It has been so long since I did any of this sort of thing (and wasn't even any good at it back then!), so apologies if I'm way off the mark.I'd go for a proof by contradiction. We know that X>Y with probability zero. Assume that X<Y with probability greater than zero. Then, the expectation of X would have to be less than the expectation of Y. And the distributions would be different - hence contradiction. So thereofre X=Y with probability 1 (almost surely).