Suppose the joint pdf $(X,Y)$ is $$f(x,y) = 24y(x-y) \ \ \text{for} \ \ 0 < y < x < 1$$ (the support of this joint pdf is a triangular region bounded by the lines $y=0$,$x=1$, and $y=x$)
How would you find the cdf $F(x,y)$
Attempted solution: I know that $$F(x,y) = \int \int f(x,y) dxdy$$
I just don't know what the bounds for the integrals would be and why? Also, I can't seem to calculate the CDF.
Any suggestions are greatly appreciated.
You're almost there. Integrating pdf is the way to go.
Now, to find the CDF, you need to compute $$ Pr(X\leq x,Y \leq y) = \int \int_R f(X,Y)dXdY$$ where R is the appropriate region of integration.
Now, the region of integration will depend on the values of $x$ and $y$. We consider two cases: (1) $y>x$, and (2) $y \le x$. The best is to draw each of the cases and see what restrictions do these inequalities impose on $R$.
Case 1 ($y>x$): Here $Y \in [0,x]$ and $X \in [Y,x]$.
Case 2 ($y \le x$): Here we need to split $R$ into two regions of integration. First region is $Y \in[0,y]$ and $X \in [Y,y]$. Second region is $Y \in [0,y]$ and $X \in [y,x]$.
Computing the above integral using the regions of integration derived above we get
$$ Pr(X\leq x,Y \leq y)= x^4 \text{ if } x\leq y \\ Pr(X\leq x,Y \leq y)= y^4 + 2y^2\left( 3 x^2 - 4 xy +y^2\right) \text{ if x>y } $$