Abstract
One of central problems in the theory of conditionals is the construction of a probability space, where conditionals can be interpreted as events and assigned probabilities. The problem has been given a technical formulation by van Fraassen (23), who also discussed in great detail the solution in the form of Stalnaker Bernoulli spaces. These spaces are very complex – they have the cardinality of the continuum, even if the language is finite. A natural question is, therefore, whether a technically simpler (in particular finite) partial construction can be given. In the paper we provide a new solution to the problem. We show how to construct a finite probability space \(\mathrm {S}^\#=\left(\mathrm\Omega^\#,\mathrm\Sigma^\#,\mathrm P^\#\right)\) in which simple conditionals and their Boolean combinations can be interpreted. The structure is minimal in terms of cardinality within a certain, naturally defined class of models – an interesting side-effect is an estimate of the number of non-equivalent propositions in the conditional language. We demand that the structure satisfy certain natural assumptions concerning the logic and semantics of conditionals and also that it satisfy PCCP. The construction can be easily iterated, producing interpretations for conditionals of arbitrary complexity.
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If \(\mathrm r\leq\mathrm k+1\) then \(\mathrm X\cap\mathrm Y=\varnothing\), and \(\mathrm P^\#\left(X\rightarrow Y\right)=\mathrm P\left(\mathrm Y\left|\mathrm X\right.\right)=0\) , which completes the proof.
This procedure of reducing the number of worlds can be continued, but the details are not relevant for the general claim.
References
Adams, E. W. (1965). On the logic of conditionals. Inquiry, 8, 166–197. https://doi.org/10.1080/00201746508601430
Adams, E. W. (1970). Subjunctive and indicative conditionals. Foundations of Language, 6, 89–94. https://doi.org/10.2307/2272204
Adams, E. W. (1975). The Logic of Conditionals. D. Reidel.
Adams, E. W. (1998). A Primer of Probability Logic. CLSI, Stanford University.
Egré, P., & Rott, H. (2021).The Logic of Conditionals. In E. N. Zalta (Ed.), The Stanford Encyclopedia of Philosophy (Winter 2021 ed.). URL = https://plato.stanford.edu/archives/win2021/entries/logic-conditionals/
Hájek, A. (1989). Probabilities of conditionals—revisited. Journal of Philosophical Logic, 18, 423–428.
Hájek, A. (2011). Triviality pursuit. Topoi, 30(1), 3–15. https://doi.org/10.1007/s11245-010-9083-2
Hájek, A. (2012). The fall of “Adams’ Thesis”? Journal of Logic, Language and Information, 21, 145–161. https://doi.org/10.1007/s10849-012-9157-1
Hájek, A., & Hall, N. (1994). The Hypothesis of the Conditional Construal of Conditional Probability. In E. Eells & B. Skyrms (Eds.), Probabilities and Conditionals: Belief Revision and Rational Decision (pp. 75–110). Cambridge University Press.
Hall, N. (1994). Back in the CCCP. In E. Eells & B. Skyrms (Eds.), Probabilities and Conditionals: Belief Revision and Rational Decision (pp. 141–160). Cambridge University Press.
Kaufmann, S. (2004). Conditioning against the grain: Abduction and indicative conditionals. Journal of Philosophical Logic, 33(6), 583–606.
Kaufmann, S. (2005). Conditional predictions: A probabilistic account. Linguistics and Philosophy, 28(2), 181–231.
Kaufmann, S. (2009). Conditionals right and left: Probabilities for the whole family. Journal of Philosophical Logic, 38, 1–53.
Kaufmann, S. (2015). Conditionals, conditional probability, and conditionalization. In H. Zeevat & H. C. Schmitz (Eds.), Bayesian Natural Language Semantics and Pragmatics (pp. 71–94). Springer.
Khoo, J. (2016). Probabilities of conditionals in context. Linguistics and Philosophy, 39, 1–43.
Khoo, J., & Santorio, P. (2018). Lecture Notes: Probability of Conditionals in Modal Semantics. http://paolosantorio.net/teaching.html
Lewis, D. (1976). Probabilities of conditionals and conditional probabilities. Philosophical Review, 85, 297–315.
Lewis, D. (1986). Philosophical Papers (Vol. II). Oxford: Oxford University Press.
McGee, V. (1989). Conditional probabilities and compounds of conditionals. Philosophical Review, XCVIII(4), 485–541. https://doi.org/10.2307/2185116
Stalnaker, R. (1968). A theory of conditionals. Studies in Logical Theory, American Philosophical Quarterly, 2, 98–112.
Stalnaker, R. (2009). Conditional Propositions and Conditional Assertions. In A. Egan & B. Weatherson (Eds.), Epistemic Modality (pp. 227–248). Oxford University Press. https://doi.org/10.1093/acprof:oso/9780199591596.003.0008
Stalnaker, R., & Jeffrey, R. (1994). Conditionals as Random Variables. In E. Eells & B. Skyrms (Eds.), Probabilities and Conditionals: Belief Revision and Rational Decision (pp. 31–46). Cambridge University Press.
Van Fraassen, B. C. (1976). Probabilities of conditionals. In W. L. Harper, R. Stalnaker, & G. Pearce (Eds.), Foundations of Probability Theory, Statistical Inference, and Statistical Theories of Science (pp. 261–308). D. Reidel.
Węgrecki, J., & Wroński, L. (2022). Revisiting the conditional construal of conditional probability. Logic and Logical Philosophy. https://doi.org/10.12775/LLP.2022.024
Wójtowicz, A., & Wójtowicz, K. (2021a). A stochastic graphs semantics for conditionals. Erkenntnis, 86, 1071–1105. https://doi.org/10.1007/s10670-019-00144-z
Wójtowicz, A., & Wójtowicz, K. (2021b). Dutch Book against Lewis. Synthese, 199, 9185–9217. https://link.springer.com/article/10.1007/s11229-021-03199-0
Wójtowicz, K., & Wójtowicz, A. (2022). A graph model for probabilities of nested conditionals. Linguistics and Philosophy, 45(3), 1–48. https://doi.org/10.1007/s10988-021-09324-z
Acknowledgements
For helpful discussions and critical comments on the probabilistic model, we would like to thank the audiences in MCMP in Munich, Madrid, Warsaw and Dubrovnik.
The preparation of this paper was supported by the National Science Centre (Narodowe Centrum Nauki) grant 2020/39/B/HS1/01866.
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The preparation of this paper was supported by the National Science Centre (Narodowe Centrum Nauki) grant 2020/39/B/HS1/01866.
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Appendices
Appendix 1: Proof of PCCP
In this section, we present a proof of PCCP. Performing a direct computation of P#([X → Y]#) is a rather tedious task, so we present an inductive proof. The proof proceeds by induction on the size of the sample space S (i.e., on n).
Base Case
Take n = 1. We need to prove that in this space, for any conditional X → Y we have \(\mathrm {P}^\#(\mathrm{X}\rightarrow \mathrm{Y})=\mathrm {P}\left(\left.\mathrm{Y}\right|\mathrm{X}\right).\)
In this space, we have only two conditionals, i.e. Ω → Ω and Ω → ∅.
In the case of Ω → Ω, obviously, \(\mathrm P^\#\left(\mathrm\Omega\hspace{0.17em}\rightarrow\hspace{0.17em}\mathrm\Omega\right)=1=\mathrm P\left(\left.\mathrm\Omega\right|\mathrm\Omega\right).\)
In the case of Ω → ∅, we know that [Ω → ∅]# = ∅, which means that P#(Ω → ∅) = 0. But this is exactly \(\mathrm {P}\left(\left.\varnothing\right|\mathrm\Omega\right).\)
Induction Step
In the induction step we assume that in sample spaces of size n, PCCP holds. This means that if Ω = {1,2,…,n}, then the probability \(\mathrm P^\#\left(X\rightarrow Y\right)=\mathrm P\left(\left.\mathrm Y\right|\mathrm X\right)\). In other words, the probability that in a permutation ω the first X is a Y, is \(\mathrm {P}\left(\left.\mathrm {Y}\right|\mathrm {X}\right).\)
We need to prove that the same applies to spaces of size n + 1. This means that if Ω = {1,2,…,n,n + 1}, and X → Y is a conditional with X,Y ⊆ {1,2,…,n,n + 1}, then \(\mathrm {P}^\#\left(X\hspace{0.17em}\rightarrow\hspace{0.17em}Y\right)=\mathrm {P}\left(\left.\mathrm {Y}\right|\mathrm {X}\right)\). To be exact, this means that if Ω = {1,2,…,n,n + 1}, the probability that in a permutation ω the first X is a Y, is \(\mathrm {P}\left(\left.\mathrm {Y}\right|\mathrm {X}\right).\)
Let pi = P(i) for i = 1,…,n + 1. To simplify the notation and to avoid double indices, assume that X = {k + 1,…,n + 1} and Y = {m,…,r}, with k + 1 ≤ r.Footnote 1
This means that X ∩ Y = {k + 1,…,r} and X ∩ Yc = {r + 1,…,n + 1}. Let:
Our Ω# is S(n + 1), i.e.,the group of permutations of {1,2,…,n, n + 1}. We shall divide Ω# into two disjoint sets, depending on whether the given permutation begins with an element of X or not:
Informally, this means that if ω ∈ WX, then already the first move decides whether X → Y is true or not.
Obviously:
WX and Wnon-X are disjoint, so SX ∩ Snon-X = ∅. This means that:
Computing P#(SX) is easy. Indeed:
This is the set of permutations which begin with an element of X ∩ Y. Therefore:
That was the easy part, now we need to compute \(\mathrm P^\#({\mathrm S}_{\mathrm{non}-\mathrm X})\).
\({\mathrm {S}}_{\mathrm{non}-\mathrm {X}}=\left[X\rightarrow Y\right]^\#\cap{\mathrm W}_{\mathrm{non}-\mathrm X}=\{{\mathrm\omega\hspace{0.17em}\in\hspace{0.17em}\mathrm\Omega}^\#:\;\mathrm\omega(1)\hspace{0.17em}\not\in\hspace{0.17em}\mathrm X\) and \(\left.\mathrm{\omega}\hspace{.4em}\right\vert\hspace{-.4em}=\hspace{0.17em}X\rightarrow Y\}\). So Snon-X consists of permutations which satisfy two conditions:
-
(i)
they begin with an element of Xc = {1,2,…,k};
-
(ii)
they satisfy X → Y, i.e., the first element of ω which is X is also Y.
Any permutation ω ∈ Wnon-X begins with one of the numbers from {1,2,…,k}. We can therefore define a partition of Wnon-X into k disjoint sets W1,…, Wk, depending on the first element, i.e.:
This means that:
and this is a union of pairwise disjoint sets. Therefore,
\(\mathrm {W}^{\mathrm {j}}\cap{\lbrack X\rightarrow Y\rbrack}^\#\), for j = 1,…k consist of the permutations \(\mathrm\omega\in\mathrm {S}(\mathrm {n+1})\) such that \(\mathrm\omega(1)=\mathrm j\) and the first element of ω which is X, is also Y.
Consider \(\mathrm {W}^\mathrm{1}\cap{\lbrack X\rightarrow Y\rbrack}^\#\). Obviously:
We want to compute \({\mathrm{P}}^{\#}\left({\left[X\hspace{0.17em}\to \hspace{0.17em}Y\right]}^{\#}{\left|\mathrm{W}\right.}^1 \right)\), i.e., the probability that the first element of the permutation ω (of length n + 1) which has property X also has property Y – conditioned on W1, i.e., on the assumption that ω(1) = 1.
This means that we want to compute the probability that we generate a permutation from [X → Y]#, knowing that ω(1) = 1 (i.e., that it begins with 1). But that is just to say that we have to generate an appropriate tail of length n of ω, i.e., a permutation of length n satisfying the condition that the first X to appear is a Y. We have to generate this permutation using the numbers from the set {2,3,…n,n + 1} with their probabilities rescaled (as we have already “used up” object 1). The probability distribution P* on the n-element set {2,3,…n,n + 1} is:
By construction:
This is the right moment to make use of the induction hypothesis. We know that for any probability space S* = (Ω*, Σ*, P*) of size n, the probability that a permutation ω fulfills X → Y, i.e., that the first element of ω with property X also has property Y, is \(\mathrm P^\ast\left(\left.\text{Y}\right|\text{X}\right)\). We apply the induction hypothesis to the tails of the permutations, and see that:
The same argumentation applies to W2,..,Wk, i.e.,
Finally:
But this means, that:
which completes the proof.
Appendix 2: Proof of the Minimality Result
Consider the sample space consisting of n worlds 1,2,3…n. By 1,2,3,…,n we shall denote the corresponding propositions of the language L( →). Let SPACE be a probability space in which L( →) is interpreted. Throughout the Appendix we assume that SPACE satisfies van Fraassen’s conditions. The probabilities of 1,2,3,…,n in the initial sample space S are P(k) = pk, for k = 1,…,n.
In order to prove the minimality of the space, we need to prove two relevant facts from Section 5.
2.1 Fact 1. No World W ∈ SPACE can Satisfy Two Different Formulas φσ and φσ*
Observation
Observe that if k ≠ m, then the interpretations of the propositions k and m within SPACE are distinct, i.e. \(\mathrm {P}(\mathit k\wedge\hspace{-.1em}\mathit m)\) = 0. But this means in particular, that if P(A) > 0, then P(A → (\(\mathit k\wedge\mathit m\)) = 0 i.e. A → \((\mathit k\wedge\mathit m)\) cannot be satisfied in any world. In a still different formulation: if for some world \(\textbf W,\textbf W\vert=A\rightarrow(\mathrm k\wedge\mathrm m)\), then k = m.
Consider two permutations of {1,2…,n}: σ = k1k2…kn and σ* = m1m2…mn, and the corresponding formulas φσ and φσ*. Our task is to show, that if in some world W ∈ SPACE both formulas φσ and φσ* are true, then σ = σ*, i.e. φσ = φσ*. In other words:
Assume that \(\textbf {W}\vert={{\varphi}}_{{\sigma}}\) and \(\textbf {W}\vert={{\varphi}}_{{\sigma}^{\ast}}\). As k1 is a conjunct in φσ and m1 is a conjunct in φσ*, this means that \(\textbf W\vert={\mathrm {k}}_1\) and \(\textbf {W}\vert={\mathrm {m}}_1\) i.e. \(\textbf W\vert={\mathrm {k}}_1\wedge{\mathrm {m}}_1.\)
But this is only possible if \({\mathrm k}_1={\mathrm m}_1.\)
Similarly, ¬k1 → k2 is a conjunct in φσ and ¬m1 → m2 is a conjunct in φσ*. This means, that \(\textbf W|=\neg{\mathit k}_1\rightarrow{\mathit k}_2\) and \(\textbf W|=\neg{\mathrm m}_1\rightarrow{\mathrm m}_2\) i.e. \(\textbf W|=(\neg{\mathit k}_1\rightarrow{\mathit k}_2)\wedge(\neg{\mathit m}_1\rightarrow{\mathit m}_2).\)
We know already that \({\mathrm k}_1={\mathrm m}_1.\) Now we make use of van Fraassen’s condition (I), i.e. ((A → C)∧(A → B)) ⇔ (A → (C∧B)). It implies that (¬k1 → k2)∧(¬m1 → m2) ⇔ (¬k1 → (k2∧m2)). But, according to the Observation, this means that \({\mathrm k}_2={\mathrm m}_2.\)
By analogous reasoning, considering the remaining conjuncts of φσ and φσ* we obtain ki = mi for i = 1,…,n. But this means exactly that φσ = φσ*.
We have therefore proved, that any world W \(\in\) SPACE is capable of satisfying at most one formula of the form φσ.
2.2 Fact 2. Every Formula φω ∈ L( →) has an interpretation in SPACE (provided SPACE has IND(n-3))
For simplicity of notation, we shall consider the permutation ω = 123…n, and the corresponding formula φ123…n. We will show, that in any space SPACE fulfilling van Fraassen’s condition and IND(n-3), it has an interpretation in SPACE. In fact, we will prove a stronger claim, and show that the probability of φ123…n within SPACE is
2.2.1 Toy Example
In order to present the idea of the proof, we start with the toy example taking n = 4. Here we shall only need IND(1), i.e. P(C∧(A1 → B1)) = P(C)⋅P(A1 → B1) (with C and A1 exclusive).
Consider the formula φ1234:
The last conjunct is a tautology (there are four worlds 1,2,3,4), so we can skip it i..e. \({\mathrm\varphi}_{1234}=1\wedge(\neg\mathit 1\rightarrow2)\wedge((\neg\mathit 1\wedge\neg\mathit 2)\rightarrow\mathit 3\). We shall show that
Consider the formula \({\alpha }=(\neg \mathit{1}\to \mathit{2})\wedge (( \neg \mathit{1}\wedge \neg \mathit{2})\to \mathit{3})\), i.e. the tail of φ1234. It is a general fact that \({\alpha }\iff ((\mathit{1}\wedge {\alpha }) \vee (\neg \mathit{1}\wedge {\alpha }))\) – and the disjunction is disjoint. This means that:
By IND(1), we have P(1∧α) = P(1)⋅ P(α), which means that
We need to compute \(\mathrm P(\neg\mathit{1}\wedge\mathrm\alpha)\; i.\,\mathrm e.\;\mathrm P((\neg\mathit{1}\wedge(\neg\mathit{1}\rightarrow\mathit{2})\wedge((\neg\mathit{1}\wedge\neg\mathit{2})\rightarrow\mathit{3}))).\)
By van Fraassen’s condition (III), i.e. \((A\wedge (A\to B)) \iff (A\wedge B)\) we have \((\neg \mathit{1}\wedge (\neg \mathit{1}\to \mathit{2})) \iff (\neg \mathit{1}\wedge \mathit{2})\), which is equivalent to 2. This means that
By IND(1):
By PCCP, \(\mathrm P((\neg\mathit{1}\wedge\neg\mathit{2})\rightarrow\mathit{3})=\frac{p_3}{1-\left(p_1+p_2\right)}\), which means, that
We have the equation for P(α):
After solving it, we obtain:
But \({\mathrm\varphi}_{1234}=\mathit{1}\wedge\mathrm{\alpha}\), which means that
2.2.2 The Proof in the General Case
This is an inductive proof. Consider ω = 123…n, and the corresponding formula φ123…n i.e.
We shall examine the tails of φ123…n. Let Tk be the fragment of φ123…n (the tail):
By definition, \({\mathrm T}_{\mathrm k-1}=((\neg{\mathit 1}\wedge\neg{\mathit 2}\wedge\dots\wedge\neg(\mathit {k-1}))\rightarrow\mathit k)\wedge{\mathrm T}_{\mathrm k}\). We also see that:
which means that T0 = φ123…n.
Our claim is that
which – in particular – implies that
The proof goes by backward induction from k = n-1 to k = 0.
BASE CASE. k = n-1
\({\mathrm T}_{\mathrm n-1}=((\neg\mathit 1\wedge\neg\mathit 2\wedge\dots\wedge\neg(\mathrm n-1))\rightarrow\mathrm n)\), so by PCCP.
\(\mathrm P\left({\mathrm T}_{\mathrm n-1}\right)=\frac{p_n}{1-(p_1+p_2+\dots+p_{n-1})}=1\), as required.
INDUCTION STEP
Assume that for a certain k we already have:
We need to show that
i.e. that
We shall use a similar procedure as the one used in the toy example. It is a general fact that for any formula γ:
\({\mathrm T}_{\mathrm k-1}\Leftrightarrow\lbrack(\mathrm\gamma\wedge{\mathrm T}_{\mathrm k-1})\vee(\neg\mathrm\gamma\wedge{\mathrm T}_{\mathrm k-1})\rbrack\), and this disjunction is disjoint. This means that
We take \(\upgamma =\neg \mathit 1\wedge \neg \mathit 2\wedge \dots \wedge \neg ({k}-1)\) and compute both terms of this sum separately.
The term P(γ∧ T k-1 )
The tail Tk-1 is
This means that
By principle (III), i.e. \((A\wedge (A\to B)) \iff (A\wedge B))\), we have:
which is equivalent to k. This means that
We apply IND: ⋅
The term P(¬γ∧ T k-1 )
\(\mathrm P(\neg\mathrm\gamma\wedge{\mathit T}_{\mathit {k-1}})=\mathrm P(\neg(\neg\mathit 1\wedge\neg\mathit 2\wedge\dots\wedge\neg(\mathit {k-1}))\wedge{\mathit T}_{\mathit {k-1}})\). By IND, this is equal to:
By de Morgan law, \(\mathrm P(\neg(\neg\mathit 1\wedge\neg\mathit 2\wedge\dots\wedge\neg(\mathit {k-1}))=\mathrm P(\mathit 1\vee\mathit {2}\vee\mathit 3\dots\vee(\mathit {k-1}))={\mathrm p}_1+\dots{\mathrm p}_{\mathit {k-1}}\), which means that
We have computed both needed terms \(\mathit P\left(\mathrm\gamma\wedge{\mathit T}_{\mathit {k-1}}\right)\) and \(\mathit P\left(\neg\mathrm\gamma\wedge{\mathit T}_{\mathit {k-1}}\right)\) and obtained the equation
After solving it, we have
as was needed in the inductive step.
2.3 IND(n-3) is an Essential Assumption
We shall show that IND is an essential assumption, i.e. without it we are able to construct a van Fraassen-like space of size less that n! This is not possible for n = 3, as PCCP is sufficient to generate 6 worlds. But already for n = 4, we shall construct a space which satisfies van Fraassen’s conditions but which has less than 24 elements. This is obtained by violating IND(1).
We start with S(4), i.e. with the set of permutations of length 4, but we shall eliminate some of them. Intuitively, some of the permutations will absorb other permutations – so the structure will be “less fine-grained” but still fine enough to conform to PCCP. Intuitively, PCCP is too weak to “provide insight” into the fine-grained structure of the worlds.
This is the set of 24 permutations (worlds):
1-worlds | 2-worlds | 3-worlds | 4-worlds |
---|---|---|---|
1234 | 2134 | 3124 | 4123 |
1243 | 2143 | 3142 | 4132 |
1324 | 2314 | 3214 | 4213 |
1342 | 2341 | 3241 | 4231 |
1423 | 2413 | 3412 | 4312 |
1432 | 2431 | 3421 | 4321 |
Initially, each world has probability 1/24. Observe that within the 1-worlds we can distinguish three parts, corresponding to \(\mathit 1\wedge \left(\neg\mathit 1\to\mathit 2\right), \mathit 1\wedge \left(\neg\mathit 1\to \mathit 3\right)\) and \(\mathit 1\wedge \left(\neg\mathit 1\to\mathit 4\right)\). Indeed:
The same observation applies to the 2-words, 3-words and 4-worlds.
We shall make the necessary rearrangements within the 1-worlds and 2-worlds. Intuitively, some worlds absorb other worlds (and their probabilities), and – in parallel – another compensating absorption occurs, so that PCCP is preserved.
Let the world 1243 “absorb” the world 1234, so that:
In parallel the world 2134 absorbs the world 2143, i.e.
After the absorptions the probabilities of the formulas ¬1 → 2 and (¬1∧¬2) → 3 are preserved. This is the new set of worlds after the absorption:
1-worlds | 2-worlds | 3-worlds | 4-worlds |
---|---|---|---|
1243 | 2134 | 3124 | 4123 |
3142 | 4132 | ||
1324 | 2314 | 3214 | 4213 |
1342 | 2341 | 3241 | 4231 |
1423 | 2413 | 3412 | 4312 |
1432 | 2431 | 3421 | 4321 |
The probabilities within 1-world and the 2-world have been transferred in such a way, that \(\mathrm P\left(\neg\mathit 1\rightarrow\mathit 2\right)\) and \(\mathrm P\left(\left(\neg\mathit 1\wedge\neg\mathit 2\right)\rightarrow\mathit 3\right)\) have not been affected. What is different is that now \(\mathrm P\left(1\wedge\left(\neg\mathit 1\rightarrow\mathit 2\right)\wedge\left(\left(\neg\mathit 1\wedge\neg\mathit 2\right)\rightarrow\mathit 3)\right)\right)=0\) – but this fact cannot be prohibited just by PCCP alone.
We have produced a space with 22 worlds.Footnote 2 This was possible as IND(1) is violated.
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Wójtowicz, A., Wójtowicz, K. A Minimal Probability Space for Conditionals. J Philos Logic 52, 1385–1415 (2023). https://doi.org/10.1007/s10992-023-09710-x
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DOI: https://doi.org/10.1007/s10992-023-09710-x