Write
$$
q_n=\mathbb P_p(0\leftrightarrow\partial B(n)),
\qquad a_n=\mathbb P_p(0\leftrightarrow e_n).
$$
Since $e_n\in\partial B(n)$, $q_n\geq a_n$, and therefore
$$
\limsup_{n\to\infty}-\frac1n\log q_n\leq\phi(p).
$$
For the opposite inequality, choose $x=(n,x_2,\ldots,x_d)$ as in part (ii). Reflection about $x$ sends $0$ to $2ne_1$ and preserves the lattice. Thus the increasing events $\{0\leftrightarrow x\}$ and $\{x\leftrightarrow2ne_1\}$ have the same probability. The <FKG inequality> yields
$$
a_{2n}\geq\mathbb P_p(0\leftrightarrow x)^2
\geq\left(\frac{q_n}{|\partial B(n)|}\right)^2.
$$
Consequently
$$
-\frac1n\log q_n
\geq-\frac1{2n}\log a_{2n}
-\frac1n\log|\partial B(n)|.
$$
The polynomial boundary-size estimate makes the last term tend to zero, while the first tends to $\phi(p)$. Hence
$$
\boxed{\lim_{n\to\infty}-\frac1n\log q_n=\phi(p).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 214 / 1 / b / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-214/1/b/iii/solution