Editing CHSH inequality (section)

=== Bell's 1971 derivation ===
The following is based on page 37 of Bell's ''Speakable and Unspeakable'',<ref name="Bell-1971"/> the main change being to use the symbol ‘''E''’ instead of ‘''P''’ for the expected value of the quantum correlation.  This avoids any suggestion that the [[quantum correlation]] is itself a probability.

We start with the standard assumption of independence of the two sides, enabling us to obtain the joint probabilities of pairs of outcomes by multiplying the separate probabilities, for any selected value of the "hidden variable" λ.  λ is assumed to be drawn from a fixed distribution of possible states of the source, the probability of the source being in the state λ for any particular trial being given by the density function ρ(λ), the integral of which over the complete hidden variable space is 1.  We thus assume we can write:
<math display="block">E(a, b) = \int \underline{A}(a, \lambda) \underline{B}(b, \lambda) \rho(\lambda) d\lambda</math>
where <u>''A''</u> and <u>''B''</u> are the outcomes.  Since the possible values of ''A'' and ''B'' are &minus;1, 0 and +1, it follows that:
{{NumBlk|:|<math>\left| \underline{A} \right| \leq 1 \quad \left| \underline{B} \right| \leq 1</math>|{{EquationRef|4}}}}

Then, if ''a'', ''a''&prime;, ''b'' and ''b''&prime; are alternative settings for the detectors,
:<math>\begin{align}
      &E(a, b) - E\left(a, b'\right) \\

  ={} &\int \left[ \underline{A}(a, \lambda) \underline{B}(b, \lambda) - \underline{A}(a, \lambda) \underline{B}\left(b', \lambda\right) \right] \rho(\lambda) d\lambda \\

  ={} &\int \left[ \underline{A}(a, \lambda) \underline{B}(b, \lambda) - \underline{A}(a, \lambda) \underline{B}\left(b', \lambda\right) \pm \underline{A}(a, \lambda) \underline{B}(b, \lambda) \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \mp \underline{A}(a, \lambda) \underline{B}(b, \lambda) \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \right] \rho(\lambda) d\lambda\\

  ={} &\int \underline{A}(a, \lambda) \underline{B}(b, \lambda) \left[1 \pm \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \right] \rho(\lambda) d\lambda - \int \underline{A}(a, \lambda) \underline{B}\left(b', \lambda\right) \left[1 \pm \underline{A}\left(a', \lambda\right) \underline{B}(b, \lambda) \right] \rho(\lambda) d\lambda
\end{align}</math>

Taking absolute values of both sides, and applying the [[triangle inequality]] to the right-hand side, we obtain

:<math>\left| E(a, b) - E\left(a, b'\right) \right| \leq \left| \int \underline{A}(a, \lambda) \underline{B}(b, \lambda) \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \right] \rho(\lambda) d\lambda \right| + \left| \int \underline{A}(a, \lambda) \underline{B}\left(b', \lambda\right) \left[1 \pm \underline{A}\left(a', \lambda\right) \underline{B}(b, \lambda) \right] \rho(\lambda) d\lambda \right|</math>

We use the fact that <math> \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \right] \rho(\lambda) </math> and <math> \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}(b, \lambda) \right] \rho(\lambda) </math> are both non-negative to rewrite the right-hand side of this as
<math display="block">
  \int \left| \underline{A}(a, \lambda) \underline{B}(b, \lambda) \right|
       \left| \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right)\right] \rho(\lambda) d\lambda \right| +
  \int \left| \underline{A}(a, \lambda) \underline{B}(b', \lambda) \right|
       \left| \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}(b, \lambda) \right] \rho(\lambda) d\lambda \right|
</math>

By ({{EquationNote|4}}), this must be less than or equal to
<math display="block">\int \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \right] \rho(\lambda) d\lambda + \int \left[ 1 \pm \underline{A}\left(a', \lambda\right) \underline{B}(b, \lambda) \right] \rho(\lambda) d\lambda</math>
which, using the fact that the integral of {{math|''ρ''(''λ'')}} is 1, is equal to
<math display="block">2 \pm \left[ \int \underline{A}\left(a', \lambda\right) \underline{B}\left(b', \lambda\right) \rho(\lambda) d\lambda + \int \underline{A}\left(a', \lambda\right) \underline{B}(b, \lambda) \rho(\lambda) d\lambda \right]</math>
which is equal to <math>2 \pm \left[ E\left(a', b'\right) + E\left(a', b\right) \right]</math>.

Putting this together with the left-hand side, we have:
<math display="block">\left| E(a, b) - E\left(a, b'\right) \right| \; \leq 2 \; \pm \left[ E\left(a', b'\right) + E\left(a', b\right) \right]</math>
which means that the left-hand side is less than or equal to both <math>2 + \left[ E\left(a', b'\right) + E\left(a', b\right) \right]</math> and <math>2 - \left[ E\left(a', b'\right) + E\left(a', b\right) \right]</math>. That is:
<math display="block">\left| E(a, b) - E\left(a, b'\right) \right| \; \leq \; 2 - \left| E\left(a', b'\right) + E\left(a', b\right) \right|</math>
from which we obtain
<math display="block">2
  \;\geq\; \left| E(a, b) - E\left(a, b'\right) \right| + \left| E\left(a', b'\right) + E\left(a', b\right) \right|
  \;\geq\; \left| E(a, b) - E\left(a, b'\right) + E\left(a', b'\right) + E\left(a', b\right) \right|
</math>
(by the [[triangle inequality]] again), which is the CHSH inequality.