Re: Programming is the Engineering Discipline of the Science that is Mathematics
From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Sun, 11 Jun 2006 19:26:48 GMT
Message-ID: <Y7_ig.20702$A26.476055_at_ursa-nb00s0.nbnet.nb.ca>
>
>
> It does in the frequentist probability interpretation, yes.
>
>
>
> In the Bayesian interpretation the product rule is a derivation form
> Cox's postulates, but even there P(B|A)P(A) is meaningful only when
> P(A) > 0.:
>
>
> "
> In our formal probability symbols (those with a capital P)
>
>
> P(A|B)
> ....
>
>
> We repeat the warning that a probability symbol is undefined and
> meaningless if the condi-
> tioning statement B happens to have zero probability in the context of
> our problem ...
> "
>
> Please see the book for details.
>
> Please see above or the book.
>
>
Date: Sun, 11 Jun 2006 19:26:48 GMT
Message-ID: <Y7_ig.20702$A26.476055_at_ursa-nb00s0.nbnet.nb.ca>
vc 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.
>>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 - 21:26:48 CEST