Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

this is an homework but i really tried hard before surrender, and i think that or I'm very close to the end or I'm as far as possibile.. That's the text:

Let $X_1, X_2, ..., X_n$ be independent and identically distributed random variables. And N is a nonnegative integer valued random variable (indipendent to any $X_i$).

Let $Z = \Sigma_{i=1}^NX_i$ calculate $Cov(N,Z)$.

What I have done:

I know that $Cov(N,Z) = E[NZ] -E[N]E[Z]$

What i've done is try to get $E[NZ] = E[\Sigma_{i=1}^N X_i N] =$

$= \Sigma_{n=0}^{\infty} E[\Sigma_{i=1}^N X_i N | N=n] P(N=n)$ =

$= \Sigma_{n=0}^{\infty} E[n\Sigma_{i=1}^n X_i]P(N=n) = $

$= \Sigma_{n=0}^{\infty} n E[\Sigma_{i=1}^n X_i]P(N=n) = $

As $X_i$ is iid with any other $X_j$ i use only $X_1$

$=\Sigma_{n=0}^{\infty} nE[\Sigma_{i=1}^n X_1]P(N=n) = $

$=\Sigma_{n=0}^{\infty} n^2 X_1 P(N=n) = X_1\Sigma_{n=0}^{\infty} n^2 P(N=n)$

I know that $\Sigma_{n=0}^{\infty} n P(N=n) = E[N]$ but what about $=\Sigma_{n=0}^{\infty} n^2 P(N=n)$.

Thank you

share|cite|improve this question
$E[N^2]=\sum_{n=0}^{\infty} n^2 P(N=n)$. By the way, your end result for $E[NZ]$ should contain $E[X_1]$ not just $X_1$. – Raskolnikov Oct 22 '11 at 10:04
O was just THAT simple! How stupid I am.. Thank you!!! – Fabio F. Oct 22 '11 at 10:26
+1 for showing your work. // As @Raskolnikov said, $E[\sum\limits_{i=1}^nX_i]=nE[X_1]$ and not $nX_1$. – Did Oct 22 '11 at 10:27
Also see the law of total covariance on the wiki. – Sasha Oct 22 '11 at 14:06
up vote 3 down vote accepted

After you have made Raskolnikov's correction and simplification, you can check your result with

$Cov(N,Z) = E[NZ] -E[N].E[Z]$

$= E[N.E[Z|N]] - E[N].E[E[Z|N]] $

$= E[N^2.E[X_1]] - E[N].E[N.E[X_1]] $

$= (E[N^2]- E[N]^2 ). E[X_1] $

$= Var(N) E[X_1]$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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