# Speaking conceptually about the tail $\sigma$-algebra

I have trouble giving negative answers when asked whether some event belongs to the tail $\sigma$-algebra or not. Solutions often argue conceptually, that is answer something along the lines: "This event depends on the first outcomes, so it is not part of the tail sigma algebra". Can we make this precise?

Let $(X_n)$ be a sequence of real random variables. The tail $\sigma$-algebra is $\mathcal{T}=\bigcap_{k=1}^\infty \sigma(X_k,X_{k+1},\dots)$. I'd like some Lemma that goes something like this: If $(\sigma(X_n))_{n\in\mathbb{N}}$ suffices some necessary conditions (independency, non-inclusion,... ) and there are outcomes $\omega_1,\omega_2$ such that $\omega_1\in A$, $\omega_2\notin A$, and $(X_n(\omega_2))_{n\in\mathbb{N}}=(X_n(\omega))_{n\geq k}$ for some $k\geq 2$, then $A\notin\mathcal{T}$. Can someone give a such statement and proof?

Example: If $(X_n)$ i.i.d. and $P([X_1=1])=p$, $P([X_1=-1])=1-p$, $S_n=\sum_{k=1}^n X_k$, determine whether $\limsup_n[S_n=0]\in\mathcal{T}$. We could use $(X_n(\omega_1))=(1,-1,-1,1,-1,1,-1,\dots)$ and $k=2$ in the eventual lemma above to conclude that $S_n\notin\mathcal{T}$. (Bonus question: How can we argue existence of suitable $\omega_1,\omega_2$ in this case?)

• I believe $\mathcal{A}$ above should be $\mathcal{T}$.
– user81375
Commented Jul 8, 2018 at 17:44
• @fourierwho Yes, thank you. Commented Jul 8, 2018 at 17:57

You actually already have the statement you're looking for; there are no extra hypotheses needed.

Theorem: Suppose that there exist outcomes $\omega_1$ and $\omega_2$ such that $\omega_1\in A$, $\omega_2\not\in A$, and there exists $k\in\mathbb{N}$ such that $X_n(\omega_1)=X_n(\omega_2)$ for all $n\geq k$. Then $A\not\in\mathcal{T}$.

Proof: We will prove more specifically that $A\not\in\Sigma$ where $\Sigma=\sigma(X_k,X_{k+1},\dots)$. To prove this, let $\Sigma'$ be the collection of sets $B$ which do not have this property of $A$: for any $\omega_1,\omega_2$ such that $X_n(\omega_1)=X_n(\omega_2)$ for all $n\geq k$, either $\omega_1,\omega_2\in B$ or $\omega_1,\omega_2\not\in B$. Observe that $\Sigma'$ is a $\sigma$-algebra. Indeed, it is trivial to check that for each pair $\omega_1,\omega_2$, the property above is preserved by any Boolean operations. Moreover, for any $S\subseteq\mathbb{R}$ and any $n\geq k$, $[X_n\in S]\in \Sigma'$. But $\Sigma$ is by definition the smallest $\sigma$-algebra containing all such sets $[X_n\in S]$ where $S$ can be any Borel subset of $\mathbb{R}$. Since $\Sigma'$ is another such $\sigma$-algebra, we have $\Sigma\subseteq \Sigma'$. Since $A\not\in\Sigma'$ by hypothesis, $A\not\in\Sigma$, as desired.

One might add to Eric Wofsey's answer that it's sometimes possible to argue using symmetries of the underlying space.

Consider the case when the underlying probability space is $(\mathbb{R}^{\mathbb{N}},\mathscr{B}^{\otimes \mathbb{N}},\mathbb{P})$ and $\mathbb{P}$ is arbitrary. In this case, I identify the sequence $\{X_{j}\}_{j \in \mathbb{N}}$ with the coordinate maps via $X_{j}(x_{1},x_{2},\dots) = x_{j}$.

The space $\mathbb{R}^{\oplus \mathbb{N}} = \{x \in \mathbb{R}^{\mathbb{N}} \, \mid \, x_{i} = 0 \, \, \text{for all but finitely many} \, \, i\}$ acts measurably on $\mathbb{R}^{\mathbb{N}}$ via \begin{equation*} x \cdot y = (x_{1} + y_{1},x_{2} + y_{2},\cdots) \quad x \in \mathbb{R}^{\oplus \mathbb{N}}, y \in \mathbb{R}^{\mathbb{N}}. \end{equation*} I'll write $R_{x} : \mathbb{R}^{\mathbb{N}} \to \mathbb{R}^{\mathbb{N}}$ for the map $R_{x}(y) = x \cdot y$.

The following claim identifies elements in the tail $\sigma$-algebra:

In the present set-up, $A \in \mathcal{T}$ if and only if $R_{x}^{-1}(A) = A$ independently of $x \in \mathbb{R}^{\oplus \mathbb{N}}$.

One direction (the "only if") follows from (the contrapositive of) Eric Wofsey's answer. To see the "if" direction, observe that $R_{x}^{-1}(A) = A$ (independently of $x$) implies that $(x_{1},x_{2},\dots) \mapsto 1_{A}(x_{1},x_{2},\dots)$ does not depend on $(x_{1},x_{2},\dots,x_{N})$ (irrespective of $N$). In particular, $1_{A}(x_{1},x_{2},\dots) = 1_{A}(0,\dots,0,x_{N},x_{N+1},\dots)$.

Observe that the map $G_{N} : (\mathbb{R}^{\mathbb{N} \setminus \{1,2,\dots,N -1\}}, \mathscr{B}^{\otimes \mathbb{N} \setminus \{1,2,\dots,N - 1\}}) \to (\mathbb{R}^{\mathbb{N}}, \mathscr{B}^{\otimes \mathbb{N}})$ given by $G(x_{N},x_{N+1},\dots) = (0,\dots,0,x_{N},x_{N + 1},\dots)$ is measurable. Letting $\pi_{N}(x_{1},x_{2},\dots) = (x_{N},x_{N + 1},\dots)$, we see that $1_{A} = 1_{A} \circ G \circ \pi_{N}$ and, thus, we conclude that $A$ is, in fact, $\sigma(X_{N},X_{N + 1},\dots)$-measurable.

Since $N$ was arbitrary, $A \in \mathcal{T}$.

The above also works if we are working in $(E^{\mathbb{N}},\mathcal{F}^{\otimes \mathbb{N}},\mathbb{P})$, where $(E,\mathcal{F})$ is a measurable space satisfying: for each $x,y \in E$, there is a measurable transformation $T_{x,y} : (E,\mathcal{F}) \to (E,\mathcal{F})$ such that $T_{x,y}(x) = y$. One then argues by considering maps obtained as finite products of the transformations $\{T_{x,y}\}_{(x,y) \in E^{2}}$.

I guess this falls into the category of what is "obvious" to people who actually work in probability, but I imagine it could be helpful in the study of percolation, spin systems, and the like, where one actually works in the set-up of $(G^{\mathbb{N}},\mathscr{B}_{G}^{\otimes \mathbb{N}})$ or $(G^{\mathbb{Z}^{d}},\mathscr{B}_{G}^{\otimes \mathbb{Z}^{d}})$ for some topological group $G$ and its Borel $\sigma$-algebra $\mathscr{B}_{G}$.

• This looks very interesting, I’ll need some time to go through that... Thank you! Commented Jul 9, 2018 at 23:00