TITLE: Why does the sum of $N$ Bernoulli random variables have a Poisson distribution if $N$ is Poisson distributed?
QUESTION [6 upvotes]: Assume those $N$ Bernoulli random variables are i.i.d. with probability $p$, where $N \sim \operatorname{Poisson}(\lambda)$. The finite sum of them is Poisson distributed with $p\lambda$. I read a book claims that but without proof.

REPLY [5 votes]: Here's a pedestrian way:
Let $j\in \mathbb N$,
$$\begin{align}
P\left( \sum_{k=1}^N X_k = j \right) &= \sum_{l=j}^\infty P\left((N=l)\;\cap \left(\sum_{k=1}^l X_k = j\right)\right)\\
&=\sum_{l=j}^\infty P\left(N=l\right)P\left(\sum_{k=1}^l X_k = j\right) \quad \text{by independence of $N$ and $X_k$}\\
&=\sum_{l=j}^\infty e^{-\lambda}\frac{\lambda^l}{l!} \binom lj p^j(1-p)^{l-j}\\
&=\frac{e^{-\lambda}p^j\lambda^j}{j!} \sum_{l=j}^\infty \frac{\lambda^{l-j}(1-p)^{l-j}}{(l-j)!}\\
&= \frac{e^{-\lambda p}(\lambda p)^j}{j!}
\end{align}$$
Hence $\sum_{k=1}^N X_k$ follows a Poisson distribution with parameter $\lambda p$.