Distributions of $U=\max\{X-Y,0\}$ and $V=\max\{X,Y\}-\min\{X,Y\}$ where $X,Y$ exponential

Question

Suppose $X\sim \text{Exp}(\lambda)$ and $Y\sim \text{Exp}(\mu)$ are independent.

I've shown that $Z=\min\{X, Y\}$ is independent of the event $\{X<Y\}$. I want to find the following:

1. $P(X=Z)$

2. Find the distribution of $U = \max\{X-Y,0 \}$ and of $V= \max\{X,Y\}-\min\{X,Y\}$

3. $P(X\leq t \leq X+Y)$ for $t>0$.

I'm pretty sure $P(X=Z) = P(X=\min\{X,Y\}) = \frac{\lambda}{\lambda+\mu}$. But I'm unsure how to tackle the others.

For 2. I have for $t>0$

$\begin{eqnarray*} P(U=t) &=& P(U=t|X\leq Y)P(X\leq Y) + P(U=t|X>Y)P(X>Y)\\ &=& P(U=t,X>Y)\\ &=& \int_{t/2}^t P(X=x)P(Y=x-t) \end{eqnarray*}$

but I get a very messy answer. Am I doing this right? Any help would be greatly appreciated.

Answer 1

First, you should be careful with writing $\mathbb{P}(X=x)$ with continuous RVs $X$ that have no point masses, because $\mathbb{P}(X=x)=0$ for all reals $x$ and all such continuous RVs, $X$. If you want to use the law of total probability in such a case, you apply it to the probability density function: $f(x)=\int f(x|y)f_Y(y)\mathrm{d}y$, or the cumulative distribution function: $\mathbb{P}(X\leq x)=\int \mathbb{P}(X\leq x | Y=y)f_Y(y)\mathrm{d}y$.

Now for the problem at hand. You have computed $1.$ correctly. To compute the distribution of $U$ as given, we first note that $U=0$ whenever $X\leq Y$. This has probability $\frac{\lambda}{\lambda+\mu}$, by part $1.$ Next, note that for $u>0$ we have $U>u$ if and only if $X>Y+u$. The complement of this event has probability $$\mathbb{P}(X\leq Y+u)=\int_0^\infty \mathbb{P}(X\leq y+u | Y=y)f_Y(y)\mathrm{d}y$$ $$=1-\frac{\mu}{\lambda+\mu} e^{-\lambda u}.$$ So $U$ has CDF equal to $$F_U(u)=\frac{\lambda}{\lambda+\mu} \mathbb{1}_{u=0}+\left(1-\frac{\mu}{\lambda+\mu}e^{-\lambda u}\right)\mathbb{1}_{u>0}$$ where $\mathbb{1}_A:=\mathbb{1}_A(u)$ is the indicator function of $A$, i.e. $=1$ if $u\in A$ and $0$ otherwise. (Sanity check, as $u\to \infty$, $F_U(u)\to 1$, etc)

For the second part of $2.$ one can find the distributions of the max and min individually and then in principle use convolution to find the distribution of the difference. Let me know if you have trouble working this out. Part $3.$ should also be doable via LTP, say applied to $\mathbb{P}(Y\geq t-X \geq 0)$. Let me know for further clarifications or if you spot any potential errors.