OPERATIONS RESEARCH http://dxdoorg/0287/opre007ec e - c o m p a n o n ONLY AVAILABLE IN ELECTRONIC FORM 202 INFORMS Electronc Companon Generalzed Quantty Competton for Multple Products and Loss of Effcency by Jonathan Kluberg and Georga Peraks, Operatons Research, http://dxdoorg/0287/opre007
Appendx A Proof of Lemma Lemma In a market wth dfferentated substtute products, a sngle product per frm and separate capacty constrants for each product, colludng frms always sell less quantty of each product than f they compete freely: Proof To prove ths lemma we frst formulate the olgopoly problem OP under capacty constrants It can be wrtten as: max d d st { p B d 0 d C d where B denotes the row of matrx B correspondng to frm } Usng notaton Γ = dagb, the correspondng OP KKT condtons are: λ OP C = 0 p B Γ λ OP + µ OP = 0 λ OP 0 C d µ OP = 0 µ OP 0 0 Smlarly, we wrte down the monopoly problem MP under capacty constrants max d st d { p B d} 0 d C d The correspondng MP KKT condtons are: λ MP T C = 0 p 2B λ MP + µ MP = 0 λ MP 0 C d µ MP T = 0 µ MP 0 0 Step : We wll prove that µ OP = 0 Let us consder the problem that gnores the constrant 0 Ths suggests we gnore µ OP and the KKT condtons of problem OP become: p B + Γ λ OP = 0 or = B + Γ p λ OP
wth B + Γ beng an nverse M-Matrx see [] There are two cases to dstngush Ether λ OP > 0, n whch case: = C > 0 Or λ OP = 0, = B + Γ p λ OP = + }{{} p λ OP p p n λ OP n B + Γ B d = I + MΓ + p d > 0 Snce M s an M-matrx, so s I +MΓ see [] Hence I +MΓ has non-negatve elements, and the last nequalty follows from d > 0 Hence, t s always the case that that 0 As a result, µ OP = 0 0 even wthout ncludng ths constrant e the constrant Step 2: Smlarly, we now show that µ MP = 0 Followng a smlar thought process as before, we frst consder the problem that gnores µ MP s, gnores the constrant 0 Then the KKT condtons of problem MP become: that p 2B λ MP = 0 or = /2 M p λ MP Ether λ MP > 0, n whch case: = C > 0 Or λ MP = 0, = /2 M p λ MP = + }{{} p λ MP p p n λ MP n /2 M p /2 d > 0 Step 3: Characterzaton of Let K = {Set of actve constrants for the olgopoly problem} = { =,, n, λ OP > 0} We denote by K c the complement set of K and by H AB and u A the restrctons of matrx H and
vector u to rows ndexed by A and columns ndexed by B Snce K s the set of actve capacty d OP constrants for problem OP, K ck = = K c K c Snce µ OP = 0, the olgopoly KKT condtons become: p B + Γ λ OP = 0 Restrctng attenton to the set K c n off-dagonal block matrces: of nactve constrants λop K c = 0 and notng that Γ dsappears p K c B K c K c K B + Γ K c K c dop K c = 0 Usng the relaton p K c = B K c d, we get: B + Γ K c K c dop K c = B K c K d K + B K c K c B + Γ K c K c dop K c = B K c K d K c K + B K c K c d K c B K c K c K d K c 2 Clearly, on K we have: K = c K K Hence, to prove the lemma above, we only need to show: K c dmp K c Step 4: Characterzaton of Let K 2 = {Set of actve constrants for the monopoly problem} = { =,, n, λ MP > 0} We denote by K2 c the complement set of K 2 Snce K 2 s the set of actve capacty constrants for problem MP, = ck2 K c 2 Snce µ MP = 0, the monopoly KKT condtons become: p 2 B λ MP = 0 Restrctng attenton to the set K c 2 of nactve constrants λmp K c 2 = 0: p K c 2 2 B K c 2 = 0 3 Wthout loss of generalty, we now assume K 2 K and hence K2 c Kc If there were constrants n K 2 \ K, we smply remove them We show that wthout these constrants K c whch proves that capacty constrants cannot be actve on K c dop K c as they are not actve on dop K c
Restrctng further 3 to K c Kc 2 and splttng varables accordng to K K c, we get: ck2 p K c 2 B K c K Usng the relaton p K c = B K c d, we get: K \K 2 2 B K c K c dmp K c = B K c K d K + B K c K c 2 B K c K c dmp K c = B K c K 2 B K c K c d K 2 c K 2 2 K \K 2 dmp K c = 0 d K c 2 B K c K + B K c K c ck2 K \K 2 d K c 4 Step 5: As shown n, for all K c 2, dmp /2 d In partcurlar: 2 K \K 2 d K \K 2 c K \K 2 5 2 K c d K c 6 On the other hand, combnng 2 and 4, we have: B + Γ K c K c dop K c B K c K d K c K = 2 B K c K c B + Γ K c K c dop K c = 2 B K c K c dmp K c + B K c K B + Γ K c K c dop K c 2 B K c K c dmp K c B K ck d K 2 c K 2 2 2 ck2 c K2 2 K \K 2 c K \K 2 dmp K c 0 usng 5 K \K 2 7 Fnally, let s assume there exst K c K c, let s expand the -th row of 7: such that dop < Denotng {s,, s f } the ndces of b s 0 b sf K c }{{} d K c + 2 b OP d }{{} < b s 0 b sf 2 K c }{{} d K c + 2 b usng 6 Snce all the coeffcents b are non-negatve, ths s a contradcton We ust showed that K c dop K c, leadng to dmp
B Proof of Step for Theorem 3 Ignorng µ SP, the KKT condtons of problem SP become: p B d SP λ SP = 0 or d SP = M p λ SP Ether λ SP > 0, n whch case: d SP = C > 0 Or λ SP = 0, d SP = M p λ SP = + }{{} d SP d > 0 p λ SP p p n λ SP n M p C Calculatons for Theorem 4 In the unform case, matrx M can be wrtten as: α α α M = = + αi αh α α α + α nα 0 0 0 + α = 0 0 0 + α T Invertng M, we get matrx B: B = = = + α I α + α H [ I + + α + α [ I + α + α + α α + α nα H ] + α n + H ] Ths allows us to compute: Γ = dagb = + 2 α nα + α + α nα I
On the other hand, dagonalzng B as we dd wth M: +α nα 0 0 0 +α B = 0 0 0 +α We are now able to compute the dverse component of the surplus rato 2+3α nα +α 0 0 2+3α 2nα 0 +α nα I + MΓ = 0 2+3α 2nα 0 0 +α nα I + MΓ = T +α 2+3α nα 0 0 0 +α nα 2+3α 2nα 0 0 0 +α nα 2+3α 2nα Let s call d the vector whose components are the egenvectors of M, and [ ρ, ρ 2 ] the two egenvalues of: I + ΓM Γ I + MΓ ρ = +α+2α nα 2+3α nα 2 +α nα ρ 2 = +α nα+2α nα 2+3α 2nα 2 +α The rato of profts becomes: D Proof of Lemma ΠOP ΠMP = 4 ρ d 2 + ρ n 2 d =2 2 d +α nα 2 + n d +α =2 2 Lemma For a symmetrc nverse M-matrx B and a vector d wth all component postve, the followng nequalty holds: where r s the market power d 2 B + r nm d 2 B Bdag Proof Snce B s an nverse M-matrx, Ostrowsk shows n [3] that: T T B kl r kl B and B kl = B kl r Bkl kl
Introducng r = max kl r kl, we have: B kl r B Bkl kl Hence, we can wrte: d 2 B dt = d T r B r B Bkl B Bkl kl kl Bnm nm rb r r B Bkl kl rbnm nm B Bkl kl d d + dt rb 0 We denote the dagonal matrx correspondng to the dagonal of matrx B by: Γ = dagb,, B nm nm We obtan: d 2 B r d T Γ Snce H = Γ d + r d T Γd 0 rb nm nm d has two egenvalues 0 and nm, we have dt Hd nm d 2 for all d d 2 B r nm d T Γd + r d T Γd + r nm d 2 B Bdag E Dervaton of olgopoly varatonal nequalty At a Nash equlbrum soluton, the optmzaton problem facng a sngle frm s: B max d p d d B m 8 st d K
Ths problem s a maxmzaton of a concave obectve functon over a convex set, t s a convex problem A general convex problem of the form: max x st F x x K wth a concave obectve F x s equvalent see [2], [4] to the varatonal nequalty problem: Fnd x 0 K : F x 0 x x 0 0 x K Applyng ths to 8, we obtan for each frm : Fnd K : { p + B + B } T d 0 d K where B denotes the rows of matrx B correspondng to frm Now, snce the constrant set of each frm s ndependent of the quanttes chosen by other frms, t s equvalent to satsfy every one of these varatonal nequalttes for frm or to satsfy the sum of these nequaltes Clearly, f satsfes all these nequaltes t satsfes the sum of the nequaltes On the other hand f satsfes the sum of the nequaltes, by choosng d = d, for all d K, t s easy to check that t wll satsfy every varatonal nequalty separately as well The sum of these nequaltes s exactly the varatonal nequalty used n ths paper: Fnd K : { p + B + B Bdag } T d d OP 0 d K References [] C R Johnson, Inverse M-Matrces, Lnear Algebra and ts Applcatons, vol 47, 95-26, 982 [2] O Mancno and G Stampaccha, Convex programmng and varatonal nequaltes, Journal of Optmzaton Theory and Applcaton, vol 9, 3-23, 972 [3] A M Ostrowsk, Note on Bounds for Determnants wth Domnant Prncpal Dagonal, Proceedngs of the Amercan Mathematcal Socety, vol 3, No, pp 26-30, 952 [4] R M Rockafellar, Convex Functons, Monotone Operators, and Varatonal Inequaltes, Theory and Applcatons of Monotone Operators, Proceedngs of the NATO Advanced Study Insttute, Vence, Italy, 968 Edzon Oders, Gubbo, Italy, 968