Probabilistic measures of coherence: from adequacy constraints towards pluralism

Abstract

The debate on probabilistic measures of coherence flourishes for about 15 years now. Initiated by papers that have been published around the turn of the millennium, many different proposals have since then been put forward. This contribution is partly devoted to a reassessment of extant coherence measures. Focusing on a small number of reasonable adequacy constraints I show that (i) there can be no coherence measure that satisfies all constraints, and that (ii) subsets of these adequacy constraints motivate two different classes of coherence measures. These classes do not coincide with the common distinction between coherence as mutual support and coherence as relative set-theoretic overlap. Finally, I put forward arguments to the effect that for each such class of coherence measures there is an outstanding measure that outperforms all other extant proposals. One of these measures has recently been put forward in the literature, while the other one is based on a novel probabilistic measure of confirmation.

Appendices

Appendix 1: Proof of Theorem 1

Theorem 1

There cannot be a probabilistic measure of coherence that satisfies both (Independence) and (Agreement).

Proof

Consider the following probability distributions over the set $\{A_1, A_2\}$:

Straightforward calculations yield that $A_1$ and $A_2$ are probabilistically independent on both distributions $\Pr$ and $\Pr '$. Thus, a measure satisfying (Independence) will assign equal degrees of coherence to $\{A_1,A_2\}$ on both probability distributions. On the other hand, $\Pr (A_1|A_2)=1/2>1/4=\Pr '(A_1|A_2)$ and $\Pr (A_2|A_1)=1/2>1/4=\Pr '(A_2|A_1)$. Hence, a measure satisfying (Agreement) will assign a higher degree of coherence to the set $\{A_1,A_2\}$ on $\Pr$ than on $\Pr '$, i.e. it will assign different degrees of coherence to the set depending on which distribution is chosen. Thus, there cannot be a measure that satisfies both constraints. $\square$

Appendix 2: Proof of Theorem 2

Theorem 2

There are two maximally consistent subsets of the set of all constraints:

• $\Sigma _1$={(Dependence), (Independence), (Equivalence), (Inconsistency)}

• $\Sigma _2$={(Equivalence), (Inconsistency), (Agreement)}

Proof

In order to prove theorem 2, we have to show that (i) $\Sigma _1$ and $\Sigma _2$ are maximally consistent subsets of the set of all constraints and that (ii) there is no other maximally consistent subset.

• (i) Obviously, $\Sigma _1$ is a subset of all considered constraints. Furthermore, the fact that the coherence measure $\mathcal{C}_{\lambda }$ satisfies all constraints in $\Sigma _1$ establishes the consistency of $\Sigma _1$. The only constraint that is not a member of $\Sigma _1$ is (Agreement). By theorem 1, we know that (Agreement) and (Independence) are logically inconsistent. Hence, there is no consistent superset $\Sigma _1\subsetneq \Sigma '$ of constraints. Therefore, $\Sigma _1$ is a maximally consistent subset of all considered constraints. Now consider $\Sigma _2$: consistency is established by measure $\mathcal{C}_f$, and the fact that (Independence) and (Agreement) are inconsistent shows that we cannot add (Independence) to $\Sigma _2$ without loosing consistency. But the same holds for (Dependence) as there cannot be a measure that satisfies both (Dependence) and (Agreement). To see this, consider the following probability distributions:

  

  Straightforward calculations yield the following probabilities: $\Pr (A_1|A_2)=1/2>1/4=\Pr (A_1)$ and $\Pr (A_2|A_1)=1/2>1/4=\Pr (A_2)$. On the other hand, $\Pr '(A_1|A_2)=399/655<655/912=\Pr '(A_1)$ and $\Pr '(A_2|A_1)=399/655<655/912=\Pr '(A_2)$. Thus, for each possible coherence measure $\mathcal{C}$ satisfying (Dependence) with neutrality threshold $\beta _\mathcal{C}$, $\mathcal{C}_{\Pr }(A_1,A_2)>\beta _\mathcal{C}>\mathcal{C}_{\Pr '}(A_1,A_2)$. However, since $\Pr (A_1|A_2)<\Pr '(A_1|A_2)$ and $\Pr (A_2|A_1)<\Pr '(A_2|A_1)$, we get the opposite coherence ordering for each measure $\mathcal{C}'$ satisfying (Agreement), i.e. $\mathcal{C}'_{\Pr }(A_1,A_2)<\mathcal{C}'_{\Pr '}(A_1,A_2)$. Thus, there cannot be a coherence measure that satisfies (Agreement) and (Dependence). Hence, there is no consistent superset $\Sigma '$ of $\Sigma _2$, so that $\Sigma _2$ is a maximally consistent subset of all considered constraints.

• (ii) Assume that $\Sigma$ is a maximally consistent subset of the set of all considered constraints, and assume furthermore that (Independence) is an element of $\Sigma$. Than we can consistently add all constraints except (Agreement). Hence, by being maximally consistent, $\Sigma$ must be identical to $\Sigma _1$. The same holds if (Dependence) is assumed to be an element of $\Sigma$. On the other hand, if we assume (Agreement) to be in $\Sigma$, then we can consistently add (Equivalence) and (Inconsistency). Hence, both should be in $\Sigma$, turning $\Sigma$ identical to $\Sigma _2$. Finally, if either (Equivalence) or (Inconsistency) are in $\Sigma$, then by consistency, both of them should be in $\Sigma$. Furthermore, we can either add (Independence) and (Dependence) yielding $\Sigma _1$ or (Agreement), thereby yielding $\Sigma _2$. Hence, $\Sigma _1$ and $\Sigma _2$ are the only maximally consistent subsets of considered constraints.

Appendix 3: Proofs for Table 3

We prove the satisfaction and dissatisfaction of considered constraints by cases:

(Independence) Being based on probabilistic measures of confirmation, each of the coherence measures $\mathcal{C}_{d1}, \mathcal{C}_{d2}, \mathcal{C}_{r1}$ and $\mathcal{C}_{r2}$ satisfies (Independence). This is due to the fact that each confirmation measure $\mathfrak {s}$ assigns a neutral degree of coherence to probabilistically independent pairs of propositions. Accordingly, the assigned degree of coherence by measure $\mathcal{C}_{\mathfrak {s}}$ will, again, be equal to the confirmation measures neutrality threshold $\beta _{\mathfrak {s}}$. Furthermore, if a set ${\Delta }=\{\varphi _1, \ldots ,\varphi _n\}$ satisfies mutual independence, then $\Pr (\varphi _1\wedge \ldots \wedge \varphi _n)=\prod _{i\le n}\Pr (\varphi _i)$. Hence, $\mathcal{D}(S)=1$ and $\mathcal{D}^{*}(S)=0$. Since both are the neutrality thresholds for these measures, both satisfy (Independence). On the other hand, it is not totally clear what should be considered the neutrality threshold for any of the measures $\mathcal{O}, \mathcal{O}^{*}$ or $\mathcal{C}_f$ (see footnote 18). However, it can be shown that whatever threshold is chosen, it is not the case that probabilistically independent sets of propositions are always assigned a coherence degree in line with this threshold. To see this, consider again the probability distributions given in the proof of theorem 1. The corresponding degrees of coherence are as follows: $\mathcal{O}_{\Pr }(A_1,A_2)=\mathcal{O}^{*}_{\Pr }(A_1,A_2)=1/3$, $\mathcal{C}_{f,\Pr }(A_1,A_2)=1/2$, and $\mathcal{O}_{\Pr '}(A_1,A_2)=\mathcal{O}^{*}_{\Pr }(A_1,A_2)=1/7$, $\mathcal{C}_{f,\Pr '}(A_1,A_2)=1/4$. Thus, whatever the chosen threshold, all these measures will violate (Independence).

(Dependence) If each pair of non-empty, disjoint subsets is mutually confirmatory/disconfirmatory, then each confirmation measure $\mathfrak {s}$ will assign a degree of confirmation above/ below $\beta _{\mathfrak {s}}$. Accordingly, $\mathcal{C}_{d1}, \mathcal{C}_{d2}, \mathcal{C}_{r1}$ and $\mathcal{C}_{r2}$ satisfy (Dependence). Furthermore, let ${\Delta }=\{\varphi _1,\ldots ,\varphi _n\}$ be a set of mutually confirmatory/disconfirmatory propositions. The deviation measure $\mathcal{D}$ can equivalently be represented as follows (proof omitted):

$$\begin{aligned} \mathcal{D}(\varphi _1,\ldots ,\varphi _n)=\frac{\Pr (\varphi _2|\varphi _1)}{\Pr (A_2)} \times \frac{\Pr (\varphi _3|\varphi _1\wedge \varphi _2)}{\Pr (\varphi _3)}\times \cdots \times \frac{\Pr (\varphi _n|\varphi _1 \wedge \ldots \wedge \varphi _{n-1})}{\Pr (\varphi _n)} \end{aligned}$$

Thus, given that ${\Delta }$ is mutually confirmatory/disconfirmatory, the assigned degree of coherence, i.e. $\mathcal{D}({\Delta })$ will exceed/ undercut 1, and analogously, $\mathcal{D}^{*}(S)$ will exceed/ undercut 0. On the other hand, assume that $\mathcal{C}$ is either $\mathcal{O}, \mathcal{O}^{*}$ of $\mathcal{C}_f$. Now consider the following probability distribution:

Given probability distribution $\Pr$, we calculate that $\Pr (A_i|A_j)=5/22>11/85=\Pr (A_i)$ for $1\le i\not =j\le 2$, so that $A_1$ and $A_2$ are mutually confirmatory. However, $\mathcal{O}(A_1,A_2)=\mathcal{O}^{*}(A_1,A_2)=5/39\approx 0.13$ and $\mathcal{C}_f(A_1,A_2)=5/22\approx 0.23$. Thus, both sets are assigned incoherent (given that the threshold separating coherence and incoherence should be equal to or above $1/2$). On the other hand, on $\Pr '$ we get: $\Pr (A_i|A_j)=27/29<29/31=\Pr (A_i)$ for $1\le i\not = j\le 2$, so that $A_1$ and $A_2$ are mutually disconfirmatory. However, $\mathcal{O}(A_1,A_2)=\mathcal{O}^{*}(A_1,A_2)=27/31\approx 0.87$ and $\mathcal{C}_f(A_1,A_2)=27/29\approx 0.93$. Thus, both sets are assigned coherent.

(Equivalence) A proof for all considered measures except $\mathcal{C}_{r2}, \mathcal{D}^{*}, \mathcal{O}^{*}$ and $\mathcal{C}_f$ is given by Siebel and Wolff (2008). Now assume that $\varphi _1$ and $\varphi _2$ are logically equivalent, then

$$\begin{aligned} \mathcal{C}_{r2}(\varphi _1,\varphi _2)=1/2\cdot \left( \frac{\Pr (\varphi _1| \varphi _2)}{\Pr (\varphi _1|\lnot \varphi _2)}+\frac{\Pr (\varphi _2|\varphi _1)}{\Pr (\varphi _2|\lnot \varphi _1)}\right) \end{aligned}$$

is simply undefined since $\Pr (\varphi _1|\lnot \varphi _2)=\Pr (\varphi _2|\lnot \varphi _1)=0$. For $\mathcal{D}^{*}$ we get the following result:

$$\begin{aligned} \log \mathcal{D}(\varphi _1,\varphi _2)=\log [1/\Pr (\varphi _1)]=-\log \Pr (\varphi _1) \end{aligned}$$

Thus, the assigned degree of coherence is sensitive to $\varphi _1$’s prior probability and a fortiori not maximal for every set of equivalent propositions. Next we turn to $\mathcal{O}^{*}$. It is easy to see that $\mathcal{O}^{*}$ satisfies (Equivalence) given that $\mathcal{O}$ does since it is only an average of $\mathcal{O}$ values. Eventually, given that ${\Delta }=\{\varphi _1,\ldots ,\varphi _n\}$ is a set of logically equivalent propositions, each posterior probability that enters the calculation for $\mathcal{C}_f({\Delta })$ equals 1. Hence, $\mathcal{C}_f({\Delta })$ is maximal for every set of logically equivalent propositions.

(Inconsistency) Let $\varphi _1$ and $\varphi _2$ be contingent propositions such that the set $\{\varphi _1, \varphi _2\}$ is inconsistent. Then we get the following result for $\mathcal{C}_{d1}$:

$$\begin{aligned} \mathcal{C}_{d1}(\varphi _1,\varphi _2)&=1/2\cdot \left[ \Pr (\varphi _1|\varphi _2)-\Pr (\varphi _1)+\Pr (\varphi _2| \varphi _1)-\Pr (\varphi _2)\right] \\&=-1/2\cdot \left[ \Pr (\varphi _1)+\Pr (\varphi _2)\right] \end{aligned}$$

Thus, the assigned degree of coherence is sensitive to the propositions’ prior probabilities, and so $\mathcal{C}_{d1}$ violates (Independence). Now let us turn to $\mathcal{C}_{d2}$. By Bayes’ theorem we calculate:

$$\begin{aligned} \mathcal{C}_{d2}(\varphi _1,\varphi _2)&=1/2\cdot \left[ \Pr (\varphi _1| \varphi _2)-\Pr (\varphi _1|\lnot \varphi _2)+\Pr (\varphi _2|\varphi _1) -\Pr (\varphi _2|\lnot \varphi _1)\right] \\&=1/2\cdot \left[ {\Pr (\varphi _1)}/{\Pr (\lnot \varphi _2)}+{\Pr (\varphi _2)}/{\Pr (\lnot \varphi _1)}\right] \end{aligned}$$

Hence, $\mathcal{C}_{d2}$ is also sensitive to prior probabilities. Next we consider the ratio-based measures $\mathcal{C}_{r1}$ and $\mathcal{C}_{r2}$. Let ${\Delta }=\{\varphi _1,\ldots ,\varphi _n\}$ be a set of contingent propositions such that all subsets with at least two propositions are unsatisfiable. Then we get:

$$\begin{aligned} \mathcal{C}_{r1}({\Delta })=(m)^{-1}\sum _{({\Delta '},{\Delta ''}) \subseteq [{\Delta }]}r1({{\Delta '}}, {\Delta ''})=(m)^{-1}\sum _{({\Delta '},{\Delta ''}) \subseteq [{\Delta }]}\frac{\Pr ({{\Delta '}}|{\Delta }'')}{ \Pr ({\Delta }')} \end{aligned}$$

$$\begin{aligned} \mathcal{C}_{r2}({\Delta })=(m)^{-1}\sum _{({\Delta '},{\Delta ''}) \subseteq [{\Delta }]}r2({{\Delta '}}, {\Delta ''})=(m)^{-1}\sum _{({\Delta '},{\Delta ''})\subseteq [{\Delta }]}\frac{\Pr ({{\Delta '}}|{\Delta }'')}{ \Pr ({\Delta }'|\lnot {\Delta }'')} \end{aligned}$$

These latter terms are undefined for all $S'$ with at least two propositions (since in this case $\Pr (S')=\Pr (S'|S'')=0$ and therefore we would have to divide by $0$). Now we turn to the $\mathcal{D}$ and $\mathcal{O}$ and their refined versions $\mathcal{D}^{*}$ and $\mathcal{O}^{*}$. To do so, let again ${\Delta }=\{\varphi _1,\ldots ,\varphi _n\}$ be a set of contingent propositions such that all non-singleton subsets of ${\Delta }$ are unsatisfiable. Then

$$\begin{aligned} \mathcal{D}({\Delta })=\frac{\Pr (\varphi _1\wedge \ldots \wedge \varphi _n)}{\Pr (\varphi _1)\cdot \ldots \cdot \Pr (\varphi _n)} =0=\frac{\Pr (\varphi _1\wedge \ldots \wedge \varphi _n)}{\Pr (\varphi _1\vee \ldots \vee \varphi _n)}=\mathcal{O}({\Delta }) \end{aligned}$$

On the other hand, $\mathcal{D}^{*}$ is not even defined in these cases. However, if we stipulate that $log(0)=-\infty$, then $\mathcal{D}^{*}$ passes the inconsistency test. $\mathcal{O}^{*}(S)$ is an average of $\mathcal{O}$ values for all subsets with at least two propositions. Since it is assumed that all those subsets are unsatisfiable, we get

$$\begin{aligned} \mathcal{O}^{*}({\Delta })=\frac{1}{l}\sum _{2\le k\le n} \sum _{{\Delta }_i\in [{\Delta }]^k}\mathcal{O}({\Delta }_i)=0 \qquad \qquad l=\sum _{k=2}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \end{aligned}$$

Finally, we consider the firmness-based coherence measure $\mathcal{C}_f$:

$$\begin{aligned} \mathcal{C}_f({\Delta })=(m)^{-1}\sum _{({\Delta '},{\Delta ''}) \subseteq [{\Delta }]}\Pr ({{\Delta '}}| {\Delta ''}) \end{aligned}$$

Since all posterior probabilities are equal to zero (remember that we stipulated that $\Pr (\varphi |\psi )=0$ for all inconsistent $\psi$), the assigned degree of coherence is equal to the minimum possible degree $0$.

(Agreement) By Theorem 1 we know that there cannot be a probabilistic measure of coherence that satisfies both (Independence) and (Agreement). Thus, given that $\mathcal{C}_{d1}, \mathcal{C}_{d2}, \mathcal{C}_{r1}, \mathcal{C}_{r2}$, $\mathcal{D}$ and $\mathcal{D}^{*}$ satisfy the former, they necessarily have to violate the latter. The case for $\mathcal{O}$ and $\mathcal{O}^{*}$ is so far unsettled, i.e. we were neither able to prove that these measures satisfy (Agreement) nor to find a counterexample.

Next, given that $\mathcal{C}_f$ is an average of posterior probabilities, it must necessarily be the case that if all relevant posterior probabilities for $\Pr$ exceed their counterparts on $\Pr '$, then the assigned degree of $\mathcal{C}_f$-coherence on $\Pr$ exceeds the one on $\Pr '$.

Appendix 4: Proofs for Table 6

Being based on probabilistic measures of confirmation, each of the coherence measures $\mathcal{C}_{z}, \mathcal{C}_{\lambda }$ satisfies (Independence) and (Dependence). Now let ${\Delta }=\{\varphi _1,\ldots ,\varphi _n\}$ be a set of equivalent propositions, then given that $\Pr ({\Delta }'|{\Delta }'')=1\ge \Pr ({\Delta '})$ we get:

$$\begin{aligned} \mathcal{C}_{z}({\Delta })=(m)^{-1}\sum _{({\Delta '},{\Delta ''})\subseteq [{\Delta }]}z({{\Delta '}}, {\Delta ''})=(m)^{-1}\sum _{({\Delta '},{\Delta ''})\subseteq [{\Delta }]} \frac{\Pr ({{\Delta '}}|{\Delta }'')-\Pr ({\Delta '})}{1- \Pr ({\Delta }')}=1 \end{aligned}$$

and accordingly

$$\begin{aligned} \mathcal{C}_{\lambda }({\Delta })=(m)^{-1}\!\!\! \sum _{({\Delta '},{\Delta ''}) \subseteq [{\Delta }]}\!\!\!\!\!\lambda ({{\Delta '}}, {\Delta ''})=(m)^{-1}\!\!\!\!\!\sum _{({\Delta '},{\Delta ''})\subseteq [{\Delta }]}\!\!\! \frac{\Pr ({{\Delta '}}|{\Delta }'')-\Pr ({\Delta '}|\lnot {\Delta }'')}{1- \Pr ({\Delta }'|{\Delta }'')}=1 \end{aligned}$$

Thus, both measures satisfy (Equivalence). Now we turn to the inconsistency-constraint. First, consider $z$. If $\Pr ({\Delta }'|{\Delta }'')=0<\Pr ({\Delta }')$, then

$$\begin{aligned} z({\Delta }',{\Delta }'')=\frac{\Pr ({\Delta }'|{\Delta }'')-\Pr ({\Delta }')}{\Pr ({\Delta }')}=-1 \end{aligned}$$

On the other hand, if $\Delta '\models \bot$, then ${\Delta }'\models \lnot {\Delta }''$ and therefore by means of the underlying case distinction we get $z({\Delta }',{\Delta }'')=-1$. Thus, $\mathcal{C}_z$ satisfies (Inconsistency). Finally consider $\lambda$. If $\Pr ({\Delta }'|{\Delta }'')=0<\Pr ({\Delta }')$, then

$$\begin{aligned} \lambda ({\Delta }',{\Delta }'')=\frac{\Pr ({\Delta }'|{\Delta }'')-\Pr ({\Delta }'|\lnot {\Delta }'')}{\Pr ({\Delta }'|\lnot {\Delta }'')}=-1 \end{aligned}$$

On the other hand, if $\Delta '\models \bot$, then ${\Delta }'\models \lnot {\Delta }''$ and therefore we get $\lambda ({\Delta }',{\Delta }'')=-1$ by definition. Thus, $\mathcal{C}_{\lambda }$ satisfies (Inconsistency). Finally, given the inconsistency of (Independence) and (Agreement), we conclude that $\mathcal{C}_z$ and $\mathcal{C}_{\lambda }$ violate (Agreement).