Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 37 / 2 / b / Solution

We use the Van den Berg-Kesten inequality: for increasing events $A,B$ under independent bond percolation, their disjoint occurrence $A\square B$, meaning that they have disjoint sets of open edges witnessing the two events, satisfies ${\mathbb{P}}_p(A\square B)\le {\mathbb{P}}_p(A){\mathbb{P}}_p(B)$. The finite-edge statement applies here; connection to a finite-box boundary can always be witnessed before the first exit from that box.

On the event $0\leftrightarrow \partial {\Lambda }_{m+n}$ choose a simple open path to that boundary, and let $y$ be its first point on $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow \partial {\Lambda }_m$ through $y$, inside ${\Lambda }_m$. The remaining segment eventually reaches maximum-norm distance at least $n$ from $y$, because its endpoint has norm $m+n$ while $\Vert y{\Vert }_{\infty }=m$. Stop that segment when it first reaches $y+\partial {\Lambda }_n$. These two segments use disjoint edges.

For fixed $y$, let $A_y$ be connection from zero to $y$ inside ${\Lambda }_m$, and $B_y$ be connection from $y$ to $y+\partial {\Lambda }_n$ inside $y+{\Lambda }_n$. Then ${\mathbb{P}}_p(A_y)\le {\beta }_m$ and, by translation invariance, ${\mathbb{P}}_p(B_y)={\beta }_n$. The union bound and the stated disjoint-occurrence inequality give

Set $x_n=\log {\beta }_n$ and ${\alpha }_n=\log |\partial {\Lambda }_n|$. The previous part applies, sinceMoreover a straight path of $n$ open bonds has probability $p^n$, so $p^n\le {\beta }_n\le 1$. Thus the limit is finite andEquivalently, for the nonnegative decay rate $\kappa =-\gamma$, the bound is ${\beta }_n\ge |\partial {\Lambda }_n{|}^{-1}{e}^{-n\kappa }$.

The printed final exponent has the opposite sign while retaining the definition of $\gamma$ as the limit of $n^{-1}\log {\beta }_n$. With that definition it cannot be correct in general. Indeed, if $(2d-1)p<1$, count self-avoiding paths of length $\ell \ge n$ to getHence $\gamma \le \log ((2d-1)p)<0$. A lower bound with $e^{-n\gamma }$ would eventually exceed one, because the boundary factor grows only polynomially. The valid bound is the boxed expression with $e^{n\gamma }$, or the equivalent expression using $\kappa =-\gamma$.