How to continue this problem to find distribution function of $|X-Y|$ The probability density function of 
$$f(x,y) =
\begin{cases}
1/x^2,  & \text{if }0 \le x \le a\text{ if }0 \le y \le a \\
0, & \text{otherwise}  \\
\end{cases}
$$
How can you prove that $|X-Y|$ and $\min(X,Y)$ have the same distribution function?
I tried:
Firstly $f(x)=1/a$, $f(y)=1/a$, so CDF for $f(x)$ is $x/a$ CDF for $f(y)$ is $y/a$ right??  
$Z=\min(X,Y)$
The CDF of $\min(X,Y)= \Pr(Z < z)=1-(1-\Pr(X < z))(1-\Pr(Y < z))$ because at least one must be smaller than $z$, to be included in the CDF of $\min(X,Y)$
$=1-(1-x/a)(1-y/a)$,
But I am stuck with finding the CDF for $|X-Y|$. And finding the CDF for $|X-Y|$, will I have to derivate both CDFs in terms of both $x$ and $y$, and get to the same function? I am really stuck, I hope someone can help. Thanks
 A: Let $W=|X-Y|$. We find the cdf of $W$. So we want to find $\Pr(W\le w)$. The only interesting part is for $0\le w\le a$.
Draw the square where our joint density for $X$ and $Y$ lives. Now draw the two lines $x-y=w$ and $x-y=-w$. Then the probability that $W\le w$ is the area of the part of the square between these two lines, divided by $a^2$. 
It is easier to find the area of the rest of the square. The rest of the square is made up of two right-angled isosceles triangles. Together they make up a little square. Find the side of that square. 
Note that the right-angled isosceles triangles are the north-west and south-east corners of the $a\times a$ square. Let's find the leg length of the one at the south-east corner. The line $y=x-w$ meets the
 $x$-axis at $x=w$. So the leg length is $a-w$. Thus the combined area of the two triangles is $(a-w)^2$, and therefore 
$$\Pr(W\le w)=1-\left(\frac{a-w}{a}\right)^2.$$ 
You will find that this bears a very nice resemblance to the cdf you calculated for $\min(X,Y)$: the distributions of the two random variables are the same.
Remark: Instead of using geometry, we could have integrated. But why bother?
There is undoubtedly a way of seeing that the distributions are the same with no calculation. Maybe tomorrow!
A: For every $t$ in $(0,a)$, consider the square $S_t=(t,a)^2$, the triangle $U_t$ with vertices $(0,t)$, $(a-t,a)$ and $(0,a)$, and the triangle $L_t$ with vertices $(t,0)$, $(a,0)$ and $(a,a-t)$. 
Each triangle $U_t$ and $L_t$ is isocele rectangle with base $a-t$ hence $|L_t|=|U_t|=\frac12(a-t)^2$. Since $L_t\cap U_t=\varnothing$, $|L_t\cup U_t|=(a-t)^2=|S_t|$.
Now, $[|X-Y|\geqslant t]=[(X,Y)\in L_t\cup U_t]$ and $[\min(X,Y)\geqslant t]=[(X,Y)\in S_t]$ hence $|X-Y|$ and $\min(X,Y)$ are identically distributed and their common density at $t$ in $(0,a)$ is
$$
\frac1{a^2}\frac{\mathrm d |S_t|}{\mathrm dt}=\frac2{a^2}(a-t).
$$
