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For a strictly positive law $\mu$ satisfying the FKG lattice condition on a finite Boolean lattice, and an increasing function $g$, tilt to ${\mu }_t(\omega )\propto e^{tg(\omega )}\mu (\omega )$. For $t\ge 0$, the lattice inequality and $g(\omega \vee \eta )\ge g(\omega )$ give the Holley condition for ${\mu }_t{\ge }_{\mathrm{st}}\mu$. For any increasing $f$, differentiate ${\mathbb{E}}_{{\mu }_t}f\ge {\mathbb{E}}_{\mu }f$ at $t=0$ from the right. The derivative is ${\mathrm{Cov}}_{\mu }(f,g)$, proving positive association of random variables.

For $H(\sigma )={\sum }_{xy}{J}_{xy}{\sigma }_x{\sigma }_y+{\sum }_x{h}_x{\sigma }_x$ on a finite graph, assume $J_{xy}\ge 0$. Let $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. The displayed difference is $4{\sum }_{xy:x\in U,y\in W\text{ or conversely}}{J}_{xy}$: all other edge terms and all field terms cancel. Exponentiating proves the FKG lattice condition and hence positive association of random variables for this Ising model.

A fixed Ising boundary condition pins the Ising spins on a specified boundary, leaving the interior distributed with the original interaction weights. The boundary can be the inner graph vertex boundary of a finite box, or an exterior neighboring set if that convention is explicitly chosen. Conditioning on pinned Ising spins preserves the FKG lattice condition of a ferromagnetic Ising model. The FKG inequality then makes interior expectations of increasing observables nondecreasing in the pinned boundary Ising spins.

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.

Work edge by edge. Let $\omega ,\eta$ have two-bit restrictions to an edge. If these restrictions are comparable, taking their meet in a lattice and join in a lattice only reorders the two pairs, so their total equality indicators do not change. If they are incomparable, they are $01$ and $10$, while their meet in a lattice and join in a lattice are $00$ and $11$. In that case the sum of equality indicators increases from zero to two. ConsequentlyThe partition functions cancel from the required ratio, leavingThis is the ferromagnetic Ising lattice inequality in binary-spin normalization, so the measure satisfies the FKG lattice condition and is positively associated. If $\sigma (x)=2\omega (x)-1$, then ${\delta }_{\omega (x),\omega (y)}=(1+\sigma (x)\sigma (y))/2$: the conventional $\{-1,1\}$ coupling is $\beta /2$, which explains the factor two in the edge calculation.

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.

Order $\Omega =\{0,1{\}}^E$ coordinatewise. The statement ${\mu }_1{\ge }_{\mathrm{st}}{\mu }_2$ meansfor every real-valued order-preserving function $f$; equivalently, ${\mu }_1(A)\ge {\mu }_2(A)$ for every increasing event $A$. This is stochastic domination of probability measures. For strictly positive laws, the Holley condition is the sufficient conditionHere join and meet are coordinatewise maximum and minimum. The Holley inequality states that this condition implies the stated stochastic domination.

For a single positive law, the FKG lattice condition isWe prove that it gives positive association of random variables, namely ${\mathrm{Cov}}_{\mu }(f,g)\ge 0$ for all increasing $f,g$. Fix such a $g$, and for $t\ge 0$ form the strictly positive tilted lawThe lattice condition and $g(\omega \vee \eta )\ge g(\omega )$ implyThus the Holley condition applies to ${\mu }_t,\mu$, and ${\mathbb{E}}_{{\mu }_t}f\ge {\mathbb{E}}_{\mu }f$. Equality holds at $t=0$, so the right derivative there is nonnegative. Differentiating the finite sums givesHencewhich proves the requested positive association of random variables. Neither $f$ nor $g$ needs to be nonnegative; the finite space makes all differentiations justified. This is the exponential-tilt proof of positive association.

The law is a ferromagnetic Ising model with unit couplings and no field. Its masses are strictly positive. Put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, so $\pi (\sigma )=Z^{-1}{e}^{H(\sigma )}$. The order is coordinatewise with $-1<1$, which identifies the configuration space with a Boolean lattice.

Use the permitted reduction to pairs that disagree at no more than two graph vertices. If such a pair is comparable, its join and meet are the two original configurations, so the lattice inequality is equality. In the incomparable case, let the two differing graph vertices be $x,y$, with one configuration taking $(1,-1)$ there and the other $(-1,1)$. Their join takes $(1,1)$ and their meet $(-1,-1)$.

Every edge not joining $x$ to $y$ cancels ina fixed edge has the same contribution throughout, while an edge from a changed graph vertex to a fixed graph vertex has the same sum of its two endpoint-spin products before and after. An edge between $x$ and $y$ contributes $1+1-(-1)-(-1)=4$. Thus the difference is nonnegative, whether or not the two changed graph vertices are adjacent. Exponentiation, with the normalizing constants cancelling, proves the FKG lattice condition for these pairs, and hence for all pairs by the given reduction.

Equivalently one can see the full ferromagnetic Ising lattice inequality directly. Let $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. Exactly the edges between $U$ and $W$ contribute, each by four. ThereforeThe preceding part now givesso the measure is positively associated. The sign of the coupling is essential to this argument: positive couplings favor agreement and give the nonnegative lattice difference.

Order configurations coordinatewise. The FKG lattice condition on the masses of a finite Boolean lattice isThe FKG inequality states that under this condition, increasing real-valued functions $f,g$ satisfy $\mathbb{E}[fg]\ge \mathbb{E}[f]\mathbb{E}[g]$. In particular increasing events satisfy $P(A\cap B)\ge P(A)P(B)$. For a product measure with coordinate weights $w_e(0),w_e(1)$, the pair $(\min ({\omega }_e,{\eta }_e),\max ({\omega }_e,{\eta }_e))$ has the same two entries as $({\omega }_e,{\eta }_e)$. Multiplying over coordinates gives equality in the lattice condition, including degenerate Bernoulli parameters without division by zero.

For independent bond percolation on the nearest-neighbor cubic lattice, write ${\theta }_d(p)={\mathbb{P}}_p(|C(0)|=\infty )$. Define the percolation critical probability and the connective constant bywhere $c_n(d)$ counts rooted $n$-step self-avoiding walks. Translation invariance makes the root irrelevant, and part (i) proves existence of the latter limit.

First work on the square lattice, whose dual is another translated square lattice. Put $q=1-p$ and assume $q\mu (2)<1$. If the open cluster at the origin is finite, its exterior edge boundary contains a closed dual simple circuit surrounding the origin. To see this planar fact, surround its finitely many vertices by their unit square cells and follow the exterior boundary: its crossing primal edges are closed. Resolving repeated boundary vertices into simple circuits leaves a circuit separating the origin from infinity.

Let $a_n$ count such dual circuits of length $n$. Every one crosses the positive horizontal ray at distance at most $n$, because it surrounds the origin and its horizontal span is at most its length. Choosing a ray-crossing edge, an orientation and all but the last edge encodes it by one of at most $Cn$ rooted length-$n-1$ self-avoiding walks. Thus $a_n\le Cn{c}_{n-1}(2)$. Choose $b>\mu (2)$ with $qb<1$. The root-count limit gives $c_{n-1}(2)\le C_b{b}^n$, soThis summable tail alone need not make the probability of every enclosing circuit less than one. Choose a large $R$ and let $A_R$ be the absence of enclosing closed dual circuits of length at least $R$, with $P(A_R)>1/2$. Let $B_R$ require every primal edge inside $[-R,R{]}^2$ to be open. It has positive probability. Both events are increasing. The finite-measure Harris-FKG inequality extends to $A_R$ by decreasing limits over finitely many circuit exclusions, so $P(A_R\cap B_R)\ge P(A_R)P(B_R)>0$.

On $B_R$ the origin is connected to every graph vertex of that box. Any closed dual circuit enclosing the origin must then enclose the whole box, hence have length at least $R$. On $A_R\cap B_R$ no such circuit exists, so the origin cluster is infinite. This proves the connective-constant Peierls bound, $p_c(2)\le 1-1/\mu (2)$. Finally the lattice in dimension $d\ge 2$ contains a coordinate copy of the square lattice, whose edge law is unchanged. Percolation in that subgraph implies percolation in the full graph, and hence