Let $U=\operatorname{min}\{X,Y\}$ and $V=\operatorname{max}\{X,Y\}$. Show that $V-U$ is independent of $U$.

Question

Let $X$ and $Y$ be exponentially distributed random variables with parameter $1$ and let $U=\operatorname{min}\{X,Y\}$ and $V=\operatorname{max}\{X,Y\}$. Show that $V-U$ is independent of $U$.

We have shown that $U$ is distributed exponentially with parameter $2$.

I am surprised to find that I don't actually know how to do this. I know of no other way than to show that $\mathbb{P}(U<x,V-U<y)=\mathbb{P}(U<x)\space\mathbb{P}(V-U<y)$ and I don't think I know how to compute the left hand side.

Can we do $$\int^{\infty}_0f_V(v)\mathbb{P}(x>U>v-y)\operatorname{dv}=\int^{\infty}_0\left(\left(\int F_X(t)F_Y(t)\operatorname{dt}\right)\left(\int^x_{v-y}2e^{-2u}\operatorname{du}\right)\right)\operatorname{dv}?$$

As $F_V(v)=F_X(v)F_Y(v)$ where $F_X$ and $F_Y$ are the distribution functions of $X$ and $Y$ respectively and $\int^x_{v-y}2e^{-2u}\operatorname{du}=\mathbb{P}(v-y<U<x)$.

I think I've seen this before but I really don't think this is what I'm meant to do, is this correct in general and is there a better way in this specific case?

Any guidance would help me out a lot, thanks!

Answer 1

This follows from properties of the Poisson process. Let two independent Poisson processes, both with rate 1, start at 0. Then $X$ is waiting time before first event in process 1, and $Y$ is waiting time before first event in process 2. If we view the two processes together, combined, it is one Poisson process with rate 2.

Then $U=\min(X,Y)$ is the waiting time before first event in the combined process, and $V-U$ is the interarrival time between first and second event in the combined process. We know that different interarrival times in a poisson process are independent, which gives the result, without any calculations.

Answer 2

Well, iff $X,Y$ are independent as well as identically exponentially distributed, then we have the following, by the Law of Total Probability:

$$\def\P{\operatorname{\mathsf P}}\begin{align}\P(U<u, V-U<w) ~=~&{ \P(X<Y,X<u,Y-X<w)+\P(Y\leqslant X,Y<u,X-Y<w)}\\[1ex] =~& \P(X<Y<w+X, X<u)+\P(Y\leqslant X<w+Y, Y< u) \\[1ex] =~& \int_0^u e^{-x}\P(x<Y<w+x\mid X=x)\operatorname d x+\int_0^u e^{-y}\P(y\leqslant X<w+y\mid Y=y)\operatorname d y\\[3ex] \overset{\text{iid}}=~& 2\int_0^u e^{-x}(e^{-x}-e^{-(w+x)})\operatorname d x\\[1ex] =~& 2(1-e^{-w})\int_0^u e^{-2x}\operatorname d x \\[1ex] =~& (1-e^{-w})(1-e^{-2u})\end{align}$$

Do similarly for $\P(U<u)$ and $\P(V-U<v)$