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Math Help - P.D.F of X-Y when X Y are independent

  1. #1
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    P.D.F of X-Y when X Y are independent

    In general, how do you solve for X-Y?

    e.g X and Y have p.d.f f(x)=3x^2 for 0<x<1
    I know how to solve for X+Y
    but for X-Y, C.D.F is effectively the region above a line Y=X-Z bounded by the square 0 to 1.

    I think I am stuck with a non standard multiple integral from top left to right bottom.

    any idea?
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  2. #2
    Moo
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    Re: P.D.F of X-Y when X Y are independent

    Hello,

    X-Y=X+(-Y)
    you can easily get the pdf of -Y, with the region that is different. You can then use the convolution or the transform of variables method.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Re: P.D.F of X-Y when X Y are independent

    Note that -1\le a\le 1

    where P(Z\le a)=P(X-Y\le a), which has two cases.

    If -1\le a\le 0 use \int_{-a}^1\int_0^{a+y}9x^2y^2dxdy

    and if 0< a\le 1 use 1-\int_{a}^1\int_0^{x-a}9x^2y^2dydx

    It's 2am, I hope my bounds are correct, but the idea is certainly right.
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