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Bob Badour wrote:
>
> Are you seriously suggesting that true and false are trivial and
> uninteresting? Should we all pack up and go home?
>
I am suggesting that a meaningful PT statement reduction to a propositional logic statement is trivial and uninteresting. Also, there is a problem with the conditional probability not being equal the probability of the conditional (see Lewis's result) which makes conditional probabily untranslatable to modus ponens in principle.
Going in the opposite direction, generalization, PT is not truth functional , that is the probability of a compound statement is not determined solely by its components probabilities (see my trivial puzzle). Also, importantly, it appears impossible to find an axiom system for any known probabilistic logic that would be sound and complete (except some special cases). Obviously, lack of such axiom system makes a formal derivation (a hallmark of any logic) impossible.
Apparently, despite obvious similarities and profound connections, both had better be used what they are best at and attempts to merge them do not seem very productive (see abundant literature on probabilistic logics).
> >
> > What Jaynes did in his derivation of the sum/product rules has got
> > nothing to do with your mindless playing with formulas. See the
> > argument from authority in my previous messages.
>
> Your argument from authority was flawed. I will reply in the other thread.
The argument from authority was a quote from Jaynes' book , not mine.
> >>
> >>that should have been P(p1|p2) != P(p1).
> >
> > That's assuming that P(p1|p2) even makes sense. More general
> > formulation of such independence is just P(p1 and p2) = P(p1)* P(p2).
>
> The formulation is neither more general nor less general. It is, in
> fact, a simple substitution of the equation describing independence:
>
> (1) P(p1|p2) = P(p1)
Again, the substitution is possible only when P(p2) > 0. (See the Jaynes book).
>
> into the formula for conditional probability:
>
> (2) P(p1 and p2) = P(p1|p2)*P(p2)
>
> Substitute (1) into (2) gives P(p1 and p2) = P(p1)*P(p2)
Received on Sun Jun 11 2006 - 08:36:16 CDT