SUPPLEMENT TO LARGE CONTESTS (Econometrica, Vol. 84, No. 2, March 2016, )

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1 Econometrca Supplementary Materal SUPPLEMENT TO LARGE CONTESTS (Econometrca, Vol. 84, No. 2, March 2016, ) BY WOJCIECH OLSZEWSKI AND RON SIEGEL WE CHOOSE AN EQUILIBRIUM for each contest and refer to the sequence n whch the nth element s the equlbrum of the nth contest as the sequence of equlbra. For both our theorems, we wll show that every subsequence of ths sequence contans a further subsequence that satsfes the statement of the theorem. Ths suffces, because the followng observaton can be appled wth Z n beng the set of equlbra of contest n. Subsequence Property. Gven a sequence of sets {Z n : n = 1 2 }, suppose that for every subsequence {Z n : = 1 2 }, every sequence {z n : = 1 2 } wth z n Z n contans a subsequence {z nl : l = 1 2 } such that every element z nl has some property. Then there exsts an N such that for every n N, every element n Z n has ths property. 1 S1. PROOF OF THEOREM 1 We begn wth an outlne of the proof. Gven a subsequence of equlbra, each equlbrum n the subsequence nduces for each player a mappng from bds to expected percentle ranngs. We consder the average of those mappngs, and G 1 composed wth ths average gves a mappng T n from bds to przes. As n ncreases, ths mappng approxmates the equlbrum mappngs from bds to przes of all players n the nth contest. We then use Helly s (1912) selecton theorem to fnd a subsequence of T n that converges to some lmt mappng T from bds to przes. We show that T s contnuous and the subsequence of T n converges unformly to T. Then, for each agent type, we defne the set of optmal bds when T s treated as an nverse tarff, and defne the correspondence from agent types to sets of optmal bds. We consder a small neghborhood of the graph of ths correspondence and show that for large n, every player s best responses n the nth contest are n the x n slce of ths neghborhood. Such a slce could, n prncple, be large, even f the set of optmal bds of the correspondng agent type s small (ths would happen f the set of optmal bds of a nearby agent type s large). We show, however, that under strct sngle crossng, each agent type has a sngle optmal bd, whch s contnuous and wealy ncreases n the agent type. Ths mples that every player s best response set and, therefore, the support of her equlbrum strategy, s bounded wthn an arbtrarly small nterval as n ncreases. We then conclude 1 Otherwse, there would be a sequence {z n : = 1 2 } wth z n Z n of elements wthout the property The Econometrc Socety DOI: /ECTA11696

2 2 W. OLSZEWSKI AND R. SIEGEL that the unque mechansm nduced by T mplements the assortatve allocaton. Ths demonstrates part (b) n the statement of the theorem; part (a) then follows easly. For the proof, we tae the subsequence of equlbra to be the sequence of equlbra (ths smplfes notaton and has no effect on the proofs). We denote the equlbrum of the nth contest by σ n = (σ n 1 σn),whereσn n s player s equlbrum strategy; a strategy of player s a random varable tang values n X B whose margnal dstrbuton on X concdes wth the dstrbuton of player s types F n. By referrng to player bddng wth some probablty n a subset S of B, we mean the probablty of the set X S, that s, the probablty of S measured by the margnal dstrbuton of player s strategy on B. We denote by R n (t) the random varable that s the percentle locaton of player n the ordnal ranng of the players n the nth contest f she bds slghtly above t and the other players employ ther equlbrum strateges. 2 That s, R n (t) = 1 ( 1 + ) 1 {σ n n X [0 t]} where 1 {σ X [0 t]} s 1 f σ X [0 t] and s 0 otherwse. Let A n (t) = 1 ( 1 + Pr ( σ n n X [0 t])) be the expected percentle ranng of player. Then, by Hoeffdng s nequalty, for all t n B,wehave (S1) Pr ( R n (t) An (t) >δ ) < 2exp { 2δ 2 (n 1) } Fnally, let A n (t) = 1 n n A n (t) =1 be the average of the expected percentles ranngs of the players n the nth contest f they bd t and the other players employ ther equlbrum strateges. Let T n be the mappng from bds to przes nduced by A n. That s, T n (t) = (G n ) 1 (A n (t)), where(g n ) 1 (z) = nf{y : G n (y) z} for z>0and (G n ) 1 (0) = nf{y : G n (y) > 0}. (Inwords,(G n ) 1 (z) s the prze of an agent wth percentle ranng z when przes are dstrbuted accordng to G n.) Snce 2 Ths s the nfmum of her ranng f she bds above t, whch s equvalent to bddng t and wnnng any tes there. If tes happen wth probablty 0, then ths s equvalent to bddng t.

3 LARGE CONTESTS 3 every T n s (wealy) ncreasng, by Helly s (1912) selecton theorem for monotone functons, the sequence T n contans a subsequence that converges pontwse to a functon T : B Y. For the rest of the proof, denote ths subsequence by T n. We frst descrbe some propertes of nverse tarff T : () Inverse tarff T s (wealy) ncreasng, because every T n s (wealy) ncreasng. () We have T(0) = 0; otherwse players bddng 0 would have proftable devatons. 3 () We have T(b max ) = 1, snce A n (b max ) = 1 and thus T n (b max ) = 1. In addton, we wll use the followng property of dscrete contest equlbra. No-Gap Property. In any equlbrum, there s no nterval (a b) B of postve length n whch all players bd wth probablty 0 and some player bds n [b b max ] wth postve probablty. Proof: Suppose to the contrary and consder such a maxmal nterval (a b). A player would only bd b or slghtly hgher than b f some other player bds b wth postve probablty. But the player who bds b wth postve probablty would be better off ether by slghtly ncreasng her bd (f another player bds b and wnnng the te leads to a hgher prze) or by decreasng her bd (n the complementary case). Our frst lemma shows that T s contnuous. (So as not to obscure the structure of the proof, we relegate the proofs of all lemmas to the end of the secton.) LEMMA S1: For any t B and any sequences q m t and r m t n B, we have lm T(q m ) = lm T(r m ) = T(t). The dea of the proof s that f T were dscontnuous at some t, then for large n, t would be better to bd slghtly above t than slghtly below t.butfno player bds slghtly below t, then by the No-Gap Property, no player bds t or above. Contnuty and monotoncty of T mply the followng result. LEMMA S2: Sequence T n converges to T unformly on B. We now relate the nverse tarff T to players behavor n the equlbra that correspond to the sequence T n.denotebybr x type x s set of optmal bds gven T, that s, the bds t that maxmze U(x T(t) t).denotebybr(ε) the 3 Indeed, suppose to the contrary that T(0) >0. Ths means that for some δ>0andlarge enough n, A n (0) >G n (0) + δ. Thus, a fracton of at least G n (0) + δ players bd 0 n the nth contest wth postve probablty. Any one of them would be better off bddng slghtly above 0 and wnnng aganst all other players who bd 0 than bddng 0 and wth postve probablty losng to all other players who bd 0.

4 4 W. OLSZEWSKI AND R. SIEGEL ε-neghborhood of the graph of the correspondence that assgns to every type x the set BR x. 4 Denote by BR x (ε) the set of bds t such that (x t) BR(ε). Note that BR(ε) s a two-dmensonal open set, whle each BR x (ε) s a onedmensonal slce of BR(ε). Usng sets BR x (ε), we can characterze players equlbrum behavor. LEMMA S3: For every ε>0, there s an N such that for every n N, n the equlbrum of the nth contest, every best response of every type x n of every player belongs to BR x n (ε). Strct sngle crossng mples several propertes of BR x. LEMMA S4: For every x, the set BR x s a sngleton. In addton, the functon br that assgns to x thesngleelementofbr x s contnuous and wealy ncreasng. Lemma S4 mples that for every ε>0, there s a δ>0 such that BR x (δ) [br(x) ε br(x) + ε] for every type x. We therefore have the followng corollary of Lemmas S3 and S4. COROLLARY S1: For every ε>0, there s an N such that for every n N, n the equlbrum of the nth contest, every best response of every type x n of every player belongs to (br(x n ) ε br(xn ) + ε). To prove part (b) of the theorem, we need to show that T br s the assortatve allocaton. Ths s done by the followng lemma. LEMMA S5: We have G 1 (F(x)) = T(br(x)) for all types x. Thus, the mechansm that prescrbes for type x prze T(br(x)) and bd br(x) s a tarff mechansm that mplements the assortatve allocaton. Moreover, every type can get at least 0 by bddng 0, and type 0 gets no more than 0 (because T(br(0)) = 0). To complete the proof, t remans to show part (a) of the theorem. In short, ths part follows from Corollary S1 and Hoeffdng s nequalty (see Secton S1.6). S1.1. Proof of Lemma S1 Suppose frst that the lemma s false for some t (0 b max ] and q m t. Let y = lm T(q m ) and y = T(t) (the lmt exsts by the monotoncty of T ), and let γ = (y y )/2. 4 That s, BR(ε) s the unon over all types x and bds t BR x of the open balls of radus ε centered at (x t).

5 LARGE CONTESTS 5 Suppose that U(0 y t) strctly ncreases n y. (Recall that we assumed U(x y t) strctly ncreases n y only for x>0.) Then, by unform contnuty of U, there exst δ Δ > 0 such that every type x gans at least Δ from obtanng a prze hgher by γ at a bd hgher by δ. More precsely, (S2) U(x y + γ t + δ) U(x y t) Δ for all x, y, andt such that y + γ and t + δ belong to the doman of U. Ths mples, as U s bounded, that every type x strctly prefers bddng t + δ and obtanng wth suffcently hgh probablty a prze suffcently close to y + γ to bddng t and obtanng wth suffcently hgh probablty a prze suffcently close to y, ndependently of the przes obtaned wth the remanng probablty. Choose t = q m such that t t <δ. Next, choose n large enough so that T n (t) T(t) <γ/2and T n (t ) T(t ) <γ/2. Ths mples that T n (t) T n (s) > γ for any bd s t. By choosng n largeenough, we guarantee (see (S1), whch apples unformly to all bds) that R n (s), the percentle ranng of player who bds s n the nth contest, s close to A n (s) wth hgh probablty, and R n(t) s close to An (t) wth hgh probablty. Thus, every type x obtans a prze suffcently close to T n (t) wth a suffcently hgh probablty by bddng (slghtly above) t, and obtans a prze that s wth a suffcently hgh probablty at most slghtly hgher than T n (t ) by bddng (slghtly above) any s t. 5 Therefore, because t t <δ, no player bds any s (t δ t ] wth postve probablty, so by the No-Gap Property, T n (t ) = 1. But T n (t ) T(t ) y <y 1, a contradcton. When U(0 y t) only wealy ncreases n y, the argument above shows that for any ε>0, there exst δ Δ > 0forwhch(S2)holdsforeverytypex [ε 1]. There also exsts t = q m such that t t <δand T n (t) T n (t )>3γ/2 for large enough n. Lettng t = nf{s : T n (s) T n (t) γ} (t t], we see that only players wth types lower than ε can bd n [t t ). Thus, for small enough ε (by contnuty of G 1 and convergence of (G n ) 1 to G 1 ), to ncrease T n (t ) to T n (t ) + γ/2, multple players wth types ε or hgher must bd t wth postve probablty and, therefore, te there. But then any one of these players could proftably devate to bddng slghtly above t. 6 The argument s analogous f we suppose that the lemma s false for some t (0 b max ) and r m t. Ift = 0, then the above proof shows that for large n, no player bds t = 0 wth postve probablty. Ths means, n turn, that suffcently small bds gve lower payoffs than t = r m such that t t<δ.thus, no player bds close to t = 0 wth postve probablty, whch contradcts the No-Gap Property. 5 For t = b max, bddng slghtly above b max s mpossble. But by bddng b max,aplayerwns wth probablty 1, because b max s strctly domnated by 0 for all players. 6 By dong so, such a player would obtan wth hgh probablty a prze of at least T n (t ) + γ/2 nstead of losng the te wth postve probablty and then obtanng wth hgh probablty a prze of at most T n (t ) + ε.

6 6 W. OLSZEWSKI AND R. SIEGEL S1.2. Proof of Lemma S2 Suppose to the contrary. Then there s some δ>0and a sequence of ntegers n 1 n 2 such that for every n, there s some bd t wth T n (t ) T(t ) >δ. Passng to a subsequence f necessary, we assume that t t. Consder numbers q and q such that q <t<q and T(q ) T(q )<δ/2; such numbers exst because T s contnuous. 7 For large enough values of,we have that T n (q ) T(q ) <δ/2and T n (q ) T(q ) <δ/2. For any t [q q ], ether (a) T n (t ) T(t ) or (b) T n (t ) T(t ). By monotoncty of T and T n,wehave T ( ) n t T ( t ) ( T ) n q T ( q ) n case (a) and T ( t ) T ( ) n t T ( q ) ( T ) n q T n ( q ) T ( q ) + T ( q ) T ( q ) <δ T ( q ) T ( q ) + T ( q ) T n ( q ) <δ n case (b). Snce t [q q ] for large enough, we obtan a contradcton to the assumpton that T n (t ) T(t ) >δfor all such. S1.3. Proof of Lemma S3 Suppose to the contrary that for arbtrarly large n, n the equlbrum of the nth contest, some type x n of some player has a best response that belongs to the complement of BR x n (ε). Passng to a convergent subsequence f necessary, we assume that x n x. Note that for every x, there s a δ x > 0 such that (under the nverse tarff) any bd from the complement of BR x (ε) gves type x a payoff lower by at least δ x than any element of BR x does. Let δ = δ x. We have the followng observatons: () The maxmal payoff of type x, attaned at any bd from BR x, s contnuous n x. Ths follows from Berge s theorem. () For every ρ>0, for suffcently large n, the hghest payoff that type x n can obtan by bddng n the complement of BR x n (ε) cannot exceed by ρ the hghest payoff that type x can obtan by bddng n the complement of BR x (ε). Indeed, suppose that for a sequence n dvergng to, typex n obtans by bddng some t n the complement of BR n x (ε) apayoffatleastρ hgher than 7 If t = 0, set q = 0, and f t = b max,setq = b max.

7 LARGE CONTESTS 7 the hghest payoff that type x can obtan by bddng n the complement of BR x (ε). Passng to a convergent subsequence f necessary, we assume that t t. Snce every (x n t ) belongs to the complement of BR(ε), sodoes (x t); thus, (x t)belongs to the complement of BR x (ε). However, by contnuty of the payoff functons, bddng t gves type x a payoff by at least ρ hgher than the hghest payoff that type x can obtan by bddng n the complement of BR x (ε), a contradcton. By observatons () and (), for suffcently large n, any bd n the complement of BR x n (ε) gves type x n a payoff lower by at least δ/2 than any bd n BR x n. Indeed, by observaton () appled to ρ = δ/4, any bd n the complement of BR x n (ε) gves type x n a payoff at most δ/4 hgher than the hghest payoff that type x can obtan by bddng n the complement of BR x (ε). Ths last payoff s, n turn, lower than the payoff that type x obtans by bddng n BR x by at least δ. And by observaton (), the payoff that type x n obtans by bddng n BR x n cannot be lower by more than δ/4 than the payoff that type x obtans by bddng n BR x. By unform convergence of T n to T, the analogous statement, wth δ/2 replaced wth some smaller postve number and T replaced wth T n,salsotrue. Ths means, however, that for suffcently large n,player would be strctly better off bddng slghtly above any bd n BR x n when her type s x n than bddng n the complement of BR x n (ε). Ths s because (S1) mples that for suffcently large n, by bddng slghtly above t, the player obtans a prze arbtrarly close to T n (t) wth probablty arbtrarly close to 1. S1.4. Proof of Lemma S4 Observe that for any x <x, strct sngle crossng mples that f t BR x and t BR x, then t t. Suppose that BR x contaned two bds, t 1 <t 2,for some type x. The frst observaton and Lemma S3 mply that for any 0 <ε< (t 2 t 1 )/4, for suffcently large n, only players wth types n I =[max{x ε 0} mn{x + ε 1}] may bd n the nterval [t 1 + (t 2 t 1 )/4 t 2 (t 2 t 1 )/4]. Consder obtanng a prze that s Δ hgher n the lmt prze dstrbuton 8 by ncreasng the bd from t 1 + (t 2 t 1 )/4 tot 2 (t 2 t 1 )/2. If Δ s suffcently small, then by contnuty of G 1, the ncrease n the prze s small as well, so the assocated ncrement n utlty s negatve for all types and unformly bounded away from 0. Therefore, tang Δ/2 = F(mn{x + ε 1}) F(max{x ε 0}), fε>0s suffcently small, then for suffcently large n, every type of every player s better off bddng t 1 + (t 2 t 1 )/4 than bddng t 2 (t 2 t 1 )/2. Ths s because wth hgh probablty, the hgher bd leads to a prze that s approxmately only 8 More precsely, gven an ntal prze y, the prze that s Δ hgher n the lmt prze dstrbuton s the prze y such that Δ = G(y ) G(y ). The prze that s Δ hgher n the prze dstrbuton G n s defned smlarly.

8 8 W. OLSZEWSKI AND R. SIEGEL Δ/2 hgher n the prze dstrbuton G n. 9 By convergence of (G n ) 1 to (G) 1, for suffcently large n, ths prze s not much more than Δ/2 hgher n the lmt prze dstrbuton. Moreover, every type of every player s better off bddng t 1 + (t 2 t 1 )/4 than bddng any bd n nterval (t 2 (t 2 t 1 )/2 t 2 (t 2 t 1 )/4), because such bds are even more costly than t 2 (t 2 t 1 )/2, and enable a player to obtan a prze that s wth hgh probablty not much more than Δ/2 hgher n the lmt prze dstrbuton than the prze the player obtans by bddng t 2 (t 2 t 1 )/2. Therefore, no player bds n the nterval ((t 2 (t 2 t 1 )/2 t 2 (t 2 t 1 )/4)) wth postve probablty, so by the No-Gap Property, T n (t 2 (t 2 t 1 )/4) = 1 for suffcently large n. Thus,T(t 2 (t 2 t 1 )/4) = 1, so t 2 cannot be n BR x, because bddng slghtly above t 2 (t 2 t 1 )/4 gves type x a hgher payoff. Consequently, BR x s a sngleton for any x, and by strct sngle crossng, br s wealy ncreasng. An argument analogous to the argument used to show that BR s a sngleton also shows that br s contnuous. 10 S1.5. Proof of Lemma S5 Consder an arbtrary type x. Letx mn = mn{z : br(z) = br(x)} and x max = max{z : br(z) = br(x)} (x mn and x max are well defned because br s contnuous). Frst, observe that G 1 (F(x mn )) = G 1 (F(x max )). Indeed, by Corollary S1, for suffcently large n, all types n the nterval [x mn x max ] bd n the nth contest close to br(x). Suppose that G 1 (F(x mn )) < G 1 (F(x max )), and consder the players whose types belong to [x mn x max ] wth postve probablty. 11 Among these players, the one whose expected prze s the lowest contngent on havng a type n ths nterval can proftably devate to bddng slghtly above br(x), thereby outbddng the other players wth a type n ths nterval and obtanng a dscretely hgher prze. Suppose that x mn > 0. By Corollary S1 for any δ>0, there s an N such that f n N, then the equlbrum bds of every player wth type lower than x mn δ are lower than br(x mn ), and the equlbrum bds of every player wth type hgher than x mn are hgher than br(x mn δ). Therefore, a player who bds br(x mn ) outbds all players wth types lower than x mn δ, sot n (br(x mn )) 9 Ths follows from the convergence of F n to F and Hoeffdng s nequalty appled to random varables: { { Z n = 1 f mn x + ε 1 } x n max { x ε 0 }, 0 otherwse, for = 1 n. 10 More precsely, suppose that br s dscontnuous at some x, and apply the argument to t 1 = br(x 1 ) and t 2 = br(x 2 ),wherex 1 and x 2 are slghtly lower and hgher, respectvely, than x. 11 For large n, at least a fracton of players close to F(x max ) F(x mn ) have types that belong to [x mn x max ] wth postve probablty.

9 LARGE CONTESTS 9 (G n ) 1 (F n (x mn δ)), and a player who bds br(x mn δ) s outbd by all players wth types hgher than x mn,sot n (br(x mn δ)) (G n ) 1 (F n (x mn )). Snce T n converges to T, T and br are contnuous, (G n ) 1 converges to (G 1 ), F n converges to F, andf and G 1 are contnuous, we obtan T(br(x mn )) = G 1 (F(x mn )). Smlarly, f x max < 1, we obtan that T(br(x max )) = G 1 (F(x max )). Thus, snce br(x) = br(x mn ) = br(x max ) and G 1 (F(x)) = G 1 (F(x mn )) = G 1 (F(x max )), we have that T(br(x)) = G 1 (F(x)) when x mn > 0 or x max < 1. Fnally, t cannot be that x mn = 0 and x max = 1, because 0 = G 1 (F(0)) < G 1 (F(1)) = 1. S1.6. Proof of Theorem 1(a) Consder a type x X. Lett = br(x), andlett and t be such that T(t ) = T(t) ε/3andt(t ) = T(t)+ε/3. Fnally, let x and x be such that t = br(x ) and t = br(x ).(Taex = 0andt = 0fT(t) ε/3 < 0, and x = 1and t = br(1) f T(t)+ ε/3 > 1.) By Lemma S3, for suffcently large n, every player wth type no hgher than x bds less than the player wth type x, and every player wth type no lower than x bds more than the player wth type x. By Hoeffdng s nequalty, the player of type x outbds wth hgh probablty at least a fracton of players close to F n (x ). Snce F n converges to F, she outbds wth hgh probablty at least a fracton of players close to F(x ). So, snce (G n ) 1 converges to G 1, she obtans (wth hgh probablty) a prze no lower than G 1 (F(x )) ε/3 = T(t) 2ε/3. Smlarly, for suffcently large n, a player of type x outbds wth hgh probablty at most a fracton of players close to F(x ), and so she obtans (wth hgh probablty) a prze no hgher than G 1 (F(x )) = T(t)+ 2ε/3. (These bounds are mmedate f y ε/2 < 0orfy + ε/2 > 1.) Thus, type x obtans (wth hgh probablty) a prze that dffers from G 1 (F(x)) by at most 2ε/3. Ths proves part (a) for a sngle x, but we must show that there s an N such that for any n N, part (a) holds for all x smultaneously. Such an N can be obtaned by tang a fnte grd of types x and the correspondng grd of bds br(x) such that T(t 1 ) T(t 2 ) <ε/3 for any par of neghborng elements t 1, t 2 of the grd, and tang the largest N among the N s correspondng to x s from the grd. S2. PROOF OF THEOREM 2 Recall that G 1 (z) = nf{y : G(y) z} for z>0 and G 1 (0) = nf{y : G(y) > 0}, and note that G 1 may be dscontnuous (but s left-contnuous). Dscontnutes requre modfyng almost all the arguments used n the proof of Theorem 1. As n Secton S1, we relegate to the end of the secton the proofs of all ntermedate results.

10 10 W. OLSZEWSKI AND R. SIEGEL Let I 0 = (y l 0 yu 0 ) be a longest nterval n [0 G 1 (1)] to whch G assgns measure 0, let I 1 = (y l 1 yu 1 ) be a longest such nterval dsjont from I 1, and so on. Then every open nterval of przes that has measure zero s contaned n one of the ntervals I 0 I 1 and for any ε>0, there s a K such that the lengths of I K+1, I K+2 sum to less than ε. The defntons of R n, An, An,andT n are as n Secton S1. The defnton of T, however, must be changed. Frst, by Helly s selecton theorem, we tae a convergng subsequence of the sequence A n ; denote ts lmt by A : B [0 1]. Ths functon s wealy ncreasng (because each A n s). For the rest of the proof, denote ths convergng subsequence by A n (wth the correspondng sequence T n = (G n ) 1 A n ). Let T = G 1 A. SnceG may not have full support, we now have that T(0) = nf{z : G(z) > 0} and T(b max ) = G 1 (1); n addton, T s stll (wealy) ncreasng (compare to propertes () () from Secton S1). In addton, F n converges pontwse to F, but(g n ) 1 may not converge pontwse to G 1. It s, however, easy to chec that lm n (G n ) 1 (r) = G 1 (r) unless r s the value of G on an nterval I = (y l yu );moreover,lm n(g n ) 1 (r) G 1 (r) for every r that s the value of G on an nterval I = (y l yu ),buttcan happen that lm n (G n ) 1 (r) = y u and G 1 (r) = y l. The dscontnutes n G 1 mply that T may not be contnuous, so Lemma S1 does not hold. Ponts of dscontnuty, however, correspond to open ntervals of przes that have measure zero. More precsely, we have the followng result. LEMMA S6: For any t>0 n B, one of the followng condtons holds: 1. Mappng T s contnuous at t, that s, for any sequences q m t and r m t, we have lm T(q m ) = lm T(r m ) = T(t). 2. There s some = 1 2 such that for any sequences q m t and r m t, we have lm T(q m ) = y l and lm T(rm ) = y u. Moreover, lma(qm ) = G(y l ) and lm A(r m ) = G(y u ). Usng Lemma S6, we defne another functon T on B by settng T (t) = lm T(r) for some sequence r t (and T (b max ) = G 1 (1)). The monotoncty of T guarantees that T (t) s well defned. In addton, t s easy to chec that T s (wealy) ncreasng, rght-contnuous, and contnuous at every bd t such that condton () from Lemma S6 holds. Note that T may not be an extenson of T, because when lm T(r) T(t), we have that T (t) = lm T(r) T(t). Consder now a bd t>0 such that condton () from Lemma S6 holds. Denote ths bd t by t,where s descrbed n condton (). Then there s a bd t <t such that A(t ) = A(t) = G(y l ),soa s constant on an nterval below t. Indeed, f A(t )<G(y l ) for all t <t, then, as n the proof of Lemma S6, for large n, no player would bd any t slghtly below t. Ths would be so, because bddng slghtly above t would almost certanly gve a prze no lower than y u, whereas bddng t would almost certanly gve a prze no hgher than y l.let t l = nf{ t : A ( t ) = G ( y l )} <t

11 LARGE CONTESTS 11 It s also true that every maxmal nterval on whch T s constant wth a value lower than G 1 (1) s [t l t ) for some. Indeed, consder a maxmal nontrval nterval wth lower bound t l and upper bound t u on whch the value of T s y<g 1 (1). It suffces to show that T (t u )>y, because then condton () from Lemma S6 apples to t u, whch mples that t u = t for some, and the maxmalty of [t l t u ) yelds t l = t l. Suppose that T (t u ) = y. Then, for large enough n, bddng t u almost certanly gves a prze at most slghtly hgher than y, whereas bddng slghtly above t l almost certanly gves a prze not much lower than y. But then, for large enough n, noplayerbdsnsome neghborhood of t u, because bddng slghtly above t l leads to a hgher payoff. Ths contradcts the No-Gap Property, because y<g 1 (1). Because G 1 may be dscontnuous, T n need not converge unformly to T or T, even on the set of ponts at whch they are contnuous. In partcular, for a t [t l t ), t may be that T n (t) = (G n ) 1 (A n (t)) y u for arbtrarly large n, whereas T(t)= T (t) = y l. Nevertheless, T n converges unformly except on some neghborhoods of a fnte number of ntervals [t l t ]. More precsely, we say that T n converges unformly to T up to β on a set C f there exsts an N such that for every n N and t C, we have that T n (t) T (t) <β We then have the followng modfcaton of Lemma S2. LEMMA S7: For every β>0, there exsts a number K such that for every γ>0, T n converges unformly to T up to β on the complement of O γ = K ( t l γ t + γ ) =1 We now relate players equlbrum behavor n large contests to the nverse tarff T.DefneBR x,br(ε),andbr x (ε) as n Secton S1,wthT nstead of T (the maxmal payoff s acheved because T s ncreasng and rght-contnuous, so s upper semcontnuous). Defne the mass expended (n the nth contest) n an nterval of bds I by players wth type x S as ( n Pr(σ n =1 S I))/n. We then have the followng result, whch we use n provng the remanng results. LEMMA S8: For all, and any ε>0 and L>0, there exsts γ>0 such that for suffcently large n, we have that the followng statements hold: () The mass expended n (t l γ tl + γ) by players wth types x for whch t l / BR x(ε) s less than ε/3l. () The mass expended n (t γ t ] by players wth types x for whch t / BR x (ε) s less than ε/3l. () In addton, for any α>0, for suffcently large n, we have that the mass expended n [t l + α t α] by all players s less than ε/3l.

12 12 W. OLSZEWSKI AND R. SIEGEL Lemma S3 must also be modfed. LEMMA S9: For every ε>0, there exst K such that for every γ>0, there s an N such that for every n N n the equlbrum of the nth contest, every best response of every type x n of every player belongs to BR x n (ε) K ( t l γ t ) =1 Strct sngle crossng no longer mples that BR x s a sngleton. Instead, we have the followng result. LEMMA S10: If strct sngle crossng holds, then for all but a countable number of types, the set BR x s a sngleton. For those types for whch t s not a sngleton, BR x contans precsely two elements: t l and t for some. The correspondence that assgns to type x the set BR x s wealy ncreasng (.e., for any x <x, f t BR x and t BR x, then t t ) and upper hemcontnuous. Let br(x) = mn BR x, and note that br s ncreasng and left-contnuous, and s not rght-contnuous precsely at types x for whch BR x s not a sngleton. We then have the followng corollary of Lemmas S8, S9, ands10, whchsa modfcaton of Corollary S1. COROLLARY S2: For every ε>0, there s an N such that for n N, at least a fracton 1 ε of players bd n (br(x n) ε br(xn ) + ε) wth probablty at least 1 ε. To prove part (b) of the theorem, t remans to show that T br s the assortatve allocaton. Ths s done by the followng lemma, whch s a modfcaton of Lemma S5 that accommodates the dscontnutes n T and br. LEMMA S11: We have G 1 (F(x)) = T (br(x)) for any type x>0. To complete the proof, t remans to show (a) n the statement of the theorem. To do so, we use the followng result. LEMMA S12: For every ε δ > 0, there s an N such that for n N, each type x from a set whose F n measure s at least 1 ε bds a t wth probablty at least 1 ε, and obtans a y wth probablty at least 1 ε such that for some r n BR x, t r <δ and y T (r) <δ

13 LARGE CONTESTS 13 To see that Lemma S12 mples (a) n the statement of the theorem, choose some ε>0. Lemma S12 shows that for every δ>0, there s an N such that for n N and for a fracton 1 ε of players, the F n measure of ther types x n that satsfy the condton of Lemma S12 s at least (1 ε). Ths means that each such player obtans wth probablty at least 1 ε aprzey that dffers by at most δ from the prze T (r) for some optmal bd r of the player s type. For types x n > 0 such that br(x n ) s a unque optmal bd, ths yelds (a) by Lemma S11. However, by Lemma S9 and strct sngle crossng, there s only a countable number of other types x n, and the F measure of such types s 0 snce F has no atoms, so the F n measure of such types s arbtrarly small for suffcently large n. S2.1. Proof of Lemma S6 Let lmt(q m ) = y and lm T(r m ) = y. Both lmts y and y exst, and y y by monotoncty. Suppose that y <y.ifg assgns a postve measure to (y y ), then t assgns a postve measure to any nterval wth endponts suffcently close to y and y. In such a case, we obtan a contradcton by arguments smlar to those used n the proof of Lemma S1. Indeed, for suffcently large n, no bdder would bd slghtly below t, because bddng slghtly above t would almost certanly gve a better prze. Thus, G assgns measure zero to (y y ). Ths mples that (y y ) (y l yu) for some. By defnton, T taes values n [0 y l ] [yu 1], soy = y l and y = y u. Moreover, the monotoncty of T mples that s the same for any sequences q m t and r m t. It remans to show that lm A(q m ) = G(y l ) and lm A(r m ) = G(y u). For ths, note that f lm A(q m )>G(y l ), then lm T(qm )>y l. Smlarly, f lm A(r m )>G(y u), then lm T(rm )>y u. The nequaltes lm A(qm )<G(y l ) and lm A(r m )<G(y u ) can be ruled out smlarly f G does not have atoms at y l or yu. Suppose that G has an atom at yu and lm A(rm )<G(y u).snce lm T(r m ) = y u, A(rm )>G(y l ) for suffcently large m. Tae two numbers rm such that G(y l )<A(rm )<G(y u); denote them by t <t. Then, for suffcently large n, any player obtans a prze close to y u wth arbtrarly hgh probablty by bddng any t [t t ]. Thus, for suffcently large n, no player would bd n the nterval [(t + t )/2 t ] wth postve probablty. Ths contradcts the No-Gap Property. Suppose that G has an atom at y l and lm A(qm )<G(y l ). Then, for suffcently large n, bddng q m almost certanly gves a prze at most slghtly better than y l. In contrast, bddng rm almost certanly gves a prze at least as good as y u. Ths follows drectly from (S1) fg has an atom at yu.ifg does not, then ths agan follows from (S1) for large enough n, because A(r m )>G(y u) for any m. For large enough n, a contradcton wth the No-Gap Property s obtaned smlarly to the last part of the proof of Lemma S1 that deals wth U(0 y t)strctly ncreasng n y.

14 14 W. OLSZEWSKI AND R. SIEGEL S2.2. Proof of Lemma S7 The proof s analogous to the proof of Lemma S2. TaeaK such that the lengths of I K+1 I K+2 sum up to less than β/2. Tae any γ>0, and suppose to the contrary that there s an ncreasng sequence of ntegers n 1 n 2 n m such that for every n m, there s some bd t m / O γ wth T nm (t m ) T (t m ) β. Passng to a subsequence f necessary, we assume that the sequence t m t.taeq and q such that q <t<q and T (q ) T (q )< β/2, 12 and [ q q ] B K [ t l t ] =1 Ths s possble, snce the lengths of I K+1, I K+2 sum to less than β/2. In addton, for large enough, we have that T n (q ) T (q ) <β/2and T n (q ) T (q ) <β/2, snce the length of each I K+1, I K+2 s less than β/2. The rest of the proof concdes wth the proof of Lemma S2. S2.3. Proof of Lemma S8 Frst, observe that the maxmal payoff of type x, attaned at any bd n BR x, s stll contnuous n x. Indeed, upper semcontnuty of T s all that s needed for the contnuty of the maxmal payoff. Ths observaton mples that there exsts a δ>0 such that for any type x, any bd n the complement of BR x (ε) gves type x a payoff lower by at least δ than any bd n BR x. For (), suppose to the contrary that for any γ>0, there are arbtrarly large n such that the mass expended n (t l γ tl + γ) by players wth types x for whch t l / BR x(ε) s at least ε/3l.taeγ small enough so that the payoff that such players obtan by bddng slghtly more than any bd n BR x s hgher by δ/2 than the payoff that they would obtan by bddng t l γ and gettng yl.13 Suppose frst that t l > 0. By monotoncty of A and the defnton of tl,we have that A(t l γ) < G(yl ).Taeapostveα<ε/6L such that A(tl γ) < G(y l ) α. Foranyt tl γ and suffcently large n, fan (t) < G(y l ) α/2, then no player of type x such that t l / BR x(ε) bds t, because by bddng t, such a player would obtan wth hgh probablty a prze no hgher than y l and, 12 If t = 0, tae q = To see why bddng slghtly above any t BR x gves at least a payoff close to U(x T (t) t), consder the followng two cases: (a) We have G 1 (A(r)) > T (t) for all r>t: In ths case, snce lm n (G n ) 1 (A(r)) G 1 (A(r)) for any r>t,fn s suffcently large, then (G n ) 1 (A(r)) > T (t). Ths mples that a player obtans a prze hgher than T (t) wth arbtrarly hgh probablty by bddng r. (b) We have G 1 (A(r)) = T (t) for r>tclose enough to t: Inthscase,T (r) = T (t) for all such r. Ths mples that t = t l for some. The clam now follows from left contnuty of G 1 and the fact that lm n (G n ) 1 (q) G 1 (q) for any q.

15 LARGE CONTESTS 15 therefore, would obtan a hgher payoff by bddng slghtly more than any bd n BR x. Let γ n be defned by t l γ n = nf{t : A n (t) G(y l ) α}. SnceA(tl γ) < G(y l ) α and An (t) s rght-contnuous, we have that γ n <γ(for suffcently large n). And snce for every t<t l γ n,wehavea n (t) < G(y l ) α/2 (by defnton of γ n ), players wth types x for whch t l / BR x(ε) must expend the mass of at least ε/3l n [t l γ n t l + γ). If more than half of ths mass s expended n (t l γ n t l + γ), then we have that A n (t l +γ) > An (t l γ n)+ε/6l G(y l ) α+ε/6l>g(yl ). Ths cannot happen for suffcently large n, because for t [t l t ), wehavea(t) = G(y l ). Thus, the players wth types x for whch t l / BR x(ε) bd precsely t l γ n wth probablty at least ε/6l. Snce these players te wth each other at t l γ n,by bddng t l γ n, they must obtan a prze of a specfc type y wth probablty 1, even f they lose all tes at t l γ n. (Otherwse, each of them could obtan a hgher payoff by bddng slghtly above t l γ + γ n and wnnng the tes at t l γ n.) But a player who loses all tes at t l γ n has ran order no hgher than G(y l ) α, by defnton of γ n,soy y l. Therefore, such a player would obtan a strctly hgher payoff by bddng slghtly more than any bd n BR x. Now suppose that t l = 0. Then A(tl ) G(yl ).ThecaseA(tl )<G(yl) s handled as n the case t l > 0 above. Suppose that A(tl ) = G(yl ). Then, for any γ>0 such that t l + γ<t, for suffcently large n, the mass expended n (t l tl + γ) by all players s smaller than ε/6l, because A(t) = G(yl ) for any t (t l + γ t ). Thus, f () does not hold, for suffcently large n, the mass expended precsely at t l by the players wth types x for whch tl / BR x(ε) s at least ε/6l, and so the ranng of a player who tes at t l and loses s at most G(y l ) ε/12l. But n ths case, each player of type x for whch tl / BR x(ε) would strctly prefer bddng slghtly more than any bd n BR x to bddng t l, a contradcton. To show (), note that f t = t l for some, then () follows from (). Thus, suppose that t t l for any. Suppose to the contrary that for any γ>0, there s an arbtrarly large n such that the mass expended n (t γ t ] by players wth types x for whch t / BR x (ε) s at least ε/3l.taeγ small enough so that the payoff that such players obtan by bddng slghtly more than any bd n BR x s hgher by δ/2 than the payoff that they would obtan by bddng t γ and gettng y u. Observe that for suffcently large n, by bddng t,any player almost certanly obtans a prze at most slghtly better than T (t ) = y u. Ths s so, because t t l and so A(t ) G(y l ) for any. Therefore, for large enough n, a player wth type x for whch t / BR x (ε) would be better off bddng slghtly above any t BR x than bddng n (t γ t ]. Part () follows mmedately from the fact that the value of A on [t l t ) s G(y l ), by the defnton of tl.

16 16 W. OLSZEWSKI AND R. SIEGEL S2.4. Proof of Lemma S9 Tae a δ>0 such that for any type x, any bd n the complement of BR x (ε) gves type x a payoff lower by at least δ than any bd n BR x.taeβ>0such that for any type x, bdt, andprzesy and y wth y y β, wehave U ( x y t ) U ( x y t ) δ 3 Next, tae a K guaranteed by Lemma S7 for ths β. In addton, tae K large enough so the lengths of I K+1, I K+2 sum to less than β/2. Fnally, for any λ>0, tae an N λ that satsfes the defnton of unform convergence up to β on the complement of O λ. (Note that K s the same for all λ.) Suppose, to the contrary of the statement of the lemma, that there s a γ>0 and a subsequence of contests such that a type x n of player n the nth contest has a best response t n to the strateges of the other players that does not belong to BR x n (ε) K =1 (tl γ t ). As usual, we assume that the subsequence s the entre sequence; moreover, we assume that x n x and t n t. Consder the followng two cases: A. We have t t for any = 1 K: In ths case, for some λ>0, there s a neghborhood of t that s dsjont from O λ. By unform convergence of T n to T up to β on the complement of O λ, U ( x n Tn( t n) t n) U ( x n T ( t n) t n) δ 3 for n N λ. And because t n / BR x n (ε),foranyt BR x n,wehave Thus, we obtan U ( x n T (t) t ) U ( x n T ( t n) t n) δ U ( x n T (t) t ) U ( x n Tn( t n) t n) 2δ 3 Observe that any bd t hgher than t guarantees, for suffcently large n, a prze not much worse than T (t) wth arbtrarly hgh probablty. 14 We wll now show that by bddng t n, for suffcently hgh n, typex n obtans wth arbtrarly hgh probablty a prze no better than T n (t n )+β. Indeed, snce t does not belong to [t l t ] for any K, we have that A(t ) s bounded away from G(y l) 1 G(yl ) for K t suffcently close to t. Therefore, A n (t n ) s also bounded away from G(y l) 1 G(yl K ) for suffcently large n. Andforsuffcently large n, bddng t n gves wth arbtrarly hgh probablty a ran order arbtrarly close to A n (t n ). Snce the lengths of I K+1, I K+2 sum to less than 14 To see why, see the prevous footnote.

17 LARGE CONTESTS 17 β/2, and for any r other than G(y l) 1 G(yl K ) and suffcently large n the dfference between (G n ) 1 (r) and G 1 (r) s no larger than the length of I K+1,by bddng t n, a player obtans wth arbtrarly hgh probablty a prze no better than (G n ) 1 (A n (t n )) + β. Therefore, by defnton of β, we have that by bddng t n,typex n obtans a payoff that s hgher than U(x n Tn (t n ) t n ) by at most slghtly more than δ/3. Consequently, for suffcently large n, player would obtan by bddng some t >ta payoff strctly hgher than by bddng t n, a contradcton. B. We have t = t for some = 1 K:Thenconsderat slghtly hgher than t, such that t does not belong to [t l t ] for = 1 K, and such that () for suffcently large n, the payoff (of any player) n the nth contest of bddng t s not much lower than the payoff of bddng t n ; () for suffcently large n, we have that the dfference between U(x n T (t) t) for any t n BR x n and U(x n T (t ) t ) s not much lower than δ. Ths latter condton s possble because, by defnton, (x t )/ BR(ε), and by rght contnuty of T at t. Now, usng (), apply an argument analogous to that from case A wth t playng the role of t n, wth a contradcton obtaned by referrng to (). S2.5. Proof of Lemma S10 Monotoncty of the correspondence follows from strct sngle crossng; upper hemcontnuty follows from standard arguments. 15 Suppose that BR x contans a par of bds t 1 <t 2. Below we wll show that for any ε>0 and any nterval [a b] such that t 1 <aand b<t 2, for suffcently large n, the mass expended n [a b] by all players s at most ε. Ths mples that the functon A, and therefore T, s constant on every such nterval [a b] and, therefore, on (t 1 t 2 ).ButT (t 2 )>T (t 1 ) because t 1 <t 2 are n BR x,soby defnton of the dscontnuty ponts t of T,wemusthave(t 1 t 2 ) (t l t ) for some. And because BR x B \ =1 (tl t ), we have that t 1 = t l and t 2 = t. It remans to show that for any ε>0, for suffcently large n, the mass expended n [a b] by all players s at most ε. We wll show ths for ε/2 andplayers of types lower than x (a smlar argument apples to types hgher than x). Choose x <x such that F(x) F(x )<ε/3. For suffcently small λ>0, sup z x BR z (λ) < a. (Ths s because x <xand t 1 BR x, so every bd n BR z s at most t 1 <a.) Therefore, by Lemma S9, there s some K such that for every γ>0and suffcently large n, anybdn[a b] made by a player of type z x n the nth contest s n K =1 (tl γ t ). Consder one of these K ntervals for whch (t l γ t ) [a b]. Snce sup z x BR z (λ) < a t, t s not n BR z (λ) for any z x.ift l > sup z x BR z (λ), then by () of Lemma S8, there exsts a γ such that for suffcently large n, the mass expended n (t l γ t ) by players of 15 More precsely, ths follows from the fact that BR x s the set of all t such that (t T (t)) maxmzes type x s utlty over the closure of the graph of T, whch s a compact set.

18 18 W. OLSZEWSKI AND R. SIEGEL type z x s less than ε/6k. Ift l sup z x BR z (λ), then by () and () of Lemma S8, for suffcently large n, the mass expended n [a t ) by players of type z x s less than ε/6k. Therefore, for large enough n, the mass expended n [a b] by players of type z x s smaller than ε/6, and because F(x) F(x )<ε/3, the mass expended n [a b] by players of type z x s smaller than ε/2. S2.6. Proof of Corollary S2 Choose ε>0. Lemma S10 mples that there s a fnte number of ntervals of types wth total F mass ε/2, such that for every type x not n one of these ntervals, BR x (br(x) ε br(x) + ε). 16 Consder the F mass 1 ε/2oftypes x wth the last property, and let K be the one n the statement of Lemma S9. Then, by Lemma S9 and Lemma S8 for L = K, for suffcently large n,atmost an F mass ε/2 of those types bd outsde of (br(x) ε br(x) + ε). S2.7. Proof of Lemma S11 The proof s analogous to that of Lemma S5. Consder an arbtrary type x. Defne x mn = mn{z : br(x) BR z } and x max = max{z : br(x) BR z }.Bystrct sngle crossng, BR z has only one element br(z) = br(x) for all z (x mn x max ); t may have two elements for z = x mn or x max, n whch cases br(x) s the hgher one and the lower one of the two, respectvely. The clam that G 1 (F(x mn )) = G 1 (F(x max )) s obtaned by the same argument as n the proof of Lemma S5. The rest of the proof requres the followng mnor changes when BR x mn has two elements (and analogous changes when BR x max has two elements): () Instead of x mn, we consder x mn = mn{z : br(x mn ) BR z }, and compare the equlbrum bds of every player wth type lower than x mn δ to br(x mn ) and the equlbrum bds of every player wth type hgher than x mn to br(x mn δ). Ths change does not affect the arguments, snce G 1 (F(x mn )) = G 1 (F(x mn )). () It may not be true that the equlbrum bds of every player wth type lower than x mn δ are lower than br(x mn ) or that the equlbrum bds of every player wth type hgher than x mn are hgher than br(x mn δ), because players may bd n K =1 (tl γ t ) BR(ε) (see Lemma S9). However, ths happens only wth vanshng probablty as n grows large, so the arguments are agan not affected. 16 There s a K>0suchthat >K (t t l )<ε. For each K such that BR x ={t l t } for some type x, consder the nterval of types [x λ x + λ] [0 1], whereλ s such that (contnuous) F ncreases by no more than ε/2k on any nterval no larger than 2λ. Thesumof the F mass of these ntervals s no larger than ε/2, and the sum of the jumps of br on the complement of these ntervals s smaller than ε.

19 LARGE CONTESTS 19 S2.8. Proof of Lemma S12 Tae any λ>0. By Lemma S9, there s a large K such that for any γ>0, f n s suffcently large, the equlbrum bd of every player n the nth contest belongs wth probablty 1 to BR x n (λ) K ( t l γ t ) =1 Assume that Ks, n addton, large enough so that the lengths of I K+1, I K+2 sum to less than δ/2. We frst clam that for any t/ (t l γ t ) for all = 1 K, there exsts an N t such that for every n N t, a player who bds t n the nth contest obtans (wth hgh probablty) a prze y such that y T (t) <δ/2. We wll also show that there exsts an N = N t thatscommonforallsuchbdst. Suppose frst that t t for any = 1 K.SnceA(t) dffers from G(y l ) and G(y u ) for any = 1 K, any ran order close to A(t) also dffers from G(y l ) and G(yu ).By(S1), for suffcently large n, a player who bds t has (wth hgh probablty) a ran order close to A(t); n partcular, ths ran order dffers from G(y l ) and G(yu). By the assumpton that the lengths of I K+1, I K+2 sum to less than δ/2, ths mples that the dfference between T (t) and the prze obtaned by a player who bds t s lower than δ/2 (wth hgh probablty). Suppose that t = t for some = 1 K. By an argument analogous to the one used n the prevous case, the prze obtaned by a player who bds t cannot, as n ncreases, exceed T (t) by δ/2 wth a probablty that s bounded away from 0. And T (t) cannot exceed ths prze by δ/2 wth a probablty that s bounded away from 0 as n ncreases, because the player would proftably devate by bddng slghtly above t, whch would guarantee a prze no worse than T (t) wth arbtrarly hgh probablty. Now note that the number N t that was chosen for any bd t also has the requred property for all bds close enough to t; n the case of t = t for some = 1 K, we mean bds close enough and hgher than t. That s, for every t, there s a neghborhood W t of that t wth N t thatscommonforallbdsfrom ths neghborhood. The famly of sets W t s an open coverng of the compact set of bds t that satsfy t/ (t l γ t ) for = 1 K. Thus, t contans a fnte subcoverng, and any number N that exceeds numbers N t for all elements of ths fnte subcoverng has the requred property. Ths yelds the lemma for bds t/ (t l γ t ) for all = 1 K. Indeed, note that BR x s a sngleton and br(x) s ts only element for all except a countable number of types x.sncef has no atoms, the set of such types has F measure 1, and for such types x, equlbrum bds t/ (t l γ t ) for all = 1 K belong to (br(x) λ br(x) + λ). Ifλ s suffcently small and x s bounded

20 20 W. OLSZEWSKI AND R. SIEGEL away from 0, then t r <δfor r = br(x) and T (t) T (r) δ/2; 17 f λ s suffcently small, then also T (r) T (t) δ/2, because t/ (t l γ t ) for all = 1 K and the lengths of I K+1, I K+2 sum to less than δ/2. Fnally, by our frst clam, the prze y obtaned by bddng t must satsfy y T (t) <δ/2, so y T (r) <δ. Now consder bds t such that t s n (t l γ t ) for some = 1 K.By () of Lemma S8, we can dsregard bds t n [t l + γ t γ]. Suppose that t s n (t γ t ) and t t l for all other = 1 K. By () of Lemma S8, one can assume that t BR x n. 18 We wll show that for suffcently small γ and for suffcently large n, player obtans by bddng t (wth arbtrarly hgh probablty) a prze y n (T (t ) δ T (t )+δ). Frst, note that player cannot obtan by bddng t a prze lower than T (t ) δ (wth probablty bounded away from 0), because for small enough γ, t would be proftable to devate to bddng slghtly above t andobtanaprzenotmuchlowerthant (t ) wth hgh probablty. Player cannot obtan by bddng t a prze hgher than T (t ) + δ (wth probablty bounded away from 0), because by (S1), for any r>t and suffcently large n, the ran order of player s wth arbtrarly hgh probablty bounded above by A(r). Thus, the upper bound on the prze follows from the assumpton that t t l for all other = 1 K, and the lengths of I K+1 I K+2 sum to less than δ/2. Fnally, suppose that t s n (t l γ tl + γ) for some = 1 K.By()of Lemma S8, one can assume that t l BR x n. We wll show that for suffcently small γ and for suffcently large n, equlbrum bddng n (t l γ tl + γ) leads (wth arbtrarly hgh probablty) to a prze y (T (t l ) δ T (t l ) + δ), except a small probablty event. Indeed, by an argument smlar to that from the prevous case, such a bd cannot lead to a prze lower than T (t l ) δ (wth probablty bounded away from 0). To obtan a prze hgher than T (t l ) + δ wth a nonvanshng probablty, a player s expected ran order when bddng t cannot be lower than G(y l ) by a nonvanshng constant. But f a nonvanshng fracton of players wn a prze hgher than T (t l ) + δ wthanonvanshng probablty by bddng n (t l γ tl + γ), then the ncrease n expected ran order on the nterval (t l γ tl + γ) s bounded away from 0 for all n, whch contradcts the fact that A n (t l + γ) approaches G(yl ) as n ncreases. Dept. of Economcs, Northwestern Unversty, Evanston, IL 60208, U.S.A.; wo@northwestern.edu and 17 Indeed, for types bounded away from 0 and for suffcently small λ,wehavethatu(x y t)> U(x y t ) whenever y y >δ/2andt t <λ. (The assumpton that types are bounded away from 0 s essental, because we dd not assume that U(0 y t) strctly ncreases n y.) However, snce r = br(x),wecannothaveu(x T (t) t) > U(x T (r) r). 18 The lemma says only that the mass expended n (t γ t ] by types x for whch t / BR x (λ) for some small λ>0 s small. However, f λ>0 s suffcently small, then the mass of types x such that t / BR x but t BR x (λ) s small.

21 LARGE CONTESTS 21 Dept. of Economcs, The Pennsylvana State Unversty, Unversty Par, PA 16801, U.S.A.; Co-edtor Matthew O. Jacson handled ths manuscrpt. Manuscrpt receved June, 2013; fnal revson receved Aprl, 2015.