Joint distribution of $X_1/(X_1+X_2)$ and $X_2/(X_1+X_2)$ for independent exponential random variables $X_1$ and $X_2$ If $X_i$, $i=1,2$, are independent $gamma(\alpha_i,1)$ random variables.
Find the joint distribution of $X_1/(X_1+X_2)$ and $X_2/(X_1+X_2)$.
I am trying to use transformation method to solve it. 
Let $U=X_1/(X_1+X_2)$ and $V=X_2/(X_1+X_2)$, but I cannot represent $X_1=g_1(U,V)$ or $X_2=g_2(U,V)$
 A: As I said in my comments, $W = \frac{X_1}{X_1+X_2}$ and 
$Z = \frac{X_2}{X_1+X_2}$ do not have a joint density function
(joint pdf),
and all the probability mass lies on the line segments with end
points $(0,1)$ and $(1,0)$.  You can work out the joint distribution function (joint CDF) $F_{W,Z}(w,z)$ (which,
incidentally, is what the question asks for) quite easily.
Some values can be put down by inspection:
$$F_{W,Z}(w,z) = 
\begin{cases}
0, & w \leq 0 ~\text{or}~ z \leq 0 ~\text{or}~ w+z \leq 1,\\
1, & w \geq 1 ~\text{and}~ z \geq 1.
\end{cases}$$
For $0 < w < 1, z > 1$,
\begin{align}
F_{W,Z}(w,z) &= P\{W \leq w, Z \leq z\}\\
&= P\left\{\frac{X_1}{X_1+X_2} \leq w, 
\frac{X_2}{X_1+X_2} \leq z\right\}\\
&= P\left\{\frac{X_1}{X_1+X_2} \leq w\right\}\\
\end{align}
and you say you know how to compute that last probability since you say
in a comment on your question that "It is not hard to find the
pdf of one of them". 
Similarly,
for $w > 1, 0 < z < 1$,
$$F_{W,Z}(w,z) = P\left\{\frac{X_2}{X_1+X_2} \leq z\right\}
= P\left\{1-z \leq \frac{X_1}{X_1+X_2}\right\}$$
which you say you can compute.  Finally, for the
region $0 < w,z < 1, w+z > 1$, we have that
$$F_{W,Z}(w,z) 
= P\left\{1 - z \leq \frac{X_1}{X_1+X_2} \leq w\right\}$$
which also can be computed.  Thus, we have found the joint CDF
for $W$ and $Z$. Note that only in the last case above is
$F_{W,Z}(w,z)$ a function of both $w$ and $z$, but even here,
$\displaystyle \frac{\partial^2F_{W,Z}(w,z)}{\partial w\partial z}$
is always $0$ thus adding credibility to the claim that $W$ and $Z$
do not have a joint density function.
A: you can find out the distribution of one of them using Jacobian, and if you chose $U$ it will follow $\beta(\alpha_1,\alpha_2)$ distribution. And as you can cee $U$ and $V$ adds upto 1. $V$ will just be equal to $1-U$.So they are not jointly continuous. so you will only be able to find out probabilities like:
$P(U<t,V>1-t)$ and this will be equal to $P(U<t)$.
