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Math Help - Obtaining f_x for E(X)

  1. #1
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    Obtaining f_x for E(X)

    I'm need some assistance with intermediate information for a problem I'm attempting to solve. For

    f(x,y)=\left\{ \begin{array}{cc} K(x^{2}+y^{2}) & 20\leq x\leq30,20\leq y\leq30\\0 & \textrm{otherwise}\end{array}\right.

    I'm attempting to solve for E(X) . As per the definition,

    E(X)=\int_{20}^{30}x\cdot f_{x}(x)dx

    Where my confusion lies is, based on my calculations,

    f_{x}=2Kx

    But, according to my solutions manual (corroborated by other sources),

    f_{x}=10Kx^{2}+.05

    I don't understand how they obtained these values. What am I missing here?
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  2. #2
    Member Abu-Khalil's Avatar
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    Re: Obtaining f_x for E(X)

    By definition, for 20\le x\le 30

    f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)dy=\int_{20}^{30}K(x^2+y^2)dy=10Kx^2+ \frac{500}3K

    So

    \mathbb E(X)=\int_{-\infty}^\infty xf_X(x)dx=\int_{20}^{30}x\left(10Kx^2+\frac{500}3K  \right)dx

    Maybe you should find the value of K.
    Last edited by Abu-Khalil; July 10th 2011 at 10:42 PM.
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  3. #3
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    Re: Obtaining f_x for E(X)

    Wait, I missed the definition of the MPDFs in the last section. So

    \begin{array}{cc}f_{X}(x)=\int_{-\infty}^{\infty}f(x,y)dy & \textrm{for }-\infty<x<\infty\end{array}

    Therefore,

    f_{X}(x)=K\int_{20}^{30}(x^{2}+y^{2})dy=K\left[yx^{2}+\frac{1}{3}y^{3}\right]_{20}^{30}=10Kx^{2}+.05

    where K=\frac{3}{380,000}.

    NOW it makes sense!
    Last edited by DenJansen; July 10th 2011 at 10:49 PM.
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