How to find the joint PDF? Ley X,Y two random variables with density
$f(x,y)=8xy \  \text{if} \  
0<x<y<1$
Find the joint distribution F(x,y).
I find that $F(x,y)=2x^2y^2-x^4\ \text{  if}  \ 0<x<y<1$ but I dont know how to find the didtribution in other points.
 A: Draw a picture. We sketch how to find the cdf $F(s,t)$ in other cases. Let $T$ be the triangle where our joint density "lives." It has corners $(0,0)$, $(1,1)$, and $(0,1)$. We will be referring to it several times.
Suppose first that $t\gt 1$. Draw the point $P=(s,t)$. We want the probability of falling in the region below and to the left of $P$.
If $s\le 0$ this probability is $0$. If $s\ge 1$ this probability is $1$. Now we deal with the more interesting $0\lt s\lt 1$. This probability is 
$$\int_{x=0}^s\left(\int_{y=x}^1 8xy\,dy\right)\,dx.$$
We have just integrated the density function over the part of $T$ that is below and to the left of $P$.
Now we deal with $0\lt t\lt 1$ but $P$ not in $T$. If $s\le 0$ the cdf is $0$. If $s\ge 1$ it is $1$. Now we deal with $0\lt s\lt 1$, but $s\gt t$.  Draw a point $P$ satisfying this.
We want to integrate the joint density over the part of $T$ that is below and to the left of $P$. Looking at the picture, we see that $y$ goes from $0$ to $t$, and then $x$ goes from $0$ to $t$.
Finally, if $t\lt 0$ the cdf is $0$.
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A: The five cases are:
$$F(x,y) \;=\; \begin{cases} 0 & : x<0 \vee y<0\\[1ex]\hdashline\displaystyle \int_0^y \int_0^x f(s,t)\operatorname d s\operatorname d t & : 0\leq x< y< 1 \\[1ex]\hdashline\displaystyle \int_0^y \int_0^y f(s,t)\operatorname d s\operatorname d t & : 0\leq y\leq 1, x\geq y \\[1ex]\hdashline\displaystyle \int_0^1 \int_0^x f(s,t)\operatorname d s\operatorname d t  & : 0\leq x < 1 \leq y \\[1ex]\hdashline 1 & : 1\leq x \wedge 1\leq y\end{cases}$$
