Show that $X-Y\mid X \sim \exp(\lambda)$

Question

I'm clueless with this one:

Let $X$ and $Y$ be the minimum and the maximum of two independent exponentially distributed random variables with parameter $\lambda>0$. Show that $Y-X\mid X \sim \exp(\lambda)$.

I will appreciate any hint.

Answer 1

Here is a sketch. It requires some simple computations. No doubt there is a more elegant argument, but here it is.

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Suppose $X_1,X_2$ are iid with exponential distribution with parameter $\lambda$. $Y=X_1\vee X_2$ and $X=X_1\wedge X_2$. It is easy to check that $X$ has exponential distribution with parameter $2\lambda$: \begin{align} P[X\leq t]&=P[X_1\leq t, X_2\leq X_1]+ P[X_2\leq t, X_1< X_1]\\ &=\lambda^2\int^\infty_0 e^{-\lambda x_1}\mathbf{1}_{(x_1\leq t)}\int^\infty_{x_1}e^{-\lambda x_2}\,dx_2\,dx_1 + \lambda^2\int^\infty_0 e^{-\lambda x_2}\mathbf{1}_{(x_2\leq t)}\int^\infty_{x_2}e^{-\lambda x_1}\,dx_1\,dx_2\\ &= 2\lambda\int^t_0e^{-\lambda u}\,du \end{align}

Notice that $Y-X=|X_1-X_2|$. Then for any Borel set $A\subset(0,\infty)$ \begin{align} P[Y-X\leq t,X\in A]&=\lambda^2\int^\infty_0\int^\infty_0\mathbf{1}(X_1\wedge X_2\in A)\mathbf{1}{(|X_1-X_2|\leq t)}e^{-\lambda(x_1+x_2)}\,dx_1dx_2\\ \end{align} Splitting the integral where $X_1\leq X_2$ and $X_2< X_1$ gives \begin{align} P[Y-X\leq t,X\in A]&=(1-e^{-\lambda t} )\,2\lambda\int^\infty_0\mathbf{1}_A(x)e^{-2\lambda x}\,dx \end{align} Incidentally, this also shows that in fact $Y-X=|X_1-X_2|$ and $X=X_1\wedge X_2$ are independent. This is also a manifestation of the memoryless property of the exponential distribution.

Answer 2

Let the original variables be $X_1$ and $X_2$. Then

\begin{eqnarray*} P(Y-X\gt a\mid X=x) &=& P(X_1-X_2\gt a\mid X_2=x,X_1\ge X_2)\quad\quad\;\text{(by symmetry)} \\ &=& P(X_1-x\gt a\mid X_1\ge x)\\ &=& P(X_1\gt x+a\mid X_1\ge x)\\ &=& P(X_1\gt a)\;.\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(by memorylessness)} \end{eqnarray*}

Thus $Y-X\mid X$ is distributed as $X_1$, and thus $\sim\exp(\lambda)$.