Bivariate Transformation of Random Variables

[Yagoda]

Homework Statement

Two RVs X₁ and X₂ are continuous and have joint pdf <br /> f_{X_1,X_2}(x_1, x_2) = \begin{cases} x_1+x_2 &amp;\mbox{for } 0 &lt; x_1 &lt; 1; 0 &lt; x_2 &lt; 1<br /> \\<br /> 0 &amp; \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 &lt; z &lt; 1, 0 &lt; y &lt; 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.

 

[statdad]

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 &lt; z &lt; 1 and 0 &lt; y &lt; 1.

 

[Yagoda]

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 &lt; y &lt; 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.