# Van Den Berg-Kesten-Reimer inequality

Statement of Van Den Berg-Kesten-Reimer inequality:

Let $n$ be a positive integer. For $i\in[n]$, let $\mu_i$ be a probability measure on a finite set $\Omega_i$. Let $\Omega=\Omega_1\times\ldots\times\Omega_n$ and $\mu=\mu_1\times \ldots\times\mu_n$.

For an event $A\subset\Omega$, and an index subset $\Sigma\subset[n]$, let $A_{|\Sigma}=\lbrace\omega\in A:\forall\psi\in\Omega,(\forall i\in\Sigma,\psi_i=\omega_i)\implies\psi\in A\rbrace$. This means that, as an event, the occurrence of $A_{|\Sigma}$ is controlled by the random variables indexed by $\Sigma$ only.

For any two events $A,B\subset \Omega$, let $A\circ B=\lbrace\omega\in A\cap B:\exists\Sigma,\Lambda\subset[n],(\Sigma\cap\Lambda=\emptyset)\wedge(\omega\in A_{|\Sigma}\cap B_{|\Lambda})\rbrace$. $A\circ B$ describes the disjoint occurrence of $A$ and $B$.

The Van Den Berg-Kesten-Reimer inequality states that for all events $A,B\subset\Omega$, $\mu(A\circ B)\le\mu(A)\cdot\mu(B)$.

Question: What (non-trivial) events turn the inequality into an equality?

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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.