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Bob Badour wrote:
> vc wrote:
>
> > 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 condi-
> > tioning statement B happens to have zero probability in the context of
> > our problem ...
> > "
> >
> > Please see the book for details.
>
> And since Keith never relied on any meaningful value for P(A|B) in his
> proof, I wonder what point you are trying to make.
>
Consider a partial function f(x) defined on the set N of natural numbers as:
if x > 10 f(x) = 2*x
Now, what would be the value of x*f(x) given x = 0 ? It's not zero, it's undefined, it simply does not exist, there is no such thing as 'unmeaningful' values of f(x) given x outside the function domain. Likewise, P(A|B) is defined as probability of proposition A given proposition B is true, so if B is false P(A|B) is undefined (see Jaynes for details). Received on Sun Jun 11 2006 - 15:31:19 CDT