P(X>Y) Probability Double Integral

Question

$f(x,y) = \frac{12}{7(x^2 + xy)}$ $ 0 \le x \le 1$ and $0 \le y \le 1 $

I want to know the $P(X>Y)$. I believe the correct solution to this is integrating from 0 to 1 for dy and y to 1 for dx on f(x,y). But I would think you would want to integrate from 0 to 1 for dy and from 1 to y for dx. Can someone explain the reasoning behind this question?

Thanks.

Answer 1

I'm going to assume that the distribution $f(x,y)$ is defined on $x \in [0,1]$, $y \in [0,1]$.

If the outer integral is w.r.t. $y$, then $0 \le y \le 1$ and for any fixed $y$, we have $y \le x \le 1$.

So, the integral would be $P(X > Y) = \displaystyle\int_{0}^{1}\int_{y}^{1}f(x,y)\,dx\,dy$.

Keep in mind that $y \le 1$, so if the integral was $\displaystyle\int_{0}^{1}\int_{1}^{y}f(x,y)\,dx\,dy$, you would get a negative answer.