= Solution
Order $\Omega=\{0,1\}^E$ coordinatewise. The statement $\mu_1\ge_{\mathrm{st}}\mu_2$ means
$$
\int f\,d\mu_1\ge\int f\,d\mu_2
$$
for 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 condition
$$
\boxed{\mu_1(\omega\vee\eta)\mu_2(\omega\wedge\eta)\ge\mu_1(\omega)\mu_2(\eta)\quad\text{for all }\omega,\eta.}
$$
Here 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> is
$$
\boxed{\mu(\omega\vee\eta)\mu(\omega\wedge\eta)\ge\mu(\omega)\mu(\eta).}
$$
We prove that it gives <positive association of random variables>, namely $\operatorname{Cov}_\mu(f,g)\ge0$ for all increasing $f,g$. Fix such a $g$, and for $t\ge0$ form the strictly positive tilted law
$$
\mu_t(\omega)=\frac{e^{tg(\omega)}\mu(\omega)}{Z_t},\qquad Z_t=\sum_\omega e^{tg(\omega)}\mu(\omega).
$$
The lattice condition and $g(\omega\vee\eta)\ge g(\omega)$ imply
$$
\mu_t(\omega\vee\eta)\mu(\omega\wedge\eta)
\ge\frac{e^{tg(\omega)}}{Z_t}\mu(\omega)\mu(\eta)
=\mu_t(\omega)\mu(\eta).
$$
Thus 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 gives
$$
\left.\frac{d}{dt}\mathbb E_{\mu_t}f\right|_{t=0}
=\mathbb E_\mu(fg)-\mathbb E_\mu f\,\mathbb E_\mu g.
$$
Hence
$$
\boxed{\operatorname{Cov}_\mu(f,g)\ge0,}
$$
which 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>.
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