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If I have i.i.d. r.v.'s $X_1, X_2,...$ then the tail sigma algebra is defined as $\mathcal{T}:=\cap_n\sigma_n$ where $\sigma_n:=\sigma(X_n,X_{n+1},...)$. From this we get very nice results such as the Kolmogorov 0-1 law. I was wondering if it makes sense to consider limsup and liminfs in this fashion: $\bigcup_n\bigcap_{k\ge n}\sigma_k$ and $\bigcap_n\bigcup_{k\ge n}\sigma_k$. Do these have a sensible meaning and are there similar laws such as the 0-1 law pertaining to each?

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1 Answer 1

Let us define $\mathcal{T}_n=\bigcap_{k=n}^\infty\sigma_k$, so that the tail $\sigma$-algebra is just $\mathcal{T}=\mathcal{T}_1$.

Now, the first thing to notice is that $\sigma_{n+1}\subset\sigma_n$ for all $n$. This means that $\mathcal{T}_n=\bigcap_{k=n}^\infty\sigma_k$ does not actually depend on $n$. So $\mathcal{T}_n$=$\mathcal{T}$ for all $n$. Therefore, $$ \bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\sigma_k = \bigcup_{n=1}^\infty \mathcal{T} = \mathcal{T}. $$

On the other hand, $\sigma_{n+1}\subset\sigma_n$ implies that $\bigcup_{k=n}^\infty \sigma_k=\sigma_n$. Therefore, $$ \bigcap_{n=1}^\infty\bigcup_{k=n}^\infty\sigma_k = \bigcap_{n=1}^\infty \sigma_n = \mathcal{T}. $$

So both of these suggested generalizations are, in fact, just the tail $\sigma$-algebra, $\mathcal{T}$.

Edit:

Here is some additional information. As you may know, the intersection of $\sigma$-algebras is a $\sigma$-algebra, but the union of $\sigma$-algebras is not, in general. It is therefore common to adopt the following notation. If $\{\mathcal{F}_i\}_{i\in I}$ is a collection of $\sigma$-algebras, then $$ \bigvee_{i\in I}\mathcal{F}_i = \sigma\bigg(\bigcup_{i\in I}\mathcal{F}_i\bigg). $$ With this notation, we may write $$ \mathcal{T} = \bigcap_{n=1}^\infty\bigvee_{k=n}^\infty\sigma(X_k). $$ So in this sense, the tail $\sigma$-algebra is already a limsup. We might then wonder about the $\sigma$-algebra $$ \mathcal{S} = \bigvee_{n=1}^\infty\bigcap_{k=n}^\infty\sigma(X_k). $$ It can be shown that $\mathcal{S}\subset\mathcal{T}$, so that Kolmogorov's $0$-$1$ law is still valid on $\mathcal{S}$. To see this, suppose $A\in\bigcup_n\bigcap_{k\ge n}\sigma(X_k)$. Then there exists $n$ such that $A\in\sigma(X_k)\subset\sigma_k$ for all $k\ge n$. This implies $A\in\mathcal{T}$ and shows that $\bigcup_n\bigcap_{k\ge n}\sigma(X_k)\subset\mathcal{T}$. Since $\mathcal{T}$ is a $\sigma$-algebra, it follows that $$ \mathcal{S} = \sigma\bigg(\bigcup_n\bigcap_{k\ge n}\sigma(X_k)\bigg) \subset\mathcal{T}. $$ The reverse inclusion does not necessarily hold. For an example, let $P$ be any Borel probability measure on $\mathbb{R}$ such that $P(\mathbb{Q})=0$. For $n\in\mathbb{N}$, let $X_{2n}=0$ and $X_{2n+1}=1_{\mathbb{Q}}$. Then $X_j=0$ a.s. for every $j$, so this is indeed an iid sequence. (In general, if $U$ is constant a.s. and $V$ is any random variable, then $U$ and $V$ are independent.)

In this example, $\sigma(X_{2n})=\{\emptyset,\mathbb{R}\}$ and $\sigma(X_{2n+1})=\{\emptyset,\mathbb{Q},\mathbb{Q}^c,\mathbb{R}\}$. Hence, $\mathcal{S}=\{\emptyset,\mathbb{R}\}$ and $\mathcal{T}=\{\emptyset,\mathbb{Q},\mathbb{Q}^c,\mathbb{R}\}$.

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