Re: Programming is the Engineering Discipline of the Science that is Mathematics
Date: Mon, 12 Jun 2006 03:09:47 GMT
> Bob Badour wrote: > [..] >
>>I don't see how you draw the inference you draw. Clearly, P(B|A) and
>>P(A|B) are constituent probabilities as well.
>> Show how to derive P(A and B) given P(A) and P(B). If you
>>>can show that, you can claim that "probability depends only on the
>>>constituent probabilities". Are you unable to do that ?
>>P(A and B) = P(B|A)P(A)
>>P(A and B) = P(A|B)P(B)
>>To derive P(A and B) one must know either P(B|A) or P(A|B) just as one
>>must know the angle between two sides to use the cosine law. You fail to
>>demonstrate anything useful or meaningful by your challenge.
> > Did I ? You have infomation P(A) and P(B), just as you might have had > truth values for similar propositions in logic. However, in PT, be > it frequentist or Bayesian, knowing P(A) and P(B) alone is not > sufficient to derive P(A and B).
So? In relativistic mechanics, one must know not only the mass of an object but the speed of the object's frame of reference relative to the speed of light to calculate the effects of an applied force. In classical mechanics, one need only know the force and the mass.
Are you suggesting that makes relativity any less a generalization of classical mechanics?
If you tried to impeach probability theory's generalization of logic, you failed. You merely demonstrated it's similarity to other generalizations.
Your referring to P(A|B) is not
> really an answer because you cannot determine P(A|B) based on P(A) and > P(B) alone and thus cannot solve the problem.
One cannot determine the speed of an object relative to the speed of light by it's mass alone. Thus, you cannot solve the problem of the effects of a force applied to the object. Big deal. Unless your point is that relativity is not a generalization of classical mechanics, your observation is pointless.
If you continue to repeat non sequitur even after the non sequitur is explained to you, I will have to agree with Keith's assessment that you are a VI.
You need additional
> information so that you could compute P(A|B) or just P(A and B) > directly, whichever is easier.
Generalizations require additional information. Big deal. Relativity requires the speed of the frame of reference relative to the speed of light. The cosine law needs the angle between the two sides of known length. Unless your point is that relativity is not a generalization of classicial mechanics and that the cosine law is not a generalization of the pythagorean theorem, your statements are pointless.
>>Keith already specified the argument applies to a limit.
The exact value of the sum of 1/2^i for i in [0..infinity] is 2. The
exact value of the sum of 3 times 10^(-k) for k in [1..infinity] is 1/3.
The exact value of P(A and B) is 0 give P(A) = 0 or P(B) = 0.
> And this is an incorrect answer because the question was not what the
> limit of P(A and B) is but what exact value of P(A and B) given P(A) =
> 0 (or P(B)=0). There is an easy way to derive P(false) from Cox's
> axioms directly or from the sum/product rules.
The exact value of the sum of 1/2^i for i in [0..infinity] is 2. The exact value of the sum of 3 times 10^(-k) for k in [1..infinity] is 1/3. The exact value of P(A and B) is 0 give P(A) = 0 or P(B) = 0.
One uses limits to ascertain the truth of all three of the above statements.
>>>>P(A) = 1
>>>>P(A or B) = P(~(~A and ~B))
>>>>P(A or B) = 1 - P(~A and ~B)
>>>>P(A or B) = 1 - P(~B|~A)P(~A)
>>>Since P(~A) equals zero, the above statement does not make sense.
>>But Keith stated "in the limit of". Thus, one could read the first line
>>of what he wrote as
>>lim P(A) as P(A) -> 1
> > 'Limit' is not just a magical word, "hey, presto". One has to show > that such limit indeed exists, in what sense it exists, and even then > it would be a useless exercise because there is a simple and direct > answer (see above).
The only simple and direct answer I see above is the one Keith already gave that you refuse to accept no matter how valid it is.
The limit exists. It exists in the sense that for the entire valid range the limit of P(A and B) approaches 0 when P(A) approaches 0 or when P(B) approaches 0 or both.
>>which makes the other factor
>>lim P(~A) as P(~A) -> 0.
>>Thus, in the limit as P(A) approaches 1,
>>P(A or B) = 1 - P(~B|~A)(1-P(A)) = 1
>>If he rewrote the proof using explicit limit notation, would you still
>>object to his proof?
> See above.
Your obstinate evasiveness is just going to get you ignored for intellectual dishonesty. Received on Mon Jun 12 2006 - 05:09:47 CEST