Expanding the square and using the independence of the percolation edges gives
$$
\mathbb EX_r^2
=\sum_{\gamma,\gamma'}\mu(\gamma)\mu(\gamma')
\frac{\mathbb P(\gamma,\gamma'\text{ open})}
{\mathbb P(\gamma\text{ open})\mathbb P(\gamma'\text{ open})}.
$$
The ratio equals $p^{-N_r(\gamma,\gamma')}$, where $N_r$ is the number of common edges in the two truncated paths. It is at most $p^{-N}$ for $N=|\xi\cap\xi'|$, under the convention in the hypothesis; in particular $N\geq1$ because both paths contain $o$. For every integer $N\geq1$,
$$
p^{-N}\leq\sum_{n=1}^Np^{-n}.
$$
The <tail-sum formula> and the assumed exponential intersection tail therefore yield
$$
\mathbb EX_r^2
\leq\sum_{n=1}^\infty p^{-n}(\mu\times\mu)(N\geq n)
\leq C\sum_{n=1}^\infty\left(\frac\zeta p\right)^n
=\frac{C\zeta}{p-\zeta}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 212 / 2 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-212/2/c/i/solution