A possible generalization of the Cox Theorems (boolean algebra => bayesian probabilities) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:48:55Z http://mathoverflow.net/feeds/question/16128 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16128/a-possible-generalization-of-the-cox-theorems-boolean-algebra-bayesian-probab A possible generalization of the Cox Theorems (boolean algebra => bayesian probabilities) Rorsa 2010-02-23T05:33:58Z 2011-02-01T11:13:46Z <p>This post is an attempt to gather people to solve a particular problem in mathematics, something that can actually be published and seems to me simple enough to test this mathoverflow as a problem-solving environment, in the spirit of the Polymath project.</p> <p>Bayesian probabilities are usually justified by the Cox theorems, that can be written this way:</p> <p><em>Under some technical assumptions (continuity, etc, etc...), given a set $P$ of objects $A, B, C, \ldots$, with a boolean algebra defined over it with operations $A \vee B$ (and) and $A | B$ (or) such that</em>:</p> <p>1) $A \vee B = B \vee A$ </p> <p>2) $A \vee (B \vee C) = (A \vee B) \vee C$</p> <p>3) $A | (B \vee C) = (A|B) \vee (A|C)$</p> <p><em>and a "valuation":</em></p> <p>$f : P \rightarrow \mathcal{R}$</p> <p><em>there is a strictly monotonic "regraduation function"</em> $R : \mathcal{R} \rightarrow \mathcal{R}$ <em>such that</em>:</p> <p>$R(f(A | B)) = R(f(A)) + R(f(B))$ (sum rule) </p> <p>and</p> <p>$R(f(A\vee B)) = R(f(A) ) R(f(B))$ (product rule)</p> <p>This theorem allows one to show that any system designed to "evaluate" boolean expressions consistently with a single real number redunds in the laws of classical probability (this can be seen shortly here: arxiv:physics/0403089 and more thoroughly here: arxiv:abs/0808.0012)</p> <p>Recently this has been extended for valuations of the type $f : P \rightarrow \mathcal{R}^2$ in <a href="http://arxiv.org/abs/0907.0909" rel="nofollow">http://arxiv.org/abs/0907.0909</a> and they proved that there are just 5 canonical valuations compatible with the underlying Boolean algebra (one of them giving a complex field structure to the "valuation" field). </p> <p>My question/proposal is: is it possible/interesting/feasible to classify at least a class of valuations of the type:</p> <p>$f : P \rightarrow W$ </p> <p>where W is a continuous manifold (a "valuation field", like $\mathcal{R}$ in the original Cox theorems)? If we retrict our attention to $\mathcal{R}^n$ for example, is there, for each n, a set of canonical valuations to which all others can be reduced after a regraduation?</p> <p>If this can be done, are those nice rules for inference in some sense? Are they useful as inference tools in specific situations? </p> <p>How would you start to attack this problem?</p> http://mathoverflow.net/questions/16128/a-possible-generalization-of-the-cox-theorems-boolean-algebra-bayesian-probab/19563#19563 Answer by Kevin Knuth for A possible generalization of the Cox Theorems (boolean algebra => bayesian probabilities) Kevin Knuth 2010-03-28T01:00:07Z 2010-03-28T01:00:07Z <p>Thanks for noting our work in this area. This has been worked out in even more detail here:</p> <p>Lattice duality: The origin of probability and entropy. Neurocomputing. 67C: 245-274. DOI: 10.1016/j.neucom.2004.11.039 <a href="http://knuthlab.rit.albany.edu/papers/knuth-neurocomp-05-published.pdf" rel="nofollow">http://knuthlab.rit.albany.edu/papers/knuth-neurocomp-05-published.pdf</a></p> <p>Its fundamental application to a wide array of problems is first noted and described here arXiv:physics/0403031v1</p> <p>which was, in part, the inspiration for the quantum mechanics paper you kindly note above: arXiv:0907.0909</p> <p>Last, I have a more recent derivation of Bayesian probability theory from lattices with simple graphical proofs: arXiv:0909.3684</p> <p>This work is literally a derivation of measure theory from much the more basic symmetry of associativity. It is immediately applicable to Boolean lattices of statements, which results in Bayesian probability theory. And, as we have shown, results in information theory and the complex Feynman rules of Quantum Mechanics.</p> <p>Cheers Kevin</p>