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Math Help - E(X^2) exists =>E(E(X|Y) ^2) exists?

  1. #1
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    E(X^2) exists =>E(E(X|Y) ^2) exists?

    True/False?

    Let X,Y by random variables. If X2 has an expectation (not necessarily finite), then (E(X|Y))2 has an expectation.
    In other words, if E(X2 ) exists, does E(E(X|Y) E(X|Y)) exist?
    Last edited by Mishmish; April 2nd 2012 at 11:16 PM.
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  2. #2
    Super Member girdav's Avatar
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    Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

    If X itself hasn't an expectation, the condition expectation is not well-defined.
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    Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

    But X does have an expectation in this case, since X^2 does.
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  4. #4
    Super Member girdav's Avatar
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    Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

    I meant a finite expectation (of the absolute value for $X$). If the expectation of X^2 is finite the result holds using Jensen's inequality. So we have to deal with the case E[X^2]=+\infty.
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    Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

    Could you please indicate how Jensen's inequality can be used to prove the claim under the assumption that E(X^2) is finite?
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  6. #6
    Super Member girdav's Avatar
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    Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

    By Jensen's inequality (E(X\mid Y))^2\leq E(X^2\mid Y) so E[(E(X\mid Y))^2]\leq E[ E(X^2\mid Y)]=E[X^2].
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