# Independence of $\min(X,Y)$ and $|X-Y|$ when $(X,Y)$ are i.i.d. exponential

I am preparing for upcoming exam and trying to do all examples from book. I think I may call it homework. I stuck on one particular one. Already did a lot but stuck on something and cant find how to get out.

So we got two independent random variables X and Y and their laws.
\begin{align*}
P_X(dx)= P_Y(dx) = e^{-x} \mathbb{1}_{\mathbb{R}_+}dx
\end{align*}
And functions $U=\min\{X,Y\}$ and $V=|X-Y|$.

I have to prove, that U and Z are independent random variables.

I use formula $\mathbb{E}(\gamma(X)\psi(Y)) = \mathbb{E}(\gamma(X)) \mathbb{E}(\psi(Y))$.

I started with calculating the density. I got that:
\begin{align*}
P_U(dx)&=2e^{-2x}\mathbb{1}_{\mathbb{R}_+}dx\\
P_V(dx)&=e^{-x}\mathbb{1}_{\mathbb{R}_+}dx
\end{align*}

Then I got stuck on the integral.
\begin{align*}
\mathbb{E}(\gamma(U)\psi(V))&=\int^{\infty}_{0} \left[ \int^{\infty}_{0} \gamma(\min\{X,Y\})\psi(|X-Y|)e^{-x}dx\right]e^{-y}dy=\\
&=\int^{\infty}_{0} \left[ \int^{\infty}_{0} \gamma(U)\psi(V)e^{-x}dx\right]e^{-y}dy
\end{align*}
In the previous example with two variable functions, prof used substitution of variables. I tried to write integral using variables $V$ and $U$ only, but since these functions have singularities I think we cant use that trick here. So maybe someone have encountered such problems before and knows way out.

#### Solutions Collecting From Web of "Independence of $\min(X,Y)$ and $|X-Y|$ when $(X,Y)$ are i.i.d. exponential"
By definition and symmetry, for every bounded measurable $\varphi$, $\mathrm E(\varphi(U,V))=(\ast)$ with
$$(\ast)=2\mathrm E(\varphi(X,Y-X):X\lt Y)=2\iint [0\lt x\lt y]\,\varphi(x,y-x)\mathrm e^{-x-y}\mathrm dx\mathrm dy.$$
The change of variable $u=x$, $v=y-x$ yields $x=u$, $y=u+v$, hence the Jacobian is $1$ and
$$(\ast)=2\iint [0\lt u,0\lt v]\,\varphi(u,v)\mathrm e^{-2u-v}\mathrm du\mathrm dv=\mathrm E(\varphi(\tfrac12X,Z)),$$
where $Z$ is standard exponential and independent of $X$. Since this holds for every $\varphi$, $U$ and $V$ are independent and exponential with $2U$ and $V$ standard exponential. Thus, $U$ is distributed like $\frac12X$ and $V$ like $X$.