= Chebyshev's inequality
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{title2=$\mathbb P(|X-\mathbb E X|\geq a)\leq\operatorname{Var}(X)/a^2$}
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= Chebyshev inequality
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{synonym}
For a real <random variable> $X$ with finite <variance> $\sigma^2$, <expected value> $\mu$, and a threshold $a>0$, <Chebyshev's inequality> gives
$$
\mathbb P(|X-\mu|\geq a)\leq\frac{\sigma^2}{a^2}.
$$
One can also cap this upper bound at one, since a <probability> cannot exceed one. If $\sigma>0$, taking $a=k\sigma$ gives $\mathbb P(|X-\mu|\geq k\sigma)\leq1/k^2$. For example, at least $3/4$ of the <probability> lies strictly within two <standard deviations> of the <mean>. This statement needs only the finite <variance> assumption. If the <variance> is zero, $X=\mu$ <almost surely> and every positive deviation threshold has zero <tail probability>.
Consider a <random variable> uniformly distributed on $\{-2,-1,0,1,2\}$. Its <expected value> is zero and its <variance> is $(4+1+0+1+4)/5=2$. At $a=2$ the actual <tail probability> is $2/5$, while <Chebyshev's inequality> gives $2/2^2=1/2$. At $a=3$ the actual <tail probability> is zero, while the bound is $2/9$. A valid bound need not describe the exact shape or bounded support of a particular <probability distribution>.
\Image[/chebyshev-bound-and-sharpness.png]
{title=A general variance bound can be loose for a normal distribution and exact for another distribution}
{description=Left: the two-sided <tail probability> of the <standard normal distribution> is compared with the capped <Chebyshev's inequality> bound. Right: the three-point <probability distribution> places <probabilities> $1/8$, $3/4$, $1/8$ at $-2$, $0$, $2$, giving <variance> one and <probability> $1/4$ at a deviation of at least two. The points at the threshold are included by $\geq$.}
For the <standard normal distribution>, the actual <tail probabilities> beyond two and three <standard deviations> are approximately $0.0455003$ and $0.0026998$, whereas the general bounds are $1/4$ and $1/9$. More knowledge of the <probability distribution> can therefore improve the estimate. The comparison does not diminish the guarantee: the three-point example in <equality in Chebyshev's inequality> attains it exactly.
The proof is a one-line use of <Markov's inequality> on the nonnegative <random variable> $Y=(X-\mu)^2$. The event $|X-\mu|\geq a$ equals $Y\geq a^2$, and $\mathbb E Y=\sigma^2$. This explains why the denominator is squared and why a finite <variance> supplies the necessary information. https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/887e96172e82a615dac16fb26702d88b_MIT18_175S14_Lecture8.pdf[Scott Sheffield, MIT 18.175, lecture 8: Markov and Chebyshev bounds]{ref}.
One important use is control of a <sample mean>. For <independent and identically distributed random variables> with finite <variance> $\sigma^2$, the <variance of a sample mean> is $\sigma^2/n$. Hence
$$
\mathbb P(|\overline X_n-\mu|\geq\varepsilon)\leq\frac{\sigma^2}{n\varepsilon^2},\qquad \varepsilon>0.
$$
For every fixed tolerance, the bound tends to zero as the <sample size> grows, proving the finite-<variance> case of the <weak law of large numbers>. For $\sigma^2=1$ and $\varepsilon=0.1$, $n=2000$ makes the bound $0.05$. This is a sufficient bound under the stated assumptions, rather than an optimal required <sample size>.
An early historical source is <Pafnuty Chebyshev>'s 1867 paper “Des valeurs moyennes”, published in French translation by N. de Khanikof. Its opening theorem bounds deviations of sums using their <expected values> and <second moments>, linking averages with control of exceptional outcomes. https://www.numdam.org/item/JMPA_1867_2_12__177_0.pdf[Chebyshev, Journal de mathématiques pures et appliquées, 2nd series 12, 177–184 (1867)]{ref}.
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