Joint PDF to CDF

Question

So I have this joint PDF: $$ f(x,y)= \begin{cases} 4xy & \text{ for } 0 \leq x \leq 1, 0\leq y \leq 1\\ 0 & \text{ otherwise} \end{cases} $$

To make this a CDF, I have tried to double integrate the PDF from $-\infty$ to $x$, $-\infty$ to $y$.

$$ F(x,y)= \begin{cases} 0 & \text{ if } x<0\ \text{or}\ y<0\\ x^2y^2 & \text{ if } 0 \leq x\leq1,0 \leq y\leq1 \\ ? & \text{ if } 0 \leq x\leq1,y>1\\ ? & \text{ if } 0 \leq y\leq1,x>1\\ 1 & \text{ if } x>1,y>1 \end{cases} $$

Can someone help me understand exactly what to do for the cases with the question marks? I am unsure of which function to apply the double integral to in order to obtain the result.

Thanks in advance.

Answer 1

An example: $$F(\tfrac13,2)=\iint_{\mathbb R^2}[x\leqslant\tfrac13,y\leqslant2]\,4xy\,[0\leqslant x\leqslant1,0\leqslant y\leqslant 1]\,\mathrm dx\mathrm dy, $$ that is, $$ F(\tfrac13,2)=\iint_{\mathbb R^2}[0\leqslant x\leqslant\tfrac13,0\leqslant y\leqslant1]\,4xy\,\mathrm dx\mathrm dy, $$ and, finally, $$ F(\tfrac13,2)=\left.x^2\right|_{x=0}^{x=1/3}\cdot\left.y^2\right|_{y=0}^{y=1}=\left(\tfrac13\right)^2$$ Thus, your $?$ are $x^2$ and $y^2$ respectively.

Answer 2

$$\begin{align}f(x,y) &= \begin{cases} 4xy & \text{ for } 0 \leq x \leq 1, 0\leq y \leq 1\\ 0 & \text{ otherwise} \end{cases} \\[2ex]F(x,y)&= \begin{cases} 0 & \text{ if }& x<0\text{ or }y<0\\ 4\int_0^x \int_0^y st~\mathrm d t~\mathrm d s & \text{ if }& 0 \leq x\lt 1~,~0 \leq y\lt 1 \\ 4\int_0^x \int_0^1 st~\mathrm d t~\mathrm d s & \text{ if }& 0 \leq x\lt 1~,~1\leq y\\ 4\int_0^1 \int_0^y st~\mathrm d t~\mathrm d s & \text{ if }& 1\leq x\qquad~,~0 \leq y\lt 1\\ 4\int_0^1\int_0^1 st~\mathrm d t~\mathrm d s & \text{ if }& 1\leq x\qquad~,~1\leq y \end{cases} \\[1ex]&=\begin{cases} 0 & \text{ if }& x<0\text{ or }y<0\\ x^2y^2 & \text{ if }& 0 \leq x\lt 1~,~0 \leq y\lt 1 \\ x^2 & \text{ if }& 0 \leq x\lt 1~,~1\leq y\\ y^2 & \text{ if }& 1\leq x\qquad~,~0 \leq y\lt 1\\ 1 & \text{ if }& 1\leq x\qquad~,~1\leq y \end{cases} \end{align}$$