# Vysochanskij–Petunin inequality

(Redirected from Vysochanskiï-Petunin inequality)

In probability theory, the Vysochanskij–Petunin inequality gives a lower bound for the probability that a random variable with finite variance lies within a certain number of standard deviations of the variable's mean, or equivalently an upper bound for the probability that it lies further away. The sole restriction on the distribution is that it be unimodal and have finite variance. (This implies that it is a continuous probability distribution except at the mode, which may have a non-zero probability.) The theorem applies even to heavily skewed distributions and puts bounds on how much of the data is, or is not, "in the middle."

## Theorem

Let X be a random variable with unimodal distribution, mean μ and finite, non-zero variance σ2. Then, for any λ > √(8/3) = 1.63299…,

$P(\left|X-\mu\right|\geq \lambda\sigma)\leq\frac{4}{9\lambda^2}.$

Furthermore, the equality is attained for a random variable having a probability 1 − 4/(3 λ2) of being exactly equal to the mean, and which, when it is not equal to the mean, is distributed uniformly in an interval centred on the mean. When λ is less than √(8/3), there exist non-symmetric distributions for which the 4/(9 λ2) bound is exceeded.

## Properties

The theorem refines Chebyshev's inequality by including the factor of 4/9, made possible by the condition that the distribution be unimodal.

It is common, in the construction of control charts and other statistical heuristics, to set λ = 3, corresponding to an upper probability bound of 4/81= 0.04938…, and to construct 3-sigma limits to bound nearly all (i.e. 99.73%) of the values of a process output. Without unimodality Chebyshev's inequality would give a looser bound of 1/9 = 0.11111….