The joint density of the max and min of two independent exponentials

Let $X=\min(S,T)$ and $Y=\max(S,T)$ for independent exponential variables $S$ and $T$. Find the joint density of $X$ and $Y$. Are $X$ and $Y$ independent? How would you suggest I approach this?

Let $F_{S},F_{T}$ and $F$ denote the distribution function of $S,T$ and $\left(X,Y\right)$.

If $y\leq x$ then $\left\{ X\leq x\wedge Y\leq y\right\} =\left\{ S\leq y\wedge T\leq y\right\} $ leading to:

$F\left(x,y\right)=P\left\{ S\leq y\wedge T\leq y\right\} =F_{S}\left(y\right)F_{T}\left(y\right)$ if $y\leq x$.

If $y>x$ then $\left\{ X\leq x\wedge Y\leq y\right\} =\left\{ \left[S\leq x\wedge T\leq y\right]\vee\left[S\leq y\wedge T\leq x\right]\right\} $

In that case $F\left(x,y\right)=P\left\{ \left[S\leq x\wedge T\leq y\right]\vee\left[S\leq y\wedge T\leq x\right]\right\} $

Practicizing the rule $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$ we find:

$F\left(x,y\right)=P\left\{ S\leq x\wedge T\leq y\right\} +P\left\{ S\leq y\wedge T\leq x\right\} -P\left\{ S\leq x\wedge T\leq x\right\} $ leading to:

$F\left(x,y\right)=F_{S}\left(x\right)F_{T}\left(y\right)+F_{S}\left(y\right)F_{T}\left(x\right)-F_{S}\left(x\right)F_{T}\left(x\right)$ if $y>x$.

Based on the distribution function you can find the density if it exists.

This method works in general for independent $S$ and $T$ so can be applied to the special case of independent exponential variables.

Hint: Remember that the joint density can be found by $$f_{X,Y}(x,y) = \frac{\partial}{\partial x} \frac{\partial}{\partial y} P(X \leq x, Y \leq y)$$

Since $\{X > x\} \iff \{ S > x \text{ and } T > x\}$ and $\{ Y \leq y\} \iff \{S \leq y \text{ and } T\leq y\}$. To find $P(X \leq x, Y\leq y)$ you can notice that $$P(Y \leq y) = P(X \leq x, Y \leq y) + P(X > x, Y \leq y)$$ since either $\{ X > x\}$ or $\{X \leq x\}$. Now, you can solve for $$P(X \leq x, Y \leq y) = P(X > x, Y \leq y) - P(Y\leq y)$$ and by the earlier remarks $$P(X>x, Y\leq y) = P(S>x, T>x, S\leq y, T\leq y) = P(x < S \leq y, x < T \leq y) \\= P(x<S\leq y)P(x < T \leq y)$$ The last equality came from the fact that $S$ and $T$ are independent. Put these pieces together and you find that $P(X \leq x, Y \leq y) = \cdots$ what? And from there $f_{X,Y}(x,y) = \cdots$ what?

If $S,T$ are identically distibuted (the question doesn't say that) and $\Pr(S>x)=e^{-x/\theta}$ then \begin{align} & \Pr(\min\{S,T\}>x) \\[8pt] = {} & \Pr(S>x\ \&\ T>x) \\[8pt] = {} & \Pr(S>x)\cdot\Pr(T>x) \\[8pt] = {} & (e^{-x/\theta})^2 = e^{-x/(\theta/2)}. \end{align} Thus $\min\{S,T\}$ is exponentially distributed with expected value $\theta/2$.

$\max$ is a bit more complicated than $\min$; maybe I'll return to it. I posted what you see above because the two answers that had appeared so far were more complicated than they needed to be.

Now let's look at the two together. We have \begin{align} & f_{S,T}(x,y) \,dx\,dy \\[8pt] = {} & e^{-x/\theta} e^{-y/\theta} \, \frac{dx} \theta \,\frac{dy} \theta \text{ for } x>0,\ y>0. \tag 1 \end{align} For any $(x,y)$, if $x<y$ we'll leave that point where it is, but if $x>y$ we want to interchange $x$ and $y$ so that the smaller one, the min, will appear first in the ordered pair. The mapping $(x,y)\mapsto(y,x)$ just takes the part of the first quadrant that's below the line $y=x$ and folds it over so that it's above that line, without changing sizes or shapes, so the Jacobian by which one multiplies is $1$. And notice from $(1)$ that interchanging the So just add together the density on the part of the first quadrant that above $y=x$ and the folded over copy of it, getting \begin{align} & f_{X,Y}(x,y)\,dx\,dy \\[8pt] = {} & 2 e^{-x/\theta} e^{-y/\theta}\, \frac{dx}\theta\,\frac{dy}\theta\text{ for } \underbrace{0<x<y}_{\text{This part is different!}} \end{align}

The asymmetry comes only from the part over the $\underbrace{\text{underbrace}}$ where it says "This part is different!".

Let's check by integrating out $y$ to get the marginal density of $X$: \begin{align} & \int_{y:=x}^{\infty} 2 e^{-x/\theta} e^{-y/\theta} \, \frac{dy}\theta \\[8pt] = {} & 2e^{-x/\theta} \int_x^\infty e^{-y/\theta} \, \frac{dy} \theta \\[8pt] = {} & 2e^{-x/\theta} \int_{x/\theta}^\infty e^{-u}\,du \\[8pt] = {} & 2 e^{-x/(\theta/2)}, \end{align} which leads to $\Pr(X>x) = e^{-x/(\theta/2)}$, as found above.

Maybe I'll add more later . . .