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Consider the measurable space $(\Omega,\mathscr{B})$ endowed with two positive measures: a "volume $\nu$" and a probability measure $\mu$. For example, one might take $\Omega=\mathbb{R}^n$ (with the usual $\sigma$-algebra) and $\nu$ as the Lebesgue measure.

I would like to quantify the notion of "a high proportion of $\mu$'s mass is concentrated on a region of small volume". To that end, let us define the function $F:[0,1]\to[0,\infty)$ by $$ F_{\mu/\nu}(x) = \inf\{\nu(A): A\in\mathscr{B},\mu(A)\ge x \}.$$

Question: Is this (or perhaps a closely related) notion known?

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  • $\begingroup$ Don’t you want sup instead of inf? Otherwise it seems that F has constant value 0. $\endgroup$
    – Ramiro de la Vega
    Commented Jul 21, 2019 at 2:45
  • $\begingroup$ @RamirodelaVega I think I do want the inf, but I did have a typo: the inf should be over all $A$ s.t. $\mu(A)\ge x$ (I previously hade $\le x$). $\endgroup$
    – Aryeh Kontorovich
    Commented Jul 21, 2019 at 5:22
  • $\begingroup$ As a sanity check: for $\mu=\nu=$ the Lebesgue measure on $[0,1]$, we have $F(x)=x$. $\endgroup$
    – Aryeh Kontorovich
    Commented Jul 21, 2019 at 5:23

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The function you propose is related to the L'evy concentration function, studied by Kolmogorov, Rogozin, Esseen and others. See the special volume [1] https://link.springer.com/chapter/10.1007/978-94-011-2260-3_70

The classic book [2] has a chapter devoted to concentration functions with many references and the paper [3] has a quite sharp estimate; connection to combinatorics are in [4]. Also related is the study of small-ball probabilities, see the survey [5].

Returning to the original question, Decompose $\mu=\mu_a+\mu_s$ where $\mu_a$ is absolutely continuous to $\nu$ with Radon-Nikodym derivative $f$, and $\mu_s$ is singular to $\nu$. Write $M_s$ for the total mass of $\mu_s$. If $M_s \ge x$ then $F(x)=0$. Otherwise consider the sets $A_c:=\{f>c\}$. If there is such a set with $\mu(A_c)=x-M_s$ then $F(x)=\nu(A_c)$. If there is no such $c$, find the infimmum $c_*$ of the constants $c$ such that $\mu(A_c)<x-M_s$. one needs to do some tie-breaking inside the level set where $f=c_*$ and take a subset there of suitable $\mu$ measure $x-M_s-\mu(A_{c_*})$. Thus $F(x)=\nu(A_{c_*})+(x-\nu(A_{c_*}))/c_*$.

[1] Kruglov, V. M. "Concentration Functions (No. 45)." Selected Works of AN Kolmogorov. Springer, Dordrecht, 1992. 571-574.

[2] Petrov, Valentin Vladimirovich. Sums of independent random variables. Vol. 82. Springer Science & Business Media, 2012.

[3] Kesten, Harry. "A sharper form of the Doeblin–Levy–Kolmogorov–Rogozin inequality for concentration functions." Mathematica scandinavica 25.1 (1970): 133-144.

[4] Halász, Gábor. "Estimates for the concentration function of combinatorial number theory and probability." Periodica Mathematica Hungarica 8.3-4 (1977): 197-211.

[5] Li, Wenbo V., and Q-M. Shao. "Gaussian processes: inequalities, small ball probabilities and applications." Handbook of Statistics 19 (2001): 533-597.

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  • $\begingroup$ Thanks, Yuval. I recognized the similarity to concentration functions, of course, but I recall seeing it defined differently -- $L(r)$ is the the sup of the complement of the measure of the $r$-blow-up over all sets of measure at least $1/2$. In other words, the notion that I'm familiar with uses a metric rather than a reference measure $\nu$. $\endgroup$
    – Aryeh Kontorovich
    Commented Jul 23, 2019 at 18:49
  • $\begingroup$ I also realize that my definition is not as useful as I originally thought. Again, take $\nu$ to be the Lebesgue measure over $\mathbb{R}^n$ and let $\mu$ to be the mixture of very highly peaked Gaussians over a very large, widely spaced, finite grid. According to my notion, $\mu$ would be highly concentrated -- but for many interesting applications, we would actually say that $\mu$ is quite dispersed. $\endgroup$
    – Aryeh Kontorovich
    Commented Jul 23, 2019 at 18:52
  • $\begingroup$ Yuval, what if I take the $\inf$ over convex sets only? (Restricing to $\mathbb{R}^n$.) I think that would be a more useful definition -- have you seen it anywhere? $\endgroup$
    – Aryeh Kontorovich
    Commented Jul 23, 2019 at 18:53

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