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P: 724
 Quote by Gordon Watson Bill, you talkin' to me? (In that you cite gill1109.) 1. Not sure about your mystical being? Purpose =? (Is something more needed to clarify the OP?) 2. The CHSH inequality formulation will be the same, imho: since the experimental outcomes are ±1, no matter the settings a, b, etc. 3. To clarify the OP (if that's your issue): Having derived the expectation E(AB) for the classical setting -- from your functions for A and B = ±1 -- what then is the related maximum value that that classical setting might yield for the CHSH inequality? That is: What a, b, c, d settings yield the maximum value in the CHSH formula, and what is that maximum? 4. Is it gill1109's +2? 5. Did you mean to say "the scenario is manifestly LOCAL"??? 6. So -- addressing your "What gives" -- just give me your answers to the OP -- or tell me why you can't. Especially as it seems that Bell might think you can; the given situation being wholly classical and involving no more than Bell's proposed (1964, etc.) analytical formulation.
Sorry for hijacking your thread Gordon, I was just responding to the portion by gill1109. To answer your questions, and more on topic.

- Without specifying the method by which the common pulse orientations are chosen, it is not possible to calculate an expectation value.Without ρ(x) we are hopeless to calculate a meaningful E(AB) even if A(a,x) and B(b,x) are clearly specified.

- The maxumum attainable is of course +2 as gill1109 calculated.

Aside:

However, note the following extremely important point

A(a1)B(b1)-A(a1)B(b2)-A(a2)B(b2)-A(a2)B(b1)

→ A(a1)[B(b1)-B(b2)] - A(a2)[B(b2)-B(b1)]
→ A(a1)[B(b1)-B(b2)] + A(a2)[B(b1)-B(b2)]
→ [A(a1) + A(a2)]*[B(b1)-B(b2)] ---> **!!!

if A(a1) = A(a2) = +1, and B(b1) = -B(b2) = +1 Or,
A(a1) = A(a2) = -1 and B(b1) = -B(b2) = -1, we obtain the maxium of 2.

If A(a1) = -A(a2) = ±1, OR B(b1) = B(b2) = ±1, we get a value of zero.

And if A(a1) = A(a2) = -1 and B(b1) = -B(b2) = +1 Or
A(a1) = A(a2) = +1 and B(b1) = -B(b2) = -1, we obtain the minimum of 2.

This may seem like a pointless way to arrive at the same result as gill1109 except it is obvious from the emphasized expresion that the original 4 terms (A(a1)B(b1), A(a1)B(b2), A(a2)B(b2), A(a2)B(b1)) of products in the inequality originate from only 4 functions (A(a1), B(b1), A(a2), B(b2)) which must be factorizable. You can not use 4 different runs of an experiment (i, j, k, l) to obtain results from 8 functions (A(a1i), B(b1i), A(a1j), B(b2j), A(a2k), B(b2k), A(a2l), B(b1l)) and expect the inequality to work. It is a simple exercise to verify that for the case where 4 different runs of the experiment are performed, the maximum of the expression will be

A(a1i)B(b1i) - A(a1j)B(b2j) -A(a2k)B(b2k) -A(a2l)B(b1l) <= 4

NOT 2.

Some naively leave out the experiment identifyers (i,j,k,l) and fool themselves into thinking the result can be factorized.

In order for the results from 4 different experiments to be factorizable the following equalities must hold
A(a1i) = A(a1j)
A(a2k) = A(a2l)
B(b1i) = B(b1l)
B(b2j) = B(b2k)

Practically, this means if the experimental results consisted of a list of numbers (+1, -1) for each function and you obtained 8 columns for 4 different experiments, the data MUST be sortable such that 4 of the columns are duplicates, not only in the numbers of +1s and -1s but also in the switching pattern.

Therefore it is not sufficient that A*B for one experiment gives you a certain expection value for the paired product. For the inequality to have a maximum of 2, rather than 4, the value of one pair must constrain the value of a different pair in some manner.