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
