Abstract
This paper calls for a re-appraisal of McGee’s analysis of the semantics, logic and probabilities of indicative conditionals presented in his 1989 paper Conditional probabilities and compounds of conditionals. The probabilistic measures introduced by McGee are given a new axiomatisation—built on the principle that the antecedent of a conditional is probabilistically independent of the conditional—and a more transparent method of constructing such measures is provided. McGee’s Dutch book argument is restructured to more clearly reveal that it introduces a novel contribution to the epistemology of semantic indeterminacy, and shows that its more controversial implications are unavoidable if we want to maintain the Ramsey Test along with the standard laws of probability. Importantly, it is shown that the counterexamples that have been levelled at McGee’s analysis—generating a rather wide consensus that it yields ‘unintuitive’ or ‘wrong’ probabilities for compounds —fail to strike at their intended target; for to honour the intuitions of the counterexamples one must either give up the Ramsey Test or the standard laws of probability. It will be argued that we need to give up neither if we take the counterexamples as further evidence that the indicative conditional sometimes allows for a non-epistemic ‘causal’ interpretation alongside its usual epistemic interpretation.
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Appendix: Proof of Observations and Theorems
Appendix: Proof of Observations and Theorems
Proof Proof of Observation 2:
The proof uses notions from Section ??. Consider a model where W contains four worlds, labelled 1, 2, 3, and 4. Let m be some probability mass on W that assigns non-zero probabilities to each world, and \(\hat {m}\) be the McGee-mass generated by m. Consider some 𝜖 > 0 such that \(\hat {m}(s) > \epsilon \) and \(\hat {m}(s) + \epsilon \leq 1\) for all sequences s. Let \(\hat {m}^{\prime }\) be a probability mass on sequences that is just like \(\hat {m}\) except for the following four reassignments:
The net change is 0, so \({\sum }_{f\in \mathcal {F}} \hat {m}^{\prime }(f) = 1\).
Assume a language where 1, 2, 3 and 4 serve as factual sentences true at (and only at) the corresponding world. A sequence like 134 will be understood as the disjunction 1 ∨3 ∨4.
The first reassignment has the effect of adding probability 𝜖 to the following base conditionals (and any base conditional entailed by them):
Combining all three in a conjunction we get a sentence that is true only at the sequence 〈1,3,2,4〉. The second reassignment has the effect of subtracting 𝜖 from the probability of the base conditionals:
So, the net effect of the two first reassignments leaves the probability of \(\mathbf {1234}\rightarrow \mathbf {1}\) and \(\mathbf {234}\rightarrow \mathbf {3}\) unchanged, while the probability of \( \mathbf {24}\rightarrow \mathbf {2}\) has been raised by 𝜖 and the probability of \( \mathbf {24}\rightarrow \mathbf {4} \) has been lowered by 𝜖.
The third reassignment has the effect of lowering the following by 𝜖:
The fourth reassignment has the effect of adding 𝜖 to the following:
So, the net effect of the third and fourth reassignments is that the probability of \( \mathbf {24}\rightarrow \mathbf {2}\) has been lowered by 𝜖 while the probability of \( \mathbf {24}\rightarrow \mathbf {4} \) has been raised by 𝜖. In effect: undoing the changes (as regards base conditionals) of the first two reassignments.
So, taken together the four reassignments will not affect the probability of any base conditional. Moreover, the new measure is regular. As \(\textup {Pr}_{\hat {m}}\) satisfies rRT, \(\textup {Pr}_{m^{\prime }}\) will satisfy rRT. But the probability of certain compounds of base conditionals have changed. For instance, the probability of
has changed; for this conjunction is true only at the sequence 〈3,1,2,4〉 to which we have added probability 𝜖.
Given that GI uniquely determines the probabilities of all complex sentences on the basis of the factual probabilities, and that our original measure satisfied GI, we can thus conclude that the new measure does not. But one can also see this directly. For the probability of the conjunction \(\mathbf {124}\rightarrow \mathbf {1} \wedge \mathbf {24}\rightarrow \mathbf {2}\) is unchanged by the reassignments (it is true at 〈1,3,2,4〉 to which we have added 𝜖 and at 〈3,1,2,4〉 from which we have subtracted 𝜖). By GI
But \(\textup {Pr}_{\hat {m}}(\mathbf {124}\rightarrow \mathbf {1} \wedge \mathbf {24}\rightarrow \mathbf {2}) = \textup {Pr}_{\hat {m}^{\prime }}(\mathbf {124}\rightarrow \mathbf {1} \wedge \mathbf {24}\rightarrow \mathbf {2})\) and \(\textup {Pr}_{\hat {m}}(\mathbf {3}) = \textup {Pr}_{\hat {m}^{\prime }}(\mathbf {3})\). Still \(\textup {Pr}_{\hat {m}}(\mathbf {3}\wedge \mathbf {124}\rightarrow \mathbf {1} \wedge \mathbf {24}\rightarrow \mathbf {2}) \neq \textup {Pr}_{\hat {m}^{\prime }}(\mathbf {3}\wedge \mathbf {124}\rightarrow \mathbf {1} \wedge \mathbf {24}\rightarrow \mathbf {2})\). So:
Thus GI is violated. Our measure \(\textup {Pr}_{\hat {m}^{\prime }}\) satisfies rRT but not GI. □
Proof Proof of Theorem 2:
Take any regular McGee-measure Pr. Consider the following consequence of GI:
Here each ai is an atomic sentence or a negated atomic sentence. Note that as Pr is regular we do not need to assume that Pr(A ∧ Bi) > 0. For if Pr(A ∧ Bi) = 0, then by regularity \(\textup {Pr}({(A\wedge B_{i})\rightarrow a_{i}}) = 1\) and so the i-th term could without any effect be removed from the conjunction (if Pr(A ∧ Bi) = 0 for all i then both sides of the identity are equal to Pr(¬A)).
From (A) we can establish:
For given regularity \(\textup {Pr}({\neg ((A\wedge C_{i})\rightarrow \perp )})\) is either 0 or 1. To see this assume that:
Then \(1-\textup {Pr}({(A\wedge C_{i})\rightarrow \perp }) > 0\) and so \(1 > \textup {Pr}({(A\wedge C_{i})\rightarrow \perp })\), By regularity it follows that Pr(A ∧ Ci) > 0 and so by rRT \(\textup {Pr}({(A\wedge C_{i})\rightarrow \perp }) = 0\). So \(\textup {Pr}({\neg ((A\wedge C_{i})\rightarrow \perp )}) = 1\).
From (B) it follows by basic probabilistic principles (i.e. if Pr(¬A ∧ ψ) = Pr(¬A) ×Pr(ψ), then Pr(A ∧ ψ) = Pr(A) ×Pr(ψ)) that:
One can show (see [3]) that any sentence of the form \( A\rightarrow \psi \) is logically equivalent to a disjunction \(D(A\rightarrow \psi ) = \delta _{1}\vee {\cdots } \vee \delta _{n}\) where the disjuncts are mutually logically incompatible and each δj has the form
Together with claim (C) this ensures that for any sentence \( A\rightarrow \psi \) (where ψ need not be factual):
And this is IA.
For the other direction assume that Pr is a regular ICL-measure satisfying IA. By IA (omitting the subscript 1 ≤ j ≤ n for \(\bigvee \), and 1 ≤ i ≤ n for \(\bigwedge \)):
Note (letting ⇔ denote logical equivalence):
So:
And so:
This implies GI. □
Proof Proof of Theorem 3:
Let m be a probability mass on worlds. Let s be some non-empty sequence with domain D(s) of cardinality k. Define an intermediary notion:
Let s − w be the sequence we get when we remove the world w from s. Trivially (when D(s) contains more than one world):
As before where P is a set of sequences let \(m^{*}(P) = {\sum }_{s\in P} m^{*}(s)\), and let P − w be the set of sequences we get when we remove w from each sequence in P (if P = {〈w〉}, then P − w = ∅). Let P[w] be the set of sequences in P that have w as its first element.
When P is a set of sequences, let P∘ be the set of sequences in P that have domain D(m). Our target construction \(\hat {m}\) is then related to m∗ by the identity:
(m∗ differs from \(\hat {m}\) in that it will assign a non-zero weight also to a sequence that has a strict subset of D(m) as its domain, and as a result \(m^{*}(\mathcal {F})\) can be greater than 1).
When s = 〈w1,…,wk〉 is a sequence such that w∉D(s) then the sequence 〈w1,…,wi,w,wi+ 1,…wk〉, for any i such that 0 ≤ i ≤ k + 1 is a w-variant of s (we get a w-variant by inserting w somewhere in the sequence). Let I(s,w) denote the set of w-variants of s. We can now prove the core lemma.
Lemma 1
If m(w) > 0 and w∉D(s), then m∗(I(s,w)) = m∗(s).
Proof
Assume that m(w) > 0. When m∗(s) = 0 it trivially follows that m∗(I(s,w)) = 0. So assume that m∗(s) > 0. Proof by induction over the cardinality of D(s). For the base case the cardinality is 1, so \(D(s) = \{w^{\prime }\}\) for some \(w^{\prime }\). There are two w-variants of s: \(\langle w, w^{\prime }\rangle\) and \(\langle w^{\prime }, w\rangle\). Trivially m∗(s) = 1. Moreover \(m^{*}(\{\langle w, w^{\prime }\rangle, \langle w^{\prime }, w\rangle\}) = m(w)/(m(w)+ m(w^{\prime })) + m(w^{\prime })/(m(w)+m(w^{\prime })) = 1\).
For the induction hypothesis assume that the claim holds for any s with domain of cardinality k. Consider some sequence s = 〈w1,…,wk+ 1〉 that has a domain of cardinality k + 1. Assume that w∉D(s). Let \(s^{\prime } = \langle w_{2}, \ldots , w_{k+1}\rangle\). \(D(s^{\prime })\) is of cardinality k so the claim holds for \(s^{\prime }\): \(m^{*}(I(s^{\prime }, w)) = m^{*}(s^{\prime })\). We can get nearly all the elements of I(s,w) by concatenating w1 as an initial world to each element \(s^{\prime \prime }\) of \(I(s^{\prime }, w)\), (in symbols \(\langle w_{1};s^{\prime \prime }\rangle\)). The only element of I(s,w) that we will miss in this way will be the sequence in which w is the initial element, that is, the sequence 〈w;s〉.
So
Focus first on the first term on the right-hand side of (A). We have:
where p = m(D(s)) + m(w) = the probability of the domain of the sequences involved. By the induction hypothesis:
Note that
(w is not in the domain of s, thereby the term p − m(w)). So \(m(w_{1})\times m^{*}(s^{\prime }) = (p-m(w))\times m^{*}(s)\). So
The right-hand side will be our new first term for (A).
Now take the second term on the right-hand side of (A). We have:
So, combining the two new right-hand terms for (A):
As \(\frac {1}{p}\times (p-m(w) ) + \frac {1}{p} \times m(w) = \frac {1}{p}(p-m(w) +m(w)) = 1\):
□
A set of sequences P is closed under w-variants if for any s ∈ P such that s≠〈w〉: \(I(s-w, w)\subseteq P\). Intuitively, a proposition is closed under w-variants if the placement of w in an ordering is irrelevant for the truth of the proposition.
Corollary 1
If P is closed under w-variants and every s ∈ P has a domain of at least two elements, then m∗(P) = m∗(P − w).
Proof
Assume that P is closed under w-variants. As each sequence has a domain of at least two elements P − w will be a set of sequences. As P is closed under w-variants: for each s ∈ P − w, \(I(s, w)\subseteq P\). Indeed {I(s,w) : s ∈ P − w} makes up a partition of P. So \(m^{*}(P) = {\sum }_{s\in P-w} m^{*}(I(s,w))\). By Lemma 1, m∗(s) = m∗(I(s,w)) for each s ∈ P − w. So \(m^{*}(P-w) = {\sum }_{s\in P-w} m^{*}(I(s,w))\). So m∗(P) = m∗(P − w). □
Corollary 2
-
1.
\(\hat {m}(\mathcal {F}) = 1\). (\(\hat {m}\) is a probability mass on sequences)
-
2.
\(\hat {m}([A]) = m(F(A))\). (\(\hat {m}\) extends m)
-
3.
\(\hat {m}([\neg A \wedge (A\rightarrow \varphi )]) = \hat {m}([\neg A])\times \hat {m}([A\rightarrow \varphi ])\) (\(\hat {m}\) satisfies IA).
Proof
-
(1)
First prove that if \(X\subseteq D(m)\) and P is the set of all sequences with domain X, then m∗(P) = 1. By induction over the cardinality of X. The case when X is a singleton {w} is trivial. So assume that the claim holds for cardinality k. Show that it holds for X of cardinality k + 1. Let P be the set of all sequences with domain X. Take any w ∈ X. Note that P − w is the set of all sequences with domain X −{w}. By the induction hypothesis m∗(P − w) = 1. By Corollary 2, m∗(P − w) = m∗(P). So m∗(P) = 1.
\(\mathcal {F}\) is the set of all sequences and \(\mathcal {F}^{\circ }\) is the set of all sequences with domain D(m). So \(m^{*}(\mathcal {F}^{\circ }) = \hat {m}(\mathcal {F}) = 1\).
-
(2)
Take any w ∈ F(A). If w∉D(m) then [A][w]∘ is empty and m∗([A][w]∘) = m(w) = 0 (where, recall, P[w] is the set of sequences of P that begin with w). So assume that m(w) > 0. Note that [A][w]∘− w is the set of all sequences that have domain D(m) −{w}. By construction m∗([A][w]∘) = (m(w)/m(D(m)))×m∗([A][w]−w) = m(w)×m∗([A][w]∘−w) (recall: m(D(m)) = 1). But by (1) of this Corollary, m∗([A][w]∘− w) = 1. So m∗([A][w]∘) = m(w). This holds for all w ∈ F(A) so \(m^{*}([A]^{\circ }) = \hat {m}([A]) = m(F(A))\).
-
(3)
Note that \([A\rightarrow \varphi]\) is closed under w-variants for each w ∈ F(¬A). Consider \(P = [A\rightarrow \varphi]^{\circ }\). P has domain D(m) and will be closed under w-variants for each w ∈ F(¬A) such that m(w) > 0.
First show that m∗(P[w]) = m(w) × m∗(P) for each w ∈ F(¬A). Assume that w ∈ F(¬A). If D(m) = {w} or if m(w) = 0 the claim follows immediately. So assume that D(m)≠{w} and that m(w) > 0. By construction m∗(P[w]) = m(w) × m∗(P − w). By Corollary 2 m∗(P) = m∗(P − w), so m∗(P[w])) = m(w) × m∗(P).
This holds for all w ∈ F(¬A) and so m∗([¬A] ∩ P) = m(F(¬A)) × m∗(P).
By construction \(m^{*}(P) = \hat {m}([A\rightarrow \varphi])\) and \(m^{*}([\neg A]\cap P) = \hat {m}([\neg A\wedge (A\rightarrow \varphi )])\). Moreover, \(m(F(\neg A)) = \hat {m}([\neg A])\). So:
□
It follows immediately that \(\textup {Pr}_{\hat {m}}\) is a regular McGee-measure. Uniqueness follows from McGee’s results. □
Proof Proof of Theorem 4:
Assume a measure m such that m(X) > 0. Assume a language where each set of worlds X has a corresponding atomic sentence X such that F(X) = X. Given such a language one can for every sequence s construct a sentence s such that [s] = s. Let PrX be the measure \(\textup {Pr}_{X}(\varphi ) = \textup {Pr}_{\hat {m}}(\mathbf {X}\rightarrow \varphi )\). As McGee demonstrated PrX thus defined is a McGee-measure. By rRT we know that for any factual A, \(\textup {Pr}_{X}(A) = \textup {Pr}_{\hat {m}}(A\wedge \mathbf {X)}/\textup {Pr}_{\hat {m}}(\mathbf {X})\). So the probability mass on worlds that would generate PrX is mX. So \(\textup {Pr}_{X} = \textup {Pr}_{\hat {m}_{X}}\) and so \(\textup {Pr}_{X}(\varphi ) = \hat {m}_{X}([\varphi])\).
Take any sequence s. We know that, for any \(s^{\prime }\) such that \(D(s^{\prime })\cap X\neq \emptyset \), \(s^{\prime }\in [\mathbf {X}\rightarrow \mathbf {s}]\) iff \(s^{\prime }/X = s\). So
\(\textup {Pr}_{X}(\mathbf {s}) = \hat {m}_{X}(s)\). Moreover, \(\textup {Pr}_{X}(\mathbf {s}) = \hat {m}([\mathbf {X}\rightarrow \mathbf {s}])\), so \(\hat {m}_{X}(s) = \displaystyle \sum \limits _{s^{\prime }/X =s}\hat {m}(s^{\prime })\). □
Proof Proof of Theorem 5:
(1) First a useful definition and an important property. Where f is a selection function and w is a world define fw as follows (for any set of worlds X):
fw is ‘just like’ f except that it has w as its preferred world. Note that, trivially, if P is determinately true at w, then fw ∈ P, and if P is determinately false at w then fw∉P.
Lemma 2
If P is a simple proposition that has an indeterminate truth value at w, then fw ∈ P iff f ∈ P.
Proof
Let P be a simple proposition that has an indeterminate truth value at w. Take any f. Consider the case when P is positive, so P = X > Y for some X and Y. If w ∈ X, then, contrary to assumption, X > Y has a determinate truth value at w (it is determinately true at w if w ∈ Y and determinately false at w if w∉Y). So w∉X. Note that as D(f) −{w} = D(fw) −{w}, D(f) ∩ X = ∅ iff D(fw) ∩ X = ∅ (as w∉X). So if D(f) ∩ X = ∅, then f ∈ X > Y iff fw ∈ X > Y. So assume that D(f) ∩ X≠∅. As w∉X, by construction, fw(X) = f(X). So fw(X) ∈ Y iff f(X) ∈ Y, and so fw ∈ X > Y iff f ∈ X > Y.
Consider the case when P is negative, so P = −(X > Y ) for some X and Y. If − (X > Y ) has an indeterminate truth value at w, then so does X > Y (= −P). By the above fw ∈−P iff f ∈−P, so fw ∈ P iff f ∈ P. □
Now, assume that \(C(\mathcal {K}) = C(\mathcal {K}^{\prime })\). Assume (i) that \(f\in {\mathord{SIC}} _{w}(\mathcal {K})\) but for reductio (ii) that \(f\not \in {\mathord{SIC}} _{w}(\mathcal {K}^{\prime })\).
From (ii) it follows that for every \(K^{\prime }\in \mathcal {K}^{\prime }\) we have \(f\not \in {\mathord{SIC}} _{w}(K^{\prime })\). Take any \(K^{\prime }\in \mathcal {K}^{\prime }\). \(\bigcap K^{\prime }\) cannot be determinately true at w, for then \({\mathord{SIC}} _{w}(K^{\prime }) = \mathcal {F}\) and so would contain f contrary to assumption. So we have two alternatives. Either \(\bigcap K^{\prime }\) is determinately false in w, and then, trivially, \(f^{w}\not \in \bigcap K^{\prime }\). Or \(\bigcap K^{\prime }\) has indeterminate truth value at w. Then there is some simple proposition \(P\in K^{\prime }\) such that P has indeterminate truth value at w and f∉P. By Lemma 2, fw∉P. Accordingly, \(f^{w}\not \in \bigcap K^{\prime }\). In either case: \(f^{w}\not \in \bigcap K^{\prime }\). This holds for all \(K^{\prime }\in \mathcal {K}^{\prime }\), so \(f^{w}\not \in C(\mathcal {K}^{\prime })\).
We know from (i) that for some \(K\in \mathcal {K}\), f ∈SICw(K). It follows that \(\bigcap K\) is not determinately false at w and so that no P ∈ K is determinately false in w. So take any P ∈ K. Either P has indeterminate truth value at w, in which case f ∈ P. By Lemma 2, fw ∈ P. Or P is determinately true at w. Then, trivially, fw ∈ P. In either case fw ∈ P. This holds for all P ∈ K, so \(f^{w}\in \bigcap K\). But then \(f^{w}\in C(\mathcal {K})\), but then \(C(\mathcal {K}) \neq C(\mathcal {K}^{\prime })\) and we have a contradiction.
-
(2)
Suppose that \(C(\mathcal {K})\) has indeterminate truth value at w. So there are selection functions f and \(f^{\prime }\) such that (i) \(w_{f} = w_{f^{\prime }} = w\), (ii) \(f\in C(\mathcal {K})\) and (iii) \(f^{\prime }\not \in C(\mathcal {K})\).
By (ii) there is some \(K\in \mathcal {K}\) such that \(f\in \bigcap K\). So \(\bigcap K\) is not determinately false at w (and \(\bigcap K\) can’t be determinately true at w for then \(C(\mathcal {K})\) would be determinately true at w). As \(\bigcap K\) has indeterminate truth value at w and \(f\in \bigcap K\), f ∈SICw(K) and so \(f\in {\mathord{SIC}} _{w}(\mathcal {K})\). So \({\mathord{SIC}} _{w}(\mathcal {K})\) is not determinately false at w.
By (iii) for all \(K\in \mathcal {K}\), \(f^{\prime }\not \in \bigcap K\). Suppose, for reductio, that there is some \(K\in \mathcal {K}\) such that \(f^{\prime }\in {\mathord{SIC}} _{w}(K)\). It follows that \(\bigcap K\) is not determinately false at w. So \(\bigcap K = Q\cap {\mathord{SIC}} _{w}(K)\), where Q is some proposition that is determinately true at w. But as \(w_{f^{\prime }} =w\), Q contains \(f^{\prime }\). So then \(f^{\prime }\in \bigcap K\), contrary to assumption. So \(f^{\prime }\not \in {\mathord{SIC}} _{w}(K)\). This holds for all \(K\in \mathcal {K}\) so \(f^{\prime }\not \in {\mathord{SIC}} _{w}(\mathcal {K})\) and so \({\mathord{SIC}} _{w}(\mathcal {K})\) is not determinately true at w. □
Proof Proof of Theorem 6:
2-to-1. Assume that Pr is coherent and that the class of fair bets for Pr satisfies the minimal requirements, bet-additivity and the weak cancellation condition. By the minimal requirements Pr(φ) = 1, for any ICL-validity φ (as any ICL-valid sentence φ is always determinately true and a bet on φ will give pay-off $1). Likewise, by bet-additivity Pr(φ ∨ ψ) = Pr(φ) + Pr(ψ), when ¬(φ ∧ ψ) is ICL-valid.
So Pr is an ICL-measure.
Consider a sentence \(\varphi = A_{1}\rightarrow B_{1}\wedge {\cdots } \wedge A_{n}\rightarrow B_{n}\). Assume that no \(A_{i}\rightarrow B_{i}\) is a logical truth or a logical contradiction (logical contradictions will have probability 0; if there are one or more logical truths in a conjunction they can be removed while retaining the same probability). Consider a sentence C that is incompatible with each Ai. Take any world w in which C is true. Take a fair bet b1 = ($ Pr(φ),p1) on φ. We will have p1(w) = $ Pr(φ) due to the weak cancellation condition. Take fair bets b = ($ Pr(C ∧ φ),p2) and b3 = ($ Pr(¬C ∧ φ),p3) on C ∧ φ and ¬C ∧ φ respectively. Due to bet additivity we have p1(w) = p2(w) + p3(w). That is, $ Pr(φ) = p2(w) + p3(w). But ¬C ∧ φ is determinately false at w so p3(w) = 0. So p2(w) = $ Pr(φ). This will hold for all w where C is true. So the expectation value of b2 is $ Pr(C) ×Pr(φ). But the cost of b2 is $ Pr(C ∧ φ). So Pr(C ∧ φ) = Pr(C) ×Pr(φ).
So we have GI.
(1-to-2) As McGee has provided a proof based on closely related assumptions only the finite case will be treated and only for regular measures. So assume, for simplicity, a finite model, and to avoid technicalities assume that the McGee-measure is defined on propositions, that propositions are sets of sequences, and that bets are made on propositions. Furthermore, let D(Pr) be the set of worlds w such that \(\textup {Pr}(\mathcal {P}(w))> 0\).
Where P is a proposition, let pP be the pay-off function:
(This is the Partial Compensation condition!) Let bP = ($ Pr(P),pP). We need to show that this is a fair standard bet on P.
Take any s = 〈w1,…,wn〉. We then have the unit proposition {s}. Let (recall: > is our conditional operator on sets of worlds)
\(\{s\} = \mathcal {P}(w_{1}) \cap P_{s} \cap Q_{s}\cap R_{s}\). Note that, due to Regularity, Pr(Qs) = 0 if there is some w ∈ D(Pr) such that w∉D(s). Meanwhile if there is some w ∈ D(s) such that w∉D(Pr), then Pr(Rs) = 0. So if D(Pr)≠D(s), then Pr({s}) = 0. Meanwhile, if D(s) = D(Pr), then Pr(Qs) = Pr(Rs) = 1. So if D(s) = D(Pr), then \(\textup {Pr}(\{s\}) = \textup {Pr}(\mathcal {P}(w_{1})\cap P_{s}) =\) (by GI) \(\textup {Pr}(\mathcal {P}(w_{1}))\times \textup {Pr}({P_{s}})\). (which is also the cost of the bet b{s}).
Note that Ps ∩ Qs ∩ Rs is the strongly indeterminate content of {s} at w1 (our \({\mathord{SIC}} _{w_{1}}(\{s\})\)), while at all other worlds w, SICw({s}) = ∅. So the expected pay-off for b{s} (when Pr({s}) > 0) is \({\sum }_{w\in W} \textup {Pr}(\mathcal {P}(w))\times \textup {Pr}({\mathord{SIC}} _{w}(\{s\}) = \textup {Pr}(\mathcal {P}(w_{1}))\times \textup {Pr}({P_{s}}) = \textup {Pr}(\{s\})\) which is also the cost of the bet and so b{s} is a fair standard bet.
This holds for unit propositions. We need to show that it holds for all propositions.
Lemma 3
For any propositional decompositions \(\mathcal {K}\) and \(\mathcal {K}^{\prime }\), if \(C(\mathcal {K}) \cap C(\mathcal {K}^{\prime }) = \emptyset \), then
-
1.
\({\mathord{SIC}} _{w}(\mathcal {K}) \cap {\mathord{SIC}} _{w}(\mathcal {K}^{\prime }) = \emptyset \).
-
2.
\({\mathord{SIC}} _{w}(\mathcal {K})\cup {\mathord{SIC}} _{w}(\mathcal {K}^{\prime }) = {\mathord{SIC}} _{w}(\mathcal {K}\cup \mathcal {K})\).
Proof
-
(1)
Assume that there is a \(K\in \mathcal {K}\) and a \(K^{\prime }\in \mathcal {K}^{\prime }\) and some f such that \(f\in {\mathord{SIC}} _{w}(K)\cap {\mathord{SIC}} _{w}(K^{\prime })\). It follows that no \(P\in K\cup K^{\prime }\) is determinately false in w. But then by Lemma 2 \(f^{w}\in \bigcap K \cap \bigcap K^{\prime }\). But then \(f^{w} \in C(\mathcal {K}) \cap C(\mathcal {K}^{\prime })\) which gives us a contradiction.
-
(2)
Follows trivially from the construction.
□
So, as the strongly indeterminate content is decomposition invariant, if P ∩ Q = ∅, then SICw(P) ∩SICw(Q) = ∅ and SICw(P ∪ Q) =SICw(P) ∪SICw(Q). So \({\mathord{SIC}} _{w}(P) = \bigcup _{s\in P} {\mathord{SIC}} _{w}(\{s\})\). When \(s\neq s^{\prime }\), \({\mathord{SIC}} _{w}(\{s\})\cap {\mathord{SIC}} _{w}(\{s^{\prime }\}) = \emptyset \). So by the additivity of Pr:
The pay-off for a bet on P in the event that w will be \(p_{P}(w) = \textup {Pr}({\mathord{SIC}} _{w}(P)) = {\sum }_{s\in P} \textup {Pr}({\mathord{SIC}} _{w}(\{s\})\), so the expected pay-off will be
So bP will be a fair standard bet for all P, and the set of fair standard bets will be additive. Moreover, the minimal requirements will hold (for if P is determinately true at w then \({\mathord{SIC}} _{w}(P) = \mathcal {F}\), and if P is determinately false at w then SICw(P) = ∅). Finally, the Weak Cancellation condition will hold trivially. □
Proof Proof of Theorem 7:
The proof was—effectively—given in the proof of Theorem 6. For it was there shown that the class of fair standard bets on φ for a McGee-measure satisfies partial compensation and the latter entails the minimal requirements, bet-additivity and the partial compensation condition. □
Proof Proof of Observation 4:
Assume a measure Pr that is coherent according to the settlement conditions. From bet-additivity: \(\textup {Pr}(A\rightarrow B) = \textup {Pr}(A\wedge (A\rightarrow B)) + \textup {Pr}(\neg A\wedge (A\rightarrow B))\). Given rRT, \(\textup {Pr}(A\rightarrow B) = \textup {Pr}(A\wedge B)/\textup {Pr}(A)\). Given the minimal requirements \(\textup {Pr}(A\wedge A\rightarrow B) = \textup {Pr}(A\wedge B)\), so it follows that \(\textup {Pr}(\neg A\wedge (A\rightarrow B)) = \textup {Pr}(\neg A)\times \textup {Pr}(A\rightarrow B)\).
Now assume factual sentences EF,E1,…,En that partition the worlds so that any fair bet (c,p) on \(\neg A\wedge (A\rightarrow B)\) will satisfy: \(p(w) = p(w^{\prime })\) if \(w,w^{\prime }\in E_{x}\). So we can write p(Ex) for the pay-off in partition Ex. Assume that EF is the outcome that makes our sentence determinately false (i.e. EF are the worlds in which A is true, and so worlds where p(w) = $0). The remaining elements E1,…,En partition possible ways in which ¬A is true. Assume that 0 < Pr(¬A) < 1. The expected value of a fair bet on \(\textup {Pr}(\neg A\wedge A\rightarrow B)\) should be \(\$\textup {Pr}(\neg A)\times \textup {Pr}(A\rightarrow B)\).
If n = 1 we get \(\textup {Pr}(E_{1})p(E_{1}) = \textup {Pr}(\neg A)\textup {Pr}(A\rightarrow B)\). But Pr(E1) = Pr(¬A). So \(p(E_{1}) = \$\textup {Pr}(A\rightarrow B)\). However, the pay-off \(\$\textup {Pr}(A\rightarrow B)\) is not identical to either of the three allowed pay-offs $0, $1 or \(\$\textup {Pr}(\neg A)\textup {Pr}(A\rightarrow B)\).
So n > 1. There are uncountably many ways of distributing the probability of ¬A over the Ei’s in a way that does not affect the probability of either ¬A or \(A\rightarrow B\) (given rRT). However, the expected pay-off on a bet on \(\neg A\wedge A\rightarrow B\) for each distribution must be the same: for any given distribution Pr we need \(\textup {Pr}(E_{1})p(E_{1})+{\cdots } +\textup {Pr}(E_{n})p(E_{n}) = \textup {Pr}(\neg A)\textup {Pr}(A\rightarrow B) = (\textup {Pr}(E_{1}) + {\cdots } +\textup {Pr}(E_{n}))\textup {Pr}(A\rightarrow B)\). But given that each pay-off p(Ei) can only take one of three values this cannot be achieved, not without imposing constraints on how the probabilities on the Ei’s are distributed. □
Proof Proof of Observation 5:
First a useful lemma.
Lemma 4
Assume that all logical and probabilistic properties stated in the Observation hold, and that rRT holds. Assume, moreover that \(\textup {Pr}(\varphi \wedge (A\rightarrow B)) = \textup {Pr}(\varphi )\), and that Pr(A) > 0. Then either Pr(A ∧¬B) = 0 or Pr(¬A ∧ C) = 0 for every factual C that logically implies φ.
Proof
Assume that \(\textup {Pr}(\varphi \wedge (A\rightarrow B)) = \textup {Pr}(\varphi )\), that Pr(A) > 0 and that C logically implies φ.
Assume for reductio that \(\textup {Pr}(\varphi \wedge (A\rightarrow \neg B))>0\). As \(\textup {Pr}(\varphi \wedge (A\rightarrow B)) = \textup {Pr}(\varphi )\), by ordinary probabilistic reasoning it then follows that \(\textup {Pr}(\varphi \wedge (A\rightarrow B) \wedge (A\rightarrow \neg B))>0\). \( \varphi \wedge (A\rightarrow B) \wedge (A\rightarrow \neg B)\) implies \((A\rightarrow B) \wedge (A\rightarrow \neg B)\) which by normality is equivalent to \( A\rightarrow (B \wedge \neg B)\). As Pr(A) > 0, we can apply rRT; but this will give \(\textup {Pr}(A\rightarrow (B \wedge \neg B)) = 0\). So \(\textup {Pr}((A\rightarrow B) \wedge (A\rightarrow \neg B)) = 0\) and so \(\textup {Pr}(\varphi \wedge (A\rightarrow B) \wedge (A\rightarrow \neg B)) = 0\), and we have a contradiction.
So \(\textup {Pr}(\varphi \wedge A\rightarrow \neg B) = 0\). As C implies φ, \(\textup {Pr}(C\wedge (A\rightarrow \neg B)) = 0\).
Let
By Sα, ψ logically implies \(((A\vee \neg C)\wedge A)\rightarrow \neg B\) and so by LLE ψ logically implies \(A\rightarrow \neg B\). As \(\textup {Pr}(C\wedge (A\rightarrow \neg B)) = 0\), Pr(C ∧ ψ) = 0.
By rMP, ¬A ∧¬C ∧ ψ and A ∧ B ∧¬C ∧ ψ are logical contradictions and so will have probability 0. The four sentences C, ¬A ∧¬C, A ∧ B ∧¬C and A ∧¬B ∧¬C form a logical partition, and the first three get probability 0 when conjoined with ψ, so by additivity we have:
However, by rCS, A ∧¬B ∧¬C logically implies ψ so
So:
By rRT:
So:
This can only hold if either Pr(A ∧¬B) = 0, or Pr(A ∨¬C) = 1 (i.e. if Pr(¬A ∧ C) = 0). □
Claim 1. Suppose that (i) \(\textup {Pr}((A\rightarrow B)\wedge (C\rightarrow B)) = \textup {Pr}(C\rightarrow B)\), (ii) Pr(A ∧¬B) > 0, (iii) Pr(¬A ∧ C ∧ B) > 0. Suppose, moreover, for reductio that Pr satisfies rRT. By rCS, C ∧ B logically implies \( (C\rightarrow B)\). By Lemma 4, either Pr(A ∧¬B) = 0 or Pr(¬A ∧ C ∧ B)) = 0. The first alternative falls with premise (ii) and the second with premise (iii) and so we have a contradiction.
Claim 2. Suppose that (i) \(\textup {Pr}(C\wedge (A\rightarrow B)) = \textup {Pr}(C)\), (ii) Pr(A ∧¬B) > 0, and (iii) Pr(¬A ∧ C) > 0. Furthermore, by way of reductio, suppose that Pr satisfies rRT. C logically implies itself so from Lemma 4 we have either Pr(A ∧¬B) = 0 or Pr(¬A ∧ C) = 0. The first alternative falls with premise (ii) and the second with premise (iii) and so we have a contradiction.
Claim (3). Suppose that (i) \(\textup {Pr}((A\rightarrow B)\wedge (\neg A\rightarrow B)) = \textup {Pr}(B)\), (ii) 0 < Pr(A) < 1, (iii) 0 < Pr(B) < 1. Suppose, moreover, for reductio that Pr satisfies rRT.
By additivity \(\textup {Pr}((A\rightarrow B)\wedge (\neg A\rightarrow B)) = \textup {Pr}(A\wedge (A\rightarrow B)\wedge (\neg A\rightarrow B)) +\textup {Pr}(\neg A \wedge (A\rightarrow B)\wedge (\neg A\rightarrow B))\). By rMP and rCS \(A\wedge (A\rightarrow B)\) is logically equivalent to A ∧ B, and \(\neg A\wedge (\neg A\rightarrow B)\) is logically equivalent to ¬A ∧ B. So \(\textup {Pr}((A\rightarrow B)\wedge (\neg A\rightarrow B)) = \textup {Pr}(A\wedge B\wedge (\neg A\rightarrow B)) +\textup {Pr}(\neg A \wedge B \wedge (A\rightarrow B)) = \textup {Pr}(B)\) (by (i)). As Pr(B) = Pr(A ∧ B) + Pr(¬A ∧ B) it follows that (1) \(\textup {Pr}(A\wedge B\wedge (\neg A\rightarrow B)) = \textup {Pr}(A\wedge B)\) and (2) \(\textup {Pr}(\neg A \wedge B\wedge (A\rightarrow B)) = \textup {Pr}(\neg A\wedge B)\).
Consider the first of these identities. A ∧ B logically implies itself, so from Lemma 4 either (a) Pr(¬A ∧¬B) = 0 or (b) Pr(¬¬A ∧ A ∧ B) = Pr(A ∧ B) = 0. Consider the second identity. ¬A ∧ B logically implies itself, so from Lemma 4 either (c) Pr(A ∧¬B) = 0 or (d) Pr(¬A ∧¬A ∧ B) = Pr(¬A ∧ B) = 0. Four possibilities. If (a)+(c) hold then Pr(¬B) = 0 and this contradicts premise (iii). If (a)+(d) then Pr(¬A) = 0 and this contradicts premise (ii). If (b)+(c) then Pr(A) = 0 and this contradicts premise (ii). If (b)+(d) then Pr(B) = 0 and this contradicts premise (iii). Those are all the alternatives, so we have a contradiction. □
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Cantwell, J. Revisiting McGee’s Probabilistic Analysis of Conditionals. J Philos Logic 51, 973–1017 (2022). https://doi.org/10.1007/s10992-022-09657-5
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DOI: https://doi.org/10.1007/s10992-022-09657-5