Conditional expectation of $\max(X,Y)$ and $\min(X,Y)$ when $X,Y$ are iid and exponentially distributed I am trying to compute the conditional expectation $$E[\max(X,Y) | \min(X,Y)]$$ where $X$ and $Y$ are two iid random variables with $X,Y \sim \exp(1)$. 
I already calculated the densities of $\min(X,Y)$ and $\max(X,Y)$, but I failed in calculating the joint density. Is this the right way? How can I compute the joint density then? Or do I have to take another ansatz? 
 A: As indicated in the comments, a useful idea when maxima and minima are involved is to consider well adapted events. Here, introducing $Z=\min\{X,Y\}$ and $W=\max\{X,Y\}$, one sees that $[z\leqslant Z,W\leqslant w]$ is $[z\leqslant X\leqslant w]\cap[z\leqslant Y\leqslant w]$ for every nonnegative $z$ and $w$ such that $z\leqslant w$. Here is a computation: since the probability that a standard exponential random variable is $\geqslant x$ is $\mathrm e^{-x}$ for every nonnegative $x$, the events $[z\leqslant X\leqslant w]$ and $[z\leqslant Y\leqslant w]$ both have probability $\mathrm e^{-z}-\mathrm e^{-w}$. Hence,
$$
\mathrm P(z\leqslant Z,W\leqslant w)=(\mathrm e^{-z}-\mathrm e^{-w})^2.
$$
Differentiating this with respect to $z$ and $w$ yields the density of $(Z,W)$ as
$$
2\mathrm e^{-z-w}\cdot[0\leqslant z\leqslant w].
$$
This formula is all right but, because of the indicator functions in it, I am afraid to make mistakes when using it, so I try to simplify it. Let $V=W-Z$, then $Z\geqslant0$, $V\geqslant 0$, and using $v=w-z$, the density becomes
$$
2\mathrm e^{-z-(v+z)}\cdot[0\leqslant z\leqslant v+z]=2\mathrm e^{-2z}\cdot[z\geqslant 0]\cdot\mathrm e^{-v}\cdot[v\geqslant0].
$$
This proves that $Z$ and $V$ are independent with $Z$ exponential of parameter $2$ and $V$ of parameter $1$ and yields at last the answer to the initial question as
$$
\mathrm E(W\mid Z)=\mathrm E(V+Z\mid Z)=\mathrm E(V)+Z=1+Z.
$$
The same technique yields that the order statistic $(X^{(k)})_{1\leqslant k\leqslant n}$ of an i.i.d. sample $(X_k)_{1\leqslant k\leqslant n}$ of standard exponential random variables, defined by the conditions that $\{X^{(1)},X^{(2)},\ldots,X^{(n)}\}=\{X_1,X_2,\ldots,X_n\}$ and that $X^{(1)}<X^{(2)}<\cdots <X^{(n)}$, is distributed like $(Z_1,Z_1+Z_2,\ldots,Z_1+Z_2+\cdots+Z_n)$ for independent exponential random variables $(Z_k)_{1\leqslant k\leqslant n}$ such that the distribution of $Z_k$ is exponential with parameter $n-k+1$. A consequence is that, for every $1\leqslant k\leqslant\ell\leqslant n$,
$$
\mathrm E(X^{(\ell)}\mid X^{(k)})=X^{(k)}+\sum\limits_{i=n-\ell+1}^{n-k}\frac1i.
$$
A: For two independent exponential distributed variables $(X,Y)$, the joint distribution is
$$
   \mathbb{P}(x,y) = \mathrm{e}^{-x-y} \mathbf{1}_{x >0 } \mathbf{1}_{y >0 } \, \mathrm{d} x \mathrm{d} y
$$
Since $x+y = \min(x,y) + \max(x,y)$, and $\min(x,y) \le \max(x,y)$ the joint distribution of 
$(U,V) = (\min(X,Y), \max(X,Y))$ is
$$
   \mathbb{P}(u,v) = \mathcal{N} \mathrm{e}^{-u-v}  \mathbf{1}_{v \ge u >0 } \, \mathrm{d} u \mathrm{d} v
$$
The normalization constant is easy to find as
$$
   \int_0^\infty \mathrm{d} v \int_0^v \mathrm{d} u \,\, \mathrm{e}^{-u-v} = 
   \int_0^\infty \mathrm{d} v \,\, \mathrm{e}^{-v} ( 1 - \mathrm{e}^{-v} ) = 1 - \frac{1}{2} = \frac{1}{2} = \frac{1}{\mathcal{N}}  
$$
Thus the conditional expectation we seek to find is found as follows (assuming $u>0$):
$$
   \mathbb{E}(\max(X,Y) \vert \min(X,Y) = u) = \frac{\int_0^\infty v \mathrm{d} P(u,v)}{\int_u^\infty \mathrm{d} P(u,v)} = \frac{\int_u^\infty  \mathcal{N} v \mathrm{e}^{-u-v} \mathrm{d} v}{\int_u^\infty  \mathcal{N} \mathrm{e}^{-u-v} \mathrm{d} v} = 1 + u
$$ 
A: If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$,
$$\begin{align*}
F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\
&= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\
&= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\
&= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z)
\end{align*}
$$ 
while for $w < z$,
$$\begin{align*}
F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\
&= P\{X \leq w, Y \leq w\}\\
&= F_{X,Y}(w,w).
\end{align*}
$$ 
Consequently, if $X$ and $Y$ are jointly continuous
random variables, then 
$$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = 
\begin{cases}
f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\
\\
0, & \text{if}~w < z.
\end{cases}
$$
The conditional density of $W$ given $Z = z$ is 
$$
f_{W \mid Z}(w \mid z) = \frac{f_{Z,W}(z,w)}{f_Z(z)}
= \begin{cases}
\frac{f_{X,Y}(z,w) + f_{X,Y}(w,z)}{\int_z^{\infty} f_{X,Y}(z,w) + f_{X,Y}(w,z)\ 
\mathrm dw}, & w > z,\\
0, & w < z,
\end{cases}
$$
and so with $f_{X,Y}(x,y) = e^{-x-y}$ for $x, y \geq 0$
$$
\begin{align*}E[W \mid Z = z] 
&= \frac{\int_z^\infty w\cdot f_{X,Y}(z,w) + w\cdot f_{X,Y}(w,z)\  \mathrm dw}{
\int_z^\infty f_{X,Y}(z,w) + f_{X,Y}(w,z)\  \mathrm dw}\\
&= \frac{\int_z^\infty w\cdot e^{-w-z} + w\cdot e^{-w-z}\  \mathrm dw}{
\int_z^\infty e^{-w-z} + e^{-w-z}\  \mathrm dw}\\
&= \frac{2e^{-2z}\int_z^\infty w\cdot e^{-w}\  \mathrm dw}{
2e^{-2z}} = \frac{2e^{-z}[\left . (-we^{-w})\right\vert_z^{\infty}
+ \int_z^{\infty}e^{-w}\ \mathrm dw]}{2e^{-2z}}\\
&= 1 + z.
\end{align*}
$$
A: You can presumably find $\Pr(X\le a)$ and $\Pr(b \lt X \le a) $ and so  $\Pr(b \lt X \le a) \Pr(b \lt Y \le a)$.
Take the derivatives of this with respect to $a$ then and $b$ (changing the sign as $b$ is a lower limit) and add an indicator such as $I_{b\le a}$, and you have the joint density which we might call $p(a,b)$ with $a$ acting as $\max(X,Y)$ and $b$ acting as $\min(X,Y)$.
You can then work out the conditional density $p(a|b) =\dfrac{p(a,b)}{\int_{a=b}^\infty p(a,b) \; da}$ and the conditional mean $E[a|b] = \int_{a=b}^\infty a\; p(a|b) \; da$, which will be a function of $b$. 
To check, remember that the exponential distribution is memoryless so if $X$ and $Y$ have mean $\mu$ then you will have $E[\max(X,Y) | \min(X,Y)] = \min(X,Y) +\mu$.
A: Observe that $\max\left(X,Y\right)=X+Y-\min\left(X,Y\right)$ so that:
$$\begin{aligned}\mathsf{E}\left[\max\left(X,Y\right)\mid\min\left(X,Y\right)\right] & =\mathsf{E}\left[X+Y-\min\left(X,Y\right)\mid\min\left(X,Y\right)\right]\\
& =2\mathsf{E}\left[X\mid\min\left(X,Y\right)\right]-\min\left(X,Y\right)
\end{aligned}
\tag1$$
Here the second equality is based on symmetry.
For a fixed $m>0$ we find:
$$\begin{aligned}\mathsf{E}\left[X\mid\min\left(X,Y\right)=m\right] & =\frac{1}{2}m+\frac{1}{2}\mathsf{E}\left[X\mid X>m\right]\\
 & =\frac{1}{2}m+\frac{1}{2}\left(m+\mathsf{E}X\right)\\
 & =m+\frac{1}{2}\mathsf EX\\
 & =m+\frac{1}{2}
\end{aligned}
\tag2$$
Here the first equality is based on symmetry and the second on the fact that exponential distribution has no memory.
Then $(2)$ leads us to the conclusion:
$$\mathsf{E}\left[X\mid\min\left(X,Y\right)\right]=\min\left(X,Y\right)+\frac{1}{2}$$
and substitution in $(1)$ results in:
$$\mathsf{E}\left[\max\left(X,Y\right)\mid\min\left(X,Y\right)\right]=\min\left(X,Y\right)+1$$
A: another way is to realize that the min and (max-min) are independent by the memoryless property of exponential distribution. Find the joint density of min and (max - min) and then apply transformation of random variables,---the jocobian. 
