How to find limit of integration for marginal densities after transformation?

Question

Here is the problem and solution for part b, which I need help with :

How do you get the highlighted limits of integration?

Four equations I have are:

$u = xy$

$v = x/y$

$x = \sqrt{uv}$

$y = \sqrt{u/v}$

I feel like u(v) and v(u) were found, graphed, then used to find the region to integrate. I really have no idea what's going on here, so any help would be greatly appreciated.

Thank you!

Answer 1

Note: I think the answer provided is wrong. For $0<v<1$, the limits of integration should be $\frac{1}{v}, \infty$ instead of $\frac{1}{2}, \infty$.

We begin by noting that $u \in [1,\infty)$ and $v \in [0, \infty)$. This is easily derived from the limits of $x$ and $y$.

This is meant to be more intuitive than rigorous (although I will do my best to not be hand-wavy), but I hope that the idea is communicated sufficiently:

We first show that the support for $f_{u,v}$ does not include any of the points $$\{ (u,v): v>u \}$$

(i.e. $v$ must always be smaller than $u$)

We do so by contradiction. Suppose $v>u$. Then $$\frac{v}{u} > 1 \implies \frac{x}{y} \frac{1}{xy} > 1 \implies \frac{1}{y^2} > 1$$

Which is a contradiction since $y \geq 1$

Note that the points $$\{ (u,v): v=u \}$$ are in the support, since $u = v$ whenever $y = 1$.

Now, we show that the support for $f_{u,v}$ does not include any of the points $$\{ (u,v): v < \frac{1}{u} \}$$

Again, we show this by contradiction. Suppose $v < \frac{1}{u}$. Then

$$uv < 1 \implies x^2 < 1 $$

Which is a contradiction, since $x \geq 1$

We also note that the points

$$\{ (u,v): v = \frac{1}{u} \}$$ are in the support, since $ v = \frac{1}{u}$ whenever $x = 1$.

So now, we know that the support $S_{u,v}$ must be a subset of the set

$$S = \{(u,v): v \leq u, v \geq \frac{1}{u}\}$$

It turns out that every point in the set $S$ is a point in the support (If you don't want to see the proof, just skip to the image below). To prove this, suppose $(u,v) \in S$, then we will have:

$$ \begin{align} &v \leq u \text{ and } uv \geq 1 \\ \implies &y^2 \geq 1 \text{ and } x^2 \geq 1 \\ \implies &y \geq 1 \text{ and } x \geq 1 \\ \implies &(x,y) \in supp(f_{x,y}) \\ \implies &(u,v) \in supp(f_{u,v}) \end{align} $$

Where the last implication follows from the fact that the mapping $(x,y) \to (xy, \frac{x}{y})$ is bijective.

Proof of bijection:

1. Injective: If there are points $(x_1,y_1),(x_2,y_2) \in supp(f_{x,y})$ such that $x_1y_1 = x_2y_2$ and $\frac{x_1}{y_1} = \frac{x_2}{y_2}$, then $$\begin{align} \implies & (x_1 = x_2 = 0 \text{ or } y_1 = y_2) \text{ or } (x_1 = x_2 \text{ or } y_1 = y_2 = 0) \end{align}$$ Either way we are done.
2. Surjective: Trivial. Just look at the range of $(xy,\frac{x}{y})$ which we are working with.

Now, you can easily conclude that the support $S_{u,v}$ is indeed given by $S$, and shade that region out (apologies for the rather ugly picture):

This gives the region of integration from which the limits that are taken should be fairly obvious.