How to show the difference of max and min of exponential random variables is exponential? Let $X \sim \exp(\lambda_1)$ and $Y \sim \exp(\lambda_2)$ be two exponential random variables. Let $M= \max(X,Y)$ and $L= \min(X, Y)$. We know that $M -L = |X-Y|$. 
How to show $M -L$ is distributed exponentially?
My try:
$$
P(M-L \leq t) = P( |X-Y| \leq t)=\int_{-\infty}^\infty \int_{x = y- t}^{x= y+t} \lambda_1 e^{-\lambda_1x}\lambda_2e^{-\lambda_2y} \, dx \, dy
$$
First I do not know if this is the double integral that leads to the solution. Second, when I try to solve it does not a convergent integral. Can you help me on that?
 A: $M-L$ is exponentially distributed iff $\lambda_1 = \lambda_2$.

Claim: If $\lambda_1 = \lambda_2$, then $M-L \sim Expo(\lambda_1)$.

This is a standard result.  Informally, if you think of $X, Y$ as the typical exponential waiting times for two different buses, then once the first bus arrived, $M-L$ is the time until the second bus arrives, but since the second bus is memoryless, it "didn't care" that time $L$ has elapsed.  Hence $M-L \sim Expo(\lambda)$.
You can find more formal proofs, some using explicit integrals, in the links provided by StubbornAtom in the comments.  Alternatively, I once wrote this proof which explicitly replicates the two-bus reasoning above.  My proof does not involve evaluating integrals but is still rigorous (IMHO).

Claim: If $\lambda_1 \neq \lambda_2$, then $M-L$ is not exponential.

Again consider the two buses example.  If the more frequent bus comes first, then $M-L \sim Expo(\min(\lambda_1, \lambda_2))$ because you must now wait for the rarer bus.  If the rarer bus comes first, the $M-L \sim Expo(\max(\lambda_1, \lambda_2))$ because you are now waiting for the more frequent bus.  So the overall distribution is a "mixture" and is not a single exponential.
E.g. both this MSE answer and these notes give the pdf $f_Z$ of $Z = X - Y$.  Since $M-L = |X - Y|$, the pdf of $M-L$ is simply:
$$f_{M-L}(v) = f_Z(v) + f_Z(-v) = {\lambda_1 \lambda_2 \over \lambda_1 + \lambda_2} (e^{-\lambda_1 v} + e^{-\lambda_2 v}) ~~\text{for } v \ge 0$$
and of course $f_{M-L}(v) = 0$ for $v < 0$.
A: Some complications arise:
$ \displaystyle \require{cancel}
\int_{\xcancel{-\infty}}^\infty \cdots \, dy \quad $  Here you need $ \displaystyle \int_0^\infty\cdots \, dy.$
Inside that integral you need $ \displaystyle \int_{\min\{y-t,0\}}^{y+t} \cdots\, dx. $
And $ \displaystyle \min\{y-t,0\} = \begin{cases} 0 & \text{if } y\le t, \\ y-t & \text{if }y>t. \end{cases}$
So here's one way:
\begin{align}
& \Pr(|X-Y|\le t) \\[8pt] = {} & \int_0^t \left( \int_0^{y+t} e^{-\lambda_1 x} (\lambda_1 \, dx) \right) e^{-\lambda_2 y} (\lambda_2 \, dy) \\[8pt]
& {} + \int_t^\infty \left( \int_{y-t}^{y+t} e^{-\lambda_1 x} (\lambda_1 \, dx) \right) e^{-\lambda_2 y} (\lambda_2 \, dy)
\end{align}
I suspect there's a more elegant way. Maybe later.
