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Math Help - Help setting integral limits for P(X1 <X3 -X2)

  1. #1
    Len
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    Help setting integral limits for P(X1 <X3 -X2)

    Let (X1 ,X2, X3) be i.i.d. r.v.'s with common probability density function

    f(x)= e^(-x), 0 <x and 0 otherwise

    Find P(X1 <X3 -X2)

    I'm having a difficult time setting up my integral

    I know f(x1,x2,x3) = f(x1)*f(x2)*f(x3)= e^(-x1-x2-x3)

    I can't figure out the limits.

    Any help would certainly be appreciated
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  2. #2
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    Re: Help setting integral limits for P(X1 <X3 -X2)

    If the pdf was supported between zero and one, say, then the space would be a unit cube, and the region satsifying X1 < X3 - X2 ...

    ... or let's call it z < x - y ...

    ... would be the tetrahedron underneath z = x - y ...



    Which would be...

     \displaystyle{\int_0^1\ \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }

    ... but extended becomes...

     \displaystyle{\int_0^{\infty} \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }
    Thanks from Len
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  3. #3
    Len
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    Re: Help setting integral limits for P(X1 <X3 -X2)

    Quote Originally Posted by tom@ballooncalculus View Post
    If the pdf was supported between zero and one, say, then the space would be a unit cube, and the region satsifying X1 < X3 - X2 ...

    ... or let's call it z < x - y ...

    ... would be the tetrahedron underneath z = x - y ...



    Which would be...

     \displaystyle{\int_0^1\ \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }

    ... but extended becomes...

     \displaystyle{\int_0^{\infty} \int_0^x\ \int_0^{x - y}\ e^{-x-y-z}\ dz\ dy\ dx }
    I really appreciate you breaking it down for me. That's what I ended up "guessing" but this was very helpful. When I solved the integral I got 1/4
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