2
$\begingroup$

I've been trying to understand Cox's Theorem and the problems surrounding it. There's so much information on this topic that I've become confused as to the exact state of the theorem. I've gathered there are three main issues, but since they cover a broad variety of topics I've split it up into multiple questions. I hope this is sufficiently narrowed down (the original question).

My main reference is K. Van Horn, A Guide to Cox's Theorem, 2003.

The assumption that a plausibility $(A|B)$ (A given B) can always be represented by a real number is often a point of controversy for various reasons (see here for the other two issues with this assumption).

The first is that this assumes universal comparability between propositions, which some find unwarranted. I know Jaynes (Probability Theory as Extended Logic, 2003, Appendix A) argues for universal comparability, though I'm not entirely sure that his argument is sufficient.

The second issue with this assumption is something that I think ties into the first, namely that representing belief in proposition $(A|B)$ as a real number, says nothing about our doubt of $(A|B)$; $(\neg A|B)$. This objection is often raised in the context of belief-function theory (which I'm not really familiar with) (G. Shafer, A Mathematical Theory of Evidence, 1976).

However, further on in his proof Cox postulates that a function $S$ exists such that $(\neg A|B) = S(A|B)$, which seems to show that we can express doubt in terms of belief. For that reason, I don't really understand the objection. Is this just a matter of which axioms we should find reasonable?

To summarise:

  1. Is Jaynes' argument for universal comparability considered sufficient?

  2. What is the problem with one-dimensional theory not saying anything about doubt?

$\endgroup$
1
$\begingroup$

"which seems to show"

The objection is exactly the opposite of what you seem to think: in probability theory, the belief in notA is a function of the belief in A. That is what other people think should not be assumed.

Within Shafer's framework for evidence modelling, for instance, even if "If A is true, then notA must be false" it is not necessarily the case that the available evidence that A is true will simultaneously show that all forms of notA are false. For instance, imagine you want to learn a counting number but you can only receive evidence in the form of upper bounds for that number. There's no way in this scheme to get any evidence that 1 is not the number. Thus it's reasonable to set Bel(not1)=0 regardless of what you think Bel(1) is. In Shafer's language, Bel(not1)=0 is equivalent to the plausibility Pl(1) being 1, which is consistent with the fact that you simply can't receive any information that positively makes 1 implausible.

-- I wouldn't say Jaynes argues for universal comparability (or anything). Jaynes "knows" what "the" solution "has" to be, so he quite arbitrarily discards all the paths leading to alternative solutions he "knows" to be wrong, i.e. different from his. If you look attentively you may realize that he actively tries to make it sound that all those paths are not worth exploring, or, in typical Jaynes rhetoric, if somebody thinks they're worth, at some point "they"'ll report their unpromising findings back to us so "we" need not waste time on them.

-- One problem about doubt is the following. Imagine that I let you test one coin, and you toss it 1000 times. Although you may initially have thought it a fair coin, the tossing results in heads 20 times and tails 980 times. You decide to assign Heads a probability P(H) close to .02.

Now another person totally unrelated to me has a coin which you don't have the opportunity to test it (for instance you just saw them put the coin into the slot of a vending machine). Since you know nothing about the coin, it looks like you may set P(H)=P(T)=.5.

The problem is that you have more evidence that the first coin actually produces heads (you have no evidence that the second coin ever lands heads at all), but your give it a much smaller P(H). It is solved in Shafer's framework by making the natural assignment Bel(H)=Bel(T)=0 for the second coin.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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