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 $V=\frac{X}{X+Y}$
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 $V=\frac{X}{X+Y}$
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 $b > 0$, $\int_0^\infty \exp(-bx)\,\mathrm dx = b^{-1}$.
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 $(0,1)$.