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CHSH inequality requires both locality and realism. I will equate here realism with counterfactual definiteness.

Now counterfactual definiteness tells us that given two different measurements on the same object, described by random variables $C_1$ and $C_2$, there exists a joint probability distribution for $C_1$ and $C_2$ (this is not always the case, search for the marginal problem and indeed we know that outcomes of measurements of noncommuting observables do not posses a joint probability distribution). Now if we can assume the existence of a joint probability distribution, then expectation values $E(C_1) + E(C_2)$ may be joined together to have $E(C_1 + C_2)$.

So suppose that we now have four random variables $A_1$ and $B_1$ local to Alice and $A_2$ and $B_2$ for Bob, which can take values $\pm 1$. The expression in the CHSH inequality is $$ |E(A_1 A_2) + E(A_1 B_2) + E(B_1 A_2) - E(B_1 B_2)| $$ Now if we can assume realism (counterfactual definiteness), there exists a joint probability distribution for the outcomes of all four random variables, and we can join the expectation values in the above together to get $$ \Bigl\lvert E\bigl(A_1 (A_2 + B_2) + B_1 (A_2 -B_2)\bigr)\Bigr\rvert $$ So now we do the standard trick where either $A_2$ and $B_2$ are equal or they are opposite, so that the first term is either $\pm 2$ and same for the second term.

Now to my question. I obviously used the realism assumption in the above. I assume locality means that the marginal distributions for $A_1$ and $B_1$ are independent of the choice of random variables Bob makes, $A_2$ or $B_2$ (but I otherwise allow for the correlation of outcomes, as long as it's independent of measurement choice, as there may be hidden variables that were encoded at the source of the state that produce correlations). Where did I use the locality assumption in this derivation? I would like if you either point out precisely where this assumption is needed in this calculation or argue that it is not needed with a convincing justification.

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Suppose that:

1) $A_2$ and $B_2$ have some fixed joint probability distribution.

2) When $A_2$ is measured, we have $A_1=B_1=A_2$. This gives a well-defined joint probability distribution for $(A_1,B_1)$.

3) When $B_2$ is measured, we have $A_1=-B_1=B_2$. This also gives a well-defined joint probability distribution for $(A_1,B_1$).

So these assumptions seem to satisfy your notion of "realism". But your expression now becomes $$E(A_1A_2)+E(A_1B_2)+E(B_1A_2)-E(B_1B_2)=4$$ which violates the desired inequality. So your realism assumption can't be enough.

The problem is that you are treating $A_1(A_2+B_2)$ as a well-defined classical random variable, which (implicitly) assumes that the probability distribution of $A_1$ is independent of which of $A_2,B_2$ is measured.

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  • $\begingroup$ @Steven I had a feeling it had to there and your answer demonstrates that realism is not enough. I gave you a +1, but I am still a bit confused about why I can't treat $A_1(A_2 + B_2)$ as a classical random variable if there exists a joint probability distribution $p(a_1, a_2, b_2)$. $\endgroup$
    – Sebastian Meznaric
    Commented Sep 12, 2012 at 14:45
  • $\begingroup$ Also, just going through your example again this reminds me very much of the three coins problem in the marginal problem. Are you sure there is a joint probability distribution $p(a_1, b_1, a_2, b_2)$ for this case? Perhaps merely assuming the existence of such a distribution implies the assumption of locality. But if this were so, stating it might be superfluous. $\endgroup$
    – Sebastian Meznaric
    Commented Sep 12, 2012 at 14:56
  • $\begingroup$ Sebastian: The existence of a joint probability distribution for all four variables is definitely sufficient to get the inequality. The existence of joint probability distributions for $(A_1,B_1)$ and $(A_2,B_2)$ (i.e. "realism") is definitely <i>not</i> sufficient to imply a joint probability distribution for all four. Locality is a sufficient additional assumption. $\endgroup$
    – Steven Landsburg
    Commented Sep 12, 2012 at 15:02
  • $\begingroup$ Now I see how I confused myself. I took counterfactual definiteness to mean that the joint distribution $p(a_1, b_1, a_2, b_2)$ exists. However, it seems the assumption is local counterfactual definiteness so that $p(a_1, b_1)$ exists and $p(a_2, b_2)$ exists. Then adding locality implies the existence of $p(a_1, b_1, a_2, b_2)$. This is making a lot more sense now. $\endgroup$
    – Sebastian Meznaric
    Commented Sep 12, 2012 at 15:07
  • $\begingroup$ Either way though, our discussion here seems to imply to me that perhaps realism in the global sense (i.e. existence of $p(a_1, b_1, a_2, b_2)$) already contains in itself locality. I find this very neat. $\endgroup$
    – Sebastian Meznaric
    Commented Sep 12, 2012 at 15:10

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