Just a heads-up: This is for a homework assignment so feel free to point in the right direction rather than answer it.
Suppose that $ X $ and $ Y $ are random variables whose joint p.d.f. $ f $ is defined by $$ \forall (x,y) \in [0,1]^{2}: \qquad {f_{X,Y}}(x,y) \stackrel{\text{df}}{=} 4 x y. $$
I’ve found that $ X $ and $ Y $ are independent (with the marginal p.d.f.’s $ {f_{X}}(x) = 2 x $ and $ {f_{Y}}(y) = 2 y $). Now, I’m asked to find $ \mathsf{P}(2 X > Y) $.
I’m not even sure where to start with this. At this point, I’m thinking (because of an example in our text) that I have to use the identity $$ \mathsf{P}(2 X > Y) = \mathsf{P}(2 X - Y > 0) = 1 - \mathsf{P}(2 X - Y \leq 0) $$ to integrate $ f_{X,Y} $ over the region $ \{ (x,y) \mid 0 \leq x \leq 1 ~ \text{and} ~ 0 \leq y \leq 2 x \} $. However, I’m not sure why or if I’m allowed to use the joint p.d.f. when we’ve already done stuff to $ X $ and $ Y $.
Am I allowed to? Or am I completely on the wrong course (I feel like it’s probably this one, haha)?
Thanks in advance.
Edit: Fixed up the specification of the region. -_-
Yes, almost. You want to integrate over the region within the joint support of the random variables where the constraint is holds.
That is: $\{(x,y)\mid 0\leq x\le 1, 0\le y\le 1, 2x>y\}\\ = \{(x,y)\mid 0\leq x\le 1/2, 0\le y< 2x\}\cup\{(x,y)\mid 1/2< x\le 1, 0\le y\le 1\}$
$$\begin{align}\mathsf P(2X>Y) ~=~& \iint_{2x>y} f_{X,Y}(x,y)~\mathsf d(x, y)\\[1.25ex]=~& \int_0^1\int_0^{\min(1, 2x)} 4xy~\mathsf d y~\mathsf d x\\[1ex]=~& \int_0^{1/2}\int_0^{2x}4xy~\mathsf d y~\mathsf d x+\int_{1/2}^1\int_0^1 4xy~\mathsf d y~\mathsf d x\end{align}$$