Talk:Probability axioms
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Would it be possibvle to rejig this in terms of Borel sets? Just a more elegant way of expressing it IMHO; here it's as if the defintion of sigma algebra comes out of probability theory. It might be moe confusing for the general reader however.
It's more confusing for me, anyway. The universe S is not necessarily a topological space, so what is a Borel set in this context?
Fool 23:42 Mar 11, 2003 (UTC)
"Fool" is right. Borel sets are by definition members of the sigma-algebra generated by a topology. But there need not be any topology on a probability space. Or, at least, no topology is explicitly contemplated by the conventional Kolmogorovian definition.
I have moved this article to "Probability axioms" (plural!). Usually it is better to use the singular than the plural in the title of an article; "zebra" is better than "zebras". But in the case of this article, it is colossally silly. This is not about axioms as individual things; it is about systems of axioms, or, at least, about one particular system of axioms --- the one formulated by Kolmogorov. Michael Hardy 00:41 Mar 12, 2003 (UTC)
I think Cox's axioms should be stated here in addition to Kolmogorov's; if I'm not mistaken, the Kolmogorov axioms are derivable as theorems from Cox's -- which, again IINM, was Cox's point, that the accepted laws of probability are derivable from more basic assumptions. The one bit that doesn't carry over, if we start from Cox's axioms, is countable additivity -- IINM finite additivity derives from Cox's axioms but not countable additivity. Wile E. Heresiarch 15:30, 27 Dec 2003 (UTC)
I believe Axiom 1 should be stated simply as P(E) >= 0 rather than 0 <= P(E) <=1, since P(E) <=1 is actually a consequence of P(S) = 1, P(E) >= 0 and countable additivity.