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.
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.
Not a sole function of its complexity, but if A and B have the same complexity, and you have no further initial reason to place more belief in one or the other, then would you agree that you should assign P(A) = P(B)?
Complexity is a function of the hypothesis. Other functions can be made. In fact, complexity isn’t even a specific function. What language are we using?
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)?
No, because you can’t say anything about the relationship of P(A) in comparison to P(C|D)
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?)
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.
How do you know that? Why must P(A) be a function of the complexity of A?
Also, this is only sufficient to yield a bound on Occam’s razor. How do you know that the universe doesn’t favor a given complexity?
Not a sole function of its complexity, but if A and B have the same complexity, and you have no further initial reason to place more belief in one or the other, then would you agree that you should assign P(A) = P(B)?
Complexity is a function of the hypothesis. Other functions can be made. In fact, complexity isn’t even a specific function. What language are we using?