Given the probability that X > Y and X If $P[X>Y]=\frac{1}{2}$, and $[X<Y]=\frac{1}{2}$, then is $E[X]=E[Y]$?
How can I visualize this problem? 

Assume that $X$ is standard normal and consider
$$Y=2X+X^+.$$
Then $E[X]=0$, $[Y\gt X]=[X\gt0]$ hence $P[Y\gt X]=\frac12$, and $[Y\lt X]=[X\lt0]$ hence $P[Y\lt X]=\frac12$.
Furthermore, $E[Y]=2E[X]+E[X^+]=2\cdot0+\frac1{\sqrt{2\pi}}$. Thus, $E[Y]\ne E[X]$.
