Find and identify the distribution of $W = \frac{Y}{X+Y}$ Suppose the joint PDF of X and Y is $f_{x,y}(x,y)=e^{-x-y},x,y\geq0$.
Find and identify the distribution of $W = \frac{Y}{X+Y}$. 
I know that since $x,y\geq0$, then $0\leq w\leq 1$. I want to do this by first finding $F_W (w)=P(W\leq w)=P(\frac{Y}{X+Y}\leq w)$. I need to figure out what region is represented by the inequality $\frac{Y}{X+Y}\leq w$, but I'm not sure how to represent this in the x,y plane. Any suggestions? Once I do that, I know how to do the integration. 
 A: $X$ and $Y$ are strictly positive, and thus too is $W$.   (Further, the support for $W$ is $(0;1]~$. ) 
So when $w\in(0;1]$ then $Y/(X+Y)\leq w$ is equivalent to $Y\leq w(X+Y)$ or $Y\leq wX/(1-w)$.
So taking it one step at a time, this is simply:
$$\mathsf P(W\leq w)~{=~\mathsf P\big(Y\leq wX/(1-w)\big) \\ = \mathbf 1_{w\in(0;1]}\cdot\iint\limits_{y\leqslant wx/(1-w)} f_{X,Y}(x,y)\operatorname d (x, y)\\= \mathbf 1_{w\in(0;1]}\cdot\iint\limits_{y\leqslant wx/(1-w)} e^{-(x+y)}\mathbf 1_{0\leqslant x,0\leqslant y}\operatorname d (x, y)\\ = \mathbf 1_{w\in(0;1]}\cdot\int\limits_0^\infty e^{-x}\int\limits_0^{wx/(1-w)} e^{-y}\operatorname d y\operatorname d x}$$
Then of course, you don't have to integrate that all, if you just want the pdf.$$f_W(w) ~{= \mathbf 1_{w\in(0;1]}\cdotp\tfrac{\mathrm d ~~}{\mathrm d~w}\int_0^\infty\int_0^{g(w,x)} f_{X,Y}(x,y)\operatorname d y\operatorname d x \\ = \mathbf 1_{w\in(0;1]}\cdotp\int_0^\infty g^{(1,0)}(w,x)\cdot f_{X,Y}(x, g(w,x))\operatorname d x  } $$
Where $g(w,x)= wx/(1-w)$ and $g^{(1,0)}(w,x)=x/(1-w)^2$.
A: $$\frac{Y}{X+Y} \leq w \implies Y \leq \frac{wX}{1-w} \equiv Y \leq kX$$
where $k = \frac{w}{1-w}$. It is the region below and including $Y = kX$ and the integration will be $\int_{0}^{\infty}\int_{0}^{kx}f_{X,Y}(x,y)dydx$. The value of the integral will be a function of $k$ (equivalently of $w$) and to get the pdf you need to differentiate with respect to $w$.
