In two recent essays, Roche and Sober (Reference Roche and Sober2013, Reference Roche and Sober2014) argue that the proposition that a hypothesis H would explain an observation O is evidentially irrelevant to H. Where E says that, were H and O true, H would explain O, Roche and Sober’s thesis is that

(1)$\begin{array}{}P\left(H\right|\mathrm{O\&E})=P(H\left|O\right)\mathrm{.}\end{array}$

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 Reference Roche and Sober2014, 193).

In endorsing SOT, Roche and Sober are presumably not making a claim about subjective probabilities, for an agent could virtually always assign coherent subjective probabilities on which the above equality is false. I will instead 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 inference to the best explanation. For example, van Fraassen (Reference van Fraassen1989, 166) famously 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. Van Fraassen does consider a precise version of inference to the best explanation, but it is an uncharitable one, on which inference to the best explanation is understood as a non-Bayesian updating rule on which good explanations get higher probabilities than Bayesian conditionalization would give them.Footnote ¹ Roche and Sober are thus to be commended for stating a precise antiexplanationist 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 proposition being conditioned on.Footnote ² So we can rewrite Roche and Sober’s claim as

(2)$\begin{array}{}P\left(H\right|\mathrm{O\&E\&K})=P(H\left|\mathrm{O\&K}\right)\mathrm{.}\end{array}$

Here H, O, and K are variables that could be filled in with various propositions 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 | O&E&K) = P(H | O&K). However, this interpretation is uncharitable because it is trivially false. For example, suppose K includes the material conditional [(O&E) ⊃ H] but does not include the material 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 | O&E&K) = 1, but P(H | O&K) < 1. Hence,Footnote ³

(3)$\begin{array}{}P\left(H\right|\mathrm{O\&E\&K})>P(H\left|\mathrm{O\&K}\right)\mathrm{.}\end{array}$

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 explanation 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 (Reference Roche and Sober2013, 660) and I agree that in this case

(4)$\begin{array}{}P\left(\right[\mathrm{S\; gets\; cancer}\left]\right|\left[\mathrm{S\; smokes}\right]\mathrm{\&K})>P(\left[\mathrm{S\; gets\; cancer}\right]\left|K\right)\mathrm{,}\end{array}$

which implies

(5)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&K})>P(\left[\mathrm{S\; smokes}\right]\left|K\right)\mathrm{.}\end{array}$

In a response to Roche and Sober (Reference Roche and Sober2013), McCain and Poston (Reference McCain and Poston2014, 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 cancer:Footnote ⁴ “The data indicated that there was some causal process—albeit unknown at that time—that explains the correlation between smoking and lung cancer. … Exactly this feature—a justified belief in an unknown explanatory story—plays a crucial role in using the data from observation to get justified beliefs about the relevant frequencies. Apart from a general justified belief in some explanatory story accounting for that data, the observational data would not justify beliefs about the relevant frequencies.” Although McCain and Poston are talking about objective frequencies and not epistemic probabilities in this quote, it is similarly true that [S smokes] supports [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 reason 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 confirmation” (Reference McCain and Poston2014, 150).Footnote ⁵ But it does not show this. Rather, it shows that explanatory 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 [Reference Roche and Sober2014] response to McCain and Poston.)

Roche and Sober (Reference Roche and Sober2013, 661) hold that, as SOT implies in this case,

(6)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E\&K})=P(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&K}\right)\mathrm{,}\end{array}$

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 immediately is that (6) involves the counterfactual [If (S smokes) and (S gets cancer) were true, (S smokes) would explain (S gets cancer)]. In ordinary circumstances we would only get evidence for a counterfactual about a particular case like this by getting evidence for broader explanatory claims like [In general, 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.Footnote ⁶ Let C₁ be the hypothesis that smoking causes cancer, 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 irrelevant. In this context, SOT* says that

I will now show that SOT* is false in this context.

First, note that

(7)$\begin{array}{}P\left(\right[\mathrm{S\; gets\; cancer}\left]\right|\left[\mathrm{S\; smokes}\right]\&~\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})=P(\left[\mathrm{S\; gets\; cancer}\right]\left|~\right[C_{1}v{C}_{2}v{C}_3\left]\mathrm{\&K}\right)\mathrm{.}\end{array}$

This is because on the (extremely unlikely) hypothesis that there is no explanatory 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 argument. Equation (7) is an instance of the more general principle

However, this principle is subject to counterexample.Footnote ⁷ Sober (Reference Sober2001) observes that although there is no explanatory connection between the price 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 (Reference Steel2003, 313) observes, “British bread prices provide information about Venetian tides (and vice versa) only in virtue 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 principle 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 (Reference Sober2001, 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 explanation 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

(8)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\&~\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})=P(\left[\mathrm{S\; smokes}\right]\left|~\right[C_{1}v{C}_{2}v{C}_3\left]\mathrm{\&K}\right)\mathrm{.}\end{array}$

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 irrelevant to each other. Hence,

(9)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|~\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})=P(\left[\mathrm{S\; smokes}\right]\left|K\right)\mathrm{.}\end{array}$

From (8) and (9) it follows that

(10)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\&~\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})=P(\left[\mathrm{S\; smokes}\right]\left|K\right)\mathrm{.}\end{array}$

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 S smokes. However, according to (5), that S gets cancer raises the probability 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

(11)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\&\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})>P(\left[\mathrm{S\; smokes}\right]\left|K\right)\mathrm{.}\end{array}$

From (10) and (11) it follows that

(12)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\&\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})>P(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\&~\right[C_{1}v{C}_{2}v{C}_3\left]\mathrm{\&K}\right)\mathrm{,}\end{array}$

which implies that

(13)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\&\left[C_{1}v{C}_{2}v{C}_3\right]\mathrm{\&K})>P(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&K}\right)\mathrm{.}\end{array}$

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

(14)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\&C_1\mathrm{\&K})>P(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&K}\right)\mathrm{.}\end{array}$

Equation (14) contradicts SOT*. So SOT* is false.Footnote ⁸

This example shows two other things. First, it shows that Roche and Sober are wrong to claim (Reference Roche and Sober2013, 662) that the asymmetry of explanation suggests that explanatory facts like C₁ cannot be confirmatory. As they observe, confirmation is symmetric whereas explanation is not. If X explains Y, Y does not explain X, but if P(X | Y&K) > P(X | K), P(Y | X&K) > P(Y | K). But this does not mean that [X explains Y] should not make a difference to the degree 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 [Y explains X] but by ruling out [There is no explanatory connection between X and Y].Footnote 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 smoking 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 explanatory claim from our hypothesis, the negation of our explanatory claim (~[C₁vC₂vC₃]) screens off our observation from our hypothesis. Our explanatory claim thus mediates the move from observation to hypothesis.Footnote 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

(15)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E}\mathrm{\&K})>P(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&K}\right)\mathrm{.}\end{array}$

We can break down the left-hand and right-hand sides of (6) and (15) as follows:

(16)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E\&K})=P\left(C_1\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E}\mathrm{\&K}\left)P\right(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&E}\&C_1\mathrm{\&K}\right)+P\left(~C_1\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E}\mathrm{\&K}\left)P\right(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&E}\&~C_1\mathrm{\&K}\right)\mathrm{,}\end{array}$

(17)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&K})=P\left(C_1\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&K}\left)P\right(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\&C_1\mathrm{\&K}\right)+P\left(~C_1\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&K}\left)P\right(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\&~C_1\mathrm{\&K}\right)\mathrm{.}\end{array}$

Plausibly,

(18)$\begin{array}{}P\left(C_1\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E}\mathrm{\&K})>P(C_1\left|\right[\mathrm{S\; gets\; cancer}\left]\mathrm{\&K}\right),\end{array}$

and

(19)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E}\&C_1\mathrm{\&K})\ge P(\left[\mathrm{S\; smokes}\right]\left|\right[\mathrm{S\; gets\; cancer}\left]\&C_1\mathrm{\&K}\right)\mathrm{.}\end{array}$

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.Footnote 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

(20)$\begin{array}{}P\left(\right[\mathrm{S\; smokes}\left]\right|\left[\mathrm{S\; gets\; cancer}\right]\mathrm{\&E}\&~C_1\mathrm{\&K})=0\mathrm{,}\end{array}$

and so the second summand in (16) equals 0, and hence is less than the second 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 cancer, 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 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 inference to the best explanation have had in mind when disagreeing about whether explanation is relevant to confirmation. And if we build into K information 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 | [S gets cancer]&K) = 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)’s first 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 explanatory 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 confirmation; it guides confirmation.