Week 11 Worksheet Solutions

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1. The partition $\xi =([0],[1],[2])$ is generating for the shift map so we may just compute $$\begin{array}{rl}\mathsf{H}(T)& =\mathsf{H}(T,\xi )=\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\mathsf{H}(\xi \vee T^{-1}\xi \vee \cdots \vee T^{-(N-1)}\xi )\\ & =\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}{3}^N\cdot {\displaystyle \frac{1}{3^N}}\log (3^N)\end{array}$$ giving an entropy of $\log 3$.
2. With respect to normalized counting measure, every partition of $\mathbb{Z}/p$ has entropy at most $\log p$ and therefore, for any partition $\xi$ we have $$\mathsf{H}(T,\xi )\le \underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{\log p}{N}}$$ which is zero.
3. It is directly verified that $0$, $1$ and $$000110$$ are the only string of lengths 1 and 2 respectively that do not contain consecutive ones. Fix $n\ge 3$ and let $s_1\cdots s_r$ be a string of zeroes and ones with no consecutive ones. If $s_1=0$ then there are $w_{n-1}$ possibilities for the repaining strings as any string of length $n-1$ not containing consecutive ones can appear. If $s_1=1$ then $s_2=0$ but then there are $w_{n-2}$ possibilities for the remaining $n-2$ strings. Thus $w_n=w_{n-1}+w_{n-2}$.
4. Fix $0<p\le 1$ and let $\mu$ be the $(1-p,p)$ fair coin measure. As $T$ is ergodic for $\mu$ we can deduce that $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{1}_{[11]}(T^{n}x)=\int 1_{[11]}\mathsf{d}\mu =p^2$$ for $\mu$ almost every $x\in \{0,1{\}}^{\mathbb{N}}$. But no sequence in $Y$ has consecutive ones in it, so we must have $\mu (Y)=0$.
5. The $(1-p,p)$ coin measure on $\{0,1{\}}^{\mathbb{N}}$ has entropy $$-p\log p-(1-p)\log (1-p)$$ which, for every $0\le p\le 1$ is less than $\log 3$. As isomorphic systems have the same entropy, no coin measure on $\{0,1{\}}^{\mathbb{N}}$ gives a system isomorphic to the fair coin measure on $\{0,1,2{\}}^{\mathbb{N}}$.
6. The corresponding transition matrix is $$B=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}\right]$$ and since all entries of $B^2$ are positive we may apply Parry's theorem to calculate the entropy. The eigenvalues of B are $$1-\sqrt{3},0,1+\sqrt{3}$$ and we set $\lambda =1+\sqrt{3}$. The entropy is $\log (1+\sqrt{3})$.
   
   The vectors $$U=\left[\begin{array}{ccc}\lambda & 2& \lambda \end{array}\right]V=\left[\begin{array}{c}\lambda \\ 2\\ \lambda \end{array}\right]$$ are left and right eigenvectors respectively of $B$ with eigenvalue $\lambda$. Parry's theorem tells us to take $$Q=\left[\begin{array}{ccc}{\displaystyle \frac{1}{\lambda }}& {\displaystyle \frac{2}{{\lambda }^2}}& {\displaystyle \frac{1}{\lambda }}\\ {\displaystyle \frac{1}{2}}& 0& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{\lambda }}& {\displaystyle \frac{2}{{\lambda }^2}}& {\displaystyle \frac{1}{\lambda }}\end{array}\right]$$ and $$p=\left[\begin{array}{ccc}{\displaystyle \frac{{\lambda }^2}{4+2{\lambda }^2}}& {\displaystyle \frac{4}{4+2{\lambda }^2}}& {\displaystyle \frac{{\lambda }^2}{4+2{\lambda }^2}}\end{array}\right]$$