Manfred comments on [Link] The Bayesian argument against induction.

Manfred 19 Jul 2011 8:00 UTC
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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.
- Zack_M_Davis 19 Jul 2011 18:54 UTC
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  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.
  - Manfred 19 Jul 2011 19:29 UTC
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    Okay. Yay, a use for the retract button where it’s still good to have the text visible!
- leekelly 19 Jul 2011 12:58 UTC
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  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?