If I have that $X_i$ is an independent random variable in $S_N=X_1+X_2+...+X_N$ where $E[X_i]=m^i$, $m\neq 1$, $S_0=0$ and $N \sim Po(\lambda)$ is independent of the RV's, how can I show that $E[S_N]=\frac{m}{m-1}(e^{\lambda(m-1)}-1)$?

I can only figure out how to do this for iid RV's using $g_{S_N}(t)=\{ X_1,X_2,... $ are iid$\}=g_N(g_X(t))$, any help appreciated...


Just use the Law of Total Expectation.

$$\begin{align}\mathsf E(S_N)~=~& \mathsf E(\mathsf E(S_N\mid N))\\~=~&\mathsf E(\sum_{k=1}^N \mathsf E(X_k))\\ \vdots\quad&\end{align}$$

  • $\begingroup$ Thank you! Also, as a comment for future readers remember the mean is a linear operator. $\endgroup$ – litmus Oct 5 '16 at 22:38
  • 1
    $\begingroup$ Oh, yes, indeed, and also that independence or otherwise or the random variables does not affect the Linearity of Expectation. $\endgroup$ – Graham Kemp Oct 5 '16 at 23:10

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