# Maximize Worst Case Bayes Score

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.

Now, let’s move on to the formal definitions and proofs.

Fix some language $L$, for example the language of first order set theory. Fix a consistent theory $T$ of $L$, for example ZFC. Fix a nowhere zero probability measure $\mu$ on $L$, for example $\mu(\phi)=2^{-\ell(\phi)}$, where $\ell(\phi)$ is the number of bits necessary to encode $\phi$.

A probability assignment of $L$ is any function from $L$ to the interval $[0,1]$. Note that this can be any function, and does not have to represent a probability distribution. Given a probability assignment $P$ of $L$, and a model $M$ of $T$, we can define the Bayes Score of $P$ with respect to $M$ by

${\displaystyle \mbox{Bayes}(M,P)=\sum_{M\models \phi}\log_2(P(\phi))\mu(\phi)+\sum_{M\models\neg \phi}\log_2(1-P(\phi))\mu(\phi). }$

We define the Worst Case Bayes Score $\mbox{WCB}(P)$ to be the infimum of $\mbox{Bayes}(M,P)$ over all models $M$ of $T$.
Let $\mathbb{P}$ denote the probability assignment that maximizes the function $\mbox{WCB}$. We will show that this maximum exists and is unique, so $\mathbb{P}$ is well defined.

In fact, $\mathbb{P}$ also coherent, meaning that there exists a probability distribution on the set of all models of $T$ such that $\mathbb{P}(\phi)$ is exactly the probability that a randomly chosen model satisfies $\phi$. Since the natural definition of a measurable subset of models comes from unions and intersections of the sets of all models satisfying a given sentence, we can think of $\mathbb{P}$ as an actual probability distribution on the set of all models of $T$.

First, we must show that there exists a probability assignment $P$ which maximizes $\mbox{WCB}$.

Note that $\mbox{Bayes}(M,P)$ either diverges to $-\infty$, or converges to a non-positive real number. If $P$ is the identically $1/2$ function, then $\mbox{WCB}(P)=-1$, so there is at least one $P$ for which $\mbox{WCB}(P)$ is finite. This means that when maximizing $\mbox{WCB}(P)$, we need only consider $P$ for which $\mbox{Bayes}(M,P)$ converges to a number between $-1$ and $0$ for all $M$.

Assume by way of contradiction that there is no $P$ which maximizes $\mbox{WCB}$. Then there must be some supremum value $m$ such that $\mbox{WCB}$ can get arbitrarily close to $m$, but never equals or surpasses $m$. Consider an infinite sequence probability assignments $\{P_i\}$ such that $\mbox{WCB}(P_i)\rightarrow m$. We may take a subsequence of $\{P_i\}$ in order to assume that  $\{P_i(\phi)\}$ converges for every sentence $\phi$. Let $P$ be such that $P_i(\phi)\rightarrow P(\phi)$ for all $\phi$.

By assumption, $\mbox{WCB}(P)$ must be less than $m$. Take any model $M$ for which $\mbox{Bayes}(M,P). Then there exists a finite subset $S$ of $L$ such that $\mbox{Bayes}_S(M,P), where

${\displaystyle \mbox{Bayes}_S(M,P)=\sum_{\phi\in S, M\models \phi}\log_2(P(\phi))\mu(\phi)+\sum_{\phi\in S, M\models\neg \phi}\log_2(1-P(\phi))\mu(\phi). }$

Note that in order to keep the Bayes score at least $-1$, any $P_i$ must satisfy $2^{-1/\mu(\phi)}\leq P_i(\phi)\leq 1$ if $M\models \phi$, and $0\leq P_i(\phi)\leq 1-2^{-1/\mu(\phi)}$ if $M\models\neg\phi$. Consider the space of all functions $f$ from $S$ to $[0,1]$ satisfying these inequalities. Since there are only finitely many values restricted to closed and bounded intervals, this space is compact. Further, $\mbox{Bayes}_S(M,f)$ is a continuous function of $f$, defined everywhere on this compact set. Therefore,

${\displaystyle \lim_{i\rightarrow\infty}\mbox{Bayes}_S(M,P_i)=\mbox{Bayes}_S(M,P)

However, clearly $\mbox{WCB}(P_i)\leq\mbox{Bayes}(M,P_i)\leq\mbox{Bayes}_S(M,P_i)$, so

${\displaystyle \lim_{i\rightarrow\infty}\mbox{WCB}(P_i)

contradicting our assumption that $\mbox{WCB}(P_i)$ converges to $m$.

Next, we will show that there is a unique probability assignment which maximizes $\mbox{WCB}$. Assume by way of contradiction that there were two probability assignments, $P_1$ and $P_2$ which maximize $\mbox{WCB}$. Consider the probability assignment $P_3$, given by

${\displaystyle P_3(\phi)=\frac{\sqrt{P_1(\phi)P_2(\phi)}}{\sqrt{P_1(\phi)P_2(\phi)}+\sqrt{(1-P_1(\phi))(1-P_2(\phi))}}.}$

It is quick to check that this definition satisfies both

${\displaystyle \log_2(P_3(\phi))\geq \frac{\log_2(P_1(\phi))+\log_2(P_2(\phi))}{2}}$

and

${\displaystyle \log_2(1-P_3(\phi))\geq \frac{\log_2(1-P_1(\phi))+\log_2(1-P_2(\phi))}{2},}$

and in both cases equality holds only when $P_1(\phi)=P_2(\phi).$

Therefore, we get that for any fixed model, $M$,

${\displaystyle \mbox{Bayes}(M,P_3(\phi))\geq \frac{\mbox{Bayes}(M,P_1(\phi))+\mbox{Bayes}(M,P_2(\phi))}{2},}$

By looking at the improvement coming from a single sentence $\phi$ with $P_1(\phi)\neq P_2(\phi),$ we see that

${\displaystyle \mbox{Bayes}(M,P_3(\phi))-\frac{\mbox{Bayes}(M,P_1(\phi))+\mbox{Bayes}(M,P_2(\phi))}{2},}$

is actually bounded away from $0$, which means that

${\displaystyle \mbox{WCB}(P_3(\phi))\geq \frac{\mbox{WCB}(P_1(\phi))+\mbox{WCB}(P_2(\phi))}{2},}$

which contradicts the fact that $P_1$ and $P_2$ maximize $\mbox{WCB}$.

This means that there is a unique probability assignment, $\mathbb{P}$, which maximizes $\mbox{WCB}$, but we still need to show that $\mathbb{P}$ is coherent. For this, we will use the alternate definition of coherence given in Theorem 1 here. Namely that $\mathbb{P}$ is coherent if and only if $\mathbb{P}$ assigns probability 0 to every contradiction, probability 1 to every tautology, and satisfies $\mathbb{P}(\phi)=\mathbb{P}(\phi\wedge\psi)+\mathbb{P}(\phi\wedge\neg\psi)$ for all  $\phi$ and  $\psi$.

Clearly $\mathbb{P}$ assigns probability 0 to every contradiction, since otherwise we could increase the Bayes Score in all models by the same amount by updating that probability to 0. Similarly $\mathbb{P}$ clearly assigns probability 1 to all tautologies.

If $\mathbb{P}(\phi)\neq\mathbb{P}(\phi\wedge\psi)+\mathbb{P}(\phi\wedge\neg\psi)$, then we update all three probabilities as follows:

$\mathbb{P}(\phi)\mapsto \frac{1}{1+\frac{1-\mathbb{P}(\phi)}{\mathbb{P}(\phi)}(2^{-x/\mu(\phi)})},$

$\mathbb{P}(\phi\wedge\psi)\mapsto \frac{1}{1+\frac{1-\mathbb{P}(\phi\wedge\psi)}{\mathbb{P}(\phi\wedge\psi)}(2^{x/\mu(\phi\wedge\psi)})},$

and

$\mathbb{P}(\phi\wedge\neg\psi)\mapsto \frac{1}{1+\frac{1-\mathbb{P}(\phi\wedge\neg\psi)}{\mathbb{P}(\phi\wedge\neg\psi)}(2^{x/\mu(\phi\wedge\neg\psi)})},$

where $x$ is the unique real number such that the three new probabilities satisfy $\mathbb{P}(\phi)=\mathbb{P}(\phi\wedge\psi)+\mathbb{P}(\phi\wedge\neg\psi)$. This correction can increases Bayes Score by the same amount in all models, and therefore increase $\mbox{WCB}$, contradicting the maximality of $\mbox{WCB}(\mathbb{P})$. Therefore $\mathbb{P}$ is coherent as desired.

Finally, we show that any given probability assignment $P$ is coherent if and only if it is impossible to simultaneously improve the Bayes Score by an amount bounded away from 0 in all models. The above proof that $\mathbb{P}$ is coherent actually shows one direction of this proof, since the only fact it used about $\mathbb{P}$ is that you could not simultaneously improve the Bayes Score by an amount bounded away from 0 in all models. For the other direction, assume by way of contradiction that $P$ is coherent, and that there exists a $Q$ and an $\epsilon>0$ such that $\mbox{Bayes}(M,Q)-\mbox{Bayes}(M,P)>\epsilon$ for all $M$.

In particular, since  $P$. is coherent, it represents a probability distribution on models, so we can choose a random model  $M$ from the distribution  $P$. If we do so, we must have that

$\mathbb{E}(\mbox{Bayes}(M,Q))-\mathbb{E}(\mbox{Bayes}(M,P))>0.$

However, this contradicts the well known fact that the expectation of Bayes Score is maximized by choosing honest probabilities corresponding the actual distribution $M$ is chosen from.

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. In other words, show that $\mbox{Bayes}(M,\mathbb{P})$ is independent of $M$. I believe it is true, but have not yet found a proof.