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Math Help - Probability Question on jointly distributed function

  1. #1
    DCU
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    Probability Question on jointly distributed function

    I'm having problems figuring out the limits for this. The exponential is e^-(x + y) apologies about the lack of clarity on it.
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    Moo
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    Hello,

    You must have \iint_{\mathbb{R}^2} f_{X,Y}(x,y) ~dxdy=1

    That is to say, by writing the boundaries:

    \int_0^\infty \int_0^y cxe^{-(x+y)} ~dxdy=1

    maybe reversing the order of integration will help :
    =\int_0^\infty \int_x^\infty cxe^{-(x+y)} ~dydx=c\int_0^\infty x \int_x^\infty e^{-(x+y)} ~dydx=1

    and that's all
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  3. #3
    DCU
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    For part (c) here would the limits be 0,inf for y and 0, 2-y for x dxdy?
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    Moo
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    Quote Originally Posted by DCU View Post
    For part (c) here would the limits be 0,inf for y and 0, 2-y for x dxdy?
    integral of the pdf, yup that's correct
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  5. #5
    DCU
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    Thanks, I think I might pass this exam!
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    Moo
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    Quote Originally Posted by DCU View Post
    Thanks, I think I might pass this exam!
    Sure ! Just follow your intuition, it doesn't look so bad

    and if you have any other question, you know where you can get answers/help...
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  7. #7
    MHF Contributor matheagle's Avatar
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    I'd go one further and also pull out the e^{-x}, as I did below.

    Quote Originally Posted by Moo View Post
    Hello,

    You must have \iint_{\mathbb{R}^2} f_{X,Y}(x,y) ~dxdy=1

    That is to say, by writing the boundaries:

    \int_0^\infty \int_0^y cxe^{-(x+y)} ~dxdy=1

    maybe reversing the order of integration will help :
    1=\int_0^\infty \int_x^\infty cxe^{-(x+y)} ~dydx=c\int_0^\infty x e^{-x}\int_x^\infty e^{-y} ~dydx

    and that's all
    1=c\int_0^\infty x e^{-x}e^{-x} dx=c\int_0^\infty x e^{-2x}dx

    Which can either be solved by parts or by recognizing the constant in the gamma density
    with \alpha=2 and what I call \beta =.5.
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