# Can one realize this as an ergodic process?

Consider the lattice $$\mathbb Z^2$$ and take iid random variables $$Y_e$$ on all edges $$e$$ of the graph.

We then define random variables $$X_i:=\sum_{e \text{ adjacent to } i}Y_e.$$

In other words: For every vertex we define a random variable $$X_i$$ that is the sum of the $$Y_e$$ on its neighbouring edges.

I ask: Is there a probability space $$(\Omega,\mathcal A, \mathbb P)$$ such that this stochastic process $$(X_i)$$ is ergodic with respect to the standard shift $$(T_i)_{i \in \mathbb Z^2}$$ where $$(T_i f)(z)=f(z-i)$$? If the $$X_i$$ were independent this would be easy as the product measure would do it. But they are only almost independent in the sense that $$X_i$$ and $$X_j$$ are independent if $$\vert i-j\vert \ge 2$$.

I recall for you:

A stochastic process is called ergodic if there is an ergodic family of measure preserving transformations $$(T_i)$$ such that $$X_i(T_j\omega)= X_{i-j}(\omega).$$

Now, the shift is a reasonable candidate to be measure preserving and it satisfies also the property $$X_i(T_j\omega)= X_{i-j}(\omega)$$

Can we also realize it as an ergodic family?

Let me add that the situation does not seem to be that bad by noting that due to independence on edges we have that

$$\frac{\sum_{\vert i \vert \le n} X_i}{n^2} \rightarrow 4\mathbb E(Y_e).$$

• This follows from a well known fact in ergodic theory: a factor of an ergodic process is ergodic. – Anthony Quas Feb 12 '19 at 16:39

## 2 Answers

Let $$\mu$$ be the common law of the $$Y_e$$. I call $$(\Omega,\mathcal F,\mathbb P)$$ the underlying probability space: $$\Omega = \mathbb R^{\mathbb Z^2}$$, $$\mathcal F = \mathcal B(\mathbb R)^{\otimes\mathbb Z^2}$$, $$\mathbb P$$ is the image of $$\mu^{\otimes\text{edges}}$$ under the application that maps the $$Y_e$$ to the $$X_i$$. Let $$S$$ be the set of events depending only on a finite number of $$X_i$$'s. Set $$\mathcal A:=\{A\in\mathcal F\text{ such that }\inf_{B\in S}\mathbb P(A\Delta B)=0\}.$$ (In what follows, what we will be interested in is that $$|\mathbb P(A)-\mathbb P(B)|\leq\mathbb P(A\Delta B)$$.) It is not hard to prove that $$\mathcal A$$ is a $$\sigma$$-algebra, once you've realised that $$\big(\bigcup_nA_n\big)\Delta\big(\bigcup_nB_n\big)\subset\bigcup_n(A_n\Delta B_n);$$ in particular, $$\mathcal A$$ is in fact just $$\mathcal F$$.

Now let $$A\in\mathcal F$$ such that $$T_i^{-1}A=A$$ for any $$i\in\mathbb Z^2$$. For $$\varepsilon>0$$, I can choose $$A_\varepsilon$$ such that $$\mathbb P(A\Delta A_\varepsilon)\leq\varepsilon$$ and $$A_\varepsilon$$ depends only on a finite number of $$Y_e$$, say only those contained in a ball of radius $$N_\varepsilon$$. Because $$(C\cap D)\Delta(C'\cap D')\subset (C\Delta C')\cup (D\Delta D')$$, we have \begin{align*}|\mathbb P(A)^2-\mathbb P(A)| & = |\mathbb P(A)\mathbb P(T^{-1}_iA)-\mathbb P(A\cap T^{-1}_iA)| \\ & \leq |\mathbb P(A_\varepsilon)\mathbb P(T^{-1}_iA_\varepsilon)-\mathbb P(A_\varepsilon\cap T^{-1}_iA_\varepsilon)| + 4\varepsilon,\end{align*} and the first term is zero provided $$|i|>2N_\varepsilon$$. This implies $$\mathbb P(A)^2=\mathbb P(A)$$, hence $$\mathbb P(A)\in\{0;1\}$$.

(It is a classical argument, although I don't remember seeing it outside a probability lesson.)

As mentioned in the comment of Anthony Quas, the process you define is a factor of a Bernoulli shift on $$\mathbb{Z}^2$$. I will explain in more detail what this means. Let $$\mu \in \mathrm{Prob}(\mathbb{R})$$ be the law of the $$Y_e$$. Consider the product space $$(\mathbb{R}^{\mathbb{Z}^2},\mu^{\otimes \mathbb{Z}^2})$$. The group $$\mathbb{Z}^2$$ acts on this space by shifting coordinates. This is the Bernoulli shift of $$\mathbb{Z}^2$$ over $$\mu$$.

Now, define a map $$\phi:\mathbb{R}^{\mathbb{Z}^2} \to \mathbb{R}^{\mathbb{Z}^2}$$ by setting $$\phi(\overline{a})(e) = \sum_{i \sim e} \overline{a}(i)$$. The law of the process you describe is $$\phi_\ast \mu^{\otimes \mathbb{Z}^2}$$, so you are asking about ergodicity of the process $$(\mathbb{R}^{\mathbb{Z}^2},\phi_\ast \mu^{\mathbb{Z}^2})$$ where again $$\mathbb{Z}^2$$ acts by shifting coordinates. It is easy to see that if $$A$$ is a nontrivial invariant set in the factor process then $$\phi^{-1}(A)$$ is a nontrivial invariant set in the Bernoulli process. Hence there do not exist nontrivial invariant sets in the factor process.

It follows from Ornstein theory that any factor of a Bernoulli process of $$\mathbb{Z}^2$$ is actually measurably conjugate to a Bernoulli process. Intuitively, this means there is a way to recode the factor process as a family of iid random variables. However, the encoding might be hard to describe explicitly. The property that factors of Bernoulli processes are conjugate to Bernoulli processes holds more generally when the acting group is amenable. It fails for certain nonamenable acting groups and the nature of this failure is a topic of current research.