# In a Bell scenario, why can correlations be nonlocal only if there are at least two measurement settings to choose from?

In (Brunner et al. 2013), the authors mention (end of pag. 6) that a set of correlations $p(ab|xy)$ can be nonlocal only if $\Delta\ge2$ and $m\ge2$, where $\Delta$ is the number of measurement outcomes for the two parties (that is, the number of different values that $a$ and $b$ can assume), and $m$ the number of measurement settings that one can choose from (that is, the number of values of $x$ and $y$).

A probability distribution $p(ab|xy)$ is here said to be nonlocal if it can be written as $$p(ab|xy)=\sum_\lambda q(\lambda) p(a|x,\lambda)p(b|y,\lambda).\tag1$$ This means that if either there is only one possible measurement outcome, or only one possible measurement setting, then all probability distributions can be written as in (1).

The $\Delta=1$ (only one measurement outcome) case is trivial: if this is the case, denoting with $1$ the only possible measurement outcome, we have $p(11|xy)=1$ for all $x,y$. Without even needing the hidden variable, we can thus just take $p(1|x)=p(1|y)=1$ and we get the desired decomposition (1).

The $m=1$ case, however, is less trivial. In this case the question seems equivalent to asking whether an arbitrary probability distribution $p(a,b)$ can be written as $$p(a,b)=\sum_\lambda q(\lambda)p(a|\lambda)p(b|\lambda).$$ The paper does not mention any reference to support this fact. How can this be proven?

Make a $\lambda_{a,b}$ for every pair $(a,b)$.
Then make $q(\lambda_{a,b}) = p(a,b)\,$, and
$p(a|\lambda_{a,b}) = p(b|\lambda_{a,b}) = 1.$
The Bell's nonlocality scenario with $$N$$ parties and where each party can choose from $$M$$ measurements and each measurement outcomes $$K$$ values is known as the $$N$$-$$M$$-$$K$$ Bell's nonlocality scenario. There are several kinds of arguments for the reason that the simplest nontrivial Bell's nonlocality scenario is $$2-2-2$$ case. Here, I would like to point out an argument based on a theorem by A. Fine (Fine's theorem), which says that the existence of the local hidden variable model is equivalent to the existence of the joint probability distribution for all measurements involved in the Bell test and all the experimentally accessible statistics can be reproduced from this joint probability distribution by taking marginals.
Since quantum theory is a non-signaling theory ($$\sum_{a}p(a,b|A,B)=\sum_{a'}p(a',b|A,B)$$ for arbitrary measurements $$A,A'$$ chosen by Alice and $$B$$ chosen by Bob, similar constraints also need to set for Bob). If Alice and Bob can only choose one measurement, then we always have an experimentally accessible joint probability distribution $$p(a,b|A,B)$$, which is guaranteed by the assumption that compatible measurements can be measured jointly, and since two measurements are carried out by two space-like separated parties the measurement statistics must satisfy the non-signaling condition. Since only $$A,B$$ are involved in this Bell test and we have a joint probability distribution of them, then there must be a local hidden variable model for this $$2-1-K$$ scenario.
For the case where Alice can choose from two measurements $$A,A'$$ and Bob only can choose measurement $$B$$, by the assumption of non-signaling and compatible measurements have joint probability distribution, the experimentally accessible probability distribution is $$\{p(a|A),p(a'|A'),p(b|B),p(a,b|A,B),p(a',b|A',B)\}$$, a joint probability distribution can be consructed as $$p(a,a',b)=\frac{p(a,b|A,B)p(a',b|A',B)}{p(b|B)}$$, it satisfies the non-signaling condition, thus this also guarantees the existence of the local hidden variable model. The case where Bob can choose from two measurements and Alice can only choose one measurement is similar. Thus the simplest nontrivial case must be $$2-2-2$$ scenario, for wich the Bell inequality is CHSH inequality.