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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.

This lattice condition on configuration masses is the hypothesis of the FKG inequality. A product measure on a finite Boolean lattice satisfies it with equality, since in each coordinate the minimum and maximum retain the original two coordinate values. It implies positive association of random variables for increasing functions.

A real-valued function on a partially ordered set is increasing when it preserves its order. Indicators of increasing events are examples. In spin or edge configuration spaces the order is coordinatewise, so increasing observables grow when a zero coordinate is replaced by one. Positive association of random variables concerns pairs of such observables, and stochastic domination compares their expectations.

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:

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.

Under the intended additional assumption of mutually independent, continuously calibrated null p-values, each test fails to reject with probability $0.95$. The familywise error rate is thereforeThere is a qualification to the printed use of “not correlated”: zero pairwise correlation coefficients do not generally imply mutual independence. A multivariate normal distribution with zero off-diagonal covariances does give independence, but an arbitrary biomarker distribution need not. The formula above requires independent rejection events, not just uncorrelated measurements.

The usual intuition for positively dependent tests is that they reject together, reducing the effective number of opportunities for a false positive. A rigorous sufficient assumption is positive association of random variables for the rejection indicators $I_1,\dots ,I_{10}$. Products of their non-rejection indicators are decreasing functions; association and induction giveThus under this stronger positive-dependence assumption the probability is no larger, and perfect dependence gives $0.05$ instead of $0.4013$. Strict inequality is not forced by every form of positive dependence.

Pairwise positive correlation coefficients alone, as printed, is insufficient. For a counterexample choose a vector of ten exchangeable random variables that are rejection indicators as follows: with probabilities $0.545,0.450,0.005$, respectively, reject none, reject one uniformly selected test, or reject all ten. Thenbut the familywise error rate is $0.455>1-0.95^{10}$. This can be realized with valid null p-values: conditional on the indicators, independently draw $p_i$ uniformly on $(0,0.05)$ if $I_i=1$ and on $(0.05,1)$ otherwise. Every $p_i$ is uniform on $(0,1)$, and $\mathrm{Cov}(p_i,p_j)=0.25\mathrm{Cov}(I_i,I_j)>0$. Taking biomarker statistics $X_i=1-p_i$ gives pairwise positively correlated null statistics with exactly these one-sided tests. Therefore the unconditional larger-or-smaller claim needs a specified dependence model.

Positive association of random variables and translation invariance give $C_{m+n}\ge C_m{C}_n$ for axis connection probabilities. The Fekete lemma applied to $-\log C_n$ proves the displayed limit of a sequence and the bound $C_n\le e^{-n\alpha }$. The lower single-edge conditional bound gives $C_n\ge [p/(p+q(1-p)){]}^n$, so the decay rate is finite and nonnegative. It may be zero.

Choose a graph vertex $x=(n,x_2,\dots ,x_d)$ on the boundary of a lattice box. Translation and reflection invariance identify the probabilities of $0\leftrightarrow x$ and $x\leftrightarrow 2n{e}_1$. Their intersection implies $0\leftrightarrow 2n{e}_1$, so positive association of random variables gives $C_{2n}\ge \varphi (0\leftrightarrow x{)}^2$. A union bound and lattice symmetries choose such an $x$ with probability at least the box-boundary connection probability divided by the number of boundary graph vertices. This polynomial factor disappears in the exponential rate, identifying boundary and axial inverse correlation lengths.