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Math Help - independent variables

  1. #1
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    independent variables

    IF X \;and\; Y are independent continuous random variables with identical \;but\; unknown distributions.how can I calculate P(X>Y).

    I tried using this concept from the independence and conditional distribution:

    P(X>Y)= E[P(X>Y)|Y] = \displaystyle \int P(X>Y) \; f(y)\;dy = \displaystyle \int (1-F_{X}(y))\; f(y)\;dy

    is this correct?
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  2. #2
    MHF Contributor matheagle's Avatar
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    \int_{-\infty}^{\infty}\int_{-\infty}^xf(x)f(y)dydx

    =\int_{-\infty}^{\infty}f(x)\int_{-\infty}^xf(y)dydx

    =\int_{-\infty}^{\infty}f(x)F(x)dx

    ={[F(x)]^2\over 2}\Bigr|^{\infty}_{-\infty}

    =.5-0=.5
    Last edited by matheagle; November 19th 2010 at 11:57 PM.
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  3. #3
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    Thank you so much. I have one more thing to ask. The second integral has limit going from -infinity to x. Is it because we are finding P(X>Y)?
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  4. #4
    MHF Contributor matheagle's Avatar
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    because y is bounded above by x, this is just basic calculus
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  5. #5
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    Quote Originally Posted by chutiya View Post
    IF X \;and\; Y are independent continuous random variables with identical \;but\; unknown distributions.how can I calculate P(X>Y).

    I tried using this concept from the independence and conditional distribution:

    P(X>Y)= E[P(X>Y)|Y] = \displaystyle \int P(X>Y) \; f(y)\;dy = \displaystyle \int (1-F_{X}(y))\; f(y)\;dy

    is this correct?
    1=P(X>Y)+P(Y>X)+P(X=Y)

    but P(X>Y)=P(Y>X) by symmetry and for any continuous distribution P(X=Y)=0

    So the result follows without doing any integrals.

    CB
    Last edited by CaptainBlack; November 20th 2010 at 12:58 AM.
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  6. #6
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    Quote Originally Posted by matheagle View Post
    because y is bounded above by x, this is just basic calculus


    thank you.. i am feeling so dumb!

    lastly, i am getting confused on the third step of your second post..

    isn't \int_{-\infty}^{\infty}}f(x)F(x)dx  ={[F(x)]^2}\Bigr|^{\infty}_{-\infty}?
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  7. #7
    MHF Contributor matheagle's Avatar
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    F(infty)=1
    F(-infty)=0
    those are really limits, but it's after 2AM
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by chutiya View Post
    thank you.. i am feeling so dumb!

    lastly, i am getting confused on the third step of your second post..

    isn't \int_{-\infty}^{\infty}}f(x)F(x)dx  ={[F(x)]^2}\Bigr|^{\infty}_{-\infty}?
    NO

    Calc 1, let u=F(x), then du/dx=F'(x)=f(x)

    AND \int udu=u^2/2+c
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