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harrylin
#15
Mar1-12, 02:06 PM
P: 3,187
Quote Quote by jtbell View Post
I'm pretty much a spectator in these discussions, but I'd like to point out that there was a long thread here about three years ago, about Jaynes's objections to Bell:

http://www.physicsforums.com/showthread.php?t=283519

This was before you joined PF, so you may not have seen this. It may or may not fit in with the direction you were planning to go.

It was split off from another thread, by the way, which is why it appears to start in the middle of a discussion.
I now had a better look at it, and I think that in particular posts #26 and #31 are important. Anyway I'll give a short summary of how I now see it.

If Jaynes' criticism focuses on Bell's equation no.11 in his "socks" paper, it was perhaps due to a misunderstanding about what Bell meant (his comments were based on an earlier paper).

P(AB|a,b,x) = P(A|a,x) P(B|b,x) (Bel 11)

Here x stands for Bell's lambda, which corresponds to the circumstances that lead to a single pair correlation (in contrast to my earlier X, which causes the overall correlation for many pairs).

According to Jaynes it should be instead, for example:

P(AB|a,b,x) = P(A|B,a,b,x) P(B|a,b,x)

Perhaps Jaynes thought that Bell meant:

P(AB|a,b,X) = P(A|a,X) P(B|b,X)

in which case Jaynes claimed that:

P(AB|a,b,X) = P(A|B,a,b,X) P(B|a,b,X)

This is really tricky.

However, he really was disagreeing with the integral equation.
According to him, it should not be:

P(AB|a,b) = ∫ P(A|a,x) P(B|b,x) p(x) dx

but:

P(AB|a,b) = ∫ P(AB|a,b,x) P(x|a,b) dx

and thus:

P(AB|a,b) = ∫ P(AB|a,b,x) p(x) dx = ∫ P(A|B,a,b,x) P(B|a,b,x) p(x) dx

Is my summary of the disagreement correct?

What is the significance of little p(x) instead of P(x)?