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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)
>> > the requirement for such definition being that P(A) <>0, naturally. >> >>Keith used the equivalent definition:
>>From the Jaynes book:
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.
>>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 Sun Jun 11 2006 - 14:26:48 CDT