Let be $S$ a r.v. define as $S=\sum_{i=1}^N X_i$, where:
$-$ $X_i$ has mean $\mu_x$ and variance $\sigma_x^2$;
$-$ $N \sim Poisson(\lambda)$;
$-$ $X_i$ are i.i.d. and $X_i$ and $N$ are independent.
I've found a topic where is shown that $E[S]=E[N]E[X_j]$; but what about $var(S)$?
Could be $var(S)=var(\sum_{i=1}^N X_i)prob(N=n)=\sum_{i=1}^N var(X_i)prob(N=n)=Nprob(N=n)var(X_i)=n(\frac{\lambda^n}{n!})\sigma_x^2=\frac{\lambda^n}{n-1!}\sigma_x^2$?
Thanks for your help.
KB
In this situation one can show that $V(S) = E(N)V(X) + V(N)[E(X)]^2.$ One proof of this uses a conditioning argument. Proofs are shown in many beginning and intermediate level probability texts (perhaps including yours).
So in your case $V(S) = \lambda\sigma_x^2 + \lambda\mu_x^2.$
Here is a simulation in R for a million repetitions of such an experiment, where $\lambda = 10$ and $X_j \sim Norm(\mu = 100, \sigma=15)$. In this case, we should have $E(S) = 10(100) = 100$ and $V(S) = 10(225) + 10(10000) = 102250$ or $SD(S) = 319.7655.$ I chose these values of the parameters to illustrate the importance of the second term in the expression for the variance.
The simulation is accurate to about the nearest integer.
m = 10^6; s = numeric(m)
for (i in 1:m) {
n = rpois(1, 10)
s[i] = sum(rnorm(n, 100, 15)) }
mean(s); sd(s)
## 999.7464
## 319.9339
The histogram of simulated values of $S$ shows a skewed distribution. (A sum of a fixed number $n=10$ of these iid normal terms would, of course, be normal--and with much smaller variance.)
