Online Appendix to Position Auctions with Budget-Constraints: Implications for Advertisers and Publishers

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1 Onlne Appendx to Poston Auctons wth Budget-Constants: Implcatons fo Advetses and Publshes Lst of Contents A. Poofs of Lemmas and Popostons B. Suppotng Poofs n the Equlbum Devaton B.1. Equlbum wth Low Resevaton Pce ( < mn { B, B }) B.2. Equlbum wth Hgh Resevaton Pce ( mn { B, B }) C. Numecal Smulatons wth Moe than Two Stategc Advetses D. Poston Auctons wth Endogenous Budgets E. Poston Auctons wth Qualty Scoes

2 A. Poofs of Lemmas and Popostons We fst outlne the logc behnd ou devaton of the UNE and then pesent a set of suppotng lemmas (Lemmas A1-A5), whch lead to the poofs fo the lemmas and popostons n secton 3. The detaled poofs fo Lemmas A1-A5 can be found n secton B.1 below. To help follow the poofs, we pesent a summay of notatons n Table A1 and a chaactezaton of advetses payoff functons n Table A2. We summaze how to deve the UNE n the whole budget space. As shown n Table A2, advetses pofts can take dffeent functonal foms based on the postons and duatons of beng anked. On the bass of these dffeng functonal foms, we ntally patton the budget space so that each advetse s poft takes the same fom n each egon. To do ths, we fst deve an advetse s wllngness-to-pay (pe clck) ({w, w, w, w } n Table A1), defned as the CPC such that the focal advetse s ndffeent between ntally stayng at the fst and the second postons. We then patton the budget space nto smalle egons based on the coespondng budget thesholds ({B, B, B, B } n Table A1), defned as the smallest budget that s nexhaustble when payng the wllngness-to-pay (pe clck) at the fst poston. Unde such a patton, an advetse s poft takes the same functonal fom wthn each budget egon, whch allows us to chaacteze the UNE case by case. We solve fo the UNE n each egon n the followng steps: (1) chaacteze the best-esponse coespondences of advetses; (2) elmnate weakly domnated bds; (3) fnd the UNE as the ntesecton of the best-esponse coespondences. Fnally, gven the UNE n these fne budget egons, we goup the esults and classfy them nto dffeent types, Type-I, Type-II, Type-III, o BI, based on equlbum outcomes. Ths leads to the patton of budget space n Fgues 1(a) and 1(b) n the pape. 1

3 Notaton π B Π ρ w = (1 1 ) π + ; B = w w = π B π +B ; B = w w = ( 1)(π )B 2 ; B = w w = π B +(1 1 )(π )B B +(π ) w = ( 1)B ( B ) ; B = w Explanaton Advetse s value-pe-clck Advetse s budget Advetse s poft Clck ato between the 1 st and the 2 nd poston Resevaton value CPC pad by the thd hghest bdde s wllngness-to-pay when b B & b B budget theshold unde w. s wllngness-to-pay when B b < B budget theshold unde w. s wllngness-to-pay when B b < B budget theshold unde w. s wllngness-to-pay when B b & B b ; budget theshold unde w. A bd b s.t. s budget s depleted by exactly at the end of the day and s budget s depleted soone. B A budget B s.t. w = w. Notes. The subscpt o of each notaton ndcates whethe the functon s - o -specfc. Table A1. Summay of Notatons. Scenao Condtons on Budgets Advetses Pofts Both ae non-depleted (t, t = 1) Only B s depleted (t < 1 = t ) Only B s depleted (t < 1 = t ) Both ae depleted and B s depleted soone (t t 1) Both ae depleted and B s depleted soone (t < t 1) B > b B > B < b B B b + (1 B b ) B b B + (1 B ) B < B b, B B b B b + B B b 1 B > B b, B < B + B B b 1 Π = (π )B b Π = (π b )B Π = B (π b ) ; ; Π = (π b ) Π = π Π = (π b )B b ; + (1 B b ) (π ) + (1 B ) (π ) Π = (π )B Π = (π b )B b Π = (π )B + B Bb (π ) Π = (π )B Table A2. Chaactezaton of Advetses Pofts When b > b 2

4 We fst consde a useful benchmak n whch advetses have unlmted budgets. Clam 1. Let B = B = +, then the UNE s b k = w k, whee k =,. Poof of Clam 1. We pove that any bd b w s weakly domnated by w. Fo a bd b > w, advetse obtans the same poft between b and w when ethe b (0, w ] [b, + ). When b (w, b ), s poft of bddng w s π, whch s geate than (π b ), the poft of bddng b. To see ths, we have (π ) > (π b ) s equvalent to b > w. Smlaly, we can pove any bd b < w s also weakly domnated by w. The same poof can be appled to advetse. Next we pesent Lemmas A1-A5, whch ae used to chaacteze the UNE when < mn { B, B }. Lemma A1. The best-esponse coespondence of k s bd, denoted by b k (b k ), s as follows: 1. When b k B k 2. When B k 3. When B k 4. When B k & b k B k, b k(b k ) = { {x x > b k} f b k < k {x x b k } f b k w ; k b k < B k, b k(b k ) = { {x x > b k} f b k < w k {x x b k } f b k w k ; b k < B k, b k(b k ) = { {x x > b k} f b k < w k, k b k ε f b k w k, k ; b k & B k b k, b k (b k ) = { {x x > b k} f b k < w k, k b k ε f b k w k, k, whee k =,, and ε > 0 epesents a penny the smallest unt of payment. We specfy the best-esponse coespondences fo advetses based on the condtons of budget exhaustblty n Lemma A1. Ths gves se to fou dstnct values of an advetse s wllngnessto-pay n dffeent stuatons. Note that the wllngness-to-pay n case 1 n Lemma A1 s the same as the UNE when advetses have unlmted budgets, whch suggests that budgets mght not play a stategc ole n bds unde cetan condtons. Next we nvestgate the odnal elatonshp of the budget thesholds assocated wth these wllngness-to-pay n ode to fnd the budget egon n whch advetses best-esponse coespondences ae unchanged. 3

5 Lemma A2. Fou sets of nequalty equvalences hold among fou budget thesholds. 1. B k B k B k B k B k B k, 2. B k B k, k B k B k, k B k, k B k, k, 3. B k B k B k B k, k B k, k B k, 4. B k B k B k B k, k B k, k B k, whee B k = w k, B k = w k, B k, k = w k, k, B k, k = w k, k, and k =,. Lemma A2 mples that B k always les between B k and B k, B k, k les between B k and B k, k, B k, k les between B k and B k, and B k, k les between B k and B k. These elatonshps allow us to patton the budget space nto smalle egons and smplfy ou analyss by educng the numbe of specfcatons of advetses payoffs. Lemma A3. The UNE n all possble cases s descbed n Table A3. Case 1 b, b = Case 2 b, b = When B B B B B B B [B, B ) B [B, B ) B [B, + ) { w ε, w, f w w w, w ε f w > w w ε, w w, b [w, B ] B B B B B [B, B ) B [B, B ) B [B, + ) { w ε, w, f w w w, w ε f w > w w, w ε w, w ε Case 3 B B B B B B [B, B ) B [B, + ) b, b = w, w ε w, w ε Case 4 B B b, b = b [w, B When B > B Case 5 b, b = Case 6 B B B B B (B, B ) B [B, B ) B [B, + ) { w ε, w, f w w w, w ε f w > w w, w ε b [w, B ], w, B B B B B (B, B ) B [B, B ) B [B, + ) 4

6 b, b = Case 7 { w ε, w, f w w w, w ε f w > w w ε, w w ε, w B (B, B ) B B B B B B [B, + ) b, b = w, w ε b [w, B ], w Case 8 B B b, b = b [w, B ], w Table A3. Chaactezaton of the UNE n All Possble Cases We fst deve the UNE n the half budget space whee B B and then extend the esults to the othe half. We enumeate all possble odengs fo the budget thesholds and patton the budget space accodngly. The advantage of ths patton s that n each egon, fo ethe advetse, a bd below the wllngness-to-pay s weakly domnated by a bd equal to that wllngness-to-pay (see fomal poof n secton B.1). Ths efnement helps us elmnate unntutve Nash equlba n each budget egon. Yet, the condtons to dentfy some cases n Table A3 ae stll unclea. Fo example, we need to undestand the condtons unde whch B B to sepaate case 1 fom 2. Also, we need the compason between w and w to dentfy the equlbum bd n the fst subcase of cases 1, 2, 5, and 6. Lemmas A4 and A5 pesent the esults egadng these compasons. Lemma A4. If B B, then B B. Futhemoe, f B B, then B B s equvalent to B B, whee B = π (π ) (1 1 4( 1)(π )(π ) ) π 2( 1)(π ) (π ) 2 +. Lemma A5. If B B B, then w w. Lemma A4 and A5 smplfy some ambguous condtons n Table A3. Lemma A4 suggests that we do not need to consde case 6 n Table A3 because B B wheneve B B. Lemma A5 ndcates that the UNE n the fst subcase of case 5 n Table A3 should be (b, b ) = (w, w ε). These two obsevatons help us chaacteze the UNE n the whole budget space ncludng egons 1 to 9 shown n Fgue A1(a). The exact UNE n each egon s epoted n Table A4. 5

7 Poof of Lemma 1. The mathematcal expesson fo each budget egon n Fgue A1(a) s gven below. Regon 1 coesponds to whee B [B, mn{b, B }). Regon 2 coesponds to whee B B & B [B, B ). Regon 3 coesponds to whee B < B & B B. Regon 4 coesponds to whee B [B, B ) & B [max {B, B }, B ). Regon 5 coesponds to whee B [B, B ) & B > B. Regon 6 coesponds to whee B, B B. Regon 7 coesponds to whee B B & B [B, B ). Regon 8 coesponds to whee B B & B < B. Regon 9 coesponds to whee B [B, B ). The equlbum analyss leads to the followng coespondence between budget egons n Fgue 1(a) and n Fgue A1(a): Θ I = {4,5,9 & 1 whee w > w }, Θ I = {2 &1 whee w w }, Θ II Θ L BI = {8}, Θ II = {3}, and = {6 & 7}. Statements egadng the descptons of Type-I and Type-II equlbum ae dect fom the UNE descbed n Table A4. 6

8 Regon n Fg. 1(a) Θ I Regon n Fg. A1(a) 1 st Poston Intally b b 1 (whee w w ),4 & 9 w w ε 5 w w ε Θ I 1 (whee w > w ) & 2 w ε w Θ II 8 b [w, B Θ II L Θ BI Regon n Fg. 1(b) Θ I Θ I 3 w b [w, B 6 b [w, B 7 b [w, B ] Regon n Fg. A1(b) 13 & 15 whee w w & w w 16 & 15 whee w > w & w w ] ] w w w w w ε w ε w Θ II 14 b [w, B Θ II Θ III Θ III 17 w b [w, B 13 & 15 whee w w & w < w w w 16 & 15 whee w > w & w < w w w 11 whee B < b [w, w ] 11 whee B b [w, + ) 12 b [w, + ) ] w ] ] S Θ BI 18 whee B < b [w, w ] 18 whee B b [w, + ) 19 b [w, + ) 10 o wth equal pob. Table A4. Chaactezaton of the UNE 7

9 (a) when < mn { B, B } (b) when mn { B, B } Fgue A1. Depcton of Budget Regons 8

10 Poof of Lemma 2. When mn { B, B }, we deve the UNE n a smla way as done fo Lemma 1. We epot the UNE n Table A4, whee the budget egons ae demonstated n Fgue A1(b). The detaled poofs ae avalable n secton B.2 n ths onlne Appendx. We epot the coespondence between budget egons n Fgue 1(b) and n Fgue A1(b): Θ I = {13 &15 whee w [ w, w ]}, Θ I = {16 &15 whee w (w, w ]}, Θ II Θ II = { 14}, Θ III w & w > w }, and Θ S BI = {13 & 15 whee w w & w > w }, Θ III = {17}, = {16 &15 whee w > = {10,11,12,18 & 19}. Statements egadng the descptons of a Type-III equlbum ae dect fom the UNE descbed n Table A4. To help undestand the Type-III bddng equlbum n Lemma 2, we povde a smple numecal example below. Example of a Type-III equlbum. Assume = 1, = 3, π = 3, π = 2, B = 2.5, and B = 1.5. Lemma 2 pedcts a Type-III equlbum outcome n whch b > b = ( 1)B ( B ) = We fst check whethe advetse has ncentve to devate. When bddng b, advetse stays at the fst poston untl he budget uns out at t = B = 3. Hence, he poft s B (π b 4 b ) = If advetse devates and bds below b, advetse stays at the second poston fst and takes the fst poston afte s budget s depleted. In ths devaton, advetse can do no bette than to bd one penny below b, whch bngs he less poft than at b : B b (π ) + mn {(1 B b B B b ), ( )} (π ) = 21 < b Next we dscuss advetse s bd stategy. The hghest possble poft fo advetse equals the maxmum clck volume B s B (π ) = 3 2. By bddng b = 10 9, advetse spends B tmes the hghest magnal value pe clck (π ), whch 9 = 3 b 4 tme at the second poston and the emanng 1 4 tme at the fst poston so that he total spendng s = 3 2, whch exactly equals he budget. Any bd below b esults n stctly less poft due to unused budget and theefoe s suboptmal fo. It s also unpoftable fo to bd above b : f b > b, s poft

11 s B (π b ) < B (π b ) snce b >. Fnally, we pove that any bd b > b s weakly domnated by b. Note that b and b lead to the same poft when b s ethe above b o below b. Howeve, when b les between b and b, advetse s poft of bddng b s stll 3, whch s 2 stctly hghe than the poft of bddng b because B b (π b ) < 3 2 when b > b >. We show n ths example that the second advetse s unque UNE bd should ensue he own budget depleton pecsely at the end of the peod. Ths mples that the second advetse s tme of stay at the second poston t, should satsfy t + (1 t ) = B so that the total spendng fo at the fst and the second postons s equal to he budget. Hence, n equlbum, the second advetse s tme of ascenson to the fst poston, t = B, should be equal to t, whch depends only on B and (t = t = 3 n ths example). To deplete the fst advetse s 4 budget at a cetan tme pont, the second advetse has to bd popotonally to the fst advetse s budget, as we descbed n the Type-III equlbum. b Poof of Poposton 1. Gven the UNE deved n Lemmas 1 and 2, we calculate each advetse s poft n dffeent budget egons and epot the esult n Table A5. We descbe the elatonshp between advetses pofts and budgets n two steps. Fo each advetse, say advetse, we fst show that wthn each budget egon, Π weakly nceases wth B except fo the egons 13 and 15 whee w < w (.e., Θ III ). As shown n Table A5, Π s ndependent of B n egons 3, 5, 6, 7, 8, 12, 14, 17. Π takes fou dffeent foms n the emanng egons. In egon 2 and pat of 1, 15, 16, Π = (π ) B w + (π )(1 B w ), whch nceases wth w. Snce w nceases wth B, Π nceases wth B as well. In egons 4, 9 and pat of 1, 13, 15, Π = π B w B = (π 1 )[π (1 )B ], whch nceases wth B π +(π )(1 1. In egons 10, 18, 19 and pat of )B /B 11, 15, 16, Π = π B B, whch nceases wth B because π >. In egons 13 and 15 whee w < w (.e., Θ III ), Π = π B B w = π ( B ) B ( 1), whch deceases wth B. The second step s to pove that Π weakly nceases wth B at the bounday of two adacent egons. We can vefy that Π emans the same on all boundaes except the boundaes of 5/6, 9/7, 9/8, 13/14, at 10

12 whch Π stctly nceases wth B due to the decease n CPC. So fa, we have fnshed the poof egadng the elatonshp between Π and B. The poof fo the elatonshp between Π and B s smla and theefoe omtted. Poof of Poposton 2. We epot the publshe s evenue Π P n each budget egon n Table A5 and then pove the statements made n Poposton 2 n thee steps. Fst, we show that Π P weakly nceases wth B and B n each budget egon except 1 and 15. We pove seveal less obvous cases. Fo example, n egon 2, Π P = B + (1 B w ) ( + ρ) + B w = ( + ρ) + B (1 (+ρ 1) w ). Because w nceases wth both B and B, Π P nceases wth both B and B hee. In egon 5, Π P = ( + ρ) + B (1 (+ρ 1) ), whch nceases wth B w because w nceases wth B n ths egon. Second, gven any B >, Π P stctly deceases wth B wheneve an ncease n B causes a cossng ove mn {B, B } because b deceases wth B. Smlaly, gven any B (, B ), Π P stctly deceases wth B wheneve an ncease n B cause a cossng ove B. Thd, we pove that Π P stctly deceases wth B wheneve such an ncease causes advetse s ntal equlbum poston to change fom the second to the fst. Such a change occus unde thee stuatons: 1) B nceases fom egon 3 to 5; 2) B nceases so that w = w n egon 1; 3) B nceases so that w = w n egon 15. In the fst case, Π P (egon 5) Π P (egon 3) = (B B ) ( (+ρ 1) 1) < 0 because B B > 0 and B > B > > ( + B ρ 1). In the second case when w = w n egon 1, the evenue change s B (1 (+ρ 1) w ) B (1 (+ρ 1) w ) = (B B ) (1 (+ρ 1) ) < 0 because B w < B n egon 1 whee w = w, as ndcated by Lemma A5. In the thd case when w = w n egon 15, we can vefy that w w = w and w w = w. Hence, the evenue change s the same as n egon 1. Fnally, we can manually check that Π P always weakly nceases when a budget ncease causes a cossng of the bounday of a egon othe than what was dscussed above. 11

13 Regon n Fg. 1(a) Θ I Regon n Fg. A1(a) 1 (whee w w ),4 & 9 Θ I 1 (whee w > w ) & 2 Advetse s Poft (Π ) π B w B 5 π (π ) B w + (π )(1 B w ) Advetse s Poft (Π ) (π ) B w + (π )(1 B ) w (π ) B + w (π )(1 B ) w π B w B Publshe s Revenue (Π P ) B B + (1 ) ( + w ρ) + B w B + (1 B ) ( + w ρ) + B w B + (1 B w ) ( + ρ) + B w Θ II 8 (π w ) π B + Θ II Θ BI 3 π (π w ) B + L 6 & 7 (π w ) π B + Regon n Fg. 1(b) Θ I Θ I Regon n Fg. A1(b) 13 & 15 whee w w & w w 16 & 15 whee w > w & w w π B w B (π ) B w + (π )(1 B w ) (π ) B w + (π )(1 π B w B B ) w B B + (1 ) ( + w ρ) + B w B + (1 B w ) ( + ρ) + B w Θ II 14 (π w ) π B + Θ II Θ III Θ III S Θ BI 17 π (π w ) B + 13 & 15 whee w w & w < w 16 & 15 whee w > w & w < w 12 & 11 whee B 19 & 18 whee B 10, 11 whee B <, & 18 whee B < π B w π B B B (π ) π B π B π B π B w π B B B B B + B + (1 B ) ρ w B + B + (1 B ) ρ w + B + (1 B ) ρ B (π ) + B + (1 B ) ρ B π B B (1 ρ)(b + B ) + ( + 1)ρ Table A5. Advetses Pofts and Publshe s Revenue n Equlbum 12

14 B. Suppotng Poofs n the Equlbum Devaton We pove suppotng lemmas fo the UNE chaactezaton n Lemmas 1 and 2 n sectons B.1 and B.2 espectvely. Fo the ease of exposton, we popose seveal defntons, whch ae epettvely used n ou poofs. Defnton 1. The UBRC denotes the best-esponse coespondence wth an undomnated bd. Defnton 2. An advetse s budget s suffcent f t s nexhaustble unde he hghest wllngness to bd, that s, B B = w. B.1. Equlbum wth Low Resevaton Pce ( < mn { B, B }) We pove suppotng Lemmas A1-A5 pevously descbed n secton A. Poof of Lemma A1. Wthout loss of genealty, we assume k = n ths poof. When < mn { B, B }, an advetse s poft functons can be smplfed as (π b ) mn { B, 1} f b b > b Π (b, b ) = {. (π ) mn { B, 1} + (π b ) max {0,1 B } f b b < b In case 1 whee b B & b B, advetse s poft s Π (b, b ) = { (π b ) f b > b π f b < b. Because (π b ) π s equvalent to b w, advetse s best-esponse coespondence s b (b ) = { {x x > b } f b < w {x x b } f b w. In case 2 whee B b < B, π B B Π (b, b ) = { b f b > b. π f b < b Because π B b s B π s equvalent to b w, advetse s best-esponse coespondence b (b ) = { {x x > b } f b < w {x x b } f b w. The best-esponse coespondences n cases 3 and 4 ae deved n a smla way. 13

15 Poof of Lemma A2. Wthout loss of genealty, we assume k = n ths poof. Fo the fst set of nequaltes, B B B ( 1)π + B π B π +B B B π B π +B ( 1)π + B B. We can manually vefy othe nequalty equvalences n a smla way based on the defntons of budget thesholds shown n Table A1. The detaled poofs ae qute tedous and can be povded by the authos upon equest. The ntuton behnd these nequaltes s the followng. When both advetses budgets ae nsuffcent, the wllngness-to-pay (pe clck) unde whch an advetse s ndffeent between two postons s always hghe when the advetse does not have the theat of budget exhauston by the val s bd than when she does. Ths explans the fst two sets of nequalty equvalence as B k and B k, k ae budget thesholds fo advetses who have the theat of budget exhauston, and B k and B k, k ae budget thesholds fo advetses who do not. Smlaly, the wllngness-to-pay (pe clck) s always lowe when the advetse has a chance to deplete the val s budget than when she does not. Ths explans the last two sets of nequalty equvalences. Next we chaacteze the UNE n all possble cases. To do so, we fst focus on the halfbudget space whee B B and then extend the esults to the othe half budget space. The odnal elatonshps among budget thesholds n Lemma A2 shed lght on how to sepaate the half-budget space nto sub-egons. 1 In patcula, we fst specfy the ange of B by odeng {B, B, B, B, B } nto dffeent cases and then specfy the ange of B by odeng {B, B, B } wthn each case. We demonstate the devaton of UNE n one case n the followng example. The UNE n othe cases ae deved n a smla way. Example. Equlbum devaton when B B B B B. Step 1: Chaacteze the best-esponse coespondences of both advetses as b (b ) = { {x x > b } f b < w b ε f b ; b (b ) = { {x x > b } f b < w w b ε f b. w Step 2: Elmnate all weakly domnated bds. 1 Lemma A2 suggests that the ode of {B, B, B, B, B } n the half-budget space whee B B can only have sx possbltes: 1. B B B B B ; 2. B B B B B ; 3. B B B B B ; 4. B B B B B ; 5. B B B B B ; 6. B B B B B. Smlaly, fo advetse, the ode of {B, B, B, B, B } also only have sx possbltes: 1. B B B B B ; 2. B B B B B ; 3. B B B B B ; 4. B B B B B ; 5. B B B B B ; 6. B B B B B. 14

16 Fo advetse, we pove that any bd b < w s weakly domnated by a bd of w. To see ths, fst consde a b < b, s best-esponse coespondence suggests that s poft s the same between bddng w and b. If b [b, w ), the best-esponse coespondence ndcates that s poft s stctly hghe by bddng w than bddng b. Fnally, f b w, Lemma A2 suggests that s budget s depletable unde w because B B B B. Hence, s poft s stctly hghe by bddng w because bddng w allows to take the fst poston soone and theeby eceve moe clcks than bddng b. Smlaly, any bd b < w s weakly domnated by a bd of w fo advetse hee. Thus, the best-esponse coespondences wth undomnated bds, o the UBRCs ae b (b ) = { {x x > w } f b < w b ε f w b ; b (b ) = { {x x > w } f b < w. b ε f w b Step 3: The UNE s detemned by the ntesecton of the best-esponse coespondences. (b, b ) = { (w, w ε) f w w (w ε, w ) f w < w. 2 Fgue A2. The UNE When B B B B B Notes. We assume w < w n Fgue A2. The sold and the dashed pat demonstate the UBRC of advetse and espectvely. The UNE s detemned by the ntesectons of two pats. 2 Thee ae actually two ntesectons of advetses UBRC n Fgue A2. Howeve, we teat two equlba (b, b + ε) and (b ε, b) as the same because the equlbum outcome fo all pates ae the same povded that ε 0. We mpose ths equlba equvalence n the est of the pape. 15

17 Poof of Lemma A3. We pove only Case 1. The othe cases follow smlaly. In Case 1 whee B B B B, the subcase 1 whee B [B, B ) has been dscussed n the example above. In subcase 2 whee B [B, B ), the UBRC fo advetses s {x x > w } f b w b (b ) = { w f w < b B b ε f B < b. To see ths, we pove that any bd b (, w ) (w, B ) s weakly domnated by a bd of w fo advetse. It can be vefed that fo any b (, w ), advetse obtans the same poft wth b and w when b b o b w but stctly hghe poft wth w when b (b, w ). Smlaly, fo any b (w, B ), advetse obtans the same poft wth b and w when b w o b b but stctly hghe poft wth w when b (w, b). Meanwhle, the UBRC fo advetse s b (b ) = { {x x > w } f b < w b ε f w b. Based on Lemma A2, w les between w and B. Thus, the UNE n ths subcase 2 s b = w ε and b = w. In subcase 3 whee B [B, + ), the UBRC fo advetse s the same as n subcase 2, whle the UBRC fo advetse becomes b (b ) = { {x x > w } f b < w. Because w w B b ε f w b n ths subcase, the UNE s b = w and b [w, B ]. The poofs of the emanng cases ae smla and theeby omtted. Poof of Lemma A4. The lemma s poved by establshng two clams. Fst, we show that B B holds when B (, B ). Second, we pove that B and B ntesect only once at B = B when B (, B ). We stat wth the poof of the fst clam. Note that B B s equvalent to f(x) 0, whee f(x) = (π )( 1)x 2 + π ( π )x + (π )π 2 and x = π + B (π, π ), whch coesponds to B (, B ). Hee f(x) s a convex functon. At the nteval s boundaes, we have f(π ) = ( 1)π 2 π > 0 and f(π ) = ( 1)π 2 (π π + ) > 0. Next we show that f(x) 0 fo any x (π, π ) by consdeng thee cases. 16

18 Case 1: π. Suppose thee exsts an x (π, π ), s.t. f(x ) < 0. Snce f(π ) and f(π ) > 0, thee exsts two eal-valued oots of f(x), denoted by x 1 and x 2, whch must be located wthn the nteval (π, π ). Thus, x 1 + x 2 should be postve. Howeve, t can be vefed n ths case that x 1 + x 2 = (π )π < 0. Ths contadcton mples f(x) 0 fo all (π )( 1) x (π, π ). Case 2: π > & > 2. Followng the pevous logc, suppose agan f(x ) < 0 fo some x (π, π ). Then we should have x 1 + x 2 = (π )π (π )( 1) > 2π, whch s equvalent to (π ) > 2( 1)(π ). Because > 2 & π π, the evese s tue. Ths contadcton mples f(x) 0 fo all x (π, π ). Case 3: π > & 2. Suppose f(x ) < 0 fo some x (π, π ). Then we should have x 1 + x 2 = (π )π (π )( 1) < 2π, Howeve, we pove that x 1 + x 2 > 2π. To see ths, fst notce that the necessay and suffcent condton fo f(x) to have eal-valued oots s to have a postve dscmnant, whch s equvalent to π (π ) 2 4π (π ) π 4( 1)(π ). Thus, x 1 + x 2 = (π )π (π )( 1) > 2π, the last nequalty of whch s equvalent to 2(π ) > (π ). Fo the second clam, note that B (B = ) > = B (B = ) and B (B = B ) < B < B (B = B ), so thee s at least one ntesecton of B and B when B (, B ). Based on the poof above, thee ae at most two ntesectons of B and B fo B (, + ). Snce t s mpossble fo these two ntesectons to locate wthn (, B ) (othewse the odnal elatonshp between B and B should be the same at two boundaes), we have poved that thee s only one ntesecton of B and B wthn ths nteval. Usng the same logc, we can futhe pove that ths ntesecton must be wthn B (, B ) because B (B = B ) = B < B (B = B ). Poof of Lemma A5. We show that gven any B, w (B ) w (B ) when B [B, B ]. Fst, we pove that t s tue at the two boundaes. When B = B, w (B ) = B [(2 1 )π (1 1 )]. Its B +(π ) devatve w..t. π s w π [(2 1 ) B ] > 0 when B >. Thus, w (B ) > 17

19 B [(2 1 )π (1 1 )] B +(π ) = w (B ) when B = B. When B = B, fom Lemma A2, we have w (B ) = B < w (B ) because B B fom Lemma A4. Next we show that w (B ) > w (B ) wthn the nteval. Note that w (B ) > w (B ) s equvalent to f(b ) > 0, whee f(b ) = [B π + (π ) (1 1 ) B ] [B + (π )] [B π + (π ) (1 1 ) B ] [B + (π )], whch s a concave functon of B. Fom above we know f(b ) and f(b ) ae postve, whch leads to f(θb + (1 θ)b ) > θf(b ) + (1 θ)f(b ) > 0, fo any θ (0,1). Thus, we have poved w w when B B B. B.2. Equlbum wth Hgh Resevaton Pce ( mn { B, B }) We fst pesent seveal useful lemmas. Lemma B1. When B k, the UBRC of advetse k s b k (b k ) =, whee k =,. Poof of Lemma B1. Wthout loss of genealty, we assume k = n ths poof. Advetse s payoff functon s (π b ) B b Π (b ) = { (π ) B f b > b. f b < b Snce advetse s budget wll defntely be depleted egadless of he poston, he weakly undomnated bd s the esevaton value. Lemma B2. w k, k w k, k w k, k w k, k B k ( B k), whee w 1 k, k = ( 1)B k ( B k ) and k =,. Poof of Lemma B2. Wthout loss of genealty, we assume k = n ths poof. w ( 1)B B ( B ) ( B ). Futhemoe, 1 w π B +(1 1 )(π )B B +(π ) B B ( B ). Fnally, w 1 w ( 1)B > π 1 B +(1 ( B ) )(π )B B +(π ) > (1 1 ) B > ( 1)B B + ( 1)(π )B > ( B )π B + ( 1)(π )B ( B ) ( 1)B B + 18

20 ( 1)(π )[B B ( B )] > ( B )π B ( 1)π B B > ( B )π B B ( B ). 1 Lemma B3. When < B k, the UBRC b k (b k ) s descbed below: whee k =,. 1. When B k < ( B k), b 1 k (b k ) = ; 2. When ( B k) 1 {x x w k, k } f b k < w k, k B k < B k, b k (b k ) = { b k ε f w k, k b k w k, k; w k, k f b k > w k, k {x x w k} f b k < w k w k f w k b k B k 3. When B k B k, b k (b k ) = b k ε f B k < b, k < w k, k { w k, k f b k > w k, k Poof of Lemma B3. Wthout loss of genealty, we assume k = n ths poof. We fst deve the poft functon fo advetse, that s, (π b ) B f b b > b Π (b, b ) = {, (π ) mn{t 1, 1} + (π ) max{0, mn {1 t 1, t 2 }} f b < b whee t 1 = B s s duaton n the lstng and t b 2 = B t 1 s the maxmum emanng tme fo advetse to stay at the fst poston afte leaves. We clam that the hghest possble poft at the second poston can be ewtten as follows: Π o (b b < b ) = { π f b B (π ) B b + (π ) (1 B b ) f B < b w (π ) B f b > w whee w efes to the bd such that advetse s budget s used up ght at the end of the game peod gven that s budget s exhaustble. In othe wods, t 1 (w ) + t 2 (w ) = 1. We now show that Π o (b b < b ) mght take thee functonal foms. When b B s so small that advetse has no way to deplete B, advetse s poft takes the lowest fom π. When b exceeds B but s modeate, advetse s optmal esponse s to bd b ε and s own budget wll not be 19

21 depleted by dung the day. Howeve, when b > w, advetse can only obtan a facton of clcks that he budget can affod, whch equals B, fo any b [w, b ε]. Theefoe, s maxmum poft s (π ) B. By defnton, we have w > B povded that B >. Next we deve advetse s UBRC. In case 1 whee B < ( B ), Lemma B2 ndcates 1 > w. Snce b > w, advetse s poft functon s the same as the case n whch B <, whch mples b (b ) =. To show case 2, we fst must establsh that B > ( B ). We have 1 B = π B > π +B because B nceases wth π and π >. Snce B >, we also have ( B ) whch poves that B > ( B ) 1 1 < ( ) 1 =, as long as B >. Next we show that any b [, w ) s weakly domnated by w and any b (w, + ) s weakly domnated by w. Fo the fome statement, we consde b [, w ), then Π s the same wth b and w when b < b. When b > w, snce B < B, Lemma A2 ndcates B < w, whch suggests Π (w ) = (π ) B w + (π ) (1 B w ) > (π ) B b + (π ) (1 B b ) Π (b ). When b les between b and w, Π (w ) = (π b ) B b (π w ) B w = (π ) B w + (π ) (1 B w ) > (π ) B b + (π ) (1 B b ) Π (b ). Thus, we have poved that any b [, w ) s weakly domnated by the bd of w. Followng smla pocedues, we can pove that any b (w, + ) s weakly domnated by the bd of w. Now we ae eady to vefy the UBRC n case 2. When b < w, the poft of stayng at the fst poston s (π b ) B b (π w ) B w = (π ) B w + (π ) (1 B w ) > (π ) B b + (π ) (1 B b ), whch s the hghest possble poft of stayng at the second poston. Thus, advetse should choose to bd above b when b < w. Gven that any bd below w s weakly domnated by the bd of w, the UBRC n ths case s b (b ) = {x x w }. When w b w, the poft at the second poston equals (π ) B b + (π ) (1 B b ), 20

22 whch exceeds the poft at the fst and nceases wth b. Thus, b (b ) = b ε. When b > w, the poft at the second poston s Π (b b < b ) = whch acheves ts maxmum (π ) B { π f b B (π ) B + (π b ) (1 B ) f B < b b w, (π ) B f w < b < b when b w. Because any bd above w s weakly domnated by a bd of w, the UBRC n ths case s theefoe b (b ) = w. Fnally, fo case 3 n whch B B, Lemma A2 ndcates that advetse s budget s nexhaustble by a bd of w. Hence, the theshold below whch all bds ae weakly domnated changes fom w to w. Futhemoe, as shown n the poof of Lemma A3, any bd b (w, B ] s also weakly domnated by w. Followng the smla pocedue n the poof fo case 2, we can vefy the UBRC n case 3. Lemma B4. If B [max {B, ( B ) ( 1) left pat of egon 13 n Fgue A1(b)), then w w. }, B ] and B [max {, ( 1)B }, mn{b, }] (the Poof of Lemma B4. Followng the logc n the poof of Lemma A5, we only need to pove that gven any B wthn n ths ange, w (B ) w (B ) holds at the lowe- and uppe-boundaes of B. We fst look at the case n whch B 2 (.e., B 2 1 ( B ) ). Followng exactly the same ( 1) pocedues n the poof of Lemma A5, we can show that w (B ) w (B ) when B = B o B = B. As fo the case n whch B < 2 2 1, the uppe bound of B emans the same whle the lowe bound changes fom B to ( B ) ( 1). Next we show w (B ) w (B ) when B = ( B ). When B = ( B ), ( 1) w = and B ( B ) (because the cuve B ( 1) = ( B ) les above ( 1) B = ( B ) ( 1) Lemma B4. n ths egon), whch ndcates w = w. Ths completes ou poof of ( 1) 21

23 Omtted Detals of the Poof of Lemma 2. We vefy the UNE n budget egons descbed n Table A4. In egon 10 whee B ( B ) 22 1 that both advetses bd n equlbum. Regon 11 whee B [ & B ( B ), Lemma B3 ndcates 1 1 B, B ] & B (0, max {, 1 B }] actually conssts of two subcases. Coespondng to Fgue A1(b), n the left pat of egon 11, whee B, snce B B 1 s equvalent to B ( B ), 1 t mples b = based on Lemmas B1 and B3. Applyng Lemma B3 to advetse futhe ndcates that any b [w, w ] and b = s the UNE n ths left subcase. As fo the ght pat whee B (, B ] & B (0, ], we know b = fom Lemma B1. On the othe hand, snce B s also not too lage, advetse has smultaneously both the theat of budget depleton by at the fst poston and the ncentve to deplete s budget at the second poston. Ths mples that any bd above w s weakly undomnated fo advetse. Hence, any b [w, + ) and b = s the UNE n the ght subcase. Smla analyss can be appled to egons 12 and 14 to deve the UNE. Next we deve the UNE n a moe complcated case, egon 13, whch can be futhe sepaated nto thee pats: the left pat whee B <, the mddle pat whee B [, B ) and the ght pat whee B [B, B ]. We fst pove that the advetse wth a hghe w (w ) wns the fst poston n any UNE. Lemma B4 says w w n the left and mddle pat of egon 13 and Lemma A2 says w B w n the ght pat because B [B, B ]. Thus, we have w w n egon 13. Suppose advetse bds hghe at equlbum, then b > b w because w s the lowest weakly undomnated bd fo advetse. Snce w w, Lemma B3 ndcates that advetse s best esponse s to bd below b, whch contadcts wth b > b. Thus, we have poved that advetse bds hghe n any UNE hee. In the mddle and ght pat of egon 13, s UBRC s b (b ) = { {x x > w } f b < w b ε f w b. Combned wth s UBRC descbed n Lemma B3, we can see that the UNE must be b = w and b = mn{w ε, w }. Actually, we can pove that gven any B [, B ], thee must exst a B # such that b = w when B (, B # ) and b = w ε when B [B #, ] n the UNE. Too see ths, fst note that gven B, w > w s equvalent to f(b ) = ( 1)(π

24 )B 2 + [π B ( 1)(π )]B + ( 1)B 2 [( 1) + π ]B > 0. The devatve s f (B ) = 2( 1)(π )B + π B ( 1)(π ), whch nceases wth π because B. Hence, we have f (B ) 2( 1)( )B + π B ( 1)( ) = π B > 0. Next we pove w > w when B = and w w when B =. When B =, w = + > w. When B =, w = B w because B B and the esult fom Lemma A2. Thus, we have shown that f() 0, f() > 0 and f > 0. Hence, thee exsts a B # [, ) s.t. f(b # ) = 0. The smla analyss can be appled to the left pat of egon 13 whee B < & B [ ( 1)B, B ]. So fa, we have completed the UNE devaton n the half budget space whee B B. The UNE n the othe half space (.e., egon 15 to 19) can be obtaned n a smla way. 23

25 C. Numecal Smulatons wth Moe than Two Stategc Advetses We demonstate that the man effects of budgets n Popostons 1 and 2 stll exst n the pesence of moe than two stategc advetses. In ou man analyss, we fnd that an advetse s budget may have negatve mpacts on both an advetse s own poft and on the publshe s evenue n a poston aucton wth two budget-constaned bddes. Howeve, t s analytcally ntactable to fnd an UNE n a settng wth moe than two bddes. Theefoe, we esot to numecal analyss by consdeng thee stategc bddes. To llustate the fndngs fom ths numecal execse, assume thee bddes have values-pe-clck π = 3, π = 2, π k = 1, espectvely. We also assume = 2 and = 0.5, whee epesents the fouth hghest bd. We seach fo the equlbum bd as follows. Specfcally, we set advetses ntal bds at the values-pe-clck and update the best-esponse bds teatvely. In each teaton, we allow each advetse s upwad devaton to be 0.01 above the val s bd and the downwad devaton to be ethe 0.01 below the val s bd o the lowest bd that acheves the maxmum poft. We epeat the pocess untl eachng an equlbum unde whch no one has ncentve to devate. To depct the elatonshp between an advetse s budget and poft n a two dmensonal space, we vay one focal advetse s budget fom 0 to 6, whle fxng the othe two advetses budgets at a cetan level denoted by {B 1, B 2 }. Fo smplcty, we allow B 1 and B 2 to take thee possble values fom {1, 1.5, 2}. Ths leads to 3*3=9 smulatons fo each focal advetse. We conduct these nne smulatons fo each advetse,, and k espectvely. We fnd that addng an addtonal stategc advetse does not elmnate the possblty of the negatve effects of budgets on advetses pofts and on the publshe s evenues. As an llustaton, the thee plots at the top n Fgue A3 show that an advetse s poft can decease wth he own budget and the thee plots at the bottom show that the publshe s evenue can decease when an advetse s budget nceases. 24

26 Fgue A3. Relatonshp between Budget and Payoff 25

27 D. Poston Auctons wth Endogenous Budgets In ths extenson, we elax the assumpton of fxed budget and allow the advetses to set budgets befoe onng the aucton. By Poposton 1, f advetses ae not n egon Θ III and can costlessly ncease budgets, they should always set daly budgets above the suffcent theshold,.e., B and B. Howeve, an expanson of an advetsng budget typcally eques the fm to dvet spendng fom othe nvestments (Che and Gale 1998). The oppotunty cost fo the advetsng budget expanson s defned as the hghest net etun fom outsde optons such as nvestments n fnancal makets o othe advetsng channels. We consde a two-stage game to model budget decsons made by advetses. In the fst stage, advetses smultaneously choose the advetsng budgets befoe patcpatng n the poston aucton. In the second stage, both advetses budgets become common knowledge and advetses compete on bds as shown n ou pevous model. The atonale fo ths two-stage settng s due to the much hghe fequency of the change n bds made by advetses than the fequency of change n budgets. We assume that the magnal cost of budget expanson s symmetc and constant, that s, C(B) = cb whee c > 0 efes to the magnal cost of the budget expanson. We assume that advetses now face a softe budget constant: they mght ethe choose a suffcent budget B H = B to obtan the geatest budget powe n the poston aucton o a smalle budget B L = αb, fo some postve α < 1. We assume the esevaton value to be zeo fo tactablty. Unde ths settng, advetses need to balance the tade-off between the hghe cost of settng a suffcent budget n the fst stage and the mpoved advetsng poft n the second stage. We solve the game backwads and epot ou fndngs n the followng poposton. Poposton D1. When α s elatvely small, advetse s budget weakly deceases n c, wheeas advetse s budget and the publshe s evenue ae n an nveted-u elatonshp wth c. In othe wods, a decease n c can lead to a decease n the publshe s evenue despte an ncease n advetse s choce of budget. We pove Poposton D1 based on Lemmas D1-D3. Let Π mn k denote advetse k s total advetsng poft when chooses B m and chooses B n mn, whee k =, and m, n = H, L. All Π k ae aleady descbed n Table A5. 26

28 Lemma D1. When π π (1, δ] and α mn { B B, α 1, α 2 }, the budget choces of advetses at equlbum ae descbed below. 1. When c < c 1, B = B H and B = B L ; 2. When c 1 c c 2, B = B L and B = B H ; 3. When c > c 2, B = B L and B = B L, whee c 1 = (+α α)π απ π +( 1)π (1+α α)π 1 α, c 2 = (+α α)π απ π +( 1)π (1+α α)π 1 α, c 3 = π π, δ s the theshold s.t. (1 α)π c 1 = c 3 as functons of π π, α 1 s the theshold s.t. c 1 = c 2 as functons of α and α 2 s the theshold s.t. c 2 = c 3 as functons of α. Poof of Lemma D1. When both advetses choose hgh budgets, advetse stays at the fst poston and pays w fo each clck. Theefoe, we have Π HH = (π w ) cb and Π HH = π cb. Smlaly, we have Π HL = (π w (αb )) cb, Π HL = π cαb. When advetse chooses low budget and chooses hgh budget, the condton α < B B ndcates that the UNE s n egon 3 n Fgue A1(a). Thus, we have Π LH = π cαb and Π LH = (π w (αb )) cb. When both choose low budgets, we have Π LL = (π w ) αb w cαb, and Π LL αb = π + π w (1 αb ) cαb w, whee w (αb, αb ) = π αb and w (αb ) = π αb π +αb. 1 +(1 )π αb αb +π By defnton, Π HH < Π HL always holds. Besdes, Π HL > Π LL s equvalent to c < c 1, Π LH > Π LL s equvalent to c < c 2, and Π HH > Π LH s equvalent to c < c 3. Next we pove that when α s elatvely small, thee must be c 1, c 3 < c 2. To see ths, consde an exteme case n whch α = 0, then c 1 = π π +( 1)π < π = c π +( 1)π 2, and c 3 = π π = 1 π = 1 π π ( 1)π < 1 ( 1)π = c ( 1 )π π +( 1)π 2. Thus, as long as α s not too lage, we have c 1, c 3 < c 2. Howeve, the elatonshp between c 1 and c 3 s uncetan. We can check that thee exsts a theshold δ such that c 1 c 3 f and only f π π δ. 27

29 Now we ae eady to vefy the equlbum. When c < c 1, we know Π HL > Π LL and Π HH < Π HL, whch ndcates that (B H, B L ) s the only equlbum. When c 1 c < c 2, we have Π HL Π LL Π LH > Π LL, whch suggest (B L, B H ) n equlbum. The equlbum when c c 2 s both advetses selectng the low budget. Lemma D2. When π π > δ and α mn { B B, α 1, α 2 }, the budget choces of advetses at equlbum ae descbed below. 1. When c < c 1, B = B H and B = B L ; 2. When c 1 c < c 3, only mxed-stategy equlbum exsts and p H 3. When c 3 c < c 2, B = B L and B = B H ; 3. When c c 2, B = B L and B = B L, whee p H (p H ) denotes advetse s ( s) pobablty of choosng B H (B H ). c < 0 whle p H > 0; c Poof of Lemma D2. The equlbum devaton s smla to that n Lemma D1. The only dffeence s the case n whch c 1 c < c 3. In ths case, we can check that no pue-stategy equlbum exsts and theefoe only mxed-stategy equlbum exsts. The mxed-stategy equlbum (p H, p H ) s detemned by the followng condtons: both advetses ae ndffeent between hgh and low budgets gven the val s stategy. Thus, p H s detemned by p H Π HH + Π LH Π LL (1 p H )Π LH = p H Π HL + (1 p H )Π LH, whch mples that p H = Π LH Π LL +Π HL ΠHH. Snce the numeato of p H deceases wth c and the denomnato s a postve constant, we have poved that p H < 0. Smlaly, we can deve that p c H LL HL Π = Π Π LL Π HL +Π HH ΠLH. Because the numeato of p H nceases wth c and the denomnato s a postve constant, we have poved that p H > 0. c Lemma D3. The publshe s evenue has an nveted-u elatonshp wth the magnal budget cost. 28

30 Poof of Lemma D3. When π π (1, δ], the publshe s evenue fst nceases fom B L to B (B L ) when c exceeds c 1, and then dops to B L when c exceeds c 2. Thus, the publshe s evenue fst nceases and then declnes wth c. When π π > δ, we only need to show that the publshe s evenue s concave n c n the egon wth mxed-stategy equlbum. When c 1 c < c 3, the publshe s expected evenue s E(Π P ) = p H p H B + p H p L (αb ) + p L p H (αb ) + p L p L (αb ) = p H p H (B αb ) p H α(b + B ) 2p H αb. By defnton, we have 2 p H p H Lemma D2 ndcates that 2 p H c p H c c 2 = 2 H p c 2 p H + 2p H c 2 p H + 2 p c H p H c. < 0. Futhemoe, based on the functonal fom of p H and p H, we know that they ae both lnea functons of c, whch mples that 2 p H we have poved 2 E(Π P ) c 2 when π π δ. c 2 = 2 H p = 0. Thus, c 2 < 0 when c 1 c < c 3. The est of the poof s smla to that n the case Poof of Poposton D1. Lemmas D1-D3 dectly lead to Poposton D1. 29

31 E. Poston Auctons wth Qualty Scoe We assumed that the publshe anks ads based only on advetses bds n ou man model. Nevetheless, as mentoned n footnote 2 n the man text, some publshes (e.g., Google) ank ads based on the poduct of bds b k and qualty scoes q k, k =,. The use of qualty scoes helps the publshe adust the ankng to eflect an advetse s qualty. Unde ths system, an advetse s pe-clck payment s equal to the smallest amount to keep he cuent ank,.e., CPC = b q, whee s the next hghest bdde. In ths settng, the esevaton pce can be e- q ntepeted as the hghest weghted bd of an unanked advetse. Followng the method used n pevous lteatue (Edelman et al. 2007, Vaan 2007), we show n the clam below that the UNE deved n ou basc model can be easly tansfeed to the UNE of a poston aucton wth qualty scoes, assumng q k s commonly known. Hence, ou fndngs egadng the mpact of budgets on advetses pofts and on the publshe s evenue stll hold when qualty scoes ae consdeed. Clam 2. Suppose a poston aucton, A, uses qualty scoes and a poston aucton, B, does not. The pa of bds {b k} s the UNE of poston aucton A wth value-pe-clck {π k }, budget {B k }, and qualty scoe {q k }, f and only f the pa of bds {b kb } s the UNE of poston aucton B wth value-pe-clck {π k B } and budget {B k B }, whee k =,, b kb = q k b k, π k B = q k π k, and B k B = q k B k. Poof of Clam 2. We pove Clam 2 by showng that both advetses pofts n a poston aucton A ae the same as the pofts n a poston aucton B deved n Table A2 afte the epaametezaton. Take the fst scenao whee both advetses budgets ae non-depleted as an example. Fo advetse, non-depleton means he budget B s geate than the numbe of clcks at the fst poston tmes he CPC q b, whch suggests B q > q b B q q > q b B B > b B. Smlaly, snce advetse pays q pe clck n aucton A, non-depleton means B > q B B >. Thus, the condtons fo budget depleton n aucton A s the same as the condtons n aucton B descbed n the second column n Table A2 afte the e-paametezaton. The poof fo the depleton condtons unde othe scenaos s smla. Next we pove that advetses pofts n aucton A ae {Π B k = 1 Π q k (b B k, π B k, B B k, B B k )}, k whee Π k s the poft functon descbed n the thd column n Table A2. Stll take the fst 30

32 scenao as an llustaton. In ths case, s poft n aucton A s (π q b q ) = (q π q b ) q = 1 q Π (b B, π B ); s poft s π q = 1 q Π (π B ). The poof fo othe scenaos s smla and theefoe omtted. Snce advetses bd choces only depend on the ato of pofts between postons, the UNE of a poston aucton A wth {π k, B k, q k } wll be {b k = b kb q k }, whee {b kb } s the UNE of a poston aucton B wth {π k B, B k B }. 31