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Let $A$ be a non-empty measurable subset of euclidean $\mathbb R^n$ and $x_0 \in \mathbb R^n\setminus A$. Define $d(x_0,A) := \inf_{a \in A} d(x_0,a)$, the distance of $x_0$ from $A$. Finally, let $\sigma > 0$.

Question 1. Are there any interesting relationships (functional equalities, inequalities, etc.) between $d(x_0,A)$ and $p(x_0,A) := \mathbb P_{x \sim \mathcal N(x_0,\sigma^2I)}(x \in A)$ ?

Notes

  • I'm fine with partial answers to Question 1 which make "reasonable" assumptions on $A$ (convex, closed, smoothness, curvature, etc.)
  • Ultimately, I'm interested in estimating $p(x_0,A)$. The sought-for connection with $d(x_0,A)$ is just a rough guess (inspired by the case of half-spaces worked-out below). It may turn out that other quantities are petinent for the estimation are in fact, not $d(x_0,A)$ in general.

Motivating example: half-spaces

Say $b \in \mathbb R$, $u \in \mathbb R^n$ with $\|u\| = 1$, and $A:=\{x \in \mathbb R^n \mid x^Tu \le b\}$, a half-space, and let $x_0 \in \mathbb R^n\setminus A$. Note that $d(x_0,A)=x_0^Tu-b$. Let $\Phi$ be the CDF of the standard 1D Gaussian distribution. Then one easily computes $$ \begin{split} p(x_0,A) &= \mathbb P_{x \sim \mathcal N(x_0,\sigma^2I)}(x \in A) = \mathbb P_{x \sim \mathcal N(x_0,\sigma^2I)}(x^Tu \le b)\\ &= \mathbb P_{z \sim \mathcal N(0,I)}(z^Tu \le (b-x_0^Tu)/\sigma)=\Phi(-d(x_0,A)/\sigma), \end{split} \tag{*} $$ or equivalently, $d(x_0,A) = -\sigma\Phi^{-1}(p(x_0,A))=\sigma\Phi^{-1}(1-p(x_0,A))$.

Edit: rough upper-bound on $p(x_0,A)$ via closed convex hulls

Suppose $A$ is closed convex nonempty subset of $\mathbb R^n$. Let $H_0$ be the half-space containing $A$, bordered by the hyper-plane with outward normal pointing in the direction of $x_0-a_0$, where $a_0$ is the (unique!) point of $A$ such that $d(x_0,A) = d(x_0,a_0)$. In other words,

$$ H_0 := \{x \in \mathbb R^n \mid (x-a_0)^Tu_0 \le 0\} = \{x \in \mathbb R^n \mid x^Tu_0 \le b_0\}, $$ where $u_0 := (x_0-a_0)/d(x_0,A)$, a unit vector in $\mathbb R^n$ and $b_0 := a_0^Tu_0 \in \mathbb R$. Note that $d(x_0,A) = d(x_0,H_0) = x_0^Tu_0-b_0$. Now, if $x \sim \mathcal N(x_0,\sigma^2 I)$, we have

$$ \mathbb P(x \in A) \le \mathbb P(x \in H_0) \overset{*}{=} \Phi(-d(x_0,A)/\sigma), $$

i.e $p(x_0,A) \le \Phi(-d(x_0,A)/\sigma)$, or equivalent, $d(x_0,A) \ge \sigma\Phi^{-1}(1-p(x_0,A))$.

The following theorem follows.

Lemma. Let $A$ be a nonempty measurable subset of $\mathbb R^n$, and let $\overline{co}(A)$ denote is the closure of it's convex hall. Let $x_0 \in \mathbb R^n\setminus \overline{co}(A)$. Then $$ d(x_0,A) \ge d(x_0,\overline{co}(A)) \ge \sigma\Phi^{-1}(1-p(x_0,\overline{co}(A))). $$

dohmatob
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    (1) $d(x_0, A)$ tells you that the open ball $B$ of that radius around $x_0$ does not intersect $A$. It tells you nothing else. $A$ could be a tiny blob of positive measure arbitrarily close to $0$, in which case $P(x \in A) \to 0$. Or $A$ could be the entire complement $\mathbb{R}^n - B$, in which case $P(x \in A) = 1 - P(x \in B)$ I don't know the formula for that but I presume this has been solved... In between these two trivial bounds, almost anything goes. (2) If you have the volume of $A$, then a better bound is possible, esp. since pdf of $x$ is radially symmetric. – antkam Jan 07 '20 at 16:56
  • @antkam (1) Thanks for the input. Indeed, it seems beyond the linear case, $d(x_0,A)$ isn't much information. (2) Concerning your second remark, I'm fine with bounds which depend on the volume of $A$, or any other geometric aspec of $A$. Are you thinking along the lines of a coarea-type inequalities like Eilenberg's inequality ? Would like to hear more about this. (3) You "don't know formula" of what ? – dohmatob Jan 07 '20 at 17:19
  • Actually, I take back my comment about volume. It might be that volume is infinite (e.g take the example of half-space). Could you kindly clarify your remark about "if you have the volume of $A$" ? Thanks in advance. N.B.: I understand that by "volume", you meant Lebesgue measure. – dohmatob Jan 07 '20 at 18:13
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    (2) I don't know any of the inequalities you named. :) I was just thinking, since PDF of $x$ is purely a function of distance (and not of direction) and is decreasing in distance, $P(x \in A)$ is maximized when the (finite) volume is spread as close to $x_0$ as possible subject to $d(x_0, A) =r$ constraint, i.e. in a concentric shell around $x_0$ with inner radius $r$ and outer radius $R>r$ s.t. the volume is $A$. (3) I dont know the formula for $P(x \in B)$. I'm sure its (easily?) derivable from PDF of $x$; I just don't know it off hand. – antkam Jan 07 '20 at 18:16
  • @antkam This is interesting. We are saying the same thing more or less. Actually, the symmetry of the density of $x$ can be used in the coarea formula to get (the rough bound): $P(x \in A) \le P(d(x,x_0) \ge d(x_0,A))=1-\gamma(n/2,d(x_0,A)^2/(2\sigma^2)) / \Gamma(n/2)$. *What's this ?* I've computed $P(d(x,x_0) \le r)$ in another unrelated post here https://math.stackexchange.com/a/3496441/168758. For the finite-volume case, your "concentric" construction you help me get tighter bounds (as, I'll integrate between a finite range of distances, instead of $[0, \infty]$). Thanks again. – dohmatob Jan 07 '20 at 18:47
  • More precisely, using your idea, a for the finite-volume case should be something like $P(x \in A) \le P(r \le d(x,x_0) \le R) = \frac{\gamma(n/2,R^2/(2\sigma^2))-\gamma(n/2,r^2/(2\sigma^2))}{\Gamma(n/2)}$, where $r:=d(x_0,A)$, and $R > d(x_0,A)$ is such that the $Annulus(x_0; r, R)$ has same volume as $A$. N.B $\gamma(\cdot,\cdot)$ is the lower-incomplete gamma function https://en.wikipedia.org/wiki/Incomplete_gamma_function. It's the unnormalized density of the generalized gamma distribution (aka Amoroso distribution). – dohmatob Jan 07 '20 at 18:58
  • $R = (d(x_0,A)^n + vol(A)/\omega_n)^{1/n}$ where $\omega_n$ is the volume of the unit sphere in $\mathbb R^n$. – dohmatob Jan 07 '20 at 19:06
  • I am not good at the actual integrals and stuff, so I'm glad you know how to expand on my ideas! :) The concentric shell maximizes $P(x \in A)$ for given $d$ and volume. To minimize $P(x\in A)$, a given volume can be made as "infinitely" long and thin and pointing away from $x_0$, so that doesnt provide a useful lower bound. However, if you also know e.g. the diameter of $A$, i.e. the max distance between $2$ points in $A$, or if you assume $A$ is a sphere, then again this becomes a puzzle in geometry / integrals. Good luck! – antkam Jan 07 '20 at 19:41

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