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

$$left{(x,y)colon frac 12 leq x leq 1, 0 leq y leq x-frac 12right} bigcup left{(x,y)colon 0 leq x leq frac 12, frac 12 leq y < x + frac 12right}$$ 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.

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

$$left{(x,y)colon frac 12 leq x leq 1, 0 leq y leq x-frac 12right} bigcup left{(x,y)colon 0 leq x leq frac 12, frac 12 leq y < x + frac 12right}$$ 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.