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

Question

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 $q(a,b|x,y)$.

Answer 1

Upon some further reflection, I found a possible answer. The idea is that yes, one can reformulate the CHSH inequalities without having to use expectation values, but rather just dealing with the relevant conditional probability distributions. I'll use the notation also used for the more in-depth discussion in this other answer of mine.

The (a?) standard way to formulate the CHSH inequalities is via $$S = \sum_{a,b} (-1)^{ab} \mathbb{E}_{ab}[XY] = \sum_{a,b,x,y} (-1)^{x+y+ab} p(ab|xy),\\ \mathbb{E}_{ab}[XY]\equiv \sum_{x,y\in\{0,1\}} (-1)^{x+y} p(ab|xy).$$ One can show that $-2\le S\le 2$ for any local realistic probability distribution.

The above writing for $S$ is in itself, technically, a way to write $S$ without necessarily invoking expectation values (talking about the RHS of the first equation). However, there is also a more "natural" way to understand this quantity without dealing with random variables at all. In fact, one can observe that $$\frac12 S = \sum_{ab} (-1)^{ab} P_{\rm same}(a,b) - 1.$$ In other words, $S$ can be seen as a linear combination of $P_{\rm same}(a,b)$ factors, where $P_{\rm same}(a,b)\equiv p(00|ab)+p(11|ab)$ is the probability of Alice and Bob finding the same outcome when measuring in the bases $a$ and $b$, respectively.