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For anyone still following this, I have tried to restate my arguments in a new way here:
http://www.criticalrationalism.net/2011/07/20/more-on-inductive-probability/
I'm back and there's been no response, so I'll be specific. Starting from
p(A v B|B) – p(A v B) + p(A v ~B|B) – p(A v ~B) = .15
Using p(X v Y) = p(X) + p(Y) - p(XY), we get
.15 = p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)
= p(A|B) + 1 - p(A) - p(A) - p(B) + p(AB) + p(A|B) - p(A) - p(~B) + p(A~B)
= 2 p(A|B) - 2 p(A)
= twice the thing you started from, which is bad.
Manfred,
I calculated the result for about three different sets of probabilities before making the original post. The equation was correct each time. I could have just been mistaken, of course, but even Zack (the commenter above) conceded that the equation is true.
EDIT: Oh, I see now. You have changed all my disjunctions into conjunctions. Why?
In the very next equation after "A = (A v B) & (A v ~B)", you write:
p(A v B|B) – p(A v B) + p(A v ~B|B) – p(A v ~B) = .15
This is the equation where you put in the plus signs. Additionally, you can break things down like that inside the P() operator, but you can't just move that to outside the P() operator, because things might be correlated (and, in the case of B and ~B, certainly are).
Well, it wasn't actually an equation. That's why I used the =||= symbol. It was a bientailment. It asserts logical equivalence (in classical logic), and it means something slightly different than an equals symbol. The equation with the plus signs and the logical equivalence shouldn't be confused.
In response to
[Link] The Bayesian argument against induction.
Whoops - v is apparently intended to mean mean "or" or "union." It's just that the author incorrectly translates P(A v B) & P(A v ~B) into P(A v B) + P(A v ~B).
Hi,
I am the author. It wasn't a mistranslation. The logical equivalence was not translated into anything. It was merely intended to break down A according to its logical consequences shared with B. I never wrote "P(A v B) + P(A v ~B)," because that would be irrelevant.
In response to
[Link] The Bayesian argument against induction.
A =||= (A v B) & (A v ~B)
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.
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There's no abstract. It is not obvious what is the point of the article is.
I linked to the previous post. That begins with something like an abstract: a statement of intent, at least.
This isn't an article or a paper: it's a blog post.