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Math Help - Conditional Expectations

  1. #1
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    Conditional Expectations

    Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1
    Prove that E(x|Y=y) = y/2 and E(Y|X=x) = x + (1-x)/2

    So, I found the density function f(x,y) to be equal to 2, as long as x and y are within the bounds.
    I set up E(x|Y=y) = integral from 0 to y of x*f(x|y) dx
    f(x|y) = f(x,y)/f(y)
    f(y) = integral from 0 to 1 of 2 dx, therefore 2.
    Then f(x|y) = 2/2 = 1
    And E(x|Y=y) = integral from 0 to y of x * dx, but then that would give (y^2)/2, which doesn't appear to be correct...can anyone help me please? I'm slightly befuddled. Also, would the second one go the same way?
    Thank you!
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  2. #2
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    Quote Originally Posted by mistykz View Post
    Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1
    Prove that E(x|Y=y) = y/2 and E(Y|X=x) = x + (1-x)/2

    So, I found the density function f(x,y) to be equal to 2, as long as x and y are within the bounds.
    I set up E(x|Y=y) = integral from 0 to y of x*f(x|y) dx
    f(x|y) = f(x,y)/f(y)
    f(y) = integral from 0 to 1 of 2 dx, therefore 2.
    Then f(x|y) = 2/2 = 1
    And E(x|Y=y) = integral from 0 to y of x * dx, but then that would give (y^2)/2, which doesn't appear to be correct...can anyone help me please? I'm slightly befuddled. Also, would the second one go the same way?
    Thank you!
    I assume X and Y are independent (the answers you quote suggest that they must be).

    I will set one up for you:

    E(X \, | \, Y = y) = \int_0^y x \, f(x \, | \, y) \, dx where f(x \, | \, y) = \frac{f(x, y)}{f(y)} = \frac{1}{y} since f(x, y) = (1) (1) = 1 for 0 \leq x \leq y \leq 1 (assuming independence) and f(y) = \int_0^y 1 \, dx = y.

    Do the calculation and you will get the answer you mention. You should now attempt finding E(Y \, | \, X = x).
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    I got kind of stuck on the second one. I used E(Y|X=x) = integral from x to 1 of y*f(y|x) dy, with f(y|x) = f(y,x)/f(x)
    i got f(x) = integral x to 1 of 2dy = 2(1-x)
    problem here is that this puts the (1-x) in the denominator of my original function, and it needs to be in the numerator...
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    Quote Originally Posted by mistykz View Post
    I got kind of stuck on the second one. I used E(Y|X=x) = integral from x to 1 of y*f(y|x) dy, with f(y|x) = f(y,x)/f(x)
    i got f(x) = integral x to 1 of 2dy = 2(1-x) Mr F says: See first comment below.
    problem here is that this puts the (1-x) in the denominator of my original function, and it needs to be in the numerator...
    f(x, y) \neq 2. f(x, y) = 1. I said this in my first post. So f(x) = 1 - x.

    Then E(Y \, | \, X = x) = \int_{y=x}^{y=1} \frac{y}{1 - x} \, dy = \frac{1+x}{2} = x + \frac{1-x}{2}.
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  5. #5
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    Based on...
    'Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1'
    the random variables are NOT independent.
    P(X>.5|Y=.5)\not= P(X>.5|Y=.3)
    If they are indep, then the region must be a rectangle.
    This is a triangle.
    f(x,y)=2 on 0\le x\le y \le 1.
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  6. #6
    MHF Contributor matheagle's Avatar
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    The joint density is f(x,y)=2 on 0<x<y<1.
    So the marginal of Y is \int_0^y 2 dx=2y on 0<y<1.
    Hence f_{X|Y}=1/y on 0<x<y<1.
    Giving us E(X|Y)=\int_0^y  x/y dx=y/2.
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  7. #7
    MHF Contributor matheagle's Avatar
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    The second aswer is correct, but it reduces to (x+1)/2.
    Since the joint density is 2 on that triangle, we have

    f(x)=2\int_x^1 dy=2(1-x).

    Thus the conditional density of Y given X is 1/(1-x) on that same triangle.
    Therefore
    E(Y|X)=\int_x^1 {y\over 1-x}dy={1-x^2\over 2(1-x)}={1+x\over 2}.
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  8. #8
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    Quote Originally Posted by matheagle View Post
    Based on...
    'Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1'
    the random variables are NOT independent.
    P(X>.5|Y=.5)\not= P(X>.5|Y=.3)
    If they are indep, then the region must be a rectangle.
    This is a triangle.
    f(x,y)=2 on 0\le x\le y \le 1.
    That's what happens when my mind is on the tour .....
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