6
$\begingroup$

Let $T$ be a measure preserving transformation of a standard probability space $(X,\mathcal{B},\mu)$. A partition $\alpha$ of $X$ is said to be a generator for $T$ if the smallest $T$ invariant $\sigma$-algebra of measurable sets containing the pieces of $\alpha$ is all of $\mathcal{B}$.

Suppose there is a generator $\alpha$ for $T$ and a finite set $F \subseteq \mathbb{Z} \setminus \{0\}$ such that $\alpha$ is measurable with respect to $\bigvee_{n \in F} T^n \alpha$. Does this imply that $T$ has zero entropy?

$\endgroup$
3
$\begingroup$

No, this is not true. This is true if $F \subset \mathbb{N}$, but not in the general case. I am lost with the objects of ergodic theory so I start by rephrasing in the probabilistic language.

Let ${(Z_n)}_{n \geq 0}$ be the stationary (shift-invariant) process defined by $Z_n(x)=\alpha(T^n(x))$ where $\alpha(x)$ denotes the element of the partition $\alpha$ to which $x$ belongs. The (usual) definition of the entropy of $T$ with respect to $\alpha$ is $h(T,\alpha)=\lim \frac{H(Z_1,\ldots,Z_n)}{n}$, and it is well known that $h(T)=h(T,\alpha)$ when $\alpha$ is a generating partition. Moreover, introducing the decreasing sequence of $\sigma$-fields ${\cal F}_n=\sigma(Z_m; m \geq n)$, it is well known that $h(T,\alpha)=H({\cal F}_0 \mid {\cal F}_1)$.

Assume your $F$ is included in $\mathbb{N} \setminus \{0\}$. That means that $Z_0$ is measurable with respect to $\sigma(Z_k; k \in F) \subset {\cal F}_1$. Therefore ${\cal F}_0 \subset {\cal F}_1$ (then ${\cal F}_0={\cal F}_1)$ because ${\cal F}_0 = {\cal F}_1 \vee \sigma(Z_0)$. Hence $H({\cal F}_0 \mid {\cal F}_1)=0$.

If your $F$ contains some negative integers, then your assumption is equivalent to $\sigma(Z_{n_0}) \subset \sigma(Z_k; k \in F)$ where $F \subset \mathbb{N}\setminus\{n_0\}$. It is easy to get a counter-example for $n_0=1$ and $F=\{0,2\}$.

Take a stationary process ${(Z_n)}_{n \geq 0}$ with non-zero entropy, taking its values in a finite alphabet, and define a new stationary process ${(Y_n)}_{n \geq 0}$ by setting $Y_n = (Z_n, Z_{n+1})$. Then $Y_1 \subset \sigma(Y_0,Y_2)$. But it is clear from the formula defining $h(T,\alpha)$ that ${(Y_n)}_{n \geq 0}$ has the same entropy as ${(Z_n)}_{n \geq 0}$. This is also clear from the other formula because the process ${(Y_n)}_{n \geq 0}$ has the same decreasing sequence of $\sigma$-fields as ${(Z_n)}_{n \geq 0}$.

For example, if you take for ${(Z_n)}_{n \geq 0}$ a sequence of independent symmetric Bernoulli variables, then you can see ${(Y_n)}_{n \geq 0}$ as a stationary Markov chain on the vertices of a square $ABCD$, where $Y_n$ has the uniform distribution and it jumps from $A$ to $A$ or to the "next" point $B$ with equal probabilities, it jumps from $B$ to $B$ or to the "next" point $C$ with equal probabilities, etc. Its entropy is $\ln 2$ and it is easy to see that $Y_1 \subset \sigma(Y_0,Y_2)$.

So the answer is no, and it is still no with additional assumptions such as ergodicity, $K$, Bernoullicity.

I seize the opportunity to give a try to the markovchain R package:

library(markovchain)

Y <- new("markovchain", states = c("A", "B", "C", "D"),
         transitionMatrix = 0.5*matrix(c(c(1, 1, 0, 0), 
                                     c(0, 1, 1, 0), 
                                     c(0, 0, 1, 1), 
                                     c(1, 0, 0, 1)), 
                                   byrow=TRUE, nrow = 4),
         name = "Counter-example")

plot(Y)

enter image description here

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.