Quarkster comments on Where Recursive Justification Hits Bottom - Less Wrong

41 Post author: Eliezer_Yudkowsky 08 July 2008 10:16AM

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Comment author: Psy-Kosh 01 December 2009 12:35:09AM 0 points [-]

Not sure why you were silently voted down into negatives here, but if I understand your meaning correctly, then you're basically saying this:

P(A)*P(B|A)

vs

P(C)

aren't automatically comparable because C, well, isn't A?

I'd then say "if C and A are in "similar terms"/level of complexity... ie, if the principle of indifference or whatever would lead you to assign equivalent probabilities to P(C) and P(A) (suppose, say, C = ~A and C and both have similar complexity), then you could apply it.

(or did I miss your meaning?)

Comment author: Quarkster 01 December 2009 01:00:42AM *  1 point [-]

You got my meaning. I have a bad habit of under-explaining things.

As far as the second part goes, I'm wary of the math. While I would imagine that your argument would tend to work out much of the time, it certainly isn't a proof, and Bayes' Theorem doesn't deal with the respective complexity of the canonical events A and B except to say that they are each more probable individually than separately. Issues of what is meant by the complexity of the events also arise. I suspect that if your assertion was easy to prove, then it would have been proven by now and mentioned in the main entry.

Thus, while Occam's razor may follow from Bayes' theorem in certain cases, I am far from satisfied that it does for all cases.