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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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    Wea$k$ly related but wit$h$ a different transliteration from Russian: $h$ttp://math.stackexchange.com/quest$i$ons/64945/is-there-one-tailed-version-of-vysochanskiipetunin-inequality-like-chebyshev2012-05-25

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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.