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Math Help - Expectation

  1. #1
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    Expectation

    I have attached the question on this thread.

    I have managed to attain the following:

    P_x(x)=\alpha^2xe^{-\alpha x}

    P_{y|x}(y|x)=xe^{-xy}

    Yet I cannot obtain the fact that

    E[XY]=1

    I'm not sure how to derive this using my results.
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  2. #2
    Moo
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    Re: Expectation

    Hello,

    Consider that E[XY]=E[E[XY|X]]

    But E[XY|X]=X*E[Y|X].

    Since you know the pdf of Y|X, you'll easily get the expectation. Then you'll get E[something with respect to X], which is again easy because you have the pdf of X.
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  3. #3
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    Re: Expectation

    E[Y|X]=\int^\infty_0 yxe^{-xy} dy=[-ye^{-xy}]^\infty_0-\frac{1}{x}[e^{-xy}]^\infty_0=\frac{1}{x}

    And is follows that:

    xE[Y|X]=1

    and then we just take the expectation

    E[XE[Y|X]]=E[1]=1=E[XY]

    Am I correct?
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  4. #4
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    Re: Expectation

    Yes, it's correct.

    A few remarks :
    - are you aware of the properties of the conditional expectation that have been used ?
    - in the pdf of X and of Y|X, you should respectively add that it's for x>0 and y>0
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