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Math Help - Proving question

  1. #1
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    Proving question

    Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]? Therefore i conclude that:

    Cov(X, E[Y/X])
    = E[(X - E[X])(E[Y/X] - E[E[Y/X]])]
    = E[(X - E[X])(Y- E[Y])]
    = Cov(X, Y)

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]?
    No you can't. But, it is true that E[f(X)E[Y|X]]=E[f(X)Y] for any function f (that's a characterization of the conditional expectation).

    Therefore i conclude that:

    Cov(X, E[Y/X])
    = E[(X - E[X])(E[Y/X] - E[E[Y/X]])]
    = E[(X - E[X])(Y- E[Y])]
    = Cov(X, Y)

    Thanks in advance!
    You can use the above equality with f(X)=X-E[X] : it gives E[(X-E[X])E[Y|X]]=E[(X-E[X])Y]. Since additionaly E[E[Y|X]]=E[Y], this justifies your second equality.
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