Identify a line with its unique parameter pair in $E=(0,\infty )\times [0,2\pi )$, usingA Poisson line process means a Poisson point process on the Borel parameter space $E$: for each measurable $B$ of finite mean measure, the number of its parameter points is Poisson with mean $\mu (B)$, and counts in finitely many disjoint sets are independent. Infinite-measure sets are handled by a countable finite-measure exhaustion. Here $\mu (dp,d\theta )=dpd\theta$ is sigma-finite and nonatomic, so there are countably many distinct lines almost surely. It is not an assertion that line counts are Poisson with respect to planar area.

Fix $r>0$ and let $D_r=\{z:|z|<r\}$. Exactly the lines with $0<p<r$ hit this disk, apart from the zero-probability tangent case. Thus the count of disk-hitting lines isEvery intersection in $D_r$ uses two of these lines. Writing $N_r=\#(\Phi \cap D_r)$, we haveThis also proves local finiteness of the intersection point process. Initially it gives a non-strict bound $\mathbb{P}(N_r\ge 1)\le 1-(1+2\pi r)e^{-2\pi r}$.

To establish strictness, choose the two parameter boxesEach has positive measure, and both lie among the disk-hitting lines. If one line from each box intersected at $z\in D_r$, thenBut their distance parameters differ by more than $r/4$, a contradiction. The event of exactly one line in each box and no other disk-hitting lines has probabilityOn this event $M_r=2$ but $N_r=0$. ThereforeThe following original diagram illustrates why two disk-hitting lines need not produce an intersection inside the disk.

To decide whether $\Phi$ is Poisson, compute its intensity measure and compare its void probability. Use the equivalent signed line coordinates $L(t,\theta )$ with $t\in \mathbb{R}$ and $0\le \theta <\pi$. If $t>0$ this is the original pair $(t,\theta )$; if $t<0$ it is $(-t,\theta +\pi )$. The measure is $dtd\theta$, with no extra factor of two. Translation by $a$ sends $t$ to $t+a\cdot n_{\theta }$, preserving this measure. Thus both the line and intersection processes are stationary.

Almost surely no two lines are parallel and no three are concurrent. In each finite-intensity parameter region, conditional line parameters have continuous joint densities. Parallel pairs and concurrent triples satisfy lower-dimensional equations and have probability zero; a countable exhaustion proves the assertion for all lines. Hence different unordered pairs have distinct intersections almost surely. For a bounded Borel set $A$, the second Poisson factorial moment measure identity givesFor nonparallel normals, their intersection $z$ satisfies $t=z\cdot n_{\theta }$, $s=z\cdot n_{\varphi }$. The absolute Jacobian determinant of this transformation is $|\sin (\varphi -\theta )|$. Consequentlysince the inner integral is $2$ for every $\theta$. In particular $\mathbb{E}{N}_r={\pi }^2{r}^2$.

If $\Phi$ were a Poisson point process, the count $N_r$ would therefore be Poisson with that mean, giving $\mathbb{P}(N_r=0)=e^{-{\pi }^2{r}^2}$. But the earlier necessary-two-lines argument givesAt $r=2/\pi$ this lower bound is $5e^{-4}$, whereas the supposed Poisson void probability is $e^{-4}$. This contradiction proves $\Phi$ is not a Poisson point process. Its points share parent lines, and the explicit void/mean comparison demonstrates the resulting dependence rather than merely asserting it.

The intensity function of a point process is positive and continuous inside the interval, so its measure is finite on every compact subinterval. Consequently the Poisson point process is locally finite there. Its nonatomic intensity makes it simple and gives no point at $0$ almost surely. At the endpoints,Both half-intervals therefore have infinite intensity. More explicitly, exhaust either half by increasing compact intervals with means $t_j\to \infty$. For any fixed integer $k$,Taking the countable union over $k$ proves infinitely many points on each side almost surely.

Local finiteness excludes accumulation in the interior, including at $0$. Hence there is a largest negative point and a smallest positive point. For example, given any negative point, the compact interval between it and $0$ contains finitely many points and therefore a largest one. Call it $X_0$; call the smallest positive point $X_1$, and enumerate consecutively in both directions. Infinite counts give every integer index, and the lack of interior accumulation implies $X_n\uparrow 1$ as $n\to \infty$ and $X_n\downarrow -1$ as $n\to -\infty$. Thus the specified two-sided labeling exists almost surely.

Use the cumulative-intensity transformationIt has $f(0)=0$, positive derivative, and limits $-\infty ,+\infty$ at the two endpoints, so it is an increasing bijection onto $\mathbb{R}$. One explicit expression isFor any bounded interval $(a,b)$ in the target, its preimage has intensityDisjoint target sets have disjoint preimages, so their counts are independent. This proves directly, or by the Poisson mapping theorem, that the transformed points form a unit-rate Poisson process on the real line.

Put $Y_n=f(X_n)$. On the positive half-line, the successive spacings are independent exponential random variables of mean $1$, so $Y_n=E_1+\cdots +E_n$. The strong law of large numbers gives $Y_n/n\to 1$ almost surely. On the negative half-line, independently, the reflected points satisfy $-Y_{-k}=E_1^{\prime }+\cdots +E_{k+1}^{\prime }$, because $Y_0$ is the first point to the left of zero. Thus $-Y_{-k}/k\to 1$, or $Y_n/n\to 1$ also as $n\to -\infty$. The shift by one has no effect on these limits.

From the explicit formula, or by integrating the endpoint intensity asymptotics,Substitute $x=X_n$ and use the strong law for Poisson arrival times. At the right endpoint,soAt the left endpoint the corresponding result isEquivalently $8|n|(1+X_n)\to 1$. The different powers of $|n|$ reflect the different endpoint singularity orders, an instance of power-law boundary accumulation of a Poisson point process.

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.

Define the total as the nonnegative sum $W={\sum }_{(z,w)\in \Pi }w$, or equivalently as the increasing limit of its finite-intensity truncations. the Campbell theorem givesSince $M<\infty$, the nonnegative random variable $W$ cannot be infinite on an event of positive probability. Thus $W$ is finite almost surely, even if the number of plants is infinite.

Separate the square into its diagonal and ordered distinct-pair terms. Nonnegativity allows the use of Tonelli's theorem and increasing limits, givingThe assumed second integral is finite, so $W$ is square-integrable. Subtracting $(\mathbb{E}W{)}^2$ yieldsThe term $v$ accounts for random weights at fixed locations; the term $m^2$ accounts for the Poisson fluctuations in the locations and count. Nonnegativity is natural for weights. For signed marks, absolute first-moment integrability, rather than just a conditionally convergent integral of the signed mean, would be needed for the ordinary sum.

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

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.

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 $p_c$:

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.