Suppose that A and B are 2 r.v with joint probability distribution given by $f_{AB}(a,b) = -\frac{1}{2}(ln(a)+ln(b))$, if $0 \lt a \lt 1$ and $0 \lt b \lt 1$, and 0 otherwise.
Define $X = A+B$ and $Y = \frac{A}{B}$.
How do I derive the joint distribution of $f_{XY}(x,y)$?
Progress: I know that X will take values $0 \lt x \lt 2$, and the Jacobian = $2y$ . Any help will be much appreciated.
The jacobian results to me
$$|J|=\frac{x}{(y+1)^2}$$
and after standard calculations the joint density results to me
$$f_{XY}(x,y)=\frac{-x}{2(y+1)^2}\log\frac{x^2y}{(y+1)^2}\cdot\left[\mathbb{1}_{(0;1]}(x)\mathbb{1}_{(0;+\infty)}(y)+\mathbb{1}_{(1;2)}(x)\mathbb{1}_{\left(x-1;\frac{1}{x-1}\right)}(y)\right]$$
This because when $x\in(0;1]$ there are no problems but when $1<x<2$, in order to have
$$0<\frac{xy}{y+1}<1$$
And
$$0<\frac{x}{y+1}<1$$
You need to have
$$x-1<y<\frac{1}{x-1}$$
Alternative notation
Joint support can alternatively be expressed by
$$0<x<\min\left(\frac{y+1}{y};y+1\right)$$
$$0<y<+\infty$$
To understand this alternative notation it is enough to do a drawing of support region