Let $X$ and $Y$ be independent, exponentially distributed with mean 1. Show $\dfrac{X}{X+Y}$ is uniformly distributed in the interval $(0, 1)$. I suspect that to solve this we make a substitution like $U=X+Y$, $V=Y$ or something like this. Any help on getting started?
 A: This type of transformation of random variables is known as a "two-variable to one-variable" transformation. I highly recommend not using the Jacobian method for this transformation.
Instead, set $Z = \dfrac{X}{X+Y}$. Then
$$\begin{align*}
\Pr\left(Z \leq z\right) = \Pr\left(\dfrac{X}{X+Y}\leq z\right) = \Pr\left(\dfrac{X+Y}{X} \geq \dfrac{1}{z}\right) &= \Pr\left(1+\dfrac{Y}{X} \geq \dfrac{1}{z}\right) \\
&= \Pr\left(\dfrac{Y}{X} \geq \dfrac{1}{z} - 1\right) \\
&= \Pr\left(Y \geq X\left(\dfrac{1}{z}-1\right)\right)\text{.}
\end{align*}$$
Fix $z$. Then 
$$\Pr\left(Y \geq X\left(\dfrac{1}{z}-1\right)\right) = \int_{0}^{\infty}\int\limits_{x(1/z-1)}^{\infty}e^{-x}e^{-y}\text{ d}y\text{ d}x = \int\limits_{0}^{\infty}e^{-x}e^{-x(1/z-1)}\text{ d}x = \int\limits_{0}^{\infty}e^{-x/z}\text{ d}x = z\text{.}$$
Thus $F_{Z}(z) = z$ for $z \in (0, 1)$. Since $f_{Z}(z) = F^{\prime}_{Z}(z) = 1$ for $z \in (0, 1)$, $Z$ is uniformly distributed in $(0, 1)$.
A: \begin{align}
v & = \frac x {x+y}, & u & = x+y, \\[10pt]
x & = uv, & y & = u(1-v).
\end{align}
So you have $u\ge0$ and $0\le v\le 1$ and $e^{-(x+y)}= e^{-u}$.
And $dx\,dy = (\text{jacobian}\cdot du\,dv)$.
