FKG Inequality - A Special Case: The Harris Inequality

A Special Case: The Harris Inequality

If the lattice 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 and, then (we may assume that the measure μ is uniform)

More generally, for any probability measure μ on and increasing functions ƒ and g,

which follows immediately from

The lattice condition is trivially satisfied also when the lattice is the product of totally ordered lattices, and 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 can be found, e.g., in Section 2.2 of Grimmett (1999).

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