I am interested in an upper bound on the largest possible gap between the mean and median of a probability distribution on [0,1]. I am only interested in continuous density functions. My conjecture is that the upper bound is 0.5 but I can’t see how to prove it. Any help much appreciated.


2 Answers 2


If $X$ has a finite second moment, then one can prove that $$|\text{Median}(X) - \mathbb{E}(X)| \leq \sqrt{\operatorname{Var}(X)}.$$ Then, you can also prove that if $X$ is bounded in $[0,c]$ then $$\operatorname{Var}(X) \leq \frac{c}{4}.$$ In your case $c=1$ so that $\operatorname{Var}(X) \leq1/4. $ Combining these results, $$|\text{Median}(X) - \mathbb{E}(X)| \leq \sqrt{1/4} = 0.5.$$

  • $\begingroup$ Can you say something about how to prove the first inequality? $\endgroup$
    – Simd
    May 20, 2023 at 8:03
  • $\begingroup$ Take a look at this question. $\endgroup$
    – Bergson
    May 20, 2023 at 8:27

This replaces a since-deleted incorrect answer of mine.

Let $F$ be the distribution function of a random variable $X$ taking values in $[0,1]$, let its median be $m$ and its mean be $\mu$. We have $F(m)=1/2$ and (by integration by parts) $\mu=\int_0^1(1-F(x))\,dx$.

For given value of $m$, what $F$ maximizes $\mu$? Clearly $1-F(x)\le1$ on $[0,m)$ and $1-F(x)\le 1/2$ on $[m,1]$, so $\mu\le m + (1-m)/2 = (1+m)/2$.

And for given $m$, what $F$ minimizes $\mu$? Clearly $1-F(x)\ge 1/2$ on $[0,m)$ and $1-F(x)\ge1$ on $[m,1]$, so $\mu\ge m/2$.

Summarizing, $$ -\frac{m}2\le \mu-m\le \frac{ 1-m}2.$$

Since $m\in[0,1]$ this delivers the sought-after bound $|\mu-m|\le 1/2$. The values $\mu-m=\pm1/2$ cannot be attained by distributions with continuous density functions, but all in-between values can


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