Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 226 / 1 / b

The displayed identity is false with the printed non-strict inequality. For example, take $A=B=\{y\}$ and let $r=P_y({\tilde{H}}_y<\infty )\in (0,1)$. The event $L_y\le {\tilde{H}}_y$ says that the walk returns to $y$ at most once, so its left side is ${\mu }_y(1-r^2)$, whereas its right side is $e_{\{y\}}(y)={\mu }_y(1-r)$.

The standard and evidently intended last-exit decomposition for a transient random walk has $L_B<{\tilde{H}}_A$. Decompose that corrected event according to $n=L_B$ and $z=X_n\in B$. The Strong Markov property at time $n$ givesReversibility of the random walk on a graph gives the path-reversal identitySince $e_B(z)={\mu }_z{P}_z({\tilde{H}}_B=\infty )$ is the equilibrium measure of a finite set, summing first over $n$ and then over $z$ yields

Solved by gpt-5.6-sol high.

Assume $A\subset B$ and use the corrected strict identity from part (i). If a walk starts at $y\in A$ and never returns to $A$, then transience makes $L_B<\infty ={\tilde{H}}_A$. ThusThis proves monotonicity of the capacity of a finite set for a transient random walk.

Solved by gpt-5.6-sol high.

The inequality need not be strict. Start with the nearest-neighbour weighted graph ${\mathbb{Z}}^3$, attach a new leaf $b$ only to the origin $o$, and give its edge positive weight. Take $A=\{o\}$ and $B=\{o,b\}$. A walk starting from $b$ must move to $o$ at its first non-killed step, while any walk reaching $b$ from elsewhere must first pass through $o$. Hence the equilibrium potential of a finite set is the same for the two sets:Using $\mathrm{cap}(C)=\mathcal{E}(h_C,h_C)$ for a finite set $C$ givesdespite $A\ne B$.

Solved by gpt-5.6-sol high.