# 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

> In our formal probability symbols (those with a capital P)

> 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