# If $X, Y$ ~ $\exp{\alpha}$ determine the PDF of $\frac {X}{X+Y}$

Let $$X,Y$$ be independent random variables the are distributed $$\exp{(\alpha)}$$ whereby $$\alpha > 0$$

Determine the Distribution Density of $$\frac {X}{X+Y}$$

My idea:

I believe it is too simple to suggest:

$$f_{ \frac {X}{X+Y}}(x,y)=\frac{\alpha\exp{(-\alpha x)}}{\alpha\exp{(-\alpha x)}+\alpha\exp{(-\alpha y)}}=\frac{\exp{(-\alpha x)}}{\exp{(-\alpha x)}+\exp{(-\alpha y)}}$$

But I fail to see another way of finding the joint PDF.

Any ideas?

Make the change of variables $$W = X$$ and $$Z = \displaystyle\frac{X}{X+Y}$$, where $$W\geq 0$$ and $$0\leq Z\leq 1$$. Since $$X$$ and $$Y$$ are independent, we obtain the following result \begin{align*} f_{W,Z}(w,z) = f_{X,Y}\left(w,\frac{w(1 - z)}{z}\right)|\det J(w,z)| = f_{X}(w)f_{Y}\left(\frac{w(1 - z)}{z}\right)|\det J(w,z)| \end{align*}
Once you have the expression of $$f_{W,Z}(w,z)$$ at hand, you determine its marginal distribution related to $$Z$$, which is the random variable whose distribution you are interested in. Can you proceed from here?