Savanah Boone

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 $ϵ>0$, then there is a compact set ${K}_{ϵ}\subseteq X$ such that $\mu \left({K}_{ϵ}\right)>1-ϵ$.
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, $\mu \left(\prod {A}_{i}\right)=\prod \nu \left({A}_{i}\right)$ 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 $\left\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\right\}$ is infinite almost surely under $\mu$ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{ϵ}$ intuitively.

bap1287dg

Expert

Here is one concrete example of such a set ${K}_{ϵ}$. Let ${M}_{n}$ for $n\in \mathbb{N}$ be numbers such that $\nu \left(\left[-{M}_{n},{M}_{n}{\right]}^{c}\right)<ϵ{2}^{-n}$. Then, consider the set
${K}_{ϵ}=\left\{\left({a}_{n}{\right)}_{n\in \mathbb{N}}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall }n,\phantom{\rule{thinmathspace}{0ex}}|{a}_{n}|\le {M}_{n}\right\}\subset {\mathbb{R}}^{\mathbb{N}}.$
We have
$\mu \left({K}_{ϵ}^{c}\right)\le \sum _{n=1}^{\mathrm{\infty }}\nu \left(\left\{|{a}_{n}|>{M}_{n}\right\}\right)<ϵ\sum _{n=1}^{\mathrm{\infty }}{2}^{-n}=ϵ.$
Also, ${K}_{ϵ}=\prod _{n=1}^{\mathrm{\infty }}\left[-{M}_{n},{M}_{n}\right]$ and so ${K}_{ϵ}$ is the product of compact sets, and hence compact itself by Tychonoff's theorem.

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