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

From: vc <boston103_at_hotmail.com>
Date: 10 Jun 2006 20:28:36 -0700
Message-ID: <1149996515.976096.54170@f6g2000cwb.googlegroups.com>


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

Original text of this message

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