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Gauss's inequality is for unimodal distributions, concerning distance from the mode.

A similar result is Vysochanskiï–Petunin inequality, which is for the distance from the mean rather than the mode. Chebyshev's inequality generalizes Vysochanskiï–Petunin inequality by concerning distance from the mean without requiring unimodality.

I wonder if there is generalization of Gauss's inequality for distributions not necessarily unimodal?

Thanks!

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Chebyshev's inequality is crude: The probability of being more than $k$ standard deviations away from the mean is at most $1/k^2$. A similar coarse estimate applies to the distance from the mode or any other point. Let $r_x^2$ be the average of the square of the distance from $x$, which can be computed as $(x-\mu)^2 + \sigma^2$. Then the probability of being more than $kr_x$ away from $x$ is at most $1/k^2$. This is true for all $x$, including a mode. The same works if we let $x$ denote a set, including a set of modes, although computing $r_x^2$ becomes harder.

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Thanks! How is $r_x^2$ defined when x is a set? –  Tim May 25 '12 at 1:01
    
The distance between the point $y$ and the set $X$ is $d(y,X) = \inf_{x \in X} d(y,x)$. Then $r_X^2$ can be defined as the average of the square of the distance to $X$. This is no longer a quadratic polynomial so it can't be computed as easily as $(x-\mu)^2 + \sigma^2$, but the inequality still holds. –  Douglas Zare May 25 '12 at 1:30
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