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Chebyshev Inequality

Apply Markov's inequality with a=k^2 to obtain

P[(x-mu)^2>=k^2]<=(<(x-mu)^2>)/(k^2)=(sigma^2)/(k^2).
(1)

Therefore, if a random variable x has a finite mean mu and finite variance sigma^2, then for all k>0,

P(|x-mu|>=k)<=(sigma^2)/(k^2)
(2)
P(|x-mu|>=ksigma)<=1/(k^2).
(3)

See also

Chebyshev Sum Inequality

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References

Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 11, 1972.Hardy, G. H.; Littlewood, J. E.; and Pólya, G. "Tchebychef's Inequality." §2.17 and §5.8 in Inequalities, 2nd ed. Cambridge, England: Cambridge University Press, pp. 43-45 and 123, 1988.Papoulis, A. Probability, Random Variables, and Stochastic Processes, 2nd ed. New York: McGraw-Hill, pp. 149-151, 1984.

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Chebyshev Inequality

Cite this as:

Weisstein, Eric W. "Chebyshev Inequality." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ChebyshevInequality.html

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