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