vc wrote:
> Bob Badour wrote:
>
>>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
>
> There is a much simpler argument for P( false and B) = zero. First,
> you prove that P(false) = zero (you have to prove that because you do
> not have any rules yet) and then the answer is obvious.
You are assuming that the literal, false, is substitutable for A. While
that's a valid assumption in logic, I am unsure whether it is valid in
probability theory. I have seen nothing to indicate the substitution is
valid and the method seems to beg the question.
>>Is the above not true regardless of P(B|A) ?
>
> P(B|A) does not exist if P(A) = 0, and we need not lim(P(A and B)
> (even if it exists) but just P(A and B) which does exist (see above).
In other words, you are now changing your position. Instead of arguing
that Keith cannot prove what he proved, you are now saying that of all
the perfectly valid proofs, you prefer a different one. That seems like
a real waste of everybody's time.
Received on Sun Jun 11 2006 - 22:21:44 CDT