The question:

$X_1$, $X_2$, etc. are independent and identically distributed non-negative integer valued random variables. $N$ is a non-negative integer valued random variable which is independent of $X_1$, $X_2$ etc.., and $Y$ = $X_1 + X_2 + X_3 + … + X_N$ . (We take $Y = 0$ if $N = 0$).

Prove that $\mathbb{E}[Y] = \mathbb{E}[X_1]\mathbb{E}[N]$.

My attempt:

I know that the probability generating $G_Y(s)$ of $Y$ is equal to $G_N(G_X(s))$... but I'm not sure how that's helpful here.

My intuition leads me in this direction:

$\mathbb{E}[Y] = \sum\limits_{n=0}^{\infty} \mathbb{E}[Y|N=n]\mathbb{P}(N=n)$

$= \sum\limits_{n=0}^{\infty} \mathbb{E}[nX_1|N = n]\mathbb{P}(N=n)$

$= \sum\limits_{n=0}^{\infty} n \mathbb{E}[X_1|N = n]\mathbb{P}(N=n)$ (is this step valid??)

But I don't know where to go from here.

  • 1
    $\begingroup$ Don't you mean that $N$ is a positive integer random variable? You start with $X_1$. Also don't you mean that $Y=X_1+\cdots+X_N$? $\endgroup$ – drhab Nov 26 '17 at 13:04
  • $\begingroup$ @drhab No to your first question - Y is taken to be 0 if N=0. Yes to your second question, thanks! Have edited it to make both points clearer. $\endgroup$ – StackExchanger10293848 Nov 26 '17 at 13:12
  • $\begingroup$ I suspect not $Y=X_1+\cdots +X_n$ but $Y=X_1+\cdots +X_N$ $\endgroup$ – drhab Nov 26 '17 at 13:13
  • $\begingroup$ D'oh! Third time's a charm... $\endgroup$ – StackExchanger10293848 Nov 26 '17 at 13:16
  • 1
    $\begingroup$ drhab's answer is the correct continuation of your work. But to be clear, your work is good and you just need to drop the condition to get $E(X_1\mid N=n)=E(X_1)$. Then factor this out of the summation. The step you have a question about validity for is indeed valid because $n$ is just a constant at the point (relative to that particular term in the summation). You just need to keep $n$ inside the summation over $n$. $\endgroup$ – jdods Nov 26 '17 at 13:34

$$\begin{aligned}\mathsf{E}Y & =\sum_{n=0}^{\infty}\mathsf{E}\left[Y\mid N=n\right]\mathsf{P}\left(N=n\right)\\ & =\sum_{n=1}^{\infty}\mathsf{E}\left[Y\mid N=n\right]\mathsf{P}\left(N=n\right)\\ & =\sum_{n=1}^{\infty}\mathsf{E}\left[X_{1}+\cdots+X_{n}\mid N=n\right]\mathsf{P}\left(N=n\right)\\ & =\sum_{n=1}^{\infty}\mathsf{E}\left[X_{1}+\cdots+X_{n}\right]\mathsf{P}\left(N=n\right)\\ & =\sum_{n=1}^{\infty}\left[\mathsf{E}X_{1}+\cdots+\mathsf{E}X_{n}\right]\mathsf{P}\left(N=n\right)\\ & =\sum_{n=1}^{\infty}n\mathsf{E}X_{1}\mathsf{P}\left(N=n\right)\\ & =\mathsf{E}X_{1}\sum_{n=1}^{\infty}n\mathsf{P}\left(N=n\right)\\ & =\mathsf{E}X_{1}\mathsf{E}N \end{aligned} $$

second equality: because the first term is $0$ since $\mathsf E[Y\mid N=0]=0$.

fourth equality: because $N$ is independent wrt the $X_i$

fifth equality: linearity of expectation.

seventh equality: factor $\mathsf EX_1$ does not depend on index $n$ so can be taken out of the summation and placed before summation symbol.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.