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Homework Help: Show that the expectation is zero

  1. Sep 12, 2006 #1
    In one of the proofs I'm reading, I need to show that the
    [tex]\phi(X)[/tex]
    that will minimize [tex]E[Y - \phi(X)]^{2}[/tex] is
    [tex]\phi(X) = E(Y|X)[/tex]

    the proof is shown to proceed as follows:
    [tex]E[Y - \phi(X)]^{2}[/tex]
    [tex]=E[(Y - E(Y|X)) + (E(Y|X) - \phi(X))]^{2}[/tex]

    expanding, we have

    [tex]=E[(Y - E(Y|X))^{2} + 2(Y - E(Y|X))(E(Y|X) - \phi(X)) + (E(Y|X) - \phi(X))^{2}][/tex]
    [tex]=E[(Y - E(Y|X))^{2}] + 2E[(Y - E(Y|X))(E(Y|X) - \phi(X))] + E[(E(Y|X) - \phi(X))]^{2}][/tex]

    the proof then states that the middle term is equal to zero. How is this true?
    I got
    [tex]2E[(Y - E(Y|X))(E(Y|X) - \phi(X))] [/tex]
    [tex]=2E[YE(Y|X) - (E(Y|X))^{2} - Y\phi(X) + E(Y|X)\phi(X)][/tex]
    [tex]=2E[YE(Y|X)] - 2E[(E(Y|X))^{2}] - 2E[Y\phi(X)] + 2E[E(Y|X)\phi(X)][/tex]
    [tex]=2E(Y|X)E(Y) - 2(E(Y|X))^{2} - 2E[Y\phi(X)] + 2E(Y|X)E(\phi(X))[/tex]

    I'm not sure on how to proceed next.
    I'm thinking that the first 2 terms will cancel as will the last two...but how exactly I'm not sure.

    help please. thanks.
     
  2. jcsd
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