Intuitively, median and average are identical when the distribution of values is symmetrical. But is that a necessary condition? If not, what is?
2026-03-26 06:06:53.1774505213
Under what conditions are the median and average identical?
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No, it is not a necessary condition.
Consider for instance the discrete probability distribution $p$ supported on $\{-1,0,\frac{4}{5}\}$ such that $$ p(-1)=\frac{1}{3}, \quad p(0) = \frac{1}{4}, \quad p\left(\frac{4}{5}\right) = \frac{5}{12} $$ Then $\mathbb{E}_{X\sim p}[X] = -1\cdot\frac{1}{3}+\frac{4}{5}\cdot\frac{5}{12} = 0$, and the median of $p$ is also $0$. But the distribution is not symmetric.
As far as I know, there is no simple necessary condition besides what is essentially a restatement of the definitions.