minkwe wrote:Jochen wrote:OK, so how do you justify the existence of a joint PD? If I am allowed nonlocal signalling, then I can always alter the PD of one part of the system due to measurements done on the other, and no joint PD exists.

If you review my response carefully, you will see examples of how to do it. For example, the your state-machine with non-local disturbance, in which the measurements are always done in sequence A,B,C,D, would produce a joint PD. So long as the 4 numbers are jointly measured, it matters not one iota what physical mechanism is generating them, you will get a joint PD of outcomes and Bell's inequalities will apply.

The problem is that the sequential measurement you proposed does not generate the PD of the outcomes---in particular, it will be impossible to derive the correlations between different variables, as you effectively only measure single-observable marginals. Take a simple example where A and B are always perfectly correlated, but each +1 or -1 with 50% probability. Then, measuring sequences of the form A, B, A, B, ... will never tell you anything about the correlation, and you'll only recover the statement that A=+/-1 with 50% probability, and B=+/-1 with 50% probability, which is the same you would get if, say, both were perfectly anticorrelated, or not correlated at all. And again, the measurement of all quantities at once may be physically impossible.

minkwe wrote:original bell inequalities would have been violated by 3 spin-half particles, the goal-posts would not have needed to be shifted. The original Bell inequalities involve 3 simultaneous measurements at settings (a,b,c). That can be done for 3 spin-half particles directly, and the original Bell's inequalities will apply, shouldn't they?

Each Bell inequality is derived for a specific setting; thus, that Bell inequalities can't be violated in a setting they are not apt to is no great mystery, and in particular does not imply any shifting of goalposts. It's not any more mysterious than that in one system, observables A,B,C may be correlated, while in a different one, they might not be.

Otherwise, why would you expect a joint PD (A,B,C) to be present in this case, but absent in the case of only 2 particles, if you refuse to acknowledge that it is the joint measurement that determines whether you have a joint PD, so that when you measure the 3 simultaneously, you get a joint PD P(A,B,C), and when you measure only 2 you do not get a joint PD P(A,B,C). It does not matter what physical mechanism is generating the outcomes.

The reason is simply that in the two-particle case, not all of these observables are compatible, while in a three-particle setting, they are. But for compatible observables, a joint PD can always be found, and they can always be measured jointly.

minkwe wrote:Jochen wrote:Take three coins, , , and . If you flip the first and second together, they will always agree, giving both H = heads or T = tails with 50% probability. If you flip the second and third together, they will always land oppositely (again with 50% probability). If you flip the second and third together, they will always agree (50 % HH, 50% TT). Clearly, there does not exist a joint PD . Nevertheless, flipping each coin on its own is a perfectly ordinary stochastic process---for which, however, BIs can't be derived.

But your contrived example does not translate into a general truth.

No, but it suffices to refute your assertion that every statistical process always can be ascribed a joint probability distribution by exhibiting a stochastic process in which this is not the case.

The point was, that contrary to your claim that pre-determination of values is required, a random number generator which generates the 4 numbers (A,B,C,D) without any pre-determination, must obey the CHSH inequality because the outcomes represent a joint PD P(A,B,C,D). Therefore pre-determination is not required to obtain a joint PD. Do you agree or disagree?

I disagree: a random process does not always produce something which can be described by a joint PD, as is shown by the coin example.

You have a bigger burden than me. You are trying to argue that it is ALWAYS possible to reconstruct a joint PD for ALL local HV theories. Using specific contrived examples do not serve your purpose. I'm arguing that in some cases where you have a local HV theory, it is not always possible to reconstruct a joint PD from measured pairs. So I can use simple contrived examples to make my point effectively, you cannot.

I've given a general argument why any local HV theory can be described by a joint PD: the PD is simply a convex combination of all the value-assignments the theory makes, which exists if there are such value-assignments and there is no disturbance. You've tried to refute this by proposing that for a class of systems, a joint PD can always be found even though there is no value-assignment; I have against this general statement provided an example which refutes it, a system in this class of stochastic systems for which no joint PD can be found.

minkwe wrote:You do not say what you mean by "at variance" so I don't know how to answer a claim that is not substantiated. I use degrees of freedom it in the same way it is usually used. So please explain what you mean by "at variance".

I asked you to elaborate on how you used a term, because I would use it differently---to me, degrees of freedom are something like, for instance, the allowed number of independent motions of a molecule, the number of parameters to describe an object's motion, etc. You seem to apply the term to the freedom the experimenters have in a Bell test. Hence, I asked to clarify; no need to get all defensive.

minkwe wrote:So you misunderstood me. The fact that averages actually measured on separate sets of particles are always reproducible does not and cannot reasonably be taken to mean that those averages are the same ones represented by the terms in the inequality which were not measured on the same set of particles. It is a fact that the measured averages are reproducible. But that has nothing to do with the fact that the actually measured correlations are not the same as the unmeasured counterfactual ones from the same set of particles.

This is exactly what the assumption of hidden variables amounts to: if there are such hidden variables, then the statistical properties of the unmeasured ones are the same as the measured ones, since the particle or whatever system could not have known beforehand which ones would be measured, and hence, could not have arranged for the unmeasured ones to differ in any relevant way.

minkwe wrote:In the expression , for the particle pair , three of the paired terms are counterfactual and one is actual.

But only at the measurement, by the free choice of the experimenter, is it decided which one is made actual. Hence, the statistics must agree for every choice they could make; but then, the inference from the above being below 2 for every single particle to it being below two on average even if not all terms are measured on every particle is valid.

The terms are not independent from each other, there are 4 independent values which are free to vary in that expression, and those 4 values are cyclically shared by the paired terms such that they are not independent.

Come again? The 4 independent values are not independent? I don't know how to parse this.

Once three of the paired terms are given, the fourth one is determined automatically. That is the origin of the relationship . The problem is that you are trying to take away the feature that gives you that relationship while at the same time claiming that the relationship continues to hold.

Again, the logic is: it holds for every particle; therefore, it holds statistically. But since the measurement is only decided after the HV-assignment, there is no way for things to be arranged such that the 'counterfactual' outcomes differ from the observed ones---since I could just as well have measured C and D instead of A and B, I can infer that it must be the case that C and D must have the same statistical properties whether I actually measure them or not; because otherwise, had I instead chosen to measure them, I would have observed statistics incompatible with the predictions of QM.

minkwe wrote:I already have written two local realistic simulations of EPRB experiments, which reproduce the QM correlations without any disturbance, and without non-locality. You can find them https://github.com/minkwe/epr-simple/, and https://github.com/minkwe/epr-clocked. But you are not going to understand what is going on there, if you do not understand my arguments.

I think I understand quite well what's going on there, and hence you should be able to anticipate my response: get rid of the nondetections in the first, or of the failures to produce particle pairs in the second, and then we can talk. A local realistic strategy should work just as well in the case of ideal detectors and sources.