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For a nonnegative integer-valued random variable $X$, its probability generating function isinitially defined at least for $|z|\le 1$. A general real-valued random variable need not have a probability generating function in this sense. For a binomial distribution, the binomial theorem givesDifferentiating the generating function produces factorial moments:Since $X^2=X(X-1)+X$, the mean and variance areThese formulas include the degenerate endpoint probabilities and $n=0$.

For independent random variables, their sum has the product of their probability generating functions:Comparing polynomial coefficients identifies a binomial distribution with parameters $n+m,p$:and the probability is zero at all other values. The common success probability is essential to this binomial form; independent binomial counts with different probabilities do not in general add to a binomial count.

The transformation is strictly increasing on $0\le x<1$, has range $[0,\infty )$, and its inverse is $x=y/(y+3)$. Therefore the cumulative distribution function isDifferentiation gives the probability density functionIts value at zero may be assigned arbitrarily without changing the distribution. There is no atom at zero or infinity. The expression is undefined at $X=1$, but that event has probability zero under the continuous uniform distribution, so any convention there leaves the law unchanged. Finally ${\int }_0^{\infty }3/(y+3{)}^{2}dy=1$, as required.

The inclusion-exclusion principle givesFor the coat problem, all $n!$ permutations are equally likely. If a specified set of $j$ people gets its own coats, the other coats can be assigned in $(n-j)!$ ways, so that intersection has probability $(n-j)!/n!$. Taking the complement of the union of correct-coat events gives the derangement probabilityTo have exactly $m$ fixed coats, choose their recipients and derange all remaining coats. Thus the fixed point count of a uniform random permutation hasand zero probability outside this range. The empty permutation has $D_0=1$, so the formula includes $m=n$; it also gives zero for $m=n-1$, since a lone remaining coat cannot be misplaced.

For fixed $m$, $n-m\to \infty$, and the exponential series givesThese are the probabilities of a Poisson distribution of mean one. The limit is for fixed $m$, not for $m$ increasing with $n$.

Let $X_t$ count $t$-cliques, and put ${\mu }_t=\left(\genfrac{}{}{0ex}{}{n}{t}\right)p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Since ${\mu }_{r+1}\to 0$, the Markov inequality gives $\mathrm{Pr}(X_{r+1}>0)\to 0$. Any larger clique contains an $(r+1)$-clique, so it remains to prove $X_r>0$ with high probability.

We give the required growing-clique overlap bound explicitly. Set $\mu ={\mu }_r$ and, for $2\le s\le r$,The quantity $a_s$ is the probability that a uniformly chosen $r$-set overlaps a fixed $r$-set in exactly $s$ vertices. Two clique indicators share exactly $\left(\genfrac{}{}{0ex}{}{s}{2}\right)$ edge requirements, so their joint-to-product ratio is $p^{-\left(\genfrac{}{}{0ex}{}{s}{2}\right)}$. Overlap zero or one gives independence. Hence the overlap formula for the variance of a clique count impliesWe must bound this sum with $r$ allowed to grow.

Eventually ${\mu }_r\ge 1$ and ${\mu }_{r+1}\le 1$. Using $(n/t{)}^t\le (\genfrac{}{}{0ex}{}{n}{t})\le (en/t{)}^t$, these inequalities giveFor $2\le s\le r/2$, a union bound over the $s$ selected shared vertices gives $a_s\le \left(\genfrac{}{}{0ex}{}{r}{s}\right)(r/(n-r){)}^s$. Put $a=(s-1)/(r-1)<1/2$. Since $n\ge r^3$ and $n-r\ge n/2$,For each fixed $s$, the first bound tends to zero if $r$ stays bounded, because its power of $n$ is negative. If $r\to \infty$, the middle bound tends to zero because $a\to 0$. Thus each fixed-$s$ term tends to zero along arbitrary admissible sequences. The bounding series ${\sum }_{s\ge 2}(2e^2/s{)}^s$ converges. Splitting at a fixed large index, then taking $n\to \infty$ and the index to infinity, proves that the entire lower-half sum is $o(1)$.

For the upper half write $s=r-j$. The exact diagonal comparison isFor $1\le j\le r/2$, the second bound on $p$ givesThe last inequality uses $n\ge r^3$: the exponent of $n$ in the bracket is negative, so substituting $n=r^3$ bounds it above. Uniformly for this range of $j$, $r^{2(j+1)/r}\le 9j$. Indeed $r^{2/r}\le 3$, and concavity of $\log (3j)-(2j/r)\log r$, checked at $j=1$ and $j=r/2$, gives $r^{2j/r}\le 3j$. It follows thatThus $\mathrm{Var}{X}_r=o({\mu }^2)$. The second moment method gives $\mathrm{Pr}(X_r=0)\to 0$. Combining this with the absence of larger cliques provesThe restriction on $r$ is used in both overlap ranges; divergence of a clique count in a binomial random graph alone would not suffice for arbitrary growing $r$.

Couple the uniform random graph process by giving every possible edge an independent uniform label in $[0,1]$ and revealing them in increasing order. The graph at label threshold $p$ is a binomial random graph. Define ${\tau }_0$ as the first step with no isolated vertices and ${\tau }_c$ as the first connected graph. Deterministically ${\tau }_c\ge {\tau }_0$. We prove the reverse inequality with high probability by tracking the same process through a short interval.

Put $w=\frac{1}{4}\log \log n$, $p_{\pm }=(\log n\pm w)/n$, and $L=n{p}_{-}$. At $p_{-}$ the isolated vertex count $Z$ hasThe variance estimate follows by comparing two isolated-vertex indicators, whose joint-to-product probability ratio is $(1-p_{-}{)}^{-1}$. Thus $Z>0$ with high probability. Also the Markov inequality gives $Z\le \log n$ with probability tending to one.

At this same threshold, exclude every graph component with order $2\le s\le n/2$. For $2\le s\le n/\log n$, any such component contains a spanning tree. The Cayley formula and a union bound therefore giveIn this range the bracket is at most $A=e^{2}L{e}^w/n=o(1)$. Summing over $s\ge 2$ gives $O(n{A}^2/L)=O(L{e}^{2w}/n)=o(1)$. For $n/\log n<s\le n/2$, even the cut-absence bound is enough:The bracket tends to zero, so this range also has total probability $o(1)$. Consequently the graph at $p_{-}$ consists of one component of order greater than $n/2$ and its isolated-vertex set $S$, where $1\le |S|\le \log n$ with high probability.

Conditional on the entire graph at $p_{-}$, every unrevealed edge label remains independently uniform on $(p_{-},1)$. In particular,There are no isolated vertices at $p_{+}$ with high probability, because their expected value is asymptotic to $e^{-w}\to 0$. On the intersection of these events, each vertex of $S$ loses isolation by joining the original large component, never by forming a separate small component with another vertex of $S$. At every step between these two thresholds there is therefore one nontrivial component and some isolated vertices. The final isolated vertex disappears precisely when the graph becomes connected. This proves the connectivity hitting time equals disappearance of isolated vertices result:

For the critical-window probability, let $p=(\log n+\alpha (n))/n$ with $|\alpha (n)|\le B$. For each fixed $j$, the falling factorial moment of the isolated-vertex count isuniformly for such shifts. Apply the Bonferroni inequalities to the zero-count event. Its odd and even truncations bracket $\mathrm{Pr}(Z=0)$ by partial sums of ${\sum }_{j\ge 0}(-1{)}^j{e}^{-j\alpha (n)}/j!$. Since $e^{-\alpha (n)}\le e^B$, their remainders tend uniformly to zero. Hence $\mathrm{Pr}(Z=0)=e^{-e^{-\alpha (n)}}+o(1)$. The process equality just proved changes this probability by at most its $o(1)$ failure probability. ThereforeThe ratio follows because the denominator is bounded below by $e^{-e^B}>0$; no convergence of $\alpha (n)$ is needed.

The printed second conclusion has the complementary denominator and is false as written. For $\alpha (n)=0$, the printed ratio tends to $e^{-1}/(1-e^{-1})=1/(e-1)\ne 1$. Its denominator $1-e^{-e^{-\alpha (n)}}$ is correct for the probability of being disconnected. Thus either remove the leading $1-$ in the denominator or replace “connected” by “disconnected”; the proof supplies both corrected versions.

Put $M=2m$ and $a_i=\left(\genfrac{}{}{0ex}{}{d_i}{2}\right)$. Use the configuration model: give vertex $i$ exactly $d_i$ labelled half-edges and choose a uniform perfect matching of all $M$ half-edges. There arepairings, with $(2m{)}_m$ denoting a falling factorial. Each allowed simple graph has exactly ${\prod }_i{d}_i!$ representations: at every vertex, assign its incident graph edges bijectively to its labelled half-edges. Therefore its enumeration reduces to the probability that a pairing is simple and avoids $G_0$.

Count three kinds of prescribed bad patterns. A loop pattern consists of a pair of half-edges at one vertex. A double-edge pattern consists of two half-edges at each of two distinct vertices and one of their two cross-pairings. A forbidden-edge pattern consists of one half-edge at each endpoint of an edge of $G_0$. Let their respective counts in the random pairing be $Z_L,Z_D,Z_F$. Simplicity and avoidance are equivalent to $Z=Z_L+Z_D+Z_F=0$, including exclusion of triple edges because those contain double-edge patterns.

A prescribed collection of $t$ disjoint half-edge pairs occurs with probabilityConsequently the bad-pattern expected values areHere $\lambda =({\sum }_i{a}_i)/m$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$, as in the PDF. Since $1\le d_i\le \Delta$ and $G_0$ has bounded maximum degree, $M=\Theta (n)$ and all three means are uniformly bounded. For the double-edge estimate, ${\sum }_i{a}_i^2=O(n)$, so its contribution after division by $M^2$ is $o(1)$.

We now give the moment-counting step rather than assume independence. For each fixed $h$, expand the factorial moment $\mathbb{E}(Z{)}_h$ as the sum over ordered distinct bad patterns. Collections on disjoint vertex sets have, by the displayed pairing probability, joint probabilities equal to the product probabilities times $1+O(1/n)$. Their total contribution isDeleting collections with a shared vertex changes the independent-product sum by $O(1/n)$, since all ordinary and forbidden vertex degrees are bounded.

For actual joint probabilities of overlapping patterns, incompatible uses of a half-edge give zero. Compatible collections with no shared paired edge lose at least one free vertex choice, without reducing the number of required pairs, so contribute $O(1/n)$. A shared paired edge can occur only between two double-edge patterns or between a double-edge pattern and a forbidden-edge pattern. In the first case their union requires at least three parallel paired edges: there are $O(n^2)$ endpoint choices and probability $O(n^{-3})$. In the second case the endpoint pair must be an edge of $G_0$: there are only $O(n)$ choices and at least two required pairs, giving $O(n^{-1})$. Adding further patterns in a fixed collection does not remove this deficit: a disjoint addition contributes bounded order, while an addition to an existing cluster has only bounded half-edge choices unless it supplies new free vertices, compensated by its new required pairs. Summing the finitely many overlap types for fixed $h$ therefore gives $o(1)$. Thus, with $\Lambda =\lambda /2+{\lambda }^2/4+\mu$,The Bonferroni inequalities bracket $\mathrm{Pr}(Z=0)$ by the odd and even partial sums of ${\sum }_h(-1{)}^h\mathbb{E}(Z{)}_h/h!$. The means $\Lambda$ are bounded, so the corresponding exponential-series remainders vanish uniformly as the truncation increases. Taking $n\to \infty$ and then the truncation to infinity proves the bounded-degree pairing avoidance estimate $\mathrm{Pr}(Z=0)=e^{-\Lambda }+o(1)$. This does not require $\lambda$ or $\mu$ to converge. The exponential is bounded away from zero, so the additive error is also a relative $o(1)$ error.

Multiplying by the pairing count and dividing by the representation factor givesThe factor $(2m{)}_m$ multiplies the exponential; it is not part of its exponent. The TeX aid misgroups this factor, whereas the original PDF is unambiguous. The bounded-degree proof also covers the stated additional assumption $2m-n\to \infty$.

Again put $Y=\mathbb{E}[X\mid \mathcal{G}]$. For any fixed real $c$, the random variableis nonnegative and integrable. The sign function of $Y-c$ is bounded and $\mathcal{G}$-measurable, so conditioning and equality of the probability distributions giveThus $D_c=0$ almost surely. On $\{Y>c\}$ this forces $X\ge c$; on $\{Y<c\}$ it forces $X\le c$; on $\{Y=c\}$ it forces $X=c$. In particular $\{Y=c\}\subseteq \{X=c\}$ up to a null set. These two events have equal probability, again by equality of the probability distributions, so they agree up to a null set. ConsequentlyTaking $c=0$ establishes the suggested sign assertion, but this alone would not exclude unequal positive values. Apply the same argument simultaneously to the countable set of rational $c$. Outside the union of their null sets, unequal $X$ and $Y$ would have a rational strictly between them, giving opposite signs. This proves the conditional expectation preserving a distribution result:Only integrability was used; no finite second moment or hidden square-integrability assumption is needed.

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

Use $T_y=\inf \{n\ge 0:X_n=y\}$ and the positive return time $T_x^{+}=\inf \{n\ge 1:X_n=x\}$. The walk has transitions $p_{uv}=c_{uv}/d(u)$, $d(u)={\sum }_v{c}_{uv}$; its transition matrix defines a reversible Markov chain, although the transition probabilities need not be symmetric.

The voltage $h(u)=P_u(T_x<T_y)$ equals one at $x$, zero at $y$, and is a harmonic function on a graph away from the terminals: $h(u)={\sum }_v{p}_{uv}h(v)$. For this unit electric potential difference the electric current isThe first-step analysis therefore proves the return probability from effective resistance identity:Finite irreducibility ensures eventual hitting of one of the terminals, so no missing escape event appears in this finite-network calculation.

For an infinite locally finite connected graph, exhaust it by finite sets and wire their exteriors to one sink. The event of exiting before the first return decreases to the event of never returning. Thuswith $1/\infty =0$. In particular recurrence is equivalent to infinite effective resistance to infinity.

After deleting an edge, consider any infinite component and choose a root $o$ away from that edge's endpoints. Its vertex degree $d(o)$ is unchanged. On every finite wired exhaustion the Rayleigh monotonicity principle makes the resistance no smaller; the limit of a sequence does too. A recurrent original network consequently leaves a recurrent component. Finite components give recurrent walks as well, with an isolated graph vertex viewed as an absorbing state. This proves that edge deletion preserves recurrence. The phrase “no less recurrent” refers to this recurrence comparison: at a deleted-edge endpoint the vertex degree factor also changes, so monotonicity of its first-return probability alone should not be inferred.

A self-avoiding walk of length $n$ is a sequence of $n+1$ distinct graph vertices joined consecutively by edges. Split a rooted self-avoiding walk of length $m+n$ after $m$ steps. There are ${\kappa }_m$ possible prefixes; each remaining segment is a self-avoiding $n$-step walk from its current graph vertex, and forbidding intersections with the prefix can only reduce its count below ${\kappa }_n$. Translation invariance therefore givesApply the Fekete lemma, proved in Question 4(a), to $\log {\kappa }_n$. It gives $\log \mu ={\inf }_n{n}^{-1}\log {\kappa }_n$ and hence existence of the connective constant. Walks using only the $d$ positive coordinate directions never revisit a graph vertex, so there are at least $d^n$. A walk has $2d$ choices for its first step and at most $2d-1$ subsequently because immediate reversal is forbidden. Thus

For independent bond percolation, let $C(0)$ be the percolation cluster of the origin, $\theta (p)=P_p(|C(0)|=\infty )$, and $p_c=\inf \{p:\theta (p)>0\}$. An infinite percolation cluster in a locally finite graph contains an open self-avoiding path of every length. The union bound givesWhen $p\mu <1$, the right-hand side tends to zero, since its $n$th root tends to $p\mu$. Hence the connective-constant lower bound for percolation is

For the upper bound in two dimensions, use the square planar dual graph. Call a dual edge available when its crossing primal edge is closed. Each dual edge is available with probability $q=1-p$, independently. The planar graph boundary lemma says that a finite primal percolation cluster is enclosed by a graph cycle consisting of such edges: its external edge boundary gives this graph cycle. We show that $q\mu <1$ guarantees positive percolation probability.

Let $N_n$ count enclosing graph cycles of length $n$. Because the origin lies inside their coordinate bounding box and their diameter is at most $n$, every such graph cycle has a graph vertex within a box of side $2n+3$. From each possible root, the first $n-1$ edges give a self-avoiding walk; the final closing edge is then determined. Dropping the closure restriction gives the convenient boundChoose $\rho$ with $q\mu <\rho <1$. The connective-constant limit of a sequence implies ${\kappa }_{n-1}{q}^n\le C{\rho }^n$. Consequently the large-circuit tailis summable.

Condition all primal edges inside a large square $[-L,L{]}^2$ to be open; this event has positive probability. The origin percolation cluster then contains that square. If it is finite, its enclosing graph cycle has length at least $L$. A proposed graph cycle using a pinned-open edge now has conditional probability zero; every other proposed graph cycle still has probability $q^n$ by independence of random variables. For $L$ large enough the union bound for the remaining graph cycles is less than one. There is therefore positive conditional probability that the origin percolation cluster is infinite. This proves the connective-constant Peierls bound without needing the exact value of $p_c$:

Order configurations coordinatewise: $\omega \le \eta$ when $\omega (e)\le \eta (e)$ for every coordinate. An increasing event $A$ is an upward-closed set: $\omega \in A$ and $\omega \le \eta$ imply $\eta \in A$.

The finite-lattice FKG inequality states that a probability mass function satisfying the FKG lattice conditionis positively associated. Here meet in a lattice and join in a lattice are coordinatewise minimum and maximum. Thus for increasing observables $f,g$,Taking indicator functions gives $\mu (A\cap B)\ge \mu (A)\mu (B)$ for all increasing events. The strictly positive version suffices for the Ising weights here; the theorem also holds for nonnegative log-supermodular weights of positive total mass.

With the inner boundary specified in the PDF, the Ising boundary condition pins $\omega {|}_{\partial \Lambda }=\zeta$. Explicitly ${\mu }_{\Lambda }^{\zeta }$ is ${\mu }_{\Lambda }$ conditioned on those Ising spins. Its partition function sums over unpinned interior Ising spins only; edges between two pinned sites contribute a constant that cancels on normalization.

Pinning coordinates preserves the FKG lattice condition, since meet in a lattice and join in a lattice still have the same pinned values. It also gives monotonicity in the pins. For any increasing observable $f$, positive association of random variables with an unpinned Ising spin $\omega (z)$ givesBoth conditional probabilities are positive. The same inequality holds after other Ising spins have been pinned. Changing pins one at a time proves stochastic domination of the measures as $\zeta$ increases.

Now let $\Lambda \subset {\Lambda }^{\prime }$ and take an increasing observable depending on a fixed set inside the smaller box. Given the Ising spins on $\partial \Lambda$, the larger-box measure induces precisely the smaller-box interior law with those boundary values: nearest-neighbor interactions cannot see farther once this boundary is fixed. Its conditional law is therefore between the all-zero and all-one smaller-box laws. Averaging yieldsThese are bounded monotone sequences along any increasing box exhaustion.

For every finite set $S$, the event that all Ising spins in $S$ are one is increasing, so its probability converges in both boundary sequences. Probabilities of arbitrary finite zero/one patterns are finite inclusion-exclusion combinations of these all-one probabilities, and hence converge too. The limits are consistent finite-dimensional distributions; they define measures on $\{0,1{\}}^{{\mathbb{Z}}^d}$ and give weak convergence of probability measures in the product topology. Thus both extremal infinite-volume Ising measures ${\mu }^0,{\mu }^1$ exist, with ${\mu }^0$ below ${\mu }^1$ in stochastic domination. This argument proves existence even when the two limits are different.

Write $\varphi ={\varphi }_{p,q}^1$ and $C_n=\varphi (0\leftrightarrow e_n)$. We use these standard facts about the infinite-volume wired random-cluster measure for $q\ge 1$: it is translation invariant and invariant under coordinate reflections/permutations; it has positive association of the random-cluster model; and every conditional edge-open probability is at leastThe last bound follows because an edge either leaves its component count unchanged on opening, giving probability $p$, or merges two components, giving $p/(p+q(1-p))$. These are standard infinite-volume properties of the limit obtained from the finite wired laws by weak convergence of probability measures. Positive association of random variables applies to connection events by increasing approximation with finite-path events.

Opening the $n$ edges of the straight segment gives $C_n\ge r^n$, so its logarithm is finite. The events $A=\{0\leftrightarrow e_m\}$ and $B=\{e_m\leftrightarrow e_{m+n}\}$ are increasing, and their intersection implies connection to $e_{m+n}$. ThusThe sequence $a_n=-\log C_n$ is a subadditive sequence, with $0\le a_n\le -n\log r$. Part (a) proves the random-cluster inverse correlation length exists and is finite:Since $a_n/n\ge \alpha$ for every $n$, exponentiating yields the requested bound

Let $B_n=\varphi (0\leftrightarrow \partial {\Lambda }_n)$ and $s_n=|\partial {\Lambda }_n|$. This connection event is the union over $x\in \partial {\Lambda }_n$ of $\{0\leftrightarrow x\}$. The union bound therefore supplies a graph vertex with connection probability at least $B_n/s_n$.

Every boundary graph vertex has some coordinate equal to $n$ or $-n$. By applying a permutation and a graph automorphisms, which preserve both the law and the boundary, choose such a graph vertex in the form $x=(n,x_2,\dots ,x_d)$ while retaining its probability bound. Now $e_{2n}-x=(n,-x_2,\dots ,-x_d)$ is a reflection of $x$. Translation and reflection invariance givePositive association of random variables applied to the two connection events, whose intersection implies $0\leftrightarrow e_{2n}$, proves the reflection comparison of random-cluster connections:Also $C_n\le B_n$, since any path to $e_n$ meets the box boundary. Consequentlyand henceHere $s_n=(2n+1{)}^d-(2n-1{)}^d=O(n^{d-1})$, so $\log s_n/n\to 0$. Both outside logarithmic connection rates converge to $\alpha (p,q)$ by part (b). Squeezing gives the rate for every box size, and therefore for the requested even subsequence:

In bond percolation on ${\mathbb{Z}}^d$, edges are independently open with probability $p$ and closed otherwise. An open percolation cluster is a connected component using only open edges. In site percolation it is the graph vertices that are independently declared open. The percolation probability $\theta (p)=P_p(0\leftrightarrow \infty )$ measures the chance that the origin lies in an infinite percolation cluster, and the percolation critical probability is $p_c=\inf \{p:\theta (p)>0\}$. Monotone coupling of Bernoulli percolation by independent uniform distribution makes $\theta$ nondecreasing.

The sharpness of the percolation transition says that below $p_c$ connection probabilities on ${\mathbb{Z}}^d$ decay exponentially with distance, while above $p_c$ the percolation probability is positive. There is almost surely a unique infinite percolation cluster in the supercritical regime. On the square lattice, the Harris-Kesten theorem gives bond threshold $1/2$, and there is no infinite percolation cluster at percolation critical probability. The emergence of connections on arbitrarily large scales is the geometric meaning of the phase transition.

Several observables describe this singular behavior. The percolation susceptibility is $\chi (p)=E_p|C(0)|$ for $p<p_c$. A directional correlation length can be defined through exponential two-point decay, $\xi (p{)}^{-1}=-{\lim }_n{n}^{-1}\log P_p(0\leftrightarrow n{e}_1)$. Percolation critical exponents describe leading powers, conveniently defined logarithmically rather than requiring exact asymptotic amplitudes:The first limit of a sequence approaches from above, the other two from below. At percolation critical probability, percolation two-point connection probability is expected to have leading power $|x{|}^{-(d-2+\eta )}$, and the number per site of percolation clusters of size $s$ has leading power $s^{-\tau }$. This cluster-count probability distribution differs from the size-biased probability distribution seen by choosing a site. Scaling relations for critical exponents such as $\gamma =(2-\eta )\nu$ and $2\beta +\gamma =d\nu$ connect the exponents in their applicable scaling regime; hyperscaling relation is not valid above the upper critical dimension of percolation in ordinary mean-field regimes.

The percolation universality hypothesis predicts that dimension and large-scale symmetry, rather than microscopic graph details, determine these exponents and scaling functions. Thus square-lattice bond and triangular-lattice site models are expected to share the planar graph universality class despite different microscopic definitions. Universality does not mean identical critical probabilities, amplitudes or finite-lattice distributions. Exact values and existence of exponents require model-specific theorems, not merely this prediction.

Two dimensions have an additional structure: conformal maps preserve angles, and critical crossing probabilities are expected to be unchanged under conformal transport of the domain and its marked boundary arcs. A crossing probability then depends only on the conformal shape of that marked quadrilateral. The Cardy boundary crossing formula gives this dependence in terms of its boundary cross ratio. For critical triangular-lattice site percolation, conformal invariance and the resulting exponents are established; examples are $\beta =5/36$, $\gamma =43/18$, $\nu =4/3$, with the rigorous statements in leading-power form. These values are summarized in Smirnov and Werner's original paper. Their universality extension to arbitrary planar lattices should not be presented as the same theorem.

An exploration path separating the two colors, with opposite boundary colors on two arcs, has a domain Markov property of a percolation exploration: conditional on the explored portion, the unexplored colors retain their independent laws with the newly exposed boundary colors. Conformal invariance, this domain Markov property and suitable regularity lead to Schramm–Loewner evolution. In the upper half-plane its chordal equation iswhere $B_t$ is standard Brownian motion and time is normalized by half-plane capacity. Critical percolation corresponds to $\kappa =6$. Thus the complicated discrete interface becomes a conformally natural random curve driven by one real Brownian motion. Its crossing and arm probabilities supply critical exponents and, through scaling relations for critical exponents, near-critical singularities. This makes conformal methods predictive tools for the transition, rather than only descriptions of an interface's appearance.

In the voter model on ${\mathbb{Z}}^2$, each site has a rate-one Poisson process clock; when it rings, that site copies one of its four neighbors, chosen uniformly. States are zero or one. For a cylinder function $f$, the local infinitesimal generator iswhere ${\eta }^{x\leftarrow y}$ agrees with $\eta$ except that the voter model opinion at $x$ is replaced by the voter model opinion at $y$. Only sites in the support of $f$ contribute, so this infinitesimal generator sum is finite.

The voter-model duality comes from independent Poisson processes of copying arrows of rate $1/4$ on each ordered neighboring pair. Trace the ancestors of specified sites backwards from time $t$: each ancestral line is a rate-one continuous-time random walk with uniform nearest-neighbor steps, and lines coalesce when they meet. The voter model opinions at time $t$ are the initial voter model opinions at those ancestral locations. This construction also explains the general duality theorem being used.

Two ancestral walks on ${\mathbb{Z}}^2$ meet almost surely. Until their meeting, their difference is a rate-two symmetric nearest-neighbor walk, and simple random walk in two dimensions is recurrent. If ${\tau }_{xy}$ is their meeting time, this givesfor every initial law $\nu$. If $\nu$ is invariant, the left side is the fixed probability $\nu (\eta (x)\ne \eta (y))$, so it is zero. This holds for every pair. Countability then implies that $\nu$ is concentrated on configurations with all voter model opinions equal. Both constant configurations are absorbing state. Thus the invariant measures of the two-dimensional voter model are exactlyNo translation-invariance assumption on $\nu$ was used.

For the last assertion, let the initial set of one-valued opinions $F$ be finite. Single-line duality givesEach transition probability tends to zero. One direct verification is the Fourier transform for the rate-one walk:The integrand has modulus at most one and tends to zero except at the single point $k=0$; dominated convergence proves the claim. For any finite observation set $K$, the union bound now givesTherefore every finite voter model opinion pattern converges in law to the all-zero pattern. These cylinder distributions determine weak convergence of probability measures in the compact product topology, proving finite-seed local extinction in the voter model and

Write $z=(x,y)$ and $\mu (dz)=\lambda (z)dz$. Treat the nonnegative plant weights as independent marks conditional on the locations, with a measurable mark law $K_z(dw)$ depending only on $z$. Its first and second moments areThis is the independent-marking interpretation appropriate for location-dependent weights.

The general results needed are the following. The Independent marking theorem for Poisson point processes says that such marks turn a Poisson point process of intensity $\mu$ into a Poisson point process on location-mark space with intensity $\nu (dz,dw)=\mu (dz)K_z(dw)$. The Campbell first-moment formula says that for any nonnegative measurable $g$, $\mathbb{E}{\sum }_{u\in \Pi }g(u)=\int gd\nu$, allowing infinite values. The second Poisson factorial moment measure identity says that for nonnegative measurable $h$,These statements apply to sigma-finite intensities and therefore do not require finitely many plants in the entire field. We use these general results without proving them.