Properties of exponential random variables: memoryless property and sums/differences

Solution 1:

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:

$$P(M < a) = P(X<a, Y<a) = P(X<a)P(Y<a) = (1-e^{-\lambda a})^2$$