Price Competition under Linear Demand and Finite Inventories: Contraction and Approximate Equilibria

Transcription

1 Prce Competton under Lnear Demand and Fnte Inventores: Contracton and Approxmate Equbra Jayang Gao, Krshnamurthy Iyer, Huseyn Topaogu 1 Abstract We consder a compettve prcng probem where there are mutpe frms wth mted nventores of substtutabe products. Each frm chooses the prces that t charges for ts product over a fnte seng horzon. The demand of each frm onty depends on the prces of a frms n a determnstc fashon through a near demand functon. The goa of each frm s to choose ts prces to maxmze ts tota revenue. We study two types of equbrum. In equbrum wthout recourse, each frm chooses ts entre prce traectory at the begnnng of the seng horzon. In equbrum wth recourse, each frm adusts ts prce at each tme perod as a functon of the current nventores of a of the frms. Athough the demand of each frm s a determnstc functon of the prces so that there s no uncertanty n the responses of the frms, there s a stark dfference between equbra wth and wthout recourse. Consderng the commony studed dagonay domnant regme, where the demand of a frm s affected more by ts prce than the prces of the other frms, we show that the equbrum wthout recourse exsts and t s unque. In contrast, we demonstrate that the equbrum wth recourse may not exst or may not be unque. Motvated by ths resut, we ook for approxmate equbrum wth recourse. Consderng a ow nfuence regme, where the effect of the prce of each frm on the demand of the others s dmnshng, we show that the equbra wthout recourse s an approxmate equbrum even when we aow the frms not to commt to a prce traectory at the begnnng of the seng horzon. 1 Introducton In many practca stuatons, mutpe frms seng substtutabe products set ther prces compettvey to se mted nventores over a fnte seng horzon, when the demand of each frm onty depends on the prces charged by a frms. For exampe, arnes compettvey set the prces for ther mted seat nventores n a partcuar market. Frms seng eectronc products take the prces of ther compettors nto consderaton when settng ther prces. In ths paper, we consder mutpe frms wth mted nventores of substtutabe products. Each frm chooses the prces that t charges for ts product over a fnte seng horzon. The demand that each frm faces s a determnstc functon of the prces charged by a of the frms, where the demand of a frm s neary decreasng n ts prce and neary ncreasng n the prces of the other frms. Each frm chooses ts prces over a fnte seng horzon to maxmze ts tota revenue. Man Contrbutons. We study two types of equbrum for the compettve prcng settng descrbed n the prevous paragraph. In equbrum wthout recourse, at the begnnng of the seng horzon, each frm chooses the prces that t charges over the whoe seng horzon and commts to ths prce traectory, under the assumpton that the other frms do the same. In equbrum 1 Jayang Gao and Krshnamurhy Iyer are wth Schoo of Operatons Research and Informaton Engneerng, Corne Unversty, Ithaca, NY Huseyn Topaogu s wth Schoo of Operatons Research and Informaton Engneerng, Corne Tech, New York, NY The ema addresses of the authors are g838@corne.edu, kryer@corne.edu and topaogu@ore.corne.edu. 1

2 wth recourse, at each tme perod n the seng horzon, each frm observes the nventores of a of the frms and chooses ts prce at the current tme perod, agan under the assumpton that the other frms do the same. Despte the fact that the demand of each frm s a determnstc functon of the prces so that there s no uncertanty n the responses of the frms, we show a cear contrast between the equbrum wthout recourse and the equbrum wth recourse. We consder the dagona domnant regme, where the prce charged by each frm affects ts demand more than the prces charged by the other frms. In other words, f a of the compettors of a frm decrease ther prces by a certan amount, then the frm can decrease ts prce by the same amount to ensure that ts demand does not decrease. Ths regme s rather standard n the exstng terature and t s used n, for exampe, Aon and Federgruen (2007) and Gaego and Hu (2014). Focusng on the equbrum wthout recourse, we show that the best response of each frm to the prce traectores of the other frms s a contracton mappng, when vewed as a functon of the prces of the other frms. In ths case, t mmedatey foows that the equbrum wthout recourse aways exsts and t s unque. To our knowedge, ths contracton property n the settng of prce competton wth mted nventores s new. We gve counterexampes to show that the equbrum wth recourse may not exst or may not be unque. Motvated by ths observaton, we ook for an approxmate equbrum that s guaranteed to exst. In partcuar, we consder the settng where a frm can adust ts prce at each tme perod based on the current nventores of a of the frms. We ca a strategy profe for the frms an ɛ-equbrum wth recourse f no frm can mprove ts tota revenue by more than ɛ by devatng from ts strategy profe. We consder a ow nfuence regme, where the effect of the prce of a frm on the demand of another frm s dmnshng, whch naturay hods when the number of frms s arge. We show that the equbrum wthout recourse, whch we know to unquey exst, s an ɛ-equbrum wth recourse even when the frms may adust ther prces. So, ntutvey speakng, an ɛ-equbrum wth recourse s expected to exst when the number of frms s arge. Our resuts fs a gap n a fundamenta cass of revenue management probems. Athough there s no uncertanty n the responses of the frms, the equbra wth and wthout recourse are not the same concept and they can actuay be quatatvey rather dfferent. Whe the equbrum wthout recourse unquey exsts, one cannot say the same thng for the equbrum wth recourse. Aso, as far as we are aware, our contracton argument for showng the exstence and unqueness of the equbrum wthout recourse does not appear n the terature. Ths argument becomes surprsngy effectve when deang wth near demand functons, but t s st an open queston whether smar arguments hod for other demand functons. Lasty, our resuts ndcate that the equbrum wthout recourse n a ow nfuence regme turns out to be an approxmate equbrum wth recourse, even when the frms are aowed to not to commt to a prce traectory. Lterature Revew. Smar to us, Gaego and Hu (2014) consder prce competton among mutpe frms wth mted nventores over a fnte seng horzon. In the dagonay domnant 2

3 regme, they show the unque exstence of the equbrum wthout recourse, but ther argument does not show the contracton of the best response. Havng the contracton property aows fndng the equbrum wthout recourse by successvey computng the best response of each frm to the others. Furthermore, for the settng we consder, we show that the equbrum wth recourse can be dfferent from the equbrum wthour recourse and the former equbrum may not exst or may not be unque, but the equbrum wthout recourse may be a good approxmaton to the equbrum wth recourse n the ow nfuence regme. There are a number of papers that study prce competton over a snge perod. Mgrom and Roberts (1990) show that pure Nash equbrum exsts for a wde cass of so caed supermoduar demand modes. Gaego et a. (2006) provde suffcent condtons for unqueness of equbrum n the Bertrand game when the demands of the frms are nonnear functons of the prces, there s a cost assocated wth satsfyng a certan voume of demand and each frm s nterested n maxmzng ts expected proft. In partcuar, the authors use a sghty more genera verson of the mutnoma demand mode to capture the reatonshp between prce and demand. The cost s a nonnear functon of the demand voume. Person et a. (2013) dentfy the condtons for exstence and unqueness of pure Nash equbrum when the demands are characterzed by a mxture of mutnoma ogt modes and the cost of satsfyng a certan voume of demand s near n the demand voume. Gaego and Wang (2014) consder the prce competton between mutpe frms when the reatonshp between demand and prce s characterzed by the nested ogt mode and provde condtons to ensure the exstence and unqueness of the equbrum. Nazerzadeh and Peraks (2015) prove the exstence of pure strategy equbrum n a prce competton game between two suppers when capacty s prvate nformaton. Consderng the papers on prce competton over mutpe tme perods, Levn et a. (2009) study a stochastc game when there are strategc consumers choosng the tme to purchase. Ln and Sbdar (2009) study a compettve prcng probem when the reatonshp between demand and prce s captured by the mutnoma ogt mode and nventory eves are pubc nformaton. Martnez-de Abenz and Taur (2011) study the prcng game between two frms wth mted nventores facng stochastc demand. The authors characterze the unque subgame perfect equbrum. Lu and Zhang (2013) show that there exsts a unque pure strategy Markov perfect equbrum n a prcng game between two frms offerng vertcay dfferentated products. Organzaton. In Secton 2, we descrbe the compettve prcng settng, formuate the optmzaton probem that computes the best response of a frm when the frms commt to prce traectores at the begnnng of the seng horzon and show that the best response s a contracton mappng when vewed as a functon of the prces of the other frms, whch aows us to concude that the equbrum wthout recourse unquey exsts. In Secton 3, we defne the equbrum wth recourse and provde counterexampes to show that the equbrum wth recourse may not exst or may not be unque. In Secton 4, we show that the equbrum wthout recourse s an ɛ-equbrum wth recourse n a ow nfuence regme. In Secton 5, we concude. 3

4 2 Equbrum wthout Recourse There are n frms ndexed by N = {1,..., n}. cannot be repenshed over the seng horzon. Frm has c unts of nta nventory, whch There are τ tme perods n the seng horzon ndexed by T = {1,..., τ}. We use p t to denote the prce charged by frm at tme perod t. Usng p t = (p t 1,..., pt n) to denote the prces charged a of the frms at tme perod t, the demand faced by frm at tme perod t s gven by D t (pt ) = α t βt pt + γt, pt, where αt > 0, βt > 0 and γ, t > 0. We assume that the prce charged by each frm affects ts demand more than the prces charged by the other frms, n the sense that γt, < βt for a N, t T. Aso, usng p t = (pt 1,..., pt 1, pt +1,..., pt n) to denote the prces charged by frms other than frm at tme perod t, to avod negatve demand quanttes, we restrct the strategy space of the frms such that each frm charges the prce p t at tme perod t that satsfes αt βt pt + γt, pt 0, gven the prces p t charged by the other frms. If the frms other than frm commt to the prce traectores p = {p t : t T }, then we can obtan the best response of frm by sovng the probem { ( max α t β t p t + ) ( γ, t p t p t : α t β t p t + ) γ, t p t c, t T t T } α t β t p t + γ t, p t 0 t T, p t 0 t T. (1) Snce β t > 0, probem (1) has a strcty convex obectve functon and near constrants, whch mpes that the best response of frm s unque. Usng the non-negatve dua mutpers v and {u t : t T } for the frst and second constrant n probem (1), the Karush-Kuhn-Tucker (KKT) condtons for ths probem are ( ( α t β t p t + ) ) ( γ, t p t c v = 0, α t β t p t + ) γ, t p t u t = 0 t T, (2) t T α t 2 β t p t + γ t, p t + β t (v u t ) = 0 t T. Snce probem (1) has a concave obectve functon and near constrants, the KKT condtons above are necessary and suffcent at optmaty; see Boyd and Vandenberghe (2004). In other words, for a feasbe souton {p t : t T } to probem (1), there exst correspondng non-negatve dua mutpers v and {u t : t T } that satsfy the KKT condtons n (2) f and ony f {pt : t T } s the optma souton to probem (1). Note that we do not assocate dua mutpers wth the constrants p t 0 for a t T n probem (1) snce t s never optma for frm to charge a negatve prce. Therefore, we can actuay vew the constrants p t 0 for a t T as redundant constrants. We use the KKT condtons n (2) extensvey to characterze the best response of frm to the prce traectores p of the other frms. In the rest of ths secton, we excusvey focus on the strateges wthout recourse, where each frm commts to a prce traectory {p t : t T } at the begnnng of the seng horzon and does not adust these prces durng the course of the seng horzon. If the prce traectory {p t : t T } chosen by each frm s the best response to the prce 4

5 traectores p chosen by the other frms, then we say that the prce traectores {p t : t T } chosen by the frms s an equbrum wthout recourse. We show that there exsts a unque equbrum wthout recourse. Furthermore, f we start wth any prce traectory {p t : t T } for the frms and successvey compute the best response of each frm to the prce traectores of the other frms, then the best response of each frm forms a contracton mappng when vewed as a functon of the prces charged by the other frms. Usng ths resut, we show that there exsts a unque equbrum wthout recourse. To capture the best response of frm to the prces charged by the other frms, we defne the set of tme perods T (ν, p ) = { t T : αt + γt, pt β t } > ν. In the next emma, we use T (ν, p ) to gve a succnct characterzaton of the souton {p t : t T } and the correspondng dua mutpers v and {u t : t T } that satsfy the KKT condtons. Lemma 1 If a feasbe souton {p t : t T } to probem (1) and the correspondng non-negatve dua mutpers v and {u t : t T } satsfy the KKT condtons n (2), then we have α t + γt, pt p t 2 β = t + v f t T (v, p ) 0 f t T (v, p ) 2 α t + u t γt, pt = β t f t T (v, p ), v αt + γt, pt β t f t T (v, p ). Proof. Snce the souton {p t : t T }, aong wth the dua mutpers v and {u t : t T }, satsfes the KKT condtons n (2), sovng for u t n the thrd KKT condton, we have u t = αt + γt, pt β t 2 p t + v. (3) For notatona brevty, we et t = (α t + γt, pt )/βt. Therefore, we can wrte (3) as u t = t 2 pt + v. Furthermore, notng that β t > 0 and dvdng the second KKT condton n (2) by β t, we observe that ( t pt ) ut = 0 for a t T. Consder any t T (v, p ). By the defnton of T (v, p ), we have t > v. In ths case, by (3), t foows that u t = t 2 pt + v < 2 ( t pt ). Mutpyng the ast chan of nequates by ut and notng that ( t pt ) ut = 0, we get (u t )2 0, whch mpes that u t = 0. Usng ths vaue of u t n (3) and sovng for p t, we have pt = t /2 + v /2. Therefore, the desred resut hods for any t T (v, p ). Consder any t T (v, p ). By the defnton of T (v, p ), we have t v. In ths case, usng (3), t foows that u t = t 2 pt + v 2 ( t pt ). Mutpyng the ast chan of nequates by t pt and notng that ( t pt ) ut = 0, we have ( t pt )2 0, whch mpes that p t = t. Usng ths vaue of pt n (3) and notng the defnton of t, we get u t = v t. Therefore, the desred resut hods for any t T (v, p ). By Lemma 1, we can characterze the souton {p t : t T } and the dua mutpers v and {u t : t T } that satsfy the KKT condtons n (2) ony by usng the vaue of v. If we know the 5

6 vaue of v, then we can compute the set of tme perods T (v, p ), n whch case, we can choose the vaues of {p t : t T } and {ut : t T } as gven n Lemma 1. Throughout the rest of ths secton, we ndeed choose the vaues of {p t : t T } and {ut : t T } as gven n Lemma 1, snce we are nterested n souton thats satsfy the KKT condtons. Naturay, we do not know the vaue of v that aows us to obtan the souton {p t : t T } and the dua mutpers v and {u t : t T } that satsfy the KKT condtons. In the next emma, we gve a characterzaton of the vaue of v that corresponds to the souton {p t : t T } and the dua mutpers v and {u t : t T } satsfyng the KKT condtons n (2). In partcuar, we consder the functon ( α t β t ν + γ, t p t t T (ν,p ) { G (ν, p ) = ( max α t + γ, t p t t T (ν,p ) ) 2 c f ν > 0 } ) 2 c, 0 f ν = 0. In the appendx, Lemma 7 shows that G (, p ) s strcty decreasng over some [0, ν ] and has a unque root. In the next emma, we use ts root to characterze a souton to the KKT condtons. Lemma 2 If a feasbe souton {p t : t T } to probem (1) and the correspondng non-negatve dua mutpers v and {u t : t T } satsfy the KKT condtons n (2), then we have G (v, p ) = 0. Proof. As n the proof of Lemma 1, we et t = (αt + γt, pt )/βt for notatona brevty. By Lemma 1, we have p t = ( t + v )/2 for a t T (v, p ) and p t = t for a t T (v, p ). Frst, we assume that v = 0. Snce α t > 0, we have T (v, p ) = T by the defnton of T (v, p ), whch mpes that p t = ( t + v )/2 = t /2 for a t T. In ths case, we obtan 1 2 t T βt t = t T βt ( t pt ) = t T (αt βt pt + γt, pt ) c, where the second equaty uses the defnton of t and the nequaty foows from the fact that {pt : t T } s a feasbe souton to probem (1). The ast chan of nequates mpy that t T βt t 2 c 0. Notng the defnton of t and the fact that T (v, p ) = T, we obtan t T (v,p ) (αt + γt, pt ) 2 c 0, whch mpes that G (v, p ) = G (0, p ) = 0. Therefore, the desred resut hods when v = 0. Second, we assume that v > 0. Usng the fact that p t = ( t + v )/2 for a t T (v, p ) and p t = t for a t T (v, p ), we have t T (αt βt pt + γt, pt ) = t T βt ( t pt ) = t T (v,p ) βt ( t v )/2. Snce v > 0, by the frst KKT condton n (2), we aso have c = t T (αt βt pt + γt, pt ). In ths case, by the ast chan of equates, we get c = 1 2 t T (v,p ) β t ( t v ). By the defnton of t, the equaty above s equvaent to t T (v,p ) (αt β v + γt, pt ) 2 c = 0, whch mpes that G (v, p ) = 0. Therefore, the desred resut hods when v > 0. By Lemma 2, f a feasbe souton {p t : t T } to probem (1) and the correspondng nonnegatve dua mutpers v and {u t : t T } satsfy the KKT condtons n (2), then v must be 6

7 the unque root of G (, p ). Aso, by Lemma 1, the vaues of {p t : t T } and {ut : t T } must be gven as n Lemma 1. In the next theorem, we use these resuts to show that the best response of frm s a contracton mappng when vewed as a functon of the prces of the other frms. Theorem 3 Let {p t (p ) : t T } be the optma souton to probem (1) as a functon of the prces charged by the frms other than frm. For any two prce traectores ˆp = { ˆp t : t T } and p = { p t : t T } adopted by the frms other than frm, we have { p t ( ˆp ) p t ( p ) max γt, ˆpt pt } t T β t. Proof. For notatona brevty, we et ˆp t = p t ( ˆp ) and p t = p t ( p ). In other words, {ˆp t : t T } s the optma souton to probem (1) when we sove ths probem after repacng p wth ˆp. Smary, { p t : t T } s the optma souton to probem (1) when we sove ths probem after repacng p wth p. Aso, we et ˆv and ṽ be such that G (ˆv, ˆp ) = 0 and G (ṽ, p ) = 0. Wthout oss of generaty, we assume that ˆv ṽ. Otherwse, we nterchange the roes of ˆv and ṽ. Fnay, we et ˆ t = (αt + γt, ˆpt )/βt and t = (αt + γt, pt )/βt for notatona brevty. Note that ˆ t t γt, ˆpt pt /βt. In ths case, usng M = max t T { γt, ˆpt pt /βt }, we have ˆ t t M for a t T. We proceed to examnng four cases to show that ˆp t pt 1 2 M max{m, ˆv ṽ } for a t T. Frst, we assume that t T (ˆv, ˆp ) and t T (ṽ, p ). Usng Lemma 1, we have ˆp t pt = 1 2 ˆ t + ˆv t ṽt 1 2 ˆ t t (ˆv ṽ ) 1 2 M (ˆv ṽ ) 1 2 M max{m, ˆv ṽ }, as desred. Second, we assume that t T (ˆv, ˆp ) and t T (ṽ, p ). Usng Lemma 1 once more, we have ˆp t pt = ˆ t t M 1 2 M max{m, ˆv ṽ }, as desred. Thrd, we assume that t T (ˆv, ˆp ) and t T (ṽ, p ). Snce t T (v, ˆp ), we have ˆ t > ˆv, whch mpes ˆv t < ˆ t t M. Aso, snce t T (ṽ, p ), we have t ṽ, whch mpes ˆv t ˆv ṽ. Notng the ast two nequates, t foows that ˆv t max{m, ˆv ṽ }. In ths case, usng Lemma 1 one ast tme and usng the fact that ˆv t max{m, ˆv ṽ }, we obtan ˆp t pt = 1 2 ˆ t ˆv t 1 2 ˆ t t ˆv t 1 2 M ˆv t 1 2 M max{m, ˆv ṽ }, as desred. Fourth, we assume that t T (ˆv, ˆp ) and t T (ṽ, p ), n whch case, we can foow the same argument n the thrd case to obtan ˆp t pt 1 2 M max{m, ˆv ṽ }. The precedng dscusson shows that ˆp t pt 1 2 M max{m, ˆv ṽ }. If ˆv M, then notng that ṽ 0, the ast nequaty mpes that ˆp t pt M, whch s the resut we want to show! In the rest of the proof, we proceed under the assumpton that ˆv > M. Consder the functon G (, p ). By Lemma 7 n the appendx, the functon G (, p ) s strcty decreasng over the nterva [0, ν ) for some ν and constant over the nterva [ν, ). By the same emma, we aso have G (ν, p ) = 2 c < 0. In the rest of the proof, we show that G (ˆv M, p ) 0. Aso, we have G (ṽ, p ) = 0 by Lemma 2. In ths case, snce G (ν, p ) < 0 and G (, p ) s strcty decreasng over the nterva [0, ν ) and constant over the nterva [ν, ), havng G (ˆv M, p ) 0 and G (ṽ, p ) = 0 mpes that ˆv M ṽ. Therefore, we have 7

8 ˆv ṽ M, so that we get ˆp t pt 1 2 M max{m, ˆv ṽ } = M, whch s the resut we want to show. It remans to show that G (ˆv M, p ) 0. Usng 1( ) to denote the ndcator functon, snce ˆv > M, by the defnton of G (, p ), we have G (ˆv M, p ) = ( α t β t (ˆv M ) + γ, t p t t T (ˆv M, p ) = ( α t β t (ˆv M ) + γ, t p t t T (ˆv, ˆp ) + t T t T 1(t T (ˆv M, p ) \ T (ˆv, ˆp )) ) 2 c ) 2 c ( α t β t (ˆv M ) + ) γ, t p t ( 1(t T (ˆv, ˆp ) \ T (ˆv M, p )) α t β t (ˆv M ) + ) γ, t p t. (4) We consder each one of the three terms on the rght sde above one by one. For the frst term, by Lemma 2, we have G (ˆv, ˆp ) = 0. By the defnton of M, we aso have γt, ˆpt pt /βt M γt, ˆpt pt M t T (ˆv, ˆp ) βt. Thus, we get for a t T, so that t T (ˆv, ˆp ) t T (ˆv, ˆp ) ( α t β t (ˆv M ) + γ, t p t ( α t β t ˆv + ) γ, t ˆp t = t T (ˆv, ˆp ) = G (ˆv, ˆp ) + G (ˆv, ˆp ) t T (ˆv, ˆp ) t T (ˆv, ˆp ) ) 2 c 2 c + γ t, ( p t ˆp t ) + M γ, t p t ˆp t + M t T (ˆv, ˆp ) β t t T (ˆv, ˆp ) t T (ˆv, ˆp ) γ, t ( p t ˆp t ) + M β t 0, β t t T (ˆv, ˆp ) where the second equaty uses the fact that ˆv > M 0 so that we have G (ˆv, ˆp ) = t T (ˆv, ˆp ) (αt βt ˆv + γt, ˆpt ) 2 c. Therefore, the frst term on the rght sde of (4) s non-negatve. For the second term, by the defnton of T (ˆv M, p ), we have α t + γt, pt > β t (ˆv M ) for a t T (ˆv M, p ). Therefore, we have 1(t T (ˆv M, p ) \ T (ˆv, ˆp )) (α t βt (ˆv M ) + γt, pt ) 0, whch mpes that the second term on the rght sde of (4) s non-negatve. For the thrd term, we have α t + γt, pt β t (ˆv M ) for a t T (ˆv M, p ). Therefore, we have 1(t T (ˆv, ˆp ) \ T (ˆv M, p )) (α t βt (ˆv M ) + γt, pt ) 0, ndcatng that the thrd term on the rght sde of (4) s non-postve. So, the frst and second terms on the rght sde of (4) s non-negatve, whereas the thrd term s non-postve, n whch case, we have G (ˆv M, p ) 0. For the vector y = {y t : t T }, we defne the norm on R τ as y = max t T y t. By the assumpton that γt, < βt for a N and t T, Theorem 3 mpes that the best response of frm s a contracton mappng wth respect to the norm, when vewed as a functon of the prces of the other frms. Therefore, t mmedatey foows that f the prce charged by each frm affects ts demand more than the prces charged by the other frms, then there aways exsts a unque equbrum wthout recourse. 8

9 3 Equbrum wth Recourse In ths secton, we consder strateges wth recourse, where each frm can change ts prce at the current tme perod as a functon of ts nventory and the nventores of the other frms. In other words, the frms do not commt to a prce traectory at the begnnng of the seng horzon. We et x t be the nventory of frm at the begnnng of tme perod t. Focusng on Markovan strateges wthout oss of generaty, as a functon of the nventores x t = (x t 1,..., xt n) of a of the frms, we use P t(xt ) to denote the prce charged by frm at tme perod t. It s usefu to vew P t ( ) as a functon that determnes the strategy of frm at tme perod t as a functon of the nventores of a of the frms. We use P t = (P t 1 ( ),..., P t n( )) to capture the strateges of a of the frms at tme perod t and P t = (P t 1 ( ),..., P t 1 ( ), P t +1 ( ),..., P t n( )) to capture the strateges of the frms other than frm at tme perod t. If the frms other than frm use the strateges P = {P t : t T }, then we can fnd the best response strategy of frm by sovng the dynamc program { ( V t (x t ) = max α t β t p t + ) γ, t P t (x t ) ( = x t α t β t p t + ) γ, t P t (x t ), ( x t+1 = x t α t βt P t (xt ) + γ, t pt + x t+1 p t 0, x t+1 0 N p t + V t+1 (x t+1 ) : α t β t p t + γ, t P t (x t ) 0, }, {,} ) γ, t P t (x t ) N \ {}, wth the boundary condton that V τ+1 ( ) = 0. An optma souton to the probem above characterzes the best response strategy of frm at tme perod t. If the strategy {P t ( ) : t T } chosen by each frm s the best response to the strategy P chosen by the other frms, then we say that the strateges {P t : t T } chosen by the frms s an equbrum wth recourse. In the prevous secton, we show that there aways exsts a unque equbrum when we focus on strateges wthout recourse. We gve two numerca exampes to show that f we focus on strateges wth recourse, then there may not exst an equbrum or there may be mutpe equbra. Consder the case where there are two frms and the seng horzon has two tme perods. For gven nventores of the two frms at the second tme perod, the probem of computng the equbrum strategy at the second tme perod s dentca to fndng an equbrum wthout recourse. So, there exsts a unque equbrum strategy for the frms at the second tme perod for gven nventores. Note that the prces charged by the frms n an equbrum wthout recourse at the second tme perod depends on the nventores of the frms at the second tme perod, whch, n turn, depends on the prces charged by the frms at the frst tme perod. To obtan an equbrum wth recourse, we compute the best response strategy of each frm at the frst tme perod as a functon of the prce of the other frm at the frst tme perod. Reca that f we fx the prces of the frms at the frst tme perod, then we fx the nventores at the second tme perod, 9

10 Frst Perod Prce of Frm Best Response of Frm 2 to Frm 1 Best Response of Frm 1 to Frm 2 Tota Revenue for Frm Frst Perod Prce of Frm Frst Perod Prce of Frm 1 Fgure 1: Best response of each frm when equbrum wth recourse does not exst and revenue of the frst frm. n whch case, we can compute the equbrum strateges at the second tme perod. We pot the best response of each frm at the frst tme perod as a functon of the prce of the other frm. An equbrum wth recourse corresponds to the ntersecton of the two best response curves. Consder the parameters α t = 4, β1 = 4, β2 = 2, γ1, = 16/5, γ2, = 1, c = 3 for a {1, 2}, {1, 2} \ {} and t {1, 2}, whch satsfy γt,t < βt for a {1, 2, } and t {1, 2}, so that we know that there exsts a unque equbrum wthout recourse. On the eft sde of Fgure 1, the sod ne pots the best response of second frm at the frst tme perod on the vertca axs, as a functon of the prce of the frst frm on the horzonta axs, whereas the dashed ne pots the best response of the frst frm at the frst tme perod on the horzonta axs as a functon of the prce of the second frm on the vertca axs. The two best response functons do not ntersect. Therefore, there does not exsts an equbrum wth recourse. The man drver of the ack of equbrum s the dscontnuty n the best response functon. The dscontnuty s due to the fact that the revenue of each frm s a mut-moda functon of ts prce at the frst tme perod. On the rght sde of Fgure 1, we show the revenue of the frst frm as a functon of ts prce at the frst tme perod, when the prce of the second frms s fxed at 2.2. So, frm 1 can ump from one mode to another based on the prce of the second frm. Consderng the parameters α t = 4, β1 = 5, β2 = 2, γ1, = 0.1, γ2, = 1 and c = 5 for a {1, 2}, {1, 2} \ {} and t {1, 2}, Fgure 2 shows the best response of each frm at the frst tme perod as a functon of the prce charged by the other frm. The best response functons ntersect at two ponts, ndcatng mutpe equbra wth recourse. 4 An Approxmate Equbrum If the strategy {P t ( ) : t T } chosen by each frm cannot ncrease the revenue of frm by more than ɛ gven the the other frms use the strateges P, then we say that the prce strateges {P t : t T } chosen by the frms s an ɛ-equbrum wth recourse. Snce there may not exst an 10

11 Frst Perod Prce of Frm Best Response of Frm 2 to Frm 1 Best Response of Frm 1 to Frm Frst Perod Prce of Frm 1 Fgure 2: Best response of each frm when there are mutpe equbra wth recourse. equbrum wth recourse or there may be mutpe equbra wth recourse, we focus on ɛ-equbra wth recourse. We consder a ow nfuence regme, where, roughy speakng, the prce charged by a frm affects ts demand more than the prces charged by each of the other frms. In partcuar, ettng M = max N,t T γt, /βt and notng the assumpton that γt, < βt for a N, t T, we have M < 1. We consder the regme where the prce charged by a frm affects ts demand so much more than the prces charged by each of the other frms such that we have γ, t /βt < (1/M) 1. When γt, /βt < 1 and the number of frms s arge, we expect ths assumpton to hod. For exampe, f we have a symmetrc settng, where the parameters reated to each frm are the same, then under the assumpton that γt, < βt, we have γt, /βt < 1/(n 1), n whch case, the ow nfuence regme naturay hods as the number of frms gets arge. In the ow nfuence regme, we show that the equbrum wthout recourse studed n the prevous secton s an ɛ-equbrum wth recourse. Intutvey speakng, ths resut uses the fact that f γ, t /βt s sma, then any devaton of a frm from a gven prce traectory has tte nfuence on the prces of the other frms n the subsequent tme perods. In the next emma, we formaze ths dea. Throughout the rest of ths secton, we use µ = max N, N\{}, t T γ, t /βt and β = max N, t T β t/ mn N, t T β t. Note that µ s expected to be sma n the ow nfuence regme. Lemma 4 Fxng the prces ˆp 1 charged by the frms at the frst tme perod, et the prces { ˆp t : t T \ {1}} form the equbrum wthout recourse n the remanng porton of the seng horzon. Defne the prces p 1 at the frst tme perod as p 1 = ˆp1 + δ and p1 = ˆp for a N \ {} for some δ 0. Fxng the prces p 1 charged by the frms at the frst tme perod, et the prces { p t : t T \ {1}} form the equbrum wthout recourse n the remanng porton of the seng horzon. If we have µ < (1/M) 1, then max, t T \{1} ˆp t pt 2 µ β δ 1 M Mµ. Proof. We consder the probem over the tme perods T \ {1} where the nventory for each frm at the second tme perod s gven by c D 1( p1 ). By the dscusson that foows Theorem 3, f we start wth any set of prces for the frms at the nta teraton and teratvey we compute the best 11

12 response of each frm to prces at the prevous teraton, then we reach the equbrum wthout recourse. Therefore, to compute { p t : t T \ {1}}, we consder the probem over the tme perods T \ {1} wth the nventory of each frm at the second tme perod gven by c D 1 ( p1 ) and startng wth the prces { ˆp t : t T \{1}} at the nta teraton, we teratvey compute the best response of each frm to the prces at the prevous teraton. Lettng { p t,k : t T \ {1}} be the prce traectores for the frms at teraton k, we know that m k p t,k = p t for a N, t T \ {1}. Therefore, for a t T \ {1} and N, we have ˆp t pt = ˆpt pt,1 + k=1 ( pt,k p t,k+1 ) k=1 pt,k p t,k+1, where the nequaty uses the fact that p t,1 = ˆp t. In ths case, to bound ˆp t pt, we can bound pt,k p t,k+1 for a k = 1, 2,... and add up the bounds on the atter quantty. We proceed to boundng p t,k p t,k+1. By defnton, the prce traectory { p t,k+1 : t T \ {1}} of frm at teraton k + 1 s the best response of frm to the prce traectores { p t,k : t T \ {1}} of the other frms at teraton k. In ths case, Theorem 3 mpes that p t,k+1 p t,k max t T \{1} { γt, pt,k p t,k 1 /β t } for a N, t T \ {1}, k = 2, 3,.... For notatona brevty, we et Φ k = max t T \{1} { p t,k+1 p t,k } so that the ast nequaty yeds Φ k max t T \{1} γt, Φk 1 /β t for a N, k = 2, 3,.... Usng the nequaty Φ k max t T \{1} for frm = wth and notng the defntons of M and µ, t foows that Φ k γt, Φk 1 /β t max t T \{1} { N\{,} γt, Φk 1 β t } + γt, Φk 1 { β t M max N\{,} Φ k 1 } + µ Φ k 1 (5) for a and k = 2, 3,.... Usng the nequaty Φ k max t T \{1} γt, Φk 1 /β t agan for frm =, we get Φ k max t T \{1} γt, Φk 1 /β t M max Φ k 1 for a k = 2, 3,.... If we use the ast nequaty n (5), then for a and k = 3, 4,..., we have Φ k M max N\{,}{Φ k 1 }+ Mµ max {Φ k 2 }. So, ettng Θ k = max Φ k, the ast nequaty yeds Θ k MΘ k 1 + Mµ Θ k 2 for a k = 3, 4,.... Addng the nequaty above over a k = 3, 4,..., we obtan k=3 Θk M k=2 Θk + Mµ k=1 Θk, whch s equvaent to k=1 Θk M k=2 Θk + Mµ k=1 Θk + Θ 1 + Θ 2 = (M + Mµ) k=1 Θk + (1 M) Θ 1 + Θ 2. Rearrangng the terms n the ast chan of nequates, we get k=1 Θk ((1 M) Θ 1 + Θ 2 )/(1 M Mµ). Therefore, f we can bound Θ 1 and Θ 2, then we can bound k=1 Θk. When we ncrease the prce of frm at the frst tme perod by δ, the nventory of frm at the second tme perod changes by at most β 1 δ and the nventory of frm at the second tme perod changes by at most γ1, δ. In the appendx, Lemma 8 shows that f we fx the prce traectores of the frms other than frm, then the best response of frm, when vewed as a functon of ts nta nventory, s Lpschtz wth constant 1/β mn, where we et β mn = mn N, t T β t. Note that the best response of frm does not depend on the nventores of the other frms, snce the prce traectores of the other frms s fxed. By defnton, f we consder the probem over the tme perods T \{1} wth the nventory of each frm at the second tme perod gven by c D 1( ˆp1 ), by defnton, { ˆp t : t T \ {1}} s the best response 12

13 to the prce traectores { ˆp t : t T \ {1}}. Aso, f we consder the probem over the tme perods T \{1} wth the nventory of each frm at the second tme perod gven by c D 1( p1 ), by defnton, { p t,2 : t T \ {1}} s the best response to the prce traectores { p t,1 : t T \ {1}}. Snce the prce traectores { ˆp t : t T \ {1}} and { pt,1 : t T \ {1}} are the same, Lemma 8 n the appendx mpes that p t,2 p t,1 = pt,2 ˆp t (c D 1( p1 )) (c D 1( ˆp1 )) /β mn for a t T \ {1}. As dscussed at the begnnng of ths paragraph, the expresson on the rght sde of the ast nequaty s bounded by β 1 δ when = and bounded by γ1, δ when = wth. Therefore, we obtan p t,2 p t,1 β 1δ/β mn β δ and p t,2 p t,1 γ1, δ/β mn β µ δ for a. The second one of the ast two nequates yeds Θ 1 = max, t T \{1} { p t,2 p t,1 } µ β δ. The frst one of the ast two nequates yeds Φ 1 = max t T \{1}{ p t,2 p t,1 } β δ, n whch case, notng (5), we get Θ 2 = max {Φ 2 } M Θ1 + µ Φ 1 Mµ β δ + µ β δ. Thus, Θ 1 and Θ 2 are respectvey bounded by µ β δ and (1 + M) µ β δ. In ths case, for a and t T \ {1}, we have ˆp t p t k=1 p t,k+1 p t,k max,t T \{1} k=1 (1 M) Θ1 + Θ 2 1 M Mµ { } p t,k+1 p t,k = k=1 Θ k (1 M) µ β δ + (1 + M) µ β δ 1 M Mµ 2 µ β δ 1 M Mµ, where the frst nequaty foows from the dscusson at the begnnng of the proof and the equaty s by the defnton of Θ k and Φ k. Consder the probem over the tme perods κ,..., τ when the nventores of the frms at tme perod κ are gven by x = (x 1,..., x n ). We use p N,t (κ, x) to denote the prce charged by frm at tme perod t n the equbrum wthout recourse. We consder the foowng strategy wth recourse for frm. If the nventores of the frms at tme perod t s gven by x, then frm charges the prce p N,t (t, x). In other words, ettng P R,t ( ) be the strategy functon of frm under ths strategy wth recourse, we have P R,t (x) = p N,t (t, x). Usng P R,t = (P R,t 1 ( ),..., Pn R,t ( )) to capture the strateges of a of the frms at tme perod t and c = (c 1,..., c n ) to denote the nventores of the frms at the frst tme perod, note that f a frms use the strateges {P R,t : t T } over the seng horzon, then the prce charged by each frm at each tme perod t s gven by p N,t (1, c), whch s precsey the prces correspondng to the equbrum wthout recourse when we consder the probem over the tme perods T wth the nventores of the frms at the frst tme perod gven by c. However, f one of the frms devate from the strateges {P R,t : t T } at a tme perod, then the prces charged by the frms w be dfferent from those n the equbrum wthout recourse. Therefore, t s not generay true that the strateges {P R,t : t T } correspond to an equbrum wth recourse. In the remander of ths secton, we show that the strateges {P R,t : t T } correspond to an ɛ-equbrum wthout recourse n the ow nfuence regme. In the next emma, we show that f frm unateray devates from the strategy {P R,t ( ) : t T }, but the other frms use the strateges {P R,t : t T }, then frm does not ncrease ts revenue by more than a smpe functon of µ. In the proof of the next emma, we make use of the fact that there s a natura upper bound on the prces that can be used by each frm. 13 As dscussed n Secton 2, we restrct

14 the strategy space of the frms such that α t βt pt + γt, pt 0 for a N. Notng the defnton of M, for a N, we get p t α t/βt + γt, pt /βt max N {α t/βt } + max N { γt, /βt } max N{p t } max N{α t/βt } + M max N{p t }, whch mpes that max N {p t } max N{α t/βt } + M max N{p t }. In ths case, we obtan the upper bound on the maxmum prce gven by max N {p t } max N{α t/βt }/(1 M). Lemma 5 Assume that the strateges of a of the frms are {P R,t : t T }. Let Π N be the revenue of frm under these strateges. Aso, assume that the strateges of the frms other than frm are {P R,t : t T }, but frm devates to charge an arbtrary prce at the frst tme perod and uses the strategy {P R,t ( ) : t T \ {1}} at the other tme perods. Let Π D be the revenue of frm under ths strategy. Lettng P max = max N {α t/βt }/(1 M) and β max = max N,t T β t, we have Π D Π N 2 β Mβ max Pmax 2 (τ 1) µ. 1 M Mµ Proof. We et ˆp t be the prce charged by frm at tme perod t n the equbrum wthout recourse. As dscussed rght before the emma, gven that a of the frms use the strategy {P R,t : t T }, the reazed prces are {ˆp t : N, t T }. We use ˆq1 to denote the arbtrary prce charged by frm at the frst tme perod. Gven that frm uses the strategy {P R,t ( ) : t T \{1}} at the other tme perods and the other frms use the strategy {P R,t : t T }, we et {ˆq t : N, t T } be the reazed prces. For each frm, note that ˆq 1 = P R,1 (c) = p N,1 (1, c) = ˆp 1. Aso, by Lemma 4, for a and t T \ {1}, we have ˆp t ˆqt 2 µ β ˆp 1 ˆq1 /(1 M Mµ) 2 µ β P max /(1 M Mµ). We use π t(pt, pt ) to denote the revenue of frm at tme perod t when frm charges the prce p t and the other frms charge the prce pt = (pt 1,..., pt 1, pt +1,..., pt n). We have Π N = t T πt (ˆpt, ˆpt ) and ΠD = t T πt (ˆqt, ˆqt ). In ths case, we get Π N = π(ˆp t t, ˆp t ) π(ˆq t, t ˆp t ) = Π D π(ˆq t, t ˆq ) t + t T t T t T \{1} t T \{1} π t (ˆq t, ˆp t ), where the nequaty uses the fact that {ˆp t : t T } s the best response of frm to the prces { ˆp t : t T } and the second equaty uses the fact that ˆq1 = ˆp1. Usng Dt (pt, pt ) to denote the demand of frm at tme perod t when frm charges the prce p t and the other frms charge the prces p t, by the nequaty above, we get ΠD Π N t T \{1} πt (ˆqt, ˆqt ) πt (ˆqt, ˆpt ) = t T \{1} ˆqt Dt (ˆqt, ˆqt ) ˆqt Dt (ˆqt, ˆpt ). Snce Dt (ˆqt, ˆqt ) Dt (ˆqt, ˆpt ) = (αt +βt ˆqt γt, ˆqt ) (α t + βt ˆqt γt, ˆpt ) = γt, (ˆpt ˆqt ), the ast chan of nequates yeds ΠD Π N t T \{1} ˆqt γt, ˆqt ˆpt t T \{1} P max Mβ t max ˆq t ˆpt, where we use the fact that ˆq t P max and M γt, /βt. At the begnnng of the proof, we show that ˆpt ˆqt 2 µ β P max /(1 M Mµ) for a and t T \ {1}. In ths case, we obtan Π D Π N t T \{1} P max Mβ t max { ˆq t ˆpt } = 2 βmβ max Pmax 2 (τ 1) µ/(1 M Mµ). In the next theorem, we show that the strategy {P R,t : t T } s an ɛ-equbrum wth recourse, when the number of frms s arge so that µ s expected to be sma. 14

15 Theorem 6 Assume that the strateges of a of the frms are {P R,t : t T }. Let Π N be the revenue of frm under these strateges. Aso, assume that the strategy of the frms other than frm are {P R,T : t T }, but frm uses an arbtrary strategy over the whoe seng horzon. Let Π A be the revenue of frm under these strateges. Lettng Γ µ = β Mβ max P 2 max/(1 M Mµ), we have Π A Π N Γ µ τ (τ 1) µ. Proof. Consder the probem over the tme perods κ,..., τ. We use Rev κ (P κ( ),..., P τ ( ), P, x) to denote the revenue of frm over the tme perods κ,..., τ, when the frm uses the strategy {P κ( ),..., P τ ( )}, the other frms use the strategy P and the nventores at tme perod κ are gven by x. We et {Q t ( ) : t T } be an arbtrary strategy used by frm. We use nducton over the tme perods to show that Rev κ (Q κ ( ),..., Qτ ( ), P R, x) Rev κ (P R,κ ( ),..., P R,τ ( ), P R, x) Γ µ (τ κ + 1) (τ κ) µ. In ths case, the resut foows by notng that Π A = Rev 1 (Q 1 ( ),..., Qτ ( ), P R, x), ΠN = Rev 1 (P R,1 ( ),..., P R,τ ( ), P R, x) and usng the ast nequaty wth κ = 1. Consder the case κ = τ. We have Rev τ (Q τ ( ), P R, x) Rev τ (P R,τ ( ), P R R,τ, x) 0, where the nequaty foows form the fact that P (x) s the best response of frm to the prces P R,t (x). Therefore the resut hods for κ = τ. Assumng that the resut hods for κ = t + 1, we show that the resut hods for κ = t. Usng D t (p t, pt ) to denote the vector of demands for the frms when the prces are (p t, pt ) and ettng x = x D t (Q t (x), P t (x)), observe that Rev t (Q t ( ), Qt+1 ( ),..., Q τ ( ), P R, x) Revt (Q t R,t+1 ( ), P ( ),..., P R,τ ( ), P R, x) = Rev t+1 (Q t+1 ( ),..., Q τ ( ), P R, x ) Rev t+1 (P R,t+1 ( ),..., R R,τ ( ), P R, x ), snce a frms make the same prcng decsons at tme perod t n the two revenue expressons on the eft sde of the equaty. By the nducton assumpton, the rght sde of the ast equaty s bounded by Γ µ (τ t) (τ t 1) µ. Aso, consderng Rev t (Q t R,t+1 ( ), P ( ),..., R R,τ ( ), P R, x) Rev t (P R,t ( ), P R,t+1 ( )..., P R,τ ( ), P R, x), ths expresson s the change n the revenue of frm when frm devates from the strategy {P R,t ( ) : t T } ony at the nta perod and there are τ t + 1 tme perods n the probem. By Lemma 5, ths expresson s bounded by 2 Γ µ (τ t) µ. In ths case, notng that Rev t (Q t ( ),..., Q τ ( ), P, R x) Rev t (P R,t ( ),..., P R,τ ( ), P, R x) = Rev t (Q t ( ), Q t+1 ( ),..., Q τ ( ), P, R x) Rev t (Q t ( ), P R,t+1 ( ),..., R R,τ ( ), P, R x) ( ),..., R R,τ ( ), P, R x) Rev t (P R,t ( ),..., P R,τ + Rev t (Q t ( ), P R,t+1 ( ), P R, x), the two dfferences on the rght sde above are bounded by Γ µ (τ t) (τ t + 1)µ and 2 Γ µ (τ t) µ. Snce Γ µ (τ t) (τ t 1)µ + 2 Γ µ (τ t) µ = Γ µ (τ t + 1) (τ t) µ, the resut hods for κ = t. We observe that as µ approaches zero, Γ µ τ (τ 1) µ approaches zero as we. Therefore, by the theorem above, f we are n the ow nfuence regme, then no frm can mprove ts revenue sgnfcanty by devatng from the pocy {P R,t : t T }, whch mpes that {P R,t : t T } s an ɛ-equbrum wth recourse. As dscussed earer, the prce traectory reazed under the strategy {P R,t : t T } s precsey the prce traectory of the equbrum wthout recourse. 15

16 5 Concusons We studed a compettve prcng probem. In equbrum wthout recourse, each frm commts to a prce traectory, whereas n equbrum wth recourse, each frm can adust ts prce at the current tme perod based on the nventores of a of the frms. Athough the demand s a determnstc functon of the prces so that there s no uncertanty n the responses of the frms, we showed that the two equbrum concepts can be qute dfferent. Whe the equbrum wthout recourse aways unquey exsts, the equbrum wth recourse may not exst or may not be unque. Our unqueness proof for the equbrum wthout recourse uses a contracton property. A natura research drecton s to extend such contracton propertes to demand modes other than the near demand mode. Aso, t woud be usefu to see whether an anaogue of equbrum wthout recourse can be defned under stochastc demand and check whether t unquey exsts. References Aon, G. and Federgruen, A. (2007), Competton n servce ndustres, Operatons Research 55(1), Boyd, S. and Vandenberghe, L. (2004), Convex optmzaton, Cambrdge unversty press. Gaego, G. and Hu, M. (2014), Dynamc prcng of pershabe assets under competton, Management Scence 60(5), Gaego, G., Huh, W. T., Kang, W. and Phps, R. (2006), Prce competton wth the attracton demand mode: Exstence of unque equbrum and ts stabty, Manufacturng & Servce Operatons Management 8(4), Gaego, G. and Wang, R. (2014), Mutproduct prce optmzaton and competton under the nested ogt mode wth product-dfferentated prce senstvtes, Operatons Research 62(2), Levn, Y., McG, J. and Nedak, M. (2009), Dynamc prcng n the presence of strategc consumers and ogopostc competton, Management scence 55(1), Ln, K. Y. and Sbdar, S. Y. (2009), Dynamc prce competton wth dscrete customer choces, European Journa of Operatona Research 197(3), Lu, Q. and Zhang, D. (2013), Dynamc prcng competton wth strategc customers under vertca product dfferentaton, Management Scence 59(1), Martnez-de Abenz, V. and Taur, K. (2011), Dynamc prce competton wth fxed capactes, Management Scence 57(6), Mgrom, P. and Roberts, J. (1990), Ratonazabty, earnng, and equbrum n games wth strategc compementartes, Econometrca: Journa of the Econometrc Socety pp Nazerzadeh, H. and Peraks, G. (2015), Non-near prcng competton wth prvate capacty nformaton, Soca Scence Eectronc Pubshng. Person, M. P., Aon, G. and Federgruen, A. (2013), Prce competton under mxed mutnoma ogt demand functons, Management Scence 59, 8. 16

17 A Appendx: Omtted Resuts In the foowng emma, we show eementary propertes for the functon G (, p ). We use these propertes throughout the paper. Lemma 7 For fxed p, et ν = nf{ν R + : T (ν, p ) = }. The functon G (, p ) satsfes the foowng propertes. (a) The functon G (ν, p ) s contnuous n ν (0, ). (b) The functon G (ν, p ) s strcty decreasng n ν [0, ν ) and constant n ν [ν, ) satsfyng G (ν, p ) = 2 c for a ν [ν, ). (c) There exsts unque ˆν [0, ) satsfyng G (ˆν, p ) = 0. Proof. Frst, we show Part a. Fx ν > 0 and ɛ > 0 sma enough that ν ɛ > 0. The defnton of T (ν, p ) mpes that T (ν ɛ, p ) T (ν, p ). Aso, f t T (ν ɛ, p ) \ T (ν, p ), then we have ν αt + γt, pt β t > ν ɛ, whch mpes that we have 0 α t βt ν + γt, pt > βt ɛ. For = T (ν ɛ, p ) and U = T \ T + so that T = notatona brevty, we et T + = T (ν, p ), T U. Notng the defnton of G (ν, p ), we have T + G (ν, p ) G (ν ɛ, p ) { = t T + ( α t β t ν + γ, t p t } { ) 2 c = ( α t β t ν + ) γ, t p t β t ɛ, t U t T + U t T ( α t β t (ν ɛ) + } ) γ, t p t 2 c Snce 0 α t βt ν + γt, pt > βt ɛ for a t U, the equaty above yeds t T + U β t ɛ G (ν, p ) G (ν ɛ, p ) t T + β t ɛ, so that G (ν, p ) s contnuous n ν (0, ). Second, we show Part b. Fx ν (0, ν ), n whch case, by the defnton of ν, we have T (ν, p ). In the proof of Part a, we show that G (ν, p ) G (ν ɛ, p ) t T (ν,p ) βt ɛ for a ɛ > 0 sma enough that ν ɛ > 0. Snce β t > 0 for a t T and T (ν, p ), the ast nequaty mpes that G (ν ɛ, p ) > G (ν, p ) for a ν (0, ν ) and ɛ > 0 sma enough that ν ɛ > 0. Aso, notng that α t > 0 and βt > 0, by the defnton of T (ν, p ), we have T (ɛ, p ) = T for sma enough ɛ > 0. In ths case, by the defnton of G (ν, p ), we obtan G (0, p ) t T (αt + γt, pt ) 2 c > t T (αt βt ɛ + γt, pt ) 2 c = G (ɛ, p ), whch mpes that G (0, p ) > G (ɛ, p ) for sma enough ɛ > 0. Therefore, we have G (ν ɛ, p ) > G (ν, p ) for a ν (0, ν ) and ɛ > 0 sma enough that ν ɛ > 0. Aso, we have G (0, p ) > G (ɛ, p ) for sma enough ɛ > 0. The ast two statements estabsh that G (ν, p ) s strcty decreasng n ν [0, ν ). Lasty, fx ν (ν, ). By the defnton of ν, we have T (ν, p ) =, n whch case, by the defnton of G (ν, p ), we obtan G (ν, p ) = 2 c. Snce G (ν, p ) = 2 c for a 17

18 ν (ν, ) and G (ν, p ) s contnuous n ν (0, ), t must be the case that G (ν, p ) = 2 c as we. Therefore, we have G (ν, p ) = 2 c for a ν [ν, ). Thrd, we show Part c. Assume that G (0, p ) > 0. Snce α t > 0 and β t > 0, we have T (0, p ) = T by the defnton of T (ν, p ). In ths case, by the defnton of G (ν, p ), we get G (0, p ) = t T (αt + γt, pt ) 2 c > 0. Smary, snce α t > 0 and β t > 0, we have T (ɛ, p ) = T for sma enough ɛ > 0. In ths case, by the defnton of G (ν, p ), we have G (ɛ, p ) = t T (αt βt ɛ + γt, pt ) 2 c. Therefore, we have m ɛ 0 G (ɛ, p ) = G (0, p ), ndcatng that G (ν, p ) s contnuous at ν = 0. Notng Part a, t foows that G (ν, p ) s contnuous n ν [0, ). Snce G (ν, p ) s strcty decreasng n ν [0, ν ) and G (v, p ) < 0 for a ν [ν, ) by Part b and G (ν, p ) s contnuous n ν [0, ) wth G (0, p ) > 0, there exsts unque ˆν such that G (ˆν, p ) = 0. Next, assume that G (0, p ) = 0. Ceary ˆν = 0 satsfes G (ˆν, p ) = 0. Aso, snce G (ν, p ) s strcty decreasng n ν [0, ν ) and constant at a negatve vaue for ν [ν, ) by Part b, there cannot be another ˆν such that G (ˆν, p ) = 0. In the next emma, we show that f we fx the prce traectores of the frms other than frm, then the best response of frm, when vewed as a functon of ts nta nventory, s Lpschtz. Lemma 8 Fx the prces p charged by the frms other than frm and et {p t (c ) : t T } be the optma souton to probem (1) as a functon of the nta nventory of frm. β mn = mn N, t T β t, for any two nta nventory eves ĉ and c, we have { } max p t (ĉ ) p t ( c ) 1 ĉ c. t T β mn Lettng Proof. Snce the prces p charged by the frms other than frm are fxed and we work wth two dfferent nta nventory eves, we drop the argument p from G (ν, p ) and make the dependence of G (ν, p ) on c expct. Thus, we use G (ν, c ) to denote G (ν, p ) throughout the proof. For notatona brevty, we et ˆp t = pt (ĉ ) and p t = pt ( c ). Notng Lemma 2, we et ˆv and ṽ be such that G (ˆv, ĉ ) = 0 and G (ṽ, c ) = 0. Wthout oss of generaty, we assume that ˆv ṽ. Snce G (ν, c ) s non-ncreasng n ν [0, ) by Lemma 7 and G (ṽ, c ) = 0, we have G (ˆv, c ) 0. Repeatng the same argument n the frst paragraph of the proof of Theorem 3, we aso get ˆp t pt 1 2 (ˆv ṽ ) for a t T. The ony dfference s that we have M = 0 n ths context snce the prces charged by the frms other than frm are fxed. If ˆv 2 ĉ c /β mn, then the ast nequaty mpes that ˆp t pt ĉ c /β mn, whch s the resut that we want to show! In the rest of the proof, we proceed under the assumpton that ˆv > 2 ĉ c /β mn. Notng that G (ˆv, ĉ ) = 0 and G (ν, ĉ ) < 0 for a ν [ν, ) by Lemma 7, the defnton of ν mpes that T (ˆv, p ). If, otherwse, T (ˆv, p ) =, then we obtan ν ˆv by the defnton of ν, whch contradcts the fact that G (ˆv, ĉ ) = 0, G (ν, ĉ ) < 0 and G (, ĉ ) s decreasng. Aso, snce ˆv > 2 ĉ c /β mn 0, by the defnton of G (ν, c ), we obtan G (ˆv, c ) G (ˆv, ĉ ) = 2 ( c ĉ ). Notng that G (ˆv, ĉ ) = 0, the ast equaty yeds G (ˆv, c ) = 2 ( c ĉ ). In the proof 18

19 of Part a of Lemma 7, we show that G (ν, c ) G (ν ɛ, c ) t T (ν,p ) βt ɛ for a ν (0, ) and ɛ > 0 sma enough that ν ɛ > 0. Usng ths nequaty wth ν = ˆv and c = c, we obtan G (ˆv ɛ, c ) G (ˆv, c )+ t T (ˆv,p ) βt ɛ. Usng the ast nequaty wth ɛ = 2 ĉ c /β mn, snce T (ˆv, p ) and G (ˆv, c ) = 2 ( c ĉ ), we get G ( ˆv 2 β mn ĉ c, c ) 2 ( c ĉ ) + β mn 2 β mn ĉ c 0 = G (ṽ, c ). Notng that G (ν, c ) s strcty decreasng n ν [0, ν ) and constant at a negatve vaue for ν [ν, ) by Part b of Lemma 7, havng G (ˆv 2 β mn ĉ c, c ) 0 = G (ṽ, c ) mpes that ˆv 2 β mn ĉ c ṽ. So, we obtan ˆp t pt 1 2 (ˆv ṽ ) 1 β mn ĉ c for a t T. 19