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$X$ and $Y$ have density given by $$f_{X, Y}\left(x, y\right) = \frac{1}{x^2y^2}$$

where $x \geq 1$, $y \geq 1$. Let $U = 4XY$ and $V = \frac{X}{Y}$. The joint density of $U, V$ is given by

$$f_{U, V}\left(u, v\right) = \mid J\left(x(u, v), y(u, v)\right)\mid^{-1} f_{X, Y}\left(x(u, v), y(u, v)\right)$$

The Jacobian \begin{vmatrix}4y&4x\\\frac{1}{y}&\frac{-x}{y^2}\end{vmatrix} evaluates to $\frac{-4x}{y} - \frac{4x}{y} = \frac{-8x}{y}$. Then, we solve for $u$ and $v$ in order to plug into the expression for the density. Solving the system of equations for $x$ and $y$, we get

$$x = \frac{(uv)^{\frac{1}{2}}}{2}$$ $$y = \frac{1}{2} \left(\frac{u}{v}\right)^{\frac{1}{2}}$$

Then we plug into the equation for the density for $u$ and $v$.

$$f_{U, V}\left(u, v\right) = \frac{-8x}{y} \frac{1}{x^2y^2} = \frac{-8 \frac{(uv)^{\frac{1}{2}}}{2}}{\frac{1}{2} \left(\frac{u}{v}\right)^{\frac{1}{2}}} \frac{1}{\left(\frac{(uv)^{\frac{1}{2}}}{2}\right)^{2} \left(\frac{1}{2} \left(\frac{u}{v}\right)^{\frac{1}{2}}\right)^{2}} = \frac{2}{u^2v}$$

Now, the domain of $u$ and $v$ is what really confuses me. I know that $x, y \geq 1$. So my thought was that since $u = 4xy$ and $v = \frac{x}{y}$, then $u \geq 4$ and $0 \leq v \leq \infty$. This is correct for $u$, but it is incorrect for $v$, and I am very confused why?

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$u>4$ is okay, but you also need $(\sqrt{uv})/2>1$ and $(\sqrt{u/v})/2>1$.

That is $v>4/u$ and $v>4u$.

So your domain is $$\{\langle u,v\rangle: 4<u~,~ 4/u<v<u/4\}$$

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  • $\begingroup$ How would you find the marginal density of for V. I know that the way to do it is by integrating the joint density with respect to $u$, but I am getting a wrong answer for this. Any ideas? What should the bounds be? $\endgroup$ – Eoin S Mar 26 at 16:23

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