The Problem of Coherence and Truth Redux

Abstract

In “What price coherence?” (Analysis 54:129–132, 1994), Klein and Warfield put forward a simple argument that triggered an extensive debate on the epistemic virtues of coherence. As is well-known, this debate yielded far-reaching impossibility results to the effect that coherence is not conducive to truth, even if construed in a ceteris paribus sense. A large part of the present paper is devoted to a re-evaluation of these results. As is argued, all explications of truth-conduciveness leave out an important aspect: while it might not be the case that coherence is truth-conducive, it might be conducive to verisimilitude or epistemic utility. Unfortunately, it is shown that the answer for both these issues must be in the negative, again. Furthermore, we shift the focus from sets of beliefs to particular beliefs: as is shown, neither is any of the extant probabilistic measures of coherence truth-conducive on the level of particular beliefs, nor does weakening these measures to quasi-orderings establish the link between coherence and truth for an important amount of measures. All in all, the results in this paper cast a serious doubt on the approach of establishing a link between coherence and truth. Finally, recent arguments that shift the focus from the relationship between coherence and truth to the one between coherence and confirmation are assessed.

Appendices

Appendix 1: Proof of Observation 3.1

Consider the following probability distributions, where $x=1-\Pr (A_1\vee A_2\vee A_3)$:

$A_1$	$A_2$	$A_3$	Probability	$A_1$	$A_2$	$A_3$	Probability
T	T	T	1/5	F	T	T	11/65
T	T	F	2/41	F	T	F	7/44
T	F	T	3/37	F	F	T	1/18
T	F	F	$\frac{122786261713}{1248935336885}$	F	F	F	x

$A_1^{\prime }$	$A_2^{\prime }$	$A_3^{\prime }$	Probability	$A_1^{\prime }$	$A_2^{\prime }$	$A_3^{\prime }$	Probability
T	T	T	34/171	F	T	T	13/71
T	T	F	2/13	F	T	F	1/52
T	F	T	1/25	F	F	T	3/20
T	F	F	$\frac{15124126187369}{13521705438129525}$	F	F	F	$\hbox {x}^{\prime }$

The table only gives a small sample of the complete distribution over six variables. However, note that $\Pr (\bigwedge _{i\le 3}A_i)=0.200>0.199\approx \Pr (\bigwedge _{i\le 3}A_i')$ and $\prod _{i\le 3}\Pr (A_i)=\prod _{i\le 3}\Pr (A_i')=0.125$. Straightforward calculations yield the following results:

	${\mathcal {O}}$	${\mathcal {D}}^{*}$	${\mathcal {O}}^{*}$	${\mathcal {C}}_d$	${\mathcal {C}}_r$	${\mathcal {C}}_s$	${\mathcal {C}}_l$	${\mathcal {C}}_k$	${\mathcal {C}}_z$	${\mathcal {C}}_f$
$\{A_1,A_2,A_3\}$	0.246	1.292	0.381	0.119	1.276	0.212	1.792	0.246	0.228	0.572
$\{A_1^{\prime },A_2^{\prime },A_3^{\prime }\}$	0.266	1.367	0.425	0.125	1.281	0.230	2.176	0.263	0.247	0.587

All mentioned measures assign a lower degree of coherence to the set $\{A_1,A_2,A_3\}$ which is in conflict with the requirements of Definition 3.2.

Appendix 2: Proof of Observation 5.1

For the vast majority of measures it is possible to show that Definition 5.1 is already violated for pairs of propositions. For these measures consider two sets of propositions $\{A_1,A_2\}$ and $\{A_1,A_3\}$ and the following probability distribution, where $x=1-\Pr (A_1\vee A_2\vee A_3)$:

$A_1$	$A_2$	$A_3$	Probability	$A_1$	$A_2$	$A_3$	Probability
T	T	T	7/62	F	T	T	5/51
T	T	F	14/65	F	T	F	5/52
T	F	T	6/73	F	F	T	1/182
T	F	F	1/220	F	F	F	x

The coherence values for the relevant measures are given in the following table.

	${\mathcal {O}}$	${\mathcal {D}}^{*}$	${\mathcal {O}}^{*}$	${\mathcal {C}}_d$	${\mathcal {C}}_s$	${\mathcal {C}}_l$	${\mathcal {C}}_k$	${\mathcal {C}}_z$	${\mathcal {C}}_f$
$\{A_1,A_2\}$	0.539	1.574	0.539	0.241	0.453	2.921	0.480	0.463	0.710
$\{A_1,A_3\}$	0.376	1.514	0.376	0.205	0.316	2.370	0.402	0.326	0.562

Given that $\Pr (A_1|A_2)\approx 0.628<0.653\approx \Pr (A_1|A_3)$, all these measures are not truth-conducive in the sense of Definition 5.1. Note that in order to prove the analogous result for the two missing coherence measures ${\mathcal {D}}$ and ${\mathcal {C}}_{r}$, we have to consider at least one set of propositions with more than two elements. This is because for the considered pairs of sets, $\Pr (A_1|A_2)<\Pr (A_1|A_3)$ already entails ${\mathcal {D}}(A_1,A_2)<{\mathcal {D}}(A_1,A_3)$ and the same holds for ${\mathcal {C}}_r$.

Therefore, let $S=\{A_1,A_2\}$ and $S'=\{A_1,A_2,A_3\}$ and consider the following probability distribution, where $x=1-\Pr (A_1\vee A_2\vee A_3)$:

$A_1$	$A_2$	$A_3$	Probability	$A_1$	$A_2$	$A_3$	Probability
T	T	T	12/49	F	T	T	12/37
T	T	F	2/25	F	T	F	3/44
T	F	T	9/73	F	F	T	1/72
T	F	F	6/53	F	F	F	x

Against the background of this distribution we get the desired result that even though $A_1$’s posterior probability given $A_2$ (approx. 0.453) exceeds its posterior probability given both $A_2$ and $A_3$ (approx. 0.430), we have

$$\begin{aligned} {\mathcal {D}}(S) \approx 0.807&< 0.861 \approx {\mathcal {D}}(S'')\\ {\mathcal {C}}_{r}(S) \approx 0.807&< 0.936 \approx {\mathcal {C}}_r(S'') \end{aligned}$$

Hence, these measures are not truth-conducive in the sense of Definition 5.1, too.

Appendix 3: Proof of Observation 5.2

Consider again the former probability distribution:

$A_1$	$A_2$	$A_3$	Probability	$A_1$	$A_2$	$A_3$	Probability
T	T	T	12/49	F	T	T	12/37
T	T	F	2/25	F	T	F	3/44
T	F	T	9/73	F	F	T	1/72
T	F	F	6/53	F	F	F	x

We can easily extend the calculated coherence values to all considered measures. The following table contains additional values for all measures but ${\mathcal {O}}$.

	${\mathcal {D}}^{*}$	${\mathcal {O}}^{*}$	${\mathcal {C}}_d$	${\mathcal {C}}_s$	${\mathcal {C}}_l$	${\mathcal {C}}_k$	${\mathcal {C}}_z$	${\mathcal {C}}_f$
$\{A_1,A_2\}$	0.807	0.341	−0.124	−0.350	0.594	−0.256	−0.193	0.516
$\{A_1,A_2,A_3\}$	0.930	0.417	−0.036	−0.082	0.940	−0.064	−0.028	0.565

As the table shows, these measures agree with ${\mathcal {D}}$ and ${\mathcal {C}}_r$ in that the extended set $\{A_1,A_2,A_3\}$ is more coherent than its subset $\{A_1,A_2\}$. Taking into account that nonetheless $A_1^{\prime }$ posterior probability is lower for this extended set, this result shows that all considered measures (except ${\mathcal {O}}$) are not truth-conducive in the sense of Definition 5.2.

Appendix 4: Proof of Observation 5.3

First of all, we can utilize the following probability distribution in order to show that the orderings induced by Definition 5.6 are not truth-conducive for all confirmation measures but f:

$A_1$	$A_2$	$A_3$	Probability	$A_1$	$A_2$	$A_3$	Probability
T	T	T	1/30	F	T	T	5/44
T	T	F	4/59	F	T	F	10/61
T	F	T	1/153	F	F	T	53/105
T	F	F	6/55	F	F	F	x

The confirmation values for the relevant measures that are currently of interest are given in the following table:

Confirmation	d	r	s	l	k	z
${\xi (A_1,A_2)}$	0.050	1.232	0.081	1.317	0.137	0.064
${\xi (A_1,A_2|A_3)}$	0.166	3.745	0.214	4.550	0.640	0.177
${\xi (A_2,A_1)}$	0.088	1.232	0.112	1.435	0.179	0.141
${\xi (A_2,A_1|A_3)}$	0.613	3.745	0.652	17.743	0.893	0.789

As the table indicates, all considered confirmation measures agree in that the there is a larger degree of confirmation between $A_1$ and $A_2$ when $A_3$ is taken for granted. This, however, is in sharp contrast with the relevant conditional probabilities: as was mentioned before, $A_1$’s conditional probability given $A_2$ exceeds its conditional probability given $A_2$ and $A_3$. Accordingly, these measures are not truth-conducive in the sense of Definition 5.3.

The latter distribution also suffices to show that the refined deviation measure ${\mathcal {D}}^{*}$ is not truth-conducive in this sense. This is due to the fact that

$$\begin{aligned} {\mathcal {D}}^{*}(A_1,A_2) \approx 1.232 < 3.745 \approx {\mathcal {D}}^{*}(A_1,A_2|A_3) \end{aligned}$$

To show that the refined overlap measure ${\mathcal {O}}^{*}$ is not truth-conducive in the sense of Definition 5.3 we utilize the following distribution involving four propositions with $x=1-\Pr (A_1\vee A_2\vee A_3\vee A_4)$:

$A_1$	$A_2$	$A_3$	$A_4$	Probability	$A_1$	$A_2$	$A_3$	$A_4$	Probability
T	T	T	T	1/25	F	T	T	T	1/43
T	T	T	F	1/23	F	T	T	F	1/36
T	T	F	T	3/56	F	T	F	T	1/44
T	T	F	F	5/69	F	T	F	F	2/39
T	F	T	T	4/39	F	F	T	T	1/59
T	F	T	F	1/57	F	F	T	F	6/71
T	F	F	T	1/56	F	F	F	T	2/53
T	F	F	F	27/94	F	F	F	F	x

According to the refined overlap measure all non-singleton subsets of $\{A_2,A_3,A_4\}$ are assigned a higher degree of coherence conditional on $A_1$. However, $A_2$’s conditional probability given $A_1$, $A_3$ and $A_4$ is lower than its conditional probability given only $A_3$ and $A_4$. Hence, ${\mathcal {O}}^{*}$ also violates Definition 5.3.

Now we turn to the remaining confirmation measure f. In order to show that this measure is truth-conducive in the sense of Definition 5.6, note that for each pair $(S',S'')\in [S]$ the following claim holds by definition:

($\dagger _f$):

If $f(S',S''|A)>f(S',S'')$, then $\Pr (S'|S'',A)>\Pr (S'|S'')$.

Hence, let $S'=\{x\}$ for some $x\in S$ and $S''=S\setminus \{x\}$, then the fact that $f(S',S''|A)>f(S',S'')$ by definition together with $(\dagger _f)$ entails the desired claim.

Appendix 5: Proof of Observation 6.2

If ${\mathcal {D}}({\mathbf {E}},H)>1$ for some set ${\mathbf {E}}=\{E_1,\ldots ,E_n\}$, then we get the following derivation:

$$\begin{aligned} {\mathcal {D}}({\mathbf {E}},H)>1&\Rightarrow \frac{\Pr (H|E_1,\ldots ,E_n)\Pr (H)^{n-1}}{\Pr (H|E_1)\cdot \ldots \cdot \Pr (H|E_n)}>1\\&\Rightarrow \Pr (H|E_1,\ldots ,E_n)>\underbrace{\frac{\Pr (H|E_1)}{\Pr (H)}}_{>1} \cdot \ldots \cdot \underbrace{\frac{\Pr (H|E_{n-1})}{\Pr (H)}}_{>1}\cdot \Pr (H|E_n)\\&\Rightarrow \Pr (H|E_1,\ldots ,E_n)>\Pr (H|E_n)\\&\Rightarrow \Pr (H|E_1,\ldots ,E_n)>\Pr (H) \end{aligned}$$

This latter fact means that $\xi (H,{\mathbf {E}})>\theta$ for all relevance-sensitive $\xi$. $\square$

Appendix 6: Proof of Observation 6.4

Keeping in mind that by assumption $\Pr (H|E_2)=\Pr (H|E_3)$, we get

$$\begin{aligned} {\mathcal {D}}({\mathbf {E}},H)>{\mathcal {D}}({\mathbf {E}}^{\prime },H)&\Leftrightarrow \frac{\Pr (H|E_1,E_2)\cdot \Pr (H)}{\Pr (H|E_1)\Pr (H|E_2)} >\frac{\Pr (H|E_1,E_3)\cdot \Pr (H)}{\Pr (H|E_1)\Pr (H|E_3)}\\&\Leftrightarrow \Pr (H|E_1,E_2)>\Pr (H|E_1,E_3) \end{aligned}$$

Thus, if $\xi$ satisfies (FPI), then $\Pr (H|E_1,E_2)>\Pr (H|E_1,E_3)$ entails that $\xi (H,{\mathbf {E}})>\xi (H,{\mathbf {E}}^{\prime })$. $\square$

Appendix 7: Proof of Observation 6.5

H	$E_1$	$E_2$	$E_2$	Probability	H	$E_1$	$E_2$	$E_2$	Probability
T	T	T	T	1/495	F	T	T	T	1/33
T	T	T	F	3/46	F	T	T	F	1/47
T	T	F	T	6/41	F	T	F	T	1/11
T	T	F	F	1/41	F	T	F	F	401/182172
T	F	T	T	1663/39606	F	F	T	T	1/68
T	F	T	F	5099/39606	F	F	T	F	4621/58938
T	F	F	T	1/21	F	F	F	T	1/114
T	F	F	F	1/177	F	F	F	F	x

Straightforward calculations yield the following results: $\Pr (H|E_1)\approx .622>.462\approx \Pr (H)>.363\approx \Pr (H|\lnot E_1)$ and $\Pr (H|E_i)=\Pr (H|E_j)$ as well as $\Pr (H|\lnot E_i)=\Pr (H|\lnot E_j)$ for all $1\le i,\,j\le 3$. Hence, ${\mathbf {E}}\cup {\mathbf {E}}^{\prime }$ is an equal positive evidence set for H. Furthermore, ${\mathcal {D}}({\mathbf {E}},H)\approx .676>.657\approx {\mathcal {D}}({\mathbf {E}}^{\prime },H)$; however, $n(H,{\mathbf {E}})\approx .050<.096\approx n(H,{\mathbf {E}}^{\prime })$, $m(H,{\mathbf {E}})\approx .027< .052\approx m(H,{\mathbf {E}}^{\prime })$ and $s(H,{\mathbf {E}})\approx .118<.121\approx s(H,{\mathbf {E}}^{\prime })$.

Appendix 8: Proof of Observation 6.6

H	$E_1$	$E_2$	$E_2$	Probability	H	$E_1$	$E_2$	$E_2$	Probability
T	T	T	T	1/12	F	T	T	T	1/31
T	T	T	F	1/205	F	T	T	F	1/306
T	T	F	T	1/1406	F	T	F	T	1/651
T	T	F	F	1/150	F	T	F	F	431/12852
T	F	T	T	3293/817950	F	F	T	T	1/918
T	F	T	F	6763/2017610	F	F	T	F	4513/132804
T	F	F	T	1/133	F	F	F	T	1/28
T	F	F	F	20/67	F	F	F	F	x

Given this probability distribution we calculate: $\Pr (H|E_1)\approx .575>.409\approx \Pr (H)>.376\approx \Pr (H|\lnot E_1)$ and $\Pr (H|E_i)=\Pr (H|E_j)$ as well as $\Pr (H|\lnot E_i)=\Pr (H|\lnot E_j)$ for all $1\le i,\,j\le 3$. Hence, ${\mathbf {E}}\cup {\mathbf {E}}^{\prime }$ is an equal positive evidence set for H. Furthermore, ${\mathcal {O}}({\mathbf {E}},H)\approx 1.444>1.428\approx {\mathcal {O}}({\mathbf {E}}^{\prime },H)$; however, $\Pr (H|{\mathbf {E}})\approx .7129<.7132\approx \Pr (H|{\mathbf {E}}^{\prime })$ and therefore all (FPI)-measures will agree in that $\xi (H,{\mathbf {E}})<\xi (H,{\mathbf {E}}^{\prime })$.

Appendix 9: Proof of Observation 6.7

The proof of Observation 6.7 utilizes Lemma 6.1:

$$\begin{aligned} {\mathcal {C}}_f({\mathbf {E}},H)&>{\mathcal {C}}_f({\mathbf {E}}^{\prime },H)\\&\Leftrightarrow \frac{\Pr (E_1|E_2, H)+\Pr (E_2|E_1,H)}{\Pr (E_1|E_2)+\Pr (E_2|E_1)} >\frac{\Pr (E_1|E_3, H)+\Pr (E_3|E_1,H)}{\Pr (E_1|E_3)+\Pr (E_3|E_1)}\\&\Leftrightarrow \Pr (H|E_1,E_2)\cdot \frac{\sum _{i=1,2}\Pr (H\wedge E_i)^{-1}}{\sum _{i=1,2}\Pr (E_i)^{-1}} > \Pr (H|E_1,E_3)\cdot \frac{\sum _{j=1,3}\Pr (H\wedge E_j)^{-1}}{\sum _{j=1,3}\Pr (E_j)^{-1}}\\&\Leftrightarrow \frac{ \Pr (H|E_1,E_2)}{ \Pr (H|E_1,E_3)} > \frac{\sum _{i=1,2}\Pr (H\wedge E_i)^{-1}}{\sum _{i=1,2}\Pr (E_i)^{-1}}\cdot \frac{\sum _{j=1,3} \Pr (E_j)^{-1}}{\sum _{j=1,3}\Pr (H\wedge E_j)^{-1}}\\&\Leftrightarrow \frac{ \Pr (H|E_1,E_2)}{ \Pr (H|E_1,E_3)} > 1 \quad\quad\quad ({\text {Lemma}}\; 6.1) \end{aligned}$$

This completes the proof of Observation 6.7.

Appendix 10: Proof of Lemma 6.1

If ${\mathbf {E}}\cup {\mathbf {E}}^{\prime }$ is an equal positive evidence set for H, then (i) $\Pr (H|E_i)>\Pr (H)>\Pr (H|\lnot E_i)$ and (ii) $\Pr (H|\pm E_i)=\Pr (H|\pm E_j)$ for all $1\le i,\,j\le 3$. Now we get:

$$\begin{aligned} \Pr (H)&= \Pr (H\wedge E_i) + \Pr (H\wedge \lnot E_i)\\&= \Pr (H|E_i)\cdot \Pr (E_i) + \Pr (H|\lnot E_i)\cdot \Pr (\lnot E_i)\\&\mathop {=}\limits ^{\text {(ii)}} \Pr (H|E_j)\cdot \Pr (E_i) + \Pr (H|\lnot E_j) \cdot \Pr (\lnot E_i) \end{aligned}$$

and also

$$\begin{aligned} \Pr (H)&= \Pr (H|E_j)\cdot \Pr (E_j) + \Pr (H|\lnot E_j)\cdot \Pr (\lnot E_j) \end{aligned}$$

Hence we get

$$\begin{aligned} \Pr (H|E_j)\cdot \Pr (E_i) + \Pr (H|\lnot E_j)\cdot \Pr (\lnot E_i) = \Pr (H|E_j)\cdot \Pr (E_j) + \Pr (H|\lnot E_j)\cdot \Pr (\lnot E_j) \end{aligned}$$

and thus

$$\begin{aligned} \Pr (H|E_j)\cdot (\Pr (E_i)-\Pr (E_j)) = \Pr (H|\lnot E_j)\cdot (\Pr (\lnot E_j) -\Pr (\lnot E_i)) \end{aligned}$$

from which we conclude that either $\Pr (H|E_i)=\Pr (H|\lnot E_i)$ in contradiction to (i) or $\Pr (E_i)=\Pr (E_j)$. With this latter identity and $\Pr (H|E_i)=\Pr (H|E_j)$ we conclude that also $\Pr (H\wedge \pm E_i)=\Pr (H\wedge \pm E_j)$.