Barker, Sean (UK) wrote:
> Waclaw
>
> From the point of view of practical mathematics, probability
> over continuous distributions is defined through probability density
> functions, with the probability for any interval being the integral of
> the pdf over that interval. If you reduce the interval to zero, then the
> probability goes to zero. No problem there. (01)
Depends on what you mean by 'reduce to zero'. There is the empty
interval, and infinitely many singleton intervals [x,x]. For the empty
interval, obviously P(X = x  x in the empty interval) = 0. For any
singleton interval, it is not that obvious to me that P(X = x) = 0 for
some x for which the 'reduced to zero' interval [x,x] is considered. (02)
>
> I'm not sure what you want to say with "P(exists x: > X=x) = 1"
> since the domain of X is fixed by definition, and therefore "the
> probability that X takes a value in its domain" doesn't seem to mean
> very much  probability is about random events, not things that are true
> by definition. (03)
This part comes from Kathy's post, and she should be asked: (04)
[KL] This raises some tricky mathematical issues. In finite domains, we
can equate probability zero with unsatisfiability and probability 1
with validity. But in the example I gave above, UniformRand = X is
satisfiable for any X between zero and one, yet (Prob (UniformRand =
X)) is zero for all X, and moreover, Prob(Exists X UniformRand=X))
is equal to 1. (05)
vQ (06)
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