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Math Help - probability question

  1. #1
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    probability question

    X and Y are independent random variables. If P(X>u)>=P(Y>U) for all u, then P(X>Y)>=1/2

    Thanks a lot
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  2. #2
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    Quote Originally Posted by violetsf View Post
    X and Y are independent random variables. If P(X>u)>=P(Y>U) for all u, then P(X>Y)>=1/2

    Thanks a lot
    Assume X and Y are continuous random variables with pdf f(x) and g(y) (the argument is easily modified for the discrete case). Then the joint pdf is f(x) g(y) due to independence of X and Y.

    What my argument boils down to is that in the XY-plane, the 'weighted area' below the line Y = X is greater that the 'weighted area' above the line Y = X since \Pr(X > u) \geq \Pr(Y > u):


    1. \Pr(X > Y) = \int_{y=-\infty}^{\infty} g(y) \int_{x = y}^{\infty} f(x) \, dx \, dy = \int_{u=-\infty}^{\infty} g(u) \Pr(X > u) \, du.

    2. \Pr(X < Y) = \Pr(Y > X) = \int_{x=-\infty}^{\infty} f(x) \int_{y = x}^{\infty} g(y) \, dy \, dx = \int_{u=-\infty}^{\infty} f(u) \Pr(Y > u) \, du.

    3. \Pr(X > Y) + \Pr(X < Y) = 1.

    Now use 1., 2., 3. and \Pr(X > u) \geq \Pr(Y > u) to show that \Pr(X > Y) + \Pr(X > Y) \geq 1 \Rightarrow \Pr(X > Y) \geq \frac{1}{2}.
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  3. #3
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    Thank you very much!
    how about #3, I mean where is P(X=Y)
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  4. #4
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    Quote Originally Posted by violetsf View Post
    Thank you very much!
    how about #3, I mean where is P(X=Y)
    Obviously you just replace the typos < and > with \leq and \geq.
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  5. #5
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    Another question: f(u) and g(u) are not the same, right?
    then how can you compare #1 and #2 ?
    Thanks!
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  6. #6
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    Quote Originally Posted by violetsf View Post
    Another question: f(u) and g(u) are not the same, right?

    [snip]
    Right.

    Quote Originally Posted by violetsf View Post
    then how can you compare #1 and #2 ?
    Thanks!
    I'll need to think about this! It seems less simple now than when I posted.

    I'm confident the approach is correct, but it might not be the most fruitful approach (I don't see what other approach could be taken at this moment).
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