Abstract
A new formal model of belief dynamics is proposed, in which the epistemic agent has both probabilistic beliefs and full beliefs. The agent has full belief in a proposition if and only if she considers the probability that it is false to be so close to zero that she chooses to disregard that probability. She treats such a proposition as having the probability 1, but, importantly, she is still willing and able to revise that probability assignment if she receives information that gives her sufficient reasons to do so. Such a proposition is (presently) undoubted, but not undoubtable (incorrigible). In the formal model it is assigned a probability 1 − δ, where δ is an infinitesimal number. The proposed model employs probabilistic belief states that contain several underlying probability functions representing alternative probabilistic states of the world. Furthermore, a distinction is made between update and revision, in the same way as in the literature on (dichotomous) belief change. The formal properties of the model are investigated, including properties relevant for learning from experience. The set of propositions whose probabilities are infinitesimally close to 1 forms a (logically closed) belief set. Operations that change the probabilistic belief state give rise to changes in this belief set, which have much in common with traditional operations of belief change.
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Acknowledgements
Open access funding provided by Royal Institute of Technology. I would like to thank Branden Fitelson, Wlodek Rabinowicz, and the other participants at the workshop Cognitive values: Theoretical and Practical Problems at the Royal Swedish Academy of Letters, History and Antiquities in Stockholm on June 10-11, 2019, for most valuable comments. I am also grateful to an anonymous referee for unusually helpful proposals for improvements of the text.
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Appendices
Appendix: A: A Brief Introduction to Hyperreal Numbers
Readers wanting an introduction to the arithmetic of hyperreals and infinitesimals are referred to Keisler [25].
The number system of hyperreal numbers (R∗) is an extension of the set R of real numbers. It consists of both finite and infinite numbers, but our focus will be on the finite numbers. The hyperreal finite numbers consist of (1) the real numbers and (2) other numbers on an extended number line, which are posited between the real numbers. The positive hyperreal numbers that are infinitely close to 0 (larger than 0 but smaller than all positive real numbers) are the positive infinitesimals. The negative infinitesimals are the numbers that are smaller than 0 but larger than all negative real numbers.
Two hyperreal numbers s and t are infinitesimally close (denoted s ≈ t) if and only if their difference s − t is infinitesimal. The relation ≈ is an equivalence relation. Each finite hyperreal number is infinitely close to exactly one real number, which is called its standard part. The standard part of the number t is denoted st(t). Standard parts satisfy the following rules.
st(−s) = −st(s)
st(s + t) = st(s) + st(t)
st(s − t) = st(s) −st(t)
st(s × t) = st(s) ×st(t)
If st(t)≠ 0 then st(s/t) = st(s)/st(t)
The set of hyperreal numbers satisfies the same algebraic laws as the real numbers. Let δ and 𝜖 be infinitesimals, and let let s and t be finite numbers that are not infinitesimals. Then:
The following are infinitesimals: δ + 𝜖, δ𝜖, sδ and δ/s.
The following are finite and not infinitesimal: s + 𝜖, st and s/t.
The following is finite and may or may not be infinitesimal: s + t
The following may be infinite or finite, and in the latter case it may or may not be infinitesimal: δ/𝜖.
Appendix: B: Proofs
Proof of Observation 1
If \(\mathfrak {p}(a)=0\) or \(\mathfrak {p}(a)=1\), then all three properties follow directly from clause (0) of Definition 5. If \(0\neq \mathfrak {p}(a)\neq 1\), then:
Part 1: In clause (1) of the definition, rewrite the right-hand side as follows:
Since δ, 1 − δ, \({\mathfrak {p}(a_{1}\&b)}/{\mathfrak {p}(a_{1})}\), and \({\mathfrak {p}(\neg a_{1}\&b)}/{\mathfrak {p}(\neg a_{1})}\) are all non-negative, so is the whole expression. The non-negativity of the right-hand side of clause (3) follows in the same way.
Part 2: Clause (3) of Definition 5 yields:
Part 3: Using clause (3) of Definition 5, we obtain:
□
Proof of Observation 2
Part 1: This follows directly if \(\mathfrak {p}(e_{1}\vee \dots \vee e_{n})=0\). If \(\mathfrak {p}(e_{1}\vee \dots \vee e_{n})\neq 0\), then apply clause (3) of Definition 5. We obtain:
Part 2:
□
Proof of Observation 3
Part 1: If \(\mathfrak {p}(d)=0\) or \(\mathfrak {p}(d)=1\), then the desired result follows directly from clause (0) of Definition 5. If \(0\neq \mathfrak {p}(d)\neq 1\), then we use clause (3) and obtain:
Part 2: If \(\mathfrak {p}(d)=0\) or \(\mathfrak {p}(d)=1\), then the desired result follows directly from clause (0) of Definition 5. If \(0\neq \mathfrak {p}(d)\neq 1\), then we use clause (3) and obtain:
□
Proof of Observation 4
Parts 1, 5, and 9: Let B = {b1, b2}, \(\mathfrak {p}(a\mid b_{1})=1\), \(\mathfrak {p}(a\mid b_{2})=0\), and \(\mathfrak {p}(b_{1})= \mathfrak {p}(a)\). Setting d = a and δ = 0, we obtain from clause (3) of Definition 5:
Parts 2 and 11: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(a)\), \(\mathfrak {p}(b_{2})=1-\mathfrak {p}(a)\), \(\mathfrak {p}(a\mid b_{1})=1\), \(\mathfrak {p}(a\mid b_{2})=0\), and \(\mathfrak {p}(d\& b_{1})=\mathfrak {p}(d\& b_{2})\neq 0\). It follows from clause (3) of Definition 5 that:
Parts 3 and 7: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(a)\), \(\mathfrak {p}(b_{2})=1-\mathfrak {p}(a)\), \(\mathfrak {p}(a\mid b_{1})=1\), and \(\mathfrak {p}(a\mid b_{2})=0\). Let \(0\lessapprox \delta \) and d = ¬a. It follows from clause (3) of Definition 5 that:
Parts 4, 8, and 12: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(a)\), \(\mathfrak {p}(b_{2})=1-\mathfrak {p}(a)\), \(\mathfrak {p}(a\mid b_{1})=1\), \(\mathfrak {p}(a\mid b_{2})=0\). Let d = ¬a and δ = 0. It follows from clause (3) of Definition 5 that:
Parts 6 and 10: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(a)\), \(\mathfrak {p}(b_{2})=1-\mathfrak {p}(a)\), \(\mathfrak {p}(a\mid b_{1})=1\), and \(\mathfrak {p}(a\mid b_{2})=0\). Let d = a and let \(0\lessapprox \delta \). Then:
□
Lemma 1
Let x1,…, xn and y1,…, yn be positive numbers. Then:
Proof of Lemma 1
We first prove the case when n = 2:
Next, we assume that this holds for n. We then have:
□
Proof of Observation 5
Part 1: If \(\mathfrak {p}(a)=0\) or \(\mathfrak {p}(a)=1\), then \(\mathfrak {p}*a=\mathfrak {p}\) and consequently \(\pmb {(}\mathfrak {p}*a\pmb {)}(a)=\mathfrak {p}(a)\). If \(0\neq \mathfrak {p}(a)\neq 1\), then Lemma 1 yields:
Part 2: Let \(\pmb {(}\mathfrak {p}*_{\delta } d\pmb {)}(a)\not \approx \mathfrak {p}(a)\). First we note that if \(\mathfrak {p}(a\mid b)\approx \mathfrak {p}(a)\) for all b ∈ B, then \(\pmb {(}\mathfrak {p}*_{\delta } d\pmb {)}(a)\approx \mathfrak {p}(a)\). Thus there are elements b of B such that \(\mathfrak {p}(a\mid b)\not \approx \mathfrak {p}(a)\). There are three options:
Case (i): For all b with \(\mathfrak {p}(a)\ll \mathfrak {p}(a\mid b)\) we have:
Case (ii): For all b with \(\mathfrak {p}(a)\approx \mathfrak {p}(a\mid b)\) we similarly obtain \(\mathfrak {p}(b)\approx \pmb {(}\mathfrak {p}*_{\delta } a\pmb {)}(b)\). Case (iii): For all b with \(\mathfrak {p}(a\mid b)\ll \mathfrak {p}(a)\) we similarly obtain \(\pmb {(}\mathfrak {p}*_{\delta } a\pmb {)}(b)\ll \mathfrak {p}(b)\). It now follows from clause (3) of Definition 5 that \(\mathfrak {p}(a)\ll \pmb {(}\mathfrak {p}*_{\delta } a\pmb {)}(a)\). □
Proof of Observation 6
Part 1: Let B = {b1, b2}, let a, d ∈ E and d ⊩ a, and let \(\mathfrak {p}\) satisfy the following conditions: \(\mathfrak {p}(b_{1})=0.5\), \(\mathfrak {p}(b_{1}\&a)=0.4\), \(\mathfrak {p}(b_{1}\&a\&d)=0.1\), \(\mathfrak {p}(b_{2}\&a)=0.3\), and \(\mathfrak {p}(b_{2}\&a\&d)=0.2\). Then \(\mathfrak {p}(a)=0.7\), and:
Part 2: Let B = {b1, b2}, let a, d ∈ E, and let \(\mathfrak {p}\) satisfy the following conditions: \(\mathfrak {p}(b_{1})= 0.5\), \(\mathfrak {p}(a\&b_{1})=0.375\), \(\mathfrak {p}(d\&b_{1})=0.5\), \(\mathfrak {p}(a\&b_{2})=0.125\), and \(\mathfrak {p}(d\&b_{2})=0\). Then:
and:
□
Proof of Observation 7
If \(\mathfrak {p}(a)=1\), then \(\mathfrak {p}*_{\delta } a=\mathfrak {p}\), and \(\pmb {(}\mathfrak {p}*_{\delta } a\pmb {)}(\neg a)=\mathfrak {p}(\neg a)=0\), and the desired result follows directly. The case when \(\mathfrak {p}(a)=0\) can be proved analogously. If \(0\neq \mathfrak {p}(a)\neq 1\), then:
□
Proof of Theorem 1
Part 1: By induction. Initial step:
Induction step: We assume that
We write \(\mathfrak {p}_{n-1} = \mathfrak {p}*_{\delta _{1}}a_{1}*_{\delta _{2}}\ldots *_{\delta _{n-1}}a_{n-1}\) and \(\mathfrak {p}_{n} = \mathfrak {p}*_{\delta _{1}}a_{1}*_{\delta _{2}}\ldots *_{\delta _{n}}a_{n}\). Then:
With the induction hypothesis, this yields:
Part 2:
□
Proof of Observation 8
which converges to 1 with increasing n. □
Proof of Observation 9
The following example serves to prove both parts of the observation: Let B = {b1, b2, b3}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(b_{2})= 0.5-\epsilon \), \(\mathfrak {p}(b_{3})=2\epsilon \), \(\mathfrak {p}(a_{1}\mid b_{1})=1\), \(\mathfrak {p}(a_{1}\mid b_{2})=0\), \(\mathfrak {p}(a_{1}\mid b_{3})=1\), \(\mathfrak {p}(a_{2}\mid b_{1})=0\), \(\mathfrak {p}(a_{2}\mid b_{2})=1\), and \(\mathfrak {p}(a_{2}\mid b_{3})=1\). We then have:
It follows that:
For an example, let δ1 = δ2 = 𝜖. It then follows that:
It follows that \(\pmb {(}\mathfrak {p}*_{\delta _{1}}a_{1}*_{\delta _{2}}a_{2}\pmb {)}(a_{1})\approx 0.8\) and \(\pmb {(}\mathfrak {p}*_{\delta _{2}}a_{2}*_{\delta _{1}}a_{1}\pmb {)}(a_{1})\approx 1\). □
Proof of Observation 10
It is sufficient to prove that (1) If \(a\in [{}[ \mathfrak {p}]{}]\) and a ⊩ d, then \(d\in [{}[ \mathfrak {p}]{}]\), and (2) If \(a_{1}\in [{}[ \mathfrak {p}]{}]\) and \(a_{2}\in [{}[ \mathfrak {p}]{}]\), then \(a_{1}\&a_{2}\in [{}[ \mathfrak {p}]{}]\).
For (1), let \(a\in [{}[ \mathfrak {p}]{}]\). If a ⊩ d, then d is equivalent with a ∨ (d&¬a), and it follows from the third Kolmogorov axiom that \(\mathfrak {p}(a)\leq \mathfrak {p}(a\vee (d\&\neg a))=\mathfrak {p}(d)\), hence \(\mathsf {st}(\mathfrak {p}(d))=1\).
For (2), let \(\mathfrak {p}(a_{1})=1-\delta _{1}\) and \(\mathfrak {p}(a_{2})=1-\delta _{2}\). Then \(\mathfrak {p}(\neg a_{1})= \delta _{1}\) and \(\mathfrak {p}(\neg a_{2})= \delta _{2}\), and:
□
Proof of Observation 11
Part 1: From Observations 1 and 10.
Part 2: We will prove this part in its reverse form:
\(d\notin \text {Cn}([{}[\mathfrak {p}]{}] \cup \{a\})\)
\(a\to d \notin [{}[\mathfrak {p}]{}]\)
\(\mathfrak {p}(a\to d) \not \approx 1\)
\(\mathfrak {p}(a\&\neg d) \not \approx 0\)
For some \(b^{\prime }\in B\): \(\mathfrak {p}(a\&\neg d \& b ') \not \approx 0\)
\(\mathfrak {p}(a\& b ') \not \approx 0, \mathfrak {p}(\neg d \& b ') \not \approx 0,\mathfrak {p}(a) \not \approx 0, \text { and } \mathfrak {p}(b ') \not \approx 0.\)
\(\frac {\mathfrak {p}(a\&b ')\times \mathfrak {p}(\neg d \& b ')}{\mathfrak {p}(a)\times \mathfrak {p}(b^{\prime })} \not \approx 0\)
\(\sum \limits _{b\in B}\frac {\mathfrak {p}(a\&b )\times \mathfrak {p}(\neg d \& b)}{\mathfrak {p}(a)\times \mathfrak {p}(b)} \not \approx 0\)
\(\pmb {(}\mathfrak {p}*_{\delta } a\pmb {)}(\neg d) \not \approx 0\)
\(\pmb {(}\mathfrak {p}*_{\delta } a\pmb {)}(d) \not \approx 1\)
\(d\notin [{}[\mathfrak {p}*_{\delta } a]{}]\)
Part 3: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(b_{2})=0.5, \mathfrak {p}(a\mid b_{1})=1\), and \(\mathfrak {p}(a\mid b_{2})=0.5\). Then \(\mathfrak {p}(\neg a)=0.25\), thus \(\neg a\notin [{}[\mathfrak {p}]{}]\). Furthermore:
Thus, \(a\notin [{}[\mathfrak {p}*_{\delta } a]{}]\).
Part 4: Let 0 ≪ r ≪ 1 and let B be such that \(\mathfrak {p}(a\mid b)\leq r\) for all b ∈ B. It follows that \(\pmb {(}\mathfrak {p}*_{\delta } d\pmb {)}(a)\leq r\) for all d and δ.
Parts 5-6: These results follow directly from the definitions. □
Proof of Observation 12
Part 1: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=1-\epsilon \), \(\mathfrak {p}(b_{2})=\epsilon \), \(\mathfrak {p}(a\mid b_{1})= \mathfrak {p}(c\mid b_{1})= 0\), \(\mathfrak {p}(a\mid b_{2})= 0.5\), and \(\mathfrak {p}(c\mid b_{2})= 1\).
We then have \( a \notin [{}[\mathfrak {p}*_{\delta } a]{}]\) and \(c\notin [{}[\mathfrak {p}]{}]\). However:
and thus \(c\in [{}[\mathfrak {p}*_{\delta } a]{}]\).
Part 2: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(b_{2})=\mathfrak {p}(a\mid b_{1})=0.5\), \(\mathfrak {p}(d\mid b_{1})=0\), and \(\mathfrak {p}(a \mid b_{2})= \mathfrak {p}(d\mid b_{2})=1\). It follows that for all δ:
□
Proof of Observation 13
Let \(\neg a\notin [{}[ \mathfrak {p}]{}]\) and \(d \in [{}[ \mathfrak {p}]{}]\). We are going to show that \(d\in [{}[ \mathfrak {p}*_{\delta } a]{}]\).
First step: First, we are going to show that for all \(b^{\prime }\in B\), if \(\mathfrak {p}(b^{\prime }\mid a)\not \approx 0\), then \(\mathfrak {p}(b^{\prime })\not \approx 0\). Let \(\mathfrak {p}(b^{\prime }\mid a)\not \approx 0\), i.e. \({\mathfrak {p}(a\&b^{\prime })}/{\mathfrak {p}(a)}\not \approx 0\). It follows from \(\neg a\notin [{}[ \mathfrak {p}]{}]\) that \(\mathfrak {p}(a)\not \approx 0\), thus \(\mathfrak {p}(a\&b^{\prime })\not \approx 0\), thus \(\mathfrak {p}(b^{\prime })\not \approx 0\).
Second step: We are going to show that for all \(b^{\prime }\in B\), if \(\mathfrak {p}(b^{\prime })\not \approx 0\), then \(\mathfrak {p}(d\mid b^{\prime })\approx 1\). Let \(\mathfrak {p}(b^{\prime })\not \approx 0\). It follows from \(d \in [{}[ \mathfrak {p}]{}]\) that \({\sum }_{b\in B}(\mathfrak {p}(b)\times \mathfrak {p}(d\mid b))\approx 1\). We also have \({\sum }_{b\in B}\mathfrak {p}(b)=1\) and \(\mathfrak {p}(d\mid b)\leq 1\) for all b ∈ B, from which the desired conclusion follows.
Third step:
□
Proof of Observation 14
Part 1: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(b_{2})=0.5\), \(\mathfrak {p}(a_{1}\&a_{2}\&d\mid b_{1})=1\), and \(\mathfrak {p}(a_{1}\neg \&a_{2}\&\neg d\mid b_{2})=\mathfrak {p}(\neg a_{1}\& a_{2}\&\neg d\mid b_{2})=0.5\). Then:
Thus, \(d\in [{}[\mathfrak {p}*_{\delta }(a_{1}\&a_{2})]{}]\). Furthermore:
Thus, \(a_{2}\to d\notin [{}[\mathfrak {p}*_{\delta ^{\prime }} a_{1}]{}]\) and consequently \(d\notin \text {Cn}([{}[\mathfrak {p}*_{\delta ^{\prime }} a_{1}]{}]\cup \{a_{2}\})\).
Part 2: Let B = {b1, b2}, \(\mathfrak {p}(b_{1})=\mathfrak {p}(b_{2})=0.5\) and \(\mathfrak {p}(a_{1}\&a_{2}\& b_{1})=\mathfrak {p}(\neg a_{1}\&\neg a_{2}\& b_{1})=\mathfrak {p}(a_{1}\&a_{2}\& b_{2})=\mathfrak {p}(\neg a_{1}\&\neg a_{2}\& b_{2})=0.25\). Then:
and:
Thus \(\neg a_{2}\notin [{}[\mathfrak {p}*_{\delta }a_{1}]{}]\) and \(a_{2}\notin [{}[\mathfrak {p}*_{\delta ^{\prime }}(a_{1}\&a_{2})]{}]\). □
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Hansson, S.O. Revising Probabilities and Full Beliefs. J Philos Logic 49, 1005–1039 (2020). https://doi.org/10.1007/s10992-020-09545-w
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DOI: https://doi.org/10.1007/s10992-020-09545-w