Efficiency of (Revenue-)Optimal Mechanisms

Transcription

1 Effcency of Revenue-Optmal Mechansms Gagan Aggarwal Google Inc. Mountan Vew, CA Gagan Goel Georga Tech Atlanta, GA Aranyak Mehta Google Inc. Mountan Vew, CA ABSTRACT We compare the expected effcency of revenue maxmzng or optmal mechansms wth that of effcency maxmzng ones. We show that the effcency of the revenue maxmzng mechansm for sellng a sngle tem wth k + log e k + 1 bdders s at least as much as the effcency of the effcency maxmzng mechansm wth k bdders, when bdder valuatons are drawn..d. from a Monotone Hazard Rate dstrbuton. Surprsngly, we also show that ths bound s tght wthn a small addtve constant of 4.7. In other words, Θlog k extra bdders suffce for the revenue maxmzng mechansm to match the effcency of the effcency maxmzng mechansm, whle olog k do not. Ths s n contrast to the result of Bulow and Klemperer [1] comparng the revenue of the two mechansms, where only one extra bdder suffces. More precsely, they show that the revenue of the effcency maxmzng mechansm wth k + 1 bdders s no less than the revenue of the revenue maxmzng mechansm wth k bdders. We extend our result for the case of sellng t dentcal tems and show that Θlog k + tθlog log k extra bdders suffce for the revenue maxmzng mechansm to match the effcency of the effcency maxmzng mechansm. In order to prove our results, we do a classfcaton of Monotone Hazard Rate MHR dstrbutons and dentfy a famly of MHR dstrbutons, such that for each class n our classfcaton, there s a member of ths famly that s pontwse lower than every dstrbuton n that class. Ths lets us prove nterestng structural theorems about dstrbutons wth Monotone Hazard Rate. Categores and Subject escrptors F.m [Theory of Computaton]: Mscellaneous General Terms esgn, Economcs, Theory work done whle the author was vstng Google Inc. Permsson to make dgtal or hard copes of all or part of ths work for personal or classroom use s granted wthout fee provded that copes are not made or dstrbuted for proft or commercal advantage and that copes bear ths notce and the full ctaton on the frst page. To copy otherwse, to republsh, to post on servers or to redstrbute to lsts, requres pror specfc permsson and/or a fee. EC 9, July 6 1, 29, Stanford, Calforna, USA. Copyrght 29 ACM /9/7...$5.. Keywords Aucton esgn, Effcency, VCG, Optmal Auctons 1. INTROUCTION Auctons are bd-based mechansms for buyng and sellng of goods. The two most common objectve functons n aucton desgn are effcency and revenue. Effcency s the sum of the surplus of both the seller and the buyer, whch represents the total socal welfare, whereas revenue s the surplus of the seller only. The effcency and revenue of auctons has been the subject of extensve study n aucton theory see, e.g., the survey [9], and the ctatons wthn. Ideally we would lke to smultaneously maxmze both the objectve functons. But these two goals cannot be acheved smultaneously. Thus, we have the celebrated Vckrey-Clarke- Groves VCG mechansm [8, 2, 3] whch s a truthful mechansm that maxmzes effcency on the one hand, and Myerson s Optmal Mechansm [5] that maxmzes revenue on the other. If one s nterested n both objectve functons, then one has to trade-off the two. The exact balance of how much weght to gve to each objectve s, perhaps, a dffcult queston n any real-world settng. Generally, prvate sellers would lke to maxmze revenue, but keepng long-term beneft of the busness n mnd, they mght want to keep socal welfare hgh as well. Smlarly, the prmary goal n allocatng publc goods s to maxmze socal welfare, but a secondary objectve mght be to rase revenue. In the lght of ths dlemma, a natural queston that arses s: how sub-optmal s the revenue of the VCG mechansm, and how sub-optmal s the effcency of Myerson s Optmal Mechansm. Bulow and Klemperer [1] gve a structural theorem whch characterzes the sub-optmalty of the revenue of the VCG mechansm. They show that for the case of sellng a sngle tem, the VCG mechansm wth one extra bdder makes at least as much revenue n expectaton as the expected revenue of Myerson s Optmal mechansm when bdder valuatons are drawn from a class of dstrbutons called regular dstrbutons. A seller, who s currently usng the VCG mechansm and wshes to ncrease her revenue, faces two choces: 1 ncrease the reserve prce closer to Myerson s reserve prce, or 2 attract more bdders by nvestng n sales and marketng. Bulow and Klemperer s theorem gves an nsght nto the trade-offs between these two choces. In ths paper, we characterze the sub-optmalty of the effcency of Myerson s Optmal mechansm. We show that, surprsngly, there exsts a class of dstrbutons wth monotone hazard rate for whch a constant number of extra bd-

2 ders does not suffce for Myerson s optmal mechansm to match the effcency of the VCG mechansm. In fact, we show that one needs at least ωlog k extra bdders, where k s the number of bdders partcpatng n the VCG mechansm. We match ths lower bound by showng that, for all monotone hazard rate dstrbutons, Olog k extra bdders always suffce our upper and lower bounds are tght up to a small addtve constant. Ths contradcts the followng ntuton: snce the effcency of Myerson s Optmal mechansm gets closer to the VCG mechansm as the number of bdders k ncreases, we would expect the number of extra bdders needed would go down as k ncreases. Another way of nterpretng our upper bound s that wth Olog k extra bdders, Myerson s Optmal mechansm smultaneously maxmzes both revenue and effcency. We extend our upper bound result to the case of sellng multple dentcal tems as well. In order to prove the above results, we do a classfcaton of Monotone Hazard Rate MHR dstrbutons and dentfy a famly of MHR dstrbutons, such that for each class n our classfcaton, there s a member of ths famly that s pontwse lower than every dstrbuton n that class. Ths enables us to prove certan mportant structural propertes of dstrbutons wth Monotone Hazard Rate, whch helps us prove our man theorems. 1.1 Model Our model conssts of a sngle seller and k buyers bdders. We wll consder the case of sellng a sngle tem, and also the case of sellng t dentcal tems when each bdder has unt demand. The prvate values v [k] of the bdders are drawn ndependently from a common dstrbuton. Here v represents the value of the bdder for one unt of the tem. We wll use f and F to denote the probablty densty functon and cumulatve dstrbuton functon of the dstrbuton respectvely. The hazard rate of a dstrbuton s gven by h x := f x/1 F x. A dstrbuton s sad to have a monotone hazard rate MHR f h. s a non-decreasng functon of x. For the most part, we wll assume as s common n the economcs lterature that the gven dstrbuton has a monotone hazard rate. Many common famles of dstrbutons such as the Unform and the Exponental famles have MHR. We wll assume that the support of les between [,. Also, for the ease of presentng man deas, we wll assume that F s a contnuous functon even though the results hold n non-contnuous case as well. We wll restrct our attenton to truthful auctons, those n whch bdders have no ncentve to msreport ther true valuaton. Thus we can assume that the bdders bd ther true prvate values,.e., the bd vector b [k] s same as the value vector v [k]. From here onwards, we wll use the terms bd and value nterchangeably. For a gven mechansm, ts effcency on a gven nput s defned as the sum of the valuatons of the bdders who get the good, whle ts revenue s defned as sum of the payments to the seller. Snce the prvate values of bdders are not arbtrary but rather drawn from a dstrbuton, we wll be nterested n the values of effcency and revenue n expectaton. We wll use EffM to denote the expected effcency of a mechansm M. 1.2 Optmal Auctons: Vckrey and Myerson We wll use Emak to denote the effcency maxmzng VCG aucton wth k bdders wll always be clear from context. It assgns the tem to the hghest bdder and charges t the second hghest bd. In the case of sellng t dentcal tems, Emak allocates the tems to the t hghest bdders, chargng each of them the t + 1th hghest bd. Smlarly, we wll use Rmak to denote revenue maxmzng aucton Myerson s aucton wth k bdders. Myerson [5] defned a noton of vrtual valuaton ψ of a bdder, where ψ := v 1 h v Myerson showed that the revenue maxmzng truthful aucton s the one whch maxmzes the vrtual effcency sum of the vrtual valuatons of the aucton wnners. Thus, n the sngle tem case, t assgns the tem to the bdder wth the hghest non-negatve vrtual value and does not sell the tem f all vrtual values are negatve. If a dstrbuton satsfes the regularty condton, defned as ψx beng a non-decreasng functon, then the above condton s equvalent to assgnng tem to the hghest bdder as long hs vrtual value s non-negatve note that dstrbutons wth MHR always satsfy the regularty condton. Ths cutoff value at whch ψx = s called the reserve prce r. efnton 1 Reserve prce. Reserve prce of dstrbuton s defned as: r := x, s.t. h x = 1/x We drop the subscrpt, f s clear from context. Thus n the sngle tem case wth regular dstrbutons, Rmak assgns the tem to the hghest bdder, as long as ts bd s no smaller than the reserve prce, and charges t the maxmum of the reserve prce and the second hghest bd t does not sell the tem f all bds are below the reserve prce. In the case of sellng t dentcal tems, Rmak fnds the k k bdders whose bds are above the reserve prce, allocates one tem each to the hghest mn{t, k } bdders and charges each one the maxmum of the reserve prce and the t+1th hghest bd. 1.3 Related Work Bulow and Klemperer [1] characterzed the revenue suboptmalty of Ema. They showed that Emak + 1 wth one extra bdder has at least as much expected revenue as Rmak. Ther result can be nterpreted n a b-crtera sense; VCG auctons wth one extra bdder smultaneously maxmze both revenue and effcency. For the case of t dentcal tems, they show that t addtonal bdders are needed for the result to hold. In [7], Roughgarden and Sundararajan gave the approxmaton factor of the optmal revenue that s obtaned by Emak. They show that, for t dentcal tems and k bdders wth unt demand, the revenue of Emak s at least 1 t/k tmes the revenue of Rmak. Neeman [6] also studed the percentage of revenue whch Emak makes compared to Rmak n the sngle tem case. [6] used a numercal analyss approach and assumed that the dstrbuton s any general dstrbuton not restrcted to regular or MHR as n [7] but wth a bounded support. In another related work lookng at smultaneously optmzaton of both revenue and effcency, Lkhodedov and

3 Sandholm [4] gave a mechansm whch maxmzes effcency, gven a lower bound constrant on the total revenue. 1.4 Our Results We study the number of extra bdders requred for Myerson s revenue-optmal mechansm to acheve at least as much effcency as the effcency maxmzng VCG mechansm. Let = 1 1/e. For bdder valuatons drawn from a dstrbuton wth Monotone Hazard Rate, we prove the followng: Sngle tem Upper Bound Theorem 8: In the sngle tem case, m log 1 2k + 2 extra bdders suffce to for the revenue maxmzng mechansm to acheve at least as much effcency as the effcency maxmzng mechansm wth k bdders for any k. Sngle tem Lower Bound Theorem 9: In the sngle tem case, we demonstrate a dstrbuton havng monotone hazard rate, such that for any k, f the effcencyoptmal mechansm has k bdders, and the revenueoptmal mechansm has log 1/ k extra bdders, then the effcency of the latter s strctly less than the effcency of the former. In other words, log 1/ k extra bdders do not suffce. Mult tem Upper Bound Theorem 11: In the case of sellng t dentcal tems, wth bdders havng untdemand: m + s extra bdders suffce for the revenueoptmal mechansm to acheve at least as much effcency as the effcency-optmal mechansm wth k bdders, where m = log 1 2k + 2 and s t ɛt log m, for every ɛ > and large enough k. Thus, approxmately, log k + t log log k extra bdders suffce. We also show that f both auctons have the same number of bdders k, then the rato of the effcency of the revenue-optmal aucton to the optmal effcency s at least 1 k. The proof s easy and we omt t n ths extended abstract 1. We also prove Secton 5 that our upper bound result does not hold for regular dstrbutons we show that for every k, m, there s a regular dstrbuton for whch the effcency of the revenue-optmal mechansm wth k + m bdders s strctly lower than the effcency of the effcency-optmal mechansm wth k bdders. Outlne of the paper: Secton 2 descrbes some basc setup whch s common to the rest of the paper. Secton 3 descrbes our results for the sngle tem case. We start wth the defnton of two quanttes Gan and Loss, and analyze the expresson Gan Loss n Secton 3.1. Secton 3.2 and 3.3 prove our upper and lower bound results respectvely. In Secton 4, we extend out upper bound result to the case of sellng t ndentcal tems to bdders wth unt-demand. Secton 5 deals wth the case of regular dstrbutons. 1 In fact we can also prove usng the same technques that the rato of the revenue of VCG to that of Myerson s Optmal Aucton s 1 k 1, whch mproves on the polynomal bound provded n [7]. 2. BASIC SETUP To compare the two mechansms Rmak+m and Emak, whch have a dfferent number of bdders, we wll thnk of the process of drawng ther bdder values as drawng k + m bdder values b 1, b 2,..b k+m ndependently from the dstrbuton the frst k bdders b 1, b 2,..b k partcpate n both Rmak + m and Emak, whereas the last m bdders partcpate only n Rmak + m. The followng lemma wll prove useful n the subsequent sectons. Lemma 1. For any MHR dstrbuton : F r 1 1/e, where r s the reserve prce for the dstrbuton. Proof. Let the hazard rate of dstrbuton be hx. By defnton of hazard rate, we have F x = 1 e x htdt. By defnton of reserve prce, hr = 1/r. Also snce has MHR, we get hx 1/r, x r. Thus, x r : 1 e x htdt 1 e x 1/rdt F r 1 e 1/rdt F r 1 e 1 settng x = r 3. SELLING ONE ITEM We begn by notng that when sellng a sngle tem, f the value of any of the frst k bdders s greater than or equal to the reserve prce r, then Rmak + m acheves at least as much effcency as Emak. The more challengng case for the upper bound occurs when the value drawn by all the frst k bdders s less than the reserve prce. In ths case, Emak acheves an effcency equal to the hghest value among the values of the frst k bdders, whereas the contrbuton of the frst k bdders to Rmak + m s zero as all of these bdders have a value less than the reserve prce cutoff. In other words, condtoned on the event that frst k bdders value s less than the reserve r xf prce r, the expected effcency of Emak s k, F k r where F k k and f are the c.d.f and p.d.f of the maxmum of k numbers pcked..d. from we wll drop the subscrpt whenever s clear from the context. Note that F k x = F k x, and f k x = kf k 1 xfx. Also, condtoned on ths event, the expected contrbuton of the frst k bdders b 1, b 2,..., b k to the effcency of Rmak + l s zero. We defne: Loss = k xf F k r 1 To make up for ths lost effcency, the revenue maxmzng mechansm has m extra bdders. The contrbuton to the expected effcency of Rmak + m from these m extra bdders b k+1, b k+2,..b k+m, condtoned on the event that the value of bdders b 1, b 2,..b k s less than r, s at least 1 F m r r. Ths s because of the fact that all the draws are ndependent and the probablty that at least one of the m extra bdders wll have a value hgher than the reserve prce r s 1 F m r. We defne: Gan = 1 F m r r 2

4 If for all dstrbutons, t s true that Gan Loss for some k, m, then we know that the effcency of Rmak+ m s at least as much as the effcency of Emak. Moreover, f we can demonstrate a dstrbuton s.t. the expected contrbuton to the effcency of Rmak+m from these extra m bdders b k+1, b k+2,..b k+m, s strctly less than Gan,.e., f we show that Gan Loss <, for some k, l, and that there s no addtonal gan to Rmak + m n the case when one of the frst k bdders has value equal to or greater than r, then we would have shown that m extra bdders does not suffce. Therefore, we examne ths key expresson Gan Loss n the next secton. We wll omt the subscrpt whenever t s clear from the context. 3.1 The expresson Gan Loss Recall the expressons Gan and Loss as defned n Equatons 2 and 1. For the purpose of gettng a better handle on the expresson Gan Loss, we wll partton the set of all MHR dstrbutons nto dfferent classes, accordng to ther optmal reserve prce r and the value of the cdf at the optmal reserve prce, as follows. Let r, φ be the set of MHR dstrbutons wth a fxed reserve prce r and F r = φ. Note that, by Lemma 1, r, φ s non-empty only f φ [, 1 1/e]. Also note that all these dstrbutons have the same value for the expresson Gan and dffer only n the numerator of the expresson Loss. Next, we fnd a dstrbuton n r, φ whch maxmzes the numerator of Loss. efnton 2 strbuton G φ,r. Let r, φ [, 1 1/e], and let tφ, r = r1 + ln1 φ [, r]. Then, x < tφ, r, 1 e r 1 x tφ,r tφ, r x r, G φ,r x = φ + 1 φ.x r r x r + ɛ, ɛ 1 x r + ɛ. Here ɛ s any postve number. It can be verfed that the above dstrbuton has the followng propertes: G φ,r s MHR. The optmal reserve prce for G φ,r s r. G φ,r r = φ. Therefore, G φ,r r, φ. Next, we prove that G φ,r s pont-wse no larger than every other functon n r, φ Lemma 2. For every dstrbuton r, φ, and any y [, r]: F y G φ,r y Proof. The cdf of a dstrbuton wth hazard rate h. can be wrtten as F y = 1 e hzdz. Now, by defnton, r 1 h =, mplyng hr = 1/r. r Snce h. s an ncreasng functon, therefore h z 1/r for all z r. Also note that h Gφ,r z = for z < tφ, r, and h Gφ,r z = 1/r for z [tφ, r, r]. Thus, for any y [tφ, r, r] we have y h zdz h zdz y h Gφ,r zdz h Gφ,r zdz h zdz h Gφ,r zdz Snce F r = G φ,r r = φ, and recallng from the defnton of hazard rate that F x = 1 e x h tdt for all dstrbutons we have hzdz = zdz. Therefore, hg φ,r h zdz h Gφ,r zdz 1 e h zdz 1 e h G φ,r zdz Thus, for any y [tφ, r, r], F y G φ,r y. Snce G φ,r y = for y [, tφ, r, we have proved the lemma. Next we prove that the numerator of Loss among r, φ s maxmzed at G φ,r. Lemma 3. For every dstrbuton r, φ: xf k x[g k φ,r x] dx Proof. For any r, φ, we have xf k k = rf r F k ntegratng by parts rg k φ,r r = x[g k φ,r x] dx G k φ,r The nequalty follows from Lemma 2 and the fact that r = rgk φ,r r = rφk. rf k Now that we know that, among all dstrbutons n r, φ, G φ,r maxmzes the expresson Loss, we can calculate the maxmum value of Loss n terms of r and φ. Lemma 6 examnes the numerator of Loss at G φ,r. Fact 4. For a fxed λ, t, and k: 1 e λx t k dx = 1 e λx t x 1 k λ =1 Proof. Let g k = 1 e λx t k dx. We have g k = 1 e λx t k 1 dx e λx t 1 e λx t k 1 dx g k = g k 1 1 λ.k 1 e λx t k g k = g 1 λ g k = x 1 λ k 1 e λx t =1 k 1 e λx t =1

5 Corollary 5. For a fxed r, t and k, x t 1 e r t k dx = r t r k φ =1, where φ = 1 e r t r. Lemma 6. = r φ,r x] dx φ k + ln1 φ + k =1 x[gk Proof. Integratng by parts, we have x[g k φ,r x] dx = rφ k = rφ k = rφ k [ tφ,r G k φ,r φ 1 e 1 r x tφ,r k dx r tφ, r r by Corollary 5 k =1 ] φ = rφ k + r1 + ln1 φ r + r by defnton of tφ, r k = r φ k φ + ln1 φ + Let r, φ. Then, for any gven m, from equatons 2 and 1 and Lemma 6, we have: Gan Loss Gan Gφ,r Loss Gφ,r r = 1 φ m r r = =1 φ k + ln1 φ + k =1 φ k φ k+m ln1 φ k =1 φ k =1 φ φ φ k Upper bound on the number of extra bdders requred To prove an upper bound for dstrbutons n r, φ, we need to fnd values of m as a functon of k for whch the expresson n Equaton 3 s non-negatve. To do ths, we defne a un-varate polynomal qx := x k+m + ln1 x + Snce φ [, 1 1/e] by Lemma 1, t suffces to fnd m for whch qx for all x [, 1 1/e]. Snce q =, f we fnd values of m for whch q x k =1 x, x [, 1 1/e], then qx x [, 1 1/e]. Now, q x = k + mx k+m 1 1 k 1 x + =1 x 1 = k + mx k+m x + 1 xk 1 x = k + mx k+m 1 xk 1 x = xk 1 x k + mxm 1 1 x 1 Snce xk, 1 x q x ff x m 1 1 x 1. It s easy k+m to see that, for x [, 1 1/e], x m 1 1 x s maxmzed at x = 1 1/e for m > 2. Let c = For ths choce of m, e, and m = logc2k + 2. x m 1 1 x 1 k + m 1 c logc2k+1 1 e 1 k + log c2k + 2 whch can be seen to be non-postve for all k. Ths proves the followng lemma. Lemma 7. Gven any and k, let m = log e 2k Then, log k log e Gan Loss Thus, we have establshed the followng theorem. Theorem 8 One-tem Upper Bound. For the case of sellng one tem, we have EffRmak + m EffEmak for any k and m log 1 2k + 2, where = 1 1/e. Thus, Olog k extra bdders suffce for the revenue maxmzng mechansm to acheve at least as much effcency as the effcency maxmzng mechansm wth k bdders. 3.3 Lower bound on the number of extra bdders requred In ths secton, we wll prove a lower bound on the number of extra bdders m needed for the revenue maxmzng mechansm to acheve at least as much effcency as the effcency maxmzng mechansm wth k bdders. To prove such a lower bound, t suffces to specfy a dstrbuton s.t. the contrbuton of the m extra bdders to expected effcency of Rmak + m s no more than Gan and show that Gan Loss < for ths choce of k and m. Consder the dstrbuton G φ,r for any choce of r, φ see defnton 2, and arbtrarly small ɛ. We frst show that for the dstrbuton specfed by G φ,r x, the contrbuton of m extra bdders to the effcency of Rmak + m s arbtrarly close to Gan Gφ,r. To see ths, note that the maxmum value possble under dstrbuton G φ,r x s r + ɛ; thus, the maxmum possble effcency for any draw of bdder values s r + ɛ. When all the frst k bdders draw a value below r, the m extra bdders contrbute a maxmum of r + ɛ to the effcency of Rmak+m wth probablty equal to 1 G m φ,r. Thus, the total contrbuton of the m extra bdders to the effcency of Rmak + m s arbtrarly close to r1 G m φ,r, whch s the same as Gan. Let = 1 1/e. Let mk = log 1/ k Next we ll show that Gan Gφ,r Loss Gφ,r < for some

6 choce of φ, any k and for all m mk. To do ths, recall the polynomal qx = x k+m + ln1 x + k x =1. By Equaton 3, we know that Gan Gφ,r Loss Gφ,r = rqφ. φ k Therefore, we just need to show that qx > for the above choce of m and some x we wll choose φ to be that x. q = k+mk + ln1 + = k+mk = k+mk =1 =k+1 > k+mk 1 k + 1 = k+mk + k =1 =k+1 k =1 k+1 k + 11 where the last nequalty follows from the choce of m. Ths proves the followng theorem. Theorem 9 One-tem Lower Bound. Let = 1 1/e and let mk = log 1/ k Then, f bdders are drawn..d. from the dstrbuton descrbed by G,r, EffRmak + m < EffEmak for any r, any k and any m mk. In other words, mk extra bdders do not suffce for the revenue maxmzng mechansm to acheve as much effcency as the effcency maxmzng mechansm wth k bdders. Note that mk log k+1 log e 2.2, whch dffers from the upper bound only by a small addtve constant, namely EXTENSION TO MULTIPLE ITEMS In ths secton we consder the case of sellng t dentcal tems. As before, the effcency maxmzng mechansm gets k bdders, and we want to fnd the smallest number of extra bdders that suffce for the revenue maxmzng aucton to make as much effcency. As seen n the prevous secton, m = mk = log 1 2k + 2 suffce when sellng only one tem. So clearly, mt extra bdders would suffce n the case of t dentcal tems. However, as we show below, t s possble to prove a much tghter bound. The queston s to fnd the smallest number s = sk such that the effcency of the revenue maxmzng mechansm wth k + m + s bdders, Rmak + m + s, s at least as much as the effcency of the effcency maxmzng mechansm on k bdders, Emak, where m = mk. As before, to compare the two mechansms wth dfferent number of bdders, we wll thnk of the process of drawng the values of k + m + s bdders b 1, b 2,..b k+m+s ndependently from the gven dstrbuton ; the frst k bdders b 1, b 2,..b k partcpate n both Rmak + m + s and Emak, whereas the last m + s bdders partcpate only n Rmak + m + s. We partton the space of draws of bdder values nto k +1 parts: For =,..., k, the th part, Ω, conssts of those draws n whch exactly of the frst k bdders have values greater than r. We now focus on one of these parts, say, the th part Ω, and try to determne the expected effcency of Emak and Rmak + m + s over ths restrcted space. Wlog, we may assume that < t f t, then the effcency of Rmak s already equal to that of Emak. We defne t = t and k = k. Let b max1, b max2,, b maxk be the bds of frst k bdders n the decreasng order of ther value. Also let Γ denote the sum of the hghest of these bds, condtoned on beng n Ω. The expected effcency of Emak condtoned on beng n Ω, EffEma k, equals Now, = E[ b maxj Ω ] + E[ j=1 Γ + t E[b max+1 Ω ] E[b max+1 Ω ] = k j=+1 b maxj Ω ] E[maxb +1, b +2,, b k b 1..b r & b +1..b k < r] because of the symmetry = E[maxb +1, b +2,, b k b +1..b k < r] = snce b js are ndependent xf k Therefore, F k r EffEma k Γ + t k xf F k r Condtoned on beng n Ω, the contrbuton to the expected effcency of Rmak + m + s from the frst k bdders s Γ. The contrbuton to the expected effcency of Rmak + m + s from the remanng m + s bdders depends on how many of the extra bdders have value above r. If j of these bdders have value more than r, then the contrbuton s at least mnj, t r. We have the followng lower bound on ths contrbuton note that the contrbuton from the extra m+s bdders s ndependent of the frst k bdders, and hence of Ω : Lemma 1. For any ɛ >, and for large enough m, f s t ɛt log m, then the total contrbuton to the effcency of Rmak + m + s from the remanng m + s bdders s at least rt 1 F m r. Proof. Recall that φ = F r. Let a j = m+s j φ j. The contrbuton s: But, a jt j t 1 r j= t 1 ja j + 1 j=1 t 1 = r t a jt j j= a jt m + sj φ m φ s j 1 φ j t j j! φ m m + s j 1 1/e s t j φ m 4 φ m+s j 1

7 for s t ɛt log m + log t and any ɛ > and large enough m. Thus the contrbuton s at least rt 1 φ m. Let m = log e/ 2k + 2. Then, usng Lemma 1 and the dscusson above, the effcency of Rma condtoned on beng n Ω s: EffRma k + m + s Γ + t 1 F r m r Γ + t k xf F k r by Lemma 7, and snce k k EffEma k by equaton 4 Thus we have proved the followng theorem we have not tred to optmze the constants or how large k has to be for ths result to hold. Theorem 11 Mult-tem Upper Bound. In the case of sellng t dentcal tems, we have EffRmak + m + s EffEmak for m log 1 2k +2 and s t ɛt log m, for every ɛ > and large enough k. Thus, approxmately, log k + t log log k extra bdders suffce. 5. THE CASE OF REGULAR ISTRIBUTION In ths secton, we wll show that for any gven k and m, there exsts a regular dstrbuton s.t. expected effcency of Rmak+m s less than the expected effcency of Emak. To recall, a dstrbuton s regular f and only f the functon ψx := x 1 h s non decreasng n x. Now x consder the followng dstrbuton: P ɛ,rx := { 1 ɛ x+ɛ x < r, 1 x r One can easly verfy that the above dstrbuton s regular for every choce of ɛ, r >, by evaluatng ts ψ functon. Moreover the reserve prce of ths dstrbuton equals r. Now, usng smlar arguments as used n the prevous sectons, we can show that the contrbuton of the extra m bdders to Eff[Rmak + m] s r1 1 ɛ r+ɛ k r1 1 ɛ r+ɛ m. ɛ r+ɛ m 1 Also, the extra contrbuton of the frst k bdders to Eff[Emak] over Eff[Rmak + m] when all of the frst k bdders have a value below reserve prce r s x[p k ɛ,rx] dx. Now, as we decrease the value ɛ, the term r1 1 ɛ r+ɛ m decreases and x[p k ɛ,rx] dx ncreases for a fxed k and m. Moreover, one can show that there exsts a small enough ɛ := ɛ such that x[p k ɛ,rx] dx s more than r1 1 ɛ m. r+ɛ Thus, for the dstrbuton P ɛ,r, the loss n Eff[Rmak+m] because of the reserve prce s more than the gan from extra m bdders. Acknowledgment: We thank Hal Varan for useful comments and dscussons. 6. REFERENCES [1] J. Bulow and P. Klemperer. Auctons Versus Negotatons. AMERICAN ECONOMIC REVIEW, 86:18 194, [2] E. Clarke. Multpart prcng of publc goods. Publc Choce, 111:17 33, [3] T. Groves. Incentves n teams. Econometrca, 414: , [4] A. Lkhodedov and T. Sandholm. Aucton mechansm for optmally tradng off revenue and effcency. In Proceedngs of the 4th ACM conference on Electronc commerce, pages ACM New York, NY, USA, 23. [5] R. Myerson. Optmal Aucton esgn. MATH. OPER. RES., 61:58 73, [6] Z. Neeman. The effectveness of Englsh auctons. Games and Economc Behavor, 432: , 23. [7] T. Roughgarden and M. Sundararajan. Is effcency expensve. In Thrd Workshop on Sponsored Search Auctons, 27. [8] W. Vckrey. Counterspeculaton, auctons, and compettve sealed tenders. Journal of Fnance, 161:8 37, [9] R. Zhan. Optmalty and Effcency n Auctons esgn: A Survey.