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 \[ \begin{align*} \int Y_n \intd \mu_p &{} = \int Z_n \left( - \int Z_n \intd \mu_p \right) \intd \mu_p \\ &{} = \int Z_n \intd \mu_p - \int Z_n \intd \mu_p = 0 \end{align*} \] by linearity and the fact that $\mu_p(X) = 1$.
7. From \[ \begin{align*} Y_n Y_{n+1} &{} = Z_n Z_{n+1} - p^2 Z_n - p^2 Z_{n+1} + p^4 \\ &{} = X_n X_{n+1} X_{n+2} - p^2 X_n X_{n+1} - p^2 X_{n+1} X_{n+2} + p^4 \end{align*} \] we can calculate \[ \int Y_n Y_{n+1} \intd \mu_p = p^3 - p^4 \]
8. From \[ \begin{align*} Y_n Y_{n+2} &{} = Z_n Z_{n+2} - p^2 Z_n - p^2 Z_{n+2} + p^4 \\ &{} = X_n X_{n+1} X_{n+2} X_{n+3} - p^2 X_n X_{n+1} - p^2 X_{n+2} X_{n+3} + p^4 \end{align*} \] we can calculate \[ \int Y_n Y_{n+2} \intd \mu_p = 0 \]
9. The integral \[ \begin{align*} & Y_a Y_b Y_c Y_d \\ = {}& (X_a X_{a+1} - p^2)(X_b X_{b+1} - p^2)(X_c X_{c+1} - p^2)(X_d X_{d+1} - p^2) \end{align*} \] 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 $E \g 0$ such that at most $EN^2$ 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.