Let $\{X_1, X_2, X_3,\ldots \}$ be a sequence of random variables with mean of $1$.   If $N$ is a geometric random variable with probability mass function:- $$\mathsf P(N=k) ~=~ \frac 1{2^k} ~ \big[k\in \{1,2,3,\ldots\}\big]$$ > ... and it is independent of all $X_k$, then what is ?:- $$\mathsf E(X_1+X_2+X_3+\cdots+X_N)$$

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I know that $E[aX+bY]=a E[X]+b E[Y]$

Also I know that $\sum_{n = 1}^\infty \frac{1}{2^n}=1.$

So $E[X_1+X_2+...+X_N]=E[X_1]+E[X_2]+...+E[X_N]=\sum_{n = 1}^N\frac{1}{2^n}$

As limit n-> $\inf$ above term becomes one.

Thus Answer is 1.

Am I correct ?

  • $\begingroup$ I don't understand the question: is $\;N\;$ a random variable that also indexes in $\;X_1+\ldots+ X_N\; ?$ $\endgroup$
    – DonAntonio
    Feb 15, 2016 at 10:39
  • $\begingroup$ Your answer, and more importantly the way you achieved it is not correct. Note that the LHS is a number and the RHS is a random variable. $\endgroup$
    – drhab
    Feb 15, 2016 at 10:53
  • $\begingroup$ @Joanpemo Yeah, the random variable $S$ of interest is defined as $$S(\omega)=\sum_{k=1}^{N(\omega)}X_k(\omega).$$ $\endgroup$
    – Did
    Feb 15, 2016 at 10:53
  • $\begingroup$ Thank you @did. I see this is way over my actual knowledge of probability. $\endgroup$
    – DonAntonio
    Feb 15, 2016 at 10:58

2 Answers 2


Theorem If $X_1,X_2...$ are r.v.i.i.d. with $E[X_i]=\mu<\infty$ and $N$ is a random variable with $E[N]<\infty$ independent respect to $X_i$, then $E[X_1+...+X_N]=\mu E[N]$


$\begin{eqnarray} E[X_1+...+X_N]&=&E[E[X_1+...+X_N|N]]\\ &=&\sum_{n=0}^\infty P(N=n)E[X_1+...+X_N|N=n]\\ &=&\sum_{n=0}^\infty P(N=n)E[X_1+...+X_n|N=n]\\ &=&\sum_{n=0}^\infty P(N=n)E[X_1+...+X_n]\\ &=&\sum_{n=0}^\infty P(N=n)n\mu\\ &=&\mu\sum_{n=0}^\infty nP(N=n)\\ &=&\mu E[N] \end{eqnarray}$

Now, your excercise is trivial.



  • $\mathbb E\sum_{i=1}^NX_i=\mathbb E(\sum_{i=1}^NX_i\mid N=n)P(N=n)$

  • $\mathbb E(\sum_{i=1}^NX_i\mid N=n)=\mathbb E(\sum_{i=1}^nX_i\mid N=n)=\sum_{i=1}^n\mathbb E(X_i\mid N=n)$

  • $N$ and $X_i$ are independent.


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