= FKG inequality
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The <FKG inequality> turns the <FKG lattice condition> into <positive association of random variables>. Let $\mu$ be a strictly positive <probability measure> on a finite <distributive lattice> and suppose
$$
\mu(x\wedge y)\mu(x\vee y)\geq\mu(x)\mu(y)
$$
for every $x,y$. Then for increasing real-valued <functions> $f,g$,
$$
\mathbb E_\mu[fg]\geq\mathbb E_\mu f\,\mathbb E_\mu g.
$$
Taking <indicator functions> gives $\mu(A\cap B)\geq\mu(A)\mu(B)$ for <increasing events>. The theorem applies to suitable dependent laws as well as <product measures>.<Correlation inequalities on some partially ordered sets by C. M. Fortuin and others>[Abstract and main finite-distributive-lattice association theorem]{ref}
For example, on the <Boolean lattice> $\{0,1\}^2$, give the configurations $00,01,10,11$ unnormalized masses $2,1,1,2$. The only nontrivial lattice check is $2\cdot2\geq1\cdot1$. Each coordinate-one <event> has <probability> $1/2$, while their intersection has <probability> $1/3>1/4$. Thus the coordinates are dependent but positively associated. For a <product measure>, the <FKG lattice condition> holds with equality and recovers the <Harris-FKG inequality>.
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