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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 $V=\frac{X}{X+Y}$

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Distribution in what sense? Do you mean distribution of $U$ with respect to the measure on $[0,+\infty)^2$ whose derivative is $f$? –  tomasz Dec 29 '12 at 2:34
    
or maybe just a hint ??? We know that 0<u<1 i have try it with an integral but im not sure about the limits –  Gmath Dec 29 '12 at 2:39
    
@tomasz yes find p(U<u)=p(x/x+y<u)=p(x<(u/1-u)y) so the limit to x is from o to (u/1-u)y but does y go from o to infinity ??? –  Gmath Dec 29 '12 at 2:42

2 Answers 2

up vote 1 down vote accepted

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

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yeah it did it like You buy first dx and then dy but yeah V is U(o,1) however does anybody are familiar with another method ?? one where you leave Y like a constant and then do x= (u/1-u)Y and then calculate dx and then find a function of f(V,Y) and finally calculate the marginal of V ??? because thats what i have to do –  Gmath Dec 29 '12 at 4:04

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 $(0,1)$.

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Here, I adapted one of my previous answers to a different question math.stackexchange.com/a/261378/48763 –  Learner Dec 29 '12 at 3:18
    
so fu(u) =1 right ? I did it with a double integral but i was not sure thanks a lot !!! –  Gmath Dec 29 '12 at 3:31
    
Yes, $f(v)=1$ on $[0,1]$ and 0 otherwise. Please observe that I redefined the names of the variables. –  Learner Dec 29 '12 at 3:32
    
Yeah that´s what I see,thanks again and one more question if v=x/x+y is V>y or v<Y ? how can I prove that Y is greater or not that V ??? –  Gmath Dec 29 '12 at 3:34
    
I cant vote ,I do not have 15 as reputation –  Gmath Dec 29 '12 at 3:36

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