Probability given X & Y are independent rand. variables and 2 p.d.f.s

Question

first time poster here, maybe you guys can help me out.

Given that $X,Y$ are independent random vars with these pdfs:

$$f_X(x) = \begin{cases}1,& 0 < x < 1,\\0,& \text{otherwise}\end{cases}$$ $$f_Y(y) =\begin{cases} 8y,& 0 < y < \frac12\\0,& \text{otherwise}\end{cases}$$

how would I determine the value of $Pr(X>Y)$?

What I know: I sort of understand what $Pr(X>Y)$ means from this: Finding probability P(X<Y), but I don't know how to apply anything, or the steps to take -- it would be really helpful to have a step-by-step response on how to go about this, thanks!

Answer 1

The accepted answer on the post that you linked seems to work.

If $f_X(x)$ is as you defined, then its cdf is $$F_X(x):=p(X<x)= \int_0^x f_X(t)\mathop{dt}=x.$$

Then $$P(X>Y)=\int_0^{1/2} P(X>y)f_Y(y)\mathop{dy}=\int_0^{1/2}(1-F_X(y))f_Y(y)\mathop{dy}.$$