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Consider the standard two-party CHSH scenario. Each party can perform one of two measurements (denoted with $x,y\in\{0,1\}$) and observe one of two outcomes (denoted with $a,b\in\{0,1\}$).

Let $P(ab|xy)$ be the probability of observing outcomes $a,b$ when choosing the measurements settings $x,y$. Local realistic theories are those that, for some probability distribution over some hidden variable $\lambda$, satisfy $$P(ab|xy)=\sum_\lambda q(\lambda)P_\lambda(a|x)P_\lambda(b|y).\tag1$$ Define the local polytope $\mathcal L$ as the set of theories that can be written as in (1). Note that we identify here a theory with its set of conditional probabilities: $\boldsymbol P\equiv (P(ab|xy))_{ab,xy}$.

Denote with $E_{xy}$ the expectation values given by $E_{xy}=\sum_{ab}(-1)^{a+b}P(ab|xy)$. We then know that all local realistic theories (i.e. all theories $\boldsymbol P\in\mathcal L$) satisfy the CHSH inequality: $$\Big|\sum_{xy}(-1)^{xy} E_{xy}\Big| = |E_{00}+ E_{01} + E_{10} - E_{11}| \le 2.\tag2$$ Is the opposite true? In other words, do all theories satisfying (2) admit local realistic explanations?

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Yes. As you've effectively said, all cases satisfying (2) are in a polytope and therefore convex. All the vertices of that polytope are deterministic strategies, and so every point inside the polytope can be described as a convex combination of these, and that gives you (at least) one such local realistic explanation.

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  • $\begingroup$ this means that the local polytope has exactly two nontrivial (nontrivial meaning excluding the faces of the form $P(ab|xy)\ge0$) facets, correct? One for $S=2$ and the other one for $S=-2$. I ask because this paper seems to state that the only facet is $S=2$ (see eq. 2) $\endgroup$ – glS Mar 23 at 9:50
  • $\begingroup$ I'd have said so. Certainly the statement from that paper doesn't seem entirely mathematically consistent. However, have you looked at ref 6 in that paper? It's been a while since I looked at it, but they thought a lot about uniqueness/equivalence of different facets, which may be relevant... Either that or it's a typo and he left out the modulus sign. $\endgroup$ – DaftWullie Mar 23 at 10:11

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