Is the mean of a continuous random variable $X$ equal to its median?

Question

If we assume $X$ to be a normal distributed random variable, then we can prove than $\Bbb P(X<mean) = \Phi(0) = 0.5$. This is true for any symmetric distribution.

I'm not sure how to prove/disprove it for an asymmetric (around mean) PDF. We're basically trying to see if mean is always the $50\%$ percentile.

Answer 1

Let $X \sim \operatorname{Exp}(1)$. The mean is $$E[X]=1 ,$$ while the median is $\log(2)$.

Answer 2

Hint: take the density function $f(x) = 2x$ for $0\le x\le 1$ and $0$ otherwise. Can you calculate median and mean?

Answer 3

Let $\mathsf{P}(X=-1)=\frac{n}{n+1}$ and $\mathsf{P}(X=n)=\frac1{n+1}$.

Then $\mathsf EX=0$ so:$$\mathsf{P}(X<\mathsf EX)=\frac{n}{n+1}$$

Look what happens for a large $n$.