Editing Chebyshev's inequality (section)

===Cantelli's inequality===
[[Cantelli's inequality]]<ref name=Cantelli1910>Cantelli F. (1910) Intorno ad un teorema fondamentale della teoria del rischio. Bolletino dell Associazione degli Attuari Italiani</ref> due to [[Francesco Paolo Cantelli]] states that for a real random variable (''X'') with mean (''μ'') and variance (''σ''<sup>2</sup>)

: <math> \Pr(X - \mu \ge a) \le \frac{\sigma^2}{ \sigma^2 + a^2 } </math>

where ''a'' ≥ 0.

This inequality can be used to prove a one tailed variant of Chebyshev's inequality with ''k'' > 0<ref name=Grimmett00>Grimmett and Stirzaker, problem 7.11.9. Several proofs of this result can be found in [http://www.mcdowella.demon.co.uk/Chebyshev.html Chebyshev's Inequalities] {{Webarchive|url=https://web.archive.org/web/20190224000121/http://www.mcdowella.demon.co.uk/Chebyshev.html |date=2019-02-24 }} by A. G. McDowell.</ref>

:<math> \Pr(X - \mu \geq k \sigma) \leq \frac{ 1 }{ 1 + k^2 }. </math>

The bound on the one tailed variant is known to be sharp. To see this consider the random variable ''X'' that takes the values

: <math> X = 1 </math> with probability <math> \frac{ \sigma^2 } { 1 + \sigma^2 }</math>
: <math> X = - \sigma^2 </math> with probability <math> \frac{ 1 } { 1 + \sigma^2 }.</math>

Then E(''X'') = 0 and E(''X''<sup>2</sup>) = ''σ''<sup>2</sup> and P(''X'' < 1) = 1 / (1 + ''σ''<sup>2</sup>).

==== An application: distance between the mean and the median ====
<!-- This section is linked from [[median]] and [[exponential distribution]]. -->

The one-sided variant can be used to prove the proposition that for [[probability distribution]]s having an [[expected value]] and a [[median]], the mean and the median can never differ from each other by more than one [[standard deviation]].  To express this in symbols let ''μ'', ''ν'', and ''σ'' be respectively the mean, the median, and the standard deviation. Then

:<math> \left | \mu - \nu \right | \leq \sigma. </math>

There is no need to assume that the variance is finite because this inequality is trivially true if the variance is infinite.

The proof is as follows. Setting ''k''&nbsp;=&nbsp;1 in the statement for the one-sided inequality gives:

:<math>\Pr(X - \mu \geq \sigma) \leq \frac{ 1 }{ 2 } \implies \Pr(X \geq \mu + \sigma) \leq \frac{ 1 }{ 2 }. </math>

Changing the sign of ''X'' and of ''μ'', we get

:<math>\Pr(X \leq \mu - \sigma) \leq \frac{ 1 }{ 2 }. </math>

As the median is by definition any real number&nbsp;''m'' that satisfies the inequalities

:<math>\Pr(X\leq m) \geq \frac{1}{2}\text{ and }\Pr(X\geq m) \geq \frac{1}{2}</math>

this implies that the median lies within one standard deviation of the mean. A proof using Jensen's inequality also [[Median#Inequality relating means and medians|exists]].