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

glS
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  • 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
  • It does matter whether this is a conditional distribution or a multivariate distribution: In the latter case, all marginals must be prob. distributions. In the former case, certain "marginals" are simply not marginals (if I'm not mistaken). – Norbert Schuch Jul 22 '18 at 23:04
  • If "convex cone" is direct enough, this is sth. which should directly follow from CHSH etc.. – Norbert Schuch Jul 22 '18 at 23:04
  • @NorbertSchuch If eg $p(a,b,x,y)\ge0$ and $\sum_{a,b,x,y}p(a,b,x,y)=1$, then aren't all marginals prob distributions as well? About the "direct argument" let's put it this way: we want to probe features of $p(a,b|x,y)$, and CHSH does this computing possible values of a sum of a very specific expectation value computed over it. This looks "indirect" in the sense that there should/could be a way to reach similar conclusions without introducing seemingly arbitrary sums of expectation values. Convexity arguments do seem promising in this regard – glS Jul 22 '18 at 23:27
  • $p(a,b|x,y)$ is a point in a 16-dimensional space, and for each theory element of a convex polytope. The LHV polytope is different from the quantum mechanics polytope. Is that "direct"? – Norbert Schuch Jul 22 '18 at 23:31
  • Note that for a cond. distribution $p(a,b|x,y)$, we have that $\sum_{x,y} p(a,b|x,y)=1$ for all $a,b$. This is quite different from what you say. – Norbert Schuch Jul 22 '18 at 23:32
  • @NorbertSchuch well, now I'm confused. I guess I see intuitively why $\sum_x p(a|x)=1$ should hold, but isn't the conditional probability always defined using join prob and marginals: $p(a|x)\equiv p(a,x)/p(x)$? How do you define it otherwise? The LHV polytope is different from the quantum mechanics polytope. Is that "direct"? isn't this just a more sophisticated way to state the same question? A straightforward way to see why this is true would be "direct", I would say – glS Jul 23 '18 at 13:48
  • I have to apologize: I meant $\sum_a p(a|x)=1$ for all $x$. Note that in the Bell setting, $p(x)$ is not even defined to start with. $p(a,b|x,y)$ is a description of what the experiment does when you choose certain parameters x,y. – Norbert Schuch Jul 23 '18 at 14:20
  • @NorbertSchuch if the sum is over $a$ than the notations are totally equivalent: $p(a|x)\equiv p(a,x)/p(x)$ gives $\sum_a p(a|x)=\sum_a p(a,x)/(\sum_{a'} p(a',x))=1$. Actually, I would take this to be the proof of $\sum_a p(a|x)=1$. Indeed, I don't see how you even define $p(a|x)$ otherwise. I do agree that in practice you don't need to worry about the distinction though, as in some sense $p(x)$ captures the probability that the experimenter chooses to do something. But mathematically shouldn't everything being equivalent? Replace joint probs with conditionals if you like – glS Jul 23 '18 at 21:20
  • p(x) is just not a property of the experimental setup. I don't see how it makes sense to talk about it, or use it in an argument deriving anything about the experimental outcome. – Norbert Schuch Jul 23 '18 at 21:39
  • @NorbertSchuch I did some edits to the question, using conditionals instead of joint probabilities, and attempting to make it clearer what I'm asking for. Does it help? – glS Jul 24 '18 at 11:07
  • You might be interested in Hardy's paradox. It proves local hidden variable theories cannot exist; without relying on a probabilistic distribution. – Ali Jul 24 '18 at 12:04
  • Let us consider the CHSH setting. I believe you do not need the probability interpretation for the CHSH inequality when "viewed" through the hidden variables; it should provide the bounds (the range) for the possible observed values, which can then be shown experimentally to be violated. (The other part of the CHSH, interpretation through the QM and the attainment of Cirel’son bound, is a different question.) – Mahir Lokvancic Mar 15 '22 at 20:27

1 Answers1

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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.

glS
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