MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Bayesian probabilities are usually justified by the Cox theorems, that can be written this way:

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 $A \wedge B$ (and) and $A | B$ (or) such that:

1) $A \wedge B = B \wedge A$

2) $A \wedge (B \wedge C) = (A \wedge B) \wedge C$

3) $A | (B \wedge C) = (A|B) \wedge (A|C)$

and a "valuation":

$f : P \rightarrow \mathcal{R}$

there is a strictly monotonic "regraduation function" $R : \mathcal{R} \rightarrow \mathcal{R}$ such that, for:

$R(f(A\wedge B)) = R(f(A)) + R(f(B))$ (sum rule)


$R(f(A|B)) = R(f(A) ) R(f(B))$ (product rule)

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)

Recently this has been extended for valuations of the type $f : P \rightarrow \mathcal{R}^2$ in 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).

My question/proposal is: is it possible/interesting/feasible to classify at least a class of valuations of the type:

$f : P \rightarrow W$

where W is a continuous manifold? 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?

If this can be done, are those nice rules for inference in some sense? Are they useful as inference tools in specific situations?

share|cite|improve this question
I think that all your \wedge's should be \vee's. – Joel David Hamkins Feb 23 '10 at 14:49
I see that you made the change, but now I'm confused, because you call \vee "and" and | "or", which also seems somehow backwards to me. The sum rule suggests that you want \vee to mean "or", isn't that right? This was why I had made my suggestion above. – Joel David Hamkins Feb 24 '10 at 16:40
Your notation is highly non-standard. In a Boolean algebra or lattice, $\wedge$ means "and" and $\vee$ means "or". – Joel David Hamkins Feb 1 '11 at 13:37
up vote 4 down vote accepted

Thanks for noting our work in this area. This has been worked out in even more detail here:

Lattice duality: The origin of probability and entropy. Neurocomputing. 67C: 245-274. DOI: 10.1016/j.neucom.2004.11.039

Its fundamental application to a wide array of problems is first noted and described here arXiv:physics/0403031v1

which was, in part, the inspiration for the quantum mechanics paper you kindly note above: arXiv:0907.0909

Last, I have a more recent derivation of Bayesian probability theory from lattices with simple graphical proofs: arXiv:0909.3684

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.

Cheers Kevin

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.