# Bivariate Transformation of Random Variables

Problem. If $X$ and $Y$ measure the lifetimes of two components operating independently. Suppose each has density (in unit of 100 hours)

$$f(x) = \begin{cases} \frac{1}{x^2}, & \text{if } x > 1 \\ 0, & \text{elsewhere}, \end{cases}$$

If $Z = \sqrt{XY}$ measures the quality of the system, show that $Z$ has density function

$$f(z) = \begin{cases} 4\frac{\ln(z)}{z^3}, & \text{if } z > 1 \\ 0, &\text{elsewhere} \end{cases}$$

I use the substitutions $Z = \sqrt{XY}$ and $U = Y$ to obtain that the Jacobian is $-2z/u$, but then when I try to solve for the marginal distribution of $z$, I obtain a divergent integral! My joint distribution function for $u$ and $z$ comes out to be $2 z^{-3} u^{-1}$. I'm not sure what's going wrong.

When substituting $x=z^2/u$ and $y=u$ into $$f_{Z,U}(z,u)=|J|f_X(x)f_Y(y)=|J|f_X(z^2/u)f_Y(u),$$ do not forget that both densities are non-zero only when the variable is greater than $1$. So, $$f_{Z,U}(z,u)=\frac{2z}{u}\frac{1}{(z^2/u)^2}\mathbb 1_{\{z^2/u > 1\}}\frac{1}{u^2}\mathbb 1_{\{u>1\}},$$ and the joint pdf is $$f_{Z,U}(z,u)=\begin{cases} 2z^{-3}u^{-1}, & 1<u<z^2,\cr 0, & \text{elsewhere}\end{cases}$$
• I appreciate this, but what exactly was the method used for establishing that $u < z^2$? That's the only part that I was struggling with. – JohnTravolski Nov 13 '17 at 6:50
• $x=\frac{z^2}{u} > 1$. – NCh Nov 13 '17 at 12:35