M-dependency and ergodicity

Question

Let $X_n, n \in \mathbb{N}$ be a strictly stationnary m-dependent sequence of real-valued random variables on an underlying probability space $(\Omega,\mathcal{A},\mathbb{P})$, i.e. $X_n$ and $X_{n'}$ are independent if $\vert n-n' \vert\geq m$. Strictly stationnary here means that $\{X_n,n \in \mathbb{N}\} =\{X_{n+k},n \in \mathbb{N}\} $ in distribution for all $k \in \mathbb{N}$. This already implies that $X_n, n \in \mathbb{N}$ is a measure-preserving dynamical system with respect to $\tau: \Omega \rightarrow \Omega$ with $X_{n+1}(\omega) = X_n(\tau(\omega))$. Can we also conclude ergodicity, i.e. that any $A \in \mathcal{A}$ with $\tau(A)=A$ is $\mathbb{P}$-trivial?

Answer 1

I am fairly sure you can, though please do check the following for issues/errors. Take any $A$ in the invariant sigma algebra, and invoke the ergodic theorem on $1_A$.

Then, you would hope to get

$$\frac{1}{mn} \sum_{i=0}^{mn} 1_A(X_n) \to 1_A$$ almost surely as $n \to \infty$.

Note that

$$\frac{1}{mn} \sum_{i=0}^{mn} 1_A(X_n) = \frac{1}{mn} \left(\sum_{j=0}^{m-1} \sum_{i=0}^{n} 1_A(X_{j + km}) \right) \to \mathbb{E} 1_A $$ as $n \to \infty$ as each of the subsequences $\{X_j + km\}_{k \in \mathbb{N}}$ are i.i.d and therefore ergodic, which implies $A$ is trivial.

Answer 2

I don't think the proof of user "E-A" is correct, however I came up with another one (please feel free to check): One knows that $X_{n+mj},j \in \mathbb{N}$ is i.i.d. for any $n \in \mathbb{N}$. Let $A \in \mathcal{A}$ be an invariant event, i.e. $\tau(A)=A$. Then we have that $$ A \in \bigcap_{n \geq 1} \sigma(X_n,X_{n+1},\ldots) \subset \sigma\bigg( \bigcup_{j=1}^{m-1} \bigcap_{n \geq 0} \sigma(X_{(n+k)m+j},k \geq 0) \bigg) $$ Now for each $j=0,\ldots,m-1$ we can interpret $\mathcal{A}_j=\bigcap_{n \geq 0} \sigma(X_{(n+k)m+j},k \geq 0)$ as the terminal $\sigma$-algebra of an i.i.d. process and we have with Kolmogorov's $0$-$1$ law we conclude $\mathbb{P}(A') \in\{0,1\} $ for each element $A' \in \mathcal{A}_j$ for any $j=1,\ldots,m-1$. Since $A \in \sigma\big( \bigcup_{j=1}^{m-1} \mathcal{A}_j \big)$ we can write $A$ as a finite intersection/union of elements in $\bigcup_{j=1}^{m-1}\mathcal{A}_j$. We conclude $\mathbb{P}(A) \in \{0,1\}$.