0

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\tau$ be a measurable map on $(\Omega,\mathcal A,\operatorname P)$ with $\operatorname P\circ\:\tau^{-1}=\operatorname P$, $X:\Omega\to\mathbb R$ be $\mathcal A$-measurable and $$X_n:=X\circ\tau^n\;\;\;\text{for }n\in\mathbb N.$$

Assume $(X_n)_{n\in\mathbb N_0}$ is independent. Are we able to conclude that $\operatorname P$ is $\tau$-ergodic?

My idea is as follows: We should have $$\tau^{-n}(A)=\left\{\omega\in\Omega:\tau^n(\omega)\in A\right\}\in\mathcal F_n:=\sigma(X_k,k\ge n)\tag1$$ for all $A\in\mathcal A$ and $n\in\mathbb N$. So, we should be able to conclude that $$\mathcal I:=\left\{A\in\mathcal A:\tau^{-1}(A)=A\right\}\subseteq\mathcal T:=\bigcap_{n\in\mathbb N}\mathcal F_n\tag2$$ and the independence assumption should yield the claim by Kolmogorov's 0-1 law. Am I missing something?

EDIT: A similar statement is proven in Probability Theory: A Comprehensive Course: enter image description here

0xbadf00d
  • 13,422

1 Answers1

1

This statement is not correct as currently written, but can be made true with a minor modification. Specifically, if we assume $\mathcal A=\sigma(X_n:n\in\mathbb N_0)$, then the statement is true. This is a fairly natural assumption in many applications, since the underlying $\sigma$-field need not carry more information than the lifetime of the stochastic process in consideration.

To see the statement holds under this assumption, one can use your line of reasoning, since any $A\in\mathcal A$ is of the form $A=\{(X_n)_{n\in\mathbb N_0} \in B\}$ for some $B\in\mathcal B(\mathbb R)^{\otimes\mathbb N_0}$. This also should hopefully illuminate why your argument does not work in full generality, since obviously if $\mathcal A$ contains events outside of $\sigma(X_n:n\in\mathbb N_0)$, then for such an event $A$ one does not have that $\tau^{-n}(A)$ is measurable with respect to $\sigma(X_n,X_{n+1},\ldots)$. The counterexample where $X_0$ (and hence every $X_n$) is constant demonstrates this, since in this case $\sigma(X_n:n\in\mathbb N_0)$ is trivial, and presumably your underlying $\sigma$-field $\mathcal A$ is not.

Jason
  • 15,438