Cumulative distribution function properties in terms of complements

Question

What is the expression $$\int_{\frac{1}{3}}^1\int_{\frac{1}{3}}^1f(x,y)dxdy$$in terms of CDF $F$ when $f$ is its pdf and bivariate distribution as support $[0,1]\times[0,1]$?

I thought it was $1-F(1/3,1/3)$ but someone said it is wrong.

Answer 1

$1−F(1,1/3)-F(1/3,1)+F(1/3,1/3)$

Just look at it geometrically: the square $[0,1]\times[0,1]$ is split into 4 rectangles by a horizontal line $y=1/3$ and a vertical line $x=1/3$.

$F(x,y)$ is the integral over the rectangle with vertices $(0,0)$ and $(x,y)$.

To get the integral you need, you have to take the whole square ($F(1,1)=1$) and remove two thin rectangles ($F(1,1/3)$ and $F(1/3,1)$), but now you removed the small square ($F(1/3,1/3)$) twice, so you need to re-add it.