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Math Help - [SOLVED] Expectation of a Conditional Distribution

  1. #1
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    [SOLVED] Expectation of a Conditional Distribution

    I am stuck in this problem

    Let X and Y have the joint pdf f(x,y)=1, -x<y<x, 0<x<1, zero elsewhere. Show that, on the set of positive probability density, the graph E(Y|x) is a straight line whereas that of E(X|y) is not a straight line.

    Here's what I have so far, I am not sure if it's right though

    The marginal pdf of X is
    f_{1}(x)=\int_{-x}^{x} f(x,y)dy=\int_{-x}^{x} 1 dy = y|^{-x}_{x}=x-(-x)=2x for 0<x<1

    So the conditional f(Y|x) is
    f(Y|x)=\frac{1}{2x} for 0<x<1

    E(Y|x)=\int^{x} _{-x} \frac{y}{2x}dy=\frac {y^2}{4x} |^{x} _{-x}=\frac{x^2-x^2}{4x}=0???

    What are the limits of integration for the marginal pdf of Y??

    f_2(y)=\int ^?_? f(x,y)dx
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  2. #2
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    0<x<1

    E(Y|x)=\int^{x} _{-x} \frac{y}{2x}dy=\frac {y^2}{4x} |^{x} _{-x}=\frac{x^2-x^2}{4x}=0???
    Correct

    What are the limits of integration for the marginal pdf of Y??

    f_2(y)=\int ^?_? f(x,y)dx
    if you know that Y = y, then we have the restriction -x<y<x as well as 0<x<1.
    -x<y \implies x>-y, and together with x>y this gives x>|y|.
    So the limits are |y|<x<1.
    Last edited by badgerigar; November 26th 2008 at 02:57 AM. Reason: post didnt work first time
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