Abstract
This paper is about the statics and dynamics of belief states that are represented by pairs consisting of an agent’s credences (represented by a subjective probability measure) and her categorical beliefs (represented by a set of possible worlds). Regarding the static side, we argue that the latter proposition should be coherent with respect to the probability measure and that its probability should reach a certain threshold value. On the dynamic side, we advocate Jeffrey conditionalisation as the principal mode of changing one’s belief state. This updating method fits the idea of the Lockean Thesis better than plain Bayesian conditionalisation, and it affords a flexible method for adding and withdrawing categorical beliefs. We show that it fails to satisfy the traditional principles of Inclusion and Preservation for belief revision and the principle of Recovery for belief withdrawals, as well as the Levi and Harper identities. We take this to be a problem for the latter principles rather than for the idea of coherent belief change.
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Acknowledgements
We gratefully acknowledge the generous support by the Swedish Collegium for Advanced Study in Uppsala where both authors enjoyed ideal research conditions as fellows in spring 2016. We also thank audiences at the 2017 Madeira Workshop on Belief Revision, Argumentation, Ontologies and Norms, at the 7th annual meeting of the DFG priority program New Frameworks of Rationality in Etelsen and at a workshop on New Perspectives on Conditionals and Reasoning in Regensburg, as well as two anonymouos reviewers of this journal for excellent questions and comments.
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Appendix: Proofs
Appendix: Proofs
Proof
of Observation 2
- (i)
Assume that Bel is consistent and locally threshold based. So there is some threshold t such that \(\mathit {Bel} = \bigcap \{C: Pr (C)\geq t\}\). Assume that u ∈Bel and that Pr(u) ≤ Pr(v). As u ∈Bel, \(Pr (\overline {\{u\}}) < t\). As Pr(u) ≤ Pr(v), \(Pr (\overline {\{v\}}) \leq Pr (\overline {\{u\}})\). So \(Pr (\overline {\{v\}}) < t\). As \(\overline {\{v\}}\) is the weakest proposition incompatible with v we know that for every C such that Pr(C) ≥ t, v ∈ C. So v ∈Bel.
- (ii)
Assume that Pr assumes at least two different values on U. Let D be the set of Pr-minimal worlds in U (i.e. u ∈ D iff Pr(u) ≤ Pr(v) for all v ∈ U). Let t = 1 − Pr(u), where u ∈ D. Let \(\mathit {Bel} = \overline {D}\). One can show that
$$ \mathit{Bel} = \bigcap \{C: Pr(C)\geq t\}, $$i.e. that Bel is threshold based. For note that \(Pr (\overline {\{u\}})\geq t\) for all u ∈ D. This means that for all u ∈ D, \(u\not \in \bigcap \{C: Pr (C)\geq t\}\). So \( \bigcap \{C: Pr (C)\geq t\}\subseteq \mathit {Bel}\). Assume that \(v\in \mathit {Bel} (=\overline {D})\). As Pr(v) > Pr(u) (for u ∈ D), 1 − Pr(v) < t. As \(Pr (\overline {\{v\}}) = 1-Pr (v)\), \(Pr (\overline {\{v\}})< t\). But \(\overline {\{v\}}\) is the weakest proposition excluding v, so v ∈ C for each C such that Pr(C) ≥ t. So \(v\in \bigcap \{C: Pr (C)\geq t\}\) and so \(\mathit {Bel}\subseteq \bigcap \{C: Pr (C)\geq t\}\).
- (iii)
Assume that Pr assumes at least two different positive values on U. Now let E be the set of positiveu ∈ E iff Pr(u)≠ 0 and Pr(u) ≤ Pr(v) for all v ∈ U such that Pr(v)≠ 0). Let t = 1 − Pr(u) where u ∈ E. Where D as before is the set of Pr-minimal worlds, let \(\mathit {Bel} = \overline {D \cup E}\). Clearly, 0 < Pr(Bel) < 1, and one can show that
$$ \mathit{Bel} = \bigcap \{C: Pr(C)\geq t\}, $$i.e. that Bel is threshold based. For note that \(Pr (\overline {\{u\}})\geq t\) for all u ∈ D ∪ E. This means that for all u ∈ D ∪ E, \(u\not \in \bigcap \{C: Pr (C)\geq t\}\). So \( \bigcap \{C: Pr (C)\geq t\}\subseteq \mathit {Bel}\). Assume that \(v\in \mathit {Bel} (=\overline {D \cup E})\). As Pr(v) > Pr(u) (for u ∈ E), 1 − Pr(v) < t. As \(Pr (\overline {\{v\}}) = 1-Pr (v)\), \(Pr (\overline {\{v\}})< t\). But \(\overline {\{v\}}\) is the weakest proposition excluding v, so v ∈ C for each C such that Pr(C) ≥ t. So \(v\in \bigcap \{C: Pr (C)\geq t\}\) and so \(\mathit {Bel}\subseteq \bigcap \{C: Pr (C)\geq t\}\).
□
Proof
of Observation 3
- (i)
From Observation 2 we know that the threshold based belief sets form an inclusion chain. As U is finite this inclusion chain has an inclusion minimal set: the strongest threshold based belief set.
- (ii)
Let C be the strongest threshold based belief set and let Bel be a coherent belief set such that C ⊆Bel. Let t0 be a threshold that generates C (so \( C= \bigcap \{E: Pr (E)\geq t_{0}\}\)). If Bel = U then Bel is threshold based, so assume that Bel ⊂ U. Let u be an element of \(\overline {\mathit {Bel}}\) such that for every element v of \(\overline {\mathit {Bel}}\), Pr(v) ≤ Pr(u). Let t = 1 − Pr(u). It follows that t ≥ t0 (as otherwise u ∈ C). As u∉C there is a proposition E such that u∉E and Pr(E) ≥ t0. It follows that for every proposition E such that \(E = \overline {\{v\}}\) where v∉Bel, Pr(E) ≥ t0, so C ⊆ E. Note that if Pr(E) ≥ t, then Bel ⊆ E (as every element v of Bel has a higher probability than u, we have \(Pr (\overline {\{v\}}) < t\)). So \(\mathit {Bel} = \bigcap \{E: Pr (E) \geq t\}\).
□
Proof
of Observation 4 Let \(\{u_{1}, \ldots , u_{n}\} = \overline {\mathit {Bel}}\). For any ui define \(A_{i} = \overline {\{u_{i}\}}\). As ui∉Bel, Pr(ui) < 0.5, so Pr(Ai) > 0.5. So there is some t such that t > 0.5 and Pr(Ai) > t for each i. Clearly, \(\mathit {Bel} = \bigcap _{1\leq i\leq n} A_{i}\). □
Proof
of Observation 5 Let Bel0,…,Beln be an enumeration of the coherent belief sets in increasing strength.
If n = 0, the claim is trivial. So assume that n ≥ 1, and let \(P_{\mathit {Bel}_{i}}(. ) = P(. \vert \mathit {Bel}_{i})\) for any i < n. As Beli+ 1 is a strict subset of Beli there are worlds u and v such that v ∈Beli+ 1, u∉Beli and u,v ∈Beli such that Pr(u) < Pr(v). As a result \(Pr _{\mathit {Bel}_{i}}(u) < Pr _{\mathit {Bel}_{i}}(v)\). Note that Beli+ 1 is the maximal coherent and \(Pr _{\mathit {Bel}_{i}}\)-informative belief set. From Observation 2(iii) we know that there is a \(Pr _{\mathit {Bel}_{i}}\)-informative locally threshold based belief set, and from Observations 2(i) and 3 we know that in relation to \(Pr _{\mathit {Bel}_{i}}\), Beli+ 1 is threshold based. □
Proof of Observation 6
Let Bel0,…,Beln be an inclusion chain of consistent belief sets such that Bel0 is locally threshold based and each Beli+ 1 is locally threshold based relative to \(Pr _{\mathit {Bel}_{i}}\). Proof by induction over n. For i = 0 it follows from Observation 2 that Bel0 is coherent. Assume, for the induction step, that Beli is coherent. By assumption Beli+ 1 is locally threshold based relative to \(Pr _{\mathit {Bel}_{i}}\). From Observations 2 and 3 we know that any threshold based belief set is coherent. So Beli+ 1 is coherent given \(Pr _{\mathit {Bel}_{i}}\). But as Beli is coherent relative to Pr, Beli+ 1 will also be coherent relative to Pr. □
Proof
of Observation 8
For any arbitrary threshold t such that 0 < t < 1, we can use the following model with \(W=\{w_{1}, w_{2}, w_{3}, \dots , w_{n_{2}+2}\}\):
So \(A=\{w_{1}, w_{3}, \dots , w_{n_{2}+2}\}\), Pr(A) = 1 − p1 and \(Pr (\overline {A})=p_{1}\). Let \(n_{2} = \text {floor}(\frac {2(1-t)}{t})+1 > \frac {2(1-t)}{t}\). Notice that n2 depends on t. Then we assign the following probabilities
Notice that \(p_{2} < \frac {(1-t)\cdot t}{2(1-t)} = \frac {t}{2} = p_{1}\) and that 2p1 + n2p2 = 1.
The belief set supported by (Pr,t) is Bel = Belt(Pr) = {w1,w2}.
Now we show that \(\mathit {Bel}* A = \mathit {Bel}_{t}(Pr ^{\alpha }_{A}) = A = \{w_{1}, w_{3}, \dots , w_{n_{2}+2}\}\) where α is any arbitrary Jeffrey parameter suitable for obtaining A as a belief.
Our proof strategy in choosing (13) is to have the relationship between p1 and p2 so that there is no α such that \(Pr ^{\alpha }_{A}(\{w_{1}\}) \geq t\). This implies that even when A is probable enough to be believed, the probability of w1 can’t itself pass the threshold t. So \(w_{3}, \dots , w_{n_{2}+2}\) will all be elements of the new belief set, which will be A, resulting in a violation of Preservation.
Meanwhile Bel ∩ A = {w1}≠∅. In order for a revision by A to satisfy Preservation the new belief set should be a subset of (and in fact identical to) {w1}. This means that we need an α such that
That is, we need an α between 0 and 1 such that
That is
But this is impossible since one can show that
This is so because we have \(Pr (w_{1}) = \frac {t}{2}\) and \(Pr (A) = 1-\frac {t}{2}\), and \(t(1-\frac {t}{2}) - \frac {t}{2} = \frac {1}{2}t(1-t)\) which is positive for t between 0 and 1.
So given the threshold t and the probability measure Pr (defined in dependence of t), every possible revision of Bel = {w1,w2} by A, minimal or otherwise, results in the belief set Bel ∗ A = A and violates Preservation. □
Proof
of Observation 9 For any arbitrary threshold t such that 0 < t < 1, we can use the following model with \(W=\{w_{1}, w_{2}, w_{3}, w_{4}, \dots , w_{m_{4}+3}\}\):
So A = {w2,w3}. Let \(m_{4} = \text {floor}(\frac {2(3-t)}{t})+1 > \frac {2(3-t)}{t}\). Notice that m4 depends on t. The role of the set \(\{w_{4} \dots , w_{m_{4}+3}\}\) is to store the residual probability in such a way that it spreads so thin (over so many worlds) that it is not relevant. Then we assign the following probabilities
One quickly finds that the assignments in (14) satisfy the requirement p1 > p2 > p3 > p4. Our proof strategy in choosing these assignments was to guarantee that p1 + p2 + p3 = t. So the belief set supported by (Pr,t) is Bel = Belt(Pr) = {w1,w2,w3}. Meanwhile \(Pr (A) = \frac {t}{2}\) and \(Pr (\overline {A}) = 1-\frac {t}{2}\).
The relationship between p1, p2 and p3 is made to be such that as soon as \(Pr ^{\alpha }_{A}(w_{3}) > Pr ^{\alpha }_{A}(w_{1})\), i.e., α > α3, we already have \(Pr ^{\alpha }_{A}(w_{2}) \geq t\). This implies that as soon as A is probable enough to be believed, the probability of w2 by itself passes the threshold t, leaving w3 outside of the new belief set that we get when revising by A. Now we show that \(\mathit {Bel}* A = \mathit {Bel}_{t}(Pr ^{\alpha }_{A}) = A = \{w_{2}\}\) where α is any arbitrary Jeffrey parameter suitable for obtaining A as a belief. Consider
For α = α3, \(\textit {UppA}(Pr ^{\alpha }_{A}) = \{w_{2}\}\) and \(Pr (\textit {UppA}(Pr ^{\alpha }_{A})) = \frac {\alpha }{Pr (A)}\cdot p_{2} = \frac {t(3-t)\cdot 2\cdot t}{2\cdot t\cdot (3-t)} = t\).
As long as α < α3, A is not yet believed, since then \(\textit {UppA}(Pr ^{\alpha }_{A}) \subseteq \{w_{2}\}\) and Pr({w2}) < t. If α > α3, on the other hand, Pr({w2}) > t, so Bel ∗ A = {w2} for every A-accepting revision Bel ∗ A, minimal or otherwise.
But Bel ∩ A = {w1,w2,w3}∩{w2,w3} = {w2,w3}. So the result is a failure of Inclusion. □
Proof
of Observation 10
The claim is trivial if A is not believed in (Pr,t). So suppose that A is believed in (Pr,t) in the first place, i.e., that Pr(UppA(Pr)) ≥ t, or equivalently, that \(i(t) < i(\overline {A})\) for the initial probability measure Pr.
First we note that in order to withdraw the belief A, we need to find an α such that \(\mathit {Bel}_{t}(Pr ^{\alpha }_{A})\) includes some \(\overline {A}\)-worlds. By coherence, then, it includes at least \(\overline {A}\text {-layer}_{i(\overline {A})}\).
Second, we convince ourselves that αi(t) is such a Jeffrey parameter α. Let \(Pr ^{\prime } = Pr ^{\alpha _{i(t)}}_{A}\). Clearly, UppA(Pr′) is A-spherei(t)− 1 = spherei− 1 (these are the spheres of Pr). Since i(t) < \(i(\overline {A})\), we know that αi(t) < Pr(A). So \(Pr ^{\prime }(A\text {-sphere}_{i(t)-1}) = \linebreak [4] \frac {\alpha _{i(t)}}{Pr (A)}\) ⋅ Pr(A-spherei(t)− 1) < Pr(A-spherei(t)− 1) < t. By the definition of αi(t), A-spherei(t)− 1 remains the i − 1st sphere of Pr′, and the ith layer of Pr′ is \(A\text {-layer}_{i(t)} \cup \overline {A}\text {-layer}_{i(\overline {A})}\). Since A-spherei(t) = A-spherei(t)− 1 ∪ A-layeri(t), we know that the smallest sphere of Pr′ with a probability of at least t includes A-spherei(t).
If αi(t) is the maximal Jeffrey parameter α that withdraws the belief A, then we have proved exactly what we needed to prove.
Suppose, however, that there is a larger α > αi(t) that withdraws A. Then the posterior probabilites of the elements of A-layeri(t) are \(\frac {\alpha }{Pr (A)}\cdot p_{i(t)} > \frac {\alpha _{i(t)}}{Pr (A)}\cdot p_{i(t)}\), while those of the elements of \(\overline {A}\text {-layer}_{i(\overline {A})}\) are \(\frac {1-\alpha }{Pr (\overline {A})}\cdot p_{i(\overline {A})} < \frac {1-\alpha _{i(t)}}{Pr (\overline {A})}\cdot p_{i(\overline {A})}\). Hence, by the definition of αi(t), the posterior probabilites of the elements of A-layeri(t) are higher than those of the elements of \(\overline {A}\text {-layer}_{i(\overline {A})}\). Since we noted above that \(\mathit {Bel}_{t}(Pr ^{\alpha }_{A})\) includes \(\overline {A}\text {-layer}_{i(\overline {A})}\), it follows by coherence, that \(\mathit {Bel}_{t}(Pr ^{\alpha }_{A})\) also includes A-layeri(t) and A-spherei(t)− 1, the elements of which have still higher probabilities. Since A-spherei(t) = A-spherei(t)− 1 ∪ A-layeri(t), we have proved exactly what we needed to prove. □
Proof
of Observation 11 For any arbitrary threshold t such that 0 < t < 1, we can use the following model with \(W=\{w_{1}, w_{2}, \dots , w_{n_{2}+1},\)\(w_{n_{2}+2},\dots , w_{2n_{2}+1}\}\):
where n2 = m3. So \(A=\{w_{1}, w_{2}, \dots , w_{n_{2}+1}\}\). Clearly, Pr(A) = p1 + n2p2 and \(Pr (\overline {A})=n_{2}p_{3}\). Let \(n_{2} = m_{3} = \text {floor}(\frac {2\cdot (1-t)}{3t})+1 > \frac {2\cdot (1-t)}{3t}\). Notice that this number depends on t. Then we assign the following probabilities
Since \(t = \frac {2\cdot (1-t)}{3\cdot {2\cdot (1-t)}/{3t}} > p_{2}\), one quickly finds that the assignments in (15) satisfy the requirement p1 > p2 > p3. Our proof strategy in choosing these assignments was to have p1 + n2p2 + n2p3 = 1. Clearly the belief set supported by (Pr,t) is Bel = Belt(Pr) = {w1}. We also get that \(Pr (A) = t + \frac {2}{3}\cdot (1-t) = \frac {1}{3}\cdot (2+t)\) and \(Pr (\overline {A}) = \frac {1}{3}\cdot (1-t)\).
The relationship between p1, p2 and p3 is made to be such that A gets lost as a belief exactly when \(Pr ^{\alpha }_{A}(w_{2}) = Pr ^{\alpha }_{A}(w_{n_{2}+2})\), i.e., when α is lowered to α2. This implies that as soon as A is improbable enough not to be believed, the probability of w2 by itself passes the threshold t, leaving w3 outside of the new belief set that we get when revising by A. Now we show that where α is the maximal Jeffrey parameter to remove A from the agent’s beliefs. Consider
As long as α > α2, \(\textit {UppA}(Pr ^{\alpha }_{A}) = \{w_{1}, w_{2}, \dots , w_{n_{2}+1}\} = A\) and \(Pr ^{\alpha }_{A}(\textit {UppA}(Pr ^{\alpha }_{A})) = \alpha > \frac {2+t}{4-t}\). But \(\frac {2+t}{4-t} > t\) for t between 0 and 1 (note that \(\frac {2+t}{4-t} > t\) transforms to (1 − t)(2 − t) > 0). So \(Pr (\textit {UppA}(Pr ^{\alpha }_{A})) > t\) which means that Bel = A as long as α2 < α < Pr(A).
A gets lost as a belief if α ≤ α2. To see this, note that \(\textit {UppA}(Pr ^{\alpha _{2}}_{A}) = \{w_{1}\}\) and
and this is less than t for t between 0 and 1 (note that \(\frac {3t}{4-t} < t\) transforms to t(t − 1) < 0). Thus the maximal Jeffrey parameter to eliminate A from the agent’s belief set is α2, and this means that the minimising equals W.
But then while Bel = {w1}. So the result is a failure of Recovery. □
Proof
of Observation 12 (i) For any arbitrary threshold t such that 0 < t < 1, we can use the following model with \(W=\{w_{1}, w_{2}, w_{3}, w_{4}, \dots , w_{n_{4}+3}\}\):
So \(A=\{w_{3}, w_{4}, \dots , w_{m_{4}+3}\}\). Let \(n_{4} = \text {floor}(\frac {2-t}{t})+1 > \frac {2-t}{t}\). Notice that n4 depends on t. (The number n4 of possible worlds at the fourth layer must be chosen greater than \(\frac {2-t}{t}\), in order to make sure that p4 < p3.) Then we assign the following probabilities
It is routine to check that the assignments in (16) satisfy the requirements that p1 > p2 > p3 > p4 and that p1 + p2 + p3 + n4p4 = 1. Notice also that with the probabilities of (16), Pr(A) = (1 − t)2 and \(Pr (\overline {A}) = t\cdot (2-t)\).
As our last bit of preparation, we verify that Bel = Belt(Pr) = {w1}. For this it is sufficient to show that
But (17) reduces to 2 − t(1 + t) ≥ 0 which is true for all t between 0 and 1.
Now we show that where α is the minimising Jeffrey parameter to remove \(\overline {A}\) from the agent’s beliefs, and at the same time \(\mathit {Bel}*A = \mathit {Bel}_{t}(Pr ^{\alpha }_{A}) = \{w_{3}, w_{4}, \dots , w_{n_{4}+3}\}\) where α is any Jeffrey parameter suitable for obtaining A as a belief.
We now turn to revisions. Given the definition of p3 it follows that the upper A-layer is never sufficient for accepting A, as the proportion of this layer within the proposition A is less than t: \(\frac {p_{3}}{Pr (A)} < t\). So in any revision by A, even for α = 1, the resulting belief set will include the worlds \(\{w_{4}, \ldots , w_{n_{4}+3}\}\) of the fourth layer.
Next contractions. We are assuming that when there is a minimising way of contracting \(\overline {A}\) (a maximal Jeffrey parameter yielding a contraction) this is the unique contraction method that will be used. We want to establish (a) that there exists a minimal way of contracting \(\overline {A}\), and (b) that the resulting belief set does not contain the worlds from the fourth layer. Once this is shown we will have a violation of the Levi identity.
We first establish that there is a minimal way of contracting \(\overline {A}\). We do this in two steps. First we show that as long as the world w2 of the second layer is more probable than the world w3 of the third layer the first and second layer jointly exceed the threshold (which means that \(\overline {A}\) is believed). That is, as long as \(\frac {\alpha }{Pr (A)}\cdot p_{3} < \frac {1-\alpha }{Pr (\overline {A})}\cdot p_{2}\), we have t ≤ 1 − α. Simplifying the first expression we thus need to show that as long as
we have t ≤ 1 − α. Now (18) is equivalent to
So it is enough to show that
After a few transformations, (20) reduces to t(3 − t) ≤ 2 which is quickly seen to hold for t between 0 and 1.
Next we show that as soon as the third layer becomes as probable as the second layer, \(\overline {A}\) is not believed any more; that is, as soon as \(\frac {\alpha }{Pr (A)}\cdot p_{3} = \frac {1-\alpha }{Pr (\overline {A})}\cdot p_{2}\) we have \(\frac {1-\alpha }{Pr (\overline {A})}\cdot p_{1} < t\) (when the first layer falls below the threshold we need more layers but now the worlds in the original layers 2 and 3 are equiprobable, which means we have to add A-worlds to the belief set to get above the threshold which means that \(\overline {A}\) is withdrawn). Simplifying the first expression we get
which is equivalent to
What we need to show is:
For this it is sufficient to show that
or equivalently \(\frac {1}{4+t(2-t)} \ < \ \frac {1}{4+t}\) which clearly holds for all t between 0 and 1.
Finally we show that when α is chosen to ensure a minimal contraction of \(\overline {A}\), the fourth layer is not in the resulting belief set. That is, when \(\frac {\alpha }{Pr (A)}\cdot p_{3} = \frac {1-\alpha }{Pr (\overline {A})}\cdot p_{2}\), the sum of the three top layers (that is, \(\frac {1-\alpha }{Pr (\overline {A})}\cdot p_{1} + \frac {1-\alpha }{Pr (\overline {A})}\cdot p_{2} + \frac {\alpha }{Pr (A)}\cdot p_{3} = (1-\alpha ) + \frac {\alpha }{Pr (A)}\cdot p_{3}\)) is greater or equal to t. Simplifying the expressions we need to show that
which reduces to
Now we can make use of (22) and see that it is sufficient to show that
After a few transformations, (27) reduces to t(5 − t) ≤ 4 which is easily recognised to be true for all t between 0 and 1.
(ii) To see that the Harper identity fails we use the same model as above, with the roles of A and \(\overline {A}\) reversed. Then we have Bel = {w1}⊆ A, and \(\mathit {Bel}*\overline {A} = \{w_{3}, w_{4},\ldots , w_{n_{4}+3}\}\). Thus neither nor , and the Harper identity is doubly violated. □
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Cantwell, J., Rott, H. Probability, coherent belief and coherent belief changes. Ann Math Artif Intell 87, 259–291 (2019). https://doi.org/10.1007/s10472-019-09649-3
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DOI: https://doi.org/10.1007/s10472-019-09649-3