Fix a theory $T$ over a language $L$. A coherent probability function is one that satisfies laws of probability theory, each coherent probability function represents a probability distribution on complete logical extensions of $T$.

One of many equivalent definitions of coherence is that $P$ is coherent if $P\left(s_1\right)+P\left(s_2\right)+\dots +P\left(s_k\right)=1$ whenever $T$ can prove that exactly one of $s_1,\dots ,s_k$ is true.

Another very basic desirable property is that $P\left(s\right)=1$ only when $s$ is provable. There have been many proposals of specific coherent probability assignments that all satisfy this basic requirement. Many satisfy stronger requirements that give bounds on how far $P\left(s\right)$ is from 1 when $s$ is not provable.

In this post, I modify the framework slightly to instead talk about conditional probability. Consider a function $P$ which takes in a consistent theory $T$ and a sentence $s$, and outputs a number $P\left(s|T\right)\in [0,1]$, which represents the conditional probability of $s$ given $T$.

We say that $P$ is coherent if:

1. $P\left(s_1|T\right)+P\left(s_2|T\right)+\dots +P\left(s_k|T\right)=1$ whenever $T$ can prove that exactly one of $s_1,\dots ,s_k$ is true, and

2. $P(s\wedge r|T)=P\left(r|T\right)\cdot P(s|T\cup \{r\left\}\right).$

3. If $s$ proves every sentence in $T$, then $P(s|R\cup T)\ge P\left(s|R\right)$.

Theorem: There is no coherent conditional probability function $P$ such that $P\left(s|T\right)=1$ only when $T$ proves $s$.

Proof:

This proof will use the notation of log odds $\ell \left(p\right)={\log }_2\left(\frac{p}{1-p}\right)$ to make things simpler.

Let $P$ be a coherent conditional probability function. Fix a sentence $s$ which is neither provable nor disprovable from the empty theory. Construct an infinite sequences of theories as follows:

1. $T_0$ is the empty theory.

2. To construct $T_{n+1}$, choose a sentence $r_n$ such that neither $s\to r_n$ nor $s\to \neg r_n$ are provable in $T_n$. If $P(s\wedge r_n|T_n)\le P(s\wedge \neg r_n|T_n)$, then let $T_{n+1}=T_n\cup \{s\to r_n\}$. Otherwise, let $T_{n+1}=T_n\cup \{s\to \neg r_n\}$.

Fix an $n$, and without loss of generality, assume $P(s\wedge r_n|T_n)\le P(s\wedge \neg r_n|T_n)$. Since $P$ is coherent we have $P(s\wedge r|T_n)+P(s\wedge \neg r|T_n)=P\left(s|T_n\right).$ In particular, this means that $P(s\wedge r|T_n)\le \frac{1}{2}P\left(s|T_n\right)$.

Observe that $P(s\wedge (s\to r\left)|T_n\right)=P\left(s|T_{n+1}\right)P(s\to r|T_n)$, and $P(\neg s\wedge (s\to r\left)|T_n\right)=P\left(\neg s|T_{n+1}\right)P(s\to r|T_n)$. Therefore, $P(s\wedge r|T_n)/P\left(\neg s|T_n\right)=P\left(s|T_{n+1}\right)/P\left(\neg s|T_{n+1}\right)$, so $\frac{1}{2}P\left(s|T_n\right)/P\left(\neg s|T_n\right)\ge P\left(s|T_{n+1}\right)/P\left(\neg s|T_{n+1}\right)$.

In the language of log odds, this means that $\ell \left(P\right(s|T_n\left)\right)-1\ge \ell \left(P\right(s|T_{n+1}\left)\right)$.

Let $T_{\infty }$ be the union of all the $T_n$. Note that by the third condition of coherence, $\ell \left(P\right(\neg s|T_{\infty }\left)\right)\ge \ell \left(P\right(\neg s|T_n\left)\right)$ for all $n$, so $\ell \left(P\right(s|T_{\infty }\left)\right)\le \ell \left(P\right(s|T_n\left)\right)$ for all $n$

Consider $\ell \left(P\right(s|T_0\left)\right)$ and $\ell \left(P\right(s|T_{\infty }\left)\right)$. These numbers cannot both be finite, since $\ell \left(P\right(s|T_{\infty }\left)\right)\le \ell \left(P\right(s|T_n\left)\right)\le \ell \left(P\right(s|T_0\left)\right)-n$. Therefore, at least one of $P\left(s|T_0\right)$ and $P\left(s|T_{\infty }\right)$ must be 0 or 1. However neither $T_0$ nor $T_{\infty }$ prove or disprove $s$, so this means that $P$ assigns conditional probability 1 to some statement that cannot be proven.

Open Problem: Does this theorem still hold if we leave condition 3 out of the definition of coherence?