[Math] Why does the sum of $N$ Bernoulli random variables have a Poisson distribution if $N$ is Poisson distributed

Question

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.

Answer 1

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 $\lambda p$.