Oracle FAQ | Your Portal to the Oracle Knowledge Grid |
Home -> Community -> Usenet -> comp.databases.theory -> Re: Programming is the Engineering Discipline of the Science that is Mathematics
Keith H Duggar wrote:
[Irrelevant stuff skipped]
> > > 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.
>
> No but you claimed "P(p1 and p2) is not equal P(p1)*P(p1) in
> general" which is of course pointing to the possibility of
> p1 and p2 being dependent which is of course equivalent to a
> conditional statement P(p1|p2) = P(p1).
It's getting funnier. Are claiming now that, given two propositions p1 and p2 with probabilities P(p1) and P(p2), "the possibility of p1 and p2 being dependent [is] equivalent to a conditional statement P(p1|p2) = P(p1)" ? You've got it backwards, amigo.
>
> > 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).
>
> Which the proof you called "mindless playing with formulas"
> provides. How about this, since you believe I'm wrong why
> don't YOU provide a proof? I would love to see what you
> consider is /not/ "mindless playing with formulas"
>
> > > 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.
>
> LMAO. "mindless playing with formulas"? You vc are a
> mindless idiot who has an extremely limited understanding of
> probability theory. You clearly know nothing about the Cox
> formulation or his derivation of the product rule IN THE
> FORM I gave above as p) from functional requirements
> ALONE.
Assuming Bayesian treatment (which was not specified originally, mind
you), the derivation is still meaningless. Let's try some argument
from authority:
"
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 conditioning
statement B happens to have zero probability in the context of
our problem ...
"
[chaff skipped]
> > Unfortunately, there can be no 'thus'.
>
> Unfortunately, you are acting like a total VI moron at this
> point. Read Jaynes' book or Cox's paper, comprehend them,
> educate yourself. After that come here and apologize for
> being a VI. Before that STFU.
Unfortunately, there still cannot be any 'thus'. See the argument from authority above.
>
> -- Keith --
Received on Sat Jun 10 2006 - 22:28:36 CDT