Week 7 Tutorial Solutions

1. ${\mu }_p([01])=(1-p)p$
2. ${\mu }_p([01])=(1-p)p$ because $1_{[01]}$ is simple.
3. $Z_1=1_{[11]}$ is simple so its integral is ${\mu }_p([11])=p^2$.
4. $T^{-1}([01])=[001]\cup [101]$ is a disjoint union so $${\mu }_p(T^{-1}([01]))={\mu }_p([001])\cup {\mu }_p([101])$$ and ${\mu }_p(T^{-1}([01]))=(1-p{)}^{2}p+p(1-p)p=(1-p)p$.
5. $Z_2=1_{T^{-1}[11]}$ is simple. Its integral is ${\mu }_p(T^{-1}[11])$ which is $${\mu }_p([011])+{\mu }_p([111])=p^2$$ as $[011]$ and $[111]$ are disjoint.
6. Calculate $$\begin{array}{rl}\int Y_n\mathsf{d}{\mu }_p& =\int Z_n(-\int Z_n\mathsf{d}{\mu }_p)\mathsf{d}{\mu }_p\\ & =\int Z_n\mathsf{d}{\mu }_p-\int Z_n\mathsf{d}{\mu }_p=0\end{array}$$ by linearity and the fact that ${\mu }_p(X)=1$.
7. From $$\begin{array}{rl}{Y}_n{Y}_{n+1}& =Z_n{Z}_{n+1}-p^2{Z}_n-p^2{Z}_{n+1}+p^4\\ & =X_n{X}_{n+1}{X}_{n+2}-p^2{X}_n{X}_{n+1}-p^2{X}_{n+1}{X}_{n+2}+p^4\end{array}$$ we can calculate $$\int Y_n{Y}_{n+1}\mathsf{d}{\mu }_p=p^3-p^4$$
8. From $$\begin{array}{rl}{Y}_n{Y}_{n+2}& =Z_n{Z}_{n+2}-p^2{Z}_n-p^2{Z}_{n+2}+p^4\\ & =X_n{X}_{n+1}{X}_{n+2}{X}_{n+3}-p^2{X}_n{X}_{n+1}-p^2{X}_{n+2}{X}_{n+3}+p^4\end{array}$$ we can calculate $$\int Y_n{Y}_{n+2}\mathsf{d}{\mu }_p=0$$
9. The integral $$\begin{array}{rl}& Y_a{Y}_b{Y}_c{Y}_d\\ =& (X_a{X}_{a+1}-p^2)(X_b{X}_{b+1}-p^2)(X_c{X}_{c+1}-p^2)(X_d{X}_{d+1}-p^2)\end{array}$$ will be zero if $$a\notin \{b-1,b,b+1,c-1,c,c+1,d-1,d,d+1\}$$ or the same holds for any permutation of the indices.
10. There is a constant $E>0$ such that at most $E{N}^2$ of all possible choices $1\le a,b,c,d\le N$ result in a non-zero integral in the previous question. It is in any case bounded by 1 so our proof from the notes of the strong law of large numbers works in this setting as well.