Large Contests - Online Appendix

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1 Large Cotests - Oe Appedx Wojcech Oszewsk ad Ro Sege August 2015 We choose a equbrum for each cotest, ad refer to the sequece whch the -th eemet s the equbrum of the -th cotest as the sequece of equbra. For both our theorems, we w show that every subsequece of ths sequece cotas a further subsequece that satsfes the statemet of the theorem. Ths suffces, because the foowg observato cabeappedwthz beg the set of equbra of cotest. (Subsequece Property) Gve a sequece of sets {Z : =1, 2,...}, supposethatfor every subsequece {Z k : k =1, 2,...}, every sequece {z k : k =1, 2,...} wth z k Z k cotas a subsequece {z k : =1, 2,...} such that every eemet z k has some property. The there exsts a N such that for every N every eemet Z hasthsproperty. 1 1 Proof of Theorem 1 We beg wth a oute of the proof. Gve a subsequece of equbra, each equbrum the subsequece duces for each payer a mappg from bds to expected percete rakgs. We cosder the average of those mappgs, ad G 1 composed wth ths average gves a mappg T from bds to przes. As creases, ths mappg approxmates the equbrum mappgs from bds to przes of a payers the -th cotest. We the use Hey s (1912) seecto theorem to fd a subsequece of T that coverges to some mt mappg T from bds to przes. We show that T s cotuous ad the subsequece of T coverges uformy to T. The, for each aget type we defe the set of optma bds whe T s treated as a Departmet of Ecoomcs, Northwester Uversty, Evasto, IL (e-ma: wo@orthwester.edu) ad The Pesyvaa State Uversty, Uversty Park, IL (e-ma: rus41@psu.edu). 1 Otherwse, there woud be a sequece {z k : k =1, 2,...} wth z k Z k of eemets wthout the property.

2 verse tarff, addefe the correspodece from aget types to sets of optma bds. We cosder a sma eghborhood of the graph of ths correspodece, ad show that for arge every payer s best resposes the -thcotestarethe x -sce of ths eghborhood. Such a sce coud prcpe be arge eve f the set of optma bds of the correspodg aget type s sma (ths woud happe f the set of optma bds of a earby aget type s arge). We show, however, that uder strct sge crossg each aget type has a sge optma bd, whch s cotuous ad weaky creases the aget type. Ths mpes that every payer s best respose set, ad therefore the support of her equbrum strategy, s bouded wth a arbtrary sma terva as creases. We the cocude that the uque mechasm duced by T mpemets the assortatve aocato. Ths demostrates part (b) the statemet of the theorem; part (a) the foows easy. For the proof, we take the subsequece of equbra to be the sequece of equbra (ths smpfes otato ad has o effect o the proofs). We deote the equbrum of the -th cotest by σ =(σ 1,...,σ ), whereσ s payer s equbrum strategy; a strategy of payer s a radom varabe takg vaues X B whose marga dstrbuto o X cocdes wth the dstrbuto of payer s types F. By referrg to payer bddg wth some probabty a subset S of B, we mea the probabty of the set X S,.e., the probabty of S measured by the marga dstrbuto of payer s strategy o B. We deote by R (t) the radom varabe that s the percete ocato of payer the orda rakg of the payers the -th cotest f she bds sghty above t ad the other payers empoy ther equbrum strateges. 2 That s, Ã! R (t) = 1 1+ X k6= 1 {σ k X [0,t]} where 1 {σ X [0,t]} s 1 f σ X [0,t] ad 0 otherwse. Let Ã,! A (t) = 1 1+ X Pr (σ k X [0,t]) k6= be the expected percete rakg of payer. The, by Hoeffdg s equaty, for a t B we have Pr ( R (t) A (t) >δ) < 2exp 2δ 2 ( 1) ª. (1) Fay, et A (t) = 1 X A (t) 2 Ths s the fmumofherrakgfshebdsabovet, whch s equvaet to bddg t ad wg ay tes there. If tes happe wth probabty 0, the ths s equvaet to bddg t. 2 =1

3 be the average of the expected percetes rakgs of the payers the -th cotest f they bd t ad the other payers empoy ther equbrum strateges. Let T be the mappg from bds to przes duced by A.Thats,T (t) =(G ) 1 (A (t)), where (G ) 1 (z) =f{y : G (y) z} for z>0, ad(g ) 1 (0) = f {y : G (y) > 0}. (I words, (G ) 1 (z) s the prze of a aget wth percete rakg z whe przes are dstrbuted accordg to G.) Sce every T s (weaky) creasg, by Hey s (1912) seecto theorem for mootoe fuctos the sequece T cotas a subsequece that coverges potwse to a fucto T : B Y. For the rest of the proof, deote ths subsequece by T. We frst descrbe some propertes of verse tarff T : (1) T s (weaky) creasg, because every T s (weaky) creasg. (2) T (0) = 0, otherwse payers bddg 0 woud have proftabe devatos. 3 (3) T (b max )=1, because A (b max )=1ad therefore T (b max )=1. I addto, we w use the foowg property of dscrete cotest equbra: (No-Gap Property) I ay equbrum, there s o terva (a, b) B of postve egth whch a payers bd wth probabty 0 ad some payer bds [b, b max ] wth postve probabty. Proof: Suppose the cotrary, ad cosder such a maxma terva (a, b). A payer woud oy bd b or sghty hgher tha b f some other payer bds b wth postve probabty. But the payer who bds b wth postve probabty woud be better off ether by sghty creasg her bd (f aother payer bds b ad wg the te eads to a hgher prze) or by decreasg her bd ( the compemetary case). Our frst emma shows that T s cotuous. (I order ot to obscure the structure of the proof, we reegate to the ed of the secto the proofs of a emmas.) Lemma 1 For ay t B ad ay sequeces q m t ad r m t B, wehavem T (q m )= m T (r m )=T(t). The dea of the proof s that f T were dscotuous at some t, theforarge t woud be better to bd sghty above t tha sghty beow t. But f o payer bds sghty beow t, the by the No-Gap Property o payer bds t or above. 3 Ideed, suppose to the cotrary that T (0) > 0. Ths meas that for some δ>0ad arge eough, A (0) >G (0) + δ. Thus, a fracto of at east G (0) + δ payers bd 0 the -th cotest wth postve probabty. Ay oe of them woud be better off bddg sghty above 0, ad wg agast a other payers who bd 0, tha bddg 0 ad wth postve probabty osg to a other payers who bd 0. 3

4 Cotuty ad mootocty of T mpy the foowg resut. Lemma 2 T coverges to T uformy o B. Weowreatetheversetarff T to payers behavor the equbra that correspod to the sequece T. Deote by BR x type x s set of optma bds gve T,.e., the bds t that maxmze U(x, T (t),t). Deote by BR(ε) the ε-eghborhood of the graph of the correspodece that assgs to every type x the set BR x. 4 Deote by BR x (ε) the set of bds t such that (x, t) BR (ε). Note that BR(ε) s a 2-dmesoa ope set, whe each BR x (ε) s a 1-dmesoa sce of BR (ε). Usg sets BR x (ε), we ca characterze payers equbrum behavor. Lemma 3 For every ε>0, there s a N such that for every N, theequbrumof the -th cotest every best respose of every type x of every payer beogs to BR x (ε). Strct sge crossg mpes severa propertes of BR x. Lemma 4 For every x the set BR x s a sgeto. I addto, the fucto br that assgs to x thesgeeemetofbr x s cotuous ad weaky creasg. Lemma 4 mpes that for every ε>0 there s a δ>0such that BR x (δ) [br (x) ε, br (x)+ε] for every type x. We therefore have the foowg coroary of Lemmas 3 ad 4. Coroary 1 For every ε>0, there s a N such that for every N, theequbrum of the -th cotest every best respose of every type x of every payer beogs to (br (x ) ε, br (x )+ε). To prove part (b) of the theorem we eed to show that T br s the assortatve aocato. Ths s doe by the foowg emma. Lemma 5 G 1 (F (x)) = T (br (x)) for a types x. Thus, the mechasm that prescrbes for type x prze T (br (x)) ad bd br (x) s a tarff mechasm that mpemets the assortatve aocato. Moreover, every type ca get at east 0 by bddg 0, adtype0 gets o more tha 0 (because T (br (0)) = 0). To compete the proof, t remas to show part (a) of the theorem. I short, ths part foows from Coroary 1 ad Hoeffdg s equaty (see Appedx 1.6). 4 That s, BR (ε) s the uo over a types x ad bds t BR x of the ope bas of radus ε cetered at (x, t). 4

5 1.1 Proof of Lemma 1 Suppose frst the emma s fase for some t (0,b max ] ad q m t. Let y 0 = m T (q m ) ad y 00 = T (t) (the mt exsts by the mootocty of T ), ad et γ =(y 00 y 0 ) /2. Suppose frst that U (0,y,t) strcty creases y. (Reca that we assumed U (x, y, t) strcty creases y oy for x > 0.) The, by uform cotuty of U, thereexstδ, > 0 such that every type x gas at east from obtag a prze hgher by γ at a bd hgher by δ. More precsey, U (x, y + γ,t + δ) U (x, y, t) (2) for a x, y, adt such that y + γ ad t + δ beog to the doma of U. Ths mpes, as U s bouded, that every type x strcty prefers bddg t + δ ad obtag wth suffcety hgh probabty a prze suffcety cose to y + γ to bddg t ad obtag wth suffcety hgh probabty a prze suffcety cose to y, depedety of the przes obtaed wth the remag probabty. Choose t 0 = q m such that t t 0 <δ. Next, choose arge eough so that T (t) T (t) < γ/2 ad T (t 0 ) T (t 0 ) <γ/2. ThsmpesthatT (t) T (s) >γfor ay bd s t 0. By choosg arge eough, we guaratee (see (1), whch appes uformy to a bds) that R (s), the percete rakg of payer who bds s the -th cotest, s cose to A (s) wth hgh probabty, ad R (t) s cose to A (t) wth hgh probabty. Thus, every type x obtas a prze suffcety cose to T (t) wth a suffcety hgh probabty by bddg (sghty above) t, ad obtas a prze that s wth a suffcety hgh probabty at most sghty hgher tha T (t 0 ) by bddg (sghty above) ay s t 0. 5 Therefore, because t t 0 <δ, o payer bds ay s (t δ, t 0 ] wth postve probabty, so by the No-Gap Property T (t 0 )=1.ButT (t 0 ) T (t 0 ) y 0 <y 00 1, a cotradcto. Whe U (0,y,t) oy weaky creases y, the argumet above shows that for ay ε>0 there exst δ, > 0 for whch (2) hods for every type x [ε, 1]. There aso exsts t 0 = q m such that t t 0 < δ ad T (t) T (t 0 ) > 3γ/2 for arge eough. Lettg t 00 =f{s : T (s) T (t) γ} (t 0,t] we see that oy payers wth types ower tha ε ca bd [t 0,t 00 ). Thus, for sma eough ε (by cotuty of G 1 ad covergece of (G ) 1 to G 1 ), order to crease T (t 0 ) to T (t 0 )+γ/2 mutpe payers wth types ε or hgher must bd t 00 wth postve probabty ad therefore te there. But the ay oe of these payers coud proftaby devate to bddg sghty above t For t = b max, bddg sghty above b max s mpossbe. But by bddg b max a payer ws wth probabty 1, because b max s strcty domated by 0 for a payers. 6 By dog so such a payer woud obta wth hgh probabty a prze of at east T (t 0 )+γ/2 stead of osg the te wth postve probabty ad the obtag wth hgh probabty a prze of at most T (t 0 )+ε. 5

6 Theargumetsaaogousfwesupposethattheemmasfaseforsomet (0,b max ) ad r m t. Ift =0, the the above proof shows that for arge o payer bds t =0wth postve probabty. Ths meas, tur, that suffcety sma bds gve ower payoffs tha t 0 = r m such that t 0 t<δ. Thus, o payer bds cose to t =0wth postve probabty, whch cotradcts the No-Gap Property. 1.2 Proof of Lemma 2 Suppose the cotrary. The, there s some δ>0ad a sequece of tegers 1, 2,... such that for every k there s some bd t k wth T k (tk ) T (t k ) >δ. Passg to a subsequece f ecessary, we assume that t k t. Cosder umbers q 0 ad q 00 such that q 0 <t<q 00 ad T (q 00 ) T (q 0 ) <δ/2; suchumbers exst because T s cotuous. 7 For arge eough vaues of k,wehavethat T k (q 0 ) T (q 0 ) < δ/2 ad T k (q 00 ) T (q 00 ) <δ/2. For ay t 0 [q 0,q 00 ],ether(a)t k (t 0 ) T (t 0 ),or(b)t k (t 0 ) T (t 0 ). By mootocty of T ad T k,wehave T k (t 0 ) T (t 0 ) T k (q 00 ) T (q 0 ) T k (q 00 ) T (q 00 ) + T (q 00 ) T (q 0 ) <δ case (a), ad T (t 0 ) T k (t 0 ) T (q 00 ) T k (q 0 ) T (q 00 ) T (q 0 ) + T (q 0 ) T k (q 0 ) <δ case (b). Sce t k [q 0,q 00 ] for arge eough k, we obta a cotradcto to the assumpto that T k (tk ) T (t k ) >δfor a such k. 1.3 Proof of Lemma 3 Suppose to the cotrary that for arbtrary arge, the equbrum of the -th cotest some type x of some payer has a best respose that beogs to the compemet of BR x (ε). Passg to a coverget subsequece f ecessary, we assume that x x. Note that for every x theresaδ x > 0 such that (uder the verse tarff) aybdfrom the compemet of BR x (ε) gves type x apayoff ower by at east δ x tha ay eemet of BR x does. Let δ = δ x. We have that: 7 If t =0set q 0 =0,adft = b max set q 00 = b max. 6

7 1. The maxma payoff of type x, attaed at ay bd from BR x,scotuousx. Ths foows from Berge s Theorem. 2. For every ρ>0, forsuffcety arge the hghest payoff that type x caobtaby bddg the compemet of BR x (ε) caot exceed by ρ the hghest payoff that type x ca obta by bddg the compemet of BR x (ε). Ideed, suppose that for a sequece k dvergg to type x k obtas by bddg some t k the compemet of BR x k (ε) apayoff at east ρ hgher tha the hghest payoff that type x ca obta by bddg the compemet of BR x (ε). Passg to a coverget subsequece f ecessary, we assume that t k t. Sce every (x k,t k ) beogs to the compemet of BR (ε), so does (x,t); thus,(x,t) beogs to the compemet of BR x (ε). However,bycotuty of the payoff fuctos, bddg t gves type x apayoff by at east ρ hgher tha the hghest payoff that type x ca obta by bddg the compemet of BR x (ε), a cotradcto. By 1 ad 2, for suffcety arge, aybdthecompemetofbr x (ε) gves type x apayoff ower by at east δ/2 tha ay bd BR x. Ideed, by 2 apped to ρ = δ/4, ay bd the compemet of BR x (ε) gves type x apayoff at most δ/4 hgher tha the hghest payoff that type x ca obta by bddg the compemet of BR x (ε). Thsast payoff s tur ower tha the payoff that type x obtas by bddg BR x by at east δ. Adby1,thepayoff that type x obtas by bddg BR x caot be ower by more tha δ/4 tha the payoff that type x obtas by bddg BR x. By uform covergece of T to T, the aaogous statemet, wth δ/2 repaced wth some smaer postve umber ad T repaced wth T, s aso true. Ths meas, however, that for suffcety arge,payer woud be strcty better off bddg sghty above ay bd BR x whe her type s x tha bddg the compemet of BR x (ε). Thssbecause (1)mpesthatforsuffcety arge, by bddg sghty above t the payer obtas a prze arbtrary cose to T (t) wth probabty arbtrary cose to Proof of Lemma 4 Observe that for ay x 0 <x 00,strctsgecrossgmpesthatft 0 BR x 0 ad t 00 BR x 00, the t 0 t 00. Suppose that BR x 0 cotaed two bds, t 1 <t 2,forsometypex. The frst observato ad Lemma 3 mpy that for ay 0 <ε<(t 2 t 1 )/4, forsuffcety arge oy payers wth types I =[max{x 0 ε, 0}, m {x 0 + ε, 1}] may bd the terva [t 1 +(t 2 t 1 )/4,t 2 (t 2 t 1 )/4]. Cosder obtag a prze that s hgher the mt prze dstrbuto 8 by creasg 8 More precsey, gve a ta prze y 0, the prze that s hgher the mt prze dstrbuto s the prze y 00 such that = G(y 00 ) G(y 0 ). The prze that s hgher the prze dstrbuto G s defed 7

8 the bd from t 1 +(t 2 t 1 )/4 to t 2 (t 2 t 1 )/2. If s suffcety sma, the by cotuty of G 1 the crease the prze s sma as we, so the assocated cremet utty s egatve for a types, ad uformy bouded away from 0. Therefore, takg /2 =F (m {x 0 + ε, 1}) F (max {x 0 ε, 0}), fε>0 s suffcety sma, the for suffcety arge every type of every payer s better off bddg t 1 +(t 2 t 1 )/4 tha bddg t 2 (t 2 t 1 )/2. Ths s because wth hgh probabty the hgher bd eads to a prze that s approxmatey oy /2 hgher the prze dstrbuto G. 9 By covergece of (G ) 1 to (G) 1,forsuffcety arge ths prze s ot much more tha /2 hgher the mt prze dstrbuto. Moreover, every type of every payer s better off bddg t 1 +(t 2 t 1 )/4 tha bddg ay bdterva(t 2 (t 2 t 1 )/2,t 2 (t 2 t 1 )/4), because such bds are eve more costy tha t 2 (t 2 t 1 )/2, ad eabe a payer to obta a prze that s wth hgh probabty ot much more tha /2 hgher the mt prze dstrbuto tha the prze the payer obtas by bddg t 2 (t 2 t 1 )/2. Therefore, o payer bds the terva ((t 2 (t 2 t 1 )/2,t 2 (t 2 t 1 )/4)) wth postve probabty, so by the No-Gap Property T (t 2 (t 2 t 1 )/4) = 1 for suffcety arge. Thus, T (t 2 (t 2 t 1 )/4) = 1, sot 2 caot be BR x, because bddg sghty above t 2 (t 2 t 1 )/4 gves type x ahgherpayoff. Cosequety, BR x s a sgeto for ay x, ad by strct sge crossg, br s weaky creasg. A argumet aaogous to the argumet used to show that BR s a sgeto aso shows that br s cotuous Proof of Lemma 5 Cosder a arbtrary type x. Let x m =m{z : br (z) =br (x)} ad x max =max{z : br (z) =br (x)} (x m ad x max are we defed because br s cotuous). Frst, observe that G 1 (F x m )=G 1 (F (x max )). Ideed, by Coroary 1, for suffcety arge a types the terva [x m,x max ] bd the -th cotest cose to br (x). Suppose that G 1 (F x m ) <G 1 (F (x max )), ad cosder the payers whose type beogs smary. 9 Ths foows from the covergece of F to F ad Hoeffdg s equaty apped to radom varabes ½ 1 f m {x Z 0 + ε, 1} x = max {x0 ε, 0}, 0 otherwse, for =1,...,. 10 More precsey, suppose that br s dscotuous at some x, ad appy the argumet to t 1 = br(x 1 ) ad t 2 = br(x 2 ) where x 1 ad x 2 are sghty ower ad hgher, respectvey, tha x. 8

9 to [x m,x max ] wth postve probabty. 11 Amog these payers, the oe whose expected prze s the owest cotget o havg a type ths terva ca proftaby devate to bddg sghty above br (x), thereby outbddg the other payers wth a type ths terva ad obtag a dscretey hgher prze. Suppose that x m > 0. By Coroary 1 for ay δ>0, theresan such that f N, the the equbrum bds of every payer wth type ower tha x m δ are ower tha br(x m ), ad the equbrum bds of every payer wth type hgher tha x m are hgher tha br x m δ. Therefore, a payer who bds br x m outbds a payers wth types ower tha x m δ, sot (br x m ) (G ) 1 (F x m δ ), ad a payer who bds br x m δ s outbd by a payers wth types hgher tha x m,sot br x m δ (G ) 1 (F x m ). Sce T coverges to T, T ad br are cotuous, (G ) 1 coverges to (G 1 ), F coverges to F,adF ad G 1 are cotuous, we obta T br x m = G 1 F x m. Smary, f x max < 1, weobtathatt (br (x max )) = G 1 (F (x max )). Thus, sce br(x) =br(x m )=br(x max ) ad G 1 (F (x)) = G 1 (F (x m )) = G 1 (F (x max )), we have that T (br (x)) = G 1 (F (x)) whe x m > 0 or x max < 1. Fay,tcaotbethat x m =0ad x max =1, because 0=G 1 (F (0)) <G 1 (F (1)) = Proof of Part (a) of Theorem 1 Cosder a type x X. Lett = br (x), adett 0 ad t 00 be such that T (t 0 )=T(t) ε/3 ad T (t 00 )=T(t)+ε/3. Fay, et x 0 ad x 00 be such that t 0 = br (x 0 ) ad t 00 = br (x 00 ).(Take x 0 =0ad t 0 =0f T (t) ε/3 < 0, adx 00 =1ad t 00 = br(1) f T (t)+ε/3 > 1.) By Lemma 3, for suffcety arge, every payer wth type o hgher tha x 0 bds ess tha the payer wth type x, ad every payer wth type o ower tha x 00 bds more tha the payer wth type x. ByHoeffdg s equaty, the payer of type x outbds wth hgh probabty at east a fracto of payers cose to F (x 0 ). Sce F coverges to F, she outbds wth hgh probabty at east a fracto of payers cose to F (x 0 ).So,sce(G ) 1 coverges to G 1, she obtas (wth hgh probabty) a prze o ower tha G 1 (F (x 0 )) ε/3 =T (t) 2ε/3. Smary, for suffcety arge apayerof type x outbds wth hgh probabty at most a fracto of payers cose to F (x 00 ), ad so she obtas (wth hgh probabty) a prze o hgher tha G 1 (F (x 00 )) = T (t)+2ε/3. (These bouds are mmedate f y ε/2 < 0 or f y + ε/2 > 1.) Thus, type x obtas (wth hgh probabty) a prze whch dffers from G 1 (F (x)) by at most 2ε/3. 11 For arge, at east a fracto of payers cose to F (x max ) F x m have types that beog to [x m,x max ] wth postve probabty. 9

10 Ths proves part (a) for a sge x, but we must show that there s a N such that for ay N part (a) hods for a x smutaeousy. Such a N ca be obtaed by takg a fte grd of types x, ad the correspodg grd of bds br (x) such that T (t 1 ) T (t 2 ) <ε/3 for ay par of eghborg eemets t 1, t 2 of the grd, ad takg the argest N amog the N s correspodg to x s from the grd. 2 Proof of Theorem 2 Reca that G 1 (z) =f{y : G (y) z} for z>0ad G 1 (0) = f {y : G (y) > 0}, ad ote that G 1 may be dscotuous (but s eft-cotuous). Dscotutes requre modfyg amost a the argumets used the proof of Theorem 1. As Appedx 1, we reegate to the ed of the secto the proofs of a termedate resuts. Let I 0 =(y0,y 0 u ) be a ogest terva [0,G 1 (1)] to whch G assgs measure 0; et I 1 =(y1,y 1 u ) be a ogest such terva dsjot from I 1, ad so o. The, every ope terva of przes that has measure zero s cotaed oe of the tervas I 0,I 1,... Ad for ay ε>0, theresak such that the egths of I K+1, I K+2,... sum up to ess tha ε. The deftos of R, A, A,adT are as Appedx 1. The defto of T, however, must be chaged. Frst, by Hey s seecto theorem, we take a covergg subsequece of the sequece A ; deote ts mt by A : B [0, 1]. Ths fucto s weaky creasg (because each A s). For the rest of the proof, deote ths covergg subsequece by A (wth the correspodg sequece T =(G ) 1 A ). Let T = G 1 A. SceG may ot have fu support, we ow have that T (0) = f{z : G(z) > 0} ad T (b max )=G 1 (1); addto, T s st (weaky) creasg (compare to propertes (1)-(3) from Appedx 1). I addto, F coverges potwse to F,but(G ) 1 may ot coverge potwse to G 1. It s, however, easy to check that m (G ) 1 (r) =G 1 (r) uess r s the vaue of G o a terva I k =(yk,yu k );moreover,m (G ) 1 (r) G 1 (r) for every r that s the vaue of G o a terva I k =(yk,yu k ), but t ca happe that m (G ) 1 (r) =yk u ad G 1 (r) =yk. The dscotutes G 1 mpy that T mayotbecotuous,solemma1doesot hod. Pots of dscotuty, however, correspod to ope tervas of przes that have measure zero. More precsey, we have the foowg resut. Lemma 6 For ay t>0 B, oe of the foowg codtos hods: 1. T s cotuous at t, that s, for ay sequeces q m t ad r m t, wehavem T (q m )= m T (r m )=T(t). 10

11 2. There s some k =1, 2,... such that for ay sequeces q m t ad r m t, wehave m T (q m )=yk ad m T (rm )=yk u.moreover,m A (qm )=G(yk ) ad m A (rm )= G(yk u). Usg Lemma 6, we defe aother fucto T o B by settg T (t) = m T (r) for some sequece r t (ad T (b max )=G 1 (1)). The mootocty of T guaratees that T (t) s we-defed. I addto, t s easy to check that T s (weaky) creasg, rght-cotuous, ad cotuous at every bd t such that codto 1 from Lemma 6 hods. Note that T may otbeaextesooft, because whe m T (r) 6= T (t), wehavethatt (t) = m T (r) 6= T (t). Cosder ow a bd t > 0 such that codto 2 from Lemma 6 hods. Deote ths bd t by t k,wherek s descrbed codto 2. The, there s a bd t 0 <t k such that A (t 0 )=A(t) =G yk,soa s costat o a terva beow tk. Ideed, f A(t 0 ) <G yk for a t 0 <t k, the, as the proof of Lemma 6, for arge o payer woud bd ay t 0 sghty beow t k. Ths woud be so, because bddg sghty above t k woud amost certay gve aprzeoowerthayk u, whereas bddg t0 woud amost certay gve a prze o hgher tha yk.let t k =f t 0 : A (t 0 )=G yk ª <tk. It s aso true that every maxma terva o whch T s costat wth a vaue ower tha G 1 (1) s t k,t k for some k. Ideed, cosder a maxma otrva terva wth ower boud t ad upper boud t u o whch the vaue of T s y<g 1 (1). It suffces to show that T (t u ) >y, because the codto 2 from Lemma 6 appes to t u, whch mpes that t u = t k for some k; ad the maxmaty of t,t u yeds t = t k. Suppose that T (t u )=y. The, for arge eough bddg t u amost certay gves a prze at most sghty hgher tha y, whereas bddg sghty above t amost certay gves a prze ot much ower tha y. But the, for arge eough, o payer bds some eghborhood of t u, because bddg sghty above t eads to a hgher payoff. Ths cotradcts the No-Gap Property, because y<g 1 (1). Because G 1 may be dscotuous, T eed ot coverge uformy to T or T,eve o the set of pots at whch they are cotuous. I partcuar, for a t t k,t k t may be that T (t) =(G ) 1 (A (t)) yk u for arbtrary arge, whereast (t) =T (t) =yk. Nevertheess, T coverges uformy except o some eghborhoods of a fte umber of tervas [t k,t k]. More precsey, we say that T coverges uformy to T up to β o a set C f there exsts a N such that for every N ad t C we have that T (t) T (t) <β. We the have the foowg modfcato of Lemma 2. 11

12 Lemma 7 For every β>0, there exsts a umber K such that for every γ>0, T coverges uformy to T up to β o the compemet of K[ O γ = (t k γ,t k + γ). k=1 We ow reate payers equbrum behavor arge cotests to the verse tarff T. Defe BR x, BR(ε), adbr x (ε) as Appedx 1 wth T stead of T (the maxma payoff s acheved because T s creasg ad rght-cotuous, so s upper sem-cotuous). Defe the mass expeded ( the -th cotest) a terva of bds I by payers wth type x S as ( P =1 Pr (σ S I)) /. We the have the foowg resut, whch we use provg the remag resuts. Lemma 8 For a k ad ay ε>0 ad L>0, thereexstsγ>0 such that for suffcety arge we have that: () The mass expeded (t k γ,t k + γ) by payers wth types x for whch t k / BR x (ε) s ess tha ε/3l; () The mass expeded (t k γ,t k ] by payers wth types x for whch t k / BR x (ε) s ess tha ε/3l. I addto, for ay α>0, forsuffcety arge we have that: () The mass expeded t k + α, t k α byapayerssessthaε/3l. Lemma 3 must aso to be modfed. Lemma 9 For every ε>0, thereexstk such that for every γ>0, there s a N such that for every N the equbrum of the -th cotest every best respose of every type x of every payer beogs to K[ BR x (ε) (t k γ,t k ). k=1 Strct sge crossg o oger mpes that BR x s a sgeto. Istead, we have the foowg resut. Lemma 10 If strct sge crossg hods, the for a but a coutabe umber of types the set BR x s a sgeto. For those types for whch t s ot a sgeto, BR x cotas precsey two eemets: t k ad t k for some k. The correspodece that assgs to type x the set BR x s weaky creasg (.e., for ay x 0 <x 00,ft 0 BR x 0 ad t 00 BR x 00, thet 0 t 00 )ad upper hem-cotuous. 12

13 Let br (x) =mbr x,adotethatbr s creasg ad eft cotuous, ad s ot rght cotuous precsey at types x for whch BR x s ot a sgeto. We the have the foowg coroary of Lemmas 8, 9, ad 10, whch s a modfcato of Coroary 1. Coroary 2 For every ε>0, there s a N such that for N at east a fracto 1 ε of of payers bd (br (x ) ε, br (x )+ε) wth probabty at east 1 ε. To prove part (b) of the theorem t remas to show that T br s the assortatve aocato. Ths s doe by the foowg emma, whch s a modfcatooflemma 5that accommodates the dscotutes T ad br. Lemma 11 G 1 (F (x)) = T (br (x)) for ay type x>0. To compete the proof, t remas to show (a) the statemet of the theorem. To do so, we use the foowg resut. Lemma 12 For every ε, δ > 0, theresan such that for N, eachtypex from a set whose F -measure s at east 1 ε bds at east wth probabty 1 ε a t, adobtasateast wth probabty 1 ε a y such that for some r BR x, t r <δad y T (r) <δ. To see that Lemma 12 mpes (a) the statemet of the theorem, choose some ε>0. Lemma 12 shows that for every δ>0, theresan such that for N ad for a fracto 1 ε of payers, thef -measure of ther types x that satsfy the codto of Lemma 12 s at east (1 ε). Ths meas that each such payer obtas wth probabty at east 1 ε aprzey that dffers by at most δ from the prze T (r) for some optma bd r of the payer s type. For types x > 0 such that br (x ) s a uque optma bd, ths yeds (a) by Lemma 11. However, by Lemma 9 ad strct sge crossg, there s oy a coutabe umber of other types x.adthef-measure of such types s 0 sce F has o atoms, so the F -measure of such types s arbtrary sma for suffcety arge. 2.1 Proof of Lemma 6 Let m T (q m ) = y 0 ad m T (r m ) = y 00. Both mts y 0 ad y 00 exst ad y 0 y 00 by mootocty. Suppose that y 0 <y 00. If G assgs a postve measure to (y 0,y 00 ),thet assgs a postve measure to ay terva wth edpots suffcety cose to y 0 ad y 00.I 13

14 such a case, we obta a cotradcto by argumets smar to those used the proof of Lemma 1. Ideed, for suffcety arge o bdder woud bd sghty beow t, because bddg sghty above t woud amost certay gve a better prze. Thus, G assgs measure zero to (y 0,y 00 ). Thsmpesthat(y 0,y 00 ) (yk,yu k ) for some k. By defto, T takes vaues [0,yk ] [yu k, 1], soy0 = yk ad y00 = yk u. Moreover, the mootocty of T mpes that k s the same for ay sequeces q m t ad r m t. Itremas to show that m A (q m )=G(yk ) ad m A (rm )=G(yk u). For ths, ote that f m A (q m ) >G(yk ),them T (qm ) >yk. Smary, f m A (rm ) > G(yk u),them T (rm ) >yk u. The equates m A (qm ) <G(yk ) ad m A (rm ) <G(yk u) ca be rued out smary f G does ot have atoms at yk or yu k. Suppose that G has a atom at yk u ad m A (rm ) <G(yk u). Sce m T (rm )=yk u, A (rm ) >G(yk ) for suffcety arge m. Taketwoumbersr m such that G(yk ) <A(rm ) <G(yk u);deotethembyt0 <t 00. The, for suffcety arge ay payer obtas a prze cose to yk u wth arbtrary hgh probabty by bddg ay t [t 0,t 00 ].Thus,forsuffcety arge, opayerwoudbdtheterva [(t 0 + t 00 )/2,t 00 ] wth postve probabty. Ths cotradcts the No-Gap Property. Suppose that G has a atom at yk ad m A (qm ) <G(yk ). The, for suffcety arge bddg q m amost certay gves a prze at most sghty better tha yk. I cotrast, bddg r m amost certay gves a prze at east as good as yk u. Ths foows drecty from (1) f G has a atom at yk u.ifgdoes ot, the ths aga foows from (1) for arge eough, because A (r m ) >G(yk u ) for ay m. For arge eough a cotradcto wth the No-Gap Property s obtaed smary to the ast part of the proof of Lemma 1 that deas wth U (0,y,t) strcty creasg y. 2.2 Proof of Lemma 7 The proof s aaogous to the proof of Lemma 2. Take a K such that the egths of I K+1, I K+2,... sum up to ess tha β/2. Takeayγ>0, ad suppose to the cotrary that there s a creasg sequece of tegers 1, 2,..., m,...such that for every m theressomebd t m / O γ wth T m (t m ) T (t m ) β. Passg to a subsequece f ecessary, we assume that the sequece t m t. Takeq 0 ad q 00 such that q 0 <t<q 00 ad T (q 00 ) T (q 0 ) <β/2, 12 ad K[ [q 0,q 00 ] B [t k,t k ]. Ths s possbe, sce the egths of I K+1, I K+2,... sum up to ess tha β/2. I addto, for arge eough k we have that T k (q 0 ) T (q 0 ) <β/2 ad T k (q 00 ) T (q 00 ) <β/2, sce 12 If t =0,takeq 0 =0. k=1 14

15 the egth of each I K+1, I K+2,... s ess tha β/2. The rest of the proof cocdes wth the proofoflemma Proof of Lemma 8 Frst, observe that the maxma payoff of type x, attaedataybdbr x, s st cotuous x. Ideed, upper sem-cotuty of T s a that s eeded for the cotuty of the maxma payoff. Ths observato mpes that there exsts a δ>0 such that for ay type x ay bd the compemet of BR x (ε) gves type x apayoff ower by at east δ tha ay bd BR x. For (), suppose the cotrary that for ay γ>0there are arbtrary arge such that themassexpeded(t k γ,t k + γ) by payers wth types x for whch t k / BR x (ε) s at east ε/3l. Take γ sma eough so that the payoff that such payers obta by bddg sghty more tha ay bd BR x s hgher by δ/2 tha the payoff that they woud obta by bddg t k γ ad gettg y k.13 Suppose frst that t k > 0. By mootocty of A ad the defto of t k,wehavethat A t k γ <G yk. Take a postve α<ε/6lsuch that A t k γ <G yk α. For ay t t k γ ad suffcety arge, fa (t) <G yk α/2, theopayeroftypex such that t k / BR x (ε) bds t, because by bddg t such a payer woud obta wth hgh probabty a prze o hgher tha yk, ad therefore woud obta a hgher payoff by bddg sghty more tha ay bd BR x. Let γ be defed by t k γ =f t : A (t) G ª yk α.scea t k γ <G yk α ad A (t) s rght-cotuous, we have that γ <γ(for suffcety arge ). Ad sce for every t<t k γ we have A (t) <G yk α/2 (by defto of γ ), payers wth types x for whch t k / BR x (ε) must exped the mass of at east ε/3l [t k γ,t k + γ). If more tha haf of ths mass s expeded t k γ,t k + γ, the we have that A t k + γ >A t k γ + ε/6l G y k α + ε/6l >G y k. Ths caot happe 13 To see why bddg sghty above ay t BR x gves at east a payoff cose to U (x, T (t),t), cosder the foowg two cases: (a) G 1 (A(r)) >T (t) for a r>t; ths case, sce m (G ) 1 (A(r)) G 1 (A(r)), for ay r>t,f s suffcety arge, the (G ) 1 (A(r)) >T (t). Ths mpes that a payer obtas a prze hgher tha T (t) wth arbtrary hgh probabty by bddg r. (b) G 1 (A(r)) = T (t) for r>tcose eough to t; thscase,t (r) =T (t) for a such r. Ths mpes that t = t k for some 0 k0. The cam ow foows from eft-cotuty of G 1 ad the fact that m (G ) 1 (q) G 1 (q) for ay q. 15

16 for suffcety arge, because for t t k,t k s A(t) =G y k. Thus, the payers wth types x for whch t k / BR x (ε) bd precsey t k γ wth probabty at east ε/6l. Sce these payers te wth each other at t k γ, by bddg t k γ they must obta a prze of aspecfc typey wth probabty 1, eve f they ose a tes at t k γ. (Otherwse, each of them coud obta a hgher payoff by bddg sghty above t k γ + γ ad wg the tes at t k γ.) But a payer who oses a tes at t k γ has rak order o hgher tha G yk α, bydefto of γ,soy yk. Therefore, such a payer woud obta a strcty hgher payoff by bddg sghty more tha ay bd BR x. Now suppose that t k =0. The A(t k ) G yk. The case A(t k ) <G yk s haded as the case t k > 0 above. Suppose that A(t k )=G yk. The, for ay γ>0such that t k + γ<t k,forsuffcety arge themassexpeded t k,t k + γ by a payers s smaer tha ε/6l, becausea (t) =G yk for ay t t k + γ,t k. Thus, f () does ot hod, for suffcety arge the mass expeded precsey at t k by the payers wth types x for whch t k / BR x (ε) s at east ε/6l, ad so the rakg of a payer who tes at t k ad oses s at most G yk ε/12l. But ths case each payer of type x for whch t k / BR x (ε) woud strcty prefer bddg sghty more tha ay bd BR x to bddg t k, a cotradcto. Toshow(),otethatft k = t k for some 0 k0, the () foows from (). Thus, suppose that t k 6= t k for ay 0 k0. Suppose the cotrary that for ay γ>0there s a arbtrary arge such that the mass expeded (t k γ,t k ] by payers wth types x for whch t k / BR x (ε) s at east ε/3l. Takeγ sma eough so that the payoff that such payers obta by bddg sghty more tha ay bd BR x s hgher by δ/2 tha the payoff that they woud obta by bddg t k γ ad gettg yk u. Observe that for suffcety arge, by bddg t k ay payer amost certay obtas a prze at most sghty better tha T (t k )=yk u.thssso, because t k 6= t k ad so A(t k) 6= G(y 0 k 0) for ay k0. Therefore, for arge eough apayer wth type x for whch t k / BR x (ε) woud be better off bddg sghty above ay t BR x tha bddg (t k γ,t k ]. Part () foows mmedatey from the fact that the vaue of A o t k,t k s G y k,by the defto of t k. 2.4 Proof of Lemma 9 Take a δ>0 such that for ay type x ay bd the compemet of BR x (ε) gves type x a payoff ower by at east δ tha ay bd BR x.takeβ>0such that for ay type x, bdt, ad przes y 0 ad y 00 wth y 0 y 00 β we have U (x, y 0,t) U (x, y 00,t) δ 3. 16

17 Next, take a K guarateed by Lemma 7 for ths β. I addto, take K arge eough so the egths of I K+1, I K+2,... sum up to ess tha β/2. Fay, for ay λ>0 take a N λ that satsfes the defto of uform covergece up to β o the compemet of O λ. (Note that K s the same for a λ.) Suppose to the cotrary of the statemet of the emma that there s a γ > 0 ad a subsequece of cotests such that a type x of payer the -th cotest has a best respose t to the strateges of the other payers that does ot beog to BR x (ε) [ K k=1 (t k γ,t k). As usua, we assume that the subsequece s the etre sequece; moreover, we assume that x x ad t t. Cosder the foowg two cases: A. (t 6= t k for ay k =1,..., K) I ths case, for some λ>0 there s a eghborhood of t that s dsjot from O λ.byuformcovergeceoft to T up to β o the compemet of O λ, U (x,t (t ),t ) U (x,t (t ),t ) δ 3 for N λ. Ad because t / BR x (ε), for ay t BR x we have U (x,t (t),t) U (x,t (t ),t ) δ. Thus, we obta U (x,t (t),t) U (x,t (t ),t ) 2δ 3. Observe that ay bd t 0 hgher tha t guaratees, for suffcety arge, aprzeotmuch worse tha T (t) wth arbtrary hgh probabty. 14 We w ow show that by bddg t,forsuffcety hgh type x obtas wth arbtrary hgh probabty a prze o better tha T (t )+β. Ideed, sce t does ot beog to t k,t k for ay k K, wehavethata (t 0 ) s bouded away from G y1,...,g y K for t 0 suffcety cose to t. Therefore, A (t ) s aso bouded away from G y1,...,g y K for suffcety arge. Ad for suffcety arge, bddg t gves wth arbtrary hgh probabty a rak order arbtrary cose to A (t ). Sce the egths of I K+1, I K+2,... sum up to ess tha β/2, adforayr other tha G y1,...,g y K ad suffcety arge the dfferece betwee (G ) 1 (r) ad G 1 (r) s o arger tha the egth of I K+1,bybddgt a payer obtas wth arbtrary hgh probabty a prze o better tha (G ) 1 (A (t ))+β. Therefore, by defto of β, wehavethatbybddgt type x obtas a payoff that s hgher tha U (x,t (t ),t ) by at most sghty more tha δ/3. Cosequety, for suffcety arge payer woudobtabybddgsomet 0 >tapayoff strcty hgher tha by bddg t, a cotradcto. 14 Toseewhy,seetheprevousfootote. 17

18 B. (t = t k for some k =1,..., K) The, cosder a t sghty hgher tha t,suchthat t does ot beog to [t k,t k] for k =1,..., K, ad such that: () for suffcety arge the payoff (of ay payer) the -th cotest of bddg t s ot much ower tha the payoff of bddg t ; () for suffcety arge, wehavethatthedfferece betwee U(x,T (t),t) for ay t BR x ad U(x,T (t ),t ) s ot much ower tha δ. Ths atter codto s possbe because, by defto, (x,t ) / BR(ε), ad by rght cotuty of T at t.now, usg (), appy a argumet aaogous to that from case A wth t payg the roe of t, wth a cotradcto obtaed by referrg to (). 2.5 Proof of Lemma 10 Mootocty of the correspodece foows from strct sge crossg, ad upper hemcotuty foows from stadard argumets. 15 Suppose that BR x cotas a par of bds t 1 <t 2. Beow we w show that for ay ε>0 ad ay terva [a, b] such that t 1 <aad b<t 2,forsuffcety arge the mass expeded [a, b] by a payers s at most ε. Ths mpes that the fucto A, adtherefore T, s costat o every such terva [a, b], adthereforeo(t 1,t 2 ). But T (t 2 ) >T (t 1 ) because t 1 <t 2 are BR x,sobydefto of the dscotuty pots t k of T we must have (t 1,t 2 ) t k,t k for some k. Ad because BRx B\ [ k=1 (t k,t k), wehavethatt 1 = t k ad t 2 = t k. It remas to show that for ay ε>0,forsuffcety arge the mass expeded [a, b] by a payers s at most ε. We w show ths for ε/2 ad payers of types ower tha x (a smar argumet appes to types hgher tha x). Choose x 0 <xsuch that F (x) F (x 0 ) <ε/3. For suffcety sma λ>0, sup z x 0BR z (λ) <a. (Ths s because x 0 <xad t 1 BR x, so every bd BR z s at most t 1 <a.) Therefore, by Lemma 9, there s some K such that for every γ>0ad suffcety arge ay bd [a, b] made by a payer of type z x 0 the -th cotest s [ K k=1 (t k γ,t k). Cosder oe of these K tervas for whch (t k γ,t k) [a, b] 6=. Scesup z x 0BR z (λ) < a t k, t k s ot BR z (λ) for ay z x 0.Ift k > sup z x 0BR z (λ), theby()oflemma8 there exsts a γ such that for suffcety arge the mass expeded t k γ,t k by payers of type z x 0 s ess tha ε/6k. Ift k sup z x 0BR z (λ), the by () ad () of Lemma 8, for suffcety arge themassexpeded[a, t k ) by payers of type z x 0 s ess tha ε/6k. 15 More precsey, ths foows from the fact that BR x s the set of a t such that (t, T (t)) maxmzes type x s utty over the cosure of the graph of T, whch s a compact set. 18

19 Therefore, for arge eough the mass expeded [a, b] by payers of type z x 0 s smaer tha ε/6, adbecausef (x) F (x 0 ) <ε/3, the mass expeded [a, b] by payers of type z x s smaer tha ε/ Proof of Coroary 2 Choose ε > 0. Lemma 10 mpes that there s a fte umber of tervas of types wth tota F -mass ε/2, such that for every type x ot oe of these tervas, BR x (br (x) ε, br (x)+ε). 16 Cosder the F -mass 1 ε/2 of types x wth the ast property, ad et K betheoethestatemetoflemma9. The,byLemma9adLemma8 for L = K, forsuffcety arge, atmostaf -mass ε/2 of those types bd outsde of (br (x) ε, br (x)+ε). 2.7 Proof of Lemma 11 The proof s aaogous to that of Lemma 5. Cosder a arbtrary type x. Defe x m = m {z : br (x) BR z } ad x max =max{z : br (x) BR z }. By strct sge crossg, BR z has oy oe eemet br(z) =br(x) for a z (x m,x max );tmayhavetwoeemetsfor z = x m or x max, whch cases br(x) s the hgher oe ad the ower oe of the two, respectvey. The cam that G 1 (F x m )=G 1 (F (x max )) s obtaed by the same argumet as the proof of Lemma 5. The rest of the proof requres the foowg mor chages whe BR x m has two eemets (ad aaogous chages whe BR x max has two eemets): 1. Istead of x m,wecosderx m = m z : br x m ª BR z,adcomparethe equbrum bds of every payer wth type ower tha x m δ to br(x m ), ad the equbrum bds of every payer wth type hgher tha x m to br x m δ. Ths chage does ot affect the argumets, sce G 1 (F x m )=G 1 (F x m ). 2. It may ot be true that the equbrum bds of every payer wth type ower tha x m δ are ower tha br(x m ), or the equbrum bds of every payer wth type hgher tha x m are hgher tha br x m δ, because payers may bd [ K k=1 (t k γ,t k) BR(ε) (see Lemma 9). However, ths happes oy wth vashg probabty as grows arge, so the argumets are aga ot affected. 16 There s a K>0 such that P k>k tk t k <ε. For each k K such that BRxk = t k,t ª k for some type x k, cosder the terva of types [x k λ, x k + λ] [0, 1], whereλ s such that (cotuous) F creases by o more tha ε/2k o ay terva o arger tha 2λ. The sum of the F -massofthesetervasso arger tha ε/2, ad the sum of the jumps of br o the compemet of these tervas s smaer tha ε. 19

20 2.8 Proof of Lemma 12 Take ay λ>0. By Lemma 9, there s a arge K such that for ay γ>0, f s suffcety arge, the equbrum bd of every payer the -th cotest beogs wth probabty 1 to BR x K[ (λ) (t k γ,t k ). k=1 Assume that K s, addto, arge eough so that the egths of I K+1, I K+2,... sum up to ess tha δ/2. We frst cam that for ay t/ (t k γ,t k) for a k =1,..., K, there exsts a N t such that for every N t,apayerwhobdst the -th cotest obtas (wth hgh probabty) a prze y such that y T (t) <δ/2. We w aso show that there exsts a N = N t that s commo for a such bds t. Suppose frst that t 6= t k for ay k =1,..., K. Sce A (t) dffers from G(yk ) ad G(yu k ) for ay k =1,..., K, ay rak order cose to A (t) aso dffers from G(yk ) ad G(yu k ).By(1), for suffcety arge, apayerwhobdst has (wth hgh probabty) a rak order cose to A (t); partcuar, ths rak order dffers from G(yk ) ad G(yu k ). By the assumpto that the egths of I K+1, I K+2,... sum up to ess tha δ/2, ths mpes that the dfferece betwee T (t) ad the prze obtaed by a payer who bds t s ower tha δ/2 (wth hgh probabty). Suppose that t = t k for some k =1,..., K. By a argumet aaogous to the oe used the prevous case, the prze obtaed by a payer who bds t caot, as creases, exceed T (t) by δ/2 wth a probabty that s bouded away from 0. AdT (t) caot exceed ths prze by δ/2 wth a probabty that s bouded away from 0 as creases, because the payer woud proftaby devate by bddg sghty above t, whch woud guaratee a prze o worse tha T (t) wth arbtrary hgh probabty. Now, ote that the umber N t that was chose for ay bd t has the requred property aso for a bds cose eough to t; the case of t = t k for some k =1,..., K, wemeabds cose eough ad hgher tha t. That s, for every t there s a eghborhood W t of that t wth N t that s commo for a bds from ths eghborhood. The famy of sets W t s a ope coverg of the compact set of bds t that satsfy t/ (t k γ,t k) for k =1,..., K. Thus, t cotas a fte subcoverg, ad ay umber N that exceeds umbers N t for a eemets of ths fte subcoverg has the requred property. Ths yeds the emma for bds t / (t k γ,t k) for a k =1,...,K. Ideed, ote that BR x s a sgeto, ad br(x) s ts oy eemet, for a except a coutabe umber of types x. Sce F has o atoms, the set of such types has F -measure 1. Ad for such types x, equbrum bds t/ (t k γ,t k) for a k =1,..., K beog to (br (x) λ, br (x)+λ). If λ 20

21 s suffcety sma, ad x s bouded away from 0, the t r <δfor r = br (x), ad T (t) T (r) δ/2. 17 Ad f λ s suffcety sma, the aso T (r) T (t) δ/2, because t/ (t k γ,t k) for a k =1,...,K ad the egths of I K+1, I K+2,... sum up to ess tha δ/2. Fay, by our frst cam, the prze y obtaed by bddg t must satsfy y T (t) <δ/2, so y T (r) <δ. Now cosder bds t such that t s (t k γ,t k) for some k =1,..., K. By () of Lemma 8, we ca dsregard bds t [t k + γ,t k γ]. Suppose that t s (t k γ,t k ) ad t k 6= t k for 0 a other k 0 =1,..., K. By () of Lemma 8, oe ca assume that t k BR x. 18 We w show that for suffcety sma γ ad for suffcety arge, payer obtas by bddg t (wth arbtrary hgh probabty) a prze y (T (t k ) δ, T (t k )+δ). Frst, ote that payer caot obta by bddg t aprzeowerthat (t k ) δ (wth probabty bouded away from 0), because for sma eough γ t woud be proftabe to devate to bddg sghty above t k, ad obta a prze ot much ower tha T (t k ) wth hgh probabty. Payer caot obta by bddg t a prze hgher tha T (t k )+δ (wth probabty bouded away from 0), because by (1), for ay r>t k ad suffcety arge the rak order of payer s wth arbtrary hgh probabty bouded above by A(r). Thus, the upper boud o the prze foows from the assumpto that t k 6= t k for a other 0 k0 =1,..., K, adtheegthsof I K+1,I K+2,... sum up to ess tha δ/2. Fay, suppose that t s t k γ,t k + γ for some k =1,..., K. By()ofLemma8,oe ca assume that t k BR x. We w show that for suffcety sma γ ad for suffcety arge, equbrum bddg t k γ,t k + γ eads (wth arbtrary hgh probabty) to a prze y (T (t k ) δ, T (t k )+δ), except a sma probabty evet. Ideed, by a argumet smar to that from the prevous case, such a bd caot ead to a prze ower tha T (t k ) δ (wth probabty bouded away from 0). To obta a prze hgher tha T (t k )+δ wth a ovashg probabty, a payer s expected rak order whe bddg t caot be ower tha G yk by a ovashg costat. But, f a ovashg fracto of payers w a prze hgher tha T (t k )+δ wth a ovashg probabty by bddg t k γ,t k + γ, the the crease expected rak order o the terva t k γ,t k + γ s bouded away 17 Ideed, for types bouded away from 0, adforsuffcety sma λ, wehavethatu(x, y, t) >U(x, y 0,t 0 ) wheever y y 0 >δ/2 ad t t 0 <λ. (The assumpto that types are bouded away fro 0 s esseta, becauseweddotassumethatu(0,y,t) strcty creases y.) However, sce r = br (x), we caot have U(x, T (t),t) >U(x, T (r),r). 18 The emma says oy that the mass expeded (t k γ,t k ] by types x for whch t k / BR x (λ) for some sma λ>0 s sma. However, f λ>0 s suffcety sma, the the mass of types x such that t k / BR x but t k BR x (λ) s sma. 21

22 from 0 for a, whch cotradcts the fact that A (t k + γ) approaches G(y k ) as creases. 22