Max and min on Uniform distribution

Question

$x$ and $y$ are independent uniformly distributed random variables in $(0,1)$. Let: $w=\max(x,y)$ and $z=\min(x,y)$. Find the pdf of:
(a) $r=w-z$
(b) $s=w+z$

I am not familiar with pdf of maximum and mimum of random variables.

Answer 1

Notice that you can only have the following two cases:
(i) $w=x$, $z=y$
(ii) $w=y$, $z=x$.

Since both cases have similar work due to symmetry, I shall only show work for (i). We have $r=x-y$ and $s=x+y$. Again, since the pdf calculation follows similar steps I shall only find the pdf for $r$.

$\because\min(x-y)=0,\max(x-y)=1$, for any $R\in(0,1)$, we have:
$P(r\le R)=P(x-y\le R)=\overset{1}{\underset{0}{\int}}P(x\le R+Y,y\le Y)\mathrm{~d}Y\\$
Since, $x$ and $y$ are independent Unif $(0,1)$ random variables, for every $Y\in(0,1)$ we can write
$P(x\le R+Y,y\le Y)=P(x\le R+Y)P(y\le Y)=(R+Y)Y$

$\therefore P(r\le R)=\overset{1}{\underset{0}{\int}}(R+Y)Y\mathrm{~d}Y= \frac{1}{6}(3 R + 2)$.