Can The Paradoxes Of Confirmation (ravens And Grue) Be Solved By Appeal To Degrees Of Confirmation?

Can the paradoxes of confirmation (ravens and grue) be solved by appeal to degrees of confirmation? 1. Introduction In this essay I will argue that the paradoxes of confirmation can be solved by appeal to degrees of confirmation, to a degree. I argue that a Bayesian account of confirmation can solve the raven paradox with appeal to degrees of confirmation and by relativizing confirmation to background knowledge. However I also show that appeal to degrees of confirmation can be of no help in solving the grue paradox. 2. Confirmation Confirmation is a relation between evidence and a hypothesis which reflects the support which the former lends to the latter. For example, one might say that the a positive pregnancy test result confirms that an individual is pregnant. Understanding the confirmation-relation is important in order to understand the relationship between evidence and hypotheses in scientific practice. Experimental science is the practice of gathering evidence in order to assess hypotheses, and it is therefore of interest to know how we are to understand the relationship between these two phenomena. Different accounts of confirmation describe the relation, and provide truth-conditions for statements of the form `E confirms H`. I first look at the Hempelian account of confirmation, before presenting both paradoxes and showing that they arise most worryingly for this account. I attempt to solve both paradoxes by reconstruing confirmation in a Bayesian fashion, with appeal to conditional probabilities. 3. Hempel Hempel presents an account of the confirmation-relation which has significant intuitive plausibility (Crupi 2013). (H-confirmation) E H-confirms H iff E entails devEH or E entails devES and S entails H - where deveH is an instance of H (without any quantifiers) naming only those individuals occurring in e. For example, the observation that `a is a man and a is mortal` H-confirms the hypothesis that `for all x, if x is a man, then x is mortal` (i.e. `all men are mortal`), since it entails `if a is a man, then a is mortal`, which is an instance of the hypothesis naming only a. This is a technical account which expresses an intuitive idea, namely that universal generalisations are confirmed by their instances. Specifically, confirmation holds when a piece of evidence logically entails an instance of the hypothesis.

4. Raven Paradox The Raven Paradox concerns the confirmation of universal-generalizations. Let h be our hypothesis, and e be our evidence: (h) All ravens are black (e) a is a black raven e H-confirms h, and seems to be a paradigmatic example of confirmation with regards to the sorts of generalizations which are made in scientific theories. The paradox arises from considering the following pair of propositions: (h*) All non-ravens are non-black (e*) a is a non-black non-raven h* and h are logically equivalent. Moreover, e* H-confirms h*. The paradoxical conclusion is that given the equivalence of h ¬and h*, e* confirms h. (PC) `a is a non-black non-raven` H-confirms `all ravens are black` In other words, the observation of a non-black non-raven H-confirms the hypothesis that all ravens are black. The paradoxical conclusion follows from the premises that 1. e H-confirms h, 2. h* is logically equivalent to h, and 3. if E entails H and H is logically equivalent to H` then E H-confirms H`. The third premise is clearly needed to motivate (PC). 5. Background knowledge Following Fitelson (2006) I think that (PC) itself is not paradoxical. This is because the above descri ption of the situation is ambiguous since it is under-described. The reason why (PC) seems to be false is that given our knowledge of the world as it is, it seems that seeing a non-black non-raven could tell us nothing about the colour of ravens. The conclusion which is paradoxical, which we implicitly draw, and which therefore underlies our feeling that (PC) is paradoxical, is: (PC*) e* confirms h given that we know that a is a non-raven. The apparent falsity of (PC) given our actual background knowledge arises because it is almost certain of a given object in the world that it is not a raven. If it is antecedently known that a is a non-raven, then learning that a is non-black is no confirmation for h. This is because once we know that a is a non-raven, learning about its colour can have no impact on h since whatever we learn it can be neither a counter-example to nor a positive instance of h. (PC*) is therefore obviously false. Furthermore (PC*) does not follow from the premises (1-3) in section 4. The fact that (PC*) is false, and that we might conflate (PC) and (PC*) demonstrates the first major flaw in Hempel`s theory, which is the fact that H-confirmation is not relativized to background knowledge. Given knowledge that a is a non-raven, e* cannot be confirmation of h. However, given a different state of the world, and therefore a different collection of background knowledge, it could be confirmation. For example, if our background knowledge contained the fact that there was only one non-black bird in existence, e* would obviously confirm h (it would make it certain). Consider yet another hypothetical situation in which we have different background knowledge. We antecedently know, of a sample of 1,000 birds, that either (i) there are 950 ravens, only 900 of which are black, or (ii) there are only 10 ravens and all of them are black. Relative to this background knowledge, e (i.e. the observation of a black raven) would disconfirm h, since seeing a black raven makes it more likely that (i) is true, and (i) being true makes it less likely that h is true. These considerations all show that H-confirmation is not a satisfactory confirmation-relation, since it does not take into account our background knowledge. H-confirmation is a two-place relation holding between an evidence-proposition and hypothesis-proposition. The raven paradox shows that it must be considered as a three-place relation between evidence, hypothesis, and background knowledge. 6. Degrees I have shown above that the raven paradox can be avoided by rejecting H-confirmation and considering confirmation as a 3-place relation. In doing so I showed that our intuitions regarding the paradox rest on conflating (PC) and (PC*). One important result of this discussion is that a re-constructed version of (PC) is true: (PC-a) e* confirms h relative to Ka (where Ka is our actual background knowledge) This is true because, while it is overwhelming likely of a given object a that it is a non-raven, it is not certain, and so Ka does not include the proposition `a is a non-raven`. This conclusion is still somewhat puzzling. If one knows nothing about a and then discovers it is a non-black non-raven, this does confirm h since it reduces the number of possible counter-examples to h. I think a further clarification of the confirmation-relation can explain why (PC-a) seems to be false, even though it is true. Having done this, I will have shown how a proper understanding of the confirmation-relation can avoid the raven paradox altogether. (PC=) e* confirms h relative to Ka to the same degree that e confirms h relative to Ka. My claim is that (PC=) (like PC* before it) is false claim which underlies our discomfort with (PC-a), and which we implicitly take (PC-a) as meaning. In our first case, (PC) underdescribed the confirmation-relation by not containing reference to background knowledge. (PC-a) under-describes the confirmation-relation by not containing reference to the degree of confirmation. To spell out the final missing aspect of the confirmation-relation, we need a way of determining the degree of confirmation. Bayesian accounts of confirmation are apt to provide this, since they define confirmation in terms of conditional probabilities. For example, on a standard Bayesian account of confirmation: (B) E confirms H relative to K iff P(H|E&K)>P(H|K) Which is to say loosely that a piece of evidence is confirmation of a hypothesis (relative to background knowledge) iff the probability of that hypothesis given the evidence is greater than the probability of that hypothesis without the evidence. The Bayesian account can explain degrees of confirmation since the RHS of (B) is a quantitative claim. One can give a value for the strength of confirmation by comparing the probability values on either side of the inequality. There are different ways of doing this, but two plausible measures for strength are: (D1) strength = P(H|E&K) - P(H|K) (D2) strength = P(H|E&K) / P(H|K) The relative merits/demerits of each of these measures are outside the scope of this essay. I will use (D2) as my measure. 7. Bayesianism applied Smith (n.d.) presents a simple estimate of the relevant probabilities to calculate the strength of confirmation. The tables below are assumptions about the world, R indicates `raven object` and B indicates `black object`. x is the relative frequency of ravens in the world, y is relative frequency of black things in the world. Furthermore, we assume that x is very small. The non-underlined values are estimated probabilities. Table 1 - assumptions about the world R ~R B p(RB)=xy (1-x)y ~B x(1-y) (1-x)(1-y)

Table 2 - assumptions about the world given h R ~R B p(RB|H)=x y -x ~B 0 1-y

By Bayes` Theorem we have: and (both assuming that x is tiny) These values are the degree of confirmation by D2 (here I have omitted k). The second equation shows that e* confirms h to degree 1/(1-x), which will be very close to 1. On the other hand, e confirms h to degree 1/y, which is a very large number. (Since y is the relative frequency of black-objects, 1/y is equal to the total number of objects in the world divided by the total number of black objects in the world.) These rough estimates show how a Bayesian account of confirmation shows that a black raven is much stronger confirmation that all ravens are black than is a non-black non-raven, and therefore that (PC=) is false. This makes the counter-intuitive conclusion that a non-black non-raven is some confirmation easier to swallow. Overall, in this section I have shown how the raven paradox can be avoided by using a confirmation relation which (i) relativizes confirmation to background knowledge, and (ii) gives sense to degrees of confirmation. A Bayesian account with these characteristics can solve the raven paradox in a satisfactory manner. 8. Grue paradox Goodman (1955) presents another paradox of confirmation, which I will call the grue paradox. It also concerns the confirmation of universal generalizations. I follow Fitelson`s (2008) characterization of the paradox. I will define the relevant predicates, hypotheses, and evidence statements. Ox - x is observed before t Gx - x is green Ex - x is an emerald £x - x is grue (Ox iff Gx) x is grue iff x is green and observed before t, or blue and not observed before t. H1 - All emeralds are green H2 - All emeralds are grue E1 - Ea&Ga E2 - Ea&(Oa iff Ga) E - Ea&Oa&Ga The paradox arises from considering a seemingly simple instance of confirmation, much like that in the raven paradox. It seems that observing a green emerald would be confirmation of H1. However, such an observation would also be confirmation of H2 (if t is some point in the near future). The grue predicate is clearly very strange, and we intuitively feel that H2 is false. The paradoxical conclusion, therefore is: (GC) Observing a green emerald before t (i.e. E) confirms both H1 and H2. First I will show the argument towards (GC). The fact that E¬1 and E2 confirm H1 and H2 respectively, follows from a Hempelian account of confirmation, since both propositions entail instances of the respective generalizations. E confirms H1 by the principle of monotonicity: (M) if `Ax` confirms H then `Ax & Bx` confirms H Hence since E1 confirms H¬1 and E is simply E1 with another conjunct, E confirms H1. E confirms H2 by (M), and by the principle of equivalence: (EQ) If E confirms H, then anything logically equivalent to E also confirms H E is equivalent to the sentence `(Ea & (Oa iff Ga)) & Oa)`, and this sentence is E2 with an extra conjunct, hence since E2 confirms H2, by (EQ) and (M) E confirms H2. 9. Bayesian solution A Bayesian account of confirmation avoided the raven paradox which befell Hempel`s account since it allowed for degrees of confirmation, and relativized confirmation to background knowledge. With regards to the grue paradox, the first thing to note is that Bayesians can block the above move to (GC) by denying (H-confirmation) and (M). (H-confirmation) can be shown to be false by way of a counter-example similar to that in section 5. There are possible collections of background knowledge such that E would disconfirm H1, and those in which E would disconfirm H2. The fact that an evidence-statement entails an instance of a hypothesis-generalization is not sufficient for it to confirm that hypothesis. (M) can be shown to be false with a different sort of counterexample: Bx - x is a black card Ax - x is the Ace of Spades Jx - x is the jack of clubs `Ba` confirms `Aa` relative to K since P(Aa|Ba&K)>P(Aa|K). However `Ba&Ja` does not confirm `Aa`, since P(Aa|Ba&Ja&K)=0. Hence (M) is false. This shows that if we take a Bayesian view of confirmation, (GC) does not follow in the same way which it does for Hempel-confirmation. However, it does not show that (GC) is false, which is the more pressing task, and what the Bayesian must do in order to solve the paradox. At this point we can consider how (GC) could be shown to be false. What we want to do, is to show that E does not confirm H2. For this to hold on a Bayesian account, it must be true (omitting K for simplicity) that: (B) P(H2|E)=P(H2) Clearly, however, (B) cannot be proved only by appeal to degrees of confirmation, since it is an equation not an inequality. Degrees of confirmation allow one to compare the strength of the confirmation relation between instances, but will not be sufficient to demonstrate that confirmation of a particular value holds. Since (B) cannot be proved by appeal to degrees of confirmation, (GC) cannot be shown false by appeal to degrees of confirmation. 10. Moderate Bayesian Response Degrees of confirmation helped us in the solution to the raven paradox because we showed that, though both black ravens and non-black non-ravens both confirmed h, instances of the former were much stronger confirmation. We did not show (PC) false, but diagnosed the paradox as resting partly on the conflation of (PC) and (PC=), which was false. With regards to the grue paradox there are two things to note. Firstly, I am not convinced that showing that while (GC) was true, green emeralds confirmed H2 to a much lesser degree than they did H1 would be a satisfactory solution to the paradox. It is highly counter-intuitive that such weird-generalizations could be confirmed by observations, and one could imagine infinitely many such weird-generalizations. Secondly, as Sober (1984) notes, it is not possible to reach even this conciliatory conclusion in this case. Adopting (D1) as a measure of degree of confirmation, he shows what must be the case if E confirms H1 more than H2: P(H1|E) - P(H1) > P(H2|E) - P(H2) iff P(E|H¬1)P(H1)/P(E) - P(H1) > P(E|H¬2)P(H2)/P(E) - P(H2) which holds iff P(H1)(1/P(E)-1) > P(H¬2)(1/P(E) - 1) which can be rewritten (on the assumption that P(E)<1) as: P(H1)>P(H2) which is to say that E confirms H1¬ more than H2 iff the prior probability of H1 is greater than that of H2. Sober points out that there does not seem to be a satisfactory way of assigning values to these prior probabilities, and I agree. In section 7 we were able to draw up estimates for differences in confirmation-strength because we were concerned with the prior probabilities of evidence statements. I do not see how one could assign prior probabilities to hypotheses in a way which would help solve the grue paradox. Moreover as I said above, I am not convinced that identifying a strong asymmetry between the strength of confirmation E gives to H1 and H2 would be a satisfactory solution to the paradox. The grue paradox needs a full-blooded response, not a conciliatory one, and appeal to degrees of confirmation is unable to give either. 11. Conclusion In conclusion, I believe I have shown the extent to with appeal to degrees of confirmation can solve the raven and grue paradoxes. The raven paradox can be solved by appeal to both degrees of confirmation, and the relativization of confirmation to background knowledge, both characteristics of Bayesian accounts of confirmation. The grue paradox, however, cannot be solved by appeal to degrees of confirmation. Appeal to degrees of confirmation can neither show its paradoxical conclusion to be false, nor mitigate it`s paradoxicality.