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Math Help - f(x,y)= 2e^-2e^-2y

  1. #1
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    Cool f(x,y)= 2e^-2e^-2y

    Does anyone know the solution to this type of problem?


    f(x,y)= 2e^-2e^-2y } for 0<x<inf

    P(x<a) for a constant a

    P(x<1, y<1)

    P(x<y)

    I have no idea how to approach this.
    Thanks!!!
    Toni
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  2. #2
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    Re: f(x,y)= 2e^-2e^-2y

    The joint density function has the property that for any set $D$ of points on the plane
    \[
    \operatorname{Prob}((X,Y)\in D)=\int_D f(x,y)\,dxdy
    \]
    So for each probability that you have to find determine the appropriate $D$ and calculate the required integral.
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  3. #3
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    Re: f(x,y)= 2e^-2e^-2y

    How?
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  4. #4
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    Re: f(x,y)= 2e^-2e^-2y

    Can you be more specific and provide one of the solutions? Not sure how to plug in the probabilities here.
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: f(x,y)= 2e^-2e^-2y

    Whoops! Wrong D.

    -Dan
    Last edited by topsquark; June 24th 2014 at 08:15 AM. Reason: Made an oopsie!
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  6. #6
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    Re: f(x,y)= 2e^-2e^-2y

    For example, for (c) you need to find a set $D$ such that $\operatorname{Prop}(X<Y)=\operatorname{Prop}((X,Y )\in D)$. Then you can calculate the right-hand side using the formula in post #2. Obviously, $D=\{(x,y)\in\Bbb R^2\mid x<y\}$.



    Then
    \begin{align*}
    \int_D f(x,y)\,ds&= \int_0^\infty\int_x^\infty 2e^{-x-2y}\,dy\,dx\\
    &=\int_0^\infty e^{-x}\int_x^\infty 2e^{-2y}\,dy\,dx\\
    &=\int_0^\infty e^{-x}\left(\left.-e^{-2y}\right|_x^\infty\right)\,dx\\
    &=\int_0^\infty e^{-3x}\,dx\\
    &=\frac{1}{3}
    \end{align*}
    Thanks from topsquark
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