Savanah Boone

Answered

2022-07-04

We can consider the sequence space ${\mathbb{R}}^{\mathbb{Z}}$, which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.

A classical result on measure theory in Polish spaces tells us that any probability measure $\mu $ on a Polish space $X$ is tight: given $\u03f5>0$, then there is a compact set ${K}_{\u03f5}\subseteq X$ such that $\mu ({K}_{\u03f5})>1-\u03f5$.

Hence any product probability measure on ${\mathbb{R}}^{\mathbb{Z}}$ is tight.

Say we fix an i.i.d. probability measure $\mu $ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mu (\prod {A}_{i})=\prod \nu ({A}_{i})$ for some probability measure $\nu $ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu $ is tight (independent of $\nu $)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu $ has unbounded support, then, for each $M>0$, the set $\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\}$ is infinite almost surely under $\mu $ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{\u03f5}$ intuitively.

A classical result on measure theory in Polish spaces tells us that any probability measure $\mu $ on a Polish space $X$ is tight: given $\u03f5>0$, then there is a compact set ${K}_{\u03f5}\subseteq X$ such that $\mu ({K}_{\u03f5})>1-\u03f5$.

Hence any product probability measure on ${\mathbb{R}}^{\mathbb{Z}}$ is tight.

Say we fix an i.i.d. probability measure $\mu $ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mu (\prod {A}_{i})=\prod \nu ({A}_{i})$ for some probability measure $\nu $ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu $ is tight (independent of $\nu $)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu $ has unbounded support, then, for each $M>0$, the set $\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\}$ is infinite almost surely under $\mu $ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{\u03f5}$ intuitively.

Answer & Explanation

bap1287dg

Expert

2022-07-05Added 13 answers

Here is one concrete example of such a set ${K}_{\u03f5}$. Let ${M}_{n}$ for $n\in \mathbb{N}$ be numbers such that $\nu ([-{M}_{n},{M}_{n}{]}^{c})<\u03f5{2}^{-n}$. Then, consider the set

${K}_{\u03f5}=\{({a}_{n}{)}_{n\in \mathbb{N}}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}n,\phantom{\rule{thinmathspace}{0ex}}|{a}_{n}|\le {M}_{n}\}\subset {\mathbb{R}}^{\mathbb{N}}.$

We have

$\mu ({K}_{\u03f5}^{c})\le \sum _{n=1}^{\mathrm{\infty}}\nu (\{|{a}_{n}|>{M}_{n}\})<\u03f5\sum _{n=1}^{\mathrm{\infty}}{2}^{-n}=\u03f5.$

Also, ${K}_{\u03f5}=\prod _{n=1}^{\mathrm{\infty}}[-{M}_{n},{M}_{n}]$ and so ${K}_{\u03f5}$ is the product of compact sets, and hence compact itself by Tychonoff's theorem.

${K}_{\u03f5}=\{({a}_{n}{)}_{n\in \mathbb{N}}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}n,\phantom{\rule{thinmathspace}{0ex}}|{a}_{n}|\le {M}_{n}\}\subset {\mathbb{R}}^{\mathbb{N}}.$

We have

$\mu ({K}_{\u03f5}^{c})\le \sum _{n=1}^{\mathrm{\infty}}\nu (\{|{a}_{n}|>{M}_{n}\})<\u03f5\sum _{n=1}^{\mathrm{\infty}}{2}^{-n}=\u03f5.$

Also, ${K}_{\u03f5}=\prod _{n=1}^{\mathrm{\infty}}[-{M}_{n},{M}_{n}]$ and so ${K}_{\u03f5}$ is the product of compact sets, and hence compact itself by Tychonoff's theorem.

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