# Bivariate Transformation of Random Variables

Yagoda

## Homework Statement

Two RVs X1 and X2 are continuous and have joint pdf $f_{X_1,X_2}(x_1, x_2) = \begin{cases} x_1+x_2 &\mbox{for } 0 < x_1 < 1; 0 < x_2 < 1 \\ 0 & \mbox{ } \text{otherwise}. \end{cases}$

Find the pdf of $Y = X_1X_2$.

## Homework Equations

I'm using the transformation "shortcut' that says if U = g1(x,y), V = g2(x,y) and h1(u,v) = x, h2(u,v) = y then
$f_{U,V}(u,v) = f_{X,Y}(h_1(u,v), h_2(u,v) \times |J|$ where |J| is the absolute value of the determinant of the Jacobian of h1 and h2.

## The Attempt at a Solution

In this setting I let $Y = X_1X_2$ and $Z = X_1$ so $h_1(y,z) = Z$ and $h_2(y,z) = Y/Z$. Then I calculated $|J| = \frac{1}{Z}$ so plugging in I got that
$f_{Y,Z}(y,z) = (z + \frac{y}{z})(\frac{1}{z}) = 1 + \frac{y}{z^2}$ where $0 < z < 1, 0 < y < z$.
So to find the pdf of Y, integrate with respect to z over all possible values of z, but this integral does not converge when evaluated from 0 to 1.

I've tried other choices for Z, but I've ended up with the same problem.

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Note: if both $0 \le x_1 \le 1$ and $0 \le x_2 \le 1$, how does the product
$X_1 X_2$ compare in size to $X_1$ alone?
Edited to add: I have not stepped through all of your work to simplify the transformation, simply noted that you have $0 < z < 1$ and $0 < y < 1$.
Since $X_1$ and $X_2$ are both smaller than 1, then their product will be smaller than $X_1$ alone, but I'm not seeing how this helps me find the pdf of $Y$.
Are the limits $0 < y < z$ not correct? It seems that this needs to be true in order for $Y/Z$ to be less than 1?
Now that I'm looking at it again, is this transformation shortcut even valid in this situation? I'm not sure if my $h_1, h_2$ are workable in this case.