<s>[INST] Bounds for change of variable U and V are independent and uniformly distributed on (0,1). Find the joint density of $X=\frac{\sqrt{U}}{\sqrt{U}+\sqrt{V}}$ and $Y={\sqrt{U}+\sqrt{V}}$.
I've been able to find without a hitch that $f_{XY}(x, y)=4(1-x)xy^3$ for some values of X and Y. My textbook confirmed that this is correct. My problem comes from evaluation the values of X and Y over which the joint pdf takes a positive value. I reasoned that, since $0<u<1$ and $0<v<1$, that $0<x<1$ and $0<y<2$. Integrating the pdf over these values of x and y, however, does not yield one. The textbooks says that the bounds are $0<x<1$ and $0<y<\min\left(\frac{1}{x}, \frac{1}{1-x}\right)$, which indeeds integrate to 1. How were these bounds for y obtained? I have a feeling that the answer might be connected to the fact that $X=\frac{\sqrt{U}}{Y}$, but I am unable to explain why.
 [/INST] $X=\frac{\sqrt{U}}{\sqrt{U}+\sqrt{V}}$
$X = \frac{\sqrt{U}}{Y}$
$ XY = \sqrt{U}$
$ \sqrt{U}$ implies $0\le xy \le 1$ implies $0\le y \le \frac{1}{x}\tag 1$
and
$X=\frac{\sqrt{U}}{\sqrt{U}+\sqrt{V}} = \frac{\sqrt{U}+\sqrt{V}-\sqrt{V}}{\sqrt{U}+\sqrt{V}} $
$X = 1-\frac{\sqrt{V}}{\sqrt{U}+\sqrt{V}} = 1-\frac{\sqrt{V}}{Y}$
$(1-X)Y = \sqrt{V}$
$\sqrt{V}$ implies $0\le (1-x)y \le 1$ implies $0\le y \le \frac{1}{1-x}\tag 2$
Combining (1) and (2)
We get
$0\le y \le min(\frac{1}{x},\frac{1}{1-x})$
</s>