Solution
= Solution
Let $X_1,X_2$ be the two <independent> <Poisson random variables>. For $k\geq0$, their discrete <convolution> is
$$
\begin{aligned}
\mathbb P(X_1+X_2=k)&=\sum_{j=0}^k\mathbb P(X_1=j)\mathbb P(X_2=k-j)\\
&=\frac{e^{-(\lambda+\mu)}}{k!}\sum_{j=0}^k\binom kj\lambda^j\mu^{k-j}\\
&=e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^k}{k!},
\end{aligned}
$$
where the last equality is the <binomial theorem>. Thus \b[$X_1+X_2\sim\operatorname{Poisson}(\lambda+\mu)$].