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Chebyshev's inequality (also known as the Bienaym茅鈥揅hebyshev inequality) assures that no more than a given fraction of values can be more than a certain distance from the mean for a wide class of probability distributions.

The Chebyshev's inequality is needed to be proved that states that for any probability distribution of any random variable\(X\)and any number\(k\)that is at least\(1\), the following holds 鈥�

\({\rm{P(|X - \mu |}} \ge {\rm{k\sigma )}} \le \frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}\)

The Variance of\(X\), where\(X\)is a discrete random variable\(X\)with set of possible values\(S\)and\({\rm{pmf p(x)}}\), denoted by\({\rm{V(X) (\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ or }}{{\rm{\sigma }}^{\rm{2}}}{\rm{)}}\)is 鈥�

\({\rm{V(X) = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {{{{\rm{(x - \mu )}}}^{\rm{2}}}} \cdot {\rm{p(x) = E}}\left( {{{{\rm{(X - \mu )}}}^{\rm{2}}}} \right)\)

So, by the definition, the following holds 鈥�

\(\begin{array}{c}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}\sum\limits_{{\rm{all x}}} {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{p(x)}} \ge \sum\limits_{{\rm{x:|x - \mu |}} \ge {\rm{k\sigma }}} {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{p(x)}}\\ \ge \sum\limits_{{\rm{x:|x - \mu |}} \ge {\rm{k\sigma }}} {{{{\rm{(k\sigma )}}}^{\rm{2}}}} {\rm{p(x) = (k\sigma }}{{\rm{)}}^{\rm{2}}}\sum\limits_{{\rm{x:|x - \mu |}} \ge {\rm{k\sigma }}} {\rm{p}} {\rm{(x)}}\\{\rm{ = }}{{\rm{k}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}{\rm{P(|X - \mu |}} \ge {\rm{k\sigma )}}\end{array}\)

Or equally this can be written as 鈥�

\(\begin{array}{*{20}{r}}{{{\rm{k}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}{\rm{P(|X - \mu |}} \ge {\rm{k\sigma )}} \le {{\rm{\sigma }}^{\rm{2}}}}\\{{\rm{P(|X - \mu |}} \ge {\rm{k\sigma )}} \le \frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}}\end{array}\)

For any\(k\)that is at least\(1\).

Therefore, the inequality is proved.