Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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 $b > 0$, $\int_0^\infty \exp(-bx)\,\mathrm dx = b^{-1}$.

share|cite|improve this answer
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 $(0,1)$.

share|cite|improve this answer
Here, I adapted one of my previous answers to a different question – 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.