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DanielLC comments on [Link] The Bayesian argument against induction. - Less Wrong Discussion
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Comments (27)
Was that supposed to be (A & B) v (A & ~B)? What it has is always true. Also, what is =||=?
If I can understand this correctly, they're saying that induction is false because the only accuracy in it is due to the hidden Bayes-structure. This is true of everything.
DanielLC,
Hi, I am the author.
The =||= just means bientailment. It's short for,
A |= (A v B) & (A v ~B) and (A v B) & (A v ~B) |= A
Where |= means entailment or logical consequence. =||= is analogous to a biconditional.
The point is that each side of a bientailment is logically equivalent, but the breakdown allows us to see how B alters the probability of different logical consequences of A.
Thank you, I think I understand most of it now. I don't see (at this hour) where the absolute values come from, but that doesn't seem to matter much. Let's focus on this line:
The conjunction of those two does contradict itself, and if you actually write out the probability of the contradiction -- using the standard product rule p(CD)=p(D)p(C|D) rather than multiplying their separate probabilities p(D)p(C) together -- you'll see that it always equals zero.
But each separate claim (A, and B~A), can increase in probability provided that they each take probability from somewhere else, namely from ~B~A. I see no problem with regarding this as an increase for our subjective confidence in A and a separate increase for B~A. Again, each grows by replacing an option (or doubt) which no longer exists for us. Some of that doubt-in-A simply changed into a different form of doubt-in-A, but some of it changed into confidence. The total doubt therefore goes down even though one part increases.