Does $P(X>x, Y>x)= P(X>x)P(Y>y)$ implies independence? We know, by definition, that two random variables are independent if
 $$P(X\leq x, Y\leq y)= P(X\leq x)P(Y\leq y).$$
If, insted, I have that 
$$P(X>x, Y>y)= P(X>x)P(Y>y),$$
does this also imply independence between $X$ and $Y$?
 A: Note that those two statements are equivalent. Dave's proof is great, but I would go for a simpler approach:


*

*$-X$ and $-Y$ are independent if, and only if, $X$ and $Y$ are indepedent.

*Apply the definition for $-X$ and $-Y$
WARNING: I am not sure about how tough to proof the first statement really is (probably not too much). However, from an intuitive perpective, it is reasonable to assume as true (otherwise independence sucks!). The second part is quite straight-forward. 
A: Yes!
Claim: $$P(X>x,Y>y) = P(X>x)P(Y>y) \implies P(X\le x,Y\le y) = P(X\le x)P(Y\le y)$$
Proof:
Let's start by assuming $P(X>x,Y>y) = P(X>x)P(Y>y)$. Now let's expand the right side.
$$ P(X>x)P(Y>y) = (1-P(X\le x)) (1-P(Y\le y)) $$
$$= 1- P(X\le x)-P(Y\le y)+P(X\le x)P(Y\le y)$$
Now let's expand the left side:
$$P(X>x,Y>y) =P(X>x\bigcap Y>y)$$
$$=P((X\le x \bigcup Y\le y)^C) $$
$$= 1- P(X\le x \bigcup Y\le y)$$
$$= 1-\bigg[P(X\le x) + P(Y\le y) - P(X\le x,Y\le y)\bigg] $$
$$= 1- P(X\le x) - P(Y\le y) + P(X\le x,Y\le y)$$
Let's set the sides equal and cancel terms.
$$1- P(X\le x) - P(Y\le y) + P(X\le x,Y\le y) = 1- P(X\le x)-P(Y\le y)+P(X\le x)P(Y\le y)$$
$$ \implies P(X\le x,Y\le y) = P(X\le x)P(Y\le y) $$
