From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/4830 Path: news.gmane.org!not-for-mail From: Steve Vickers Newsgroups: gmane.science.mathematics.categories Subject: Re: Axioms of elementary probability Date: Tue, 12 May 2009 16:34:07 +0100 Message-ID: References: Reply-To: Steve Vickers NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1242220999 5522 80.91.229.12 (13 May 2009 13:23:19 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 13 May 2009 13:23:19 +0000 (UTC) To: David Espinosa , Original-X-From: categories@mta.ca Wed May 13 15:23:09 2009 Return-path: Envelope-to: gsmc-categories@m.gmane.org Original-Received: from mailserv.mta.ca ([138.73.1.1]) by lo.gmane.org with esmtp (Exim 4.50) id 1M4EQO-0007aW-Fu for gsmc-categories@m.gmane.org; Wed, 13 May 2009 15:23:08 +0200 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1M4Di4-0002f5-Bm for categories-list@mta.ca; Wed, 13 May 2009 09:37:20 -0300 In-Reply-To: Original-Sender: categories@mta.ca Precedence: bulk Xref: news.gmane.org gmane.science.mathematics.categories:4830 Archived-At: Dear David, On structure: Domain L (say) just needs to be distributive lattice - not Boolean algebra. The axiom P(top) = 1 looks an obvious dual to P(bottom) = 0, but there's a lot to be gained from considering P with codomain [0,infinity] and forgetting P(top) = 1. Maps P: L -> [0,infinity] satisfying P(0) = 0 and the third (modular) law are called valuations - I believe this dates back to Birkhoff's book on lattice theory. In the case where L is a frame (complete lattice, with binary meet distributing over all joins) and P is Scott continuous, P is called a continuous valuation. These have been studied in domain theory (Jones, Plotkin: probabilistic power domain) and general locales (including by Heckmann, by Coquand and Spitters and by myself). More generally, the domain of P can fruitfully be any commutative monoid M. There is a universal valuation L -> M(L) in this generalized sense, with M(L) got by adjoining finite monoid structure to L and forcing the two laws. Coquand and Spitter cite an interesting construction of M(L) by Horn and Tarski. Let L* be the set of finite lists over L, and define a preorder on L* by [x_i]_{1 in I} <= [y_j]_{j in J} if for every natural number k, \/{x_K | K subseteq I, |K| = k} <= \/{y_K' | K' subseteq J, |K'| = k} where x_K = /\{x_i | i in K} etc. Then M(L) is isomorphic to L*/(equ reln corresponding to <=). The relations holding in M(L) are what can be proved from the theory. You give a ternary inclusion-and-exclusion for P(A u B u C). If you bring all the negative terms from right to left, it will still hold in M(L), and can be generalized from ternary to n-ary. I think you will get the dual (for P(A n B n C)) by considering L^op. Another interesting relation, which can be used in proving the Horn-Tarski result, is this: Sigma_{i = 0}^{n-1} x_i = Sigma_{k = 1}^{m} \/{x_I | I subseteq {0, ..., n-1}, |I| = k} Regards, Steve Vickers. References: Jones & Plotkin: "A probabilistic powerdomain of evaluations", LICS'89. Horn & Tarski: "Measures in Boolean algebras", Trans. Amer. Math. Soc. 64 (1948) Heckmann: "Probabilistic powerdomain, information systems and locales", MFPS VIII, Springer LNCS 802 (1994) Vickers: "A localic theory of lower and upper integrals", Math. Logic Quarterly 54 (2008) Coquand & Spitters: "Integrals and valuations", Journal of Logic and Analysis 1:3 (2009 David Espinosa wrote: > > Here's a question about elementary (naive, finitist) probability. > The proper, self-dual axioms for elementary probability are presumably > > P(0) = 0 > P(X) = 1 > P(A u B) + P(A n B) = P(A) + P(B) > > P's domain is a boolean algebra. P's codomain is [0,1]. > What kind of algebraic structure is [0,1] in this case? > > What can we prove from this theory? The best I can think of is inclusion / > exclusion: > > P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C) + P(A n > B n C) > P(A n B n C) = P(A) + P(B) + P(C) - P(A u B) - P(A u C) - P(B u C) + P(A u > B u C) > > Thanks, > > David > > > >