Suppose we have $f(x,y) = 2$ when $x>0$, $y>0$, $x+y <1$ and $0$ elsewhere. I want to find the $\Pr(Y>X)$ and I'm struggling with finding the correct limits of integration generally but especially when we want to find $\Pr(Y>X)$ or similar.
Can anyone explain the general method please, would be very grateful.
Here, the blue line is the line $x+y=1$. The region where $y>x$ is the triangle formed above the red line. You want to integrate your pdf over this region.
To find the limits, first notice that $y$ ranges from $0$ to $1$, and $x$ ranges from $0$ to $\frac 12$. You can choose which variable you want to integrate over first. Lets say you choose $y$. Then your integral will be of the form $$\color{red}{\int_\text{limits}}\int_\text{limits}\,dy\,\color{red}{dx}$$ First we look at the red limits. This is just the limits of $x$, i.e. $$\int_{0}^{\frac12} [... ]dx$$
Now is the key step - you want to see what the $y$ limits should be for some value of $x$. You can see from the graph that integrating in the $y$ direction always starts on the line $y=x$, so this is where you start to integrate $y$. A little more thought will lead you to seeing that you stop integrating at $1-x$, which is when you hit the blue line. So your limits on the $y$ integration are $$\int_{x}^{1-x}...\,dy$$ Now put it all together to get $$\int_{0}^1\int_0^{1-x} f(x,y)\,dy\,dx$$
However, it actually necessary to integrate here. This is because $f(x,y)$ does not depend on $x$ or $y$, so you can take it out of the integral - $$\int_A f(x,y) \,dx\, dy=\int_A 2 \,dx\, dy=2 \int_A \,dx\, dy$$ So you end up getting an area integral. So all you need to do is find the area of the region - here it is a triangle, so its area is $\frac{b\times h}2=\frac{1\times\frac12}2=\frac14$, and so the probability you require is $$P(Y>X)=2 \int_A \,dx\, dy=2\times \frac14=\frac12$$
This is the same result you will get by integrating.
