A A Non-Constructible Equilibrium 1

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1 A A Non-Contructbe Equbrum 1 The eampe depct a eparabe contet wth three payer and one prze of common vaue 1 (o v ( ) =1 c ( )). I contruct an equbrum (C, G, G) of the contet, n whch payer 1 bet-repone et the Cantor et. Let c 1 () =. Let F () =, and modfy F () by mmckng the contructon of the Cantor et on [0, 1] n the foowng way. For every removed nterva (a, b) and every (a, b), etf () =a +( a) 2 / (b a). Denote the reutng functon by F. Then F (0) = 0, F (1) = 1, and F contnuou, trcty ncreang, equa precey on the Cantor et, and trcty ower than on t compement. In partcuar, f payer 1 probabty of wnnng a prze when choong F (), then h bet-repone et the Cantor et. To acheve th, et G () atfy F () =1 (1 G ()) 2 for a [0, 1]. That, G () =1 q 1 F (). Then G () contnuou and trcty ncreang, wth G (0) = 0 and G (1) = 1. Todefne the cot functon of payer 2 and 3, et C be the Cantor functon, and reca that t contnuou and weaky ncreang, wth C (0) = 0 and C (1) = 1. Let µq c 2 () =c 3 () =1 (1 G ()) (1 C ()) = 1 1 F () (1 C ()). Then c 2 and c 3 are contnuou and trcty ncreang, wth c 2 (0) = c 3 (0) = 0 and c 2 (1) = c 3 (1) = 1. It traghtforward to verfy that (C, G, G) an equbrum of the contet and payer 1 bet-repone et the Cantor et. B The Eampe of Fgure 2 Payer cot functon are c 2 () = 3, f c 1 () = 100 ( ))2 f < ( 1) f 1 <, 1 I thank George Maath for encouragng me to provde th eampe. 1

2 and c 3 () = f ( ))2 f < ( ) + 25 ( ) 2 f < ( ) ( 0.85) 0.15 f 0.85 <. Thee cot functon gve power of 0, 1 and 1. Perturbng the cot functon ghty doe 4 2 not change the quatatve apect of the equbrum. C ProofofReutfromSecton3 C.1 Proof of Theorem 9 Payer m +2,...,n do not partcpate becaue the condton of Theorem 2 hod. Denote by the nfmum of the core at whch payer actve. Becaue v ( ) =V γ c ( ) and γ > 0, the condton of Theorem 5 hod, o <T for every m +1and there are no atom at the threhod. Becaue m, m +1 CP (0), wehave m = m+1 =0.Now, uppoethatfor, j m, j. Denote by P k ( ) and u k ( ) payer k probabty of wnnng and payoff n the equbrum a a functon of h core. Snce payer and j have potve power and are not actve beow, G = Gj =0.Thu,P = Pj. 2 Therefore, u u j V V j = P (1 α) c (aj a )+αc (aj a ) =(a j a ) c α + P (1 α). Ao, w /V =1 a c (T ) and w j /V j =1 a j c (T ), ow /V w j /V j =(a j a ) c (T ). Snce w /V = u /V and w j /V j u j /Vj,wehave (1) 0 µ w w à j u V V j V u! j =(a j a ) c (T ) c α + P (1 α). V j 2 If =0, conder the mt of the probabte of wnnng a the core approache 0 from above, and mary for u and uj. 2

3 Becaue α>0 and P 1, wehave α + P (1 α) > 0, o (1) hod f and ony f (a j a ) c (T ) c 0. Snce c <c(t ), thmpethataj a. Itremantohowthatfapayerbecomeactveatomecore,thenheremanactve unt the threhod. To do th, et u derve ome properte of payer em-eatcte. Frt normaze payer payoff o that the prze vaue 1 for a payer. To th end, note that the contet trategcay equvaent to a contet n whch a vauaton equa 1 and n whch payer cot a c ntead of γ c.wethenhave ³ q () = v 1 () v (T ) a v ()+c () = a m+1 c () 1 (1 α) a c () and ε () = q0 () q () = c 0 ()(a (α 1) + a m+1 ) (1 c () a m+1 )(a c ()(α 1) + 1).3 Vewed a a functon of a,weobtan ε () c 0 ()(1 α) = a (a c ()(α 1) + 1) 2 0, o at every core payer wth a hgher reach have hgher em-eatcte. Th mean that when a new payer become actve a etng actve payer reman actve, and that whether an actve payer reman actve depend ony on h em-eatcty and thoe of payer wth ower reache. In partcuar, payer m and m +1 are away actve, nce ther em-eatcte are away the owet. To how that payer 1,...,m 1 are actve on an nterva, oberve that the rato of em-eatcte of payer j> non-decreang n core, µ 0 εj () = ε () µ 0 (aj (α 1) + a m+1 )(a c ()(α 1) + 1) (a (α 1) + a m+1 )(a j c ()(α 1) + 1) = a m+1 (1 α) a j a m+1 (1 α) a (1 α)(a j a ) c 0 () (a j c ()(α 1) + 1) 2 0, 3 For epotona mpcty, throughout the proof I gnore the pobty that ome payer emeatcte may be 0 at ome core. Becaue payer cot are trcty ncreang, pecewe anaytcty mpe that there are at mot a fnte number of uch core, o ony ght modfcaton are needed to dea wth uch core. 3

4 and ao that th rato at mot 1 (nce payer wth a hgher reach have hgher emeatcte). Suppoe n contradcton that there a payer who not actve on an nterva, and et m 1 be the payer wth the hghet nde among uch payer. Suppoe that payer actve at >0. DenotebyH +1,...,m+1 ( ) the potve fed pont of the uppy functon defned ung the em-eatcte of payer +1,...,m+1, and conder ome 0 [, T ]. By defnton of payer and becaue payer wth hgher reache become actve at hgher core, payer m +1,..., +1 areactveat 0. So, becaue the em-eatcte of a payer 1,..., 1 are weaky hgher than that of payer, ε ( 0 ) H ( 0 ) f and ony f ε ( 0 ) H +1,...,m+1 ( 0 ).Letb = ε ( 0 ) /ε (). For a j>, nce ε j ( 0 ) /ε ( 0 ) ε j () /ε (), wehaveε j ( 0 ) /ε j () b. Therefore, H +1,...,m+1 ( 0 )= 1 m m+1 X j=+1 ε j ( 0 ) 1 m m+1 X j=+1 bε j () =bh +1,...,m+1 (). Becaue payer actve at, ε () H () o ε () H +1,...,m+1 (). Therefore, ε ( 0 )=bε () bh +1,...,m+1 () H +1,...,m+1 ( 0 ) o ε ( 0 ) H ( 0 ). Th how that once a payer become actve he reman actve unt the threhod. C.2 Proof of Coroary 5 Becaue j,wehaveg () G j () for any. By Theorem 9, both payer are actve on on,t. Ao, ε ε j, o becaue both payer are actve on,t the agorthm how u that h h j on,t. So payer tart beng actve ater and ha a ower hazard rate than payer j, whch mean that G FOSD G j. To ee th, reca that h () = (1 G ()) 0 / (1 G ()), oh h j mpe that (1 G ) 0 / (1 G ) (1 G j ) 0 / (1 G j ).Thmpethatfory,T, 0 Z y µ (1 G ()) 0 1 G () (1 G j ()) 0 µ 1 G () d =n y 1 G j () 1 G j () µ Ã! 1 G (y) 1 G =n n. 1 G j (y) 1 G j Becaue G Gj,wehave 1 G / 1 Gj 1, obytakngeponent the prevou nequaty mpe 1 G (y) 1, org 1 G j (y) j (y) G (y), a requred. 4

5 Th FOSD mpe that payer probabty of wnnng a prze hgher than that of payer j for any gven core, and hence ao n epectaton. To ee th, note that by choong >0 payer chooe a hgher core than payer j wth probabty G j (), wherea by choong payer j chooe a hgher core than payer wth probabty G (). Therefore, becaue G j () G (), for any gven core payer wn a prze wth at eat a hgh a probabty a payer j doe,.e., P () P j (). Therefore,P = R P () dg R Pj () dg, and becaue P j ( ) non-decreang, by FOSD R P j () dg R P j () dg j = P j. C.3 Proof of Theorem 10 Let f (α, ) = 1 Π m+1 k=+1 (1 a kc ()(1 α)) (1 a m+1 c ()) (1 a c ()(1 α)) m, and note that f dfferentabe n α < 1 and < T. Denote by (α) the owet uch that f (α, ) = Π m k=+1 a m. Note that Π m k=+1 a m > 1 (becaue a k ncreae n k) andf (α, 0) = 1 (nce c (0) = 0). Suppoe that when α ncreae to α 0 the vaue of f at (α) ncreae. Then, becaue f α 0, (α) > Π m k=+1 a m and f (α 0, 0) < Π m k=+1 a m, the ntermedate vaue theorem how that (α 0 ) < (α). Therefore, to how that decreae n α t uffce to how that f (a, ) / α > 0 for (0,T). By conderng the epreon for f (a, ) / α, one can verfy that t uffce to how that m+1 X j=+1 a j c () Y k {+1,...,m+1}\j (1 a k c ()(1 α)) (1 a c ()(1 α)) > Π m+1 k=+1 (1 a kc ()(1 α)) ((m ) a c ()). For th nequaty to hod, t uffce that for every j = +1,...m+1, a j c () >a c () and 1 a c ()(1 α) > 1 a j c ()(1 α), 5

6 and th hod becaue a k ncreae n k. Therefore, decreae n α<1 for every payer <m. Now conder what happen to a α approache 0. For<T, f (0,)= 1 Π m+1 k=+1 (1 a kc ()) (1 a m+1 c ()) (1 a c ()) m = Πm k=+1 (1 a kc ()) (1 a c ()) m 1. Therefore, by unform contnuty of f (α, ) on [0, α] [0,] for any α (0, 1), approache T a α approache 0. C.4 Proof of Coroary 6 Chooe β<1. By Theorem 10 there et <T and α >0 uch that for a α< α and <m, >. Chooeuch and α that ao atfy (2) V V m γ m+1 m + αγ γ m+1 m c ( ) >β V m (1 α) γ m c ( ) and (3) αγ m+1 c ( ) V m+1 (1 α) γ m+1 c ( ) < 1 β for every α< α. Conder the unque equbrum G of a uch a mpe contet wth α< α. Snce G ( ) =0for <mand G (0) = 1 for >m+1, Coroary 3 how that the CDF of payer m +1and m on [0, ] are gven by (2) and (3). Snce > for <m,eachof thee m 1 payer chooe a hgher core than payer m +1, and therefore wn a prze, wth probabty hgher than β. Payerm chooe core hgher than wth probabty hgher than β, and therefore wn a prze wth probabty hgher than β 2. 6