The covariance is
$$
\mathbb E[\Gamma(x)\Gamma(y)]
=\frac12G_{\{a\}}(x,y).
$$
Indeed, the killed-walk Green function satisfies
$$
G_{\{a\}}(x,y)
=\mathbf1_{\{x=y\}}+\frac1{2d}\sum_{z\sim x}G_{\{a\}}(z,y),
\qquad G_{\{a\}}(a,y)=0.
$$
On the other hand, the conditional mean from part 2 makes the covariance harmonic in $x\ne y,a$. At $x=y$, the conditional variance $1/2$ gives the same equation with source $\frac12\mathbf1_{\{x=y\}}$. Uniqueness of the <Dirichlet problem> therefore identifies the covariance with $G_{\{a\}}/2$. The factor $1/2$ comes from the coefficient $1/(2d)$ in this paper's energy density.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 209 / 3 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-209/3/3/solution