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?

$$M-L$$ 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 $$v < 0$$.

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.

• I will wait for the elegant way. Nov 18, 2019 at 20:48