Less Wrong is a community blog devoted to refining the art of human rationality. Please visit our About page for more information.

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.

Comment author: Quarkster 30 November 2009 11:58:09PM -2 points [-]

I don't see how it would be possible to draw conclusions about the future without assuming that the future will be like the past on the most stable level of organization can be identified. If this assumptions allows us to draw conclusions where they could not otherwise be drawn, isn't that winning?

Comment author: Psy-Kosh 08 July 2008 03:08:49PM 1 point [-]

Hrm... can't one at least go one step down past Occam's razor? ie, doesn't that more or less directly follow from P(A&B)<=P(A)?

Comment author: Quarkster 30 November 2009 11:15:28PM 3 points [-]

No, because you can't say anything about the relationship of P(A) in comparison to P(C|D)

In response to Belief in Belief
Comment author: Quarkster 29 November 2009 02:50:06AM 4 points [-]

"Those who find this confusing may find it helpful to study mathematical logic, which trains one to make very sharp distinctions between the proposition P, a proof of P, and a proof that P is provable"

This is a bit of a side question, but wouldn't a proof that P is provable be a proof of P? In fact, it sounds like a particularly elegant form of proof.