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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.

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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 $$ f_{Z,U}(z,u)=\begin{cases} 2z^{-3}u^{-1}, & 1<u<z^2,\cr 0, & \text{elsewhere}\end{cases} $$

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  • $\begingroup$ 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. $\endgroup$ – JohnTravolski Nov 13 '17 at 6:50
  • $\begingroup$ $x=\frac{z^2}{u} > 1$. $\endgroup$ – NCh Nov 13 '17 at 12:35

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