Finding the support for Transformation of Random Variables

Question

I have two independent random variables $X$ and $Y$ with exponential distribution $\sim E(5)$. $U=5X-Y$ and $V=4X+2Y$ I need to find the inequalities for the joint density $f(u,v)$. I was able to find the joint density which was $\frac{25}{14}\exp\left(-5\frac{-2(u+6)v}{14}\right)$ but I'm not sure how to proceed to find the support for the function.

Answer 1

The joint support of $(X,Y)$ is the first quadrant $[0,\infty) \times [0,\infty)$. Under the matrix transformation $$\begin{bmatrix}U \\ V\end{bmatrix} = \begin{bmatrix}5 & -1 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix},$$ it becomes clear that the joint support of $(U,V)$ must be a linear transformation of this region. Indeed, we see that when $Y = 0$, then $(U,V) = (5X, 4X)$; and when $X = 0$, then $(U,V) = (-Y, 2Y)$. Thus the joint support is the region satisfying $$\left( V \ge \frac{4U}{5} \right) \cap \left( V \ge -2U \right),$$ or equivalently, $$V \ge \max\left\{-2U, \frac{4U}{5}\right\}.$$ We could also write this as $$-\frac{V}{2} \le U \le \frac{5V}{4},$$ which perhaps is the simplest of all.

Note it is not sufficient to find the support of the marginal distributions of $U$ and $V$ separately: the joint support is needed to characterize the distribution, because in the $(U,V)$-plane, the support is not rectangular as it was in the $(X,Y)$-plane.