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Confirmation, transitivity, and Moore: the Screening-Off Approach

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Abstract

It is well known that the probabilistic relation of confirmation is not transitive in that even if E confirms H1 and H1 confirms H2, E may not confirm H2. In this paper we distinguish four senses of confirmation and examine additional conditions under which confirmation in different senses becomes transitive. We conduct this examination both in the general case where H1 confirms H2 and in the special case where H1 also logically entails H2. Based on these analyses, we argue that the Screening-Off Condition is the most important condition for transitivity in confirmation because of its generality and ease of application. We illustrate our point with the example of Moore’s “proof” of the existence of a material world, where H1 logically entails H2, the Screening-Off Condition holds, and confirmation in all four senses turns out to be transitive.

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Notes

  1. Cf. Carnap (1962, Preface to the Second Edition) on “concepts of increase in firmness” and “concepts of firmness”.

  2. We leave it open that t may be context-dependent (perhaps higher in higher-stakes contexts and lower in lower-stakes contexts). We are following the standard view here that whether H is rationally acceptable given E is determined solely by Pr(H | E) (and perhaps the context), though the view is not unproblematic. Cf. Shogenji (2012).

  3. Cf. Douven (2011, pp. 487–488) on “t-evidence”, and Chandler (2010, p. 337) on “sufficient evidence”.

  4. The example is taken from Shogenji (2003).

  5. This case is adapted from Dretske (1970, pp. 1015–1016).

  6. This sort of point is made in Chandler (2010, p. 337), Cohen (2005, pp. 424–425), Hawthorne (2004, pp. 73–75), Okasha (1999, Sect. 9), Silins (2005, p. 85; 2007, pp. 123–125), and White (2006, Sect. 5).

  7. We have in mind, of course, non-trivial such conditions and not, say, the condition that E confirms-IF H2 as a condition for transitivity in confirmation-IF in the general case.

  8. See Roche (2012a).

  9. That confirmation-IF is transitive under (C1*) is shown in Shogenji (2003).

  10. Cf. Kotzen (2012, p. 69), Kukla (1998, Sects. 4.2, 4.3, and 6.2), and Moretti (2002, p. 160; 2012, Sect. 5).

  11. We are following Kotzen (2012) in calling (C2) “the Dragging Condition”.

  12. Cf. Kotzen (2012, p. 66).

  13. Cf. Kotzen (2012, p. 72, n. 22).

  14. Moretti (2012, Sect. 5) establishes a principle similar to Theorem 2 where (X) is (C2). It can be put thus: if (a) Pr(H2) > t, (b) Pr(H1 | E) > Pr(H1), (c) Pr(H1 | E) > t, (d) H1 entails H2, and (e) (C2) holds, then Pr(H2 | E) > P(H2).

  15. Theorem 4 is essentially the same as “Closure*” in Chandler (2010, p. 337, n. 5).

  16. Here and throughout the paper when we speak of positive instances of transitivity, we have in mind non-vacuous positive instances.

  17. Each of (C1)–(C3) fails to hold in the zoo case. That (C1) fails to hold follows from the fact that Pr(H2 | E ∧ ¬H1) < Pr(H2 | ¬H1); E increases the probability of ¬H2 given ¬H1, and, so, decreases the probability of H2 given ¬H1. That (C2) fails to hold follows from the fact that Pr(E | H1) = Pr(E | ¬H2), thus Pr(E | H1) ≤ Pr(E | ¬H2); Kotzen (2012, pp. 81–82) shows that (where E confirms-IF H1) if Pr(E | H1) ≤ Pr(E | ¬H2), then (C2) does not hold. Pr(¬H2 ∧ ¬H1 | E) > Pr(¬H2 ∧ ¬H1) and Pr(H2 ∧ H1 | E) = Pr(H1 | E) > Pr(H1) = Pr(H2 ∧ H1), so, since Pr(¬H2 ∧ H1 | E) = 0 = Pr(¬H2 ∧ H1), it follows that Pr(H2 ∧ ¬H1 | E) < Pr(H2 ∧ ¬H1), thus (C3) fails to hold.

  18. The extant literature on transmission failure is extensive. See, e.g., Beebee (2001), Brown (2003, 2004), Cling (2002), Coliva (2011), Davies (1998, 2000, 2003, 2004), Dretske (2005a, b), Ebert (2005), Hale (2000), Hawthorne (2005), Kotzen (2012, Sect. 6), McKinsey (2003), McLaughlin (2003), Neta (2007), Peacocke (2004, Chap. 4, pp. 112–115), Pryor (2004), Sainsbury (2000), Schiffer (2004), Silins (2005, 2007), Smith (2009), Suarez (2000), Tucker (2010a, b), White (2006, Sect. 5), and Wright (1985, 2000a, b, 2002, 2003, 2004, 2007, 2011). For discussion of how to formalize the issue of transmission failure, see Chandler (2010), Moretti (2012), Moretti and Piazza (2011), and Okasha (2004). Cf. Pynn (2011).

  19. Compare Theorems 3 and 7. The latter shows that the antecedent condition that H1 entails H2 is essential to the former, but does not show that the same is true of the antecedent condition that H1 confirms-TSF H2. We noted earlier that it follows from H1 ⊢ H2 that H1 confirms-IF H2 (except for the uninteresting cases), but it does not follow that H1 confirms-TSF H2. So, (b) is not redundant. Moreover, the argument given in Appendix 5 for Theorem 7 does not involve cases where H1 entails H2, and therefore does not itself imply that there can be cases where (a) E confirms-TSF H1, (c) H1 entails H2, and (d) (C1)–(C3) all hold, and yet E does not confirm-TSF H2. It can be shown, however, that such cases are possible—regardless of the value specified for t. Due to space considerations we omit the proof.

  20. The Converse Consequence Condition is introduced and rejected in Hempel (1965).

  21. When H1 and H2 are mutually entailing, Pr(H1) = Pr(H2) and Pr(H1 | E) = Pr(H2 | E), in which case if Pr(H1 | E) > Pr(H1), it follows that Pr(H2 | E) > Pr(H2). Counterexamples to (CCC) are thus cases where H1 and H2 are not mutually entailing. For relevant discussion, see Milne (2000).

  22. (C1*), like (C1), is a condition for transitivity in confirmation-IF in the case where H2 entails H1. But there are cases of transitivity in confirmation-IF in the case where H2 entails H1 where (C1) holds but (C1*) does not.

  23. Testimonial/memorial/perceptual cases are discussed in Roche (2012a) and Shogenji (2003).

  24. Cf. Kotzen (2012, Sects. 3 and 4).

  25. In fact, it must be the case that Pr(H2 | ¬H1 ∧ E) > Pr(H2 | ¬H1), which means that (C1*) fails to hold. So, if E confirms-IF H1, confirms-IF&SF H1, or confirms-TSF H1, (C3) entails not-(C1*).

  26. This does not require that H2 also logically entails H1.

  27. It can be shown, further, that even under the condition that E confirms-IF H1, there is no logical entailment between (C2) and (C3).

  28. For ease of expression, we sometimes refer to E as an experience. Strictly speaking, of course, E is a proposition about an experience, not an experience itself.

  29. Some may point out that (C2) is satisfied for the purpose of transitivity in confirmation-TSF because Pr(H2) ≤ t from the second antecedent of transitivity in confirmation-TSF, while Pr(H1 | E) > t from the first antecedent of transitivity in confirmation-TSF. However, we are not assuming here that the second antecedent of transitivity in confirmation-TSF holds, precisely for the reason that Pr(H2) is questionable and questioned. The second antecedent of transitivity in the other three senses of confirmation is unproblematic: H1 confirms-IF H2, confirms-SF H2, and confirms-IF&SF H2, from Pr(H2 | H1) = 1 > P(H2) and Pr(H2 | H1) = 1 > t.

  30. Since confirmation-TSF is transitive in MOORE, it follows that H1 does not confirm-TSF H2. As we noted in Footnote 29 we are not making the assumption that H1 confirms-TSF H2. Our claim of transitivity in confirmation-TSF in MOORE is of the form: if E confirms-TSF H1 and H1 in turn confirms-TSF H2, then E confirms-TSF H2.

  31. We can state the point more simply: if Pr(H1 | E) > t, then Pr(H2) > t. Note that this does not imply the failure of (C2) Pr(H2) < Pr(H1 | E). (C2) can be true while Pr(H1 | E) > t and Pr(H2) > t, and thus the conditional is also true.

  32. Bear in mind here and throughout the remainder of the argument that Pr(H1 | E) and Pr(H2 | H1) are continuous monotonically increasing functions of β, and that Pr(H1) and Pr(H2) are continuous monotonically decreasing functions of β.

  33. See Roche (2012b) for a similar argument for the claim that regardless of the value specified for t the following condition is not a condition for transitivity in confirmation-IF&SF: Pr(H2 | E ∧ H1) > Pr(H2 | H1) and Pr(H2 | E ∧ ¬H1) > Pr(H2 | ¬H1). Note that this condition is stronger than (C1) and neither stronger nor weaker than (C1*), and that, like (C1) and (C1*), it is a condition for transitivity in confirmation-IF.

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Acknowledgments

We thank an anonymous reviewer for very helpful comments on a prior version of the paper.

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Correspondence to William Roche.

Appendices

Appendix 1: Proof of Theorem 2

Let (X) be any of (C1)–(C3). Suppose (a) E confirms-IF&SF H1, so (a1) Pr(H1 | E) > Pr(H1) and (a2) Pr(H1 | E) > t. Suppose (b) H1 confirms-IF&SF H2, therefore (b1) Pr(H2 | H1) > Pr(H2) and (b2) Pr(H2 | H1) > t. Suppose (c) H1 entails H2 and (d) (X) holds. By (a1), (b1), (c), (d), and Theorem 1, it follows that Pr(H2 | E) > Pr(H2). By (c) and the theorem that if H1 entails H2, then Pr(H2 | E) ≥ Pr(H1 | E), it follows that Pr(H2 | E) ≥ Pr(H1 | E). By (a2), it then follows that Pr(H2 | E) > t. So Pr(H2 | E) > Pr(H2) and Pr(H2 | E) > t. So E confirms-IF&SF H2.

Appendix 2: Proof of Theorem 3

Let (X) be any of (C1)–(C3). Suppose (a) E confirms-TSF H1, so (a1) Pr(H1 | E) > Pr(H1), (a2) Pr(H1 | E) > t, and (a3) Pr(H1) ≤ t. Suppose (b) H1 confirms-TSF H2, therefore (b1) Pr(H2 | H1) > Pr(H2), (b2) Pr(H2 | H1) > t, and (b3) Pr(H2) ≤ t. Suppose (c) H1 entails H2 and (d) (X) holds. By (a1), (a2), (b1), (b2), (c), (d), and Theorem 2, it follows that Pr(H2 | E) > Pr(H2) and Pr(H2 | E) > t. By (b3), Pr(H2) ≤ t. Hence Pr(H2 | E) > Pr(H2), Pr(H2 | E) > t, and Pr(H2) ≤ t. Hence E confirms-TSF H2.

Appendix 3: Proof of Theorem 5B

Suppose a card is randomly drawn from a standard deck of cards. Let E be the claim “The card drawn is a Heart”, H1 be the claim “The card drawn is a Red”, and H2 be the claim “The card drawn is a Diamond”. Then, E confirms-IF H1, since Pr(H1 | E) = 1 > Pr(H1) = 1/2, and H1 confirms-IF H2, given that Pr(H2 | H1) = 1/2 > Pr(H2) = 1/4, and both (C2) and (C3) hold, since Pr(H2) = 1/4 < Pr(H1 | E) = 1 and Pr(H2 ∧ ¬H1 | E) = 0 = Pr(H2 ∧ ¬H1). But E does not confirm-IF H2; Pr(H2 | E) = 0 < Pr(H2) = 1/4.

Appendix 4: Proof of Theorem 6

Consider the following schema, to be referred to (for lack of a better name) as “Schema”, where β ∈ ℝ+, β ≥ 1, and τ = 1 + (2/10)β + (1/10)β + (9/10)β + (1/10)β + (1/10)β + 10β:

Schema

E

H1

H2

Pr

E

H1

H2

Pr

T

T

T

1/τ

F

T

T

(1/10)β/τ

T

T

F

(2/10)β

F

T

F

(1/10)β/τ

T

F

T

(1/10)β/τ

F

F

T

0

T

F

F

(9/10)β/τ

F

F

F

10β/τ

On each instance of Schema, it follows that:

$$ \Pr ({\text{H}}2\,|\,{\text{E}} \wedge {\text{H}}1)-\Pr ({\text{H}}2\,|\,{\text{H}}1) = \frac{1}{{1 + \left( \frac{2}{10} \right)^{\beta } }} - \frac{{1 + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } }} > 0; $$
(1)
$$ \begin{gathered} \Pr ({\text{H}}2\,|\,{\text{E}} \wedge \neg {\text{H}}1)-\Pr ({\text{H}}2\,|\,\neg {\text{H}}1) = \frac{{\left( \frac{1}{10} \right)^{\beta } }}{{\left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} - \frac{{\left( \frac{1}{10} \right)^{\beta } }}{{\left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } + 10^{\beta } }} > 0; \hfill \\ \hfill \\ \end{gathered} $$
(2)
$$ \begin{aligned} \Pr ({\text{H}}1\,|\,{\text{E}})-\Pr ({\text{H}}2) & = \frac{{1 + \left( \frac{2}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} \\ & \quad - \frac{{1 + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + 10^{\beta } }} > 0; \\ \end{aligned} $$
(3)
$$ \begin{aligned} \Pr ({\text{H}}2 \wedge \neg {\text{H}}1\,|\,{\text{E}})-\Pr ({\text{H}}2 \wedge \neg {\text{H}}1) & = \frac{{\left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} \\ & \quad - \frac{{\left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + 10^{\beta } }} \\ & > 0. \\ \end{aligned} $$
(4)

By (1) and (2) it follows that (C1) holds. By (3) it follows that (C2) holds. By (4) it follows that (C3) holds.

The aim is to show that regardless of the value specified for t there are instances of Schema on which E confirms-IF&SF H1, H1 confirms-IF&SF H2, and yet, though E confirms-IF H2, E does not confirm-IF&SF H2 because Pr(H2 | E) ≯ t.

First, observe that each of Pr(H1 | E) and Pr(H2 | H1) approaches 1 as β tends to ∞:

$$ \mathop {\lim }\limits_{\beta \to \infty } \frac{{1 + \left( \frac{2}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} = 1; $$
(5)
$$ \mathop {\lim }\limits_{\beta \to \infty } \frac{{1 + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } }} = 1. $$
(6)

So, regardless of the value specified for t there is a value for β such that Pr(H1 | E) > t and Pr(H2 | H1) > t.

The same is true of Pr(H2 | E), since Pr(H2 | E), like each of Pr(H1 | E) and Pr(H2 | H1), approaches 1 as β tends to ∞:

$$ \mathop {\lim }\limits_{\beta \to \infty } \frac{{1 + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} = 1. $$
(7)

But, crucially, the following inequalities hold:

$$ \Pr ({\text{H}}1\,|\,{\text{E}}) = \frac{{1 + \left( \frac{2}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} > \frac{{1 + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} = \Pr ({\text{H}}2\,|\,{\text{E}}); $$
(8)
$$ \Pr ({\text{H}}2\,|\,{\text{H}}1) = \frac{{1 + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } }} > \frac{{1 + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } }} = \Pr ({\text{H}}2\,|\,{\text{E}}) > 0. $$
(9)

Next, consider the inequalities:

$$ \Pr ({\text{H}}1\,|\,{\text{E}})-\Pr ({\text{H}}1) > 0; $$
(10)
$$ \Pr ({\text{H}}2\,|\,{\text{H}}1)-\Pr ({\text{H}}2) > 0. $$
(11)

We noted above that each of Pr(H1 | E) and Pr(H2 | H1) approaches 1 as β tends to ∞. This is not true of Pr(H1) and Pr(H2)—quite the opposite in fact. Each of Pr(H1) and Pr(H2) approaches 0 as β tends to ∞:

$$ \mathop {\lim }\limits_{\beta \to \infty } \frac{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + 10^{\beta } }} = 0; $$
(12)
$$ \mathop {\lim }\limits_{\beta \to \infty } \frac{{1 + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } }}{{1 + \left( \frac{2}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{9}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + \left( \frac{1}{10} \right)^{\beta } + 10^{\beta } }} = 0. $$
(13)

With β = 1, Pr(H1 | E) = 6/11 > Pr(H1) = 7/62 and Pr(H2 | H1) = 11/14 > Pr(H2) = 3/31. So, given (5), (6), (12), and (13), and with β ≥ 1, it follows that (10) and (11) hold.Footnote 32

The argument now runs as follows. Take β = 1. Then Pr(H1 | E) = 6/11 > Pr(H1) = 7/62, Pr(H2 | H1) = 11/14 > Pr(H2) = 3/31, and Pr(H2 | E) = 1/2. If 6/11 > t > .5, we have an instance of Schema on which E confirms-IF&SF H1, H1 confirms-IF&SF H2, and yet, though E confirms-IF H2, E does not confirm-IF&SF H2 because Pr(H2 | E) ≯ t. If, instead, t ≥ 6/11, then let the value of β increase until Pr(H1 | E) > t and Pr(H2 | H1) > t but Pr(H2 | E) ≯ t; that there is such a value for β is guaranteed by (5), (6), (8), and (9). It will still be the case that Pr(H1 | E) > Pr(H1) and Pr(H2 | H1) > Pr(H2); this follows from (10) and (11). The resulting distribution will be an instance of Schema on which E confirms-IF&SF H1, H1 confirms-IF&SF H2, and E confirms-IF H2 but does not confirm-IF&SF H2 given that Pr(H2 | E) ≯ t.

The result is that none of (C1)–(C3) is a condition for transitivity in confirmation-IF&SF regardless of the value specified for t.Footnote 33

Appendix 5: Proof of Theorem 7

Consider Schema, and take β = 1. Then, as noted above, Pr(H1 | E) = 6/11 > Pr(H1) = 7/62, Pr(H2 | H1) = 11/14 > Pr(H2) = 3/31, and Pr(H2 | E) = 1/2. If 6/11 > t > .5, we have an instance of Schema on which E confirms-TSF H1, H1 confirms-TSF H2, and yet, though E confirms-IF H2, E does not confirm-TSF H2 because Pr(H2 | E) ≯ t. If t ≥ 6/11, then, as explained above, let the value of β increase until Pr(H1 | E) > t and Pr(H2 | H1) > t but Pr(H2 | E) ≯ t. Given (10) and (11), it will still be the case that Pr(H1 | E) > Pr(H1) and Pr(H2 | H1) > Pr(H2). Given (12) and (13), it will still be the case that Pr(H1) ≤ t and Pr(H2) ≤ t. The resulting distribution will thus be an instance of Schema on which E confirms-TSF H1, H1 confirms-TSF H2, but, since Pr(H2 | E) ≯ t, E does not confirm-TSF H2. Therefore, regardless of the value specified for t, none of (C1)–(C3) is a condition for transitivity in confirmation-TSF.

Appendix 6: Proof of Theorem 8

We showed above in the proof of Theorem 6 that regardless of the value specified for t there is an instance of Schema on which E confirms-IF&SF H1, H1 confirms-IF&SF H2, and E confirms-IF H2 but does not confirm-IF&SF H2 given that Pr(H2 | E) ≯ t. It follows immediately that regardless of the value specified for t there is an instance of Schema on which E confirms-SF H1, H1 confirms-SF H2, and E confirms-IF H2 but does not confirm-SF H2 because Pr(H2 | E) ≯ t. None of (C1)–(C3), therefore, is a condition for transitivity in confirmation-SF regardless of the value specified for t.

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Roche, W., Shogenji, T. Confirmation, transitivity, and Moore: the Screening-Off Approach. Philos Stud 168, 797–817 (2014). https://doi.org/10.1007/s11098-013-0161-3

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