Suppose f(x,y) = {1/2 for 0<=x<=2 and 0<= y <= 1} and 0 otherwise.
Find Pr(X>=Y). So everything to the right of x = y is where to integrate on I believe you can find the bounds two ways: 1. Integrate on x first then on y. $$\int_{0}^{1} \int_{y}^{2}1/2 \,dx\,dy$$ and consequently the answer is 3/4.
- Integrate on y first then on x. To me the bounds are intuitively supposed to be: $$\int_{0}^{2} \int_{o}^{x}1/2 \,dy\,dx$$
But that is not correct because the answer is 3/4.
Can someone explain how the to derive the bounds for the second method?
You have to split the integral, since from $0 \le x \le 1$, $y$ goes between $0$ and $x$, but for $1 \le x \le 2$, $y$ goes between $0$ and $1$.
So the correct integral should be $$\int_0^1{\int_0^x{1/2 dy dx}} + \int_1^2{\int_0^1{1/2 dy dx}}$$ $$= 1/4+1/2 = 3/4$$