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Math Help - [SOLVED] joint probability density function

  1. #1
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    [SOLVED] joint probability density function

    Given two iid Gaussian Random Variables X,Y with zero mean and variance \sigma^2

    U = \frac{X^2-Y^2}{\sqrt{X^2+Y^2}}

    V = \frac{2XY}{\sqrt{X^2+Y^2}}

    Derive the joint pdf f_{UV} (u,v)

    Can someone help me with this?
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  2. #2
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    Quote Originally Posted by akane View Post
    Given two iid Gaussian Random Variables X,Y with zero mean and variance \sigma^2

    U = \frac{X^2-Y^2}{\sqrt{X^2+Y^2}}

    V = \frac{2XY}{\sqrt{X^2+Y^2}}

    Derive the joint pdf f_{UV} (u,v)

    Can someone help me with this?
    You (should) know the joint pdf of X and Y. Now apply the Change of Variable Theorem:

    If the random variables X_1 and X_2 have joint pdf function f(x_1, x_2) then the pair of random variables Y_1 = Y_1(X_1, X_2) and Y_2 = Y_2(X_1, X_2) have joint pdf given by g(y_1, y_2) = f(x_1(y_1, y_2), x_2(y_1, y_2)) |J(y_1, y_2)| if (y_1, y_2) is in the range of (Y_1, Y_2) and zero otherwise, where J(y_1, y_2) = \left| \begin{array}{ccc} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_2}{\partial y_1} \\ & \\ \frac{\partial x_1}{\partial y_2} & \frac{\partial x_2}{\partial y_2} \end{array} \right|.
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  3. #3
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    Thanks for your help.

    f_{X,Y} (x,y) = f_{X}(x) f_{Y}(y) = \frac{1}{2 \pi \sigma^2} e^{-{\frac{x^2+y^2}{2 \sigma^2}}}

    f_{UV} (u,v)= |J(u,v)| f_{X,Y} (x(u,v),y(u,v))

     J (u,v)= \frac {\partial(x,y)}{\partial(u,v)} = \left|  \begin{array}{cc} \frac{\partial x}{\partial u}& \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right|

    To get J I should know x = x(u,v) and y = y(u,v).
    Since I couldn't derive these functions from expressions for U and V, I did the opposite thing, that is

    \frac {\partial(u,v)}{\partial(x,y)} = \left| \begin{array}{cc} \frac{\partial u}{\partial x}& \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ \end{array} \right| and obtained 2.

    So, I have this now:

     f_{UV}(u,v) \frac{\partial(u,v)}{\partial(x,y)} = f_{XY}(x(u,v),y(u,v))

    Hopefully this way is correct. Now I have

    f_{UV}(u,v) = \frac{1}{2} \frac{1}{2 \pi \sigma^2} e^{-{\frac{x^2+y^2}{2 \sigma^2}}}

    I got f_{UV} as a function of x and y, and don't know how to substitute x and y to get f_{UV} as a function of u and v.

    Assistant mentioned that we should use this substitution

     A = \sqrt {X^2 + Y^2}
     B = \tan {\frac{Y}{X}}

    I have tried everything I could think of, but I'm still not sure how does that help.
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