Finite FK association, spatial Markov property, and boundary monotonicity

Form the quotient vertex set $V_{\xi }$ by identifying each $\xi$-class. For an edge with distinct quotient endpoints $u,v$, let its incidence vector be $e_u+e_v\in {\mathbb{F}}_2^{V_{\xi }}$; a quotient loop has zero vector. In a component $C$ of the opened quotient graph $(V_{\xi },A)$, every edge vector has even coordinate sum, so their span lies in the even-sum subspace, of dimension $\mathrm{\mid }C\mathrm{\mid }-1$. Conversely choose a root $r\in C$. For every $v\in C\setminus \{r\}$, connectedness [F2] gives an open path from $r$ to $v$; summing its edge vectors over ${\mathbb{F}}_2$ cancels each internal vertex twice and gives $e_r+e_v$. These $\mathrm{\mid }C\mathrm{\mid }-1$ vectors are linearly independent by their distinct $v$-coordinates. Thus the incidence rank on $C$ is $\mathrm{\mid }C\mathrm{\mid }-1$, and summing over components gives $r(A)=\mathrm{\mid }{V}_{\xi }\mathrm{\mid }-k_{\xi }(A)$, including isolated vertices and quotient loops. If $U_A$ and $U_C$ are the spans of incidence vectors from edge sets $A$ and $C$, then $U_{A\cup C}=U_A+U_C$ and $U_{A\cap C}\subseteq U_A\cap U_C$. The subspace dimension formula therefore gives $r(A)+r(C)\ge r(A\cup C)+r(A\cap C)$, hence $k_{\xi }(A\cup C)+k_{\xi }(A\cap C)\ge k_{\xi }(A)+k_{\xi }(C)$.

Fix the exterior edge states outside a finite $F\subseteq E$. For an interior edge set $A\subseteq F$, the total number of components of the combined graph is the number of exterior-only components disjoint from the boundary of $F$, plus the number of components of the graph on $F$ after identifying its boundary vertices that are connected by the fixed exterior open edges or by $\xi$. The first summand and all exterior edge factors are constant in $A$; they cancel when the conditional weights are normalized. Hence the conditional law on $F$ is exactly the finite FK law with this induced boundary partition.

Since $\mathrm{\mid }A\cup C\mathrm{\mid }+\mathrm{\mid }A\cap C\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }C\mathrm{\mid }$, the factors $p^{\mathrm{\mid }A\mathrm{\mid }}(1-p{)}^{\mathrm{\mid }E\setminus A\mathrm{\mid }}$ have equality in the lattice condition. The component inequality in step 1.1 and $q\ge 1$ give the displayed log-supermodular inequality; every mass is positive by $0<p<1$ and $q\ge 1$.

Given all edges except $e$, let $s$ be the conditional number of components when $e$ is closed. If its endpoints are already connected, opening it leaves $s$ unchanged, so open-to-closed odds are $p\mathrm{/}(1-p)$ and the open probability is $p$. If the endpoints are not connected, opening it changes the component factor from $q^s$ to $q^{s-1}$, so the odds are $p\mathrm{/}[q(1-p)]$ and the open probability is $p\mathrm{/}(p+q(1-p))$. Both expressions increase with $p$; opening other edges or making the boundary partition coarser can only turn the second case into the first.

We prove positive association for a strictly positive log-supermodular measure on $\{0,1{\}}^n$ by induction on $n$. It is trivial at $n=0$. For the last coordinate $s\in \{0,1\}$, each conditional measure ${\mu }_s$ on the first $n-1$ coordinates remains log-supermodular, so induction makes each ${\mu }_s$ positively associated. The ratio $R(x)=\mu (X_n=1,x)\mathrm{/}\mu (X_n=0,x)$ is increasing in $x$: for $x\le y$, apply log-supermodularity to $(y,0)$ and $(x,1)$ to obtain $R(y)\ge R(x)$. Thus for every increasing $h$, ${\mathbb{E}}_{{\mu }_1}h={\mathbb{E}}_{{\mu }_0}(hR)\mathrm{/}{\mathbb{E}}_{{\mu }_0}R\ge {\mathbb{E}}_{{\mu }_0}h$ by positive association under ${\mu }_0$. For increasing $f,g$, their conditional means $F_s=\mathbb{E}[f\mid X_n=s]$ and $G_s=\mathbb{E}[g\mid X_n=s]$ therefore satisfy $F_1\ge F_0$ and $G_1\ge G_0$. The conditional association and the nonnegative covariance of two increasing functions of the single Bernoulli coordinate give $\mathbb{E}(fg)\ge {\sum }_s\mu (X_n=s)F_s{G}_s\ge \mathbb{E}f\mathbb{E}g$.

For two parameter/boundary choices ordered by $p_1\le p_2$ and by refinement ${\xi }_1\le {\xi }_2$, run random-scan heat-bath chains from the all-closed configuration, choosing the same edge and uniform random number at each update. Each update resamples one edge from its conditional FK law, so it preserves the corresponding FK law. Step 2.2 makes every coupled update preserve coordinatewise order. To prove marginal convergence, fix $j\in \{1,2\}$ and couple two copies of the chain with this same fixed kernel $(p_j,q,{\xi }_j)$, one from any configuration and one from its stationary FK law. If there are $r>0$ labels, a block of $r$ updates chooses them in a prescribed order with probability $r^{-r}$. If all its uniforms exceed $p_2<1$, every label is closed at the block end in both copies, regardless of their initial configurations. This event has probability $\delta =r^{-r}(1-p_2{)}^r>0$. Once these same-kernel copies meet they remain equal under shared updates. Independence of the update blocks bounds their disagreement probability after $k$ blocks by $(1-\delta {)}^k$, proving convergence of each marginal to its FK law. No absorbing equality is claimed between the different-kernel chains. For every increasing real $f$ on the finite configuration space, their order at every time gives $\mathbb{E}f(X_t^{(1)})\le \mathbb{E}f(X_t^{(2)})$; taking the two marginal limits proves stochastic domination. When $r=0$ there is one configuration and the conclusion is immediate.