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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|\mathrm{E}(X-m)\right|\!\!\!\!\!</math>
|<math>\leq \mathrm{E}\left(\left|X-m\right|\right)</math>
|-
|
|<math>\leq \mathrm{E}\left(\left|X-\mu\right|\right) = \mathrm{E}\left(\sqrt{(X-\mu)^2}\right)</math>
|-
|
|<math>\leq \sqrt{\mathrm{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
absolute deviation function
:<math>a \mapsto \mathrm{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.
==Alternative proof==
The one-tailed version of
Chebyshev's inequality is
:<math>P(X-\mu \geq k\sigma)\leq\frac{1}{1+k^2}.</math>
Letting ''k'' = 1 gives P(''X'' ≥ μ + σ) ≤ 1/2 and (by changing the sign of ''X'' and so μ) P(''X'' ≤ μ − σ) ≤ 1/2. So the median is within one standard deviation of the mean.
