billschnieder said:
There now appears to be two different meanings ascribed to what Bell is doing in equation (2), which I asked you earlier several times:
1) Bell is marginalizing with respect to λ.
2) Bell is calculating an expectation value for the probability P(AB|ab)
Which one is it? I see only a single integral and no summation, and you need one for each if you are doing both.
Yes, I didn't notice before that the left side of (2) was an expectation value rather than a straight probability. But it's not quite an expectation value for P(AB|ab) as you suggest, it's actually an expectation value for A*B, which is equivalent to a sum over all possible combinations of values for A and B of the quantity A*B*P(AB|a,b). Remember, though, Bell is assuming that the value of A and B is
completely determined by the values of a, b, and λ. So, the integral on the right of (2) is exactly equivalent to the following weighted sum of four integrals:
(+1)*(+1)\int P(A=+1, B=+1|a,b,\lambda)P(\lambda)\,d\lambda +
(+1)*(-1)\int P(A=+1, B=-1|a,b,\lambda)P(\lambda)\,d\lambda +
(-1)*(+1)\int P(A=-1, B=+1|a,b,\lambda)P(\lambda)\,d\lambda +
(-1)*(-1)\int P(A=+1, B=+1|a,b,\lambda)P(\lambda)\,d\lambda
The reason this works is because for any given value of λ, say λ=λ
i, three of the probabilities in the four integrals above will be equal to zero, while the other probability will be equal to 1. So by splitting up the single integral into the four above, you aren't overcounting or undercounting A*B*P(λ) for any specific value of λ, you're counting it exactly once. This is easier to see if you suppose λ can only take a discrete set of values from 0 to N, so the integral on the right side of (2) can be replaced by the sum \sum_{i=0}^N A(a,\lambda_i)*B(b,\lambda_i)*P(\lambda_i). Then if a,b,λ completely determine the values of A and B (which each take one of two values +1 or -1), that means the four-term sum (+1)*(+1)*P(A=+1,B=+1|a,b,λ
i) + (+1)*(-1)*P(A=+1,B=-1|a,b,λ
i) + (-1)*(+1)*P(A=-1,B=+1|a,b,λ
i) + (-1)*(-1)*P(A=-1,B=-1|a,b,λ
i) will always be equal to A(a,λ
i)B(b,λ
i) for each specific value of λ
i [for example, if a,b,λ
i determine that A=+1 and B=-1, then (+1)*(+1)*P(A=+1,B=+1|a,b,λ
i) + (+1)*(-1)*P(A=+1,B=-1|a,b,λ
i) + (-1)*(+1)*P(A=-1,B=+1|a,b,λ
i) + (-1)*(-1)*P(A=-1,B=-1|a,b,λ
i) = (+1)*(+1)*0 + (+1)*(-1)*1 + (-1)*(+1)*0 + (-1)*(-1)*0 = (+1)*(-1) = A(a,λ
i)B(b,λ
i)]. So, if we substitute the four-term sum in for the individual term A(a,λ
i)B(b,λ
i) in the sum over all possible values of λ I wrote above, we get:
\sum_{i=0}^N [(+1)*(+1)*P(A=+1,B=+1|a,b,\lambda_i)\,+\,(+1)*(-1)*P(A=+1,B=-1|a,b,\lambda_i)+ \,(-1)*(+1)*P(A=-1,B=+1|a,b,\lambda_i)\,+\,(-1)*(-1)*P(A=-1,B=-1|a,b,\lambda_i)]*P(\lambda_i)
Which can be split up into the following four sums:
\sum_{i=0}^N (+1)*(+1)*P(A=+1,B=+1|a,b,\lambda_i)*P(\lambda_i) +
\sum_{i=0}^N (+1)*(-1)*P(A=+1,B=-1|a,b,\lambda_i)*P(\lambda_i) +
\sum_{i=0}^N (-1)*(+1)*P(A=-1,B=+1|a,b,\lambda_i)*P(\lambda_i) +
\sum_{i=0}^N (-1)*(-1)*P(A=-1,B=-1|a,b,\lambda_i)*P(\lambda_i)
...which is just the discrete version of the four integrals I wrote before.
So, the left side is an expectation value which can be broken up into a weighted sum of four probabilities of the form P(AB|ab), and the right side can be broken up into a weighted sum of four integrals or sums over all possible values of λ of terms of the form P(AB|a,b,λ). For example, on the left side one of the four weighted probabilities is (+1)*(-1)*P(A=+1,B=-1|ab), and on the right side one of the four weighted integrals is (+1)*(-1)\int P(A=+1, B=-1|a,b,\lambda)P(\lambda)\,d\lambda. So if you take the marginalization equation P(A=+1,B=-1|a,b) = \int P(A=+1, B=-1|a,b,\lambda)P(\lambda)\,d\lambda and then multiply both sides by A*B=(+1)*(-1) and add this equation to three other marginalization equations where both sides have been multiplied by the corresponding value of A*B, you get something mathematically equivalent to equation (2) in Bell's proof.
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