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Thread: Jacobian transformation

  1. #1
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    Jacobian transformation

    Here is a second one I'm not too sure what to do next :

    Let $\displaystyle X_1$ and $\displaystyle X_2$ be independent, with respective probability density functions of the forms $\displaystyle f_{X_1}(x_1) = x_1 e^{\frac{-x_1^2}{2}}$, $\displaystyle x_1>0$, and $\displaystyle f_{X_2}(x_2) = \pi^{-1} (1-x_2^2)^{\frac{-1}{2}}$, $\displaystyle |x_2|<1$.
    Find the distribution of $\displaystyle X_1X_2$.

    Here's how it goes :
    Let $\displaystyle Y_1 = X_1X_2$ and $\displaystyle Y_2 = X_1$.
    The transformation $\displaystyle Y = g(X)$ is then specified by two functions :
    $\displaystyle g_1(t_1,t_2) = t_1t_2$
    and
    $\displaystyle g_2(t_1,t_2) = t_1$
    And the inverse transformation $\displaystyle X = g^{-1}(Y)$ is $\displaystyle X_1=Y_2$ and $\displaystyle X_2=\frac{Y_1}{Y_2}$ giving
    $\displaystyle g^{-1}_1(t_1,t_2) = t_2$
    and
    $\displaystyle g^{-1}_2(t_1,t_2) = \frac{t_1}{t_2}$

    So the Jacobian of the transformation is given by
    $\displaystyle det \begin{pmatrix} 0 & 1 \\ \frac{1}{y_2} & \frac{-y_1}{y^2_2} \end{pmatrix}$
    So $\displaystyle J(y_1,y_2) =\frac{-1}{y_2}$
    Hence using the theorem $\displaystyle f_{Y_1,Y_2}(y_1,y_2) = f_{X_1,X_2}(y_2,\frac{y_1}{y_2}) |J(y_1,y_2)|
    = y_2e^{\frac{-y^2_2}{2}}\pi^{-1}(1-(\frac{y_1}{y_2})^2)^{-1/2}\frac{1}{y_2} = \frac{y_2 e^{\frac{-y^2_2}{2}}}{\pi \sqrt{(y^2_2 -y_1^2)}}$

    In order to find $\displaystyle f_{Y_1}(y_1)$ we need to calculate $\displaystyle \int f_{Y_1,Y_2}(y_1,y_2) dy_2$ over the range of $\displaystyle y_2$.

    This is where I get stuck. I'm assuming the range of $\displaystyle y_2$ is the same as the range of $\displaystyle x_1$ (so $\displaystyle y_2 > 0$).
    But it doesn't seem to be integrable.
    Plus I'm not sure what is meant by find the distribution of $\displaystyle XY$.
    Does it mean identify the distribution of $\displaystyle XY$ (i.e Normal, Gamma..) ?
    Thanks for the help.
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  2. #2
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    Solved.
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