billschnieder said:
I don't follow. S is already a result of 4 different realizations, but then you appear to be averaging more than one S.
By "realisation" I mean what you might call "measurement on one particle pair". (Though I don't like that terminology much since Bell's theorem is about locally causal theories, which may or may not be theories about particles.) Alice and Bob each pick a measurement to do. They measure their systems. They each get a result which they record. That's one realisation. This would normally be repeated thousands of times in a Bell experiment to get a good statistical estimate of the Bell correlator.
Yes, I would expect ##-4 \le S_{k} \ge +4##. How you get from this to ##-2 \le \langle S_{k} \rangle \ge +2## is what the problem is.
I explained that in the subsequent part of my post. The point is that the estimator is defined in such a way that its expectation value, including the average taken over the choice of measurements (which is random), is exactly what's bounded in most derivations of the CHSH inequality. If you want me to do that explicitly, then start with the last line I wrote down: $$\begin{eqnarray}
\langle S_{k} \rangle &=& \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \\
&=& \sum_{xy} (-1)^{xy} \sum_{ab} (-1)^{a} (-1)^{b} P(ab \mid xy) \\
&=& E(00) + E(01) + E(10) - E(11) \,,
\end{eqnarray}$$ with $$\begin{eqnarray}
E(xy) &=& \sum_{ab} (-1)^{a} (-1)^{b} P(ab \mid xy) \\
&=& P(00 \mid xy) - P(01 \mid xy) - P(10 \mid xy) + P(11 \mid xy)
\end{eqnarray}$$ defined for convenience. For a Bell-local model, the probability distribution should have the form $$P(ab \mid xy) = \int \mathrm{d}\lambda \, \rho(\lambda) \, P_{\mathrm{A}}(a \mid x; \lambda) \, P_{\mathrm{B}}(b \mid y; \lambda) \,,$$ so the quantities ##E(xy)## can be written as $$E(xy) = \int \mathrm{d}\lambda \, \rho(\lambda) \, E(xy; \lambda)$$ with $$\begin{eqnarray}
E(xy; \lambda) &=& \sum_{ab} (-1)^{a} (-1)^{b} \, P_{\mathrm{A}}(a \mid x; \lambda) \, P_{\mathrm{B}}(b \mid y; \lambda) \\
&=& \bigl( P_{\mathrm{A}}(0 \mid x; \lambda) - P_{\mathrm{A}}(1 \mid x; \lambda) \bigr) \bigl( P_{\mathrm{B}}(0 \mid y; \lambda) - P_{\mathrm{B}}(1 \mid y; \lambda) \bigr) \\
&=& E_{\mathrm{A}}(x; \lambda) \, E_{\mathrm{B}}(y; \lambda) \,.
\end{eqnarray}$$ In the last line, I set $$\begin{eqnarray}
E_{\mathrm{A}}(x; \lambda) &=& P_{\mathrm{A}}(0 \mid x; \lambda) - P_{\mathrm{A}}(1 \mid x; \lambda) \,, \\
E_{\mathrm{B}}(y; \lambda) &=& P_{\mathrm{B}}(0 \mid y; \lambda) - P_{\mathrm{B}}(1 \mid y; \lambda) \,,
\end{eqnarray}$$ which are bounded by ##-1 \leq E_{\mathrm{A}}(x; \lambda) \leq 1## and ##-1 \leq E_{\mathrm{B}}(y; \lambda) \leq 1##. For any given ##\lambda##, $$\begin{eqnarray}
E(00; \lambda) + E(01; \lambda) + E(10; \lambda) - E(11; \lambda)
&=& E_{\mathrm{A}}(0; \lambda) \bigl( E_{\mathrm{B}}(0; \lambda) + E_{\mathrm{B}}(1; \lambda) \bigr) \\
&&+\> E_{\mathrm{A}}(1; \lambda) \bigl( E_{\mathrm{B}}(0; \lambda) - E_{\mathrm{B}}(1; \lambda) \bigr) \\
&\leq& \lvert E_{\mathrm{B}}(0; \lambda) + E_{\mathrm{B}}(1; \lambda) \rvert + \lvert E_{\mathrm{B}}(0; \lambda) - E_{\mathrm{B}}(1; \lambda) \rvert \\
&\leq& 2 \,,
\end{eqnarray}$$ so for the CHSH estimator expectation value, you get $$\begin{eqnarray}
\langle S_{k} \rangle &=& \int \mathrm{d}\lambda \, \rho(\lambda) \, \bigl( E(00; \lambda) + E(01; \lambda) + E(10; \lambda) - E(11; \lambda) \bigr) \\
&\leq& \max_{\lambda} \bigl( E(00; \lambda) + E(01; \lambda) + E(10; \lambda) - E(11; \lambda) \bigr) \\
&\leq& 2 \,.
\end{eqnarray}$$
