The Representation of Belief

Abstract

I derive a sufficient condition for a belief set to be representable by a probability function: if at least one comparative confidence ordering of a certain type satisfies Scott’s axiom, then the belief set used to induce that ordering is representable. This provides support for Kenny Easwaran’s project of analyzing doxastic states in terms of belief sets rather than credences.

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Notes

  1. 1.

    If an agent’s credences do not conform to, for example, Kolmogorov’s axioms of probability, then in certain betting scenarios, she is guaranteed to lose money. See [11] for one version of this ‘Dutch Book’ argument.

  2. 2.

    See [8] for a statement of the Preface Paradox, and see [4] for a proposed solution.

  3. 3.

    For a defense of this analysis, see [4]. For a detailed discussion of just the left-to-right direction, see [6].

  4. 4.

    See [2], p. 830.

  5. 5.

    The proof assumes that the probability function which represents B assigns a non-zero probability to every possible state of the world.

  6. 6.

    The proposition which consists of zero atoms is the empty proposition, denoted ⊥.

  7. 7.

    For justification of this consequence of Definition 2, see [2] (p. 828).

  8. 8.

    Formally, this is represented as \(\sum \limits _{i = 1}^{n}x_{i}=\sum \limits _{i = 1}^{n}y_{i}\), where each x i (and each y i ) is understood to be a characteristic function of the corresponding element in \(\mathcal {X}\).

  9. 9.

    If x 2 > y 2, and \(x_{1}\geqslant y_{1}\), then x 1 + x 2 > y 1 + y 2, which contradicts the supposition that x 1 + x 2 = y 1 + y 2.

  10. 10.

    It is not hard to show that \(\geqslant \) on the reals also satisfies an analog of (SA) for all n greater than or equal to 2.

  11. 11.

    Thanks to an anonymous reviewer for drawing my attention to this.

  12. 12.

    Depending on the belief set in question, the only possible exception to this might be D 1. For D 1 would include comparisons among A’s beliefs if for some p, A believes p and A believes ¬p.

  13. 13.

    Given the assumption that ⊥∉B and that ⊤∈ B, \((\mathcal {C}_{1})\) actually follows from \((\mathcal {C}_{2})\). I still state \((\mathcal {C}_{2})\) as a separate condition because in the proofs to come, it is perhaps the most important consequence of the assumption that ⊥∉B and that ⊤∈ B.

  14. 14.

    In fact, each ordering in \(\mathcal {K}\) induces the very same belief set.

  15. 15.

    To illustrate what this means: if B turns out to be representable by a probability function, then every proposition in B but not B must get assigned probability \(\frac {1}{2}\).

  16. 16.

    A necessary condition for b-representability is related to—but distinct from—this sufficient condition.

  17. 17.

    Thanks to an anonymous reviewer for pressing the importance of this point.

  18. 18.

    An analogy might help here: facts about beliefs (constrained by the \(\mathcal {C}\) construction) give rise to those orderings, much in the way that facts about fundamental particles (constrained by statistical mechanical laws) give rise to the law-like regularities of thermodynamics.

  19. 19.

    Note that the \(\mathcal {C}\) construction does not itself suggest anything about which (if any) of the generated orderings provide the best description of A’s doxastic state. At most, it articulates a necessary condition on the comparative confidence ordering that A may reasonably have: in order to be a suitable non-fundamental description of A’s belief set, that ordering must satisfy the relatively minor restrictions of the \(\mathcal {C}\) construction.

  20. 20.

    This is especially reasonable given the assumption, employed in the formulation of Definition 2, that the threshold for belief is \(\frac {1}{2}\). It is less reasonable for different belief thresholds. Again, this is discussed in Section 5 and in the A.

  21. 21.

    This is assumed merely in order to keep the example relatively simple.

  22. 22.

    There is a related argument for the claim that (SA) is a rationality constraint. In [5], pp. 60–62, James Hawthorne argues that a rational agent should have a comparative confidence ordering that either satisfies, or is extendible to, a condition he calls (X). Roughly put, an agent’s total comparative confidence ordering satisfies (X) just in case there is some way to partition the space of possibilities into equally plausible states such that the agent has very little confidence in any one of them. It can be shown that in conjunction with a few other plausible assumptions, (X) implies (SA). Thus, to the extent that (X) amounts to a necessary condition for rationality, (SA) does too. Note that this argument, unlike mine, assumes that agents actually have comparative confidence orderings.

  23. 23.

    For clarity of exposition, in this example, I occasionally describe Emily as actually having a comparative confidence ordering. Nothing hangs on that, however. The entire example could be reworded, in a convoluted way, so that the comparative confidence ordering is just a way of describing Emily’s belief set.

  24. 24.

    The other ways of violating (SA) are just as irrational as this one.

  25. 25.

    Regardless of which atom is taken to be actual, the number of truths in X and the number of truths in Y are each one. For example, if left is taken to be actual, then exactly one proposition in X is true (the proposition left) and exactly one proposition in Y is true (the proposition left ∨right).

  26. 26.

    Generalizations to other thresholds seem to be more convoluted than the 1 threshold generalization. Some of them may constitute mathematically and philosophically successful reductions of credence to belief, but it is beyond the scope of the present discussion to explore if that is so.

  27. 27.

    Of all the thresholds for which one might want to reduce probability to belief, 1 and \(\frac {1}{2}\) are among the most desirable. For they are among the few thresholds for belief which seem plausible, independent of context. In part, the results are simplest for these thresholds because there are known qualitative axiomatizations of the constraints that belief sets corresponding to them must satisfy. In the case of 1, those constraints derive from classical logic: the belief sets must be deductively closed, for instance. For the case of \(\frac {1}{2}\), see [12].

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Acknowledgements

Many thanks to Laura Callahan, Sam Carter, Kevin Dorst, Kenny Easwaran, Frankie Egan, Adam Elga, Danny Forman, Jimmy Goodrich, James Hawthorne, Nevin Johnson, Barry Loewer, Jonathan Schaffer, an anonymous reviewer, and the audience at NASSLLI 2017 for helpful comments. Special thanks are due to Branden Fitelson for all his help, guidance, and support.

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Correspondence to Isaac Wilhelm.

Appendix

Appendix

The results of Section 3 generalize to the case of a belief threshold of 1. First, a new definition of belief representability is required.

Definition 7 (B1-Representable)

Let \(\mathcal {X}\)be a finite Boolean algebra,and let \(B\subseteq \mathcal {X}\)be a belief set. Say that B is b 1 -representable just in casethere is a probability function P r such that

  1. (i)

    if P r(p) = 1then pB,and

  2. (ii)

    if P r(p) < 1then pB.

Second, the \(\mathcal {C}\) construction needs to be adjusted for the 1 threshold. The new construction consists of two parts. For the first, define the following two sets.

  • \({D_{1}^{1}}\) = {〈p,⊤〉∣pB}.

  • \({D_{2}^{1}}\) \(=\{\langle p,\bot \rangle \mid p\in \mathcal {X}\}\).

Then let \(\succeq _{B}^{\star _{1}}={D_{1}^{1}}\cup {D_{2}^{1}}\).

\({D_{1}^{1}}\) says that if A believes p, then A is at least as confident in p as in the tautological proposition. Given that the notion of b1-representability under scrutiny here takes the threshold for belief to be 1, \({D_{1}^{1}}\) is extremely plausible: it implies absolute certainty in all propositions believed. \({D_{2}^{1}}\) is plausible for same reasons D 3 is plausible (see Section 3).

For the second part of the construction, let \(\mathcal {K}_{1}\) be the set of total comparative confidence orderings ≽ that contain \(\succeq _{B}^{\star _{1}}\) as a subset, and that also satisfy the following conditions.

  • \((\mathcal {C}_{1})\) 〈⊥,⊤〉∉ ≽.

  • \((\mathcal {C}_{2}^{1})\) For all pB, 〈p,⊤〉∉ ≽.

As before, \((\mathcal {C}_{1})\) ensures that according to every ordering in \(\mathcal {K}_{1}\), A is strictly more confident in ⊤ than in ⊥. \((\mathcal {C}_{2}^{1})\) ensures that if p is not believed, then A is not at least as confident in p as in ⊤. Since the orderings in \(\mathcal {K}_{1}\) are total, this implies that if p is not believed, then A is strictly more confident in ⊤ than in p.

The following definition provides a succinct way to refer to this construction.

Definition 8 (Constructed from B in the manner of \(\mathcal {C}\)-1)

Let \(\mathcal {X}\)be a finite Boolean algebra, let \(B\subseteq \mathcal {X}\)be a belief set, and let \(\mathcal {K}_{1}\)be the set of total comparative confidence orderings that contain\(\succeq _{B}^{\star _{1}}\)and that satisfy \((\mathcal {C}_{1})\)and \((\mathcal {C}_{2}^{1})\).Then \(\mathcal {K}_{1}\)is the set of comparative confidence orderings constructed from B in the manner of \(\boldsymbol {\mathcal {C}-1}\).

As before, we need an account of how comparative confidence orderings can induce belief sets.

Definition 9 (Induced1 Belief Set)

Let \(\mathcal {X}\)be a finite Boolean algebra, and let \(\succeq \;\subseteq \mathcal {X}\times \mathcal {X}\)be a comparative confidence ordering. Let B be the belief set which consists of all and only the propositions p such that p ≽⊤.Call B the belief set induced 1 by .

Now to derive a sufficient condition for b1-representability that is analogous to Theorem 3.4. Proofs are included only when they are significantly different from the proofs of the corresponding lemmas and theorems in Section 3.

Lemma 2

Let \(\mathcal {X}\) be a finite Boolean algebra, let \(B\subseteq \mathcal {X}\) be a belief set, and let \(\mathcal {K}_{1}\) be the set of comparative confidence orderings constructed from B in the manner of \(\mathcal {C}\) -1. Each \(\succeq \;\in \mathcal {K}_{1}\) satisfies (A 1), (A 2), and (A 3).

Theorem A.1

Let \(\mathcal {X}\) be a finite Boolean algebra, let \(B\subseteq \mathcal {X}\) be a belief set, and let \(\mathcal {K}_{1}\) be the set of comparative confidence orderings constructed from B in the manner of \(\mathcal {C}\) -1. Then for each \(\succeq \;\in \mathcal {K}_{1}\) ,satisfies (SA) if and only ifis c-representable.

Theorem A.2

Let \(\mathcal {X}\) be a finite Boolean algebra, and let \(\succeq \;\subseteq \mathcal {X}\times \mathcal {X}\) be a comparative confidence ordering that induces 1 the belief set B .Let P r be a probability function that c-represents ≽.Then P r b 1 -represents B .

Proof

If P r(p) = 1then since P r(⊤) = 1,it follows that \(Pr(p)\geqslant Pr(\top )\).So by Definition 4, p ≽⊤.Therefore, by Definition 9, pB .

If P r(p) < 1then since P r(⊤) = 1,it follows that P r(p) < P r(⊤).So by Definition 4, p ≺⊤.Definition 9 implies that pB .

Therefore, by Definition 7, P r b1-represents B . □

Theorem A.3

Let \(\mathcal {X}\) be a finite Boolean algebra, let \(B\subseteq \mathcal {X}\) be a belief set, and let \(\mathcal {K}_{1}\) be the set of comparative confidence orderings constructed from B in the manner of \(\mathcal {C}\) -1. Suppose that some \(\succeq \;\in \mathcal {K}_{1}\) induces 1 the belief set B .Then B = B.

Proof

First, take pB . By Definition 9, p ≽⊤. By condition \((\mathcal {C}_{2}^{1})\), if pB, then p ≽ ⊤. Thus, pB.So B B.

Second, take pB. Then \(\langle p,\top \rangle \in {D_{1}^{1}}\),from which it follows that 〈p,⊤〉∈ ≽.Thus, by Definition 9, pB .So BB .

Therefore, B = B. □

Finally, here is the sufficient condition for b 1-representability.

Theorem A.4 (Sufficient Condition for B1-Representability)

Let \(\mathcal {X}\) be a finite Boolean algebra, let \(B\subseteq \mathcal {X}\) be a belief set, and let \(\mathcal {K}_{1}\) be the set of comparative confidence orderings constructed from B in the manner of \(\mathcal {C}\) -1. Suppose that some \(\succeq \;\in \mathcal {K}_{1}\) satisfies (SA). Then B is b 1 -representable.

Proof

By Theorem A.1, ≽is c-representable by some probability function P r.Let B be the belief set induced1by ≽.By Theorem A.2, P r b1-represents B .Since B = B by Theorem A.3, P r b1-represents B. □

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Wilhelm, I. The Representation of Belief. J Philos Logic 47, 715–732 (2018). https://doi.org/10.1007/s10992-017-9448-8

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Keywords

  • Formal epistemology
  • Representation theorem
  • Belief
  • Credence
  • Scott’s axiom