Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 226 (Random Walks and Phase Transitions)

For simple random walk on ${\mathbb{Z}}^d$, $d\ge 3$, the Green-function decay for simple random walk on the integer lattice and the Strong Markov property giveThe series over $k$ converges, so the first of the Borel-Cantelli lemmas says that almost surely only finitely many of the points $x_k$ are ever hit. Moreover, ${\mathbb{Z}}^d$ is a transient graph for $d\ge 3$, so each of those finitely many points is visited only finitely often. Therefore the simple random walk visits $A$ only finitely often almost surely:

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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

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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.

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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 $A\ne B$.

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The Dirichlet energy space with zero boundary at infinity iswhere $C_c(G)$ is the vector space of finitely supported real functions on the vertices. Transience makes the Dirichlet form of a Markov chain positive definite on this completion.

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The Strong Markov property at the first visit to $0$ gives the singleton equilibrium potential from the Green functionLet $U_n$ be an increasing exhaustion of $G$ by finite sets containing $0$, and let $g_{U_n}(x,0)$ be the Green function of a transient weighted graph stopped on leaving $U_n$. Each $g_{U_n}(\cdot ,0)$ has finite support. The Green identity and the Markov property give, for $m\ge n$,By the monotone convergence theorem, the right side tends to zero as $m,n\to \infty$, while $g_{U_n}(x,0)\uparrow g(x,0)$. Thus $g(\cdot ,0)$ is an $H_0$ limit of finitely supported functions. Dividing by the positive number $g(0,0)$ proves $h\in H_0$.

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For the normalized Green function of a transient weighted graph, the Green identity isTaking $f=g(\cdot ,0)$ and using $h=g(\cdot ,0)/g(0,0)$ therefore givesThis is also the capacity of a finite set for a transient random walk $\mathrm{cap}(\{0\})$.

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The field $\psi$ is centered and jointly Gaussian. Since the Gaussian free field has covariance $\mathbb{E}[{\phi }_x{\phi }_y]=g(x,y)$ and $h_x=g(x,0)/g(0,0)$,Also ${\psi }_0=0$ andJoint Gaussianity turns this zero covariance into independence. Therefore $\psi$ is a Pinned Gaussian free field at $0$, independent of ${\phi }_0$, with covariance equal to the Green function of the walk killed on hitting $0$.

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The family $(Y_x)$ is a Gaussian random field. For distinct $x,x^{\prime }$ its covariance isTwo distinct vertices of ${\mathbb{Z}}^d$ have a common neighbour exactly when their graph distance, equivalently their ${\ell }^1$ distance, is two. Since jointly Gaussian variables are independent exactly when they are uncorrelated,Thus the field has finite-range dependence, even though nearest-neighbour values are independent.

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Write $p(a)=P(Y_0\ge a)$; because $Y_0\sim N(0,1+1/(2d))$, $p(a)\to 0$ as $a\to \infty$. Every path in a graph of length $n$ contains, by a greedy selection, at least $n/M_d$ vertices at mutual graph distance greater than two, where $M_d=|B_2(0)|$. The corresponding field values are jointly independent by part (a). Hence the probability that a fixed path lies in the superlevel set is at mostThere are at most $(2d{)}^n$ length-$n$ paths from the origin. The union bound therefore givesChoose a finite $a$ for which $(2d)p(a{)}^{1/M_d}<1$ and let $n\to \infty$. There is then no unbounded component through the origin, and translation invariance rules out an unbounded component anywhere almost surely. Thus the critical threshold for level-set percolation satisfies $a_c(d)<\infty$.

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The one-arm event $A_R(a)$ depends on the field values in the finite ball $B_R$. Replace $V_x$ by $U_x+a$ for $x\in B_R$. In the new variables the event has threshold zero, while the product normal distribution density is ${\prod }_{x\in B_R}\phi (U_x+a)$. Differentiating this finite-dimensional integral under the integral sign gives the Gaussian shift identityThis is also an instance of Gaussian integration by parts. In particular, the asserted inequality holds, in fact with equality.

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Apply the OSSS inequality to the independent coordinates $(V_x,W_x)$ and to the indicator of the one-arm event $A_R(a)$. Use the randomized OSSS exploration of a one-arm event: choose $k$ uniformly from $\{1,\dots ,R\}$ and reveal the variables needed to explore the superlevel cluster meeting $\partial B_k$. A coordinate can be revealed only if a nearby vertex has an open connection over the relevant distance. Translation invariance, finite-range dependence and the union bound therefore give the revealment estimatefor both kinds of coordinates, after enlarging the explored neighbourhood by a distance depending only on $d$.

For an increasing Gaussian threshold event, the resampling influence of $V_x$ is bounded by a universal constant times $\mathbb{E}[V_x{1}_{A_R(a)}]$. The influence of $W_x$ is bounded by the sum of the corresponding $V$ influences at the $2d$ neighbours of $x$: indeed Gaussian integration by parts givesfirst for smooth increasing approximations and then by a limit. Consequently the OSSS bound becomesPart (c) identifies the final sum with $-{\theta }_R^{\prime }(a)$. Dividing and putting $c=C_d^{-1}>0$ proves

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