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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

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3  
$E[N^2]=\sum_{n=0}^{\infty} n^2 P(N=n)$. 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

1 Answer 1

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] $

$= Var(N) E[X_1]$

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