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Math Help - Joint p.d.f

  1. #1
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    Joint p.d.f

    Can anyone help with the following question please?

    The continuous random variables X and Y have the joint probability density function,

    f XY (x,y) = 6x, 0 < x < 1, 0 < y < 1-x

    Derive the marginal p.d.f for X and Y, and find the marginal expected values for both X and Y.

    For the marginal of X, I got 3x(1-x) using the limits of 0 to 1-x
    and for Y I got 3, using the limits of 0 to 1.

    And using the limits from 0 to 1, I got E(X) = 1/4

    My question is what limits, would I used for E(Y), as when I use 0 to 1-x I get E(Y) = (1-x)^3 - 3(1-x)^2
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  2. #2
    MHF Contributor harish21's Avatar
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    Firstly the marginal pdf of X that you have written is incorrect.

    You can check if your odf of X is correct by integrating it from 0 to 1. The pdf should equal to 1.

    In this case:

    \displaystyle  \int_0^1 f(x)\;dx = 1

    and f(x) = 3x(1-x) does not satisfy the above stated property of the pdf.

    Therefore, your expected value of X also happens to be incorrect.

    To find the expected value of Y, you would integrate:

    E[Y] = \displaystyle \int_0^{1-x} y\;f(y)\;dy
    Last edited by harish21; February 17th 2011 at 10:14 AM.
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  3. #3
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    Thanks, but I thought to find the marginal p.d.f of X, you have to integrate with respect to y.

    edit: Sorry my mistake, my integration was incorrect, I got f(x) = -6x(x-1) which also integrate to one.
    Last edited by axa121; February 17th 2011 at 10:10 AM.
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  4. #4
    MHF Contributor harish21's Avatar
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    Thats right.

    Lastly your calculation of E[Y] is completely wrong.

    you have f(y)= 3. Use this and info in post #2 to find E[Y]
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  5. #5
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    Sorry I meant Var(Y) = (1-x)^3 - 3(1-x)^2
    for E(Y) i got 3(1-x)^2 / 2

    Thank you.
    Last edited by axa121; February 17th 2011 at 10:30 AM.
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  6. #6
    MHF Contributor harish21's Avatar
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    You need to work on your integration.

    Work properly on figuring out the value of E[Y]
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  7. #7
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    Yes, ino, i have nt done it for a while, thanks btw.

    Finally, i've got it 3(1-x)^2 / 2
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  8. #8
    MHF Contributor harish21's Avatar
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    okay so you have"

    E[X]=\dfrac{3}{2}(1-x)^2

    Var(Y) = E[X^2]-(E[X])^2

    find E[X^2] now, and then the variance
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  9. #9
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    Yep,

    E(Y^2) = (1-x)^3

    Therefore Var(Y) = (1-x)^3 [ 1 - 9(1-x)/4 ]
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  10. #10
    MHF Contributor harish21's Avatar
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    that looks good.
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