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From Wikipedia about the conditions for the Vysochanskij–Petunin inequality

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.)

Does it mean that unimodality and finite variance imply the distribution is continuous except at the mode? Why is that?

Thanks!

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Consider $X \sim \operatorname{Binom}\left(2n, \frac{1}{2}\right)$. It is unimodal (with mode at $X=n$) and has finite variance. –  Sasha May 25 '12 at 5:33
    
Weakly related but with a different transliteration from Russian: math.stackexchange.com/questions/64945/… –  Henry May 25 '12 at 7:31

2 Answers 2

up vote 3 down vote accepted

The finiteness of the variance is irrelevant. As explained by Sellke and Sellke (see the reference on the WP page on unimodality), in this context, the unimodality of a CDF $F$ with mode $m$ means that $F$ is concave on $[m,+\infty)$ and convex on $(-\infty,m]$. This is an easy exercise to show that any CDF which is unimodal in this sense has no jumps except possibly at $m$.

Assuming for example that $F(x-)\lt F(x)$ at some $x\gt m$, one sees that for small values of $\varepsilon\gt0$, the segment between the points on the graph of $F$ with abscissae $x-\varepsilon$ and $x$ is at least partly above the graph. This contradicts the concavity hence $F(x-)=F(x)$ at every $x\gt m$.

In particular, discrete distributions $(p_k)$ on the integers which are unimodal in the sense that $k\mapsto p_k$ is increasing then decreasing, are not unimodal in this convexity/concavity sense.

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Your interpretation is correct.

If you regard some distributions on the integers as unimodal in the natural sense (such as the binomial and Poisson distributions) then the distribution with $$\Pr(X=0)=\frac{1}{2k^2}$$ $$\Pr(X=1)=1-\frac{1}{k^2}$$ $$\Pr(X=2)=\frac{1}{2k^2}$$ has a mean of $1$, a variance of $\frac{1}{k^2}$, and is unimodal when $k \gt \sqrt{\frac32}$. But it is bound by the Chebyshev inequality not the Vysochanskij–Petunin inequality.

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