Distribution of non-identical exponential order statistics This is a problem given as practice for an upcoming exam (without solution provided).
Let $X$ and $Y$ be independent exponential random variables, with $X = \theta e^{-\theta x}$ and $Y = \lambda e^{-\lambda y}$, with $\theta > 0$ , $\lambda > 0$, and $\theta \neq \lambda$. Define $S=\mathrm{min}(X, Y)$ and $T=\mathrm{max}(X, Y)$. Calculate the following:


*

*The joint distribution of $S$ and $T$

*The joint distribution of $S$ and $T$ conditional on $X \le Y$


The first question has been covered more or less on this site here and here, so my uncertainty is more about the second question.
For the first, since $P(T \le t) = P(T \le t, S \le s) + P(T \le t, S > s)$, I calculate
\begin{eqnarray}
P(T \le t, S \le s) &=& P(T \le t) - P(T \le t, S > s) \nonumber \\
\end{eqnarray}
Filling in the details and taking the derivatives with respect to t and s, I can get the joint density.
However, for the second part, we gain information that $X \le Y$ which means $S=X$ and $T=Y$. This seems to "fold over" the domain around the line $X=Y$, right? I'm not sure how to proceed, so any hints or help would be appreciated.

The conditional distribution of $(S,T)$ given that $X\le Y$ is the same as the conditional distribution of $(X,Y)$ given that $X\le Y$, and the conditional density is the same as the joint density of $(X,Y)$ restricted to the set $\{(x,y): 0 \le x\le y\}$ and then normalized by multiplying by a constant.
To prove the assertions above, suppose $A\subseteq \{(x,y): 0 \le x\le y\}$.  Then
\begin{align}
& \Pr((S,T)\in A \mid X\le Y) = \frac{\Pr((S,T)\in A\ \&\ X\le Y)}{\Pr(X\le Y)} \\[10pt]
\overset {(1)}= {} & \frac{\Pr((X,Y)\in A\ \&\ X\le Y)}{\Pr(X\le Y)} \overset {(2)} = \frac{\Pr((X,Y)\in A)}{\Pr(X\le Y)}.
\end{align}
The equality $(1)$ holds because the event $[(S,T)\in A\ \&\ X\le Y]$ is the same as the event $[(X,Y)\in A\ \&\ X\le Y]$. The equality $(2)$ holds because the event $[(X,Y)\in A\ \&\ X\le Y]$ is the same as the event $[(X,Y)\in A]$.
The numerator is $\displaystyle\iint_A f_{X,Y}(x,y)\, d(x,y)$ and the denominator is a "constant", in the sense that its value does not depend on which set $A$ is.  Hence we have
$$
\Pr((S,T)\in A\mid X\le Y) = \iint_A c f_{X,Y}(x,y)\,d(x,y)
$$
for some constant $c$.
Next we find the constant. The joint density is
$$
f_{X,Y}(x,y)=\theta\lambda e^{-(\theta x+\lambda y)}\text{ for }x,y\ge 0.
$$
The conditional density given $X\le Y$ is just that function restricted to $\{(x,y): 0 \le x\le y\}$ and then normalized.  To find the normalizing constant, integrate:
\begin{align}
& \int_0^\infty \int_0^y e^{-(\theta x+\lambda y)} \, dx\,dy = \int_0^\infty \left[ \frac{e^{-(\theta x+\lambda y)}}\theta \right]_{x:=0}^{x:=y} \,dy \\[10pt]
= {} & \int_0^\infty \left( \frac{e^{-(\theta+\lambda)y}}\theta - \frac{e^{-\lambda y}}\theta \right) \, dy = \frac 1 \theta \left[ \frac{e^{-(\theta+\lambda)y}}{\theta+\lambda} - \frac{e^{-\lambda y}}\lambda \right]_0^\infty \\[10pt]
= {} & \frac 1 \theta\left( \frac 1 {\theta+\lambda} - \frac 1 \lambda \right) = \frac 1 {(\theta+\lambda)\lambda}.
\end{align}
So
$$
f_{S,T\,\mid\, X\le Y} (x,y) = (\theta+\lambda)\lambda e^{-(\theta x+\lambda y)}\text{ for } 0\le x\le y.
$$
