Topologies on Types: Connections

Transcription

1 Topologes on Types: Connectons Y-Chun Chen y Syang Xong z May 28, 2008 Abstract For d erent purposes, economsts may use d erent topologes on types. We characterze the relatonshp among these varous topologes. Frst, we show that for any general types, convergence n the unform-wea topology mples convergence n both the strategc topology and the unform strategc topology. Second, we explctly construct a type whch s not the lmt of any nte types under the unform strategc topology, showng that the unform strategc topology s strctly ner than the strategc topology. Wth these results, we can lnearly ran varous topologes on the unversal type space, whch gves a clear pcture of the relatonshp between the mplcaton of types for belefs and ther mplcaton for behavors. Keywords: the unversal type space, the strategc topology; the unform strategc topology; the unform-wea topology; nterm correlated ratonalzable actons JEL Class caton: C70 We are grateful to Edde Deel for hs nvaluable support, gudance, and detaled comments. We also than Je Ely for helpful dscussons and Aaron Sojourner for useful comments. All remanng errors are our own. y Department of Economcs, Northwestern Unversty, 2001 Sherdan Road, Evanston, IL 60208, ychunchen2010@u.northwestern.edu z Department of Economcs, Northwestern Unversty, 2001 Sherdan Road, Evanston, IL 60208, xong@u.northwestern.edu 1

2 1 Introducton If two agents have smlar belefs on some payo -relevant uncertanty, do they always mae smlar decsons? In the sngle-person-decson-mang setup, where proxmty of belefs about uncertanty can be measured by the standard wea -topology, the answer s clearly yes. The answer s much more complcated n the mult-person-game setup where agents belefs about each other s belefs play a crucal role n determnng the outcome of the nteracton. In ths setup, do smlar belefs stll mply smlar behavors? One compact way to formulate the complcated object of belefs about belefs s to use the noton of types developed by Harsany (1967/1968). A player s type spec es a belef over the set of unnown relevant parameters and opponents type pro les. Mertens and amr (1985) shows that the set of all coherent herarches of belefs s the unversal type space and Harsany s dea su ers no loss of generalty. In practce, an deal formulaton of an economc stuaton may nvolve a very complcated type space whch s hard to analyze drectly. For reasons of tractablty, economsts may therefore replace ths deal formulaton wth some smpler type space n order to solve a smpler problem. To mae ths approxmaton meanngful, we should requre that these smpler types be "close" to the true types so that they wll exhbt "smlar" behavors n games. To measure the "closeness" of types, we need to de ne a topology on types. Two nds of nformaton are encapsulated n a type. On the one hand, de ned drectly on herarches of belefs, a type contans an agent s belef nformaton. On the other hand, as shown n Deel, Fudenberg, and Morrs (2006) [hereafter DFM], any two d erent Mertens- amr types have d erent behavors n some game, and hence, the behavors of an agent are also mplctly encoded n a type. We can therefore de ne topologes on types accordng to these two aspects. Examples of "belef" topologes nclude the product topology ntroduced by Mertens and amr (1985) and the unform-wea topology ntroduced by D Tllo and Fangold (2007). Examples of "behavor" topologes are the strategc topology and the unform strategc topology both ntroduced by DFM. A careful study of the connecton among these topologes wll reveal the relatonshp between the belef mplcaton of types and the behavor mplcaton of types, whch s the man theme of ths paper. 2

3 The product topology s used by Mertens and amr to acheve ther homeomorphsm result. A sequence of types t n converges to a type t n the product topology f and only f the th -order belef of t n converges to that of t for any. Snce proxmty of tals of the herarches of t n and t s not requred, t s now well nown that we may have two types whch are close under the product topology but exhbt very d erent strategc behavors (see Rubnsten (1989)). Ths rases the queston as to whether there exsts a topology under whch nearby types always have smlar strategc behavor (see Monderer and Samet (1989), Monderer and Samet (1996), Kaj and Morrs (1998), and Deel, Fudenberg, and Morrs (2006)). In partcular, DFM propose the strategc topology whch s just strong enough to guarantee that for every nte game, the correspondence whch maps types nto " nterm-correlated-ratonalzable ("-ICR) actons s contnuous. More precsely, DFM show that upper-hemcontnuty of the "-ICR correspondence under strategc convergence s equvalent to product convergence, and t s precsely the lower-hemcontnuty property whch maes strategc convergence a more strngent requrement. DFM also ntroduce the noton of unform strategc convergence, whch adds to the noton of strategc convergence the addtonal requrement that the degree of smlarty of "-ICR actons s unform over all bounded nte games. In contrast to the strategc topology, the unform strategc topology has ts own mportance, especally when appled to mechansm desgn. We dscuss ths mportance further n Secton 5.1. Though useful, DFM s de ntons of strategc topologes are complex snce they nvolve drect reference to best reples, whch are ted to games. In partcular, we are not able to tell whether a sequence of types converges n the strategc topology, unless we study the behavors of all these types n all nte games. To address ths ssue, D Tllo and Fangold (2007) propose the unform-wea topology. A sequence of types t n converges to a type t n the unform-wea topology f and only f the th -order belef of t n converges to that of t for any order and the rate of convergence s unform over. D Tllo and Fangold show that around any nte type, the strategc topology s fully characterzed by the unform-wea topology. To characterze the relatonshp among these varous topologes, we rst study the m- 3

4 plcaton of unform-wea convergence around general types. Our rst man result (Theorem 1) shows that unform-wea convergence always guarantees unform strategc convergence (hence also strategc convergence) around any types. Ths gves a su cent condton for unform strategc convergence (hence, also for strategc convergence), whch s easy to use. 1 Coupled wth D Tllo and Fangold s result, we see that around nte types, unform-wea convergence, strategc convergence and unform strategc convergence are all equvalent. In a recent paper, Ely and Pes (2007) propose an nsghtful partton of the unversal type space nto regular and crtcal types. Regular types are types around whch the strategc topology s equvalent to the product topology. Around crtcal types the strategc topology s strctly ner. Ely and Pes (2007) o er ths surprsngly concse characterzaton of crtcal types: a type s crtcal f and only f for some p > 0, t has common p-belef for some closed proper subset n the unversal type space. Therefore, nte type spaces as well as other type spaces typcally consdered n applcatons consst entrely of crtcal types. Our rst man result leaves open the queston as to whether D Tllo and Fangold s equvalence result on nte types can be extended to all crtcal types,.e., does strategc convergence also mply unform-wea convergence around any crtcal type? To answer ths queston, we construct an n nte crtcal type whch cannot be approxmated by any sequence of nte types under the unform strategc topology. Ths result, coupled wth our rst man result and DFM s denseness result for nte types under the strategc topology, shows that the D Tllo and Fangold s equvalence result does not hold for a general crtcal type. Moreover, n sharp contrast to DFM s denseness result, our second man result (Theorem 2) shows that nte types are nowhere dense under the unform strategc topology. 2 Hence, by our rst man result, nte types are nowhere dense under the unform-wea topology as well. Gven two topologes = and = 0, let "= = 0 " mean that = s wealy ner than = 0 1 As shown n D Tllo and Fangold (2007), the unform-wea topology s a generalzed noton of the common p-belef ntroduced by Monderer and Samet (1989). 2 DFM conjecture that nte types are not dense under the unform strategc topology. They am to use ther Proposton 2 to prove ths conjecture. However, Chen and Xong (2008) prove that ther Proposton 2 s not true by constructng a counterexample. Hence, pror to the current paper, whether nte types were dense under the unform strategc topology was an open queston. 4

5 and let "= = 0 " mean that = s strctly ner than = 0. Our man contrbuton can be summarzed as follows. Our rst man result shows that [unform-wea topology][unform strategc topology]. 3 Coupled wth DFM s denseness result, our second man result shows that [unform strategc topology][strategc topology]. Moreover, DFM show that [strategc topology][product topology]. Therefore, we show that these topologes are related as follows. [unform-wea topology] [unform strategc topology] [strategc topology] [product topology]. The next secton of ths paper contans basc de ntons and notatons. In Secton 3, we present our rst man result and a setch of the proof. We turn to the ssue of non-denseness of nte types n Secton 4. In Secton 5, we o er some dscusson about related ssues. All proofs are relegated to the Appendx. 2 Prelmnares Throughout ths paper, for any arbtrary separable metrc space Y wth metrc d Y, let (Y ) be the space of all probablty measures on the Borel -algebra of Y endowed wth the wea*- topology. It s well nown that the wea*-topology s metrzable wth the Prohorov dstance de ned as (; 0 ) = nf f > 0 : (E) 0 (E ) + for every Borel set E Y g, 8, 0 2 (Y ) where E fy 0 : nf y2e d Y (y 0 ; y) < g. Unless explctly noted, all product spaces wll be endowed wth the product topology and subspaces wth the relatve topology. Every nte or countable set s endowed wth the dscrete topology and denote the cardnalty of a nte set E by jej. Moreover, let supp denote the support of a measure de ned on a nte set. Fnally, for any E Y and y 2 Y, let 1 E be the ndcator functon on E and y be the pont mass on y. For smplcty, assume that there are two players, player 1 and player 2. Gven a player 2 f1; 2g, let denote the other player n f1; 2g. The basc uncertanty s a nte set whch 3 Whether [unform-wea topology][unform strategc topology] s true remans an open queston, for whch we do not have an answer. 5

6 s denoted by. Let Y 0 = and Y 1 = Y 0 (Y 0 ). Then, for 2 de ne recursvely Y = ; 1 ; :::; 2 Y 0 Y 0 Y : marg Y l 2 l = l 1, 8l = 2; :::;. Then, the Mertens-amr unversal type space s de ned as T = 1 ; 2 ; ::: 2 1 =0 Y : marg Y l 2 l = l 1, 8l 2. For each 1, let : T! Y be the natural projecton. For every player and 1, let T and Y denote the copes of T and Y respectvely, wrte : T! Y for, and de ne T = (T ). An element t 2 T s a type of player. For smplcty, we wll wrte t nstead of (t ) for the th -order belef of type t. 4 By the result of Mertens and amr (1985), T (endowed wth product topology) s homeomorphc to ( T ). Let denote ths homeomorphsm. In the Mertens-amr constructon, for any type t, the margnal dstrbuton of (t ) on Y agrees wth the dstrbuton t. Let 0 be the dscrete metrc on Y 0 =,.e., 0 (; 0 ) = 1 f 6= 0 and 0 (; ) = 0. For 1, let be the Prohorov metrc on Y wth respect to the metrc d on Y 1 de ned recursvely as d ; :::; ; 0 ; :::; = maxf 0 (; 0 ); ( 1 ; 1 ); :::; ( ; )g. As de ned by D Tllo and Fangold (2007), the unform-wea topology s generated by the metrc d uw (t; s) sup t ; s for types t and s n T. 1 Followng DFM, we assume that there s a xed exogenous bound M > 0 for the payo s of all nte games we consder. Let G = (A ; g ) =1;2 be a nte game where A s a nte set of actons for player and g : A A! [ wll use the followng recursve de nton of M; M] s the payo functon. For 0, we nterm-correlated-ratonalzable set whch s equvalent to the xed-pont de nton (see Deel, Fudenberg, and Morrs (2007)). Let R (G; ) (R (G; )) =1;2 R (t ; G; ) t 2T =1;2 \1 =0R (G; ) 4 Note that T = (T ) and hence when we wrte t 2 T wthout specfyng the type t, t should be understood as the th -order belef of some type t 2 T. 6

7 where R 0 (G; ) (A ) t 2T =1;2, R (G; ) R (G; ) =1;2 R (t ; G; ) t 2T, 8 1, =1;2 and a 2 R (t ; G; ) f and only f there exsts a measurable functon : T! (A ) such that 5 supp (; t ) R (t ; G; ) for (t ) almost surely (; t ) ; T [g (a ; a 0 ; ) (; t )] (t ) [(; dt )] g (a ; a 0 ; ) (g (a ; a ; ) g (a 0 ; a ; )) a 2A. for all a 0 2 A nfa g where For any 2 (A ), let g (a ; ; ) 2 < ja j denote the vector of expected payo d erence under,.e., 0 g (a ; ; X (a 0 ) [g (a ; a ; ) 1 g (a 0 ; a ; )] A a 0 2A a 2A Observe that R (t ; G; ) depends only on the th -order belef of type t. Through the followng two lemmas, we wll reach an alternatve characterzaton of the ICR set whch wll be used n the proof of Theorem 1. Ther proofs can be found n Appendx A.1. Frst, n the de nton of R (t ; G; ) above, a conjecture s a measurable functon from T to (A ). Snce we wll study the n uence of th -order belefs on R (t ; G; ), the followng de nton and lemma o er a useful alternatve de nton of R (t ; G; ). De nton 1 R 0 (t ; G; ) = A for all t and. For 1, a 2 R (t ; G; ) f and only f there exsts a measurable functon : T! (A ) such that supp ; t R (t ; G; ) for t almost surely ; t ; (1) T where T 0. ; t g (a ; a 0 ; ) t ; dt for all a 0 2 A, (2) 5 Throughout the paper stands for the nner product of two vectors wth the same dmenson. 7

8 Lemma 1 R (t ; G; ) = R (t ; G; ) for every nteger 0, every t, and every player. Hereafter, a measurable functon from T to (A ) whch sats es condton (1) s sad to be a vald conjecture. Moreover, an acton a whch sats es condton (2) s sad to be a best reply under for type t. Second, the followng result shows that provng a 2 R (t ; G; ) s equvalent to provng that for any mxed acton 2 (A nfa g), we can nd a vald conjecture (whch may vary wth ) such that playng a s -better than playng. Note that ths equvalent characterzaton reverses the quant er of (2) whch requres a vald conjecture worng for all a 0 (and hence for all 2 (A nfa g)). 6 Lemma 2 For any postve nteger, any 0, any nte game G, and any type t 2 T, a 2 R (t ; G; ) f and only f for every > 0 and 2 (A nfa g) there s a vald conjecture [] : T! (A ) for t under whch T g (a ; ; ) [] ; t t ; dt. For each t 2 T, de ne h (t ja ; G) to be the mnmal under whch an acton a s ratonalzable for t n G,.e., h (t ja ; G) = mn f : a 2 R (t ; G; )g. 7 Fx 2 (0; 1). Let G m be the collecton of all games where each player has m actons and the payo s are bounded by M. Let G be the collecton of all two-player nte games wth payo s bounded by M. Followng DFM, we de ne the strategc topology and the unform strategc topology on types to be the topologes generated by the metrcs d s and d us respectvely, where d s (t ; s ) 1X m=1 m d us (t ; s ) sup a 2A (G);G2G Clearly, d us (t ; s ) d s (t ; s ). sup jh (t ja ; G) h (s ja ; G)j for types t and s n T ; a 2A (G);G2G m jh (t ja ; G) h (s ja ; G)j for types t and s n T. 6 Ths mnmax step s conceptually d erent from the standard equvalence between the never-best reply and a strctly domnated strategy, though the techncal content s smlar. 7 DFM s Proposton 1 shows that the mnmum exsts. 8

9 3 d uw convergence ) d us convergence We are now ready to state our man result as follows. Theorem 1 For any " > 0 and two types t and s n T wth d uw (t ; s ) ", we have d us (t ; s ) 6M" and hence d s (t ; s ) 6M". Theorem 1 s an mmedate consequence of the followng proposton. Proposton 1 For any nte game G, any "; 0, and any types t and s n T wth t ; s ", we have R (t ; G; ) R (s ; G; + 6M") for every nteger 0. The formal proof of Proposton 1 s long and wll be presented n Appendx A.2. Here we o er a setch to hghlght the essental deas. We prove ths proposton by nducton on and dvde the proof of the nducton step nto ve sub-steps. We now provde a roadmap by descrbng the role of each step. Frst, for a 2 R (t ; G; ), there s some vald conjecture under whch a s a reply for type t. Our goal s to show that a 2 R (s ; G; + 6M") by ndng another vald conjecture 0 under whch a s a ( + 6M") best reply for type s. By Lemmas 1 and 2, t su ces to show that for every > 0 and 2 (A nfa g), there exsts a vald conjecture 0 such that T g (a ; ; ) 0 ; t t ; dt best 6M". (3) T To nd the desred 0, we observe rst that s must assgn a large probablty to " where T s the set of ; t s where randomzes among ratonalzable actons of player (step 1). Then, we de ne 0 = on T we de ne 0 outsde T " so long as t s vald. To de ne 0 on T and t s mmateral how " nt, we rst nd a nte partton f m g of (A ) such that n each partton cell, there s some representatve probablty vector q m whch supnorm-approxmates every q n m. We then use the pre-mage of f m g under to nduce a partton F m on T (step 2). 9

10 Second, step 3 obtans a sutable measurable extenson of 0 from F m to F m " so that the nducton hypothess can be nvoed to tae care of the valdty of 0. Thrd, n step 4, we solve the double-countng problem whch arses when we specfy 0 on the ntersecton of Fm " and F ". 0 m Here we wll see the advantage of allowng 0 0 to depend on. Gven, there s one obvous way to acheve (3) by maxmzng g (a ; ; ) 0 ; t.e., to assgn 0 accordng to the extenson on F m " obtaned n step 3 whenever g (a ; ; )q m g (a ; ; 0 ) q m0. Step 5 llustrates the dea of provng (3) under Step 1: focus on the support of t Let T ; t 2 T : supp ; t R t ; G;. Snce s vald, t T = 1. Moreover, wth t ; s ", we have s T " 1 ". Snce the payo has a unform bound M, how to de ne 0 on the complement of T " s not mportant because the payo resulted from ths regon s at most M". For ; t n T we can tae care of ts valdty by smply choosng 0 to be dentcal to. The ey ssue s to sutably de ne 0 for ; t n T " but outsde T, whch wll be dscussed n step 3., 3.2 Step 2: dscretze (A ) For the conjecture 0 that we are about to de ne, we wll prove a 2 R (s ; G; + 6M") by showng that (3) holds. However, t s not obvous how the condton t ; s " can be appled to evaluate the payo d erence. To solve ths problem, we dscretze the smplex (A ). Note that (A ) s a (ja j 1)-dmensonal compact set. Therefore, for any postve nteger h, we can nd a nte partton f m g m=1 of (A ) such that q m 2 m s a representatve element of m, and the supnorm of (q m q 2 m. q) s no more than 1 h We can partton T nto N = jj sets F m where for each and m, Fm f ; t 2 T : ; t 2 m g. for any Then, we can approxmate the expected payo of t under by replacng ; t wth 10

11 q m f ; t belongs to F m. Ths approxmaton has at most an error of 2MjA j h be arbtrarly small f h s su cently large. whch can 3.3 Step 3: measurable extenson of the conjecture We now specfy 0 for ; t n T 0 = on T. By step 2, T = [ ;m Fm and T " but outsde T. By step 1, we have de ned " ". = [;m Hence, one way to solve the problem s to extend the conjecture on F m to F m " for each (; m). We have to tae care of two requrements to guarantee the valdty of 0. Frst, 0 must be measurable. Second, for any ; t 2 F ", m 0 ; t must be close to q m so that the approxmaton n step 2 s stll vald. These requrements are handled by the followng lemma whose proof can be found n Appendx A.2.1. F m Lemma 3 Consder a separable metrc space (Y; d Y ), a Borel set F Y, and " > 0. Suppose f : F! s a measurable functon from F to another measurable space. Then, there s a measurable functon f " : F "! such that f " = f on F, and for every y 2 F " nf, f " (y) = f (y 0 ) for some y 0 2 F wth d Y (y; y 0 ) < ". The lemma guarantees that for ; t wthn to be equal to 0 ; s 0 ; t for some 0 ; s F " m but outsde F m, we can de ne n F m and stll preserve measurablty. The valdty of 0 wll then be granted by the nducton hypothess. Moreover, by dong so we reduce the comparson of the expected payo s to the much more tractable tas of evaluatng the probablty d erences on the sets F m " and F m subject to a double countng problem to be dscussed n the next step. That s, the expected payo s for t under and for s under 0 can be approxmated by P ;m [g (a ; ; ) q m ] t Fm and P ;m [g (a ; ; ) q m ] s F " m respectvely, for whch the condton t ; s " s readly appled. 11

12 3.4 Step 4: the double-countng problem The double-countng problem mentoned n step 3 arses because f Fm "g;m may not partton T Namely, for ; t ". n F " m \ F ", 0 m should we de ne 0 0 followng the extenson on Fm " or the one on F "? 0 m To solve ths problem, t s mportant to recall 0 Lemma 2 whch allows us to x 2 (A nfa g) before ndng the conjecture 0 to prove (3). Then, for every ; t n F " m \ F ", 0 m we can smply assgn 0 0 to follow the extenson on F m " f g (a ; ; )q m g (a ; ; 0 )q m0. Ths wll mae the expected value of g (a ; ; ) 0 (; t ) as large as possble and help us to ratonalze a. such that Formally, we can relabel the elements n ff mg ;m as follows. Let F m g (a ; ; 1 ) q m 1 g (a ; ; N ) q m N. ;m = ff n m n g N For notatonal smplcty, we often wrte F n nstead of Fm n n when no confuson may arse. By the way we order F n, we should assgn 0 ; t to follow the extenson on (Fn ) " wth n beng the smallest number such that (F n ) " contans ; t. Formally, ths amounts to modfyng the sets f(f n ) " g to the sets fe n g whch are de ned as follows, and de ne 0 followng the extenson n step 3 on each E n. E 1 = (F 1 ) " and E n = (F n ) " n [ n 1 l=1 E l for n 2. Wth ths mod caton, fe n g N 1 ", parttons T and we can approxmate the payo of s under 0 by P N [g (a ; ; n ) q mn ] s [E n ] wthout double-countng. 3.5 Step 5: combnng sets The last step s to show that the d erence of the two approxmated payo s s not large,.e., NX [g (a ; ; n ) q mn ] s (E n ) t (F n ) 4M". (4) Let A n = g (a ; ; n ) q mn and B n = s (E n ) t (F n ). For notatonal convenence, we also relabel fa n g N andfbn g N n a reverse order,.e., let Cn = A N n+1 and D n = B N n+1 for every n = 1; :::; N. 12

13 The followng clam wll be useful n showng (4) and s formally proved n Appendx A.2.2. It follows bascally from our constructon that [ l E n = [ l (F n ) " and the assumpton t ; s ". Clam 1 We have P l Bn " and P l Dn " for 1 l N. We now setch the dea for the proof of (4). Frst, recall from the prevous step that A 1 A 2 A N. For heurstc purpose, assume that N = 3 and A 3 0. We prove that P 3 An B n 2M" n two steps. If fb n g are all nonnegatve, then P 3 An B n 0 > 2M". If fb n g are all non-postve, then P 3 An B n 2M P 3 Bn 2M" because of Clam 1 and our assumpton that ja n j 2M for all n. The man problem s to deal wth the stuaton n whch B 1, B 2, and B 3 have d erent sgns. For example, suppose B 1 > 0, B 2 < 0, and B 3 < 0. We use the followng trc. Snce B 1 > 0 and A 1 A 2, we wll not ncrease the value of P 3 An B n f B 1 s moved from beng multpled by A 1 to beng multpled by A 2. That s, P 3 An B n A 2 (B 1 + B 2 ) + A 3 B 3. Then, we chec the sgn of B 1 + B 2. If (B 1 + B 2 ) 0, then 3X A n B n A 2 B 1 + B 2 + A 3 B 3 2M B 1 + B 2 + B 3 2M", where second nequalty follows because both (B 1 + B 2 ) and B 3 are non-postve and ja n j 2M; the last nequalty follows from Clam 1. If (B 1 + B 2 ) > 0, then the value of A 2 (B 1 + B 2 )+ A 3 B 3 decreases f B 1 +B 2 s further moved from beng multpled by A 2 to beng multpled by A 3,.e. A 2 (B 1 + B 2 ) + A 3 B 3 A 3 (B 1 + B 2 + B 3 ). Hence, 3X A n B n A 2 B 1 + B 2 3X + A 3 B 3 A 3 B n 2M", where the last nequalty follows from ja 3 j 2M and Clam 1. Our argument taes advantage of the property that A n s decreasng n n so that movng a postve B toward beng multpled by a smaller A does not ncrease the value. After all these moves are done, only the last term may have a postve coe cent B. If 13

14 the last term s negatve, we go bac to the sngle-sgned case. If the last B s postve, snce fa n g are all nonnegatve, we can smply throw out A n B n wthout ncreasng the value and also go bac to the sngle-sgned case. 8 The general proof nvolves dvdng the summaton P N An B n nto two groups. One group has all the nonnegatve A n s and the other group has all the negatve A n s. For both groups, we can nvoe a smlar trc and Clam 1 and conclude P N An B n 4M" (see the proofs of Clams 2 and 3 n Appendx A.2 for detals). 4 Non-denseness of nte types In ths secton, we rst show by an example that nte types are not dense under the unform strategc topology. We acheve ths goal by drectly constructng an (n nte) type t such that d us (t ; t ) M for any nte type t 16. Based upon the example, we go one step further to show that the set of nte types s nowhere dense n the unversal type space,.e., the complement of the unform strategc closure of nte types s open and dense. Fnally, we remar t s a crtcal type n the sense of Ely and Pes (2007) and comment on the mplcaton of our example to the relatonshp between the strategc topology and the unform-wea topology around crtcal types. Throughout ths secton, consder the case that = f0; 1g. Ths smpl caton allows us to follow Morrs (2002) to de ne the terated expectatons of a type t = ( 1 ; 2 ; :::) 2 T whch wll be used later. Let 0 : Y 0! [0; 1] be de ned as 0 = 1 f=1g and b (t) = ( ) d, 8 1. Y De ne b 1 (t) b (t) 2 [0; 1] 1. Say a sequence of terated expectatons x 2 [0; 1] 1 s =1 generated by a type t f x = b (t). We now that every x 2 [0; 1] 1 can be generated by some type t 2 T (see (Morrs, 2002, Example)). We are now ready to de ne the type t. Let t be the type whch has the followng 8 Note that we cannot smply delete all A n B n such that B n > 0. Consder the specal case we dscuss here for example. Whle we have P 3 An B n A 2 B 2 + A 3 B 3, we cannot apply Clam 1 to show that A 2 B 2 + A 3 B 3 2M". Recall that B 2 + B 3 = s (E 2 [ E 3 ) t (F 2 [ F 3 ). Snce (E 2 [ E 3 ) may not be equal to (F 2 [ F 3 ) ", we cannot get B 2 + B 3 " from t ; s ". 14

15 terated expectatons 8 b n(n 1) +1 (t ) ; :::; b < n(n+1) (t ) = 2 2 : (1; 1; ::::; 1), f n s odd; (0; 0; ::::; 0), f n s even. That s, b (t ) = (1; 0; 0; 1; 1; 1; 0; 0; 0; 0; :::). For notatonal convenence, denote a 0 = ( 1 2 ; 1 2 ; :::; 1 2 ; :::); a n = ( b n (t ) ; b n+1 (t ) ; b n+2 (t ) ; :::) for n = 1; 2; :::. Note that a n 6= a m for n 6= m. For = 1; 2; :::, let (a n ) be the th element of the sequence a n. One feature of a n, whch wll be useful later n the proof, s that (a n ) = (a n+ ) 1. Let b T b T T T be the smallest (w.r.t. set-ncluson) belef-closed subset such that t 2 b T. Observe that 0 (and 1) s the mnmum (and maxmum) that a th -order expectaton can acheve. Snce b (t ) = (1; 0; 0; 1; 1; 1; 0; 0; 0; 0; :::), there s a unque type t n T whch has the followng herarchy of belefs: rst-order belef: pont mass on = 1; second-order belef: pont mass on [player beleves = 0 wth probablty 1]; thrd-order belef: pont mass on [player beleves wth probablty 1 that player beleves wth probablty 1 that = 0], and so on. Let t (1) = t. Moreover, let t (n) denote the unque type n T that generates the terated expectatons a n, for n 2. Hence, [t (n)] [ b 1 (t (n)) ; t (n + 1)] = 1 for n 1. (5) That s, type t beleves hs opponent s t (2) wth probablty 1; type t (n) beleves hs opponent s t (n + 1) wth probablty 1 (cf. (Morrs, 2002, Example) and Mertens and amr (1985)). Therefore, b bt [ T b = fa 1 ; a 2 ; :::g. We now show that nte types are not dense under the unform strategc topology. We provde an outlne of our argument here, and the rgorous proof can be found n Appendx A.3.1. It s helpful to consder rst the followng mod ed verson of the hgher-order 15

16 expectaton (HOE) game due to Morrs (2002). 9 8 >< g (a;n ) 1 ; (a ;n) 1 ; = >: A = fa 0 ; a 1 ; a 2 ; a 3 ; :::g; 8 < M nf : 0, f (a ;n ) 1 = a0 ; 9 (a ;1 ) 2 ; (a ;2 a ;1 ) 2 ; = :::; (a ;n a ;n 1 ) 2 ; ::: ;, f (a ;n) 1 6= a0. Whle G = (A ; g ) =1;2 s a game wth n ntely many actons, we wll show how t can be mod ed to a nte game n Appendx A.3.1. In ths game, a player s actons are a 0, a 1, a 2,... The acton a 0 always generates the maxmal payo If the player chooses a n wth n 1, he gets the n mum over the quadratc losses between the rst coordnate of player s acton and, the quadratc loss between the second coordnate of player s acton and player s rst coordnate, and so on. That s, t s a coordnaton game n whch a player tres to match the state of nature and hs opponent s acton. The players can acheve the maxmal payo 0 by ether tang the safe acton a 0 or havng perfect coordnaton wth nature as well as every coordnate of hs opponent s acton. Frst, we have a 1 2 R (t ; 0) because for each each player of type t (n) 2 b T can ratonalze a n by holdng the belef that t (n + 1) wll choose the acton a n+1. Second, we show that for some postve but small enough, a 1 =2 R (t ; ) for every nte type t. Let T T T T be the smallest belef-closed subset such that t 2 T. Suppose nstead that a 1 s ratonalzable for t. Then, (A) player must beleve most of hs opponent s types (whch are n the support of [t ]) are playng a 2 so as to get almost perfect coordnaton; 9 One may wonder f we can use the orgnal HOE game, n whch the payo s de ned as, " 1 # g (a;n ) 1 ; (a j;n) 1 ; X = M 1 (a ;1 ) 2 (a ; a j; ) 2, such that > 0 and 1X = 1: =1 Consder two types t and t 0 such that ther expectatons d er only at the N th -order. On the one hand, we should choose a N large enough so that the game separates t and t 0 n the sense that the mnmal to ratonalze some acton under these types d ers. On the other hand, the player chooses a ;N b N [t 0 ] only f both players almost truthfully report all ther th -order expectaton for every up to N, whch requres those s to be large enough. Snce P 1 =1 = 1, t does not seem obvous to us how to resolve ths tenson for a large N. 10 We add ths safe acton a 0 to mae t easer to rule out certan actons as not beng ratonalzable. =2 16

17 and (B) a 2 must be ratonalzable for these opponents. Pc one t whch s beleved to play a 2 by t. Snce a 2 s ratonalzable for t, we can smlarly get (A) player must beleve most of player s types n the support of [t ] are playng a 3 ; and (B) a 3 must be ratonalzable for these player s types. Dong ths nductvely, we can get an n nte chan of ratonalzaton. t = bt (1)! bt (2)! bt (3)! bt (4)! bt (5)! such that (6) bt () 2 T f s odd and bt () 2 T f s even; bt ( + 1) s n the support of bt () ; and a s ratonalzable for bt (). Snce t s a nte type, some type n T must recur n ths n nte chan. That s, we can nd bt (n) = bt (m) = et (1) 2 T such that a n and a m are both a n 6= a m. ratonalzable for et (1) and Recall that a player can always acheve the maxmal payo by tang the safe acton a 0. Snce s small enough, n order to mae a n ratonalzable for et (1), a n+1 must be ratonalzable for most of et (1) s opponent s types n the support of et (1) ; to mae a m ratonalzable for et (1), a m+1 must be ratonalzable for most of et (1) s opponent s types n the support of et (1). Hence, there exsts some type et (2) n the support of et (1) such that a n+1 and a m+1 are both ratonalzable for et (2). Smlarly, we can pc et () n ths way for all 2. Hence, we can construct another common chan of for both a n and a m : et (1)! et (2)! et (3)! et (4)! et (5)! such that et () 2 T and a n+ ; a m+ R et () ; f s odd; et () 2 T and a n+ ; a m+ R et () ; f s even. ratonalzaton Snce the players are also tryng to coordnate wth nature n ths game, we also have the property (C) for a small enough and any acton a ratonalzable for a type t, (a) 1 = 0 b 1 (t) 2 [0; 1 ]. We then reach a contradcton by applyng property (C) to ths common chan 2 of ratonalzaton for a n and a m. Frst, a n 6= a m mples (a n ) = 1 6= 0 = (a m ) for some. Second, by the de ntons of a n and a m, (a n+ ) 1 = (a n ) = 1 6= 0 = (a m ) = (a m+ ) 1, 17

18 but property (C) mples, (a n+ ) 1 = (a m+ ) 1 = 0 f b 1 e t ( ) 2 [0; 1 2 ]; (a n+ ) 1 = (a m+ ) 1 = 1 f b 1 e t ( ) =2 [0; 1 2 ], because both a n+ and a m+ are ratonalzable for et ( ) n the common chan above. Ths s a contradcton. Therefore, a 1 =2 R (t ; ). In the rgorous proof for Proposton 2 n Appendx A.3.1, we formalze the argument above. We mae the game nte by truncatng a n nto ts rst N coordnates and proceed n four steps. In step 1, we show that f N s su cently large, we can stll nd a n and a m n the ratonalzaton chan n (6), such that the N-truncaton of a n and a m are dstnct. Step 2 proves propertes (A) and (C). Step 3 constructs the common chan of ratonalzaton for a n and a m as above. Step 4 derves the contradcton. In the proof, = M s small enough 16 to acheve our goal. Proposton 2 d us (t ; t ) M 16 for any nte type t. Hence, nte types are not dense under d us. In fact, Proposton 2 can be used to prove a stronger result. Theorem 2 Fnte types are nowhere dense under d us. Snce the unform-wea topology s ner than the unform strategc topology by Theorem 1, nte types are nowhere dense under d uw as well. The proofs of Proposton 2 and Theorem 2 are relegated to Appendx A.3.1 and A.3.2, respectvely. Recall that Ely and Pes (2007) de ne a crtcal type to be a type around whch the strategc topology s strctly stronger than the product topology. Recall also that D Tllo and Fangold (2007) show that the strategc topology s equvalent to the unform-wea topology around nte types. As shown n Ely and Pes (2007), every nte type s crtcal but not conversely. In partcular, the type t we construct s an n nte crtcal type. To see ths, recall that T b T b T T j s the smallest belef-closed set such that t 2 T b. Clearly, 18

19 bt b T s of common 1-belef under t. By monotoncty of the common 1-belef operator, the closure of b T b T (under the product topology n T T j ) s of common 1-belef at t. Consder the types t 1 ; t 2 2 T T j under whch both t 1 and t 2 have the rst-order belef Pr( = 1) = Pr( = 0) = 1 2. For all (t 1; t 2 ) 2 b T b T, the rst-order belefs of t 1 and t 2 are ether Pr( = 1) = 1 or Pr( = 0) = 1. Thus, t 1 ; t 2 does not belong to the closure of bt T b. Hence, the closure of T b T b s a proper closed subset of T T j. By (Ely and Pes, 2007, Theorem 3), we conclude t s a crtcal type. Snce nte types are dense under the strategc topology by (Deel, Fudenberg, and Morrs, 2006, Theorem 3), Proposton 2 shows that strategc convergence to an n nte crtcal type does not mply unform-wea convergence to ths crtcal type. 5 Dscusson 5.1 The unform strategc topology DFM study the unform strategc topology n contrast to the strategc topology they propose. The denseness result from Secton 4 demonstrates one d erence between the two topologes. In partcular, our example shows that t s sometmes hard to approxmate complcated types wth nte types when we requre unformty of ths approxmaton among all nte games. However, such a unform approxmaton may stll be relevant. Suppose we are facng a mechansm desgn problem where agents nformaton s modeled wth a complcated type space T. DFM s denseness results on strategc topology states that gven any xed game G, we can nd a smple type space T 0 to approxmate T n terms of strategc behavors n G. However, to solve a mechansm desgn problem s to search for a mechansm (game form) among all possble games. Hence, to ensure that the optmal soluton on T 0 ncurs approxmately no loss of accuracy, t s crucal that the strategc behavors under T 0 approxmate those under T n all mechansms nstead of merely n some G. Thus, the unform strategc topology s free of such a problem so long as the mechansm s searched wthn bounded nte games Suppose that G 0 s the optmal mechansm for the smpler type space T 0, whch s close to the true type 19

20 5.2 Comparson of our proof to that of D Tllo and Fangold (2007) D Tllo and Fangold (2007) prove that the unform-wea topology s ner than the strategc topology around nte types. Our rst man result extends ths to all types. In D Tllo and Fangold s proof, they explot the fact that n a nte type space, there s a mnmal unform-wea dstance between any two d erent types wthn ths type space. Thus, when d uw (t n ; t ) s su cently small relatve to the mnmum unform-wea dstance of the nte type space contanng t, the two types t n and t must beleve n approxmately the same set of opponents wth approxmately the same dstrbuton, and no double-countng problem s nvolved. In our proof, we tae advantage of the nteness of games and solve the doublecountng problem by applyng the mnmax argument. 5.3 In nte order mplcatons of Theorem 1 It s natural to doubt whether one can obtan nformaton about the entre herarchy of belefs of types. Therefore, types whch are close under the product topology may stll be deemed ndstngushable from a practcal vewpont. In accordance wth ths dea, extensve studes have been carred out on nte-order mplcatons of notons assocated wth a type. In partcular, Lpman (2003) shows that common-pror types are dense under the product topology, whle Wensten and Yldz (2007) show that n a xed game wthout commonnowledge restrctons on payo s, types wth unque ratonalzable actons are open and dense. As ponted out by DFM, nether the result n Lpman (2003) nor the result n Wensten and Yldz (2007) holds when we consder the strategc topology nstead of the product space T n the strategc topology. The behavors of T and T 0 n G 0 mght stll be qute d erent. Ths s especally true f G 0 has a lot of actons. Recall that the strategc dstance d s (t ; s ) for types t and s n T s a weghted sum of the d erence between the behavors of t and s n all bounded nte game. If G 0 has m actons wth m beng large, d s (t ; s ) assgns a small weght m on the d erence between the behavors of t and s n G 0. Therefore, even f d s (t ; s ) s small, the two types t and s may stll exhbt qute d erent behavors n G 0. However, ths problem can be avoded f we approxmate T by T 00 whch s close to T n the unform strategc topology. 20

21 topology. DFM show that there s an open set n the strategc topology whch conssts entrely of types wth noncommon prors or multple ratonalzable actons. That s, n these cases strategc open sets are rch enough to separate noncommon-pror types from common-pror types, or types wth multple ratonalzable actons from types wth unque ratonalzable actons. A straghtforward consequence of Theorem 1 s that these strategc open sets dent ed n DFM s examples must contan unform-wea open sets. Hence, nether Lpman s result nor that of Wensten and Yldz s holds n the unform-wea topology. 5.4 The dstance d The unform-wea topology s a natural way of strengthenng the product topology. In DFM s Proposton 2, they propose a metrc d whch also mples unform strategc convergence. Two types are close under d f they have unformly close expectatons on all bounded functons measurable wth respect to the th -order belefs for some. However, ths metrc d s too strong n the sense that even when we restrct attenton to nte-order belefs, the topology t generates s stll strctly stronger than the standard wea topology. In partcular, consder an example n Chen and Xong (2008). Suppose that = f0; 1g. Let t be a complete nformaton type under whch t s common 1-belef that " = 1." Let ft n g be a sequence of types under whch both players beleve " = 1" wth probablty 1 1 n and t s common 1-belef that both players beleve " = 1" wth probablty 1 1 n 1 (cf. Monderer and Samet (1989)). Then, t s common 1 n -belef that " = 1" under t n, and moreover, d uw (t n ; t)! 0. By Theorem 1, we have d us (t n ; t)! 0. However, as shown n Chen and Xong (2008), t n does not converge to t under d. In contrast to the result of D Tllo and Fangold (2007), ths example also demonstrates that even f the lmt type s a nte type, the (unform) strategc convergence does not mply the d convergence. 21

22 A Appendx A.1 Alternatve characterzatons of the ICR set A.1.1 Proof of Lemma 1 Recall that R 0 (t ; G; ) = R 0 (t ; G; ) = A and for 1, a 2 R (t ; G; ) there exsts a measurable functon : T! (A ) such that supp (; t ) R (t ; G; ) for (t ) almost surely (; t ) ; (7) T [g (a ; a 0 ; ) (; t )] (t ) [(; dt )] for all a 0 2 A, (8) and a 2 R (t ; G; ) there exsts a measurable functon : T that! (A ) such supp ; t R T where T 0. g (a ; a 0 ; ) ; t (t ; G; ) for t almost surely ; t t ; dt ; (9) for all a 0 2 A (10) Lemma 1 R (t ; G; ) = R (t ; G; ) for every nteger 0, every t, and every player. Proof. We prove ths clam by nducton on. For = 0, the lemma holds by de nton. Now suppose that the lemma holds for some nonnegatve nteger. (R (t ; G; ) R (t ; G; )) Suppose a 2 R (t ; G; ). Then, there exsts a measurable functon : T! (A ) such that (9) and (10) hold. Now consder : T! (A ) such that (; t ) ; t for all (; t ) 2 T. Note that s measurable because s measurable and T s a second countable space endowed wth the Borel -algebra (see (Alprants and Border, 1999, 4.43 Theorem)). Frst, (7) follows because the margnal dstrbuton of (t ) on T agrees wth t and both 22

23 (9) and the nducton hypothess hold. Second, for all a 0 2 A, = = g (a ; a 0 ; ) (; t ) (t ) [(; dt )] T g (a ; a 0 ; ) ; t T T g (a ; a 0 ; ) ; t (t ) ; dt t ; dt where the rst equalty s due to the de nton of and the second s agan because the margnal dstrbuton of (t ) on T a 2 R (t ; G; ). agrees wth t. Therefore, (8) holds and hence (R (t ; G; ) R (t ; G; )) Suppose a 2 R (t ; G; ). Hence, there exsts measurable functon : T! (A ) such that (7) and (8) hold. Snce T s a compact metrc space, t s a standard Borel space. Hence, there s a regular condtonal dstrbuton of (t ) on T (see (Dudley, 2002, Theorem)). De ne : T! (A ) as h ; t e; s (t ) e; ds j e = ; s = t, 8 ; t 2 T. T Then, snce s a measurable functon from T to < ja j, by (Dudley, 2002, Theorem), s a verson of the condtonal expectaton of condtonal on ; t. Hence, s measurable. Agan, (9) follows because the margnal dstrbuton of (t ) on T agrees wth t and both (7) and the nducton hypothess hold. Moreover, for all a 0 2 A, = = T T g (a ; a 0 ; ) ; t t ; dt T [g (a ; a 0 ; ) (; s )] (t ) T [g (a ; a 0 ; ) (; t )] (t ) [(; dt )] h (; ds ) j e = ; s = t (t ) ; dt where the rst equalty s because the margnal dstrbuton of (t ) on T agrees wth t and s also because of the de nton of, and the second equalty follows from the law of terated expectaton (see (Dudley, 2002, Theorem)). Therefore, (8) holds and hence a 2 R (t ; G; ). 23

24 A.1.2 The proof of Lemma 2 Lemma 2 For any postve nteger, any 0, any nte game G, and any type t 2 T, a 2 R (t ; G; ) f and only f for every > 0 and 2 (A nfa g) there s a vald conjecture [] : T! (A ) for t under whch T g (a ; ; ) [] ; t t ; dt. (11) Proof. Recall rst a Mnmax Theorem due to Fan (1952). Let f be a real-valued functon de ned on a product space X Y. Say f s convex-le on X f for every x, x 0 2 X and c 2 [0; 1], there s some x 00 2 X such that f (x 00 ; y) cf (x; y) + (1 c) f (x 0 ; y) for all y 2 Y. Say f s concave-le on Y f for every y, y 0 2 Y and c 2 [0; 1], there s some y 00 2 Y such that f (x; y 00 ) cf (x; y) + (1 c) f (x; y 0 ) for all x 2 X. Fan s Mnmax Theorem Let X be a compact Hausdor space and Y an arbtrary set (not topologzed). Let f be a real-valued functon on X Y such that for every y 2 Y, f(; y) s lower sem-contnuous on X. If f s convex-le on X and concave-le on Y, then mn sup x2x y2y f(x; y) = sup y2y mn f(x; y). x2x To apply ths result, de ne X = (A n fa g) ; Y = : T! (A ) : s a vald conjecture ; f (; ) = [g (a ; ; ) (; t )] t ; dt, 8 2 X, 2 Y. T Obvously, f s convex-le on X, concave-le on Y, and f(; ) s lower sem-contnuous on X for every. Hence, by Fan s Mnmax Theorem, we have mn 2X sup 2Y f(x; y) = sup 2Y mn 2X f(x; y). Frst, sup 2Y mn 2X f(x; y) f and only f for any > 0, there s a conjecture 2 Y such that f (; ) for all 2 X. By Lemma 1 and DFM s Lemma 1, sup 2Y mn 2X f(x; y) s therefore equvalent to a 2 R (t ; G; ). Smlarly, mn 2X sup 2Y f(x; y) s equvalent to for every > 0 and 2 X, there s some vald conjecture [] 2 Y such that (11) hold. 24

25 A.2 The proof of Proposton 1 A.2.1 The proof of Lemma 3 Lemma 3 Consder a separable metrc space (Y; d Y ), a Borel set F Y, and " > 0. Suppose f : F! s a measurable functon where s a measurable space. Then, there s a measurable functon f " : F "! such that f " = f on F, and for every y 2 F " nf, f " (y) = f (y 0 ) for some y 0 2 F wth d Y (y; y 0 ) < ". Proof. Snce F s a subset of a separable metrc space, t s also separable under the relatve topology (see (Dudley, 2002, p.32)). Let fy 1 ; y 2 ; :::g be a countable dense subset of F. For any y 2 Y, let B (y; ") denote the "-open ball around y. Frst, we clam that F " = [ 1 m=1b (y m ; "). Clearly, F " [ 1 m=1b (y m ; "). To see F " [ 1 m=1b (y m ; "), suppose y 2 F ". Then, there s some y 0 2 F such that d Y (y; y 0 ) < ". Hence, B (y; ") \ F 6=?. Snce B (y; ") \ F s relatvely open n F and fy 1 ; y 2 ; :::g s dense n F, there s m such that y m 2 B (y; ") \ F. Hence, d Y (y m ; y) < " and therefore y 2 B (y m ; "). We modfy the sets n fb (y m ; ")g 1 m=1 to B m 1 m=1 such that B 1 = B (y 1 ; ") and B m = B (y m ; ") n [ m 1 l=1 B l for m 2. Observe that 1 B m parttons F ". Moreover, for any m 2 m=1 +, snce the sets n fb (y m ; ")g 1 m=1 are open and hence measurable, B m s also measurable. We now de ne f " as 8 < f (y), f y 2 F, f " (y) = : f (y m ) f y =2 F and y 2 B m. Then, by de nton f " sats es the property that f " = f on F, and for every y 2 F " nf, f " (y) = f (y 0 ) for some y 0 2 F wth d Y (y; y 0 ) < ". Then, Fnally, we show that f " s measurable. Let 0 be an arbtrary measurable set. (f " ) 1 ( 0 ) = f 1 ( 0 ) [ [ fm2 + :f(y m)2 0 g where B m nf = fy 2 Y : y 2 B m and y =2 F g. Bm nf, 25

26 Snce f s a measurable functon and 0 s measurable, f 1 ( 0 ) s also measurable. For each m, the set B m nf s measurable. Hence, the set [ fm2+ :f(y m)2 0 g Bm nf s also measurable, because t s a countable unon of measurable sets. Therefore, (f " ) 1 ( 0 ) s measurable, and f " s a measurable functon. A.2.2 The proof of Proposton 1 Proposton 1 For any nte game G, any "; 0, and any types t and s n T wth t ; s ", we have R (t ; G; ) R (s ; G; + 6M") for every nteger 0. Proof. Let G be a nte game. Snce G s xed, we wll drop hereafter the explct reference of G from our notaton for expostonal ease. We prove the proposton by nducton. Suppose a 2 R (t ; ). By Lemma 1, there s a vald conjecture : T such that T g (a ; a 0 ; ) ; t Our goal s to show that a 2 R (s ; + 6M").! (A ) for t t ; dt for all a 0 2 A nfa g. (12) Consder rst the case of = 1. By De nton 1, T = and hence s a measurable functon from to (A ). Let 0 =. Snce R 0 (t ; ) = A, 0 s trvally vald. If " 1, the clam trvally holds because the payo s bounded by M and hence R (s ; G; + 6M") = A. Now suppose " < 1. Then, snce we endow wth the dscrete metrc, ( 0 ) " = 0 for any 0. Hence, js 1 ( 0 ) t 1 ( 0 )j " snce 1 (t 1 ; s 1 ) ". Let A = g (a ; a 0 ; ) 0 () and B = s 1 () t 1 (). Then, X X X A B = A B + A B 4M" 2 f2:b <0g f2:b 0g where the last nequalty s due to A 2M and js 1 ( 0 ) t 1 ( 0 )j " n partcular for 0 = 2 : B < 0 and for 0 = 2 : B 0. Therefore, by (12), P 2 A s 1 () 4M". Hence, a 2 R (s ; + 6M"). Now consder the nducton step. By Lemma 2, to prove a 2 R (s ; + 6M"), t su ces to show that for every > 0 and 2 (A nfa g), there s a vald conjecture 0 26

27 under whch we have T g (a ; ; ) 0 ; t s ; dt 6M". (F) Snce player has ja j actons, let q q 0 ja j njq max 1 q1j 0 ; :::; q ja j q 0 ja j o for any q and q 0 n (A ). Pc an arbtrary postve nteger h. We can dscretze (A ) wth a nte partton f m g m=1 such that for each m there s some qm 2 m wth q q m ja j 1=h for any q 2 m. Consder T ; t 2 T : supp ; t R t ;. We can nduce from f m g and a partton F m on T m = 1; :::;, Fm f ; t 2 T : ; t 2 m g. 12 such that for each 2 and Label F m as ff n m n g N where N = jj such that g (a ; ; 1 ) q m 1 g (a ; ; N ) q m N. Hereafter, we wrte F n nstead of Fm n n whenever no confuson may arse. Hence, T [ N F n and T " = [ N (F n ) ". = De ne E 1 = (F 1 ) " and E n = (F n ) " n [ n 1 l=1 E l for n 2. Observe that fen g N ", parttons T and moreover, for any 1 l N, we have l[ E n = N[ n=l+1 l[ (F n ) " ; (13) " l[ # E n = T " (F n ) ". (14) We now proceed to de ne the conjecture 0. We dvde T nto three areas: (I) T ; (II) T " n T ; (III) T n T ", and de ne 0 on these three areas respectvely. Frst, for area (I), let 0 =. Second, snce t 7! R (t ; + 6M") s upper hem-contnuous under the product topology on T, by Kuratows-Ryll-Nardzews 12 We can mae each m measurable, so that each F m s also measurable. 27

28 Theorem (see Alprants and Border (1999)), there s a measurable selecton r () wth r t 2 R (t ; + 6M") for all t 2 T. Then, we de ne 0 as r () on area (III). Thrd, we extend the de nton of 0 from area (I) to area (II) by Lemma 3. Recall that ff n g s a partton of T. Snce s vald, t s measurable on F n for every n. By Lemma 3, there s a measurable functon n () on (F n ) " such that n = on F n and for every ; t 2 (Fn ) " there s some 0 ; s 2 Fn such that d ; t ; 0 ; s < " and n ; t = 0 ; s. Recall also that E n (F n ) " "., and moreover, fe n g forms a partton of T In sum, we de ne the conjecture 0 : T 8 < 0 ; t = :! (A ) as n ; t r(t ), f ; t 2 En ; 1, f ; t =2 T ". Observe that 0 s vald n areas (I) and (III) by the de nton of r () and the valdty of. In area (II), by the nducton hypothess and the extenson n de ned above, 0 s also vald. It remans to show that the nequalty (F) holds under 0. It s a drect consequence of the followng three lemmas whch we wll prove later. Lemma 4 We have NX [g (a ; ; n ) q mn ] t [F n ] 2M ja j. (15) h Lemma 4 says that replacng on the left-hand sde of (12) a 0 by and by q mn each F n would nduce at most a loss 2MjA j h. on Lemma 5 We have T g (a ; ; ) 0 ; t NX [g (a ; ; n ) q mn ] s [E n ] 2M" s ; dt 2M ja j. (16) h 28

29 Lemma 5 says that by usng P N [g (a ; ; n ) q mn ] s [E n ] to approxmate the lefthand sde of (16), we may ncur two nds of losses and both are small. One s due to the error outsde T of 0 by q mn ", whch s at most 2M", and the other results from the approxmaton on E n, whch s at most 2MjA j h. Lemma 6 We have NX [g (a ; ; n ) q mn ] s (E n ) t (F n ) 4M". (17) Lemma 6 s a generalzaton of step 5 n Secton 3 whch says that we only have to compare the belefs of t and s on the probabltes of sets fe n g and ff n g. We wll show that the d erence of the two approxmated payo s s at most 4M". By addng up (15) (17), we get g (a ; ; ) 0 ; t T Snce 4MjA j h s ; dt 4M ja j 6M". h! 0 as h! 1, we can choose h large enough so that (F) holds. Hence, for every > 0 and 2 (A nfa g), there s a vald conjecture 0 under whch (F) holds. Thus, by Lemma 2, a 2 R (s ; + 6M"). We now prove Lemmas 4 6. Proof of Lemma 4 Recall that a s a best reply under for t. Hence, g (a ; ; ) ; t T t ; dt. (18) Snce ff n g N 1 s a partton of T, we can wrte = T NX g (a ; ; ) ; t t ; dt g (a ; ; ) ; t (;t )2F n t ; dt. (19) 29