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

From: vc <boston103_at_hotmail.com>
Date: 9 Jun 2006 18:04:47 -0700

Keith H Duggar wrote:
> vc wrote:
> > Erwin wrote:
> > > > This is why PT is a /generalization/ of logic. It
> > > > reduces to logic when applied to truth-valued
> > > > statements. Just as gamma reduces to factorial for
> > > > natural arguments. (Again no quibbles about offset by
> > > > 1 etc).
> > > >
> > >
> > > You mean like :
> > >
> > > AND (p1, p2) === p1*p2
> >
> > P(p1 and p2) is not equal P(p1)*P(p1) in general, so no
> > such 'generalization' is possible.
>
> Wow! vc is going off the VI deep end at the moment. "no such
> 'generalization' is possible"? Saying that PT is not a
> generalization is one thing; but, none possible??
>

Recall that the OP claimed that PT is a logic 'generalization' in the sense that 'probability depends only on the constituent probabilities'.  He failed to prove the bizzare assertion and refused/or was unable to solve the trivial puzzle that disproves his statement.

> Erwin, what vc was referring to is that
>
> P(AB) = P(A|B)P(B) -or-
> P(AB) = P(B|A)P(A)
>

Note, that I said nothing about conditional probabilities. I merely requested to compute the P(A and B) probability in terms of P(A) and P(B) which was promised by the OP (see above).

> where | means given and AB is short for "A and B". This is
> called the product rule. Something that vc seems not to know
> (given his questions in the other post) is that in the limit
> of true (0) and false (1) the conditional probability
> product rule reduces to the logical conjunction truth
> table. Here is the proof
>
> g : P(A) = 0
> p : P(AB) = P(B|A)P(A)
> u : P(AB) = 0

Unfortunately, it's no proof but just mindless playing with formulas. The conditional probability is *defined* as

P(B|A) def P(A and B)/P(A)

the requirement for such definition being that P(A) <>0, naturally. The definition can be found in any introductory PT textbook.

>
> g : P(B) = 0
> p : P(AB) = P(A|B)P(B)
> u : P(AB) = 0
>

See above.

> g : P(A) = 1
> g : P(B) = 1
> s : P(~B) = 0
> m : P(A) = P(AB) + P(A~B)
> p : P(A) = P(AB) + P(A|~B)P(~B)
> u : P(A) = P(AB)
> c : P(AB) = P(A)
> u : P(AB) = 1

This is even funnier. First, we do not know what P(A|B) is (see above) and second the question is what kind of events might A and B be if the probability of either is one ? What about P(A or B) given the respective probabilities are one ? Is it two by any chance ? (I asked the same question in another message).

>
> thus
>
> P(A) : P(B) : P(AB)
> 0 : 0 : 0
> 0 : 1 : 0
> 1 : 0 : 0
> 1 : 1 : 1
>

Unfortunately, there can be no 'thus'.

> descriptions
> g : given
> p : product rule
> s : sum rule
> m : marginalization (derived from sum rule)
> u : substitution
>
> -- Keith --
Received on Sat Jun 10 2006 - 03:04:47 CEST

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