Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 37 / 5

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.