In the above sections we have discussed causal conditionals like
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(16)
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a.
If John is nervous, he smokes.
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b.
If fire, then smoke.
In terms of our causal structure \(i \rightarrow e \leftarrow a\) they are of the form ‘If i, then e’. But many natural language conditionals are stated in the reverse form ‘If e, then i’:
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(17)
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a.
If John smokes, he is nervous.
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b.
If smoke, then fire.
The clear intuition for the above conditional is that it is appropriate because esignals that i is the case. There exists an evidential, or diagnostic, but no causal, dependence relation from e to i. Given that causation is asymmetric, and that we analyze conditionals in terms of causal powers, one wonders how we could analyze such conditionals. Before we will delve into that question, let us first remind ourselves that there is another empirical fact to be explained. Evans et al. (2007) and Douven and Verbrugge (2010) have shown experimentally that the acceptability of such, so-called, diagnostic conditionals is, although not identical to, still very close to the conditional probability of e given i.Footnote 17 Thus, we would like to investigate the following two questions:
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1.
Can we explain the appropriateness of diagnostic conditionals in terms of causal powers? And if so,
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2.
can we explain the relatively strong correlation that exists between the assertability/acceptability of such conditionals on the one hand, and the conditional probability of consequent given antecedent on the other?
In this section we will provide an explanation for both on the assumption that diagnostic conditionals should have somewhat different acceptability requirements than causal conditionals.
We will assume that the probability of evidential, or diagnostic, conditionals should be measured by what Cheng et al. (2007) determine as the probability that i caused e, or perhaps by what they determine as the probability that \(i \ alone\) caused e. To see what this comes down to, let us first determine the probability that given e, e is due to i, \(P(i \leadsto e|e)\). Given that e is caused by i with probability \(P(i) \times p_{ie}\), this can be given as follows:Footnote 18\(^,\)Footnote 19
$$\begin{aligned} P(i \leadsto e|e) = \frac{P(i) \times p_{ie}}{P(e)}. \end{aligned}$$
(18)
Recall that we have claimed in Sect. 2 that the conditional ‘If i, then e’ is appropriate only if \(\Delta ^* P^e_i\) is high, which means that \(\Delta ^* P^e_i>\!> \Delta ^* P^e_a\). Assuming a reading, or appropriateness condition, of diagnostic conditionals in terms of \(P(i \leadsto e|e)\), this means that ‘If e, then i’ is appropriate only if \(P(i \leadsto e|e)>\!> P(a \leadsto e|e)\). Now suppose that we assume that \(P(i) \approx P(a)\). Then it follows that \(P(i \leadsto e|e)>\!> P(a \leadsto e|e)\) iff \(p_{ie}>\!> p_{ae}\). Thus, from the fact that ‘If e then i’ is a good conditional, together with the commonly assumed causal structure \(i \rightarrow e \leftarrow a\), it follows that i is taken to be the best causal explanation of e, at least if \(P(i) \approx P(a)\).
Because \(p_{ie}>\!> p_{ae}\) it follows that \(P(e|\lnot i) = P(a) \times p_{ae}\) will be low, and thus that \(P(e|i) - P(e|\lnot i)\) will be close to P(e|i). Because for the same reason \(1 - P(e|\lnot i)\) will be close to 1, \(p_{ie} = \frac{P(e|i) - P(e|\lnot i)}{1 - P(e|\lnot i)}\) will be close to P(e|i). In this way we have explained the experimental results of Evans et al. (2007) and Douven and Verbrugge (2010) that the acceptability of diagnostic conditionals like (17-a)-(17-b) correlates well with the conditional probability of consequent given antecedent.
Similar conclusions follow when we account for acceptability of diagnostic conditionals in terms of the probability that given e, e is due to ialone, \(P(i \ alone \ \leadsto e|e)\), which abbreviates \(P((i \wedge \lnot (i \wedge a)) \ \leadsto e|e)\). To derive the latter notion, recall from a previous section that if i and a are independent of each other,
$$\begin{aligned} P(e) = P(i) \times p_{ie} + P(a) \times p_{ae} - P(i) \times p_{ie} \times P(a) \times p_{ae}. \end{aligned}$$
(19)
Because we have seen above that \(P(e|\lnot i)\) estimates \(P(a) \times p_{ae}\), this reduces to
$$\begin{aligned} P(e) = P(i) \times p_{ie} + P(e|\lnot i) - P(i) \times p_{ie} \times P(e|\lnot i). \end{aligned}$$
(20)
Given that e is caused by i with probability \(P(i) \times p_{ie}\) and that e is caused by \(i \wedge a\) with probability \(P(i) \times p_{ie} \times P(e|\lnot i)\), it follows that
$$\begin{aligned} P(i \ alone \ \leadsto e|e)= & {} \frac{P(i) \times p_{ie} - P(i) \times p_{ie} \times P(e|\lnot i)}{P(e)}\nonumber \\= & {} \frac{P(i) \times p_{ie} \times [1 - P(e|\lnot i)]}{P(e)}. \end{aligned}$$
(21)
Now, recall that under independence conditions we have derived the following for \(p_{ie}\)
$$\begin{aligned} p_{ie} = \frac{\Delta P^e_i}{1 - P(e|\lnot i)} = \frac{P(e|i) - P(e|\lnot i)}{1 - P(e|\lnot i)}. \end{aligned}$$
(22)
Substituting this measure in the above formula gives us
$$\begin{aligned} P(i \ alone \ \leadsto e|e) = \frac{P(i) \times \Delta P^e_i}{P(e)}. \end{aligned}$$
(23)
We want to conclude from sentences (17-a)-(17-b) of the form ‘If e then i’ that \(P(i \ alone\ \leadsto e|e)>\!> P(a \ alone\ \leadsto e|e)\). This holds iff \(\frac{P(i) \times \Delta P^e_i}{P(e)}>\!> \frac{P(a) \times \Delta P^e_a}{P(e)}\). If we now assume that \(P(i) \approx P(a)\), it follows that \(P(i \ alone \ \leadsto e|e)>\!> P(a \ alone\ \leadsto e|e)\) iff \(\Delta P^e_i>\!> \Delta P^e_a\). Thus, from the fact that ‘If e then i’ is a good conditional, together with the commonly assumed causal structure \(i \rightarrow e \leftarrow a\), it follows again that i is taken to be the best causal explanation of e, at least if \(P(i) \approx P(a)\).
Now, can we derive from this also that the acceptability of the conditional goes with conditional probability? Yes, this is the case, because
$$\begin{aligned} P(i \ alone \ \leadsto e|e)= & {} \frac{P(i) \times \Delta P^e_i}{P(e)}\nonumber \\= & {} \frac{[P(e|i) \times P(i)] - P(e|\lnot i) \times P(i)}{P(e)}\nonumber \\= & {} \frac{P(e \wedge i) - P(e|\lnot i) \times P(i)}{P(e)}. \end{aligned}$$
(24)
Because \(P(i \ alone\ \leadsto e|e)>\!> P(a \ alone\ \leadsto e|e)\) it follows that \(P(e|\lnot i) = P(a) \times p_{ae}\) will be low, and thus that \(P(i \ alone \ \leadsto e|e)\) will be close to \(\frac{P(e \wedge i)}{P(e)} = P(i|e)\). In this way we have explained the experimental results of Evans et al. (2007) and Douven and Verbrugge (2010) that the acceptability of diagnostic conditionals like (17-a)-(17-b) correlates well with the conditional probability of consequent given antecedent.
As usual, if we assume that i is the only potential cause of e, or if we assume incompatibility of i and a instead of independence, the derivations of our two desired conclusions are much easier.
If i is (taken to be) the only potential cause of e, the inference is trivial. First, assume that the diagnostic conditional is interpreted in terms of \(P(i \leadsto e|e) = \frac{P(i) \times p_{ie}}{P(e)}\). We have seen before that in case i is the only potential cause of e, \(p_{ie} = P(e|i)\). But this means that \(P(i \leadsto e|e) = \frac{P(i) \times P(e|i)}{P(e)} = \frac{P(i \wedge e)}{P(e)} = P(i|e)\). This indicates that (17-a) is good if P(i|e) is high, or significantly higher than P(e|a), which was what we had to explain.
So far so good, but what if we assume that the diagnostic conditional is interpreted in terms of \(P(i \ alone \ \leadsto e|e)\)? Recall that in general
$$\begin{aligned} P(i \ alone \ \leadsto e|e) = \frac{P(i) \times p_{ie} \times [1 - P(e|\lnot i)]}{P(e)}. \end{aligned}$$
(25)
If i is the only potential cause of e, \(P(e|\lnot i) = 0\), and thus \(P(i \ alone \ \leadsto e|e) = \frac{P(i) \times p_{ie}}{P(e)} = \frac{P(i \wedge e)}{P(e)} = P(i|e)\), because if i is the only cause of e, \(p_{ie} = P(e|i)\).
For the case that i and a are incompatible, the derivation is equally straightforward. First, notice that in that case \(P(i \ alone \ \leadsto e|e) = P(i \leadsto e|e) = \frac{P(i) \times p_{ie}}{P(e)}\). We have seen in the previous section that in case i and a are incompatible, \(p_{ie} = P(e|i)\). It follows that \(P(i \ alone \ \leadsto e|e) = P(i \leadsto e|e) = \frac{P(i) \times P(e|i)}{P(e)} = P(i|e)\). This indicates, again, that (17-a)-(17-b) are good if P(i|e) is high, or significantly higher than P(e|a).
In this section we have assumed, so far, that conditionals of the form ‘If i, then e’ are ambiguous: they are appropriate either due to high \(p_{ie} = \Delta ^* P^e_i\), or due to high \(P(e \leadsto i|i)\). But are (the appropriateness conditions of) conditionals really ambiguous in this way? Can’t we interpret them uniformly in terms of high \(\Delta ^* P^e_i\)? Given the form of the conditional, this means that we want to explain that \(P(e \leadsto i|i)\) is high on the assumption that \(\Delta ^* P^e_i\) is high. It turns out that we can.
Because the conditional is appropriate, it follows that \(\Delta ^* P^e_i = \frac{ P(e|i) - P(e|\lnot i)}{1 - P(e|\lnot i)}\) is high. Now recall from Sect. 2 that \(\Delta ^* P^e_i = \frac{ P(e|i) - P(e|\lnot i)}{1 - P(e|\lnot i)} = \frac{ P(e|i) - P(e)}{P(\lnot e \wedge \lnot i)}\). Similarly, \(\Delta ^* P^i_e = \frac{ P(i|e) - P(i)}{P(\lnot e \wedge \lnot i)}\). One can prove that \(P(e|i) - P(e) = \frac{P(e)}{P(i)} \times [P(i|e) - P(i)]\),Footnote 20 and thus that \(\Delta ^*P^e_i = \frac{P(e)}{P(i)} \times \Delta ^* P^i_e\). But recall that under suitable independence conditions \(\Delta ^* P^i_e = p_{ei}\). It follows that \(\Delta ^*P^e_i = \frac{P(e)}{P(i)} \times p_{ei} = \frac{P(e) \times p_{ei}}{P(i)} = P(e \leadsto i|i)\). Thus, under suitable independence conditions, as far as the numbers are concerned, \(\Delta ^*P^e_i = p_{ie} = P(e \leadsto i|i)\)! As a consequence, a conditional of the form ‘If i, then e’ is appropriate only if \(\Delta ^* P^e_i\) is high, which means that depending on the (assumed) causal structure, either \(p_{ie}\) is high, or \(P(e \leadsto i|i)\) is high! We can conclude that we don’t need two different types of conditions for a conditional of the form ‘If i, then e’ to be appropriate. One condition will do, and depending on the assumed causal structure, it will give rise to the desired causal reading. We have seen that only under some strong conditions high \(\Delta ^* P^e_i\) gives rise to high P(e|i) under the causal forward reading of the conditional, but also that it more naturally gives rise to high P(e|i) under an diagnostic, or evidential, reading. We take this to be in accordance with the experimental observations of Douven and Verbrugge (2010).
Of course, for a conditional of the form ‘If i then e’ to be appropriate it need not be the case that (i) i is a cause of e, or that (ii) e is the cause of i. It might simply be the case that there exists a semantic or deductive relation between i and e: if someone is a bachelor, he is an unmarried man, and if y + 2 = 6, then y = 4. Another reason why a conditional can be true might be due to what is sometimes called ‘metaphysical grounding’. The experimental data of Douven and Verbrugge (2010) show that for such cases there exists a strong correlation between acceptance of the conditional and its conditional probability. It is also quite clear why: both acceptance and conditional probabilities will be 1. Let us concentrate therefore on a more challenging type of ‘empirical’ conditionals whose acceptability is due neither to (i) or (ii): the case where there is a common cause of i and e. Suppose that we have a causal structure of the form \(i \leftarrow c \rightarrow e\), and thus a structure where i neither causes nor is caused by e. To make it concrete, let i stand for the falling barometer, e for storm, and c for low pressure. With this instantiation of the variables, the conditional ‘If i, then e’ is appropriate. We don’t know whether there exists a strong correlation between acceptance of the conditional and the corresponding conditional probability for such cases, but we can makes clear under which circumstances this could be the case.
It seems natural that the probability of the conditional ‘If i, then e’ can now be measured by \(P(c \leadsto i|i) \times p_{ce}\). This is \(\frac{P(c) \times p_{ci}}{P(i)} \times p_{ce} = \frac{P(c) \times p_{ci} \times p_{ce}}{P(i)}\). We have seen above that under natural conditions this is nothing but \(\frac{P(c \wedge i \wedge e)}{P(i)} = P(c \wedge e| i)\). If we now assume that (almost) only c can cause i, and thus that \(P(c|i) \approx 1\), \(P(c \leadsto i|i) \times p_{ce}\) comes down to P(e|i), the conditional probability.