Bayesian epistemology II: Arguments for Probabilism

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1 Bayesan epstemology II: Arguments for Probablsm Rchard Pettgrew May 9, The model Represent an agent s credal state at a gven tme t by a credence functon c t : F [0, 1]. where F s the algebra of propostons about whch the agent has an opnon. 1 If A F, then c t (A) = 0 ff the agent has mnmal credence n A at t. If A F, then c t (A) = 1 ff the agent has maxmal credence n A at t. Note: It s an emprcal assumpton that agents are capable of maxmal and mnmal credences; t s not a normatve clam. 2 The norms At any tme t n her epstemc lfe, an agent ought to have a credence functon c t such that Probablsm c t s a probablty functon. That s, c t ( ) = 0 and c t ( ) = 1. c t (A B) = c t (A) + c t (B) f A and B are mutually exclusve. Countable addtvty c t s countably addtve. That s, f F s nfnte, c t ( n A n ) = n c t (A n ) f A 1, A 2,... are parwse mutually exclusve. 1 Snce F s an algebra, t s closed under conjunctons, dsjunctons, and negatons. 1

2 3 Dutch Book arguments 3.1 Probablsm We wll assume that F s fnte. Ths assumpton s not necessary, but t smplfes proofs. We begn by gvng an alternatve formulaton of Probablsm. Defnton 1 An assgnment of truth values to the propostons n F s a functon v : F 0, 1} such that 0 f v(a) = 1 v( A) = 1 f v(a) = 0 and 0 f v(a) = 0 and v(b) = 0 v(a B) = 1 otherwse Defnton 2 Let V be the set of all assgnments of truth values to propostons n F. We mght thnk of each assgnment of truth values as a possble world. Thus, V s the set of all possble worlds. Note that snce F s fnte, V s fnte. Defnton 3 Let V + be the convex hull of V. That s, V + := λ v v : λ v > 0 and λ = 1 Another characterzaton of V + : t s the smallest convex set that contans all elements of V. Lemma 1 An agent satsfes Probablsm ff her credence functon c t s n V +. Proof. Suppose c t V +. To see that c t s a probablty functon, t suffces to note that: () Each v V s a probablty functon. () If p and p are probablty functons, then λp + (1 λ)p s a probablty functon. For the converse, suppose that c t s a probablty functon. Then, for each v V, let A v := A A v(a)=0 Thus, A v s the unque proposton n F such that v 0 f v = v (A v ) = 1 f v = v That s, A v s made true by v but by no other truth assgnment. Thus: A = A v } 2

3 And the A v s are dsjont. Now let Then, snce c t s a probablty functon, c t (A) = c t ( A v ) = λ v := c t (A v ) c t (A v ) = v(a)c t (A v ) = λ v v(a) as requred. So far, we have been treatng credence functons and truth value assgnments as functons from F nto [0, 1]. But, f F = A 1,..., A n }, then we mght just as well consder them as vectors n an n-dmensonal vector space. Thus, f c s a credence functon, we represent t as c = (c 1,..., c n ) where c = c(a ). Smlarly, f v s a truth value assgnment, we represent t as v = (v 1,..., v n ) where v = v(a ). Usng ths notaton, we can better state the Dutch Book argument. A Dutch book s a book of bets on the propostons n F and a prce for that book such that the prce s greater than the payoff of the book of bets n every possble world. We represent ths mathematcally as follows: Then a book of bets on F s represented by a vector (s 1,..., s n ). Ths s a set of n bets B 1,..., B n, where: B wll pay s f A s true; B wll pay 0 f A s false. Suppose p s the prce for bet B. Then the prce of the book (s 1,..., s n ) s p. The payoff of the book B = (s 1,..., s n ) at v V s v s Thus, a book (s 1,..., s n ) wth prces p for bet B s a Dutch Book ff for all v V. p > v s 3

4 3.1.1 The argument (1) Credences as bettng odds An agent wth credence r n proposton A should consder rs as a far prce for a bet that pays S f A s true and 0 f A s false. (2) Package prncple If an agent consders r as a far prce for bet 1 and r as a far prce for bet 2, then she should consder (r + r ) as a far prce for the book of bets that conssts of bets 1 and 2. (3) Undutchbookable An agent should not have credences that lead her to acceptng a Dutch book as far. (3) Theorem 1 (Dutch book theorem) (I) If c V +, then there s a book of bets (s 1,..., s n ) such that, for all v V, c s > v s (II) If c V +, then there s no book of bets (s 1,..., s n ) such that, for all v V, Therefore, (4) An agent ought to obey Probablsm Proof of Theorem 1 c s > v s (I) Suppose c V +. Then let p V + be the pont n V + that s closest to c. Then let s = c p. Then, by a classcal geometrcal result, we have that the angle between s and v p s not acute, for any v V. Thus s (v p) 0. Ths gves s v s p. But we also have s 2 > 0. But s 2 = s s = s (c p) = s c s p Thus, s p < s c. So s v < s c. That s, as requred. c s > v s (II) Suppose c V +. And let s be a vector. Then, ether the angle between v c and s s rght for all v V or for at least one v V, the angle between s and v c s acute. If the angle between s and v c s rght for all v V, then s (v c) = 0, so c s = v s for all v V. If the angle between s and v c s obtuse, then s (v c) > 0, so Ths completes the proof. c s < v s 4

5 4 Accuracy domnaton arguments In accuracy domnaton arguments, we treat epstemc states as epstemc acts, we ntroduce measures of epstemc utlty for those acts, and we employ the machnery of decson theory to derve norms that govern epstemc states. An epstemc utlty argument requres: An epstemc utlty functon For each credence functon c and possble world w, EU(c, w) measures the epstemc goodness of havng c at w. An example: The Brer score s the followng measure: B(c, w) = 1 (c(a) v w (A)) 2 A F where v w (A) = 0 f A s false 1 f A s true A norm of decson theory Ths tells us how an agent should choose from a range of dfferent acts on the bass of the epstemc utlty of those acts at dfferent worlds. Example: Act-Type Domnance Suppose there are two sorts of act: D 1 and D 2. Say that an act D s domnated by another act D f D has hgher utlty than D n every world. Now suppose: Every D n D 1 s domnated by some D n D 2. No D n D 2 s domnated by any D n D 1 or n D 2. Then the agent should choose an act from D The argument (1) The Brer score measures epstemc utlty (2) Act-Type Domnance (3) Theorem 2 (de Fnett) (I) If c V +, then there s p V + such that for all worlds w. B(c, w) < B(p, w) (II) If c V +, then there s no credence functon p such that for all worlds w. Therefore, (4) An agent ought to obey Probablsm. B(c, w) < B(p, w) 5