Chebyshev inequality

If \(X\) has finite mean \(\mu\) and standard deviation \(\sigma\), then for every \(k>0\) we have \(P(|X-\mu|\ge k\sigma)\le 1/k^2\). An equivalent form uses an absolute margin \(\varepsilon\): \(P(|X-\mu|\ge\varepsilon)\le\sigma^2/\varepsilon^2\). Both say that a small variance forces the variable to concentrate near its mean.

Taken through the complement, it guarantees concentration: at least \(1-1/k^2\) of the probability lies within \(k\sigma\) of the mean, that is, at least \(\tfrac34\) for \(k=2\) and \(\tfrac89\) for \(k=3\). These figures hold for any distribution with finite variance, symmetric or not, discrete or continuous.

That generality has a price: the bound is often loose. For a normal variable, the probability of straying more than \(2\sigma\) is about 0.05, while Chebyshev only ensures it does not exceed 0.25. With \(k\le1\) the bound is 1 or larger and says nothing.

The idea of the proof is to compare \(\sigma^2=E[(X-\mu)^2]\) with the contribution of the far region, where \((X-\mu)^2\ge\varepsilon^2\). A typical mistake is to use the variance where the standard deviation belongs, or to apply the bound to one side when the event is two-sided.