An inequality on location and scale parameters
For probability distributions having an expected value and a median, the mean (i.e., the expected value) and the median can never differ from each other by more than one standard deviation. To express this in mathematical notation, let μ, m, and σ be respectively the mean, the median, and the standard deviation. Then
- <math>\left|\mu-m\right| \leq \sigma.<math>
Proof
This proof uses Jensen's inequality twice. We have
- <math>\left|\mu-m\right|
= \left|E(X-m)\right|
\leq E\left(\left|X-m\right|\right)
\leq E\left(\left|X-\mu\right|\right)<math>
- <math>= E\left(\sqrt{(X-\mu)^2\,}\right) \leq \sqrt{E((X-\mu)^2)\,}=\sigma.<math>
The first inequality comes from (the convex version of) Jensen's inequality applied to the absolute value function, which is convex. The second comes from the fact that the median minimizes the function
- <math>a \mapsto E(\left|X-a\right|).<math>
The third inequality comes from (the concave version of) Jensen's inequality applied to the square root function, which is concave. Q.E.D.
