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Conditional Expectation

  1. Sep 17, 2012 #1
    1. Let the joint pdf be f(x,y) = 2 ; 0<x<y<1 ; 0<y<1
    Find E(Y|x) and E(X|y)



    2. Relevant equations

    E(Y|x) = [itex]\int Y*f(y|x)dy[/itex]
    f(y|x) = f(x,y) / f(x)

    3. The attempt at a solution
    f(x) = [itex]\int 2dy[/itex] from 0 to y = 2y
    f(y|x) = f(x,y)/f(x) = 1/2y
    E(Y|x) = [itex]\int Y/2Y dy[/itex] from x to 1 = [itex]\int 1/2 dy[/itex] from x to 1
    = -(x-1)/2
    = (1-x)/2

    The answer is supposed to be (1+x)/2
     
  2. jcsd
  3. Sep 17, 2012 #2

    jbunniii

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    Your expression for f(x) is wrong. It should be a function of x, not of y. Try drawing a picture of the region where f(x,y) is nonzero. Then answer this question: for a fixed value of x, what values of y will give you a nonzero f(x,y)?
     
  4. Sep 17, 2012 #3
    The only other way I can think of doing f(x) would be to integrate from 0 to 1 instead. f(x) is defined as the integral of the joint pdf in terms of y.

    So, we could get integral(2dy) from x to 1?
     
  5. Sep 17, 2012 #4

    jbunniii

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    Correct, from x to 1 (not as 0 to 1 as you wrote in the previous paragraph). Also be sure to state which values of x this is valid for.
     
  6. Sep 17, 2012 #5
    That will give us 2(1-x) so f(y|x) = 1/(1-x)

    I'm confused how this will give (1+x)/2 for E(y|x)
     
  7. Sep 17, 2012 #6
    Just kidding I worked it out, thanks.
     
  8. Sep 17, 2012 #7

    jbunniii

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    That's the answer I got. What integral are you calculating for E(y|x)?

    [edit] Cool, I see you got it.
     
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