How to prove these two random variables are independent? If $X$ and $Y$ are independent Gamma random variables with parameters $(\alpha,\lambda)$ and $(\beta,\lambda)$ respectively, how to show that $U=X+Y$ and $V=X/(X+Y)$ are independent?
 A: $U$ 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*}
and 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)\\
  & = & u \times f_X \left( uv \right) \times f_Y \left( u \left( 1 - v
  \right) \right)\\
  & = & u \times \frac{1}{\Gamma \left( \alpha \right) \lambda^{\alpha}}
  \left( uv \right)^{\alpha - 1} e^{- \frac{uv}{\lambda}}\\
  & \times & \frac{1}{\Gamma \left( \beta \right) \lambda^{\beta}} \left( u
  \left( 1 - v \right) \right)^{\beta - 1} e^{- \frac{u \left( 1 - v
  \right) }{\lambda}}\\
  & = & \frac{1}{\Gamma \left( \alpha + \beta \right)} u^{\alpha + \beta - 1}
  e^{- \frac{u}{\lambda}}\\
  & \times & \frac{\Gamma \left( \alpha + \beta \right)}{\Gamma \left( \alpha
  \right) \Gamma \left( \beta \right)} v^{\beta - 1} \left( 1 - v
  \right)^{\alpha - 1}\\
  & = & f_U \left( u ; \alpha + \beta, \lambda \right) f_V \left( v ; \beta,
  \alpha \right)
\end{eqnarray*}
The second line follows from the independence of $X$ and $Y$. The third equality comes from replacing the gamma distribution densities by their values. In the end you see
that the joint density factors into the product of two marginal densities
(one gamma and one beta). Because $f_{U,V}(u,v)=f_U(u) f_V(v)$ for every $(u,v)$, that implies independence of $U$ and $V$.
