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Bob Badour wrote:
> vc wrote:
>
> > Keith H Duggar wrote:
> > [Irrelevant stuff skipped]
> >
> > Assuming Bayesian treatment (which was not specified originally, mind
> > you), the derivation is still meaningless. Let's try some argument
> > from authority:
>
> [snip]
>
> Your whole dismissal, as I recall, depends on your observation:
>
> > P(B|A) def P(A and B)/P(A)
It does in the frequentist probability interpretation, yes.
> >
> > the requirement for such definition being that P(A) <>0, naturally.
>
> Keith used the equivalent definition:
In the Bayesian interpretation the product rule is a derivation form Cox's postulates, but even there P(B|A)P(A) is meaningful only when P(A) > 0.:
>From the Jaynes book:
"
In our formal probability symbols (those with a capital P)
P(A|B)
....
We repeat the warning that a probability symbol is undefined and
meaningless if the conditioning
statement B happens to have zero probability in the context of
our problem ...
"
Please see the book for details.
>
> P(A and B) = P(B|A)P(A), which places no requirements on P(A) because
> one does not divide by P(A).
Please see above or the book.
>
> In the case of P(A) = 0, P(A and B) = 0 and P(B|A) is indeterminate,
> which is to say, we don't care what it's value might be and it could be
> any real number; although, as a probability, we restrict it to real
> numbers in the range [0...1].
>
> Thus, both of Keith's proofs were entirely valid because he neither
> inferred nor concluded using the indeterminate P(B|A). He made the valid
> conclusion that P(A and B) = 0 when P(A) = 0.
Received on Sun Jun 11 2006 - 08:57:13 CDT