What is the sum of $k$ normal random variables where $k$ is sampled from a Poisson distribution?

Question

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.

Answer 1

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.

Answer 2

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