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Bob Badour wrote:
[..]
>
> I don't see how you draw the inference you draw. Clearly, P(B|A) and
> P(A|B) are constituent probabilities as well.
>
>
> Show how to derive P(A and B) given P(A) and P(B). If you
> > can show that, you can claim that "probability depends only on the
> > constituent probabilities". Are you unable to do that ?
>
> P(A and B) = P(B|A)P(A)
> P(A and B) = P(A|B)P(B)
>
> To derive P(A and B) one must know either P(B|A) or P(A|B) just as one
> must know the angle between two sides to use the cosine law. You fail to
> demonstrate anything useful or meaningful by your challenge.
Did I ? You have infomation P(A) and P(B), just as you might have had truth values for similar propositions in logic. However, in PT, be it frequentist or Bayesian, knowing P(A) and P(B) alone is not sufficient to derive P(A and B). Your referring to P(A|B) is not really an answer because you cannot determine P(A|B) based on P(A) and P(B) alone and thus cannot solve the problem. You need additional information so that you could compute P(A|B) or just P(A and B) directly, whichever is easier.
>
> Keith already specified the argument applies to a limit.
>
And this is an incorrect answer because the question was not what the limit of P(A and B) is but what exact value of P(A and B) given P(A) = 0 (or P(B)=0). There is an easy way to derive P(false) from Cox's axioms directly or from the sum/product rules.
>
> >>P(A) = 1
> >>P(A or B) = P(~(~A and ~B))
> >>P(A or B) = 1 - P(~A and ~B)
> >>P(A or B) = 1 - P(~B|~A)P(~A)
> >
> >
> > Since P(~A) equals zero, the above statement does not make sense.
>
> But Keith stated "in the limit of". Thus, one could read the first line
> of what he wrote as
>
> lim P(A) as P(A) -> 1
'Limit' is not just a magical word, "hey, presto". One has to show that such limit indeed exists, in what sense it exists, and even then it would be a useless exercise because there is a simple and direct answer (see above).
>
> which makes the other factor
>
> lim P(~A) as P(~A) -> 0.
>
> Thus, in the limit as P(A) approaches 1,
> P(A or B) = 1 - P(~B|~A)(1-P(A)) = 1
>
> If he rewrote the proof using explicit limit notation, would you still
> object to his proof?
See above. Received on Sun Jun 11 2006 - 21:46:00 CDT