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.
Is there one-tailed version of Vysochanskiï–Petunin inequality, like Chebyshev?
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probability-theory
inequality
1 Answers
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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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0It turns out the random variable in my case is not always unimodal. Nevertheless, the inequality seems to hold. Is there a sufficient rather than necessary (i.e., unimodality) condition, or relaxed condition for the inequality to hold? – 2011-10-07