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Assume we are given a probability space $(\mathbb{X}, \mathcal{X}, \mathbb Q)$ and a measurable distance function defined on it $d:\mathbb{X}\times \mathbb{X}\to \mathbb{R}^+\cup\{0\}$ that conforms to the usual definition of distance on metric spaces. I am trying to understand (in lay terms) the "expected value of the measure of a ball with fixed radius $r> 0$ centered at a randomly chosen point in $\mathbb{X}$ with measure $\mathbb{Q}$". That is more formally \begin{equation} \int_{\mathbb{X} }q(x)\mathbb{Q}(\mathrm{d}x) \end{equation} where $q(x) = \mathbb{Q}(B(x,r))$ and $B(x,r) = \{y\in \mathbb{X}: d(x,y)\leq r\} $. The first thing to ask here is if $q(x)$ is measurable but for now I assume $\mathcal{X}$ is large enough that it is.

My problem originates from this question: "find a lower bound for the probability that a randomly chosen iid set of points in a space are close-by". I am wondering what assumptions are needed from $\mathbb{Q}$ and $d$ such that this can be lower bounded. Of course I would not expect a bound valid for all $\mathbb{Q}$ and $d$ but any tip about how to treat this integral (or you know if anybody thought about this before and gave it a name so I can search) would be helpful. The trivial bound $\inf_{x\in\mathbb X} \mathbb{Q}(B(x,r))$ is not useful for my purposes.

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Let $S:=\mathbb X$. Let $X$ and $Y$ be iid random elements of $S$ each with distribution $\mathbb Q$. It is more transparent to rephrase the question as follows:

Can one give a good lower bound on $P(d(X,Y)\le r)$?

Let $(A_i)$ be any countable measurable partition of $S$ with each $A_i$ of diameter $\le r$. Then clearly $$P(d(X,Y)\le r)\ge\sum_i P(X\in A_i,Y\in A_i)=\sum_i P(X\in A_i)^2,$$ so that $$P(d(X,Y)\le r)\ge L(r):=\sup_{(A_i)}\sum_i P(X\in A_i)^2,$$ where the $\sup$ is taken over all countable measurable partitions $(A_i)$ of $S$ with each $A_i$ of diameter $\le r$. The lower bound $L(r)$ on $P(d(X,Y)\le r)$ is clearly attained when $S$ is an at most countable metric space with $d(x,y)>r$ for all distinct $x,y$ in $S$. One can hardly have anything better in general.

If $S=\mathbb R$ with the usual distance for $d$ and if $X$ has (say) a bounded continuous pdf $p$, then $d(X,Y)$ has the pdf $q$ given by $$q(r)=2\,1(r>0)\int_{\mathbb R}p(y)p(y+r)\,dy,$$ whence $$P(d(X,Y)\le r)\sim2r\int_{\mathbb R}p(y)^2\,dy$$ as $r\downarrow0$.

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  • $\begingroup$ Thank you very much, this was helpful. I actually realize now that without any assumptions, this probability can be made arbitrarily small by choosing the support of $\mathbb{Q}$ as the balls of radius $o(r)$, center of each separated by $2r$. $\endgroup$
    – eonaran
    Jan 5, 2021 at 10:28

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