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Conditional expectation in statistics

  1. Jan 26, 2013 #1
    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. jcsd
  3. Jan 26, 2013 #2

    HallsofIvy

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    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?

     
  4. Jan 26, 2013 #3
    Thanks for the reply!

    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...
     
  5. Jan 26, 2013 #4

    Ray Vickson

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    Cov(W,X) = E(W*X) - EW * EX (a standard result). Can you relate E(W*X) and EW to E(W|X)?
     
  6. Jan 26, 2013 #5
    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...
     
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