Let $(\Omega, \mathbb{F}, P)$ be a probability space and $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent and identically distributed random variables. How can one prove that every tail event $A$; i.e., $A \in \chi = \bigcap_{n} \mathbb{F}^{\infty}_{n}$, such that $\mathbb{F}^{\infty}_{n} = \sigma(X_n, X_{n+1}, \cdots)$, $\forall n$, is also symmetric?
Recall that an event $A$ is said to be symmetric if $\pi(A) = A$, where $\pi(\cdot)$ is an one-to-one mapping such that $\pi(n) = n$ for every $n$ with a finite number or exceptions.
Best Answer
Intuitively, $A \in \chi = \bigcap_{n} \mathbb{F}^{\infty}_{n}$ means $A$ is an event in the limiting random variable, i.e. $\lim_{n\rightarrow\infty}X_n$. Hence it is not affected by finite permutations, and is an exchangeable/symmetric event.
A precise proof is mostly about definition understanding and manipulation. Let $\mathcal{R}$ be Borel sets of real number $R$. $(R, \mathcal{R})$ is a measurable space.
An event $A \in \mathbb{F}^{\infty}_{n}$, by definition, is equivalent to $A\equiv\{\omega \in \Omega: X_i(w) \in B_i, B_i \in \mathcal{R} \text{ for }i \ge n\}$. Note the definition of $A$ has no requirements on $X_i(\omega)$ with $1 \le i < n$, and they can be any real values. We can equivalent write $B_i=R$ for $1 \le i < n$ in this case.
Furthermore, $\pi A \equiv \{\omega \in \Omega: X_{\pi(i)}(w) \in B_i \text{ for }i \ge 1\}. $ $B_i$s here are from the definition of the event $A$.
For any finite permutation $\pi$, there exists a $N$ so that for all $n \ge N$, $\pi_(n)=n$. Since $A \in \chi$, $A \in \mathbb{F}^{\infty}_{N}$. We have $$\pi A = \{\omega \in \Omega: X_{\pi(i)}(w) \in B_i \text{ for }i \ge 1\} \\ = \{\omega \in \Omega: X_{\pi(i)}(w) \in B_i\text{ for }1 \le i < N, X_{\pi(i)}(\omega) \in B_i \text{ for }i \ge N\} \\= \{\omega \in \Omega: X_{\pi(i)}(w) \in R\text{ for }1 \le i < N, X_i(\omega) \in B_i \text{ for }i \ge N\} \\= \{\omega \in \Omega: X_i(\omega) \in B_i \text{ for }i \ge N\} = A.$$