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