How Explanation Guides Confirmation How Explanation Guides Confirmation Nevin Climenhaga*y Where E is the proposition that [If H and O were true, H would explain O], William Roche and Elliot Sober have argued that P(H F O&E) 5 P(H F O). In this article I argue that not only is this equality not generally true, it is false in the very kinds of cases that Roche and Sober focus on, involving frequency data. In fact, in such cases O raises the probability of H only given that there is an explanatory connection between them. In two recent essays, Roche and Sober (2013, 2014) argue that the proposi- tion that a hypothesis H would explain an observation O is evidentially irrel- evant to H. Where E says that, were H and O true, H would explain O, Roche and Sober’s thesis is that P H O&Ej Þ 5 P Hð jOð Þ: (1) Once we know O, Roche and Sober claim, E gives us no further evidence that H is true. In other words, O screens off E from H. Call this claim the Screening-Off Thesis (SOT; Roche and Sober 2014, 193). In endorsing SOT, Roche and Sober are presumably not making a claim about subjective probabilities, for an agent could virtually always assign co- herent subjective probabilities on which the above equality is false. I will in- stead read them as making a claim about epistemic probabilities, which we can understand as rationally constraining subjective probabilities. Theses similar to SOT are endorsed by other Bayesians skeptical of in- ference to the best explanation. For example, van Fraassen (1989, 166) fa- mously denies that the claim that a hypothesis is explanatory can give it any probabilistic “bonus.” Often, however, such skeptics do not make clear in precisely what way they think that explanation is irrelevant to confirmation. *To contact the author, please write to: Department of Philosophy, University of Notre Dame; e-mail: nclimenh@nd.edu. yI am grateful to Robert Audi, Daniel Immerman, Ted Poston, and two reviewers for Phi- losophy of Science for very helpful comments on earlier drafts of this article. Received December 2015; revised February 2016. Philosophy of Science, 84 (April 2017) pp. 359–368. 0031-8248/2017/8402-0009$10.00 Copyright 2017 by the Philosophy of Science Association. All rights reserved. 359 This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM All use subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). 360 NEVIN CLIMENHAGA All u Van Fraassen does consider a precise version of inference to the best expla- nation, but it is an uncharitable one, on which inference to the best explana- tion is understood as a non-Bayesian updating rule on which good explana- tions get higher probabilities than Bayesian conditionalization would give them.1 Roche and Sober are thus to be commended for stating a precise anti- explanationist thesis that does not mischaracterize their opponents’ position. If SOT is true, there is a clear sense in which explanation is not relevant to confirmation. That said, we do need to clarify the scope of SOT before we can evaluate its significance and plausibility. It is widely acknowledged by Bayesians that all confirmation is relative to a context. In other words, we always have some background knowledge K, which may be left implicit but is always guiding our judgments of probability. While subjective Bayesians often think of this background as being part of the probability function P(�) itself, for epistemic probabilities, it is preferable to make K explicit as a conjunct of the propo- sition being conditioned on.2 So we can rewrite Roche and Sober’s claim as P H O&E&Kj Þ 5 P Hð jO&Kð Þ: (2) Here H, O, and K are variables that could be filled in with various proposi- tions or background knowledge, with the caveat that H is a hypothesis and O an observational statement. We can then ask: for which H, O, and K do Roche and Sober take (2) to be true? The most straightforward interpretation of SOT is as a universal claim: for all O, H, and K, P(H F O&E&K) 5 P(H F O&K). However, this inter- pretation is uncharitable because it is trivially false. For example, suppose K includes the material conditional [(O&E) ⊃ H] but does not include the ma- terial conditional [O ⊃ H]. (Perhaps an oracle who knows whether O, H, and E are true has told you that if O and E are true, so is H.) Then P(H F O&E&K) 5 1, but P(H F O&K) < 1. Hence,3 1. Some explanationists have defended this rule against van Fraassen’s criticisms of it (see, e.g., Douven 2013). However, whether or not van Fraassen’s original arguments against non-Bayesian explanationist updating rules work, I show (Climenhaga, forthcoming) that such rules lead tosynchronicprobabilisticincoherence. The argument of this articlesuggests that it would be better to think of the “probabilistic bonus” that explanation gives to a hy- pothesis H as the degree to which the proposition that H is explanatory confirms H. 2. Making K part of the probability function itself makes it impossible to “bring out” any part of K in the way one brings out the evidence in Bayes’s Theorem—i.e., where X is a conjunct of K and Y is an arbitrary proposition, the probability of X given Y will always equal 1. This leads to a version of the old evidence problem. By contrast, if K is one of the propositions conditioned on, and not part of the probability function itself, then X can be brought out in such a way that the probability of X given Y does not necessarily equal 1. (For more on this point, see sec. 3 of Climenhaga [2017].) 3. Roche and Sober consider the possibility of counterexamples to SOT if one drops the standard Bayesian assumption of logical omniscience: “Let I be the proposition that H This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM se subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). DISCUSSION 361 P H O&E&Kj Þ > P Hð jO&Kð Þ: (3) Although the above example shows SOT to not be universally true, [(O&E) ⊃ H] is not the kind of information we would ordinary have as part of our background knowledge. In endorsing SOT, then, Roche and Sober presumably mean to hold that (2) is true in ordinary or paradigm cases, in particular, the kinds of cases in which defenders of inference to the best ex- planation wish to claim that explanatoriness is evidentially relevant. I will now argue that even so restricted, SOT is false. I will do this by considering one of the paradigmatic statistical cases Roche and Sober use to argue for SOT, involving smoking and cancer. Suppose that K includes statistical data on which smoking and cancer are correlated as well as the base rate of cancer in the population. Roche and Sober (2013, 660) and I agree that in this case P S gets cancer½ � S smokes½ �&Kj Þ > P S gets cancer½ �ð jKð Þ, (4) which implies P S smokes½ � S gets cancer½ �&Kj Þ > P S smokes½ �ð jKð Þ: (5) In a response to Roche and Sober (2013), McCain and Poston (2014, 150) briefly argue that the frequency data in K make (4) true only because they support the existence of some causal connection between smoking and can- cer:4 “The data indicated that there was some causal process—albeit un- known at that time—that explains the correlation between smoking and 4. McCain and Poston’s main claim in that paper is that even if SOT is true, explanato- riness can still play an evidential role by increasing the “resiliency” of probabilities. In my view this is based on a mistaken view about epistemic probabilities. As I understand it, the epistemic probability of H given O&K, P(H F O&K), is a relation between the prop- ositions H and O&K, such that, if P(H F O&K) 5 n, then someone with O&K as their evidence ought to be confident in H to degree n. Following Keynes (1921), I take this re- lationship to be metaphysically necessary and knowable a priori, like the laws of logic or logically implies O. Then, plausibly, there can be cases in which Pr(H F O&I) > Pr(H F O) and Pr(H F O&I&E) > Pr(H F O). Perhaps there can even be cases in which Pr(H F O&E) > Pr(H F O). This would be especially plausible if E were in some way indicative of I. But then the point would be that Pr(H F O&I&E) 5 Pr(H F O&I). Explanatoriness has no con- firmational significance, once purely logical and mathematical facts are taken into ac- count” (2014, 195). However, these comments do not undermine the counterexample in the text. First, that counterexample involves knowledge of a material implication, not a logical implication. ([(O&E) ⊃ H] could equivalently be stated as [~(O&E) v H].) Thus, the assumption that there are some contexts in which one does not know [(O&E) ⊃ H] does not violate logical omniscience. Second, Roche and Sober’s example involves I describing an implication relationship between O and H on their own. This is what lets them say that P(H F O&E&I) 5 P(H F O&I). In the counterexample in the text, K says that O&E, but not O, implies H. Hence, in the circumstances I have described, P(H F O&E&K) > P(H F O&K). This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM All use subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). 362 NEVIN CLIMENHAGA All u lung cancer. . . . Exactly this feature—a justified belief in an unknown ex- planatory story—plays a crucial role in using the data from observation to get justified beliefs about the relevant frequencies. Apart from a general jus- tified belief in some explanatory story accounting for that data, the obser- vational data would not justify beliefs about the relevant frequencies.” Al- though McCain and Poston are talking about objective frequencies and not epistemic probabilities in this quote, it is similarly true that [S smokes] sup- ports [S gets cancer] relative to K only because K supports the existence of an explanatory connection between [S smokes] and [S gets cancer]. As I go on to argue, this means that SOT is false. McCain and Poston do not formalize this point and perhaps for this rea- son do not recognize this implication of it. Instead they write that it shows that “even if one grants Roche and Sober’s claim that ‘O screens-off E from H’ this doesn’t show that explanatory considerations are irrelevant to confir- mation” (2014, 150).5 But it does not show this. Rather, it shows that explan- atory considerations are relevant to confirmation precisely because Roche and Sober’s SOT (or a near cousin, as I clarify below) is false. (McCain and Poston’s confusion on this point may be part of the reason that Roche and Sober do not reply to it in their [2014] response to McCain and Poston.) Roche and Sober (2013, 661) hold that, as SOT implies in this case, P S smokes½ � S gets cancer½ �&E&Kj Þ5P S smokes½ �ð j S gets cancer½ �&Kð Þ, (6) where E is the proposition [If (S smokes) and (S gets cancer) were true, (S smokes) would explain (S gets cancer)]. Roche and Sober claim that “a good estimate of the probability on the right [of (6)] is furnished by frequency data; the same estimate is a good one for the probability on the left” (663). I claim that (6) is not in general true. One reason this is hard to see imme- diately is that (6) involves the counterfactual [If (S smokes) and (S gets can- mathematics. Learning new empirical information, like E, does not affect the value or re- silience of P(H F O&K). The value of this probability does not change, just as whether A&B entails A does not change. Rather, learning E simply makes a new probability rele- vant to what we should believe, namely, P(H F O&E&K), because now O&E&K describes our total evidence. 5. Immediately after this, McCain and Poston say, “Explanatory considerations are al- ready at work in setting Pr(H F O)—having E provide additional confirmation for H would be akin to double-counting the information about objective chances” (2014, 150). This suggests that they may be thinking of ‘Pr(H F O)’ as a frequency. But Roche and So- ber’s SOT is not about frequencies (although Roche and Sober may invite confusion on this point by moving back and forth between discussing epistemic probabilities and frequencies themselves). Or perhaps McCain and Poston are suggesting that the existence of an explan- atory connection between smoking and cancer is part of the background information K. (In this case P(H F O&E&K) wouldequalP(H F O&K) because K and E&K wouldbe logically equivalent.) However, the existence of an explanatory connection is neither directly ob- served nor entailed by facts that are directly observed. Hence, it should not be included in K. This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM se subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). DISCUSSION 363 cer) were true, (S smokes) would explain (S gets cancer)]. In ordinary circum- stances we would only get evidence for a counterfactual about a particular case like this by getting evidence for broader explanatory claims like [In gen- eral, smoking causes cancer]. As such, it will be helpful to start by considering claims of this form. Say that there is an explanatory connection between two phenomena X and Y if and only if, at least sometimes, X causes Y, Y causes X, or X and Y have a common cause.6 Let C1 be the hypothesis that smoking causes can- cer, C2 the hypothesis that cancer causes smoking, and C3 the hypothesis that they have a common cause. Then, ~[C1vC2vC3] is the claim that there is no explanatory connection between smoking and cancer. Let us now consider a revised Screening-Off Thesis, SOT*, which says that the existence of general explanatory connections is evidentially irrele- vant. In this context, SOT* says that 6. I h 7. Th All WhereC1 saysthatsometimes smoking causes cancer, P([S smokes]F [Sgets cancer]&C1&K) 5 P([S smokes] F [S gets cancer]&K). I will now show that SOT* is false in this context. First, note that Pð S gets cancer½ � S smokes½ �&∼ C1vC2vC3½ �&Kj Þ 5 P S gets cancer½ �ð j∼ C1vC2vC3½ �&KÞ:(7) This is because on the (extremely unlikely) hypothesis that there is no ex- planatory connection between smoking and cancer, the observed frequency data are a huge fluke. But we should not expect huge flukes to continue. If the observed association of smoking and cancer is merely coincidental, then we should expect future smokers that we observe to have cancer at the same rate as the rest of the population. So if we know that ~[C1vC2vC3], learning that S smokes does not raise the probability that S gets cancer above the probability given by the base rate of cancer in the population. An anonymous reviewer suggests the following objection to this argu- ment. Equation (7) is an instance of the more general principle (7) P(A F B&[there is no explanatory connection between A&B]&K) 5 P(A F [there is no explanatory connection between A&B]&K). However, this principle is subject to counterexample.7 Sober (2001) ob- serves that although there is no explanatory connection between the price ere ignore noncausal forms of explanation. is counterexample was originally applied to an analogous principle about frequencies. This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM use subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). 364 NEVIN CLIMENHAGA All u of bread in Britain and the height of the sea in Venice, they are nevertheless correlated: they both tend to increase over time. As such, if K reports the bread price in Britain and the sea level in Venice historically, B says that the sea level in Venice is x at some unspecified future time t, and A says that the bread price in Britain is y at t, then this equality is false. For example, learning that the sea level in Venice is much higher than at present raises the probability that the bread price is also much higher than at present. Assuming that time is not a common cause of A and B, then I accept the counterexample to the above principle, but I still hold that the principle is true in the current case. For, as Steel (2003, 313) observes, “British bread prices provide information about Venetian tides (and vice versa) only in vir- tue of telling us something about the time” (emphasis his). Knowledge of the time t screens off the bread prices from the sea levels. The above princi- ple is plausibly true when any relevant temporal information is built into our background K, which we can stipulate is the case in the smoking and cancer example. Even if this is not right, Sober (2001, 342–43) agrees that separate cause explanations “often” do not predict correlations, and I think he should accept that (7) is such a case. According to Sober, inference to a common cause is often rational because it is frequently the case that a common cause expla- nation predicts a correlation when a separate cause explanation does not. (Sober is considering cases in which it is obvious that neither A nor B causes the other, so a common cause is the only explanatory relation available.) However, it is clearly rational to infer a causal relationship between smoking and cancer from the frequency data in K. Hence, Sober’s reasoning would suggest that (7) is a case in which the above principle is true. It follows from (7) that Pð S smokes½ � S gets cancer½ �&∼ C1vC2vC3½ �&Kj Þ 5 P S smokes½ �ð j∼ C1vC2vC3½ �&KÞ:(8) Presumably learning ~[C1vC2vC3] on its own does not affect the probability of [S smokes] relative only to our background knowledge K: without any information about whether S has cancer, these two propositions are irrele- vant to each other. Hence, P S smokes½ � ∼ C1vC2vC3½ �&Kj Þ 5 P S smokes½ �ð jKð Þ: (9) From (8) and (9) it follows that P S smokes½ � S gets cancer½ �&∼ C1vC2vC3½ �&Kj Þ 5 P S smokes½ �ð jKð Þ: (10) In other words, learning that there is no explanatory connection between smoking and cancer and that S has cancer does not raise the probability that (8) This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM se subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). DISCUSSION 365 S smokes. However, according to (5), that S gets cancer raises the probabil- ity that S smokes. For (5) to be true, it must then be the case that learning that S smokes and that there is an explanatory connection between smoking and cancer raises the probability that S gets cancer. That is, from (5) and (10) it follows that P S smokes½ � S gets cancer½ �& C1vC2vC3½ �&Kj Þ > P S smokes½ �ð jKð Þ: (11) From (10) and (11) it follows that Pð S smokes½ � S gets cancer½ �& C1vC2vC3½ �&Kj Þ > P S smokes½ �ð j S gets cancer½ �&∼ C1vC2vC3½ �&KÞ,(12) which implies that Pð S smokes½ � S gets cancer½ �& C1vC2vC3½ �&Kj Þ > P S smokes½ �ð j S gets cancer½ �&KÞ:(13) Presumably any one of C1, C2, and C3 also licenses extrapolation from our frequency data. Consequently, we can replace C1vC2vC3 with any one of C1, C2, and C3, and (11)–(13) will remain true. In particular, it will be true that Pð S smokes½ � S gets cancer½ �&C1&Kj Þ > P S smokes½ �ð j S gets cancer½ �&KÞ:(14) Equation (14) contradicts SOT*. So SOT* is false.8 This example shows two other things. First, it shows that Roche and So- ber are wrong to claim (2013, 662) that the asymmetry of explanation sug- gests that explanatory facts like C1 cannot be confirmatory. As they observe, (12) (13) (14) 8. An anonymous reviewer suggests the following argument against (14): P([S smokes] F [S gets cancer]&K) should be equal to the frequency of cancer among smokers given by K. But then (14) implies that P([S smokes] F [S gets cancer]&C1&K) is greater than the fre- quency of cancer among smokers, which seems wrong. My response to this argument is to reject its first premise: K includes data on the correlations between cancer and smoking in previously observed cases. We are extrapolating from these data to a new case, and in gen- eral we should not follow the “straight rule” in so extrapolating (compare Roche and So- ber’s [2013, 663] coin toss example). This is clear in extreme cases: if all the people with cancer we have observed so far have been smokers, we should still not be 100% confident that the next person with cancer we observe will be a smoker. It is true that, as data accu- mulate, our new probabilities should tend to approach observed frequencies. But this is compatible with (14). The probability on the left-hand side of (14) is closer to the observed frequency of smoking among people with cancer than the probability on the right-hand side, but both values approach this frequency as the number of samples in K increases. This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM All use subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). 366 NEVIN CLIMENHAGA All u confirmation is symmetric whereas explanation is not. If X explains Y, Y does not explain X, but if P(X F Y&K) > P(X F K), P(Y F X&K) > P(Y F K). But this does not mean that [X explains Y] should not make a difference to the de- gree to which Y confirms X. In the above example, C1 raises the probability of [S gets cancer] not by ruling out C2 and C3 but by ruling out ~[C1vC2vC3]. So [X explains Y] can support Y not by ruling out [Yexplains X] but by rul- ing out [There is no explanatory connection between X and Y].9 Second, note that (10) and (11) together say that [S gets cancer] raises the probability that [S smokes] when it is conjoined with the claim that there is an explanatory connection between smoking and cancer but not when it is conjoined with the claim that there is no explanatory connection between them. In other words, the existence of an explanatory connection between cancer and smoking is precisely what licenses the inference from S’s smok- ing to S’s cancer. Moreover, (8) says that ~[C1vC2vC3] screens off [S gets cancer] from [S smokes]. The situation is thus almost the opposite of what Roche and Sober claim: not only does our observation not screen off our ex- planatory claim from our hypothesis, the negation of our explanatory claim (~[C1vC2vC3]) screens off our observation from our hypothesis. Our explan- atory claim thus mediates the move from observation to hypothesis.10 Equation (14) shows SOT* to be false. What about SOT? SOT implies equation (6) obtains, where E says that [If (S smokes) and (S gets cancer) were true, (S smokes) would explain (S gets cancer)]. If, by contrast, E is positively relevant to [S smokes], then Pð S smokes½ � S gets cancer½ �&E&Kj Þ > P S smokes½ �ð j S gets cancer½ �&KÞ:(15) We can break down the left-hand and right-hand sides of (6) and (15) as follows: P S smokes½ �j S gets cancer½ �&E&Kð Þ 5 P C1 S gets cancer½ �&E&Kj ÞP S smokes½ �ð j S gets cancer½ �&E&C1&Kð Þ 1 P ∼C1 S gets cancer½ �&E&Kj ÞP S smokes½ �ð j S gets cancer½ �&E&∼C1&Kð Þ, (16) (15) 9. It is compatible with this that the order of explanation is evidentially irrelevant, in that X confirms Y to the same degree regardless of what the explanatory relationship between them is. But even this does not follow from Roche and Sober’s observation that confirmation is symmetric. For confirmation is only qualitatively symmetric: X confirms Y if and only if Y confirms X. It is not plausibly quantitatively symmetric: in general, X does not confirm Y to the same degree that Y confirms X (Eells and Fitelson 2002). So, for all Roche and So- ber have said, X might well confirm Y more (or less) if Yexplains X than if X explains Y. 10. In Climenhaga (2017), I formalize the idea that explanatory connections mediate confirmation in terms of Bayesian networks. (16) This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM se subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). DISCUSSION 367 P S smokes½ �j S gets cancer½ �&Kð Þ 5 P C1 S gets cancer½ �&Kj ÞP S smokes½ �ð j S gets cancer½ �&C1&Kð Þ 1 P ∼C1 S gets cancer½ �&Kj ÞP S smokes½ �ð j S gets cancer½ �& ∼C1&Kð Þ: (17) Plausibly, P C1 S gets cancer½ �&E&Kj Þ > P C1ð j S gets cancer½ �&Kð Þ, (18) and Pð S smokes½ � S gets cancer½ �&E&C1&Kj Þ ≥ P S smokes½ �ð j S gets cancer½ �&C1&KÞ:(19) Equation (18) says that, given that S gets cancer, [(S smokes) would explain (S gets cancer) if (S smokes) were true] makes it more likely that there exists at least one instance of someone getting cancer because of smoking.11 Equation (19) says that, if we know that [S gets cancer]&C1&K, E does not make [S smokes] less likely. It follows from (18) and (19) that the first summand in (16) is greater than the first summand in (17). However, this does not yet show that (15) is true. This is because P S smokes½ �j S gets cancer½ �&E& ∼ C1&Kð Þ 5 0, (20) and so the second summand in (16) equals 0, and hence is less than the sec- ond summand in (17). Equation (20) is true because, if smoking never causes cancer, and S gets cancer, then it cannot be the case that S’s smoking causes S’s cancer. However, if it’s the case that, were S to smoke and get cancer, S’s smoking would be the cause of S’s cancer and it’s the case that S gets can- cer, then if S smokes, S’s smoking must cause S’s cancer. Hence, the only way for [S gets cancer]&E&~C1 to be true is for it to be the case that S does not smoke. Nevertheless, for (6) to be true the second summand in (17) would need to exactly equal the difference between the first summand in (16) and the (17) (19) 11. In an earlier version of this article, I claimed that E entails C1. However, an anony- mous reviewer pointed out to me that this is not true, even holding fixed the above back- ground knowledge. For example, suppose that smoking never has and never will cause cancer, so that C1 is false. S, for his part, does not smoke. Nevertheless, because of S’s unique physiology and the chemical properties of tobacco, it is true that were S to smoke, his smoking would cause him to have cancer. In this case E is true, but C1 remains false. However, inasmuch as a scenario like this in which E is true and C1 is false is incredibly unlikely, it remains extremely plausible that E confirms C1, even if it does not entail it. This content downloaded from 129.074.250.206 on September 05, 2017 14:04:04 PM All use subject to University of Chicago Press Terms and Conditions (http://www.journals.uchicago.edu/t-and-c). 368 NEVIN CLIMENHAGA All u first summand in (17). While this could be the case, there is no reason to expect it a priori. Hence, far from being a general truth, if (6) is true in this case it is only by fortuitous coincidence. More importantly, the negative influence of E on [S smokes] sketched above is not the kind of influence that either proponents or opponents of in- ference to the best explanation have had in mind when disagreeing about whether explanation is relevant to confirmation. And if we build into K infor- mation that screens off this influence, then (15) is true. For example, imagine that we know that nothing apart from smoking will give S cancer (and that S will not get cancer for no reason). In this case P(~C1 F [S gets cancer]&K) 5 0—if the only way for S to get cancer is from smoking, then if S gets cancer it is because of S’s smoking, and so C1 is true. Hence, the second summand in both (16) and (17) is 0, and the dominance of (16)’s first summand over (17)’sfirst summand is sufficient for it to be the case that equation (15) is true. I have argued in this article that Roche and Sober’s thesis that the explan- atory hypothesis [If H and O were true, H would explain O] is irrelevant to confirmation is not true in the kind of case they discuss, at least once we fix our background knowledge so as to screen off irrelevant information. I have also shown that when we move to more tractable propositions describing explanatory connections, such as those about general causal links between smoking and cancer, not only are these relevant to confirmation, they actually mediate the connection between observations and theories: the observation confirms the theory (and vice versa) only insofar as we have evidence that the described explanatory connection exists. Explanation not only adds confirma- tion; it guides confirmation. REFERENCES Climenhaga, Nevin. 2017. “The Structure of Epistemic Probabilities.” Unpublished manuscript, University of Notre Dame. ———. 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