Van Den Berg-Kesten-Reimer inequality

Let $n$ be a positive integer. For $i\in[n]$, let $\Omega_i$ be a finite set and $\mu_i$ a probability measure on it. Set $\Omega=\Omega_1\!\times\!\ldots\!\times\!\Omega_n$ and $\mu=\mu_1\!\times\!\ldots\!\times\!\mu_n$. For $A\!\subset\!\Omega$ and $\sigma\!\subset\![n]$, let $$ A_\sigma=\lbrace\omega\!\in\!A\!:\,\forall\upsilon\!\in\!\Omega,\,(\forall i\!\in\!\sigma,\,\upsilon_i\!=\!\omega_i)\!\implies\!\upsilon\!\in\!A\rbrace $$ In other terms, the occurrence of the event $A_\sigma$ is solely controlled by $\sigma$. Given two events $A,B\subset\Omega$, the notion of disjoint occurrence of $A$ and $B$ is defined by the following event
$$ A\!\circ\!B=\lbrace\omega\!\in\!A\!\cap\!B\!:\,\exists\,\sigma,\tau\!\subset\![n],\,\sigma\!\cap\!\tau\!=\!\emptyset \wedge\omega\!\in\!A_\sigma\!\cap\!B_\tau\rbrace $$

Van Den Berg-Kesten-Reimer inequality states that

$$ \forall A,B\!\subset\!\Omega,\,\,\mu(A\circ B)\le\mu(A)\cdot\mu(B) $$


Are there non-trivial events that turn the inequality stated above into an equality?


1 Answer 1


In the case of increasing events, the standard proof of the BK inequality works like this. Start with your set $\Omega$, and successively replace each $\Omega_i$ by a disjoint union of two copies $\Omega_i^1$ and $\Omega_i^2$. After step $i$, the event ${(A\circ{B})}(i)$ is defined by saying that $A$ and $B$ need to occur disjointly on $\Omega_{i+1}$ up to $\Omega_n$, and $A$ (resp. $B$) is allowed to use the $\Omega_k^1$ (resp. $\Omega_k^2$).

Then, $(A\circ B)(0)$ is just disjoint occurrence, and $(A\circ B)(n)$ is the occurrence of two independent events, its probability is $P(A)P(B)$. Besides, for every $i$, we relax the restrictions by allowing the events to be less and less disjoint - hence the BK inequality.

If there exists an $i$ that is pivotal for both $A$ and $B$ with positive probability, then when you split the corresponding $\Omega_i$, the inequality between the probabilities of $(A\circ B)(i-1)$ and $(A\circ B)(i)$ is strict, and so is BK. In other words, BK is an equality if and only if you can partition $[n]$ into two subsets, $A$ depending only on the first one and $B$ on the second one.

In yet other words, in the case of increasing events, there is no non-trivial (in the sense, where disjoint occurrence is not automatic) pair of events realizing the equality.

Without monotonicity, the argument does not work directly ...


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