FKG inequality

Let ${\displaystyle X}$ be a finite distributive lattice, and μ a nonnegative function on it, that is assumed to satisfy the (FKG) lattice condition (sometimes a function satisfying this condition is called log supermodular) i.e.,

${\displaystyle \mu (x\wedge y)\mu (x\vee y)\geq \mu (x)\mu (y)}$

for all x, y in the lattice ${\displaystyle X}$.

The FKG inequality then says that for any two monotonically increasing functions ƒ and g on ${\displaystyle X}$, the following positive correlation inequality holds:

${\displaystyle \left(\sum _{x\in X}f(x)g(x)\mu (x)\right)\left(\sum _{x\in X}\mu (x)\right)\geq \left(\sum _{x\in X}f(x)\mu (x)\right)\left(\sum _{x\in X}g(x)\mu (x)\right).}$

The same inequality (positive correlation) is true when both ƒ and g are decreasing. If one is increasing and the other is decreasing, then they are negatively correlated and the above inequality is reversed.

Similar statements hold more generally, when ${\displaystyle X}$ is not necessarily finite, not even countable. In that case, μ has to be a finite measure, and the lattice condition has to be defined using cylinder events; see, e.g., Section 2.2 of Grimmett (1999).

Petrov [Petrov V.V. (1995) Limit Theorems of Probability Theory. Oxford: Clarendon Press.] attributes the FKG inequality (known also as the Tchebyshev’s sum) inequality to Tchebyshev.

For proofs, see Fortuin, Kasteleyn & Ginibre (1971) or the Ahlswede–Daykin inequality (1978). Also, a rough sketch is given below, due to Holley (1974), using a Markov chain coupling argument.

The lattice condition for μ is also called multivariate total positivity, and sometimes the strong FKG condition; the term (multiplicative) FKG condition is also used in older literature.

The property of μ that increasing functions are positively correlated is also called having positive associations, or the weak FKG condition.

Thus, the FKG theorem can be rephrased as "the strong FKG condition implies the weak FKG condition".

If the lattice ${\displaystyle X}$ is totally ordered, then the lattice condition is satisfied trivially for any measure μ. For this case, the FKG inequality is Chebyshev's sum inequality: if the two increasing functions take on values ${\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}}$ and ${\displaystyle b_{1}\leq b_{2}\leq \cdots \leq b_{n}}$, then (we may assume that the measure μ is uniform)

${\displaystyle {\frac {a_{1}b_{1}+\cdots +a_{n}b_{n}}{n}}\geq {\frac {a_{1}+\cdots +a_{n}}{n}}\;{\frac {b_{1}+\cdots +b_{n}}{n}}.}$

More generally, for any probability measure μ on ${\displaystyle \mathbb {R} }$ and increasing functions ƒ and g,

${\displaystyle \int _{\mathbb {R} }f(x)g(x)\,d\mu (x)\geq \int _{\mathbb {R} }f(x)\,d\mu (x)\,\int _{\mathbb {R} }g(x)\,d\mu (x),}$

which follows immediately from

${\displaystyle \int _{\mathbb {R} }\int _{\mathbb {R} }[f(x)-f(y)][g(x)-g(y)]\,d\mu (x)\,d\mu (y)\geq 0.}$

The lattice condition is trivially satisfied also when the lattice is the product of totally ordered lattices, ${\displaystyle X=X_{1}\times \cdots \times X_{n}}$, and ${\displaystyle \mu =\mu _{1}\otimes \cdots \otimes \mu _{n}}$ is a product measure. Often all the factors (both the lattices and the measures) are identical, i.e., μ is the probability distribution of i.i.d. random variables.

The FKG inequality for the case of a product measure is known also as the Harris inequality after Harris (Harris 1960), who found and used it in his study of percolation in the plane. A proof of the Harris inequality that uses the above double integral trick on ${\displaystyle \mathbb {R} }$ can be found, e.g., in Section 2.2 of Grimmett (1999).

In statistical mechanics, the usual source of measures that satisfy the lattice condition (and hence the FKG inequality) is the following:

If ${\displaystyle S}$ is an ordered set (such as ${\displaystyle \{-1,+1\}}$), and ${\displaystyle \Gamma }$ is a finite or infinite graph, then the set ${\displaystyle S^{\Gamma }}$ of ${\displaystyle S}$-valued configurations is a poset that is a distributive lattice.

Now, if ${\displaystyle \Phi }$ is a submodular potential (i.e., a family of functions

${\displaystyle \Phi _{\Lambda }:S^{\Lambda }\longrightarrow \mathbb {R} \cup \{\infty \},}$

one for each finite ${\displaystyle \Lambda \subset \Gamma }$, such that each ${\displaystyle \Phi _{\Lambda }}$ is submodular), then one defines the corresponding Hamiltonians as

${\displaystyle H_{\Lambda }(\varphi ):=\sum _{\Delta \cap \Lambda \not =\emptyset }\Phi _{\Delta }(\varphi ).}$

If μ is an extremal Gibbs measure for this Hamiltonian on the set of configurations ${\displaystyle \varphi }$, then it is easy to show that μ satisfies the lattice condition, see Sheffield (2005).

A key example is the Ising model on a graph ${\displaystyle \Gamma }$. Let ${\displaystyle S=\{-1,+1\}}$, called spins, and ${\displaystyle \beta \in [0,\infty ]}$. Take the following potential:

${\displaystyle \Phi _{\Lambda }(\varphi )={\begin{cases}\beta 1_{\{\varphi (x)\not =\varphi (y)\}}&{\text{if }}\Lambda =\{x,y\}{\text{ is a pair of adjacent vertices of }}\Gamma$ ;\\0&{\text{otherwise.}}\end{cases}}}

Submodularity is easy to check; intuitively, taking the min or the max of two configurations tends to decrease the number of disagreeing spins. Then, depending on the graph ${\displaystyle \Gamma }$ and the value of ${\displaystyle \beta }$, there could be one or more extremal Gibbs measures, see, e.g., Georgii, Häggström & Maes (2001) and Lyons (2000).

The Holley inequality, due to Richard Holley (1974), states that the expectations

${\displaystyle \langle f\rangle _{i}={\frac {\sum _{x\in X}f(x)\mu _{i}(x)}{\sum _{x\in X}\mu _{i}(x)}}}$

of a monotonically increasing function ƒ on a finite distributive lattice ${\displaystyle X}$ with respect to two positive functions μ₁, μ₂ on the lattice satisfy the condition

${\displaystyle \langle f\rangle _{1}\geq \langle f\rangle _{2},}$

provided the functions satisfy the Holley condition (criterion)

${\displaystyle \mu _{2}(x\wedge y)\mu _{1}(x\vee y)\geq \mu _{1}(x)\mu _{2}(y)}$

for all x, y in the lattice.

To recover the FKG inequality: If μ satisfies the lattice condition and ƒ and g are increasing functions on ${\displaystyle X}$, then μ₁(x) = g(x)μ(x) and μ₂(x) = μ(x) will satisfy the lattice-type condition of the Holley inequality. Then the Holley inequality states that

${\displaystyle {\frac {\langle fg\rangle _{\mu }}{\langle g\rangle _{\mu }}}=\langle f\rangle _{1}\geq \langle f\rangle _{2}=\langle f\rangle _{\mu },}$

which is just the FKG inequality.

As for FKG, the Holley inequality follows from the Ahlswede–Daykin inequality.

Consider the usual case of ${\displaystyle X}$ being a product ${\displaystyle \mathbb {R} ^{V}}$ for some finite set ${\displaystyle V}$. The lattice condition on μ is easily seen to imply the following monotonicity, which has the virtue that it is often easier to check than the lattice condition:

Whenever one fixes a vertex ${\displaystyle v\in V}$ and two configurations φ and ψ outside v such that ${\displaystyle \varphi (w)\geq \psi (w)}$ for all ${\displaystyle w\not =v}$, the μ-conditional distribution of φ(v) given ${\displaystyle \{\varphi (w):w\not =v\}}$ stochastically dominates the μ-conditional distribution of ψ(v) given ${\displaystyle \{\psi (w):w\not =v\}}$.

Now, if μ satisfies this monotonicity property, that is already enough for the FKG inequality (positive associations) to hold.

Here is a rough sketch of the proof, due to Holley (1974): starting from any initial configuration on ${\displaystyle V}$, one can run a simple Markov chain (the Metropolis algorithm) that uses independent Uniform[0,1] random variables to update the configuration in each step, such that the chain has a unique stationary measure, the given μ. The monotonicity of μ implies that the configuration at each step is a monotone function of independent variables, hence the product measure version of Harris implies that it has positive associations. Therefore, the limiting stationary measure μ also has this property.

The monotonicity property has a natural version for two measures, saying that μ₁ conditionally pointwise dominates μ₂. It is again easy to see that if μ₁ and μ₂ satisfy the lattice-type condition of the Holley inequality, then μ₁ conditionally pointwise dominates μ₂. On the other hand, a Markov chain coupling argument similar to the above, but now without invoking the Harris inequality, shows that conditional pointwise domination, in fact, implies stochastically domination. Stochastic domination is equivalent to saying that ${\displaystyle \langle f\rangle _{1}\geq \langle f\rangle _{2}}$ for all increasing ƒ, thus we get a proof of the Holley inequality. (And thus also a proof of the FKG inequality, without using the Harris inequality.)

See Holley (1974) and Georgii, Häggström & Maes (2001) for details.

• Ahlswede–Daykin inequality

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