Is there a one-tailed version of the Vysochanskiï-Petunin inequality?


Thanks for you effort modulo_function, but I already know what the Vysochanskiï-Petunin inequality is. I am asking whether there is a one-tailed version of the ineqaulity i.e. whether the is a inequality that bounds the variation from the mean in one direction only.

Update 2:

Forgive the typos in the addition above!

1 Answer

  • 1 decade ago
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    When I first saw this question I thought that it was probably a joke. But, I searched and found out some real info:

    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. The sole restriction on the random variable is that the distribution be unimodal. (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 limit is attained (that is, the probability is equal to 4/(9 λ2)) 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 unsymmetric distributions for which the 4/(9 λ2) limit is exceeded.

    The theorem refines Chebyshëv's inequality by imposing 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. 95%) of the values of a process output. Without unimodality Chebyshev's inequality would give a looser bound of 1/9 = 0.11111...

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