Econometrica Supplementary Material

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1 Econometrica Supplementary Material SUPPLEMENT TO A DOUBLE-TRACK ADJUSTMENT PROCESS FOR DISCRETE MARKETS WITH SUBSTITUTES AND COMPLEMENTS Econometrica, Vol. 77, No. 3, May, 2009, BY NING SUN AND ZAIFU YANG PROOF OF LEMMA 1: Suppose that p is an equilibrium price vector. Then we know from Gul and Stacchetti 1999, Lemma 6 that for any efficient allocation π, p π constitutes an equilibrium. Clearly, i I ui π i = RN the market value of the objects. Furthermore, we have Lp = V i i I p β h N p = h i I ui π i β h π i p h β h N p h = RN. Note that for any p R n and i I, V i p u i π i p β h π i h.thusforanyp R n, we have Lp = i I V i p βh N p h i I u i π i = RN = Lp Hence, Lp = min p R n Lp, that is, p is a minimizer of the function L with Lp = RN. Suppose that ˆp is a minimizer of L with its value L ˆp = RN. Letρ be any efficient allocation of the model. We will show that ˆp ρ is an equilibrium. Clearly, we have V i ˆp u i ρi ˆp β h ρi h for every i I. We need to show that V i ˆp = u i ρi ˆp β h ρi h for every i I. Suppose to the contrary that V j ˆp > u j ρj ˆp β h ρj h for some bidder j. Adding the previous inequalities over all bidders leads to L ˆp > RN. This contradicts the hypothesis that ˆp is a minimizer of L with L ˆp = RN.Thus ˆp ρ must be an equilibrium. PROOF OF LEMMA 3: Let p = pt Z n. Observe that by Theorem 3i, Lp δ as a function of δ is continuous, generalized submodular, and convex on the set. So minimum is attained and its minimizers form a nonempty generalized lattice. Analogous to the proof of Theorem 3ii, one can further show that the set of minimizers is integrally convex and, consequently, both its smallest and largest elements are integer vectors. PROOF OF LEMMA 5: Sufficiency is obvious. Let us prove necessity. First, recall that Lemma 1 of Sun and Yang 2006 says a value function u i :2 N R satisfies the gross substitutes and complements GSC condition if and only if for any p R n,anyβ k S j for j = 1or2,anyδ 0, and any A D i p, there exists B D i p δek such that [A c S j ]\{β k } B c and [A S c j ] B. For any p R n and any A D i p, we consider the following three basic cases; the other cases can be proved in an analogously recursive way The Econometric Society DOI: /ECTA6514

2 2 N. SUN AND Z. YANG Case i, p = p δ k ek δ k ek, where the two different objects β k and β k are both in S j and δ k > 0 δ k > 0. By the definition of the GSC condition, there exists B D i p δ k ek such that [A S j ]\{β k } B and [A c S c] j B c.since p = p δ k ek δ k ek,forb D i p δ k ek, there is B D i p such that [B S j ]\{β k } B and [B c S c j ] Bc.Thuswehave[A S j ]\{β k β k } B and [A c S c j ] Bc, namely, {β x β x A S j and p x = p x } B and {β y β y A c S c and p j y = p y } B c Case ii, p = p δ l el δ l el, where the two different objects β l and β l are both in S c and δ j l > 0 δ l > 0. It follows from the above equivalent formulation of the GSC condition that there exists B D i p δ l el such that [A c S c]\{β j l} B c and [A S j ] B.Since p = p δ l el δ l el,for B D i p δ l el there is B D i p such that [B c S c]\{β j l } Bc and [B S j ] B. Thus we obtain that [A c S c]\{β j l β l } B c and [A S j ] B, namely, {β x β x A S j and p x = p x } B and {β y β y A c S c and p j y = p y } B c Case iii, p = p δ k ek δ l el, whereβ k S j, β l S c j,andδ k > 0, δ l > 0. By the definition of the GSC condition, there exists B D i p δ k ek such that [A S j ]\{β k } B and [A c S c j ] B c. Note that p = p δ k ek δ l el. Then it follows from the above equivalent formulation of the GSC condition that for B D i p δ k ek there is B D i p such that [B c S c j ]\{β l} B c and [B S j ] B. Sowehave[A c S c j ]\{β l} B c and [A S j ]\{β k } B, namely, {β x β x A S j and p x = p x } B and {β y β y A c S c and p j y = p y } B c PROOF OF LEMMA 6: Necessity is obvious. Let us prove sufficiency. Pick up any p R n and fix any A/ D i p, that is, any A for which V i p > v i A p. By continuity of V i and v i A, there exists ε>0 such that V i q > v i A q,whereq = p εea c εea. Then there exists B A such that it satisfies i or ii of Definition 3 and v i B q v i A q, which implies v i B p v i A p = v i B q v i A q [ A \ B B \ Aε >v i B q v i A q 0 To prove Theorem 2, we need to introduce an equivalent characterization of the generalized submodular function. This characterization can be easily

3 DOUBLE-TRACK ADJUSTMENT PROCESS 3 used to verify whether a function is a generalized submodular function or not. Property i is new but ii is a familiar property of the submodular function. LEMMA 7: A function f is a generalized submodular function if and only if the following statements hold: i For any x R n, any β k S j, any β l S c j, any δ k > 0, and δ l > 0, fx δ k ek δ l el fx δ l el fx δ k ek fx ii For any x R n, any distinct β k β l S j, any δ k > 0, and δ l > 0, fx δ k ek δ l el fx δ l el fx δ k ek fx PROOF: Suppose that f is a generalized submodular function. In the case of i, let p = x δ k ek and q = x δ l el. Thenp g q = x and p g q = xδ k ek δ l el. Clearly the part i conclusion holds. It is also easy to check the case of ii. Suppose that both i and ii hold. Take any p q R n. With respect to S 1 and S 2,let J S1 ={j p j >q j and β j S 1 } K S1 ={k p k <q k and β k S 1 } J S2 ={j p j >q j and β j S 2 } K S2 ={k p k <q k and β k S 2 } We consider the most general case, namely, all the above four sets are nonempty. So there exists a nonnegative vector δ = δ 1 δ n 0, such that p j = q j δ j for all j J S1 J S2 and p j = q j δ j for all j K S1 K S2.Let J S1 ={h 1 h s }, K S1 ={i 1 i t }, J S2 ={j 1 j u },andk S2 ={k 1 k v }. Then we have fp fp g q = fp f p = δ hl eh l δ kl ek l δ hr eh r δ hr eh r δ hr eh r δ kr ek r δ hr eh r δ kr ek r

4 4 N. SUN AND Z. YANG δ hr eh r δ i1 ei 1 δ hr eh r δ i1 ei 1 δ hr eh r δ kr ek r δ i1 ei 1 δ hr eh r δ kr ek r δ i1 ei 1 δ hr eh r δ ir ei r δ hr eh r δ ir ei r δ hr eh r δ kr ek r δ hr eh r δ kr ek r δ hr eh r δ hr eh r δ ir ei r δ ir ei r δ ir ei r δ j1 ej 1 δ ir ei r δ j1 ej 1 δ hr eh r δ kr ek r δ ir ei r δ j1 ej 1 δ hr eh r δ kr ek r

5 DOUBLE-TRACK ADJUSTMENT PROCESS 5 δ ir ei r δ j1 ej 1 u δ hr eh r δ ir ei r δ jr ej r = δ hr eh r δ ir ei r δ hr eh r δ kr ek r δ ir ei r u δ jr ej r δ hr eh r δ kr ek r δ ir ei r f p g q δ hr eh r f p g q f p g q u δ jr ej r δ hr eh r f p g q = fp g q fq δ hr eh r δ kr ek r δ hr eh r δ kr ek r u δ jr ej r Therefore we have fp g q fp g q fp fq. In the above derivation, the first two inequalities follow from case ii and the last two follow from case i.

6 6 N. SUN AND Z. YANG PROOF OF THEOREM 2 Necessity: Choose any two distinct items β k, β l N,anyp R n,anyδ k > 0, and any δ l > 0. If V i p V i p δ k ek = 0, the monotonicity of V i implies that V i p δ l el δ k ek V i p δ l el 0 = V i p δ k ek V i p and V i p δ l el δ k ek V i p δ l el 0 = V i p δ k ek V i p. We can now assume that V i p V i p δ k ek = ε k > 0. Then it follows that 0 <ε k δ k, V i p ε k ek = V i p δ k ek = V i p ε k, and there is a bundle A D i p and a bundle B D i p ε k ek for example, B = A withβ k A B. We need to consider the following two situations. Case i β l and β k are in the same set S j :WithregardtoA D i p and B D i p ε k ek, it follows from the GSC condition and β k A B that there are two bundles C D i p δ l el with β k C and D D i p δ l el ε k ek with β k D. Asaresult,wehave V i p δ l el δ k ek V i p δ l el V i p δ l el ε k ek V i p δ l el = v i D p δ l el ε k ek V i p δ l el = v i D p δ l el ε k V i p δ l el ε k = V i p δ k ek V i p Case ii β l and β k are not in the same set S j :WithregardtoA D i p and B D i p ε k ek, it follows from the GSC condition, Lemma 5, and β k A B that there are two bundles C D i p δ l el with β k C and D D i p δ l el ε k ek with β k D, whichleadsto V i p δ l el δ k ek V i p δ l el V i p δ l el ε k ek V i p δ l el = v i D p δ l el ε k ek V i p δ l el = v i D p δ l el ε k V i p δ l el ε k = V i p δ k ek V i p In summary, we see through Lemma 7 that V i is a generalized submodular function. Sufficiency: Suppose to the contrary that there are some p R n, β k S j, δ k > 0, and A D i p such that for every B D i p δ k ek we have [A S j ]\{β k } B or A c S c j Bc.Letε k = V i p V i p δ k ek. Clearly, 0 ε k δ k, V i p ε k ek = V i p δ k ek,anda D i p ε k ek.since A/ D i p δ k ek, it holds that D i p ε k ek D i p δ k ek and ε k < δ k.letq = p ε k ek and θ k = δ k ε k > 0. Then V i q = V i q θ k ek. Observe that A D i q and B/ D i q θ k ek for every bundle B satisfying [A S j ]\{β k } B and A c S c j Bc. This means that V i q θ k ek >

7 DOUBLE-TRACK ADJUSTMENT PROCESS 7 v i B qθ k ek for every bundle B satisfying [A S j ]\{β k } B and A c S c j B c. Furthermore, the continuity of V i and v i B implies that there exists a sufficiently small positive number θ so that V i q θ k ek θe[a S j ]\{β k } θea c S c j >v i B q θ k ek θe[a S j ]\{β k } θea c S c j for every bundle B satisfying [A S j ]\{β k } B and A c S c j Bc. This means that if B D i q θ k ek θe[a S j ]\{β k } θea c S c, then [A S j j]\ {β k } Bor A c S c j Bc. Then choosing a bundle B D i qθ k ek θe[a S j ]\{β k } θea c S c j yields V i q θ k ek θe[a S j ]\{β k } θea c S c j = v i B q θ k ek θe[a S j ]\{β k } θea c S c j = v i B q θ k ek p = v i B q θ k ek βk B p k = v i B q θ k ek B [A S j ]\{β k } θ B A c S c θ j <v i B q θ k ek [A S j ]\{β k }θ V i q θ k ek [A S j ]\{β k }θ where p = θe[a S j ]\{β k } θea c S c j. Therefore we have V i q θe[a S j ]\{β k } θea c S c j v i A q θe[a S j ]\{β k } θea c S c j = v i A q [A S j ]\{β k }θ = V i q [A S j ]\{β k }θ = V i q θ k ek [A S j ]\{β k }θ >V i q θ k ek θe[a S j ]\{β k } θea c S c j Let x = q and y = q θ k ek θe[a S j ]\{β k } θea c S c j. Then the above inequality leads to V i x g y V i x g y = V i q θe[a S j ]\{β k } θea c S c j V i q θ k ek

8 8 N. SUN AND Z. YANG >V i q V i q θ k ek θe[a S j ]\{β k } θea c S c j = V i x V i y contradicting the hypothesis that V i is a generalized submodular function. To expedite reading, we also give a proof for the following statement in the third paragraph of the proof of Theorem 5 in the Appendix of the paper. The argument here is the same as in the second paragraph of the proof for the Step 2 case of the global dynamic double-track GDDT procedure. STATEMENT: By the symmetry between Step 2 and Step 3, similarly we can also show that Lp g pt Lpt for all p R n. PROOF: To prove the statement, we first show that Lp Lpt for all p g pt. Suppose to the contrary that there exists some p g pt such that Lp < Lpt. By the convexity of L via Theorem 3i, there is a strict convex combination p of p and pt such that p {pt } and Lp <Lpt. Because of the symmetry between Step 2 and Step 3, Lemma 3 where is replaced by = and Step 3 of the GDDT procedure imply that Lpt δt = min δ Lpt δ = min δ Δ Lpt δ Lp <Lpt and so δt 0, contradicting the fact that the GDDT procedure stops in Step 3 with δt = 0. So we have Lp Lpt for all p g pt. Because p g pt g pt for all p R n, it follows that Lp g pt Lpt for all p R n. REFERENCES GUL,F., AND E. STACCHETTI 1999: Walrasian Equilibrium With Gross Substitutes, Journal of Economic Theory, 87, SUN, N., AND Z. YANG 2006: Equilibria and Indivisibilities: Gross Substitutes and Complements, Econometrica, 74, School of Economics, Shanghai University of Finance and Economics, Shanghai, China; nsun@mail.shufe.edu.cn and Faculty of Business Administration, Yokohama National University, Yokohama, Japan; yang@ynu.ac.jp. Manuscript received June, 2006; final revision received October, 2008.