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Chebyshev Inequality

Apply Markov's Inequality with $a \equiv k^2$ to obtain

P[(x-\mu)^2\geq k^2] \leq {\left\langle{(x-\mu)^2}\right\rangle{}\over k^2} = {\sigma^2\over k^2}.
\end{displaymath} (1)

Therefore, if a Random Variable $x$ has a finite Mean $\mu$ and finite Variance $\sigma^2$, then $\forall\ k
\geq 0$,
P(\vert x-\mu\vert\geq k)\leq {\sigma^2\over k^2}
\end{displaymath} (2)

P(\vert x-\mu\vert\geq k\sigma) \leq {1\over k^2}.
\end{displaymath} (3)


Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 11, 1972.

© 1996-9 Eric W. Weisstein