Joint PDF of $U=XY$ and $V=\frac{X}{Y}$ Given that $X$ and $Y$ are jointly distributed with pdf:
$$f(x, y)=\left\{\begin{array}{cc}
\frac{1}{x^{2} y^{2}}, & x \geqslant 1, y \geqslant 1 \\
0 & \text { else }
\end{array}\right\}$$
Find the joint pdf of $U=XY, V=\frac{X}{Y}$
My try:
First i found the Jacobian of the transformation:
$$\begin{aligned}
&\Rightarrow \quad x=\sqrt{u v} \quad y=\sqrt{\frac{u}{v}} \\
&J=\left|\begin{array}{ll}
\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\
\frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}
\end{array}\right|=\left|\begin{array}{ccc}
\frac{\sqrt{v}}{2 \sqrt{u}} & \frac{1}{2 \sqrt{u v}} \\
\frac{\sqrt{u}}{2 \sqrt{v}} & \frac{-1}{2} \sqrt{u} v^{\frac{-3}{2}}
\end{array}\right|=\frac{-1}{2v}
\end{aligned}$$
Now we have:
$$f(u, v)=|J| f\left(x\left(u, v\right), y(u, v)\right)$$
$$\begin{aligned}
&f(u,v)=\frac{1}{2 v} \frac{1}{(\sqrt{u v})^{2}} \frac{1}{\left(\sqrt{\frac{u}{v}}\right)^{2}} \\
&f(u,v)=\frac{1}{2 v u^{2}}
\end{aligned}$$
We have $u=xy \geq 1 $
But i am not able to find limits of $v$?

$x \geq 1$ and $y \geq 1$ become $\sqrt {uv} \geq 1$ and $\sqrt {\frac  u v} \geq 1$. So the limits are $\frac 1 u \leq v \leq u$ and $u \geq 1$.
