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Perplexed comments on Book Club Update, Chapter 3 of Probability Theory - Less Wrong
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Comments (19)
Isn't the equivalence of the "superstructure" implicit in that both systems satisfy (and can be derived from) the Kolmogorov axioms (Section 2.6.4 of the book)?
Of course Jaynes claims in 2.6.4 that his version of Bayesianism goes beyond Kolmogorov (I'm guessing he is talking about things like the principle of indifference and MAXENT.)
Do both systems satisfy the Kolmogorov axioms? One of them is countable additivity, right?
Of course, Kolmogorov's is hardly the only such development. My question is: Is there an isomorphism in reasoning that also serves as a proof of the equivalence?
Are you suggesting that Jaynes is only finitely additive? I have to admit that I don't know exactly how Jaynes's methodological preachments about taking the limit of finite set solutions translates into real math.
I'm not sure I understand your second paragraph either (I am only an amateur at math and less than amateur at analysis.) But my inclination is to say, "Yes, of course there is always a possible isomorphism in the reasonings upward from a shared collection of axioms. But no, there is not an isomorphism in the reasonings or justifications advanced in choosing that set of axioms. But I suspect I missed your point.
Incidentally, Appendix A-1 of the book includes much discussion, quite a bit of it over my head, of the relationship between Jaynes and Kolmogorov.
(Heh, I'm pretty sure being a college sophomore makes me an amateur too.)
Yep. Cox's theorem implies only finite additivity. Jaynes makes a big point of this in many places.
I'm not asking for an isomorphism in the reasoning of choosing a set of axioms. I'm asking for an isomorphism in the reasoning in using them.
For large classes (all?) of problems with discrete probability spaces, this is trivial -- just map a basis (in the topological sense) for the space onto mutually exclusive propositions. The combinatorics will be identical.