Why $2\frac{X_1}{X_1+X_2}-1$ and $X_1+X_2$ are independent if $X_1$ and $X_2$ are i.i.d. exponential?

Question

How to show that $2\frac{X_1}{X_1+X_2}-1$ and $X_1+X_2$ are independent, if $X_1$ and $X_2$ are i.i.d. exponential with mean $1$? Is there a simple way to see this?

Answer 1

\begin{align} u_1 & = x_1 + x_2 \\[8pt] u_2 & = \frac{x_1}{x_1+x_2} \\[10pt] x_1 & = u_1 u_2 \\ x_2 & = u_1(1-u_2) \\[12pt] du_1\,du_2 = \left| \det\begin{bmatrix} \dfrac{\partial u_1}{\partial x_1} & \dfrac{\partial u_1}{\partial x_2} \\[6pt] \dfrac{\partial u_2}{\partial x_1} & \dfrac{\partial u_2}{\partial x_2} \end{bmatrix} \right| \, dx_1\,dx_2& = \frac{dx_1\,dx_2}{x_1+x_2} = \frac{dx_1\,dx_2}{u_1} \\[10pt] u_1\,du_1\,du_2 & = dx_1\,dx_2 \\[15pt] e^{-x_1} e^{-x_2} \, dx_1\,dx_2 & = e^{-u_1 u_2} e^{-u_1(1-u_2)} u_1\,du_1\,du_2 \\[10pt] & = \Big( u_1 e^{-u_1} \, du_1 \Big) \Big( du_2 \Big) \end{align}

So $\dfrac{X_1}{X_1+X_2}$ is uniformly distributed on the interval from $0$ to $1$, and $X_1+X_2$ has a certain gamma distribution, and they are independent because the density factors as something depending only on $u_1$ times something depending only on $u_2$.