Inprobability theory,Cantelli's inequality (also called theChebyshev-Cantelli inequality and theone-sided Chebyshev inequality) is an improved version ofChebyshev's inequality for one-sided tail bounds.^[1][2][3] The inequality states that, for${\displaystyle \lambda >0,}$

${\displaystyle Pr(X-\mathbb{E}[X]\ge \lambda )\le \frac{{\sigma }^2}{{\sigma }^2+{\lambda }^2},}$

where

${\displaystyle X}$ is a real-valuedrandom variable,

${\displaystyle Pr}$ is theprobability measure,

${\displaystyle \mathbb{E}[X]}$ is theexpected value of${\displaystyle X}$,

${\displaystyle {\sigma }^2}$ is thevariance of${\displaystyle X}$.

Applying the Cantelli inequality to${\displaystyle -X}$ gives a bound on the lower tail,

${\displaystyle Pr(X-\mathbb{E}[X]\le -\lambda )\le \frac{{\sigma }^2}{{\sigma }^2+{\lambda }^2}.}$

While the inequality is often attributed toFrancesco Paolo Cantelli who published it in 1928,^[4] it originates in Chebyshev's work of 1874.^[5] When bounding the event random variable deviates from itsmean in only one direction (positive or negative), Cantelli's inequality gives an improvement over Chebyshev's inequality. The Chebyshev inequality has"higher moments versions" and"vector versions", and so does the Cantelli inequality.

For one-sided tail bounds, Cantelli's inequality is better, since Chebyshev's inequality can only get

${\displaystyle Pr(X-\mathbb{E}[X]\ge \lambda )\le Pr(|X-\mathbb{E}[X]|\ge \lambda )\le \frac{{\sigma }^2}{{\lambda }^2}.}$

On the other hand, for two-sided tail bounds, Cantelli's inequality gives

${\displaystyle Pr(|X-\mathbb{E}[X]|\ge \lambda )=Pr(X-\mathbb{E}[X]\ge \lambda )+Pr(X-\mathbb{E}[X]\le -\lambda )\le \frac{2{\sigma }^2}{{\sigma }^2+{\lambda }^2},}$

which is always worse than Chebyshev's inequality (when${\displaystyle \lambda \ge \sigma }$; otherwise, both inequalities bound a probability by a value greater than one, and so are trivial).

Various stronger inequalities can be shown.He, Zhang, and Zhang showed^[6] (Corollary 2.3) when${\displaystyle \mathbb{E}[X]=0,\mathbb{E}[X^2]=1}$ and${\displaystyle \lambda \ge 0}$:

${\displaystyle Pr(X\ge \lambda )\le 1-(2\sqrt{3}-3)\frac{(1+{\lambda }^2{)}^2}{\mathbb{E}[X^4]+6{\lambda }^2+{\lambda }^4}.}$

In the case${\displaystyle \lambda =0}$ this matches a bound in Berger's "The Fourth Moment Method",^[7]

${\displaystyle Pr(X\ge 0)\ge \frac{2\sqrt{3}-3}{\mathbb{E}[X^4]}.}$

This improves over Cantelli's inequality in that we can get a non-zero lower bound, even when${\displaystyle \mathbb{E}[X]=0}$.

• Chebyshev's inequality
• Paley–Zygmund inequality

• Probabilistic inequalities

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