We have that $N$ and $X_1, X_2, \dots$ are all independent. We also have $\operatorname{E} [X_j] = \mu$ and $\operatorname{Var}[X_j] = σ^2$.
Then, we introduce an integer–valued random variable, $N$, which is the random sum such that:
$$Z = \sum_{j=1}^{N+1}X_j.$$
Assuming that $N$ is distributed $\sim\mathrm{Poisson}(\lambda)$, what is the first moment and what is the variance of $Z$?
For a normal Poisson distribution, I know the variance is just $\lambda$, as is the mean. I'm having trouble understanding the implication of having the bounds be poisson distributed. Normally, I would just say "variance of the sum is the sum of the variance," but I don't think that's how it works with random sums. Any hints/guidance appreciated.
Best Answer
\begin{align} \operatorname{var}(Z) & = \operatorname{var}(\operatorname{E}(Z\mid N)) + \operatorname{E}(\operatorname{var}(Z\mid N)) & & (\text{This is the law of total variance.}) \\[10pt] & = \operatorname{var}(N\mu) + \operatorname{E}(N\sigma^2) \\[10pt] & = \mu^2 \operatorname{var}(N) + \sigma^2 \operatorname{E}(N) \\[10pt] & = \mu^2 \lambda + \sigma^2 \lambda = \lambda \operatorname{E}(X_1^2). \end{align}
Indeed, generally the $n$th cumulant of a compound Poisson distribution is the mean of the simple Poisson distribution times the $n$th raw moment of the distribution that gets compounded.
PS: The above applies if the sum is from $0$ to $N$; I'll leave it as an exercise to figure out whether something needs to change if it's from $1$ to $N+1$.