The chain is a random walk on the finite abelian group $(\mathbb{Z}/n\mathbb{Z}{)}^d$, so its stationary distribution is uniform and its characters diagonalize the transition operator. For one coordinate and $\theta =2\pi k/n$, the eigenvalue isUniformly in $p\in [0,1]$,for an absolute $c>0$. Hence the one-coordinate chi-squared distance after $t\ge n^2$ is at most $C{e}^{-ct/n^2}$. The coordinates evolve independently, so the product formula for chi-squared distance givesThe chi-squared divergence bound on total variation distance now yieldsuniformly in $p_1,\dots ,p_d$.

For $d=1$, the first nonconstant character has eigenvalue modulus $1-O(n^{-2})$, uniformly in $p$. Testing against its real or imaginary part gives a fixed positive total-variation distance until time $c{n}^2$. Thus

Exhaust ${\mathbb{Z}}^2$ by finite boxes $G_n$ and let $T_n$ be a uniform spanning tree of $G_n$. Every finite tree satisfies the handshaking lemma, soChoose the root uniformly from $V(G_n)$. The proportion of roots within any fixed distance of the boundary tends to zero, and the rooted trees converge locally to the uniform spanning tree $T$ of ${\mathbb{Z}}^2$. Since every degree is at most four, expectations also converge. Translation invariance therefore givesThe four edges incident to $0$ have equal inclusion probability by the rotations and reflections of the square lattice. If that common probability is $r$, then $4r=\mathbb{E}{\mathrm{deg}}_T(0)=2$. Consequentlyfor every edge $e$ by translation invariance.

For an infinite locally finite recurrent connected graph, take an increasing exhaustion by finite connected subgraphs and sample a uniform spanning tree in each. The restrictions to any fixed finite edge set converge in distribution; on a recurrent graph the free and wired limits coincide and form one tree. This infinite-volume law is called the uniform spanning tree of the recurrent graph.

It can be generated by Wilson's algorithm. Fix a root and an enumeration of the remaining vertices. Starting with the root, run a random walk from the first vertex not yet in the tree until it hits the existing tree, erase its loops chronologically, and add the resulting path. Recurrence ensures every walk hits the finite tree almost surely. Repeating this operation produces the infinite uniform spanning tree, independently of the enumeration.

A spanning tree of a finite connected graph $G$ is a connected acyclic subgraph containing every vertex. A uniform spanning tree is a random spanning tree chosen uniformly from the finite set of all spanning trees of $G$.

Let $v$ be the voltage with $v(a)=1$, $v(b)=0$, and harmonic values at every other vertex. For any other admissible $f$, write $f=v+g$, where $g(a)=g(b)=0$. Its discrete Dirichlet energy expands asDiscrete summation by parts turns the cross term intobecause $v$ is harmonic in the interior and $g$ vanishes at the boundary. Thus $v$ minimizes the energy. Under a unit voltage drop, its energy equals the total current from $a$ to $b$, namely the effective conductance $1/R_{\mathrm{eff}}(a,b)$. Therefore the Dirichlet principle gives

Put $A=\{a,b\}$ and let ${\tau }_x^{+}=\min \{t\ge 1:X_t=x\}$. The conductance-hitting identity for an electrical network iswhere the vertices of $D$ are wired together. It follows by taking the hitting probability of $D$ before returning to $x$ as a voltage and computing its total current out of $x$.

Split the walk into successive excursions from $x$. The first excursion that hits $A$ determines whether $a$ or $b$ is hit first. Its conditional probability of hitting $a$ is at most the probability that an arbitrary excursion hits $a$, divided by the probability that it hits $A$. Hence

A voltage with boundary values $u(a)$ and $u(b)$ is a function $u:V\to \mathbb{R}$ that is harmonic at every other vertex:A current flow is an antisymmetric function $i(x,y)=-i(y,x)$ satisfying Kirchhoff's node law ${\sum }_{y}i(x,y)=0$ away from its source and sink. Voltage and current are related by Ohm's law,The effective resistance is the voltage drop divided by the total current from $a$ to $b$. Equivalently, it is the voltage drop generated by a unit current flow:

By Tonelli theorem,If $p<p_c$, part (a) bounds the summand by $e^{-c\Vert x{\Vert }_{\infty }}$. There are only polynomially many vertices at each radius, so the series converges and $\chi (p)<\infty$.

Conversely, suppose $\chi (p)<\infty$. Then $p<1$. Choose $q>p$ so close to $p$ thatUse the standard sprinkling coupling for Bernoulli percolation: first expose the $p$-open clusters, then independently open each remaining edge with probability $\alpha =(q-p)/(1-p)$. Explore the $q$-cluster of the origin cluster by following sprinkled edges. Each discovered $p$-cluster has at most $2d$ times its number of vertices as many incident edges, so the exploration is dominated by a Galton-Watson process of mean at most $2d\alpha \chi (p)<1$. This process dies out almost surely, and hence there is no infinite $q$-open cluster. Thus $q\le p_c$, and $p<q$ implies $p<p_c$. Therefore

Because $p<p_c={\tilde{p}}_c$, choose a finite set $S\ni 0$ with ${\varphi }_p(S)=\rho <1$. Let $L$ exceed the ${\ell }^{\infty }$-distance from $0$ to every endpoint of an edge in $\partial S$, and putOn the one-arm event to distance $n>L$, take the first oriented boundary edge $(x,y)\in \partial S$ used by an open self-avoiding path. The connection $0\stackrel{S}{\leftarrow }x$, the open edge $(x,y)$, and the remaining connection from $y$ to $\partial B(n)$ occur disjointly. The van den Berg-Kesten inequality and translation invariance therefore giveIteration yields $q_n\le {\rho }^{\lfloor n/L\rfloor }$ up to an inessential finite-scale adjustment. Since $p<1$, each of the finitely many remaining $q_n$ is strictly below one, so reducing the exponent if necessary produces a constant $c>0$ valid for every $n\ge 1$:This is the finite-size proof of exponential decay of subcritical percolation.

PutFix $k\ge 1$ and write $n=qk+r$ with $0\le r<k$. Repeated subadditivity givessoThe finitely many values $x_0,\dots ,x_{k-1}$ are bounded, henceTaking the infimum over $k$ gives $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition of $\alpha$ gives $x_n/n\ge \alpha$ for every $n$. ThereforeThis is Fekete lemma.

The terms independent of $v$ in the substituted PDE satisfysoIntegrating backward from $T$ yieldsThus the requested constant is

Forone hasSubstitution into the PDE and collection of the coefficient of $v$ give the Riccati differential equationThe terminal condition also requires $B(T)=0$.

Apply the multidimensional Itô formula to ${\xi }_t=F(t,S_t,v_t)$. The stated PDE cancels its drift to $rFdt$, leavingConsequently $e^{-rt}{S}_t$ and $e^{-rt}{\xi }_t$ are local martingales under the physical measure $P$. Thus $P$ itself is an equivalent local martingale measure for the augmented market relative to the bank account. The continuous-time fundamental theorem of asset pricing says that existence of such a measure for locally bounded prices implies no free lunch with vanishing risk, and hence no arbitrage. The terminal condition also gives ${\xi }_T=\sqrt{S_T}$ as required.

Risk-neutral valuation givesFor $s\ge t$,ThereforeThe future ${\zeta }_j$ are independent and identically distributed under the stated model, sowith an empty product equal to one.

If $T\mapsto P_t^T$ is nonincreasing, thenso $1+r_t=(P_t^{t+1}{)}^{-1}\ge 1$ and $r_t\ge 0$.

Conversely, if every spot rate is nonnegative, then $B_{T+1}=B_T(1+r_T)\ge B_T$. Under a risk-neutral measure,Hence

The one-period spot interest rate is defined byand the bank account byA probability measure $Q$ equivalent to the physical measure is a risk-neutral measure when every discounted zero-coupon bond priceis a $Q$-martingale. Equivalently,

At maturity, $P_T^T=1$. If $P_t^T\le 0$ on an event $A\in {\mathcal{F}}_t$ of positive probability, buying one bond on $A$ has nonpositive cost and certain payoff $1$ on $A$ at $T$; any negative purchase cost can also be consumed or retained. This is an arbitrage. Therefore absence of arbitrage implies