At very least my conjecture is not true if \mu is not nowhere zero, which was enough for me to ignore that case, because (see my response to cousin_it) what I actually believe is that there are three very different definitions that all give the same distribution, which I think makes the distribution stand out a lot more as a good idea. Also if \mu is sometimes zero, we also lose uniqueness because we dont know what to do with sets that our Bayes score does not care about. The fact that we can do whatever we want with these sets also takes away coherence (although maybe we could artificially require coherence) I really don't want to do that, because the whole reason I like this approach so much is that it didnt require coherence and coherence came out for free.
Well for example, if Si only has one set A, Abram will think we are in that set, I will think we are in that set with probability 1/2. Now, you could require that every sentence has the same \mu as its negation, (corresponding to putting the sentence or its negation in with probability 1/2 in Abram's procedure) in which case partition X into 3 sets, A, B, and C for which the in or not in A question is given weight muA, and similarly define muB and muC.
Let muA=1/2, muB=1/4 and muC=1/4.
Abrams procedure will with probability 1/4 choose A first, probability 1/8 choose B first, probability 1/8 choose C first, with probability 1/8 choose not A first then choose B with probability 1/8 choose not A first then choose C with probability 1/16 choose not C first and end up with B, with probability 1/16 choose not B first then end up with C, and with probability 1/8 choose not B or not C first then end up with A. In the end P(A) is .375.
Notice that Abrams solution gives a different Bayes score when in set A than when in the other 2 sets. My Bayes score will not. My bayes score will give P(A) probability p where the bayes score is constant:
1/2 log p+1/4 log (1-(1-p)/2)+1/4 log (1-(1-p)/2)=1/2 log (1-p)+1/4 log ((1-p)/2)+1/4 log (1-(1-p)/2)
2 log p+log (1-(1-p)/2)=2 log (1-p)+ log ((1-p)/2)
p^2(1-(1-p)/2)=(1-p)^2((1-p)/2)
p^2(1+p)=(1-p)^3
p=.39661
If you check this p value, you should see that bayes score is independent of model.
In this post, I propose an answer to the following question:
Given a consistent but incomplete theory, how should one choose a random model of that theory?
My proposal is rather simple. Just assign probabilities to sentences in such that if an adversary were to choose a model, your Worst Case Bayes Score is maximized. This assignment of probabilities represents a probability distribution on models, and choose randomly from this distribution. However, it will take some work to show that what I just described even makes sense. We need to show that Worst Case Bayes Score can be maximized, that such a maximum is unique, and that this assignment of probabilities to sentences represents an actual probability distribution. This post gives the necessary definitions, and proves these three facts.
Finally, I will show that any given probability assignment is coherent if and only if it is impossible to change the probability assignment in a way that simultaneously improves the Bayes Score by an amount bounded away from 0 in all models. This is nice because it gives us a measure of how far a probability assignment is from being coherent. Namely, we can define the "incoherence" of a probability assignment to be the supremum amount by which you can simultaneously improve the Bayes Score in all models. This could be a useful notion since we usually cannot compute a coherent probability assignment so in practice we need to work with incoherent probability assignments which approach a coherent one.
I wrote up all the definitions and proofs on my blog, and I do not want to go through the work of translating all of the latex code over here, so you will have to read the rest of the post there. Sorry. In case you do not care enough about this to read the formal definitions, let me just say that my definition of the "Bayes Score" of a probability assignment P with respect to a model M is the sum over all true sentences s of m(s)log(P(s)) plus the sum over all false sentences s of m(s)log(1-P(s)), where m is some fixed nowhere zero probability measure on all sentences. (e.g. m(s) is 1/2 to the number of bits needed to encode s)
I would be very grateful if anyone can come up with a proof that this probability distribution which maximizes Worst Case Bayes Score has the property that its Bayes Score is independent of the choice of what model we use to judge it. I believe it is true, but have not yet found a proof.