The Veiled Raven: Bayesianism and the Limits of Evidential Relevance — Epoche C1
Hempel's paradox of the ravens is the claim that a white shoe, examined and found to be a white shoe, counts as evidence that all ravens are black. This essay does three things with it. It derives the standard Bayesian response completely, arriving at a definite number for how much less a shoe is worth than a raven. It corrects the diagnosis offered in the earlier version of this argument, which located the weakness of that response in the choice of prior probabilities; the comparative result is in fact independent of the prior, and the vulnerability lies elsewhere. And it identifies where the vulnerability actually lies — in assumptions about how the observed object came to be observed — using a counterexample of I. J. Good's in which a black raven is evidence against the hypothesis that all ravens are black. The Paradox and Its Two Premises The argument is short, and it must be laid out exactly, because everything afterwards is a question of which premise to give up. Nicod's criterion holds that a universal generalisation of the form 'all $F$ are $G$' is confirmed by the observation of an object that is both $F$ and $G$, and disconfirmed by one that is $F$ and not $G$. Observing a black raven therefore confirms that all ravens are black. The equivalence condition holds that if two statements are logically equivalent, whatever confirms one confirms the other; it is hard to reject, since to deny it is to make confirmation depend on how a hypothesis is phrased rather than on what it says. Now take the hypothesis $H$: all ravens are black. Its contrapositive — all non-black things are non-ravens — is logically equivalent to it, since both say that nothing is a raven and non-black. By Nicod's criterion, the second is confirmed by observing an object that is both non-black and a non-raven. By the equivalence condition, that observation confirms $H$ too. A white shoe is a non-black non-raven. Hence the white shoe confirms that all ravens are black, and one may do ornithology indoors. Hempel (1945) did not treat this as a reductio. He accepted the conclusion and located the discomfort in the reader rather than in the argument, holding that we tacitly import background knowledge — in particular, that we already know shoes are not ravens — and that stripped of such knowledge the inference is unobjectionable. The Bayesian response goes further: it accepts the conclusion and then measures it. The Calculation, Performed The earlier version reported that the Bayesian shows the shoe to confirm 'to a very small degree', without saying how small or why. The degree can be computed, and the computation is the heart of the matter. Set up a definite model. An object is drawn at random from the universe of objects and both of its relevant properties are noted. Write $R$ for 'is a raven', $B$ for 'is black', and let $$p = P(R), \qquad q = P(\neg B)$$ be the proportions of ravens and of non-black things among all objects. Assume, as the standard treatment does, that these two proportions are the same whether or not $H$ is true — an assumption that will be scrutinised below, since it is where the argument is weakest. Let $\epsilon$ be the proportion of ravens that are non-black if $H$ is false, and $\delta$ the proportion of non-black things that are ravens if $H$ is false. Consider first the black raven, $E_1 = R \wedge B$. If $H$ is true, every raven is black, so $$P(E_1 \mid H) = P(R)\,P(B \mid R, H) = p \cdot 1 = p,$$ whereas if $H$ is false a fraction $\epsilon$ of ravens fail to be black, giving $P(E_1 \mid \neg H) = p(1 - \epsilon)$. The likelihood ratio — the factor by which the observation multiplies the odds on $H$, which is what Bayes's theorem shows the evidence contributes — is therefore $$\Lambda_1 = \frac{P(E_1 \mid H)}{P(E_1 \mid \neg H)} = \frac{1}{1 - \epsilon}.$$ Now the white shoe, $E_2 = \neg B \wedge \neg R$. If $H$ is true then every non-black thing is a non-raven, so by exactly the parallel reasoning $$P(E_2 \mid H) = P(\neg B)\,P(\neg R \mid \neg B, H) = q, \qquad P(E_2 \mid \neg H) = q(1-\delta), \qquad \Lambda_2 = \frac{1}{1-\delta}.$$ One correction to the earlier version is already visible. It stated that the probability of encountering a non-black non-raven is slightly reduced if all ravens are black. The opposite is true: under $H$ every non-black object is a non-raven, so non-black non-ravens are marginally more probable, which is precisely why $\Lambda_2$ exceeds one and the shoe confirms at all. The comparison is settled by relating $\epsilon$ and $\delta$, and they are related because they count the same objects. Suppose there are $N$ objects in all and $k$ non-black ravens. Then there are $Np$ ravens and $Nq$ non-black things, so $$\epsilon = \frac{k}{Np}, \qquad \delta = \frac{k}{Nq}, \qquad \frac{\delta}{\epsilon} = \frac{p}{q}.$$ The count $k$ cancels; only the two base rates remain. For small $\epsilon$ and $\delta$ the logarithms of the likelihood ratios are approximately $\epsilon$ and $\delta$ themselves, so the black raven outweighs the white shoe by a factor of about $q/p$: the ratio of how common non-black things are to how common ravens are. Put numbers to it. Ravens are a vanishingly small fraction of the objects in the universe — take $p \approx 10^{-8}$ — while a large majority of objects are not black, say $q \approx 0.9$. Then $$\frac{\Lambda_1 - 1}{\Lambda_2 - 1} \approx \frac{q}{p} \approx 9 \times 10^{7}.$$ A black raven is worth roughly a hundred million white shoes. This is the substance of the Bayesian dissolution, and it is a satisfying one: the intuition that the shoe is negligible is not dismissed as bias but vindicated and quantified. The intuition was wrong only in treating negligible as zero. Hosiasson-Lindenbaum (1940) gave the first argument of essentially this form, five years before Hempel's paper appeared. Correcting the Diagnosis: It Is Not the Prior Here the earlier version's central criticism must be redirected, because as aimed it misses.