$X$ and $Y$ are two random variables, and $f_{X,Y}(x,y)=6(y-x)$, $0 < x < y < 1$
If $W=\frac{X}{Y-X}$, how can I find $f_W(w)$ and show that $W$ and $Y$ are independent? I've calculated
$A=Y-X$
$f_{X, A}(x,a)=6a$
$f_X(x)=3(x-1)^2$
$f_Y(y)=3y^2$
Now I can't find the support/domain of $W$ of $X$ to integrate to get $f_X$, $W(x,w)$ I've been stuck here for hours. Thank you for your help.
The $f_Y(y)$ is correct, just specify that $y \in (0;1)$
First of all observe that $w \in(0;\infty)$
the easiest way to proceed is to set the following system
$$\begin{cases} w=\frac{x}{y-x} \\ z=y \end{cases} $$
That is like $$\begin{cases} x=\frac{w}{1+w}z \\ y=z \end{cases} $$
Calculate the Jacobian: $|J|=\frac{z}{(1+w)^2}$
and finally, via Fundamental Transformation Theorem you get
$f_{WZ}(w,z)=6z(1-\frac{w}{1+w})\frac{z}{(1+w)^2}$
Now you can easy factorize this density in the following way
$f_{WZ}(w,z)=3z^2\mathbb{1}_{(0;1)}(z)\times\frac{2}{(1+w)^2}[1-\frac{w}{1+w}]\mathbb{1}_{(0;\infty)}(w)$
That is also
$f_{WY}(w,y)=3y^2\mathbb{1}_{(0;1)}(y)\times\frac{2}{(1+w)^2}[1-\frac{w}{1+w}]\mathbb{1}_{(0;\infty)}(w)$
that's all. Now you have $f_Y(y)$, $f_W(w)$ and you have proved that
$f_{WY}(w,y)=f_W(w)f_Y(y)$