Independence of $\min(X,Y)$ and $\max(X,Y)$ for independent $X$, $Y$? What's the reasoning for checking the independence of
$$\min(X,Y)$$ and $$\max(X,Y)$$ for independent r.v.s $X,Y$?
Is it possible that $\min$ and $\max$ both select the same r.v. in which case they would be dependent? No, because that would mean that $X=Y$, i.e. $X$ and $Y$ would not be independent.
Does the independence regarding functions of independent r.v.s apply here? Yes? In that case $X,Y$ independent $\implies$ $\min(X,Y),\space \max(X,Y)$ independent.
 A: If $X$ and $Y$ are independent continuous random variables, then $\max(X,Y)$ and
$\min(X,Y)$ are independent random variables if and only if one of the
following two conditions holds:


*

*$P(X > Y) = 1$

*$P(X < Y) = 1$
Note that the above conditions mean that $P(X=Y) = 0$ but this does
not mean that $(X=Y)$ is the same as the impossible event, that is,
there is no outcome $\omega$ in the sample space for which
$X(\omega) = Y(\omega)$. Those thoroughly confused by this notion
should recall that they might have been told that
for a continuous random variable $V$,
$P(V = a) = 0$ for all real numbers $a$ even though it is manifestly
true that $V$ can take on value $a$ for some particular $a$, 
and if they have
swallowed that whopper, then accepting that $P(X=Y)=0$ does not
mean that the event $(X=Y)$ will never occur is just a small
additional stretch of their credulity.
When $X$ and $Y$ are independent discrete random variables, then
the above condition needs to be relaxed slightly, and it is possible
to have $P(X=Y) > 0$. For example, if $(X,Y)$ takes on values
$(1,0), (2,0), (1,1), (2,1)$ with equal probability $\frac 14$, then
$(\min(X,Y), \max(X,Y))$ takes on values $(0,1), (0,2), (1,1), (1,2)$
with equal probability $\frac 14$ and thus $\min(X,Y)$ and 
$\max(X,Y))$ are
independent. A little thought will show that $(\min(X,Y), \max(X,Y))$
is the same as $(Y,X)$ in this case. A little further thought will
show that if $P(X=Y)>0$, then it must be that there is a unique
$a$ such that $P(X=a, Y= a) >0$ and that for all other real numbers
$b$, $P(X=b, Y= b) =0$. For independent discrete random variables
$X$ and $Y$, the probability mass function has nonzero values at
all points on a rectangular grid, and this grid must be strictly
below or strictly above the line $x=y$ or must have only one point
(the upper left corner or the lower right corner) on the line $x=y$; 
the point $(1,1)$ in the example above. 

An interesting follow-up question is:  
When  $X$ and $Y$ are dependent random variables, is it possible for
$\max(X,Y)$ and $\min(X,Y)$ to be independent random variables?
to which the answer is Yes, it is possible.  Consider the case when
$X$ and $Y$ are jointly continuous random variables uniformly
distributed on the set
$$\left\{(x,y)\colon \frac 12 \leq x \leq 1,
 0 \leq y \leq x-\frac 12\right\}
\bigcup 
\left\{(x,y)\colon 0 \leq x \leq \frac 12,
 \frac 12 \leq y < x + \frac 12\right\}$$
The joint density of the minimum and maximum can be worked out
as described here
where it is shown that if $Z = \min(X,Y)$ and $W = \max(X,Y)$,
then 
$$f_{Z,W}(z,w) =  \begin{cases}
f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\
\\
0, & \text{if}~w < z.
\end{cases}
$$
Applying this, it can be shown that the joint density of
$Z$ and $W$ is uniform on interior of the square with vertices
$(0,\frac 12), (\frac 12, \frac 12), (\frac 12, 1),
(0,1)$, and so $Z \sim U[0,\frac 12]$ and $W \sim U[\frac 12,1]$
are independent random variables.
