Learning Conditional Information by Jeffrey Imaging on Stalnaker Conditionals

Abstract

We propose a method of learning indicative conditional information. An agent learns conditional information by Jeffrey imaging on the minimally informative proposition expressed by a Stalnaker conditional. We show that the predictions of the proposed method align with the intuitions in Douven (Mind & Language, 27(3), 239–263 2012)’s benchmark examples. Jeffrey imaging on Stalnaker conditionals can also capture the learning of uncertain conditional information, which we illustrate by generating predictions for the Judy Benjamin Problem.

Appendix

1.1 A Possible Worlds Model of the Jeweller Example

Following the presentation in [5], we consider the Jeweller Example.

Example 5

The Jeweller Example ([5, p. 654]) A jeweller has been shot in his store and robbed of a golden watch. However, it is not clear atthis point what the relation between these two events is; perhaps someone shot the jewellerand then someone else saw an opportunity to steal the watch. Kate thinks there is somechance that Henry is the robber (R). On the other hand, she strongly doubts that he iscapable of shooting someone, and thus, that he is the shooter (S). Now the inspector, afterhearing the testimonies of several witnesses, tells Kate:

$$ \text{If Henry robbed the jeweller, then he also shot him.} $$

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As a result, Kate becomes more confident that Henry is not the robber, whilst her probability forHenry having shot the jeweller does not change.

We model Kate’s belief state as the Stalnaker model $\mathcal {M}_{St} = \langle W, R, \le , \le ^{\prime } V \rangle$ depicted in Fig. 8. W contains four elements covering the possible events of R,¬R,S,¬S, where R stands for “Henry is the robber”, and S for “Henry has shot the jeweller”. The example suggests that 0 < P(R) < 1 and P(S) = 𝜖 for a small 𝜖, and thus P(¬S) = 1 − 𝜖. The prescribed intuitions are that P^∗(R) < P(R)and P^∗(S) = P(S). We know about Kate’s degrees of belief before receiving the conditional information that 0 < P(w₁) + P(w₂) < 1 and P(w₁) + P(w₃) = 𝜖, as well as P(w₂) + P(w₄) = 1 − 𝜖. Note that Kate is ‘almost sure’ that ¬S, and thus we may treat ¬S as ‘almost factual’ information.

Fig. 8

A Stalnaker model for Kate’s belief state in the Jeweller Example. The blue arrow indicates the unique w_{((R>S)∧¬S)}-world under ≤. The red arrows indicate that each world is its most similar ¬((R > S) ∧¬S)-world under ≤^′. The teal arrows represent the transfer of (1 − 𝜖) ⋅ P(w), whilst the violet arrows represent the transfer of 𝜖 ⋅ P(w)

Kate receives certain conditional information. She learns the minimally informative proposition [R > S] = {w₁, w₃, w₄} such that P(R > S) = P^R(S) = 1. By the law of total probability, P(R > ¬S) = P^R(¬S) = 0. Taking her uncertain but almost factual information into account, Kate learns in total the minimally informative proposition [(R > S) ∧¬S], which is identical to {w₄}. By P(R > S) = 1, P((R > S) ∧¬S) = P(¬S) = 1 − 𝜖. Note the tension expressed in P((R > S) ∧¬S) = 1 − 𝜖. It basically says that S is almost surely not the case and, under the supposition of R, we exclude the possibility of ¬S. Intuitively, the thought expressed by this statement should cast doubt as to whether R is the case.

By ¬((R > S) ∧¬S) ≡ (R > ¬S) ∨ S, we also know that P(R > ¬S) ∨ S) = 𝜖. Note that the proposition [(R > S) ∧¬S] = {w₄}(interpreted as minimally informative) specifies a similarity order ≤ such that w_(R>S)∧¬S = w₄ for all w. In contrast, the proposition [(R > ¬S) ∨ S] is minimally informative in a strong sense, since it does not exclude any world w. Hence, the ‘maximally inclusive’ proposition [(R > ¬S) ∨ S] = {w₁, w₂, w₃, w₄} specifies a similarity order ≤^′≠ ≤according to which w_(R>¬S)∨S = w for each w.

We apply now Jeffrey imaging to the Jeweller Example, where k = 1 − 𝜖.

$$\begin{array}{@{}rcl@{}} P^{(R > S) \land \neg S}_{1 - \epsilon}(w^{\prime}) = P^{*}(w^{\prime}) &=& \sum\limits_{w} \left( P(w) \cdot \left\{ \begin{array}{ll} 1 - \epsilon & \text{if \(w_{(R > S) \land \neg S} = w^{\prime}\)} \\ 0 & \text{otherwise} \end{array} \right\} \right.\\ &&\left.+P(w) \cdot \left\{ \begin{array}{ll} \epsilon & \text{if \(w_{(R > \neg S) \lor S} = w^{\prime}\)} \\ 0 & \text{otherwise} \end{array} \right\}\right) \end{array} $$

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We obtain the following probability distribution after learning:

$$\begin{array}{@{}rcl@{}} P^{*}_{1 - \epsilon}(w_{1}) \!&=&\! P^{*}_{1 - \epsilon}(R \land S) = \epsilon \cdot P(w_{1})\quad\quad P^{*}_{1 - \epsilon}(w_{2}) \,=\, P^{*}_{1 - \epsilon}(R \land \neg S) \,=\, \epsilon \cdot P(w_{2}) \\ P^{*}_{1 - \epsilon}(w_{3}) \!&=&\! P^{*}_{1 - \epsilon}(\neg R \land S) = \epsilon \cdot P(w_{3})\quad P^{*}_{1 - \epsilon}(w_{4}) \,=\, P^{*}_{1 - \epsilon}(\neg R \land \neg S)= (1 - \epsilon) \\ && \cdot (P(w_{1}) \,+\, P(w_{2}) + P(w_{3}) \\ && +P(w_{4})) + \epsilon \cdot P(w_{4})\\ \end{array} $$

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The results almost comply with the prescribed intuitions. The intuition concerning the degree of belief in R is met: P^∗(R) < P(R), since $P^{*}_{1 - \epsilon }(w_{1}) + P^{*}_{1 - \epsilon }(w_{2}) < P(w_{1}) + P(w_{2})$. The intuition concerning the degree of belief in S is ‘almost’ met: $P^{*}_{1 - \epsilon }(S) \approx P(S)$, for P(w₁) + P(w₃) = 𝜖 and $P^{*}_{1 - \epsilon }(w_{1}) + P^{*}_{1 - \epsilon }(w_{3}) = P(w_{1}) \cdot \epsilon + P(w_{3}) \cdot \epsilon \approx \epsilon$. In words, the method gives us the result that Kate is now pretty sure that Henry is neither the shooter nor the robber.