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Math Help - Need help with a joint density function

  1. #1
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    Need help with a joint density function

    First let me apologise for my poor english, it's not my first language.

    Here is the joint density function:

    <br />
f(y,x)=\left\{\begin{array}{cc}e^{-y},&\mbox{ if }<br />
0 \leq x \leq y \leq 0,\\0, & \mbox{ elsewhere. } \end{array}\right.<br />

    I'm trying to find E(Y-X) and subsequenty V(Y-X).

    I started by finding the marginal density functions:

    f_1(Y)=  \int_{0}^{y} e^{-y} dx = xe^{-y}\Big|^{y}_{0}=ye^{-y}<br /> <br />

    and

    f_2(X)=  \int_{x}^{\infty} e^{-y} dy = -e^{-y}\Big|^{\infty}_{x}   =e^{-x}

    Integration isn't really my strong suit, but I think I'm on the right path here. Next, finding E(Y) and E(X)

    E(Y)=  \int_{x}^{\infty} y^2e^{-y}dy= -\frac{y^3}{3}e^{-y}\Big|^{\infty}_{x}=\frac{x^3}{3}

    E(X)= \int_{0}^{y} xe^{-x}dx=-\frac{x^2}{2}e^{-x}\Big|^{y}_{0}=-\frac{y^2}{2}e^{-y}

    This seems really counter-intuitive, but I'm not sure what I'm doing wrong (or right for that matter). I'd really appreciate any help. Thanks in advance.
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  2. #2
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    Haven't done this in a while, so I just looked on wikipedia. This is what it said (for expected value):

    Expected value - Wikipedia, the free encyclopedia
    result is valid even if X is not statistically independent of Y.

    So in your case E(Y - X) = E(Y) - E(X).

    To find your V(Y-X), you'll need to use the following property:

    Variance - Wikipedia, the free encyclopedia
    since in your case X and Y are not independent, their covariance is not 0.
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  3. #3
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    Thank you.

    Could you maybe look over my integrations for E(Y) and E(X). I'm not really confident I've done them correctly and would appreciate any help
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  4. #4
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    In your calculations of E(Y) and E(X), did you use integration by parts? If you didn't, then you made a mistake.
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  5. #5
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    That's the mistake I made. Now it makes sense. Thanks again for your help.
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  6. #6
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