1. HEMPEL’S RAVEN PARADOX A lacuna in the standard Bayesian solution Peter B. M. Vranas Iowa State University PSA’02, 8 November 2002

2. OVERVIEW PART 1 Hempel’s paradox, the Bayesian solution, and the disputed assumption PART 2 Two attempts to defend the assumption PART 3 Three attempts to refute the assumption CONCLUSION A possible way out for Bayesians

3. HEMPEL’S PARADOX l Nicod’s Condition: The proposition that an object x has properties F and G confirms the proposition that every F has G. l Equivalence Condition: If the proposition P confirms Q and Q is logically equivalent to Q', then P confirms Q'. ___________________________________________________________________________________________________________________________________________________________ l Paradoxical Conclusion: The proposition (E) that a is a nonblack nonraven confirms the proposition (H) that every raven is black.

4. THE BAYESIAN SOLUTION & THE DISPUTED ASSUMPTION l H   x(Rx  Bx) (every raven is black). l E  ~Ba~Ra (a is a nonblack nonraven). l P(Ra|~Ba) =  2 (positive but minute)—there are many more nonblack objects than ravens. The disputed assumption: P(Ba|H) = P(Ba). Equivalently: P(~Ba|H) = P(~Ba). Then: P(H|E) - P(H) = P(H)  2 /(1-  2 ), which is both positive and minute.

5. THE ASSUMPTION IS NEEDED FOR THE BAYESIAN SOLUTION l P(H|E) - P(H) is both positive and minute iff: 0 < P(H|E) - P(H) <  2 (minute). l Equivalently: 1-  2 < P(~Ba|H)/P(~Ba) < (1-  2 )[1+(  2 /P(H))]. l Equivalently: P(~Ba|H)/P(~Ba)  1. l Equivalently: P(~Ba|H)  P(~Ba).

6. OVERVIEW PART 1 Hempel’s paradox, the Bayesian solution, and the disputed assumption PART 2 Two attempts to defend the assumption PART 3 Three attempts to refute the assumption CONCLUSION A possible way out for Bayesians

7. FIRST ATTEMPT TO DEFEND THE ASSUMPTION Let P + be my subjective probability measure right after I learn that H is true. l Woodward’s (1985) argument: (1) P + (Ba) = P(Ba) (2) P + (Ba) = P(Ba|H) ______________________________________________________________________ (3) P(Ba) = P(Ba|H) l My reply: One person’s modus ponens is another’s modus tollens. Those who deny (3) will also deny (1).

8. SECOND ATTEMPT TO DEFEND THE ASSUMPTION l Principle of Conditional Indifference (PCI): If none of H 1, H 2 gives more support than the other to E, then P(E|H 1 ) = P(E|H 2 ). l So P(Ba|H) = P(Ba|~H), which is equivalent to the assumption P(Ba|H) = P(Ba). l My reply: Compare PCI to the Principle of Indifference (PI): If E gives no more support to one of H 1, H 2 than to the other, then P(H 1 |E) = P(H 2 |E). PI leads to inconsis- tencies, and so does PCI.

9. WEAK & STRONG VERSIONS OF THE ASSUMPTION l Weak: P(Ba|H) and P(Ba) may be equal. Strong: P(Ba|H) and P(Ba) should be equal. l The strong version is needed for the claim that E confirms H; i.e., that P(H|E) should exceed P(H). l But the strong version is hard to defend. Even if one refutes the claim that P(Ba|H) and P(Ba) should differ, all that follows is that they may be equal, namely the weak version.

10. OVERVIEW PART 1 Hempel’s paradox, the Bayesian solution, and the disputed assumption PART 2 Two attempts to defend the assumption PART 3 Three attempts to refute the assumption CONCLUSION A possible way out for Bayesians

11. FIRST ATTEMPT TO REFUTE THE ASSUMPTION l Horwich’s (1982) argument: The assumption has the counterintuitive consequence that H is slightly disconfirmed by the proposition that a is a black nonraven. l My reply: To the charge that on their account E confirms H, Bayesians reply that the degree of confirmation is minute. Similarly, to the charge that on their account Ba~Ra disconfirms H, Bayesians can reply that the degree of disconfirmation is minute.

12. SECOND ATTEMPT TO REFUTE THE ASSUMPTION Let (again) P + be my subjective probability measure right after I learn that H is true. l Swinburne’s (1971) argument: (1') P + (Ba) > P(Ba) (2) P + (Ba) = P(Ba|H) ______________________________________________________________________ (3') P(Ba|H) > P(Ba) l My reply: One person’s modus ponens is another’s modus tollens. Those who deny (3') will also deny (1').

13. THIRD ATTEMPT TO REFUTE THE ASSUMPTION l Maher’s (1999) argument: (1) P(~Ba|H) = P(~BaRa|H) +P(~Ba~Ra|H) (2) P(~Ba|~H) = P(~BaRa|~H)+P(~Ba~Ra|~H) _________________________________________________________________________________________________________________________________________________________ (3) P(~Ba|H) < P(~Ba|~H); equivalently, P(~Ba|H) < P(~Ba), so the assumption fails. l My reply: How does (3) follow, unless P(~Ba~Ra|H) = P(~Ba~Ra|~H)? But this is equivalent to P(H|~Ba~Ra) = P(H) and thus begs the question against Bayesians.

14. OVERVIEW PART 1 Hempel’s paradox, the Bayesian solution, and the disputed assumption PART 2 Two attempts to defend the assumption PART 3 Three attempts to refute the assumption CONCLUSION A possible way out for Bayesians

15. CONCLUSION: A POSSIBLE WAY OUT FOR BAYESIANS l Prescriptive question: Does E confirm H? The Bayesian answer (“yes, but marginally”) is incorrect if the disputed assumption is false. l Explanatory question: Why do people believe that P(H|E) = P(H)? The Bayesian answer (“because they mistake marginal confirmation for confirmational irrelevance”) can be supplemented with: “because they mistakenly take the assumption to be true (and this is because they reason by indifference)”.