# Van Den Berg-Kesten-Reimer inequality

Van Den Berg-Kesten-Reimer inequality

For a given positive integer $n$ and for every $i\in[n]$, denote by $\mu_i$ a probability measure on a finite set $\Omega_i$. Call $\mu$ and $\Omega$ the products $\mu_1\!\times\! \ldots\!\times\!\mu_n$ and $\Omega_1\!\times\!\ldots\!\times\!\Omega_n$ respectively.

For an event $A\!\subset\!\Omega$ and an index subset $\sigma\!\subset\![n]$, let $$A_\sigma=\lbrace\omega\!\in\!A\!:\,\forall\psi\!\in\!\Omega,\,(\forall i\!\in\!\sigma,\,\psi_i\!=\!\omega_i)\!\implies\!\psi\!\in\!A\rbrace$$ Hence, the occurrence of $A_\sigma$ is solely controlled by $\sigma$. For all events $A$ and $B$ of $\Omega$, the disjoint occurrence of $A$ and $B$ is described by
$$A\!\circ\!B=\lbrace\omega\!\in\!A\!\cap\!B\!:\,\exists\,\sigma,\tau\!\subset\![n],\,\sigma\!\cap\!\tau\!=\!\emptyset \wedge\omega\!\in\!A_\sigma\!\cap\!B_\tau\rbrace$$

The Van Den Berg-Kesten-Reimer inequality states that for all events $A$ and $B$ of $\Omega$,

$$\boxed{\mu(A\circ B)\le\mu(A)\cdot\mu(B)}$$

Question

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

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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.