Problems with extremum of two uniform random variables Here is the problem from the book:

Let $X = \min(U,V)$ and $Y = \max(U,V)$ for independent $\text{uniform}(0,1)$ variables $U$ and $V$.  Find the distributions of a) $X$; b) $1-Y$; c) $Y-X$.

I know that each has density $2(1-x)$ for $0<x<1$.  What I'm hoping for is an intuitive explanation of how to understand this problem and arrive at that solution without knowing the beta distribution.
 A: I'll work out the first one and leave the other two to you. First, find $\Pr(X \leq x)$. We have
$$\begin{align*}\Pr(X \leq x) &= 1 - \Pr(X > x) \\
&= 1 - \Pr(U > x, V > x) \\ &= 1 - \Pr(U > x)\Pr(V > x) \\ &= 1 - (1-x)^2 \end{align*}$$
The first step follows that, for $x$ to be a minimum $U$ and $V$ must be greater than that value. Independence gives us the second line. Use of the CDF of the uniform gives us the third.
To get the density, take the derivative to get $2(1-x)$.
A: An explanation for the result that $X$, $1-Y$, and $Y-X$ have the same distribution in this case is as follows.
First, consider a plane with coordinate axes $u$ and $v$ and let 
$(U,V)$ be a random point in the plane chosen according to some 
joint density function $f_{U,V}(u,v)$.  $U$ and $V$ need not be 
independent random variables.  Then,
$$\begin{align*}
P\{X > \alpha\} &= P\{\min(U,V) > \alpha\} = P\{U > \alpha, V > \alpha\},\\
P\{1-Y > \alpha\} &= P\{1-\max(U,V) > \alpha\} = P\{\max(U,V) < 1 - \alpha\}\\
&= P\{U < 1- \alpha, V < 1- \alpha\},\\
P\{Y-X > \alpha\} &= P\{\max(U,V)-\min(U,V) > \alpha\}\\
&= P\{U-V > \alpha\} + P\{V-U > \alpha\}.
\end{align*}$$
These three probabilities  can be found in the general case by
integrating $f_{U,V}(u,v)$ over the appropriate region which can be 
described in the three cases respectively as


*

*the northeast quadrant of the plane with southwest corner 
$(\alpha, \alpha)$

*the southwest quadrant of the plane with northeast corner 
$(1-\alpha, 1-\alpha)$

*the half-plane below the line $v < u - \alpha$ and the half-plane above the line $v > u + \alpha$
So much for generalities.  If the random point $(U,V)$ is uniformly
distributed on a region $A$ of the plane (that is,
$f_{U,V}(u,v)$ is nonzero and constant for $(u,v) \in A$,
$f_{U,V}(u,v) = 0$ for $(u,v) \notin A$) and $B$ is any
region of the plane, then
$$P\{(U,V) \in B\} = P\{(U,V) \in A\cap B\} 
= \frac{\mathrm{Area}(A\cap B)}{\mathrm{Area}(A)}.$$
In particular, if we can compute areas via mensuration
formulas learned in school, we do not need to integrate
formally.
Finally, in the special case when $A$ is the unit-area square with
opposite corners $(0,0)$ and $(1,1)$, and $\alpha$ is a number between
$0$ and $1$,
$$\begin{align*}
P\{X > \alpha\} &= P\{U > \alpha, V > \alpha\}\\
&= P\{(U,V) \in ~\mathrm{square~with~opposite~corners}~ (\alpha,\alpha)
~ \mathrm{and}~ (1,1)\\
&= (1-\alpha)^2,\\
P\{1-Y > \alpha\} &= P\{U < 1- \alpha, V < 1- \alpha\}\\
&= P\{(U,V) \in ~\mathrm{square~with~opposite~corners}~ (0,0)
~ \mathrm{and}(1-\alpha,1-\alpha)~ \\
&= (1-\alpha)^2,\\
P\{Y-X > \alpha\} &= P\{U-V > \alpha\} + P\{V-U > \alpha\}\\
&= P\{(U,V) \in ~\mathrm{triangle~with~corners}~ (\alpha,0),
(1,1-\alpha) ~\mathrm{and}~(1,0)\}\\
&\quad \quad 
+ P\{(U,V) \in ~\mathrm{triangle~with~corners}~ (0,\alpha),
(1-\alpha,1) ~\mathrm{and}~(0,1)\}\\
&= \frac{1}{2}(1-\alpha)^2 + \frac{1}{2}(1-\alpha)^2 = (1-\alpha)^2.\\
\end{align*}$$
So the complementary cumulative distribution of the three
random variables $X$, $1-Y$ and $Y-X$ is the same $(1-\alpha)^2$
in this case,
and so the three random variables have the same density function
$2(1-\alpha)$, $0 \leq \alpha \leq 1$.
