Maximize Worst Case Bayes Score

byCoscott5y17th Jun 201424 comments


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.