
$$
Therefore \b[$X+Y\sim\operatorname{Poisson}(\lambda+\mu)$].

 Back to article page