Why does the sum of $N$ Bernoulli random variables have a Poisson distribution if $N$ is Poisson distributed? 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. 
 A: 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 $\lambda p$.
A: You can prove by showing that the moment generating function of $\sum_{i=1}^{N}{X_i}$ is that of a Poisson.
\begin{align*}
 E\left[\exp\left(t \sum_{i=1}^{N}{X_i}\right)\right]
 &= E\left[E\left[\exp\left(t \sum_{i=1}^{N}{X_i}\right)\bigg|N\right]\right] \\
 &= E\left[\prod_{i=1}^{N}E[\exp(t X_i)]\right] \\
 &= E\left[(p\exp(t)+(1-p))^{N}\right] \\
 &= \sum_{i=0}^{\infty} \frac{(p\exp(t)+(1-p))^{i} \exp(-\lambda)\lambda^{i}}{i!} \\
 &= \exp(-\lambda) \sum_{i=0}^{\infty}\frac{(\lambda(p\exp(t)+(1-p)))^{i}}{i!} \\
 &= \exp(\lambda(p\exp(t)+(1-p)-1)) \\
 &= \exp(\lambda p(\exp(t)-1))
\end{align*}
Note that $\exp(\lambda p(\exp(t)-1))$ is the moment generation function of a Poisson($\lambda p$).
