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leekelly comments on [Link] The Bayesian argument against induction. - Less Wrong Discussion

4 Post author: Peterdjones 18 July 2011 09:52PM

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Comment author: leekelly 19 July 2011 01:18:39AM 2 points [-]

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.

Comment author: Manfred 19 July 2011 01:40:36AM *  0 points [-]

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).

Comment author: leekelly 19 July 2011 01:49:32AM 0 points [-]

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.

Comment author: Manfred 19 July 2011 08:00:16AM *  0 points [-]

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.

Comment author: Zack_M_Davis 19 July 2011 06:54:44PM *  2 points [-]

It looks like you simplified p(AB|B) as p(A), but in fact p(AB|B)=p(ABB)/p(B)=p(AB)/p(B)=p(A|B). (I made a similar mistake earlier.)

I get 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|B) - p(A) - p(B) + p(AB) + p(A|B) + 0 - 0 - p(A) - 1 + p(B) + p(A~B)

= - p(A) - p(B) + p(AB) + p(A|B) - p(A) + p(B) + p(A~B)

= - p(A) + p(AB) + p(A|B) - p(A) + p(A~B)

= p(A|B) -2p(A) +p(AB) + p(A~B)

= p(A|B) -2p(A) + p(A)

= p(A|B) - p(A)

But this is quod erat demonstrandum.

Comment author: Manfred 19 July 2011 07:29:07PM 0 points [-]

Okay. Yay, a use for the retract button where it's still good to have the text visible!

Comment author: leekelly 19 July 2011 12:58:55PM *  2 points [-]

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?

Comment author: Manfred 19 July 2011 01:59:37AM 0 points [-]

Ah, I'm sorry. Do you agree that the equation I quoted above is incorrect, though? I'm going to have to leave now, but the relevant basic equations of probability are P(X v Y) = P(X) + P(Y) - P(X&Y), and P(X&Y) = P(X) * P(X | Y)

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