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Math Help - Joint Probability Density Function/Finding expected Value

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    Joint Probability Density Function/Finding expected Value

    The Joint Probability Density Function of X and Y is
    f(x,y) = c(y^2 - 64x^2)e^-y, -y/8<= x <= y/8, 0<y<infinity

    So the integral would look something like this?
     \int_{0}^{\inf} \int_{-y/8}^{y/8} (y^2-64x^2)e^{-y} dxdy


    Find c and the expected Value of X.

    But I don't know how where to go from here or how to get the expected value

    C is supposed to be 1 and E[X] should be 0.
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    MHF Contributor harish21's Avatar
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    Re: Joint Probability Density Function/Finding expected Value

    Quote Originally Posted by bjnovak View Post
    The Joint Probability Density Function of X and Y is
    f(x,y) = c(y^2 - 64x^2)e^-y, -y/8<= x <= y/8, 0<y<infinity

    So the integral would look something like this?
     \int_{0}^{\infty} \int_{-y/8}^{y/8} (y^2-64x^2)e^{-y} dxdy


    Find c and the expected Value of X.

    But I don't know how where to go from here or how to get the expected value

    C is supposed to be 1 and E[X] should be 0.
    If you have studied probability density functions carefully, the joint pdf should integrate to 1(by definition). so solve for "c" using the following (its simple integration), and then recall the definition of expected value to find E[X]..

    \int_{0}^{\infty} \int_{-y/8}^{y/8} c \cdot (y^2-64x^2) \cdot e^{-y} dxdy\;=\;1
    Last edited by harish21; December 5th 2011 at 05:30 PM.
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