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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) > > the requirement for such definition being that P(A) <>0, naturally.
Keith used the equivalent definition:
P(A and B) = P(B|A)P(A), which places no requirements on P(A) because one does not divide by P(A).
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 Sat Jun 10 2006 - 23:14:27 CDT