TITLE: X,Y are independent exponentially distributed then what is the distribution of X/(X+Y)
QUESTION [7 upvotes]: Been crushing my head with this exercise. I know how to get the distribution of a ratio of exponential variables and of the sum of them, but i can't piece everything together.
The exercise goes as this:
If X,Y are independent exponentially distributed with beta = 1 (parameter of the exponential distribution = 1) then what is the distribution of X/(X+Y)
Any ideas?
Thanks a lot.

REPLY [9 votes]: Another way is to do it is to perform the following transformation. Let \begin{align*} U&=\frac{X}{X+Y} \\ V&=X+Y.\end{align*} Then \begin{align*} x(u,v)&=uv\\y(u,v)&=v-uv. \end{align*}
The determinant of the Jacobian of the transformation is
$$\left| \begin{matrix} v & u\\ -v & 1-u \end{matrix} \right|=v.
$$
Then, the joint pdf of $U$ and $V$ is
$$f_{U,V}(u,v)=f_{X,Y}(uv,v-uv)|v|=ve^{-v}, \hspace{2mm} \text{for } v\geq0, 0\leq u\leq1.
$$
Finally, "integrate out" the $v$ to obtain the marginal pdf of $U=X/(X+Y)$ as
$$f_U(u)=\int_0^{\infty}f_{U,V}(u,v)dv=\int_0^{\infty}ve^{-v}dv=1, \hspace{2mm} \text{for } 0\leq u\leq 1.$$
Hence, as Michael Hardy showed, $X/(X+Y)$ is uniformly distributed on $[0,1].$