I’m not sure what your objection is here. You appear to be using propositions A and B that we believe to be almost certainly true, in which case the plausibility of A given B must be very close to 1 by definition. Maybe you meant to negate B?
Sorry about that. Think it would have been clearer if chose a different A and B. But I believe the argument still holds because by (A|B), I did not mean (A|B, I) where I is the background information (I should have been clearer about this as well as I am going against the convention). So basically, we have are not conditioning on any background information so (A|B) is not close to one by definition.
In general though, yes Cox’s result is an existence proof, not a uniqueness one. It means that under given reasonable conditions you can use probability theory, but doesn’t tell you what probabilities to assign to which propositions.
Well what does it prove the existence of? Are you saying that Cox’s theorem implies the existence of a function F such that [the plausibility (A|B) = F(A, B) for any propositions A and B]?
Jaynes’ extension of this to objective probabilities is much more controversial and does not have anything like a mathematical proof.
I do not think that Professor Jaynes’ theory necessarily warrants a mathematical proof as we are only trying to formalise our intuitions about plausibilities. My contention is that Professor Jaynes’ theory contradicts our intuitions about plausibilities and hence the necessity of an “imprecise” theory of plausibility which addresses this problem.
We cannot argue that Cox’s theorem justifies Professor Jaynes’ theory if we start with axioms which are not consistent with our understanding of plausibility. This is what David Chapman argues:
Cox’s Theorem says that there is no formal system other than probability theory that is very similar to it; so if you want something like that, you’ve only got one choice.5 This is irrelevant unless you are considering using one of the dubious alternatives, none of which seems to work as well in practice. - What probability can’t do
Thanks for the reply!
Sorry about that. Think it would have been clearer if chose a different A and B. But I believe the argument still holds because by (A|B), I did not mean (A|B, I) where I is the background information (I should have been clearer about this as well as I am going against the convention). So basically, we have are not conditioning on any background information so (A|B) is not close to one by definition.
Well what does it prove the existence of? Are you saying that Cox’s theorem implies the existence of a function F such that [the plausibility (A|B) = F(A, B) for any propositions A and B]?
I do not think that Professor Jaynes’ theory necessarily warrants a mathematical proof as we are only trying to formalise our intuitions about plausibilities. My contention is that Professor Jaynes’ theory contradicts our intuitions about plausibilities and hence the necessity of an “imprecise” theory of plausibility which addresses this problem.
We cannot argue that Cox’s theorem justifies Professor Jaynes’ theory if we start with axioms which are not consistent with our understanding of plausibility. This is what David Chapman argues:
Chapman discusses this in more detail here: Probability theory does not extend logic.