Properties of exponential random variables: memoryless property and sums/differences My textbook, Introduction to Probability, by Blitzstein and Hwang, gives the following example:

Example 3
For $X, Y \stackrel{i.i.d.}{\sim}\text{Expo}(\lambda)$, find $\mathbb{E}[\max(X, Y) | \min(X, Y)]$.
Let $M = \max(X, Y)$ and $L = \min(X, Y)$.
By the memoryless property, $M - L$ is independent of $L$, and $M - L \sim \text{Expo}(\lambda)$.
$\dots$

The full example isn't important, because what my question pertains to are the two facts above:

*

*How does the memoryless property imply that $M - L$ is independent of $L$?


*How is it that $M - L \sim \text{Expo}(\lambda)$? In other words, how is it that the difference of two $\text{Expo}(\lambda)$ random variables has the same parameter?
I would greatly appreciate it if people could please take the time to clarify this.
 A: 
How does the memoryless property imply that $M−L$ is independent of $L$?
How is it that $M−L\sim Expo(\lambda)$?

For convenience I will write $Z = M-L$.
I think the authors are trying to appeal to a "proof by obviousness".  If you think of $X, Y$ as the typical exponential waiting times for two different buses, then once the first bus arrived, $Z=M-L$ 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 $Z \sim Expo(\lambda)$.
Now the obvious problem with "proof by obviousness" is that some people find them non-obvious.  :)  So if you don't buy the hand-wavy argument above, we can try to prove it.  For any $z, l > 0:$
$$
\begin{array}{rl}
P(Z > z \mid L = l, X > Y) &= P(X-Y > z \mid X>Y=l) \\
&= P(X>z+l \mid X>l, Y=l) \\
&= P(X>z+l \mid X>l) \,\,\,\,\,\text{...because $X,Y$ independent} \\
&= e^{-\lambda z}  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{...because $X$ is memoryless}\\
P(Z > z \mid L = l, Y > X) &= e^{-\lambda z}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{...similarly}\\
P(Z > z \mid L = l) &= P(X>Y)\,P(Z > z \mid L = l, X > Y) \\
&\,\,\,\,\,\,\,\,+ P(Y>X)\, P(Z > z \mid L = l, Y > X)\\
&= e^{-\lambda z}\\
P(Z > z) &= \int_0^\infty P(Z > z \mid L = l) f_L(l) \,dl = e^{-\lambda z}
\end{array}
$$
I think the above is watertight, but even if not, you get the idea.
Combining the last two equations, we have:
$$\forall l>0: P(Z > z \mid L = l) = e^{-\lambda z} = P(Z>z)$$
which directly shows $Z,L$ independent, and $Z \sim Expo(\lambda)$.

how is it that the difference of two $Expo(\lambda)$ random variables has the same parameter?

This question has two different wrong ideas hidden behind it.  $Z$ is indeed the difference of two random variables, i.e. $Z=M-L$, but neither $M$ nor $L$ is $\sim Expo(\lambda)$.  Instead we have:

*

*$X, Y \sim Expo(\lambda)$, given


*$Z=M-L=\max(X,Y) - \min(X,Y) \sim Expo(\lambda)$, shown above


*$L = \min(X,Y) \sim Expo(2\lambda)$, e.g. see here


*$M = \max(X,Y) $ is not exponential at all, because its CDF is not in the required form:
$$P(M < a) = P(X<a, Y<a) = P(X<a)P(Y<a) = (1-e^{-\lambda a})^2$$
