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Re: Programming is the Engineering Discipline of the Science that is Mathematics

From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Mon, 12 Jun 2006 00:22:34 GMT
Message-ID: <et2jg.20842$A26.479278@ursa-nb00s0.nbnet.nb.ca>


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

Is the above not true regardless of P(B|A) ? Received on Sun Jun 11 2006 - 19:22:34 CDT

Original text of this message

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