Minimal Assumptions for E ciciency in Asymmetric English Auctions.

Transcription

1 Mnmal ssumptons for E ccency n symmetrc Englsh uctons. Juan Dubra y Unversdad de Montevdeo and Unversdad Torcuato D Tella September, 2006 bstract I ntroduce a property of player s valuatons that ensures the exstence of an ex post e cent equlbrum n asymmetrc Englsh auctons. The property s weaer than the one n Echenque and Manell (2006) and s smlar to that n Bruln and Izmalov (2003). Relatve to ths latter paper and the wor of Krshna (2003), the use of ths property has the advantage of yeldng the desred results wthout assumng d erentablty of valuatons or that sgnals are drawn from a densty. 1 Introducton Ths paper gves a mnmal set of assumptons that ensures exstence of an e cent ex-post equlbrum (once sgnals are nown, players don t want to change ther behavor), n asymmetrc Englsh auctons. Ths set of assumptons does not nclude d erentablty of the valuatons or the use of a densty functon as n Krshna (2003) or Bruln and Izmalov (2003). Moreover, I generalze the result on Englsh auctons n Echenque and Manell (2006). The man reason why ths note s relevant for the study of Englsh auctons, s that nether d erentablty, nor the assumpton of denstes has a drect nterpretaton n terms of the economcs of auctons. The contrbuton of ths note s mportant because the Englsh aucton has had a promnent role n allocatng obects among a potental set of buyers, both n real lfe, and n economc theory. The I than Federco Echenque and leandro Manell for ther comments. Ths paper started as an attempt to weaen some of the assumptons n ther paper Echenque and Manell (2006) on comparatve statcs. y Correspondence address: Unversdad de Montevdeo, Prudenco de Pena 2440, Montevdeo, 11600, Uruguay. Emal: dubra@um.edu.uy and dubra@utdt.edu 1

2 Englsh was the rst aucton format, havng been used snce the tmes of the Roman empre, and t s the most commonly used form of aucton to sell goods nowadays (see Mcfee and McMllan (1987) and Cassady (1967) who clams that 75% or more of all auctons are Englsh). Its popularty stems from a varety of reasons: t yelds more revenue than the other common aucton format, the sealed bd, n a varety of contexts (see Mlgrom (1989) and Mlgrom and Weber, 1982); t allocates the obect e cently n a wder range of envronments; and t economzes on nformaton gatherng and bd preparaton costs (see Mlgrom, 1989). lso, relatve to other theoretcal constructons of reduced use, le the second-prce aucton, t does not requre the wnner to reveal hs true valuaton, thus avodng renegotaton between the seller and the hghest bdder. Fnally, n a varety of contexts the Englsh ucton has an ex post equlbrum, one whch remans an equlbrum even f bdders now each others valuatons or sgnals. If ths s the equlbrum actually played, then players have no ex-post regrets, or reasons to renegotate, mang the aucton format attractve. In order to clarfy my contrbuton, I wll now descrbe de assumptons of the three papers that have studed e cent ex post equlbra n asymmetrc Englsh auctons, and compare them to the ones n ths note. 2 The Model and Dscusson of ssumptons Let N = f1; 2:::; ng be the set of players. Each player observes a sgnal s 2 [0; b]. Ths sgnal s only nown to player : Sgnals are drawn accordng to some probablty measure over [0; b] n that need not posses a densty. We assume that v : [0; b] n! R s contnuous and strctly ncreasng n s and wealy ncreasng n s (vectors are wrtten n bold). For any s; let W (s) be the set of players such that v (s) v (s) for all (the set of wnners at s). Let W (s) be the cardnalty of W (s) : I now ntroduce the property that wll yeld the desred results. It states that the e ect of an ncrease n some sgnals s larger for one of the players whose sgnal ncreased than for all the rest of the players. De nton. The set of functons v sats es Own E ect Property (OEP) f for every s such that W (s) > 1 t happens that s 0 s s 0 > s only for a set P (s 0 ; s) W (s) ) ) max 2P (s 0 ;s) v s 0 max =2P (s 0 ;s) v s 0 Ths property s nspred n, and closely related to, the followng property used n Echenque and Manell (2006): v sats es the Domnant E ect Property f for any s 0 and s wth some s 0 > s ; max v s 0 v (s) > max v s 0 v (s) : :s 0 >s :s 0 s 2

3 The OEP s weaer than the DEP n ve dmensons: on the doman of applcaton, OEP apples more generally () OEP apples only when s 0 s; () t apples only when W (s) > 1; () t apples only when s 0 > s for players n W (s) ; the concluson (what s demanded of the functons) s weaer because (v) the nequalty n the concluson s wea and (v) OEP does not requre that the ncrements be larger (for players whose sgnals ncrease), but only that the nal values be larger. Of these weaenngs of the condton n Echenque and Manell, the only relevant one n terms of applcablty of the property, s the wea nequalty n the concluson. Of course, sets of functons v that satsfy the OEP and not the DEP (for reasons other than the wea nequalty) are easy to construct, but are not very relevant. 1 The strct nequalty s d erent however, snce t excludes, for example, the common value aucton. It s worth emphaszng that Echenque and Manell (2006) s not a paper about auctons, but about comparatve statcs, so they have de ned ther DEP n order to yeld comparatve statcs results n a wde varety of contexts, and not ust auctons. Hence, t s not surprsng that one can weaen ther property when usng t n a partcular settng. lso, my OEP property s smlar to Generalzed Sngle Crossng (GSC) property n Bruln and Izmalov (2003). Let 5v (s) be the gradent of v at s: They say that v sats es the GSC f for all s such that W (s) > 1 u 5 v (s) max :u >0 u 5 v (s) for all such that 2 W (s) and u = 0: If one assumes d erentablty of all v functons, then the OEP mples the GSC, but for small ncrements they are equvalent. The authors also mae two addtonal assumptons (other than the twce contnuous d erentablty of v ) that I do not use: that sgnals s are drawn from denstes, and that the valuatons are regular (the matrx of cross partal dervatves of v s non sngular). 2 The man reason why ths note s relevant for the study of Englsh auctons, s that nether d erentablty, nor non-sngularty of cross partals, nor the assumptons of denstes has a drect nterpretaton n terms of the economcs of auctons. Fnally the paper by Krshna, that generalzes pror wor by Masn, also assumes that valuatons are d erentable and that there s a densty. Moreover, hs two sngle crossng condtons exclude the common value case and other cases n whch one player s sgnal has the same e ect on hs valuaton than n other s. However, hs wor s more general than the rest of the lterature, n that he does not assume that valuatons are wealy ncreasng n other player s sgnals (although he assumes that the sum of player s sgnals s ncreasng n s sgnal). 1 Consder for example b = 1; v 1 = 10 + s 1=2 and v 2 = s 1 + s 2: The 10 precludes any equalty n valuatons, so OEP s trvally sats ed, whle the DEP s volated for ncreases n s 1: 2 The authors clam that ther GSC s necessary for the exstence of an e cent equlbrum, but n the proof they use crucally the addtonal assumpton of regularty. So t s not nown whether ther property s necessary wthout regularty. 3

4 2.1 The OEP s weaer than the Sngle Crossng property s an llustraton of the mportance of the OEP assumpton for auctons, we now show that t s weaer than the Sngle Crossng condton that I now ntroduce. Suppose there are two players; the functons v = (v 1 ; v 2 ) satsfy the Sngle Crossng Condton f at any s such that v 1 (s) = v 2 : Ths s the verson n Dasgupta and Masn (2000). In Masn (1992) the nequalty s wea, but apples to all s: Snce the Sngle Crossng s necessary for the exstence of e cent equlbra (the obect s allocated to one of the players wth the hghest valuatons for all pro les of sgnals s) n two player auctons, the OEP s also necessary. 3 Moreover, n addton to beng mpled by the SC, the OEP does not requre that the value functons v be d erentable, as does the Sngle Crossng Condton. Theorem 0. OEP s weaer than Sngle Crossng. Suppose that there are only two players, and that v 1 and v 2 are d erentable. If v sats es the SC condton or the (Masn) Sngle Crossng, t sats es the OEP. Proof. We wll assume that the OEP does not hold, and show that ths mples that the SC fals. Tae any s 00 wth W (s 00 ) > 1 (.e. v 1 (s 00 ) = v 2 (s 00 )) and an s 0 s 00 such that s 0 1 > s00 1, s0 2 = s00 2. We wll now show that v 1 (s 0 1 ; s 2) v 2 (s 0 1 ; s 2) : Let " = max " 2 [0; 1] : v 1 "s 0 + (1 ") s 00 v 2 "s 0 + (1 ") s 00 and notce that " s well de ned, snce 0 belongs to the set over whch the maxmum s taen. If " = 1; there s nothng to prove, so suppose " < 1; and de ne s = " s 0 + (1 " ) s 00 : We then have: v 1 (s) = v 2 (s) and for all s 1 such that s 0 1 > s 1 > s 1, v 1 (s 1 ; s 2 ) < v 2 (s 1 ; s 2 ) obtans. Ths mples that for all s 1 > s 1 v 1 (s 1 ; s 2 ) v 1 (s) < v 2 (s 1 ; s 2 ) v 2 (s) whch contradcts the SC condton, and therefore proves that f SC holds, so does the OEP. We wll now show that f the Masn Sngle Crossng holds, so does the OEP. s before, assume " < 1; so that v 1 s 0 1; s 2 < v2 s 0 1; s 2 (1) 3 D erent versons of the statement the sngle crossng s necessary for an e cent equlbrum can be found n Masn (1992), Bruln and Izmalov (2003) and Dasgupta and Masn (2000). 4

5 and de ne s = " s 0 + (1 " ) s 00 whch mples v 1 (s) = v 2 (s). Ths last equalty and equaton (1) contradct Masn s Sngle Crossng snce ; v 1 s 0 1; s 1 (s 1 ; s 2 2 (s 1 ; s 2 ) 8s 1 2 Z s 0 s0 1; s 1 ) 1 v 1 (s 1 ; s 2 ) v 2 s 0 1; s 2 s 1 v 2 (s 1 ; s 2 ), v 1 s 0 1; s 2 v2 s 0 1; s 2 Z s0 1 (s 1 ; s 2 2 (s 1 ; s 2 ) ds 1 ds 1 1 s 1 as was to be shown. 3 Man Results The followng Lemma bulds on Echenque and Manell. The method of proof s smlar, but: I do not use a border condton that they assume; I use OEP whch s a weaer verson of ther DEP property. For any set N; any player 2 ; and any y; let (s) = v s; y Nn : : [0; b]! R be de ned by Lemma 1. Fx any N; wth > 1; and x a pro le of types y Nn such that there exsts a y for whch: (a) for all ; 2 ; v (y) = v (y) and y < b (b) for all =2 ; and 2 ; v (y) v (y) : If v sats es the OEP, there exsts a p y > v (y) = y (for 2 ) and a wealy ncreasng functon ynn : h y ; p y! Y [y ; b] mappng prces nto types of actve players, such 2 that: () ynn p y = b for some ; and for all 2 ; p = p y mples ynn (p) = p: (2) () for all p < p y ; ynn (p) b = (b; :::; b) and the brea even condton (2) holds for all 2. () for all p p y ; and all 2 N; v ynn (p) ; y Nn p Proof. Fx any and y that satsfy condtons (a) and (b). Let b; y denote the vector y wth the th component replaced by a b: Snce s strctly ncreasng n s and y < b (by (a)) we get for all ; y < b; y : De nng a nonempty set X. For any 2 ; let = h y ; mn b; y (whch s ndependent of ) and let P be the set of nonempty subsets of : The set P has typcal elements P and P 0 ; and each s a set of prces. Let X be de ned by ( X = (P; ) : P 2 P, : P! Y y ; b ), ncreasng and ( (p)) = p, 8 2 ; 8p 2 P : 5

6 Notce that by condton (a): P = fp : p = v (y) for some 2 g 2 P s a sngleton; the functon de ned by (v (y)) = y sats es ( (p)) = p: Therefore, X s nonempty. De nng a partal order on X. De ne a partal order on X by (P 0 ; 0 ) (P; ) f and only f P 0 P and 0 (p) = (p) for all p 2 P: Showng that every chan n X has an upper bound. Tae any totally ordered set n X (a chan) f(p ; )g n X and de ne P [ P and : P! Q y ; b through (p) = (p) for any such that p 2 P : Notce that the de nton of does not depend on the spec c chosen, snce f p belongs to two d erent P and P 0; we stll get (p) = 0 (p) : I wll rst show that (P; ) 2 X; and then that (P; ) s an upper bound for f(p ; )g : It s easy to chec that s wealy ncreasng. lso, for any p 2 P; there s some for whch: p 2 P and (p) = (p) : Then, snce (P ; ) 2 X; we get showng that (P; ) 2 X: ( (p)) = p ) ( (p)) = p To see that (P; ) s an upper bound, note that for any P P and (p) = (p) for all p 2 P : Showng that the maxmal element mpled by Zorn s Lemma must have P = : Zorn s lemma then ensures that there exsts a maxmal element P M ; M n X: We now show that P M =. Suppose p 0 =2 P M ; and assume wthout loss of generalty that there s some p 2 P M such that p < p 0 (the converse case s treated symmetrcally). Let p = sup ep 2 P M : ep < p 0 and p = nf ep 2 P M : ep > p 0 : ep ep Case, p =2 P M : Consder rst the case n whch p =2 P M : We set P 0 = P M [ fp g and lettng fp n g be an ncreasng sequence n P M that converges to p ; de ne 0 on P 0 through 0 (p) = ( 0 (p) = M (p) for all p 6= p 0 0 (p ) = lm n M (p n ) Snce M s ncreasng, the lmt s well de ned. Moreover, t s easy to chec that 0 s ncreasng. For all p 2 P M ; we already now that also have that, by contnuty of ; 0 (p ) = ( 0 (p)) = : M (p) = p holds, and for p ; we lm M (p n ) = lm n n M (p n ) = lm p n = p n establshng that (P 0 ; 0 ) 2 X: Snce (P 0 ; 0 ) P M ; M by constructon, ths contradcts P M ; M beng maxmal. 6

7 Case B, p 2 P M and 9p 2 P M such that p > p 0 : Consder now the case n whch p 2 P M ; so that p < p 0 : If there s some p 2 P M such that p > p 0 ; one can follow the same steps as n Case to dscard the case n whch p =2 P M ; so assume that p 2 P M (we wll later consder the case n whch p 0 > p for all p 2 P M ). Let s = (p ), and s = (p ) and x any p wth p < p < mn (s ; (p )) p (3) ssume, wthout loss of generalty, that s > s for all (when they are equal, the sgnal of player ust becomes a xed parameter n the V functons, and thus plays no role). Let g : R! ( 1; 1) be any strctly decreasng functon wth g (0) = 0: Let (s) = (s) p and for 2 ; let 8 s >< + g h (s) = >: s + g (s) (s) (s s ) f (s) > 0 s f (s) = 0 (s s ) f (s) < 0 The functon h : Y [s ; s ]! Y [s ; s ] sats es hypothess of Brouwer, so there s a xed pont s f. We wll now show that for all ; 1. Suppose that for some ; s f > p: If snce h s f = s ; we must have sf = s (otherwse, g somethng from s f ). We then get V ynn s f = p: (4) s ; s f s f > p; we get s f > 0; and s f would be subtractng > p and snce equaton (3) ensures (s) = ( (p )) = p ; we must have s f > y for some : Let be the player n P s f ; s for whom s f = max 2P(s f ;s) s f : By applyng the OEP we see that that for player wth s f > p and s f = s ; s f max =2P(s f ;s) s f s f > p: Then, player s such that s f > p; but snce 2 P s f ; s ; we must have s f > s and h s f = s f = sf + g s f s f s wth g s f < 0 (because contradcts s f beng a xed pont. s f > p and ths means s f > 0) whch 7

8 2. If Vm ynn s f < p; for some m; then, snce h m s f = s f m; we must have s f m = s m ; because otherwse g V m ynn s f would be addng somethng strctly postve to s m. We would then have p > Vm ynn s f = Vm ynn s m ; s f m Vm ynn s m ; s m = V y Nn m (s m ; m (p )) mn (s ; (p )) > p whch s a contradcton. That s, we had chosen a small p; so that a large ncrease n the sgnal of m from s m to s m ncreases m above p: Items 1 and 2 have establshed that s f = p for all ; so that P 0 = P M [ fpg and 0 (ep) = ( 0 (ep) = M (ep) for all ep 6= p 0 (p) = s f satsfy (P 0 ; 0 ) P M ; M whch contradcts P M ; M beng maxmal. We conclude that P M = ; and that M maps = h y ; mn b; y nto Q y ; b, s ncreasng and M (p) = p, 8 2 ; 8p 2 : Case C, p 2 P M 2 P M such that p > p 0 : Recall s = (p ) and x any p wth Let g : R! ( and for 2 ; let p = (s) < p mn V b; y : (5) 1; 1) be any strctly decreasng functon wth g (0) = 0: Let 8 >< h (s) = >: s + g (s) (s s ) f (s) > 0 s f (s) = 0 s + g (s) (b s ) f (s) < 0 The functon h has a xed pont s f, so we wll show that for all ; s f = p: 1. Suppose that for some ; s f > p; so that s f = s. We then get (s) = (s) p s ; s f > p and snce (s) < p; we must have s > s for some : Let be the player n P (s ; s) for whom s f = max 2P(s f ;s) s f : By applyng the OEP we see that that for player wth s f > p and s f = s ; s f max =2P(s f ;s) s f s f > p: Then, player s such that s f > p; but snce 2 P s f ; s ; we must have s f > s and h s f = s f = sf + g s f s f s wth g s f < 0 whch contradcts s f beng a xed pont. 8

9 2. If s f < p; for some ; then, snce h s f = s f ; we must have sf the choce of p n equaton (5), we get p > whch s a contradcton. s f = b; s f Items 1 and 2 have establshed for all ; so that P 0 = P M [ fpg and ( 0 0 (ep) = M (ep) for all ep 6= p (ep) = 0 (p) = s f = b; but then, usng b; y mn V y Nn b; y p satsfy (P 0 ; 0 ) P M ; M whch contradcts P M ; M beng maxmal. We conclude that P M = ; and that M maps = h y ; mn b; y nto Q y ; b, s ncreasng and M (p) = p, 8 2 ; 8p 2 : So far we have establshed that for all p n pro le of sgnals ynn (p) s f such that h y ; mn b; y there exsts of a ynn (p) = s f = p for all ; for all p mn V b; y, and y Nn s ncreasng. Snce y and are xed throughout the proof, we wll let (p) stand for ynn (p) : Let p 1 = mn V b; y and x s 1 = p 1 : If s 1 = b for some ; the proof s complete by lettng p y = p 1 snce for all p < p 1 we have that (p) b; for f (p) was equal to b; we would get the followng contradcton p = V ( (p)) mn V ( (p)) mn V b; y = p1 > p: So assume s 1 < b for all : Then, we have that p 1 = mn V b; y = V p 1 = V s 1 and s 1 < b mply that p1 < mn V b; s 1 p 2 : Fx any p 1 < p p 2 : We can now repeat exactly the same steps as we have done so far (wth s 1 n place of y ) and show that n the doman h y ; mn V b; s 1 one has an ncreasng functon () such that V ( (p)) = p for all : Fx any s 2 = p 2 ; and notce agan that f p 2 = b for some ; the proof s complete by lettng p y = p 2 : Contnung n ths fashon, we get an ncreasng sequence of s t and p t wth the propertes that for all ; V In the lmt p 1 ; s 1 we obtan for all s t = p t < p t+1 = mn V b; s t : V (s 1 ) = p 1 = mn V b; s 1 9

10 and so, for some ; V (s 1 ) = p 1 = V b; s 1. Snce V s ncreasng n s ; ths means that s 1 = b; so that we can set p y = p 1 : Ths completes the proof of () and (). To establsh () set s 0 = ynn (p) ; y Nn and s = y: If s 0 = s condtons (a) and (b) yeld the desred result, so assume s 0 6= s: Note that: 2 mples p = v ynn (p) ; y ; Nn =2 mples that =2 P (s 0 ; s) so that the OEP and P (s 0 ; s) ensure p = max v 2P (s 0 ynn (p) ; y Nn max v ;s) =2P (s 0 ynn (p) ; y Nn v ynn (p) ; y Nn ;s) for all =2 as was to be shown. The prevous Lemma establshes the exstence of a functon that maps prces nto sgnals, the resultng pro le of sgnals beng the presumpton that other players wll have about a players sgnal, f he quts at a certan prce. The set s the set of actve players at a certan moment, and the pro le of sgnals y s decomposed n the set of sgnals of nactve players y Nn and the set of sgnals such that all actve players have sgnals greater than y : The next Lemma descrbes the presumpton of other players about a certan player s sgnal, when he should have qut, but he ddn t (n the sense that hs presumed sgnal s b; but he ddn t qut). The d erence wth the prevous Lemma s that we allow some elements of y B to be equal to b (whereas n Lemma 1 we had y B < b for all n B: Lemma 2. Fx any B N; wth B > 1; and x a pro le of types y NnB such that there exsts y B 6= b for whch for all 2 B, y < b mples v (y) = max 2N v (y) : If v sats es the OEP, there exsts a p B y > max v (y) and a wealy ncreasng functon ynnb : max v (y) ; p B Y y! [y ; b] 2B mappng prces nto types of actve players, such that: () ynnb p B y = b for some wth y < b and for all 2 ; p = p B y and y < b mply the brea even condton (2) holds; () for all p < p B y ; f y < b then ynnb () for all p p B y ; and all 2 N; v ynnb (p) ; y NnB p: (p) < b and the brea even condton (2) hold; Proof. Let B and y be as n the statement of ths Lemma. Consder rst the case n whch y < b for = ; 2 B; 6=. De nng = Bn f 2 B : y = bg and applyng Lemma 1 yelds the desred result. So assume there s a unque 2 B such that y < b: Let p B y = v (b; y ) and let v 1 (p; y ) be the nverse of v ; de ned by v v 1 (p; y ) ; y p: Then, t s easy to chec that ynnb ynnb (p) = de ned by ( b 2 Bn fg v 1 (p; y ) = : 10

11 sats es condtons () and (). To chec condton (), two cases must be consdered. (I) If W (y) > 1; we have that for s 0 = ynnb (p) ; y = ynnb (p) ; b; :::; b; y NnB = ynnb (p) ; y NnB and s = y, = P (s 0 ; s) the OEP mples that for all p; as was to be shown. p = v s 0 = max 2P (s 0 ;s) v s 0 max 6= v s 0 = max 6= v ynnb (p) ; y NnB (II) If W (y) = 1; we have that for p = max v (y) = v (y) ; v ynnb (p ) ; y NnB = max 2N v (y) > max 6= v (y) = max 6= v ynnb (p ) ; y NnB : (6) Suppose that contrary to what we want to show, there was some p such that for some 6= v ynnb (p) ; y NnB > p = v ynnb (p) ; y NnB : (7) Gven equatons (6) and (7), contnuty of ynnb (p) (ensured by constructon) and Bolzano s Theorem, there exsts a p such that max 6= v ynnb (p ) ; y NnB = v ynnb (p ) ; y NnB : Then, lettng s 0 = ynnb (p) ; y and s = ynnb (p ) ; y the OEP mples v s 0 max v s 0 v s 0, v ynnb (p) ; y v ynnb (p) ; y NnB 6= whch contradcts (7), and therefore completes the proof. The next Lemma gves the connecton between one set of functons B and the set of functons when = Bn flg for some l 2 B: Ths gves the relaton between the bddng strateges n a sub-aucton wth actve players B; and the one that follows after player l has dropped out. If varous players drop out at the same prce, one only needs to apply the Lemma repeatedly at the prce of the drops (ep n the Lemma). Lemma 3. Fx any B N; wth B > 2; and x a set of types y NnB such that there exsts a y B 6= b for whch for all 2 B, y < b mples v (y) = max 2N v (y) : ssume that v sats es the OEP, and x a p B y and ynnb as n Lemma 2. Fx any l 2 B and let = Bn flg. For any ep p B y, f s B = ynnb (ep) then for z y NnB ; s l there exsts p z v s B ; y NnB = V z s (for wth y < b) and a wealy ncreasng functon z : V z s ; p Y z! [s ; b] mappng prces nto 2 types of actve players, such that: () z p z = b for some wth y < b and for all 2 ; p = p y and y < b mply the brea even condton V z ( z (p)) = p: (8) 11

12 () for all p < p z ; f y < b then z (p) < b and the brea even condton (8) holds for all 2. () for all p p y ; and all 2 N; v ynn (p) ; y Nn p: (v) for all 2 z (ep) = ynnb (ep) : Proof. Items (), () and () follow as a drect applcaton of Lemma 2. Then, tem (v) follows from the fact that for all ; z (ep) s ; and f z (ep) > s ep = V z ( z (ep)) > V z whch s a contradcton. s ; z (p) V z We now show that the functon s contnuous. we would get s = v s B ; y NnB = V ynnb s B = V ynnb ynnb (ep) = ep Lemma 4. Contnuty. For every and y Nn satsfyng the condtons of Lemma 2, the functon ynn s contnuous. Proof. Suppose that s dscontnuous at p. It must be ether not contnuous from the rght, or from the left, so assume wthout loss of generalty that t s dscontnuous from the left: there s an " such that for all there s some p wth p p < but (p ) (p) " (we have used non decreasng). Fx then 1 = 1 and p 1 < p such that p p 1 < but (p ) (p 1 ) ". Pc then, by nducton, 0 < n < p p n 1 and p p n < n but (p ) (p n ) ": We then obtan: p n! p ; p n s ncreasng, (p n ) s ncreasng and therefore has a lmt (snce ts bounded above by b) s 1 and s 1 6= (p ) ; s 1 (p ) (because (p n ) (p ) for all n). Snce for all n and for all ; V ( (p n )) = p n we obtan by contnuty of V ; p = lm p n = lm V ( (p n )) = V (lm (p n )) = V (s 1 ) : But then, s 1 6= (p ) ; s 1 (p ) mply that for some ; s 1 < (p ) and, by V strctly ncreasng n s, V (s 1 ) < V ( (p )) = p : Ths s a contradcton, and shows that s contnuous. strategy for a player s a functon that determnes a prce at whch to qut, for each realzaton of the prvate nformaton, and each hstory of who left the aucton at what prce. Formally, a strategy for bdder s a collecton of functons, one for each set of (actve) players and each pro le p Nn of prces at whch bdders n Nn qut the aucton, : [0; b] R Nn +! R + where 2 ; > 1 and s ; p Nn > max fp : 2 Nng : The value s ; p Nn s the prce at whch bdder wll drop out f players n Nn dropped at prces p Nn and nobody quts before. 12

13 s long as p < s ; p Nn he stays n the aucton; he drops out when p = s ; p Nn ; n any hstory n whch p > s ; p Nn he drops out (ths part of the strategy wll never be used). The man pont of ths paper s showng the exstence of an e cent ex-post equlbrum. The nformal descrpton of the strateges that wll be used n that equlbrum are the followng: n the empty hstory, player remans n the aucton as long as b s > 0N (p) (the pro le of sgnals 0 sats es (a) and (b) of Lemma 1, so the functon exsts); all players now ths; player drops at the lowest prce p such that s = 0N (p) ; let the prce of the rst drop be p 1 ; let be the player who drops at p 1 and at the tme of hs drop, player sgnal becomes nown, so let y = 0N p1 ; let = Nn f g and y Nn y and notce that snce 0N p 1 sats es V 0N for all ; the pro le y = 0N 0N p 1 = p p1 sats es (a) and (b) of Lemma 1, so that a functon ynn satsfyng ()-() n that Lemma exsts. Then, player 2 remans n the aucton as long as s ynn (p), and drops at the lowest p such that s = ynn (p) : the process contnues n ths fashon. The formal descrpton of the strateges ust mentoned s as follows: n a subgame n whch types y Nn are nown and actve players are ; s ; y Nn o = ynn (s ) = mn np : ynn (p) s : Notce that snce s contnuous and wealy ncreasng, s strctly ncreasng and well de ned. Theorem 1. Ex-post Equlbrum. Suppose that v sats es the OEP. Then, the pro le of strateges that n any aucton wth actve players and sgnals of nactve players y Nn calls for a o player wth sgnal s to qut at a prce ynn (s ) = mn np : ynn (p) s consttutes an ex-post equlbrum. Proof. The proof follows Krshna s Lemma 1 closely, but does not use the fact that s unque or strctly ncreasng. Consder bdder 1 and suppose that all bdders > 1 are followng the strategy. We wll now show that player 1 does not have a pro table devaton. Consder rst the case n whch followng 1 player 1 wns when actve players are and sgnals are s: ths can only happen f players n n f1g drop at the same prce, say p : We wll now show that he earns a pro t, so that no devatons are pro table: quttng before earns hm 0; and he can never change the prce he pays. Wthout loss of generalty, let = f2; 3; :::; ag : Snce all strateges are ncreasng, all bdders n can nfer the sgnals s Nn of nactve bdders from the prces at whch they dropped. lso, snce player = 2; :::; a drop at p and snn o (s ) = mn np : snn (p) s = p 13

14 we obtan s mples = snn (p ) : Moreover, s 1 > snn 1 (p ) and therefore V1 snn snn (p ) v 1 (s) = v 1 s 1 ; snn 1 (p ) ; s Nn > v 1 snn (p ) ; s Nn = V1 snn snn (p ) = p whch means that player 1 maes a pro t, as was to be shown. p 1 = p s a second alternatve, consder the case n whch 1 calls for bdder 1 to drop at some prce n some sub-aucton wth actve bdders = f1; 2; :::; ag ; when the other players qut at sgnals s Nn, and suppose that bdder 1 consders stayng longer untl he wns the obect. Suppose he stays untl wnnng and that bdders qut n the order a; a 1; a 2; :::; 2 at prces p a :::; p 2 ; so that 1 wns at a prce p 2 : We wll show that by dong ths he can t mae a pro t. For p 2, the prce at whch player 2 quts, s 2 = snnf1;2g 2 (p 2 ) so () of Lemma 2 mples that p 2 v 1 snnf1;2g 1 (p 2 ) ; s 1 : (9) Then, snce for each xed par B; s NnB the functon snnb s ncreasng and when a bdder 2 B drops out at p, we get snnb (p ) = snnfbnfgg (p ) (by (v) of Lemma 3), we obtan snnf1;2g 1 (p 2 ) snnf1;2g 1 (p 3 ) = snnf1;2;3g 1 (p 3 ) snnf1;2;3g 1 (p 4 ) = snnf1;2;3;4g 1 (p 4 ) ::: snnfnfagg (p a ) = snn 1 (p a ) snn 1 (p 1) = s 1 : (10) (the last equalty follows from the fact that player 1 was supposed to qut at p 1 ). Equatons (9) and (10) mply that p 2 v 1 (s) so that player 1 can t mae a pro t by stayng longer than what hs strategy calls for. We have already shown that t s not pro table to qut when 1 calls for stayng, and t s not pro table to stay when 1 calls for quttng. We wll now show that f n some o equlbrum path, player 1 s stll actve at prce p when he should have qut at prce p 1 < p; then quttng s a best response (n partcular, t s better than wnnng at p). Let the set of actve bdders at p be J = f1; :::; g : Then, as n equaton (10), snnj 1 (p) snnj 1 (p +1 ) = snnfj[f+1gg 1 (p +1 ) ::: snn 1 (p a ) snn 1 (p 1) = s 1 so that p v 1 snnj (p) ; s NnJ mples p v 1 s 1 ; snnj 1 (p) ; s NnJ. Ths means quttng, as hs strategy prescrbes, s optmal. The next theorem shows that the equlbrum descrbed n Theorem 1 s e cent. Theorem 2. E cency. Suppose that v sats es the OEP. Then, the pro le of strateges that n any aucton wth actve players and sgnals of nactve players y Nn calls for a player wth sgnal o s to qut at a prce ynn (s ) = mn np : ynn (p) s consttutes an e cent equlbrum. 14

15 Proof. Wthout loss of generalty, suppose that at a pro le of sgnals s the wnner s player 1 and that the last to qut s player 2 at prce p 2 : Then, we have that s 1 > snnf1;2g 1 (p 2 ) ; s 2 > snnf1;2g 2 (p 2 ) and v 1 snnf1;2g (p 2 ) ; s Nnf1;2g = v 2 snnf1;2g (p 2 ) ; s Nnf1;2g : The OEP then tells us that for P = P s; snnf1;2g (p 2 ) ; s Nnf1;2g we must have establshng e cency. 4 v 1 (s) = max 2P v (s) max =2P v (s) References [1] Bruln, O. and S. Izmalov (2003) On E cency of the Englsh ucton, mmeo. [2] Cassady, R. (1967). uctons and uctoneerng. Bereley: Unversty of Calforna Press. [3] Dasgupta, P. and E. Masn (2000) E cent uctons, Quarterly Journal of Economcs, 115(2), pp [4] Echenque, F. and. Manell (2006) Comparatve Statcs, Englsh uctons and the Stolper- Samuelson Theorem, No 1178 of 2003 of Caltech. [5] Krshna, V. (2003) symmetrc Englsh uctons, Journal of Economc Theory 112, pp [6] Masn, E. (1992) uctons and Prvatzaton, n H. Sebert (Ed.) Prvatzaton, Insttut fur Weltwrtschaften dr Unverstat Kel, Kel, pp [7] Mcfee, P. and J. McMllan (1987) uctons and Bddng, Journal of Economc Lterature 25, [8] Mlgrom, P. (1989) uctons and Bddng: Prmer, Journal of Economc Perspectves 3, [9] Mlgrom, P. and R. Weber (1982) Theory of uctons and Compettve Bddng, Econometrca 50: Vay: n the proof of your Lemma 4, don t you need to ensure that after a player has qut hs value does not surpass that of the players that reman n the aucton? 15