# Tail $\sigma$-algebra

With a simple symmetric random walk such that $S_n=\sum\limits_{k=1}^n X_k$ and $\mathbb{P}[X_i=\pm1]=1/2$ with $S_0=0$ like in this post: Tail events and exchangeable events where Did answered some of my questions:

An alternative characterization of $\mathcal F_n$ is that $B\subseteq\Omega$ is in $\mathcal F_n$ if and only if property $P_n(B)$ holds:

$P_n(B):$ Let $\omega=(\omega_k)_{k\in\mathbb N}$ and $\omega'=(\omega'_k)_{k\in\mathbb >N}$ denote two elements of $\Omega$. If $\omega$ is in $B$ and if $\omega'_k=\omega_k$ for every $1\leqslant k\leqslant n$, then $\omega'$ is in $B$.

Edit: $\mathcal{F_n}=\sigma(X_{n+1},X_{n+2},\ldots)$.

1. The event $T=\{|S_n|\to\infty\}$ is a tail event and an exchangeable event. It is an exchangeable event because the sum $S_n$ is independent of finite permutations of the $X_k$ so it is in the the $\mathcal{E}$ "exchangeable" $\sigma$-algebra. Do I need to show this formally, it doesn't seem sufficient to state that the sets in $\{|S_n|\to\infty\}$ are independent of permutations of the first $n$ $X_k$ with $1\leq k\leq n$.

2. How would I say this in the notation in the other post? And how would I show that this is a tail event

3. The event $\{S_n=0 \text{ i.o.}\}= \limsup S_n = \bigcap\limits_{n\geq 1} \bigcup\limits_{k=n}^\infty \{S_n=0\}$ is also an exchangeable event for the same reason that the sum does depend on the order of the first few values of the $X_k$.

4. The event $R=\{S_n=0 \text{ i.o.}\}$ is not a tail event - take the set $\{1,0,1,0,1,0,\cdots\}$ which is an element in $R$. If it is a tail event then it is independent of the first $n$ of $X_k$, so change $X_1$ to $-1$ and then we have $\{-1,-2,-1,-2,-1,\cdots\}$ which is not in $R$ so it cannot be in $\mathcal{T}$ the tail sigma algebra. Is this enough to show that $R\notin \mathcal{T}$?

5. How do you show that events like $\bigcup\limits_{N\geq m} \bigcap\limits _{n\geq N} \left[ \omega: \left|n^{-1} \sum\limits_{k=m}^n\right|<\epsilon \right] \in \sigma (X_m,X_{m+1},\ldots)$? Just because the LHS depends on those $X_k$, with $k\geq m$ only?

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For which sigma-algebras $\mathcal F_n$? (Bis repetita.) – Did May 4 '13 at 8:27
These $\mathcal F_n$ are certainly not the sigma-algebras whose characterization you reproduce in the post. I really think your first task is to be clearer about which sigma-algebras $\mathcal F_n$ you are considering, either $\mathcal F_n=\sigma(X_n)$ or $\mathcal F_n=\sigma(X_k;k\leqslant n)$ or $\mathcal F_n=\sigma(S_n)$ or... The tail sigma-algebra depends heavily on the choice. – Did May 4 '13 at 8:50
1-4 is the $\mathcal{F}_n=\sigma(X_k; k\leq n)$, 5 was a general query but I thought that it still applied to the same tail sigma algebra. – shilov May 4 '13 at 10:29