expectation equals 0 and probability.

Question

If the expectation of a continuous random variable $X$ is $0$, then can we prove that $$ P(X>0)=\dfrac{1}{2}\,\,\,?$$ If so, how?

Answer 1

If the probability density function $f_X(x)$ is symmetric, i.e. $f_X(-x) = f_X(x)$, then clearly $\mathbb{E}\left(X\right) = 0$, provided $X$ is integrable. For such a density $$\Pr(X>0) = \Pr\left(X \leqslant 0\right) = \frac{1}{2}$$

However, the converse is not true, i.e. $\mathbb{E}\left(X\right)=0$ does not imply $f_X(-x) = f_X(x)$ and hence $\Pr(X>0) = \frac{1}{2}$ does not follow.

As a counter example consider a triangular distribution with $a=c=-1$ and $b=2$: $$ f_X(x) = \begin{cases}\frac{2}{9} \left(2-x\right) & -1 < x < 2 \cr 0 & \text{otherwise} \end{cases} $$