# Empirical Relationship Between Mean, Median and Mode

For a unimodal distribution which is moderately skewed, we have the following empirical relationship between mean, mode and median:

(Mean-Mode) ~ 3(Mean-Median)

Could someone please explain how the relationship was derived? I am really curious to know. Did Karl Pearson plot thousands of these relationships before forming this conclusion or is there a logical line of reasoning behind this relationship?

Thank you,

-

It's not a full explanation (i may come back to it depending on the comments), but here's an intuitive derivation:

Denoting $\mu$ the mean ($\neq$ average), $m$ the median, $\sigma$ one standard deviation, $M$ the mode, $sgn()$ the sign function and $X$ the (random) dataset.

It's well known that $|\mu-m|\leq\sigma$. Certainly, Chebyshev's inequality must imply that $|\mu-M|<3\sigma$. Finally, it's easy to show that $sgn(\mu-M)=sgn(\mu-m)$ (with the convention that if $sgn(\mu-m)==0$, then both sides equal 1).

All these hold for any unimodal distribution (regardless of skew).

EDIT: this is a frequent textbook exercise: $\begin{eqnarray} |\mu-m|&=&|E(X-m)| \ &\leq& E|X-m| \ &\leq& E|X-\mu| \ &=& E\sqrt((X-\mu)^2) \ &\leq& \sqrt(E(X-\mu)^2) \ &=&\sigma \end{eqnarray}$

The first equality derives from the definition of the mean, the third comes about because the mean is the unique minimiser (among all $c$'s) of $E|X-c|$ and the fourth from Jensen's inequality (i.e. the definition of a convex function).

Chebyshef's inequality states that:

$P(|X-\mu|>\alpha)\leq \frac{\sigma^2}{\alpha^2}$

-
I am sorry, I am just a first year math student. Could you please provide/recommend a link/book/paper that describes how the relationship was derived? – Sara Oct 20 '10 at 8:52
@Sara I think it dates back to Karl Pearson, which uses this empirical relationship for his "Pearson mode skewness". Aside from this, you may find interesting this online article, j.mp/aWymCv. – chl Oct 20 '10 at 9:42
Thank you chl and kwak for the link and answer you have provided. I will study them. – Sara Oct 20 '10 at 11:13
If mean is unique minimizer, shouldn't it be $E|X-\mu|\le E|X-m|$? – mpiktas Dec 9 '10 at 13:09
Various points: $E|X-k|$ is minimised when $k$ is the median of $X$. Von Hippel's article (linked above by chl) discuses exceptions and btinternet.com/~se16/hgb/median.htm shows the possible relationship between mean, median, mode and standard deviation, both for continuous and for discrete distributions. The 3 can in fact take any value: positive, negative, zero or infinite. – Henry Feb 9 '11 at 8:18