# Van Den Berg-Kesten-Reimer inequality

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

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

Question

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

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.