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 restrictions on the distribution are that it be unimodal and have finite variance; here unimodal implies that it is a continuous probability distribution except at the mode, which may have a non-zero probability.
Theorem Let X {\displaystyle X} be a random variable with unimodal distribution and finite variance, and α ∈ R {\displaystyle \alpha \in \mathbb {R} } . If we define ρ = E [ ( X − α ) 2 ] {\displaystyle \rho ={\sqrt {\mathbb {E} [(X-\alpha )^{2}]}}} then for any r > 0 {\displaystyle r>0} ,
Pr ( | X − α | ≥ r ) ≤ { 4 ρ 2 9 r 2 r ≥ 8 / 3 ρ 4 ρ 2 3 r 2 − 1 3 r ≤ 8 / 3 ρ . {\displaystyle {\begin{aligned}\operatorname {Pr} (|X-\alpha |\geq r)\leq {\begin{cases}{\frac {4\rho ^{2}}{9r^{2}}}&r\geq {\sqrt {8/3}}\rho \\{\frac {4\rho ^{2}}{3r^{2}}}-{\frac {1}{3}}&r\leq {\sqrt {8/3}}\rho .\\\end{cases}}\end{aligned}}}
Relation to Gauss's inequality Taking α {\displaystyle \alpha } equal to a mode of X {\displaystyle X} yields the first case of Gauss's inequality.
Tightness of Bound Without loss of generality, assume α = 0 {\displaystyle \alpha =0} and ρ = 1 {\displaystyle \rho =1} .
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