\[
\newcommand{\C}{\mathbb{C}}
\newcommand{\haar}{\mathsf{m}}
\newcommand{\P}{\mathcal{P}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\d}{\mathsf{d}}
\newcommand{\g}{>}
\newcommand{\l}{<}
\newcommand{\intd}{\,\mathsf{d}}
\newcommand{\Re}{\mathsf{Re}}
\newcommand{\area}{\mathop{\mathsf{Area}}}
\newcommand{\met}{\mathop{\mathsf{d}}}
\newcommand{\emptyset}{\varnothing}
\DeclareMathOperator{\borel}{\mathsf{Bor}}
\newcommand{\ent}{\operatorname{\mathsf{H}}}
\]
Week 11 Tutorial Solutions
- The only members of $Z$ are
\[
\begin{align*}
a ={} &\mathtt{0101010101010101}\cdots \\
b ={} &\mathtt{1010101010101010}\cdots
\end{align*}
\]
- The only invariant probability measure on $Z$ is
\[
\nu = \tfrac{1}{2} (\delta_a + \delta_b)
\]
- The only partitions of $Z$ are $(Z)$ and $(\{a\}, \{b\})$. As $T$ fixes the first and permutes the second, we have
\[
\xi \vee T^{-1} \xi \vee \cdots \vee T^{-(N-1)} \xi = \xi
\]
for both and all $N \in \N$. Therefore the entropy of $T$ is zero.
- The cylinders $[101]$ and $[111]$ are disjoint from $Y$ because $101$ and $111$ are forbidden for sequences in $Y$. The six other cylinders of length 3 have non-empty intersection with $Y$.
- If the measure has $\mu(Y) = 1$ then necessarily
\[
\begin{align*}
0 = \mu([101]) &{} = p(1) q(1,0) q(0,1) \\
0 = \mu([111]) &{} = p(1) q(1,1) q(1,1)
\end{align*}
\]
both hold.
It may be that $p(1) = 0$. In this case the matrix equation gives
\[
0 = p(0) q(0,1)
\]
and therefore $q(0,1) = 0$ as we would have to have $p(0) = 1 - p(1) = 1$. Forbidding $01$ gives a countable collection of sequences on which the single invariant measure is the point mass at the sequence that is constantly zero.
It may be that $p(1) \ne 0$. In this case we have $q(1,1) = 0$ and $q(1,0) q(0,1) = 0$. But $q(1,1) = 0$ forces $q(1,0) = 1$ so we must have $q(0,1) = 0$. There are only two sequences satisfying these rules and the only measure is again the point mass at the sequence which is constantly zero.
- Consider the graph with vertices labeled $0,1,2,3$ and edges as follows.
- An edge from 0 to 0.
- An edge from 0 to 1.
- An edge from 1 to 2.
- An edge from 1 to 3.
- An edge from 2 to 0.
- An edge from 3 to 0.
Let $\Omega$ be the corresponding subset of $\{0,1,2,3\}^\N$ i.e. the one defined by the matrix
\[
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix}
\]
and define $\pi$ by
\[
(\pi \omega)(n) = \begin{cases} 0 & \omega(n) \in \{0,2\} \\ 1 & \omega(n) \in \{1,3\} \end{cases}
\]
for all $\omega \in \Omega$. Note that $\pi(\Omega) = Y$. Let $\eta$ be any Markov measure compatible with this graph. As a Markov measure it is $T$ invariant. So too is the push-forward $\eta \circ \pi^{-1}$ and
\[
(\eta \circ \pi^{-1})(Y) = \eta(\Omega) = 1
\]
as desired.
Moreover, it is not the point mass at the origin because
\[
(\eta \circ \pi^{-1})([01]) = \eta([01]) + \eta([21]) + \eta([03]) + \eta([23])
\]
is positive for many possible values of the parameters defining $\eta$.