[I'm sorry, I've already posted the same question in the physics community, but I haven't received an answer yet.]
I'm approaching the study of Bell's inequalities and I understood the reasoning under the Bell theorem (ON THE EINSTEIN PODOLSKY ROSEN PARADOX (PDF)) and how the postulate of locality was assumed at the start of the demonstration.
However, I find problematic to arrive at the equivalence
$$ E(\vec{a},\vec{b}) = \int_{\Lambda}d\lambda \rho(\lambda)A(\vec{a},\lambda)B(\vec{b},\lambda),$$
starting from the point of view expressed by the Clauser and Horne definition of locality.
CH claimed that a system is local if there is a parameter $\lambda$ and a joint conditional probabilities that can be written as follows: $$p(a,b|x,y,\lambda) = p(a|x,\lambda)p(b|y,\lambda),$$ and $$p(a,b|x,y) = \int_\Lambda d\lambda \rho(\lambda) p(a|x,\lambda)p(b|y,\lambda)$$ which make sense since it affirms that the probability of obtaining the value $a$ depends only on the measument $x = \vec{\sigma}\cdot\vec{x} $ and the value of $\lambda$.
However, if I use this expression to write down the expectation value of the products of the two components $\vec{\sigma}\cdot\vec{a}$ and $\vec{\sigma}\cdot\vec{b}$, I obtain as follows:
$$ E (\vec{a},\vec{b}) = \sum_{i,j}a_ib_jp(a,b|x,y) = \\ = \sum_{ij}a_ib_j \int_\Lambda d\lambda \rho(\lambda) p(a|x,\lambda)p(b|y,\lambda) \\ = \int_\Lambda d\lambda \rho(\lambda) (\sum_{i}a_ip(a|x,\lambda))(\sum_{i}b_ip(b|y,\lambda)) $$ where in the last equivalence I've used the fact that if the measument are independent their covariance must be equal to $0$.
At this point, the terms in the RHS in the brackets are equal to: $$ (\sum_{i}a_ip(a|x,\lambda)) = E(a,\lambda) =? = A(\vec{a},\lambda)\quad \quad (\sum_{i}b_ip(b|y,\lambda)) = E(b,\lambda) =?= B(\vec{b},\lambda).$$
That is not the equivalence that I want to find.
In fact in the RHS of the first equation $A(\vec{a},\lambda)$ is, according to Bell original article, the result of measure $\vec{\sigma}\cdot\vec{a}$, and fixing both $\vec{a}$ and $\lambda$ it can assume only the values of $\pm1$. (The same is applied for $B(\vec{b},\lambda)$.)
Some of you knows, where I fail? How can I obtain the original equivalence (that then is proved to be violate in the case of an entangled system) starting from the CH definition of reality?
Edit #1:
I've noted that I obtain the wanted equivalence only if I assume that $p(ab|xy\lambda) = E(\vec{a}\vec{b})$, but is it possible? How can a conditional probability be linked to the mean value of the product of two components?
Edit #2:
Surfing the internet I found an article (https://arxiv.org/abs/1709.04260, page 2, right on the top) which reports the same CH's local condition (to be accurate, the article presents the discrete version) and then affirm that:
Blockquote "The central realization of Bell’s theorem is the fact that there are quantum correlations obtained by local measurements ($M_a^x$ and $M_b^y$) on distant parts of a joint entangled state $\varrho$, that according to quantum theory are described as: $$p_{Q}(a,b,|x,y) = \text{Tr}(\varrho(M_a^x\otimes M_b^y) $$ and cannot be decomposed in the LHV form (i.e. The CH condition for locality)"
So why $p_Q(a,b|x,y)$ is seen as a measure of quantum correlation (that for definition is the mean of the product of the possible output)? It isn't a joint probability distribution (as stating while obtaining the LHV form)? Is there a link between the classical correlation ($E(\vec{a},\vec{b})$) and the joint probability distribution $p(a,b|x,y,\lambda)$?
