Comprehensive Rationalizability

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1 Comprehensve Ratonalzablty Avad Hefetz Martn Meer y Burkhard C. Schpper z Prelmnary & Incomplete: August 12, 2010 Abstract We present a new soluton concept for strategc games called comprehensve ratonalzablty. It embodes common con dence of cautous ratonalty based on a sound epstemc characterzaton n a unversal type space. It nether re nes nor s re ned by terated admssblty but t concdes wth terated admssblty n many applcatons n the lterature. Keywords: cautousness. Iterated admssblty, ratonalzablty, lexcographc belef systems, JEL-Class catons: C72. The Economcs and Management Department, The Open Unversty of Israel. Emal: avadhe@openu.ac.l y Insttut für Höhere Studen, Venna. Emal: meer@hs.ac.at z Department of Economcs, Unversty of Calforna, Davs. Emal: bcschpper@ucdavs.edu

2 1 Introducton We ntroduce a noton of ratonalzablty based on lexcographc conjectures (full-support mutually sngular lexcographc belef systems) about the other players strateges, comprehensve ratonalzablty. It s based on ncreasng levels of mutual assumpton of ratonalty. Ths noton nether re nes nor s re ned by terated admssblty (teratve elmnaton of weakly domnated strateges). Ths new soluton concept s provded wth an epstemc characterzaton. 2 Lexcographc conjectures A lexcographc conjecture over a Hausdor space X s a nte sequence of regular Borel probablty measures over X = ( 1 ; : : : n ) 2 ( (X)) n whch are (1) mutually sngular: `?`0 for ` 6= `0 (2) the unon of whose supports cover X : [`n supp` = X If Y and Z are two dsjont measurable subsets of X; we say that the lexcographc conjecture = ( 1 ; : : : n ) deems Y as n ntely more lkely than Z f ` (Y ) > 0 for some ` 2 f1; : : : ; ng; and whenever ` (Z) > 0 t s the case that `0 (Y ) = 0 8`0 `: If Y 0 and Z 0 are two subsets of X; not necessarly dsjont, we say that the lexcographc conjecture = ( 1 ; : : : n ) deems Y 0 as n ntely more lkely than Z 0 f t deems Y 0 n Z 0 n ntely more lkely than Z 0. We say that Y s assumed by the lexcographc conjecture = ( 1 ; : : : n ) f t deems Y as n ntely more lkely than X n Y. 1

3 3 Games, lexcographc domnaton and lexcographc best reples Consder a game G wth players 2 I and nte strategy sets (S ) 2I equpped wth the dscrete topology, and utlty functons (u ) 2I. If = ( 1; : : : n) 2 ( (S )) n s a lexcographc conjecture over the other players strategy pro les S = Q j6= Sj, and s ; ^s 2 S are two strateges of player ; we say that the strategy s s `-domnated by ^s w.r.t, denoted s ` ^s f Z u s ; Z d ì < u ^s ; d ì whle Z u s ; Z d ì 0 = u ^s ; d ì 0 for all `0 < ` (f such `0 exsts,.e. f ` 6= 1). We further say that the strategy s s lexcographcally domnated by ^s w.r.t, denoted s ^s f s ` ^s for some ` n. If s s not lexcographcally domnated w.r.t by any other strategy of player ; we say that s s a lexcographc best reply to : We denote by LBR ( ) S the set of player s lexcographc best reples to. Let M be a set of lexcographc conjectures of player on the other players strateges. If s ` ^s for some strategy ^s of player 1 but there does not exst a lexcographc conjecture 1.e., gven t s the case that s s just as good as ^s at all levels `0 < `; but s strctly worse than ^s at level `: 2

4 ~ 2 M wth respect to whch s s optmal at all the levels up to and ncludng `; we say that s s `-unjust able w.r.t. M. 4 Comprehensve ratonalzablty Let C 1 = (S ) and R 1 = S. De ne nductvely C k+1 = 2 C k : R k s assumed by R k+1 = s 2 S : 9 2 C k+1 for whch s s a best reply Player s comprehensve ratonalzable strateges are Remark 1 R k R k 1 for every k 1. R 1 = Proof. Ths s clear snce C k C k 1. 1\ k=0 R k Proposton 1 C k 6= ; and R k 6= ; for any k 0. Proof. Both R 0 and C 0 are nonempty for all 2 I. Assume R k and C k are nonempty for all 2 I. Clam: There exsts a k+1 2 C k that assumes R k. Proof of the clam: To see ths, note that by the nducton hypothess k assumes R` for ` = 0; :::; k 1. Agan, by the nducton hypothess R k s nonempty and by Remark 1, R k R k 1. Fx k = ( k ;0; :::; k ;m 1) 2 N (A ) for some nteger m. Snce R k s nonempty by the nducton hypothess, and k s a lexcographc conjecture (n partcular t s full support sequence), we have a j m 1 such that k ;j(r k ) > 0. Let k ;n 1 ; :::; k ;n p be all the k ;j such that k ;j(r k ) > 0 wth n r < n r+1, for all r = 1; :::; p 1. Moreover, let k ;m 1 ; :::; k ;m q be all the k ;j such that k ;j(r k 1 n R k ) > 0. Snce k assumes R k, there exsts ` m 1 such that k ;j(r k 1 ) = 1 for all j = 0; :::; ` and k ;j(r k 1 ) = 0 for j > `. Note that we have fn 1 ; :::; n p g [ fm 1 ; :::; m q g = f0; :::; `g. 3

5 Now, de ne = k+1 ;0 ; :::; k+1 ;p+q+m 1 k+1 where k+1 ;k = 8 >< >: k ;n j+1 (jr k ) for j = 0; :::; p 1 k ;m j p+1 (jr k 1 n R k ) for j = p; :::; p + q 1 k ;j p q+`+1 for j = p + q; :::; m 2 ` + p + q By constructon k+1 assumes R k, R k 1, but also R j for j < k 1. To see ths, let j < k 1. Snce k assumes R j, there exsts `0 0 such that k ;r(r j ) = 1 for r `0 and k ;r(r j ) = 0 for all r > `0. But snce R k R j, we have `0 `. If R k = R j, there s nothng to show. If R k $ R j then, by the fact that k s a lexcographc conjecture (n partcular, a full support sequence), we must have `0 > `. Then, k+1 ;r (Rj ) = 1 for r = 0; :::; p + q 1, but also for all j = p + q; :::; p + q 1 ` + `0. And we have by constructon, k+1 ;r (Rj ) = 0 for r > p + q 1 ` + `0. So, k+1 nshes the proof of the clam. assumes also R j. Ths Snce C k+1 s nonempty and the game s nte, we must have that R k+1 s nonempty. 4.1 Comprehensve Ratonalzablty vs. Iterated Admssblty An acton a 2 A s weakly domnated wth respect to X Y A A f there exsts 2 (A ) wth (X) = 1 such that P b 2X (b )u (b ; a ) u (a ; a ) for every a 2 Y and P b 2X (b )u (b ; a ) > u (a ; a ) for some a 2 Y. Otherwse, we say that a s admssble wth respect to X Y. Let A 0 = A and de ne for k 1 A k+1 = a 2 A k : a s admssble wth respect to A k A k : (1) The set of teratvely admssble actons of player s A 1 = 1\ A k : (2) k=0 4

6 A game s solvable by terated admssblty f for every player 2 I, the payo functon u s constant on all outcomes n 2I A 1. Smlarly, a game s solvable by comprehensve ratonalzablty f for every player 2 I, the payo functon s constant on all outcomes n 2I R 1. Example 1 Surprsngly, comprehensve ratonalzablty s not a re nement of terated admssblty as the followng example demonstrates. x y z w a b c 4; 0 4; 1 0; 1 0; 0 0; 1 4; 1 3; 0 2; 2 2; 1 5; 0 0; 1 0; 2 The order of elmnaton under terated admssblty s a; w; c and then both y and z. The maxmal reducton or teratve admssble actons are f(x; b)g. For comprehensve ratonalzablty, the order of elmnaton s a and then both z and w. The set of comprehensve ratonalzable pro les s fx; yg fb; cg. Yet, there are other examples such as the next example where comprehensve ratonalzablty re nes the teratve admssble actons. Example 2 The followng example shows that comprehensve ratonalzablty may re ne terated admssblty. a b c x 4; 0 4; 1 0; 2 y 0; 0 0; 1 4; 2 z 3; 0 2; 2 2; 1 Here, only acton a can be elmnated under teratve admssblty and we reman wth fx; y; zg fb; cg as the maxmal reducton. However, under comprehensve ratonalzablty the order of elmnaton s a; z; b and x. Thus, the f(y; c)g s the unque comprehensve ratonalzable acton pro le. Every rst level admssble acton s rst level comprehensve ratonalzable and vce versa. Ths follows from Blume, Brandenburger and Dekel (1991b, Proposton 1). 5

7 What are su cent condtons for teratve admssble actons to be comprehensve ratonalzable? The reason for why comprehensve ratonalzablty does not re ne terated admssblty n Example 1 s that there s no full support belef of player 1 on the rst level admssble actons of player 2 such that z s a strct best reply. E.g., for player 1, playng both x or y s as good as playng z aganst the belef that puts equal probablty on both b and c. Lexcographc best reples re ne best reples. Lemma 1 If for all 2 I and any k 2, f for every a 2 A k there exsts a full support belef 2 (A k 1 ) for whch a s the strct best reply, then A k = R k for all 2 I. Consequently, A 1 = R 1 for all 2 I. Proof. For all 2 I, A 1 = R 1 follows from Blume, Brandenburger and Dekel (1991b, Proposton 1). We show by nducton that f for all 2 I, A k = R k and for every a 2 A k+1 there exsts a full support belef 2 (A k ) for whch a s the strct best reply, then A k+1 = R k+1. Consder rst the case A k+1 R k+1. We need to show that for 2 (A k ) there exst a lexcographc conjecture ( ;0 ; :::; ;n ) 2 C k+1 for some n 0 such that a s a lexcographc best reply. Consder any lexcographc conjecture ( ;0 ; :::; ;n ) wth ;0 =. Snce s full support on A k, ( ;0 ; :::; ;n ) s by de nton mutually sngular, and by the nducton hypothess A k = R k, we have that R k s assumed by ( ;0 ; :::; ;n ). Thus ( ;0 ; :::; ;n ) 2 C k+1. Snce a was the only best reply to among actons n A k and A k = R k, a s a lexcographc best reply to ( ;0 ; :::; ;n ) among actons n R k no matter what are the belefs ;`, ` = 1; :::; n, n the lexcographc conjecture. a s also a lexcographc best reply to ( ;0 ; :::; ;n ) among actons n A, snce f there exsts an acton a 0 2 A that lexcographcally domnates a wth respect to ( ;0 ; :::; ;n ), then a =2 R k. Snce by the nducton hypothess, A k = R k, a =2 A k, a contradcton. We conclude that a 2 R k+1. Consder now the case R k+1 A k+1, and suppose to the contrary that there exsts an acton a 2 R k+1 wth a =2 A k+1. Snce a 2 R k+1, there exsts a lexcographc conjecture ( ;0 ; :::; ;n ) 2 C k+1 for some n 0 such that a s a lexcographc best reply to ( ;0 ; :::; ;n ) and ( ;0 ; :::; ;n ) assumes R k. Thus, there s smallest ` 2 f1; :::; ng such that ;`0(R k ) = 0 for all `0 `. By the nducton hypothess, A k = R k. Thus by arguments analogous to Blume, Brandenburger and Dekel (1991b, Proposton 1), we construct from ( ;0 ; :::; ;`) a full support belef on A k such that a s a best reply to. Thus a 2 A k+1. 6

8 Snce there s a nte number of actons for each player, we also have A 1 = R 1 for every 2 I. In many economcally relevant examples the condton of Lemma 1 s sats ed and comprehensve ratonalzablty concdes wth terated admssblty. We dscuss some of the examples n sequel. The rst example concerns votng wth a presdent. Example 3 (Votng wth a presdent) Consder an example of majorty votng wth a presdent (Mouln, 1986, p ). Three players have to select one of three alternatves fa; b; cg. If a majorty votes for an alternatve, t wll be mplemented. Otherwse, the alternatve selected by player 1, the presdent, s selected. For smplcty, for each assgn payo s 3; 2 and 1 n the order of preferences. The strategc form s gven by the followng three matrxes where player 1 chooses matrces, player 2 rows and player 3 columns. a b c a b c a b c a b c 3; 2; 1 3; 2; 1 3; 2; 1 a 3; 2; 1 2; 1; 3 2; 1; 3 a 3; 2; 1 1; 3; 2 1; 3; 2 3; 2; 1 2; 1; 3 3; 2; 1 b 2; 1; 3 2; 1; 3 2; 1; 3 b 1; 3; 2 2; 1; 3 1; 3; 2 3; 2; 1 3; 2; 1 1; 3; 2 c 2; 1; 3 2; 1; 3 1; 3; 2 c 1; 3; 2 1; 3; 2 1; 3; 2 a b c At the rst round, the only admssble acton s a. For players 1 t s the set fa; cg and for player 2 t s fb; cg. Hence the game s reduced to a c b c 3; 2; 1 3; 2; 1 3; 2; 1 1; 3; 2 a Ths game can be further reduced by weak domnance to (a; c; c), the only teratve admssble pro le. Interestngly, n ths pro le the presdent faces hs lowest ranked alternatve. Note that for player 2 a s the strct best reply to a full support belef concentrated on for nstance the pro le of other players actons (c; a) and c s the strct best reply to a full support belef concentrated on the pro le of other players actons (b; c). Smlarly, for player 3. Thus Lemma 1 apples and the unque comprehensve ratonalzable acton pro le s (a; c; c) as well. 7

9 Example 4 (Dvdng Money (Brams, Klgour, and Davs, 1993, Osborne, 2004, p. 38)) Two players use the followng procedure to dvde $10 between themselves. Each person names a number of dollars (a nonnegatve nteger), at most equal to $10. If the sum s at most $10, then each person receves the amount of money she named and the remander s burned. If the sum exceeds $10 and the players named d erent amounts, then the person who named the smaller amount receves that amount and the other player receves the remanng money. If the sum exceeds $10 and the amounts named by the players are the same, then each player receves $5. In the rst round, every amount weakly lower than $5 s weakly domnated by $6. If the opponent names more than $6, then the sum exceeds $10 and the player receves $6. If the opponent names exactly $6, then the sum also exceeds $10, and both players receve $5. If the opponent names any amount strctly than $5, then the sum at most $10, and the player receves $6. Any other amount a 2 f$6; :::; $9g s a strct best response to a+1, and $10 s a strct best response to $0. In the followng rounds, the hghest remanng amount s weakly domnated by $6 and all other amounts are a strct best response to that amount plus 1. Thus, the maxmal reducton under terated admssblty s ($6; $6) yeldng a payo of $5 to each player. Note that snce every amount not elmnated at the prevous round s a strct best reply to some full support belef on the remanng amounts, Lemma 1 apples and the game s solvable by comprehensve ratonalzablty. 4.2 Prce Competton Consder a symmetrc Bertrand duopoly n whch each rm s restrcted to choose nteger prces. Let p be the prce and c the cost (also n ntegers), and D(p) = ( p f p 0 f p > and assume c + 1 <. The pro t functon of rm 6= j s gven by 8 >< (p ; p j ) = >: (p c)d(p ) f p > p j 1 2 c)d(p ) f p = p j 0 f p < p j Assume that the monopoly prce s unque. 8

10 Proposton 2 In the Bertrand duopoly, p = c + 1 s the unque comprehensve ratonalzable prce. It also the unque teratve admssble prce. Proof. We consder teratvely admssblty and then apply Lemma 1 to show that t concdes wth comprehensve ratonalzablty. In the rst round every prce n excess of the monopoly prce s weakly domnated by the monopoly prce. If the opponent sets a prce weakly hgher than the monopoly prce, then the monopoly prce s strctly better than any prce strctly hgher than the monopoly prce. If the opponent sets a prce strctly below the monopoly prce, the monopoly prce s as good as any prce strctly hgher than the monopoly prce. In fact, there s no belef about the opponent s prces for whch a prce strctly hgher than the monopoly prce s a best reply. Also n the rst round, every prce equal to at most c s weakly domnated by the prce c + 1. If the opponent sets a prce weakly larger than c+1, then a prce equal to c+1 s strctly better than a prce equal to at most c. If the opponent sets a prce strctly below c + 1, then a prce equal to c+1 s as good as a prce equal to c and strctly better than a prce strctly below c. That s, c s a best reply to the belef that the opponent sets a prce of at most c but t s not the unque best reply. c can never be a best reply to a full support belef. Every other prce p s a strct best response to p+1, so no other prce s weakly domnated. To see ths, note that for any p c+1 t s strctly better to obtan all the demand at the prce p than to obtan half of the demand at the prce p+1. That s, consder any p c+2. We need to show that 1 (p c)( p) < (p 1 c)( p+1) = (p 1 c)( p)+p 1 c. 2 We have 1 (p c) p 1 c and p 1 c > 0 because p c + 2. By the same argument, 2 at each subsequent round of teratve elmnaton, the hghest remanng prce s weakly domnated by the next hghest prce. The par of prces that remans s (c + 1; c + 1). Let p` be the hghest prce that remans after the `-th round of elmnaton of weakly domnated prces. Lemma 1 mples that at each level ` = 0; 1; :::, the set of admssble prces concde wth the set of comprehensve ratonalzable prces snce every prce n fc + 1; :::; p`g s a strct best reply to a full support belef over prces remanng from the prevous round. E.g., p 2 fc + 1; :::; p`g s a strct best reply to the belef that s concentrated on p Second Prce Common Value Auctons Consder a second prce common value aucton n whch each of the n bdders receves a prvately observed sgnal about the value of the object to be auctoned o. Ths sgnal x s ndependently and dentcally dstrbuted over some nte set of sgnals X N 9

11 (ntegers) and assume that every sgnal may be drawn wth postve probablty. Let x max denote the realzaton of the hghest of n sgnals. The common value of the object to each bdder s x max. Each bdder submts a bd n a sealed envelop. The hghest bdder wns and pays the second hghest bd. In case of a te, each hghest bdder obtans the object wth equal probablty. Let b : X! R denote the bd functon of player. Bddng your value s the unque bddng functon that s obtaned after two rounds of terated admssblty (see Harstad and Levn, 1985). Comprehensve ratonalzablty yelds the same outcome. Proposton 3 In the second prce common value sealed bd aucton, R k = fb(x ) = x g for all k 2 and 2 I. Proof. For each bdder 2 I, any bd strctly above max X and strctly below x s weakly domnated. Thus, by Blume, Brandenburger and Dekel (1991b, Proposton 1) R 1 = fb : For all x 2 X; x b(x ) max Xg. At the second level, suppose a bdder bds strctly above hs sgnal. If hs bd s below the hghest bd, then he does not obtan the object. Bddng hs sgnal would be a lexcographc best reply n ths case as he receves nothng and pays nothng. If the second hghest bd s below hs sgnal, then he stll receves the object. Bddng hs sgnal would be a lexcographc best reply n ths case snce he would stll obtan the object and hs payment would reman unchanged. Consder now the case n whch the second hghest bd s between hs bd and hs sgnal. By hs second level lexcographc conjectures he assumes that all bdders j 6= select a bd n fx j ; max Xg. Thus, n ths case x max > x. Ths means he would pay more than the value of the object. A lexcographc best reply s to bd hs sgnal nstead, n whch case he pays nothng and obtans nothng. In fact, t s the only lexcographc best reply snce opponents may have drawn any sgnals from X. Thus we have shown that at the second level, bddng the sgnal remans as the only comprehensve ratonalzable bd functon. 4.4 Relatonshp to other Iteratve Soluton Concepts Bernhem (1984) and Pearce (1984) ntroduced ratonalzablty. Let P 0 = A 0 and B 0 = f 2 (A )g and de ne for k 1, B k+1 = f 2 (P k )g and P k+1 = fa 2 A : a s a best reply to some 2 B k+1 g. The set of ratonalzable actons s 10

12 P 1 = T 1 k=0 P k. 2 Proposton 4 For every 2 I and k 0, R k ratonalzable acton s ratonalzable. P k. Moreover, every comprehensve Proof. By de nton, R 0 = P 0 = A for all 2 I. Moreover, for all 2 I, R 1 P 1 snce any lexcographc best reply to = ( 1 ; ::; n ) on A, for some nte n 1, s a best reply to 1. Suppose that R k P k for all 2 I and some k 2. We show that R k+1 P k+1. Suppose to the contrary that there exsts an acton a 2 R k+1 wth a =2 P k+1. There s no conjecture 2 (P k ) for whch a s a best reply. Yet, a s a lexcographc best reply to some lexcographc conjecture ( 1 ; :::; n ) 2 (R k ) n. Thus, t s n partcular a best reply to 1 on R k. Snce by the nducton hypothess, R k P k, a contradcton. P k, 1 s a conjecture on Snce well-known games such as Guess-the-Average are solvable by ratonalzablty, the result mples that they are solvable by comprehensve ratonalzablty as well. Dekel and Fudenberg (1990) ntroduce one round elmnaton of weakly domnated actons followed by teratve elmnaton of strctly domnated actons as soluton concept. Let W S 1 denote the maxmal reducton of ths procedure. Proposton 5 For every player 2 I, R 1 W S 1. Proof. At the rst level, for all players the set of actons that are not weakly domnated are the set of lexcographc best reples. Ths follows from Blume, Brandenburger and Dekel (1991b, Proposton 1). Next, by Pearce (1984, Lemma 3), every acton not strctly domnated s a best reply to some conjecture over opponents actons and vce versa. Thus ratonalzablty and terated strct domnance concde. Thus the result follows from Proposton Comprehensve Ratonalzable Implementaton Let X denote the set of smply lotteres (.e. wth nte support) over an arbtrary set of alternatves. Each player 2 I has now preferences over lotteres represented by 2 Best reply refers to the pure acton best reply. 11

13 u : X R! R, where s a nte set of utlty parameters for player. Dstnct parameters n assocated dstnct preferences orderngs over X. Moreover, we assume that a player s never nd erent between all lotteres n X. The functon s lnear n ts rst argument. We assume that the preference pro le 2 = 2I s common knowledge among players n I. Yet, the socal planner does not know and want to mplement some lottery over alternatves n X. A socal choce functon f :! X assocates wth each preference pro le a lottery over alternatves. We consder a nte mechansm wth transfers hm 1 ; :::; M n ; g; t de ned by a nte acton set M for each player, an outcome functon g : M! X that assocates wth each acton pro le n M = 2I M a lottery over outcomes, and a transfer rule t = (t ) 2I : M! R n that for each player assocates wth each acton pro le a ne. (The second argument n each player s utlty functon refers to the nes. Less nes are preferred to more.) A mechansm hm 1 ; :::; M n ; g; t and a preference pro le n de ne a strategc game wth complete nformaton. A mechansm exactly mplements a socal choce functon f n comprehensve ratonalzable actons wth nes bounded by t > 0 f and only f jt(m)j t for all m 2 M, and for any 2, there exsts m ( ) 2 M such that g(m ( )) = f( ), t(m ( )) = 0, and R 1 ( ) = fm ( )g. A socal choce functon f s exactly mplementable n comprehensve ratonalzable actons wth small nes f for all t > 0, there exsts a mechansm whch exactly mplements f wth nes bounded by t. Proposton 6 Suppose that there are at least three players. functon s exactly mplementable n comprehensve ratonalzable actons. Then any socal choce Proof. Abreu and Matsushma (1994) show that any socal choce functon s exactly mplementable n terated admssble actons. In ther proof, they show that n fact t s exactly mplementable wth one round elmnaton of weakly domnated actons followed by many rounds of elmnaton of strctly domnated actons. In ther mechansm, the maxmal reductons of both procedures concde. Thus the result follows from Proposton 5 above. 12

14 5 Epstemc characterzaton of comprehensve ratonalzablty Our next goal wll be to provde an epstemc characterzaton of comprehensve ratonalzable strateges. To ths e ect we de ne lexcographc conjectures type spaces, construct the unversal space n the category of lexcographc conjectures type spaces, and de ne wthn t the graded event whose prmary part corresponds to the set of comprehensve ratonalzable strateges. 5.1 Lexcographc conjectures type spaces For a gven game wth a set of players I and a Hausdor space of strategy pro les S = Q 2I S ; T 2I s a lexcographc conjectures type space f 8 2 I; T = S n1 T n and each Tn s a Hausdor space, wth contnuous mappngs : T! S specfyng each type s strategy, de ned by the contnuous mappngs n : T n! S ; n 1 and the contnuous mappngs 3 : T! [ n1 T n specfyng each type s state of mnd regardng the other players types, as de ned by the contnuous mappngs n : Tn! T n ; n 1 havng the property that the n-tuple of belefs n (t n) js of t n 2 Tn on S de ned by n t n () js n t n j j6= 1 () 3 In what follows we wll use the notatonal conventon Y = Q j2j;j6= Y j : 13

15 consttutes a lexcographc conjecture over S (.e., these belefs are mutually sngular and the unon of ther supports s S ): Some (but not all) of the T n may be empty. Notce that n ths de nton, only the belefs of the types t n on the other players strateges S are requred to form a lexcographc conjecture; n contrast, the belefs of the types t n on the other players types T are not requred to have supports whose unon cover T. The motvaton for ths dstncton s that pror to playng the game, player may potentally get surprsng ver able evdence that her prmary belef on the other players strategy pro le S was wrong (.e. that the other players strategy pro les not ruled out by that evdence was assgned probablty zero by the prmary belef), n whch case she resorts to her secondary belef, and so forth. In contrast, no drect ver able evdence s feasble regardng the other players belefs. Hence, pror to playng the game there cannot arse a necessty for player to replace her prmary belef about the other players belefs, and therefore player need not necessarly entertan an exhaustve arsenal of mutually sngular alternatve belefs on the other players types. Ths does not preclude, of course, that a swtch of player to a secondary (or ternary, etc.) belef about the other players strateges may be correlated wth a correspondng swtch n belef about the other players types. Even though the lexcographc belef system (t ) need not have full support, the notons of n ntely more lkely and assumpton are de ned for (t ) verbatm. De nton 1 t assumes E f (t ) assumes E and for every s 2 (E ), (t ) deems ft 2 E j (t ) = s g as n ntely more lkely than ft =2 E j (t ) = s g. Lemma 2 In any lexcographc conjecture type space, for any measurable sets Y and Z wth Y \ Z = ; we have ft that deem Y n ntely more lkely as Zg s measurable. Proof. Let Y \ Y = ; and Y; Z be measurable. For any n, T n s measurable. For ` n the set B ;>0 ` (Y ) := ft 2 Tnj (t )`(Y ) > 0g as well as the set ft 2 Tnj (t )`(Z) = 0g are measurable. 14

16 Hence B I;`(Y; Z)n = ft nj (t n)`(y ) > 0; (t n) k (Y ) = 0; for k > `; (t n) j (Z) = 0; for j `g s measurable, snce t s an ntersecton of measurable sets. The set BI (Y; Z)n := ft n 2 Tnj (t n) deems Y as n ntely more lkely than Zg s S `=1;:::;n B I;`(Y; Z), and therefore measurable. Hence the set S k1 B I (Y; Z)n s measurable. Lemma 3 In any lexcographc conjecture type space: For any measurable event E T, the event ft jt assumes E g s measurable n T. Proof. For all s 2 (E ) the set ft j (t ) deems (t ) 1 (fs g)\e as n ntely more lkely than (t ) 1 (fs g) \ (T n E )g s measurable by Lemma 2. The set ft jt assumes E g s an ntersecton of the measurable set ft j (t ) assumes E g and the measurable sets (t ) 1 (fs g) \ E, where s 2 (E ). 5.2 The unversal lexcographc conjectures type space For a gven game wth a set of players I and strategy pro les S = Q 2I S ; de ne, nductvely, for each player 2 I: 0 = S 1;n = n s ; 1;` `n 1 = S n1 1;n o n : 1;` `n s a lexcographc conjecture on S Suppose, nductvely, that 8n 1 we have already de ned k;n and k. De ne k+1;n = ns ; 1;`; : : : ; k;`; k+1;` k+1 = S n1 k+1;n `n 2 k;n o n k : 8` n margj k 1 k+1;` = k;` In the lmt, de ne 1;n = n s ; 1;`; : : : `n o : s ; 1;`; : : : ; k;` 2 k;n 8k 0 `n 1 = S n1 1;n 15

17 By the Kolmogorov extenson theorem (the verson of Metver or Bourbak 1969), for each 4! 1;n s ; 1;`; : : : `n 2 1;n there exsts a unque tuple of n regular Borel probablty measures n! 1;n 2 ( (S 1 )) n such that 8k 0 and 8` n marg j k n! 1;n = ` k+1;` In partcular, from ths condton for k = 0, we get that s a lexcographc conjecture over S : marg js Moreover, the mappngs n are contnuous, and so are also the mappngs n : 1;n! S n! 1;n ` `n de ned by n! 1;n = s Hence h 1 2I s a lexcographc conjectures type space. The unversalty and completeness of h 1 2I follow the lnes of by-now standard arguments. ( Unversalty here refers to termnalty n the category of jij-player lexcographc conjectures type spaces based on the strategy pro les set S: ) Type möpse: Let ht 2I and h ^T 2I be lexcographc type spaces on S = Q 2I S. Then h : T! ^T contnuous s.t. T h! ^T & S. ^ commutes,.e. ^ (h (t )) = (t ) for all 2 I, for all t 2 T, such that h T n (Tn) ^T n and s contnuous for all n js j and such that (t ) (h ) 1 ( ^E ) = ^ (h (t ))( ^E ) for all measurable ^E ^T. Unversalty: Let h ^T 2I be any lexcographc conjecture type space on S. Let ^t 2 ^T n. 4 Each 1;n s non-empty t contans e.g. the n-tuple of heraches expressng player s belef that all the players lexcographc conjectures over K are commonly known. 16

18 De ne 0;n(^t ) := h 0;n(^t ) := (^t ), h 0;n : ^T Also de ne 1;n(^t ) :=! T 0;n. Ths s obvously contnuous. ^ (^t ); (( ) 1 ()). Ths s a lexcographc conjecture on `n S snce h ^T s a lexcographc conjecture type space. Suppose j m;n for j 2 I and for 0 m k have already been de ned s.t. h k;n := 0;n; 1;n; :::; k;n : ^T! T k; n s contnuous. Then de ne k+1;n va k+1;n(^t )` := ^ (^t )` (h k ) 1 (). k+1;n s contnuous f h k s contnuous, whch holds by the nducton hypothess. Coherency of Let k 1 and let proj jt 2 I that marg T have marg T k 1 k 1 0;n(^t ); 1;n(^t ); :::; k;n(^t ); ::: k 1 : T k! T k 1 k+1(^t )` = k(^t )`. Let E k 1 k+1(^t )`(E k ) = k+1(^t )` (proj jt ) 1 (E k 1 k 1 ) = ^ (^t )` (h k ) 1 ((proj jt k 1 = ^ (^t )` (proj jt h k ) 1 (E k 1 ) The last equaton follows from the fact that we have proj T (h j k 1 ) j6=. And therefore marg T ^T! T for all 2 I. k 1 k 1 be the projecton. We have to show for be some measurable event n Tk 1. We ) 1 (E k+1 )) k 1 h k = ^ (^t )`(h k 1 (E k 1 )): = (proj T j h j k ) j6= = k 1 k+1(^t ) = k(^t ). Hence we have shown that h := ( 0; 1; :::) : Snce the nte cylnder sets form an algebra that generates the -algebra on T for all 2 I and snce (h (^t )) and ^ (^t ) ((h ) 1 ()) are two -addtve probablty measures that - by constructon - concde on these cylnder sets, by the Caratheodory extenson theorem, they must be equal. Hence, snce by constructon (h (^t )) = ^ (^t ), (h ) 2I s a type mops from ^T := h ^T 2I to the unversal type space. Unqueness: Let := h 2I denote from now on the lexcographc conjecture type space constructon n ths secton (5.2) and for any lexcographc conjecture type space T := ht 2I let (w T ) 2I be the type mops from T to constructed before. A close examnaton of the de nton shows Remark 2 (w ) 2I = (d ) 2I. 17

19 Proposton 7 Let (h ) 2I be a type mops from the lexcographc conjecture type space T = ht 2I to ^T = h ^T 2I. Then we have w ^T h = w T for all 2 I. Proof. Analogous to Hefetz and Samet (1998, add reference to correspondng proposton/lemma) wth the obvous changes. Corollary 1 Let T be a lexcographc conjecture type space. Then (w T ) 2I s the unque type mops from T to. Proof. Let (h ) 2I be a type mops from T to. Then we have w T = w h = h. Proposton Remark. Remark 3 (After the constructon of the unversal type space.) The projectons proj k+1 k Proof. : k+1! k are onto. Analogous to the proof of Theorem 6 n Hefetz (1993) or the onto part of the p n n (3) of Theorem 1.1 Chapter III n Mertens, Sorn and Zamr (2003, p. 108). De nton 2 Let ^( ) be the set of all 2 (( ) n for n js j s.t. ((( ) 1 ())`)`n consttutes a lexcographc conjecture on S, endowed wth the relatve topology nherted from the dsjont unon topology of the S njs j (( )) n. Proposton 8 For all 2 I, :! ^( ) s a homeomorphsm. Proof. Smlar to Meer (forthcomng) usng unversalty or by drect examnaton of the maps n the unversal type space. 5.3 The epstemc characterzaton Let ht 2I be a type space for a gven game wth strategy pro les S = Q 2I S : De ne the followng sequence of events of player s Ratonalty and Mutual Assumpton of degree k 0 of Ratonalty: RMA 0 R = nt 2 T : t s a best reply to t o js 18

20 and nductvely RMA k+1 R = t 2 RMA k R : t assumes RMA k R Furthermore, de ne the event of s ratonalty and common assumpton of ratonalty to be 1\ RCAR = RMA k R k=0 Lemma 4 For k 0: If t 2 RMA k R, then (t ) js 1 assumes (RMA k 1 R ). Proof. For k = 0, there s nothng to prove. Suppose the clam s shown for all j 2 I and all m k. Let t 2 RMA k+1 R. By the nducton hypothess (t ) js assumes (RMA m R ) for all m < k. We have to show that (t ) js assumes (RMA k R ). Snce (t ) assumes RMA k R, there s an ndex ` 1 such that (t )`0 (RMA`0R ) = 1 for all `0 ` and (t )`0 (RMA`0R ) = 0 for all `0 > `. Ths mples that f (t )`0jS (fs g) > 0, for some `0 `, then s 2 (RMA k R ). Conversely, let s 2 (RMA k R ). By the second condton n the de nton of assumpton and the de nton of n ntely more lkely, (t )`0 (ft 2 RMA k R ; (t ) = s g) > 0 for some `0 and snce (t )`00(RMA k R ) = 0 for all `00 > `, we have `0 `. But ths mples that (t )`0jS (fs g) > 0. Together wth the rst part of the nducton step, we have that (RMA k R ) s assumed by (t )js. Theorem 1 (Epstemc characterzaton of comprehensve ratonalzablty) In the unversal lexcographc conjectures type space, for each k 0 the strateges (RMA k R ) played by the types n RMA k R are the strateges n R k R k = RMA k R and also n the lmt, the comprehensve ratonalzable strateges of player are the strateges played by s types n the event of s ratonalty and common assumpton of ratonalty: R1 = RCAR 19

21 Proof. Clearly (RMA 0 R ) R 0; snce by de nton 8t 2 RMA 0 R ; (t ) s a best reply to some 2 C 0; namely to = (t ) js : Conversely, let s 2 R 0: By de nton, s s a best reply to some lexcographc conjecture = ( 1; : : : n) 2 (S ) : Our am s hence to nd a type t 1 n the unversal lexcographc type space T 1 whose strategy (t 1) s s and whose margnal belef (t 1) js on S s : To ths e ect, let _ j 1 2 (S j ) be an arbtrary full-support belef of player j on her rvals strateges, j 2 I, and de ne nductvely _ j k+1 = _ j 0 1 ;::: _ j0 k j 0 6=j j s belef that there s k-mutual certanty that the players belefs on ther rvals strateges are _ j 1 ( s the Drac probablty measure). j2i De ne now t 1 = s ; ; _ 2; : : : _ k; : : : where _ 2 = _ 2;1; : : : ; _ 2;n s de ned by _ 2;` = ì _ j ; ` n 1 j6= and nductvely _ k = _ k;1; : : : ; _ k;n s de ned by _ k;` = _ k 1;` _ j ; ` n: k 1 j6= t 1 = s ; ; _ 2; : : : _ k; : : : s thus a type n RMA 0 R T1 whose strategy (t 1) = s s a best reply to ts margnal belef (t 1) js = on the other players strateges as requred. Suppose, by nducton, that 8 2 I we have already proved that 8m k ts s the case that R m = (RMA m R ), and hence that R m = Q j6= j (RMA m R j ). We now have to prove that R k+1 = (RMA k+1 R ). If t 2 RMA k R, then (t ) s a best reply to (t ) js. By the nducton hypothess and Lemma 4, f t 2 RMA k+1 R, then (t ) s a best reply to (t ) js, whch assumes (RMA m R ) = R m for all m k (and hence 20

22 (t ) js 1 s n C k+1 ). Thus, (t ) 2 R k+1. Conversely, let s 2 R k+1, we want to nd a t 2 RMA k+1 R such that (t ) = s. There s a 2 C k+1 such that s s a best reply to. By the nducton hypothess, (RMA k R ) = R k. has the form = ( 1 ; :::; n ) for some n 1. Recall that s full support and mutually sngular. Moreover, recall that S s nonempty and nte. Consder s = (s j ) j6=. There s exactly one ` n such that `(s ) > 0. For each j 6= choose a t j (s j ) such that j (t j (s j )) = s j and t j (s j ) 2 RMA m R j for the maxmal m k among the ^t j wth j (^t j ) = s j. Now let t be the unque ^t 2 T such that (^t ) s a xed best reply to and such that (^t )` := P s 2S `(s ) t (s )() for all ` = 1; :::; n. Ths measure assumes RMA m R for all m k, t sats es the second condton of the de nton of RMA k+1 R for all m k and ts margnal on S s. And (t ) s a best reply to. That R 1 = (RMA 1 ) s proved n Lemma 6. Constructon: Let k be such that C k+1 = C k and R k+1 = R k for all 2 I. For each s 2 R k choose 1;n(s ) (s ) 2 C k for some n(s ) wth s a lexcographc best reply to 1;n(s ) (s ) = ( 1;1; :::; 1;n). Smplfy noton by wrtng 1(s ) for 1;n(s ). Lkewse, for any s =2 R k, chose a (s ) 2 C m wth m maxmal such that s s a lexcographc best reply to (s ), f such a (s ) exsts. Otherwse let (s ) be any lexcographc conjecture. De ne 2;n(s ) such that 2(s )() := 2;n(s )`() := P s 2S 1;`(s )(fs g) (s j ; j 1 (sj )) j6= (). By de nton marg js 2;n(s )`() = 1;`(s )(). Fx k 1. Assume for each 2 I and s 2 S, we have already de ned by nducton t ;m (s ) n(s ) = (s ; ( 1;`(s ); :::; m;`(s ))`n(s ) 2 T m;n(s ). Then de ne m+1;`(s )() := P s 2S 1;`(s )(fs g) (s j ;t j;m+1 (s j )) j6= (). By the nducton hypothess m;`(s ) = P s 2S 1;`(s )(fs g) (s j ;t ;m 1 (s j )) j6= and we have that marg jt ;m (s j ;t j;m (s j )) j6= () = (s j ;t j;m 1 (s j )) j6= () snce t j;m (s j ) extends t j;m 1 (s j ) by constructon. If we let t (s ) = (s ; ( 1;`(s ); :::)`n(s )) then we get by constructon of the unversal type space that (t (s ))` = P s 2S 1;`(s )(fs g) (s j ;t j (s j )) j6= (). By de nton: f s 2 R m, then 1;n(s ) assumes C p for all p < m. Therefore, by construc- 21

23 ton of the t (s ), (t (s )) assumes ft (s )js 2 Rp g =: t (Rp ). By also, by constructon, f s 2 (t (Rp )) = Rp, then t (s ) deems ft ( s ^ )j (^s ) = s and ^s 2 Rp g = ft (s )g as n ntely more lkely than ft =2 t (Rp )j (t ) = s g. Note, ths latter set gets by constructon probablty 0 at every level of (t (s )), whle (t (s ))`(ft (s )g) > 0 f and only f 1;`(s )(fs g) > 0. But snce 1;n(s ) s a lexcographc conjecture on S ths happens for some ` n(s ). We have just shown the followng: Lemma 5 For all 2 I, for all m 0, for all p < m, and for all s 2 R m, t (s ) assumes ft (s )js 2 R p g. Lemma 6 For all 2 I, R 1 = (RMA 1 R ). Proof. We now prove by nducton the followng clam: For all m 0: s 2 R m f and only f t (s ) 2 RMA m R. The ( drecton follows already from the rst part of the proof of the Theorem. We now prove the ) drecton. If m = 0, by constructon s s a best reply to t (s ) s rst order belef. Observe that by constructon (t (s ))` (ft (s )js 2 S g) = 1 for all ` n(s ). Now x any s 2 R m+1. By the nducton hypothess, for all j: ft j (s j )js j 2 R j mg = ft j (s j )jt j (s j ) 2 RMA m R j g. Now, by the above observaton, for all ` n(s ) we have (t (s ))` (ft (s )js 2 R m g) = (t (s ))`(RMA m R ). But by Lemma t (s ) assumes (ft (s )js 2 Rm g), and hence assumes RMA m R. Snce t also assumes RMA m 1 R by the nducton hypothess (for m 1), we have shown that t (s ) 2 RMA m+1 R. To sum up, snce Rk = R m for m k for all 2 I, and snce each Rk 6= ; and snce the clam holds we have s 2 R1 f and only f Rm for m k f and only f t (s ) 2 RMA m R for all m k f and only f t (s ) 2 RMA 1 R and snce t (s ) plays s, we have that R 1 (RMA 1 R ). On the other hand, f t 2 RMA 1 R t follows that t 2 RMA m R for all m 0 and therefore (t ) 2 R m for all m 0 by the rst part of the proof of the theorem. But (t ) 2 R m for all m 0 mples (t ) 2 R 1. Hence we have show that (RMA 1 R ) = R 1 for all 2 I. 22

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