# Domain for variables in joint distribution under change of variable.

$$X$$ 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?

$$u>4$$ 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
• 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? – Eoin S Mar 26 at 16:23