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Math Help - Joint probability distribution of function of two r.vs

  1. #1
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    Joint probability distribution of function of two r.vs

    If X and Y are independent standard random variables, determine the joint density function of
    U=X, V=\frac{X}{Y}
    Then use your result to show that \frac{X}{Y} is a Cauchy distribution.

    I get:

    f(x,y)=f_X(x)f_Y(y)
    \\\\\\= \frac{1}{2\pi}e^{-(x^2+y^2)/2}

    now u=x \ and \ v=\frac{x}{y} \rightarrow y=\frac{u}{v} \ and \ x=u

    J(x,y)= \begin{vmatrix} 1 & 0 \\ \frac{1}{y} & \frac{-x}{y^2} \\ \end{vmatrix} =\frac{-x}{y^2}

    so J(x,y)^-^1 = \frac{y^2}{x}

    now f_U_,_V(u,v)=f_X_,_Y(x,y)|J(x,y)|^-^1
    = f_X_,_Y(u,\frac{u}{v^2})
     f_U_,_V(u,v)=\frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}

    f_V(v)= \int^\infty_{-\infty} \frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}\ du

    However when you evaluate the integral the answer seems to be meaningless
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  2. #2
    Moo
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    Hello,

    The mistake is in there : |J(x,y)|=|x|/y
    The absolute value will change the boundaries.

    I trust you for the formula of the change of variables. I've never seen it this way, so I can't really check.
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  3. #3
    MHF Contributor matheagle's Avatar
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    It works

    f_V(v)= \int^\infty_{-\infty} \frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{|u|}{v^2}\ du

    f_V(v)=2 \int^\infty_0 \frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}\ du

    f_V(v)= \int^\infty_0 \frac{1}{\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}\ du

    Now integrate...

    w=(-u^2/2)(1+\frac{1}{v^2})
    Last edited by matheagle; May 24th 2011 at 09:16 AM.
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