1. ## Jacobian transformation

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

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

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

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

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

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

2. Solved.