Suppose I have two independent random variables $X$ and $Y$ with supports $[0, \infty)$. They have PDFs $f_{X}, f_{Y}$ and CDFs $F_{X}, F_{Y}$. I want to find:
$\Pr((x>y)\, \cap x\in(a,b) \cap y\in[0,b))$ where $0 < a < b$ are constants
I cannot simply multiply the probabilities of these three events together since they are not independent. Further, I understand from pages like this that the probability of $x<y$ and $y<b$ is $\int_{0}^{b}\int_{0}^{y} f_{X}(x) f_{Y}(y) dx dy$. My main issue is that I only understand how this issue applies to the larger random variable, but the smaller (or both) random variables.
These are my attempts at this problem:
$\Pr((x>y)\, \cap x\in(a,b) \cap y\in[0,b))$
Solution 1
$\displaystyle{\int_{0}^{b}\int_{y}^{b} f_{X}(x)f_{Y}(y)\, dx dy}$
My issue with this solution is that it does not restrict $x>a$, so I think this is not a correct solution.
Solution 2
$\displaystyle{\int_{a}^{b}\int_{y}^{b} f_{X}(x)f_{Y}(y)\, dx dy}$
I am not convinced this is correct either - this seems to restrict $y$ to $[a,b]$. However, a realisation such as $y=0, x=b$ should be included in this event.
I am also convinced that I cannot simply multiply the multiple integral by $[F_{X}(b)-F_{X}(a)]$ because this is not independent of the condition $x>y$.
What is a way that I can incorporate both the upper and lower bounds of the restrictions on both $x$ and $y$ into this formulation?
Similarly, I am trying to find: $\Pr((x<y)\, \cap x\in(a,b) \cap y\in[0,b))$ I am also proposing two possible solutions, $\displaystyle{\int_{0}^{b}\int_{a}^{y} f_{X}(x)f_{Y}(y)\, dx dy}$ and $\displaystyle{\int_{a}^{b}\int_{a}^{y} f_{X}(x)f_{Y}(y)\, dx dy}$. I see the restrictions being less of a problem in this case, since it cannot be that $y<a$ and $x>a$ but $x<y$ - but again I am not sure which is the correct formulation.
Any ideas about how to include these restrictions into the probability of the event?