Stability and Scepticism in the Modelling of Doxastic States: Probabilities and Plain Beliefs

Abstract

There are two prominent ways of formally modelling human belief. One is in terms of plain beliefs (yes-or-no beliefs, beliefs simpliciter), i.e., sets of propositions. The second one is in terms of degrees of beliefs, which are commonly taken to be representable by subjective probability functions. In relating these two ways of modelling human belief, the most natural idea is a thesis frequently attributed to John Locke: a proposition is or ought to be believed (accepted) just in case its subjective probability exceeds a contextually fixed probability threshold $t<1$. This idea is known to have two serious drawbacks: first, it denies that beliefs are closed under conjunction, and second, it may easily lead to sets of beliefs that are logically inconsistent. In this paper I present two recent accounts of aligning plain belief with subjective probability: the Stability Theory of Leitgeb (Ann Pure Appl Log 164(12):1338–1389, 2013; Philos Rev 123(2):131–171, 2014; Proc Aristot Soc Suppl Vol 89(1):143–185, 2015a; The stability of belief: an essay on rationality and coherence. Oxford University Press, Oxford, 2015b) and the Probalogical Theory (or Tracking Theory) of Lin and Kelly (Synthese 186(2):531–575, 2012a; J Philos Log 41(6):957–981, 2012b). I argue that Leitgeb’s theory may be too sceptical for the purposes of real life.

Appendix: Proofs for Section 2.1

Proof of the equivalence of (1) and (2). That (2) implies (1) (and even closure under infinite intersections) is trivial. For the converse, suppose that $LS_t^P$ is closed under finite intersections. Since we are assuming that the field for P is finite, we get that $\bigcap LS_t^P = LC_t^P$ is in $LS_t^P$. Thus, by the definition of $LS_t^P$, $P(LC_t^P) \underset{(-)}{>}t$. By the laws of probability, we then get for all A such that $LC_t^P \subseteq A$, $P(A) \underset{(-)}{>}t$, that is $A \in LS_t^P$. This is one direction of (2). For the other direction, suppose that A is in $LS_t^P$. Then $\bigcap LS_t^P \subseteq A$, i.e., $LC_t^P \subseteq A$, as desired.

Proof of the equivalence of (4) and (5), with identical propositions X. Assume (4), fix a suitable proposition X from (4) and let $w\in X$. By the left-to-right part of (4), $P(W-\{w\}) \underset{)-(}{<}t$, i.e., $P(w) \underset{)-(}{>}1-t$. By the right-to-left part of (4), $P(X)\underset{(-)}{>}t$, i.e., $P(\overline{X } )\underset{(-)}{<}1-t$. Taking this together, we get (5), for the same proposition X as in (4). For the converse, assume (5) and fix a suitable proposition X from (5). In order to prove (4), let first $P(A)\underset{(-)}{>}t$. Suppose for reductio that $X \not \subseteq A$. Then $A \subseteq W-\{w\}$ and thus $P(A) \le 1-P(w)$ for some w in X. So, for this w, $P(A) \le 1 - P(w) \underset{)-(}{<}1-(1-t) = t$, by (5), and we get a contradiction. Conversely, let $X\subseteq A$. Then $P(X)\le P(A)$. But by (5), $P(X)\underset{(-)}{>}t$, so $P(A)\underset{(-)}{>}t$, as desired for (4), with the same proposition X as in (5).

Proof that (5) implies (6). Take the X from (5). We first check the range of the threshold t. For version (${\hbox {LT}}_>^t$) of the Lockean thesis, (5) requires us to have $P(\overline{X } ) < 1-t \le \min _{w\in X} P(w)$ which means that t is in the interval $[1-\min _{w\in X} P(w),P(X))$. For version (${\hbox {LT}}_\ge ^t$) of the Lockean thesis, (5) requires us to have $P(\overline{X } ) \le 1-t < \min _{w\in X} P(w)$ which means that t is in the interval $(1-\min _{w\in X} P(w),P(X)]$. But this is just what the first part of (6) says. For the second part of (6), let B be such that $B\cap X \not = \emptyset$. Then, since by (5) $P(w)>0$ for every $w\in X$, $P(B)>0$. Moreover, $P(X\vert B) = \frac{P(X\cap B)}{P(B)} = \frac{P(X\cap B)}{P(X\cap B)+P(\overline{X } \cap B)}$. Now on the one hand $P(\overline{X } \cap B) \le P(\overline{X } ) \underset{(-)}{<}1-t$ by (5), and on the other hand $P(X\cap B) \ge P(w)$ for some w in X and $P(w)\underset{)-(}{>}1-t$ by (5) again, so $P(X\cap B)\underset{)-(}{>}1-t$. Thus clearly $P(\overline{X } \cap B)<P(X\cap B)$, and thus $\frac {1}{2} < P(X\vert B) \le 1$.

Proof that (6) implies (5). Take X from (6), choose an arbitrary $w \in X$ and put $B:=\overline{X } \cup \{w\}$. From (6), we then get $P(\overline{X } \cup \{w\})>0$ and $P(X\vert \overline{X } \cup \{w\}) > \frac {1}{2}$, that is $\frac{P(X\cap (\overline{X } \cup \{w\}))}{P(\overline{X } \cup \{w\})} > \frac {1}{2}$. Since $\overline{X }$ and $\{w\}$ are disjoint, this reduces to $\frac{P(w)}{P(\overline{X } )+P(w)} > \frac {1}{2}$, or just $P(w) > P(\overline{X } )$. Since w was chosen arbitrarily from X and the threshold condition is checked as in the direction from (5) to (6), this establishes (5).