Exchangeable strictly vs. almost surely

Question

Let $X$ be a standard Borel measurable space (for example, $\Bbb{R}$). Let $p$ be an exchangeable probability measure on the countably infinite product $X^\Bbb{N}$. Call a measurable set $A\subseteq X^\Bbb{N}$

• Strictly exchangeable if for any finite permutation $\sigma:X^\Bbb{N}\to X^\Bbb{N}$, $\sigma^{-1}(A)=A$;
• Almost surely exchangeable if for any finite permutation $\sigma:X^\Bbb{N}\to X^\Bbb{N}$, $p(\sigma^{-1}(A)\,\Delta\,A)=0$, where $\Delta$ denotes the symmetric difference.

Is it true that if $A$ is almost surely exchangeable, then there exists a strictly exchangeable $B$ such that $p(A\,\Delta\,B)=0$?

(I think I've found a proof, but it seems very specific to permutations, and I would like to see other proofs that might generalize to arbitrary actions.)

A reference would also be welcome.

Answer 1

The key observation is that strict exchangeability means $1_A(\sigma(X)) = 1_A(X)$ for all $X$ while almost sure exchangeability means the equality only holds almost surely.

Suppose $A$ is almost surely exchangeable. Let $f_n(X) = \frac{1}{n!}\sum_{\sigma \in S_n}1_A(\sigma(X))$. Then $f_n$ is strictly exchangeable over $S_n$ and $f_n = 1_A$ almost surely. Let $f(X) = \limsup_n f_n(X)$. Then $f(X)$ is strictly exchangeable and $f(X) = 1_A(X)$ almost surely. Let $B = \{f(X) = 1\}$. Then $1_B(X) = I(f(X) = 1)$, from which it is clear that $1_B(X) = 1_A(X)$ almost surely and that $1_B$ is strictly exchangeable.

Answer 2

Assumption 1. Let $p$ be invariant under the measurable transformation/action/dynamical system $\sigma$, i.e., $p(\sigma^{-1}(A))=p(A)$ for all $A\in\mathcal{B}(\mathcal{X})$.

Assumption 2. Let $\sigma$ be invertible with $\sigma^{-1}$ measurable.

Remark 1. Assumptions $1$ and $2$ imply $p(\sigma(A))=p(A)$ for all $A\in\mathcal{B}(\mathcal{X})$. Now,

1. $A$ is invariant (both forward and backwards) w.r.t. $\sigma$ if $\sigma^{-1}(A)=A$.
2. $A$ is almost surely invariant w.r.t. the tuple $(p,\sigma)$ if $p(A\bigtriangleup \sigma^{-1}(A))=0$.

Theorem 1. If $A$ is almost surely invariant, then $B\overset{\Delta}=\bigcup\limits_{n=-\infty}^{\infty} \sigma^{-n}(A)$ is invariant, i.e., $\sigma^{-1}(B)=B$ and $p(A\bigtriangleup B)=0$.

Proof. Indeed, $$p\left(A\bigtriangleup B\right)\leq\underbrace{p(A\setminus B)}_{=0}+p(B\setminus A)= p\left(\bigcup_n \sigma^{-n}(A)\setminus A\right)\leq \sum_n \underbrace{p\left(\sigma^{-n}(A)\setminus A\right)}_{=0}=0,$$ where $p(A\setminus B)=0$ since $A\subset B$ and $p\left(\sigma^{-n}(A)\setminus A\right)=0$ for every $n\in \mathbb{Z}$ from Lemma 1. $\square$

Remark 2. Your setting is a particular case of the above: $p$ is exchangeable when it is invariant under permutations. Each permutation is measurable and invertible with measurable inverse (hence, Assumptions 1 and 2 hold in your case).

--------------------- Auxiliary result -----------------------------

Lemma 1 We have that $p(\sigma^{-n}(A)\setminus A)=0$ for all $n\in\mathbb{Z}$.

Proof. Remark that this is true for $n=1$ as $A$ is almost surely invariant, i.e., $p(\sigma^{-1}(A)\setminus A)\leq p(A\bigtriangleup \sigma^{-1}(A))=0$. It is also true for $n=-1$. To see this $$p(\sigma(A)\setminus A)\leq p(\sigma(A)\bigtriangleup A)=p(\sigma\left(A\bigtriangleup \sigma^{-1}(A)\right))=p(A\bigtriangleup \sigma^{-1}(A))=0,$$ where the second to last identity holds in view of Remark 1.

Via induction, we can prove the result. Assume that $p(\sigma^{-N}(A)\setminus A)=0$. We have $$p(\sigma^{-(N+1)}(A)\setminus A)\leq p\left(\left(\sigma^{-(N+1)}(A)\setminus \sigma^{-N}(A)\right)\cup \left(\sigma^{-N}(A)\setminus A\right)\right)\leq p\left(\sigma^{-(N+1)}(A)\setminus \sigma^{-N}(A)\right)+ \underbrace{p\left(\sigma^{-N}(A)\setminus A\right)}_{=0}=p\left(\sigma^{-(N+1)}(A)\setminus \sigma^{-N}(A)\right).$$ where the first inequality holds from the fact that $D\setminus F \subset D \setminus E \cup E\setminus F$. Observe that $p\left(\sigma^{-(N+1)}(A)\setminus \sigma^{-N}(A)\right)=p\left(\sigma^{-N}\left(\sigma^{-1}(A)\setminus A\right)\right)=p\left(\sigma^{-1}(A)\setminus A\right)=0$, since $p$ is invariant under $\sigma$, and the result follows. $\square$