Assume we have a Borel probability measure $\mu$ in $\mathbb{R}^n$ and a sequence of $\mu$ distributed I.I.D. random variables $x_n$. Is there a limit formula for $supp(\mu)$, something like closure of limit points of $x_n$ or similar, which allows kind of asymptotic recovering of $supp(\mu)$ from observed data.
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2 Answers
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By f.i. https://encyclopediaofmath.org/wiki/Support_of_a_measure $supp(\mu)$ is the smallest closed set $C \subset \mathbb{R}^n$ such that $\mu(C^c) = 0$, $C^c$ the complement of $C$. Let $y := (y_n)_{n \in \mathbb{N}}$ be any sample of $(x_n)$ (independent realizations of $\mu$) and $A := \{y_n \colon n \in \mathbb{N}\}$, $A_N := \{y_n \colon n \leq N\}$ for $N \in \mathbb{N}$. Then by Borel-Cantelli with probability $1$ $A \subset C$, $A_N \uparrow A$ and $A$ is dense in $C$, in particular $\bar A = C$. Thus each sequence of iid random variables recovers the support (a.s.).
answered Jul 1, 2023 at 20:09
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1$\begingroup$ If I interpret your answer correctly, you essentially say that $\mu$-almost every realization will be a dense set inside the $supp(\mu)$ ? $\endgroup$ Commented Jul 1, 2023 at 21:29
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$\begingroup$ thanks, I think that is what I wanted! $\endgroup$ Commented Jul 1, 2023 at 21:34
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$\begingroup$ After some thinking I realized that even slightly stronger statement is true: any probability measure in $R^n$ is a weak limit of its averaged sum of Dirak measures on "iid realization" almost surely... $\endgroup$ Commented Jul 3, 2023 at 1:15
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that's basically the philosophy of Monte Carlo. So here is one suggestion: for $x_{n}$ iid uniform variables in the $supp(\mu)$ and generic integrable function $f$ we have by strong law of large numbers
$$\frac{1}{N}\sum_{i=1}^{N}f(x_{i})\to \int_{supp(\mu)}f(x)dx.$$
So by taking a localized $f$ eg. $f(x)=1_{B_{r}(x_{0})}(x)$ one can inspect the support of $\mu$.
answered Jul 1, 2023 at 19:19