Graduate Bell inequalities demonstration

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The correlation function C(x y) is defined in the context of Bell inequalities, with the relationship −1 ≤ C(x y) ≤ 1 derived from the principles of probability theory. This relationship stems from the positivity of probabilities, ensuring that P(+ + |x y), P(− − |x y), P(+ − |x y), and P(− + |x y) are all non-negative. Additionally, the normalization condition requires that the sum of these probabilities equals one. These fundamental properties of probability lead to the established bounds for the correlation function. Understanding this relationship is crucial for exploring the implications of Bell's theorem in quantum mechanics.
microsansfil
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I try to understand why the correlator boundaries is −1 ≤ C(x y) ≤ 1
Hello,

In this thesis https://tel.archives-ouvertes.fr/tel-01743877/document at "1.2.2 Bell inequalities" page 7-8 it's define a correlation function :

C(x y) = P(+ + |x y) + P(− − |x y) − P(+ − |x y) − P(− + |x y), with −1 ≤ C(x y) ≤ 1.

How do one get to this relationship −1 ≤ C(x y) ≤ 1 ?

Thanks
Patrick
 
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microsansfil said:
Summary:: I try to understand why the correlator boundaries is −1 ≤ C(x y) ≤ 1

Hello,

In this thesis https://tel.archives-ouvertes.fr/tel-01743877/document at "1.2.2 Bell inequalities" page 7-8 it's define a correlation function :

C(x y) = P(+ + |x y) + P(− − |x y) − P(+ − |x y) − P(− + |x y), with −1 ≤ C(x y) ≤ 1.

How do one get to this relationship −1 ≤ C(x y) ≤ 1 ?

Thanks
Patrick
It's just a consequence of positivity $$P(\pm \pm' | xy) \geq 0$$ and normalisation $$P(++|xy) + P(+-|xy) + P(-+|xy) + P(--|xy) = 1$$ of the probabilities.
 
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