# Can local hidden variables models be ruled out without invoking expectation values?

Bell's inequalities, in their standard form, are a statement about the limitations faced by a probability distribution that can be written as $$p(a,b|x,y)=\sum_\lambda p(\lambda) p(a,b|x,y,\lambda)=\sum_\lambda p(\lambda) p(a|x,\lambda)p(b|y,\lambda).\tag A$$

More specifically, in the CHSH setting, one sees that if $p(a,b|x,y)$ is like the above, then expectation values have the form $$g(x,y)\equiv\sum_{a,b}ab\,p(a,b|x,y)=\sum_\lambda p(\lambda) \sum_a a\,p(a|x,\lambda)\sum_b b\,p(b|y,\lambda).\tag B$$ Assuming binary outcomes $a,b=\pm1$ and considering two possible choices for $x$ and $y$, one then sees that the following holds: $$g(x_0,y_0)+g(x_0,y_1)+g(x_1,y_0)-g(x_1,y_1)\le 2.\tag C$$

This inequality should however be simply a trick to highlight the restrictions imposed by (A) over the function $(x,y)\mapsto g(x,y)$ defined in (B), with the purpose of witnessing features of $p(a,b|x,y)$.

Is there a more direct way to see the limitations imposed by (A) on the possible probability distributions, that doesn't involve computing expectation values over seemingly arbitrary functions of $a, b$?

This could for example be an argument showing that a more general (as in, not satisfying the locality constraint) distribution $q(a,b|x,y)$ can produce a wider set of outcome probabilities than $p(a,b|x,y)$ (if this is indeed the case).

Otherwise, if the above is not true for a fixed choice of $x,y$ (as it seems plausible), a possible answer could be an argument showing that, for two different sets of measurement choices $(x_0,y_0)$ and $(x_1,y_1)$, the locality constraint imposes a relation between $(a,b)\mapsto p(a,b|x_0,y_0)$ and $(a,b)\mapsto p(a,b|x_1,y_1)$ that is more restrictive then what is the case for a more general $q(a,b|x,y)$.

• Technical note: These are conditional probability distributions $p(x,y|a,b)$ -- or the other way round? Without you saying what denotes input and what output, it is hard to make sense of the above. (E.g., with my condition both p and q seem LHV models.) – Norbert Schuch Jul 22 '18 at 22:45
• More to the point: What do you mean by "direct way"? Apparently, expectation values are not "direct". But what would be direct? One distribution lies within a convex cone while the other has points outside of it. Is this sufficient? – Norbert Schuch Jul 22 '18 at 22:49
• @NorbertSchuch well, I've tried to use only the full probability distribution and its marginals, to keep the number of objects used to the minimum, but of course, everything can be equivalently (and maybe more naturally) be stated using conditional probabilities. For example, $p(a,b|x,y)\equiv p(a,b,x,y)/p(x,y)=\sum_\lambda p(a,b|x,y,\lambda)p(\lambda)=\sum_\lambda p(\lambda)p(a|x,\lambda)p(b|y,\lambda)$ and so on. – glS Jul 22 '18 at 22:54
• I purposefully used a notation that doesn't explicitly make a distinction between "input" and "output" because it seems to me that at the end of the day the distinction between which variable is "cause" and which is "effect" is a matter of interpretation that is applied on top of the maths, while the result I was looking for should not be dependent on that. Of course what is a "direct" argument is somewhat opinion-based here, but for example a general argument such as the one you seem to be mentioning should do it. More generally, something not requiring to compute seemingly arbitrary (...) – glS Jul 22 '18 at 22:58
• (...) expectation values of specific marginals. Finally, regarding the LHV model, the $p$ in the last definition should represent a distribution compatible with QM, while $q$ is the LHV model, unless I made some mistake somewhere. The only condition I imposed on that $p$ is the statistical independence of the measurement choices – glS Jul 22 '18 at 23:01