What is the sum of $k$ normal random variables where $k$ is sampled from a Poisson distribution? Let's say you have a sequence of i.i.d. random variables $r_1,r_2,r_3,\dots$ that are distributed as $\mathcal N(0,1)$. Now, consider the process where we sample a $k$ from $\text{Pois}(\lambda)$ and let $z$ equal $$\sum_{i=1}^{k}r_i$$
My question is what the distribution of $z$ is?
My first guess is that it would be $\mathcal N(0,\lambda)$. I do not how I would proof this, however. The variance might be slightly larger due to the uncertainty introduced by the Poisson distribution, for example. Proving the mean is $0$ is easy by symmetry.
 A: If the sum is $Z=\sum\limits_{i=1}^{K}r_i$, then $Z$ does not have a normal distribution
In particular, $\mathbb P(Z=0)=\mathbb P(K=0)=e^{-\lambda} > 0$ while a normal distribution represents a continuous random variable with no point of positive probability
A: The moments of even order of the normal distribution are:
$$
\begin{array}{c|r}
n & n\text{th moment} \\
\hline
2 & 1 \\
4 & 1\times3 \\
6 & 1\times3\times5 \\
8 & 1\times3\times5\times7 \\
\vdots & \vdots\qquad 
\end{array}
$$
The cumulants of even order of this compound Poisson distribution are consequently:
$$
\begin{array}{c|r}
n & n\text{th cumulant} \\
\hline
2 & 1\times\lambda \\
4 & 1\times3\times\lambda \\
6 & 1\times3\times5\times\lambda \\
8 & 1\times3\times5\times7\times\lambda \\
\vdots & \vdots\quad\quad\quad 
\end{array}
$$
This result can be derived by an application of the law of total cumulance.
But the cumulants of even order of the normal distribution with expectation $0$ and variance $\lambda$ are:
$$
\begin{array}{c|c}
n & n\text{th cumulant} \\
\hline
2 & \lambda \\
4 & 0 \\
6 & 0 \\
8 & 0 \\
\vdots & \vdots
\end{array}
$$
Therefore this compound Poisson distribution is not a normal distribution.
But for now I have to leave this as this partial answer.
