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The Vysochanskiï–Petunin inequality gives a tighter bound than Chebyshev for unimodal distributions . I'm just wondering if there is a one tailed version of it, like that of Chebyshev inequality? Please help.

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We are talking about continuous random variables with unimodal densities. The Wikipedia article you link to says

for any $ \lambda \gt \sqrt{8/3}=1.63299\ldots$, you have $\Pr(\left|X-\mu\right|\geq \lambda\sigma)\leq\dfrac{4}{9\lambda^2}.$

This, which I wrote many years ago and have not checked recently, says (trying to translate to a similar notation)

Unimodal two-tailed case:

  • If $\lambda \ge B $, then $\Pr(|X-\mu|\ge \lambda\sigma) \le \dfrac{4 }{\; 9 \lambda^2}$
  • If $\lambda \le B $, then $\Pr(|X-\mu|\ge \lambda\sigma) \le 1-\left(\dfrac{4 \lambda^2}{3(1+\lambda^2)}\right)^2$

where B is the largest root of $7x^6-14x^4-x^2+4=0$, about $1.38539\ldots...$

which seems similar but slightly stronger.

It also says

Unimodal one-tailed case:

  • $\Pr(X-\mu \ge \lambda\sigma) \le \max \left\{ \dfrac{4}{9(1+\lambda^2)}, \dfrac{3-\lambda^2}{3(1+\lambda^2)} \right\}$

so taking the first term if $\lambda \ge \sqrt{\frac{5}{3}}$ and the second if $0 \le \lambda \le \sqrt{\frac{5}{3}}$

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  • $\begingroup$ That's exactly what I'm looking for. Thanks so much. I'll go ahead and just use it for now and check the proof later. $\endgroup$ Sep 16, 2011 at 1:39
  • $\begingroup$ What is the definition of unimodality here? Strictly unimodal? Or simply put, it this unimodal according to your definition? $\endgroup$ Sep 16, 2011 at 17:37
  • $\begingroup$ Here unimodality means having cumulative distribution function which is convex up to a particular point and concave from that point onwards. Informally, it has a density function which weakly increases up to a point called the mode and then weakly decreases, so not a lopsided bimodal distribution as in your second link. The definition can include a uniform distribution (where any point in the support can be chosen as the mode); it can also include cases where the mode is a point of positive probability (informally, having infinite density). $\endgroup$
    – Henry
    Sep 16, 2011 at 19:19
  • $\begingroup$ Is there an efficient way to test unimodality of a population from samples only? The naive approach is to construct empirical CDF and then use your definition of unimodality, which is not very friendly for algorithmic implementation. For now, I just plot the histgram and check if there is only one peak. But depending on the granularity of the histgram, the conclusion of unimodality can be different even for the same sample. $\endgroup$ Sep 16, 2011 at 19:37
  • $\begingroup$ That's true. So is the question of the accuracy of the mean and standard deviation. This is a theorem for populations rather than samples. If you have a sample you can use order statistics to calculate confidence intervals or credible intervals for the probability of being in a given range. $\endgroup$
    – Henry
    Sep 16, 2011 at 20:05

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