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Bob Badour wrote:
> vc wrote:
>
> > 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).
>
> Are you quibbling that Keith should have expressed his proof in the
> limit as P(A) approaches zero? Would that overcome your objection?
>
> 0 = lim P(B|A)P(A) as P(A) -> 0
There is a much simpler argument for P( false and B) = zero. First, you prove that P(false) = zero (you have to prove that because you do not have any rules yet) and then the answer is obvious.
>
> Is the above not true regardless of P(B|A) ?
P(B|A) does not exist if P(A) = 0, and we need not lim(P(A and B) (even if it exists) but just P(A and B) which does exist (see above). Received on Sun Jun 11 2006 - 20:49:38 CDT