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Show Rationality and the Structure of the Self, Volume II: A Kantian Conception 137 (ii) if ai = bi for every i, then you are indifferent between A and B (DTP 7). (3) Confidence: For any hypotheses P and Q, you are more confident that P than you are that Q iff you prefer ($1 if P, $0 if ~P) to ($1 if Q, $0 if ~Q) (DTP 8). (4) Decomposition: A sequence of well-mannered states of affairs φ is a decomposition of a well-mannered state of affairs A (i.e. A is composable from φ) just in case φ is finite and it is logically impossible that the realization of φ will effect … a different net change in your fortune than the realization of A will. You place a monetary value of $a on A just if you are indifferent between $a and A. Then if (i) A is a well-mannered state of affairs; (ii) φ is a decomposition of A; and (iii) you place a monetary value on A and on each of the terms of φ; then the value you place on A is equal to the sum of the values you place on the terms of φ (DTP 10). (5) Modest Connectedness: Your preferences are characterized by a non-empty set V of assignments of monetary values to all well-mannered states of affairs such that each assignment satisfies Ordering, Dominance and Decomposition, and such that (i) you are indifferent between A and B just in case every member of V assigns A the same monetary value as it assigns B; and (ii) you prefer A to B just in case no member of V assigns B a greater monetary value than it assigns A and some member of V assigns A a greater value than it assigns B (DTP 13). From (3) and (5), Kaplan derives (6) Modest Probabilism: Any assignment of monetary values to all well-mannered states of affairs that satisfies Ordering, Dominance and Decomposition assigns a real number to con(P) for every hypothesis P such that, for any hypotheses P and Q in a non-empty set W of conassignments, (i) con(P) ≥ 0; (ii) if P is a tautology, then con(P) = 1; (iii) if P and Q are mutually exclusive, then con(P v Q) = con(P) + con(Q); (iv) you are just as confident that P as you are that Q just in case, on every member of W, con(P) = con(Q); and (v) you are more confident that P than that Q just in case, on no member of W, con(Q) > con(P) and, on at least one member of W, © Adrian Piper Research Archive Foundation Berlin |
