Bayesianism for Non-ideal Agents

Abstract

Orthodox Bayesianism is a highly idealized theory of how we ought to live our epistemic lives. One of the most widely discussed idealizations is that of logical omniscience: the assumption that an agent’s degrees of belief must be probabilistically coherent to be rational. It is widely agreed that this assumption is problematic if we want to reason about bounded rationality, logical learning, or other aspects of non-ideal epistemic agency. Yet, we still lack a satisfying way to avoid logical omniscience within a Bayesian framework. Some proposals merely replace logical omniscience with a different logical idealization; others sacrifice all traits of logical competence on the altar of logical non-omniscience. We think a better strategy is available: by enriching the Bayesian framework with tools that allow us to capture what agents can and cannot infer given their limited cognitive resources, we can avoid logical omniscience while retaining the idea that rational degrees of belief are in an important way constrained by the laws of probability. In this paper, we offer a formal implementation of this strategy, show how the resulting framework solves the problem of logical omniscience, and compare it to orthodox Bayesianism as we know it.

Appendix

This appendix contains proofs of three main results of the paper. All definitions can be found in §3. The results are repeated here for convenience.

Theorem 1

(n-preservation) If $A \vdash _R^n B$, then $Cr(A) = x \models \langle n \rangle (Cr(B) \ge x)$.

Proof

Suppose that $A \vdash _R^n B$ and consider any pointed model, (M, w), such that $M,w \models Cr(A) = x$, where $M = (W^P,W^I,f,V)$ and $w \in W^P$. We must show that $M,w \models \langle n \rangle (Cr(B) \ge x)$. We proceed by defining a suitable n-accessible pointed model from (M, w). Let $M' = (W'^P,W'^I,f',V')$ be a model such that $W'^P = W^P$, $W'^I = W^I$, and $V' = V$. Since $A \vdash _R^n B$, we can let $f'$ be an n-variation of (M, w) for which it holds that $f'(w) = (c,Pr_c)$, where $M',v \models B$, for all $v \in \{v' \in c: M',v' \models A\}$. By the definition of n-accessibility, then, $(M,w) {\mathop {\sim }\limits ^{n}} (M',w)$. Since $M,w \models Cr(A) = x$, (P4) tells us that $\sum _Q Pr(Q) = x$, where $Q = \{v \in S: M,v \models A\}$. Hence, $\sum _{Q'} Pr_c(Q') \ge x$, where $Q' = \{v \in c: M',v \models B\}$. By another application of (P4), $M',w \models Cr(B) \ge x$. So, by (P5), it follows that $M,w \models \langle n \rangle (Cr(B) \ge x)$. $\square$

Theorem 2

(n-certainty) If $\vdash _R^n A$, then $\models \langle n \rangle (Cr(A) = 1)$.

Proof

Suppose that $\vdash _R^n A$ and let (M, w) be any pointed model such that $M = (W^P,W^I,f,V)$ and $w \in W^P$. We must show that $M,w \models \langle n \rangle (Cr(A) =1)$. We proceed by defining a suitable n-accessible pointed model from (M, w). Let $M' = (W'^P,W'^I,f',V')$ be a model such that $W'^P = W^P$, $W'^I = W^I$, and $V' = V$. Since $\vdash _R^n A$, we can let $f'$ be an n-variation of (M, w) such that $f'(w) = (c,Pr_c)$, where $M',v \models A$, for all $v \in c$. Hence, $\sum _{Q'} Pr_c(Q') = 1$, where $Q' = \{v \in c: M',v \models A\}$. By (P4), $M',w' \models Cr(A) = 1$. By the definition of n-accessibility, $(M,w) {\mathop {\sim }\limits ^{n}} (M',w)$. So, by (P5), it follows that $M,w \models \langle n \rangle (Cr(A) = 1)$. $\square$

Theorem 3

(n-conditionality) If $A \vdash _R^n B$, then $\models \langle n \rangle (Cr(B \, | \, A) = 1)$.

Proof

Suppose that $A \vdash _R^n B$ and let (M, w) be any pointed model such that $M = (W^P,W^I,f,V)$ and $w \in W^P$. We must show that $M,w \models \langle n \rangle (Cr(B \,|\, A) = 1)$. Let $M' = (W'^P,W'^I,f',V')$ be a model such that $W'^P = W^P$, $W'^I = W^I$, and $V' = V$. Since $A \vdash _R^n B$, we can let $f'$ be an n-variation of (M, w) such that $f'(w) = (c,Pr_c)$, where $M',v \models B$, for all $v \in \{v' \in c: M',v' \models A\}$. Hence, $\sum _{Q} Pr_c(Q) = \sum _{Q'} Pr_c(Q')$, where $Q = \{v \in c: M',v \models A\,\text {and}\,M',v \models B\}$ and $Q' = \{v' \in c: M',v' \models q\}$. By (P8), $M',w \models Cr(B \,|\, A) = 1$. By the definition of n-accessibility, $(M,w) {\mathop {\sim }\limits ^{n}} (M',w)$. So, by (P5), it follows that $M,w \models \langle n \rangle (Cr(B \,|\, A) = 1)$. $\square$