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Chebyshev's inequality (Theorem)

Let $ X\in \textbf{L}^2$ be a real-valued random variable with mean $ \mu=\mathbb{E}[X]$ and variance $ \sigma^2=\operatorname{Var} [X]$. Then for any standard of accuracy $ t>0$,

$\displaystyle \mathbb{P}_{}\left\{\left\vert X - \mu \right\vert \ge t\right\} \le \frac{\sigma^2}{t^2}. $

Note: There is another Chebyshev's inequality, which is unrelated.



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"Chebyshev's inequality" is owned by rspuzio. [ full author list (2) | owner history (1) ]
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See Also: Markov's inequality, Chebyshev's inequality

Keywords:  variance, mean, deviation

Attachments:
proof of Chebyshev's inequality (Proof) by PrimeFan
Multidimensional Chebyshev's inequality (Theorem) by daniWk
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Cross-references: variance, mean, random variable
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This is version 3 of Chebyshev's inequality, born on 2002-06-17, modified 2006-10-23.
Object id is 3115, canonical name is ChebyshevsInequality2.
Accessed 12399 times total.

Classification:
AMS MSC60A99 (Probability theory and stochastic processes :: Foundations of probability theory :: Miscellaneous)

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