Given the probability that X > Y and X <Y, find E[X] and E[Y]

Question

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?

Answer 1

Let random variable $X$ take constant value $1$. Let random variable $Y$ have density $y$ on the interval $[0,\sqrt{2}]$ and $0$ elsewhere.

Then $\Pr(X\gt Y)=\Pr(Y\lt `1)=\frac{1}{2}$.

We have $E(X)=1$ and $E(Y)= \frac{2\sqrt{2}}{3}$.

Remark: The problem is basically the same as the problem of showing that the median and mean of a random variable are not necessarily equal. Almost any non-symmetric example works. The visualization is very similar to the visualization of how to make mean not equal to the median.

Answer 2

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, $E[Y]\ne E[X]$.