Distribution of a transformation Hi I just can´t with this problem.
We have $f(x,y)=e^{-(x+y)}$ with $x,y$ from 0 to $\infty$
Find the distribution of $V=\frac{X}{X+Y}$
 A: Define $V=\frac{X}{X+Y}$ and a second random variable $U = X + Y$ , then $U$ and $V$ are
obtained from $X$ and $Y$ by the transformation
\begin{eqnarray*}
  \left( \begin{array}{c}
    U\\
    V
  \end{array} \right) & = & \left( \begin{array}{c}
    X + Y\\
    \frac{X}{X + Y}
  \end{array} \right)
\end{eqnarray*}
giving rise to the inverse transformation
\begin{eqnarray*}
  \left( \begin{array}{c}
    X\\
    Y
  \end{array} \right) & = & \left( \begin{array}{c}
    UV\\
    U \left( 1 - V \right)
  \end{array} \right)
\end{eqnarray*}
The Jacobian of the transformation (which is the absolute value of
determinant)
$$ \left| J \right| = \left| \begin{array}{cc}
     V & U\\
     1 - V & - U
   \end{array} \right| = U $$
(because $U$ is a positive random variable).
Which implies that the joint density of $U$ and $V$ is
\begin{eqnarray*}
  f_{U, V} \left( u, v \right) & = & u \times f_{X, Y}  \left( uv, u \left( 1
  - v \right) \right)\mathbf 1_{0 < u < \infty}\mathbf 1_{0 < v < 1}\\
  & = & u \times \exp \left( - uv - u \left( 1 - v \right) \right)\mathbf 1_{0 < u < \infty}\mathbf 1_{0 < v < 1}\\
  & = & [u \exp \left( - u \right)\mathbf 1_{0 < u < \infty}]\mathbf 1_{0 < v < 1}\\
& = & f_U(u)f_V(v)
\end{eqnarray*}
This means the marginal of $U$ has density $u \exp \left( - u \right)$ on
$\left( 0, \infty \right)$ and 0 otherwise and that the marginal distribution
of $V$ is uniform $\left( 0, 1 \right)$ (and both are independent).
Thus, the answer for the distribution you are looking for is uniform on $(0,1)$.
A: More directly than @Learner's answer,  $X, Y \in (0, \infty)$, and so obviously
$\frac{X}{X+Y}$ takes on values in $(0,1)$. Now, for any $\alpha, ~0 < \alpha < 1$,
$$\begin{align*}
F_{\frac{X}{X+Y}}(\alpha) &= P\left\{\frac{X}{X+Y} \leq \alpha\right\}\\
&= P\left\{Y \geq \frac{1-\alpha}{\alpha}X\right\}\\
&= \int_{x=0}^\infty\int_{y=\frac{1-\alpha}{\alpha}x}^\infty 
\exp(-x-y)\,\mathrm dy\,\mathrm dx\\
&= \int_{x=0}^\infty\exp(-x)\exp\left(-\frac{1-\alpha}{\alpha}x\right)\,\mathrm dx\\
&= \int_{x=0}^\infty\exp\left(-\frac{1}{\alpha}x\right)\,\mathrm dx\\
&=\alpha
\end{align*}$$
and so $\frac{X}{X+Y} \sim U(0,1)$. The integrals are not hard to carry
out explicitly and can even be done by inspection and judicious use of
standard results such as $P\{Y > a\} = \exp(-a)$ for exponential random
variable $Y$ with parameter $1$, and  for
$b > 0$, $\int_0^\infty \exp(-bx)\,\mathrm dx = b^{-1}$.
