# Conditional expectation in statistics

1. Jan 26, 2013

### libragirl79

Hi,

I am trying to show that if the E[W|X]=0 then the Cov (W,X)=0.

Using the def of variance, and given that E[W] is zero,
I get that Cov is equal to: E[WX]-E[W * E(X)]

using conditional expectation:

E [E(WX|X)] -E[x]E[W]= E[X E[W|X]]-E[X]E[E(W|X)]=0

I am not sure if this transition (in red) is ok.

Thanks!

2. Jan 26, 2013

### HallsofIvy

The difficulty is that this statement is, in general, NOT true.
What is true is that if E[W|X]= E[W] then Cov(W, X)= 0.

What? Where is that gven?

3. Jan 26, 2013

### libragirl79

From what I understand the posed statement is true when E[W|X]=0 a.s.

E[W]=E[E[W|X]]=E[0]=0 and the def of cov being E[(W-E(W))(X-E(X))]

The question I have is if Expec value of two vars, WX, can be conditioned as I wrote above on X and then X extracted...

4. Jan 26, 2013

### Ray Vickson

Cov(W,X) = E(W*X) - EW * EX (a standard result). Can you relate E(W*X) and EW to E(W|X)?

5. Jan 26, 2013

### libragirl79

The only way I know how to relate the two was by doing the conditioning I did of both WX on X...It would be optimal if W and X were indep, then the stand result for Cov works perfectly, but that's not the case here...