Appendx for An Effcent Ascendng-Bd Aucton for Mutpe Objects: Comment For Onne Pubcaton Norak Okamoto The foowng counterexampe shows that sncere bddng by a bdders s not aways an ex post perfect equbrum under a ratonng rues that satsfy the monotoncty property. Exampe 3 Consder a case where there are two bdders, A and B, and three quanttes of an object. Let u A, u B be the margna vaue functons of the two bdders such that 5 f q [0, 1 u A (q = u B (q = 1 f q [1, 3]. Consder the hstory h 4 = (x t A, xt B t=0,1,2,3 = ( (3, 3, (3, 3, (3, 3, (3, 3. After h 4, sncere bddng of each bdder s 1. The resut under sncere bddng x 4 A = 1 If the bdders report x 4 A = x4 B = 1 after h4, then the aucton ends at z 5 = (h 4, (1, 1, yedng an assgnment (x A, x B such that 1 x A 3, 1 x B 3, x A + x B = 3. Wthout oss of generaty, suppose that x A 3. Snce bdder A dd not cnch 2 at t 3, A s payment s ya = 4x A. Therefore, A s utty s U A(x A 4x A at z5. 1
The resut under msreportng ˆx 4 A = 0 If bdder A reports ˆx 4 A = 0, and bdder B reports x4 B = 1 after h4, then the aucton ends at ẑ 5 = (h 4, (0, 1, yedng an assgnment (ˆx A, ˆx B. Snce 0 = ˆx 4 A < x4 A = 1 and any other condton of ẑ5 s the same as z 5, by the monotoncty property, ˆx A must be strcty ess than x A. Smary to case wth sncere bddng, A s utty at ẑ 5 s U A (ˆx A 4ˆx A. We cacuate the dfference between A s uttes at z 5 and ẑ 5, ( U A (x A 4x A (U A (ˆx A 4ˆx A ( x ( A ˆxA = u A (qdq 4x A u A (qdq 4ˆx A 0 0 ( x A ˆxA = u A (qdq u A (qdq 4 (x A ˆx A 0 0 x A = u A (qdq 4 (x A ˆx A. (1 ˆx A Case 1: ˆx A 1. We cacuate (1 such that x A ˆx A 4 (x A ˆx A = 3 (x A ˆx A < 0. Case 2: ˆx A < 1. We cacuate (1 such that ( x A 1 + 5 (1 ˆx A 4 (x A ˆx A = 3x A ˆx A + 4 < 0 ( x A 3 2. Thus, A s utty at ẑ 5 s strcty greater than that at z 5, and bdder A has an ncentve to msreport after h 4. Therefore, sncere bddng by a bdder s not an ex post perfect equbrum. 2
Proof of Lemma 1 Snce u s a weaky decreasng nteger-vaued functon, there s a partton {a 0,..., a m } X wth 0 = a 0 < < a m = λ and vaues {b 1,..., b m } {0, 1,..., u} wth b 1 > b 2 > > b m such that for each k wth 1 k m, u (x = b k f a k 1 < x < a k. Note that m T. Consder any x X. Let By the defnton of Remann Integra, U (x = x 0 k = arg mn{a : x a }. k 1 u (qdq = b (a a 1 + b k (x a k 1. (2 =1 Take any p {1,..., T }. Defne b 0 = T + 1. Let By equaton (2, we can verfy that r = arg mn{b : p 1 < b }, r = arg mn{b : p b }. Because b Z for each, a r = mn{arg max x X U (x (p 1x }, a r = max{arg max x X U (x px }. {b : p 1 < b } = {b : p b }. Therefore a r = a r, that s, mn{arg max U (x (p 1x } = max{arg max U (x px }. x X x X 3
To prove Lemma 2 and Proposton 1, we expan some notaton and a property of the Ausube aucton. Notaton Wth fu bd nformaton, a strategy σ s a functon that maps each nontermna hstory h H \ Z to a quantty x X, that s, σ : H \ Z X. For each non-termna hstory h H \Z, the set of hstores n the subgame that foows h s gven by H h = {h H : h = (h, h for some sequence h }, and the set of termna hstores n the subgame s gven by Z h = Z H h. For each non-termna hstory h H \ Z and each strategy σ, we denote σ h : H h \ Z h X the nduced strategy n the subgame that foows h. For each h H h \ Z h, σ (h = σ h (h. Let π ( be the utty of bdder at an n-tupe of strateges. Property 1 For each t 1, f there exsts a bdder N such that x t = C t 1 and C t 1 > 0, then the aucton ends at t,.e., t = L. Therefore, f the aucton does not end at t, then for each bdder N, x t C t 1 Proof. Suppose that x t = C t 1 and C t 1 or C t 1 = 0. > 0. Then, x t = M j xt 1 j. By bddng constrant for each j N, x t j x t 1 j. Therefore j N xt j M. Note that ths property hods under a ratonng rues. We use the property n proofs of Lemma 2 and Proposton 1. 4
Proof of Lemma 2 Consder any t {0, 1,..., T }, h t = (x s 1, x s 2,..., x s n s t 1 H t \ Z t, and (u j j N. For each j N, et σ j be sncere bddng whch s correspondng to u j, and σj h t be nduced sncere bddng n the subgame that foows h t. Take any N and σ Σ h t. Suppose that x t 1 Q (p t 1. We sha show that Let π ((σ j h t j N π (σ, (σ j h t j. z L+1 = (x s 1, x s 2,..., x s n s L be the termna hstory whch s reached by (σ j h t j N, and w L +1 = (ˆx s 1, ˆx s 2,..., ˆx s n s L be the termna hstory whch s reached by (σ, (σj h t j. Denote {(Cj t j N } L t=0 the cumuatve cnches of z L+1, and {(Ĉt j j N } L t=0 the cumuatve cnches of w L +1. Step 1. x L 1 Q (p L 1. If L 1 = t 1, x L 1 = x t 1 Q (p t 1 = Q (p L 1. Then, et L 1 t. By the defnton of sncere bddng, x L 1 = σ h t By Property 1, x L 1 ( (x 1,..., x n L 2 = mn{x L 2, max{c L 2, Q (p L 1 }}. C L 2 Therefore, x L 1 Q (p L 1. or C L 2 = 0. Then, x L 1 = mn{x L 2, Q (p L 1 }. Step 2. For each j and s mn{l 1, L 1}, x s j = ˆx s j. Ths mpes that for each s mn{l 1, L 1}, C s = M j x s j = M j ˆx s j = Ĉs. 5
For each s t 1, obvousy x s j = ˆx s j. For the cases wth t s mn{l 1, L 1}, we sha show by nducton. Let s = t. Because x t j = σ j h t(h t and ˆx t j = σ j h t(h t, x t j = ˆx t j. Let s = k wth t + 1 k mn{l 1, L 1}. Suppose that x j = ˆx j for a wth t + 1 k 1. By the defnton of sncere bddng, ( x k j = σj h t (x 1,..., x n k 1 = mn{x k 1 j, max{c k 1 j, Q j (p k }}, ( ˆx k j = σj h t (ˆx 1,..., ˆx n k 1 = mn{ˆx k 1 j, max{ĉk 1 j, Q j (p k }}. Snce k mn{l 1, L 1}, by Property 1, x k j C k 1 j or C k 1 j = 0. Thus, x k j = mn{x k 1 j, Q j (k}. Smary, we have ˆx k j = mn{ˆx k 1 j, Q j (k}. Snce x k 1 j = ˆx k 1 j, x k j = ˆx k j. Step 3. π ((σ j h t j N π (σ, (σ j h t j. We consder three cases; L = L, L > L and L < L. Case 1. L = L. By step 2, for a s L 1 = L 1, C s = Ĉs. We cacuate C L and ĈL for two cases wth x L Q (p L and x L < Q (p L. Case 1-1. x L Q (p L. By step 1, x L 1 Therefore, by emma 1, Q L 1. Thus, Q (p L x L C L x L 1 Q (p L 1. mn{arg max x X (U (x p L x } C L max{arg max x X (U (x p L x }. Hence, π ((σ j h t j N π (σ, (σ j h t j. Case 1-2. Let x L < Q (p L. We sha show that x L = x t 1. By the defnton of sncere bddng, x L = σ h t((x s 1,..., x s n s L 1 = mn{x L 1, max{c L 1, Q (p L }}. 6
Snce x L < Q (p L, x L = x L 1. If L 1 = t 1, x L 1 t 1 L 1. By the defnton of sncere bddng, = x t 1. Then, we assume x L 1 = σ h t((x s 1,..., x s n s L 2 = mn{x L 2, max{c L 2, Q (p L 1 }}. Snce Q (p L Q (p L 1, x L 1 = x L < Q (p L 1. Hence, we have x L 1 = x L 2. By repeatng ths procedure, x L = x L 1 = = x t 1. Thus, C L = x t 1. Snce bdder cannot bd more quantty than x t 1 after h t, ĈL x t 1. Then, Ĉ L C L < Q (p L. Hence, π ((σ j h t j N π (σ, (σ j h t j. Case 2. L > L. By step 2, for each s L 1, C s = Ĉs. Then, we cacuate {C s } L s=l and Ĉ L. Snce the aucton does not end at L n the hstory z L+1, by Property 1 for each j, x L j C L 1 j or C L 1 j = 0. Then, by the defnton of sncere bddng, for each j, On the other hand, x L j = mn{x L 1 j, Q j (p L }. ˆx L j = mn{ˆx L 1 1 j, max{ĉl j, Q j (p L }}. Snce x L 1 j = ˆx L 1 j, x L j ˆx L j. Thus, Ĉ L M j ˆx L j M j x L j = C L. By the defnton of cumuatve cnches, for each s {L,..., L 1}, C s x s and x L C L x L 1. For each s {L,..., L 1}, because s t, x s s sncere bddng. That s, x s = mn{x s 1, max{c s 1, Q (p s }}. 7
Snce the aucton does not end at s L 1 n the hstory z L+1, by Property 1, x s = mn{x s 1, Q (p s }. Therefore, for each s {L,..., L 1}, x s Q (p s. Thus, C s Q (p s s {L,..., L 1}, Ĉ L C L Q (p L, C L x L 1 Q (p L 1 = max{arg max x X (U (x p L x }. Hence, π ((σ j h t j N π (σ, (σ j h t j. Case 3. L < L. We frst show that Q (p L x L. Suppose that Q (p L > x L. Smary to case 1-2, we have x L = x t 1. By bddng constrant, ˆx L x t 1. Then, ˆx L x L. Snce L < L, the aucton does not end at L n the hstory w L +1. Therefore, by Property 1, for each j, ˆx L ĈL or ĈL = 0. By the defnton of sncere bddng x L j ˆx L j = mn{x L 1 j, max{c L 1 j, Q j (p L }}, = mn{ˆx L 1 j, max{ĉl 1 j, Q j (p L }} = mn{ˆx L 1 j, Q j (p L }. For each j, snce by step 2, x L 1 j = ˆx L 1 j, we have x L j ˆx L j. Hence for each j N, x L j ˆx L j. Snce the aucton ends at L n the hstory z L+1, j N xl j M. Therefore, j N ˆxL j M. Ths mpes the aucton ends at L n the hstory w L +1. Ths contradcts to L < L. Thus, Q (p L x L. By step 2, for each s L 1, C s = Ĉs. Smary to case 2, we have C L ĈL. Because x L C L, Q (p L C L ĈL. Moreover, for each s L, Ĉ L Ĉs and Q (p s Q (p L. Thus, for each s L, Q (p s Ĉs. Hence, π ((σ j h t j N π (σ, (σ j h t j. 8
Proof of Proposton 1 Consder any t {0, 1,..., T }, h t = (x s 1, x s 2,..., x s n s t 1 H t \ Z t, and (u j j N. For each j N, et σ j be sncere bddng whch s correspondng to u j, and σ j h t be nduced sncere bddng n the subgame that foows h t. Take any N and σ Σ h t. We sha show that π ((σ j h t j N π (σ, (σ j h t j. If x t 1 Q (p t 1, we can show by Lemma 2. Suppose that x t 1 > Q (p t 1. Let z L+1 = (x s 1, x s 2,..., x s n s L be the termna hstory whch s reached by (σ j h t j N, and w L +1 = (ˆx s 1, ˆx s 2,..., ˆx s n s L be the termna hstory whch s reached by (σ, (σj h t j. Denote {(Cj t j N } L t=0 the cumuatve cnches of z L+1, and {(Ĉt j j N } L t=0 the cumuatve cnches of w L +1. We consder three cases; L > t, L > L = t and L = L = t. Case 1. L > t. Snce L 1 t, by the defnton of sncere bddng, x L 1 = σ h t By Property 1, x L 1 ( (x 1,..., x n L 2 = mn{x L 2, max{c L 2, Q (p L 1 }}. C L 2 or C L 2 = 0. Then, x L 1 = mn{x L 2, Q (p L 1 }. Therefore, x L 1 Q (p L 1, whch s the same argument as step 1 of Lemma 2. Note that we ony use the assumpton x t 1 Q (p t 1 n step 1 of Lemma 2. Thus, we can prove ths case smary to Lemma 2. Case 2. L > L = t. For each j N and each s t 1, obvousy x s j = ˆx s j. Therefore, for each 9
s t 1 = L 1, C s = Ĉs. We w cacuate C L and {Ĉs } L s=l. We frst show that Q (p L C L. By the defnton of sncere bddng, x L = mn{x L 1, max{c L 1, Q (p L }}. Snce x L 1 C L 1 and x L 1 > Q (p L 1 Q (p L, x L = max{c L 1, Q (p L }. Therefore, Q (p L x L. Because x L C L x L 1, Q (p L C L. Next we show that C L ĈL. For each j, because t = L, x L j = σ j h t(h t and ˆx L j = σ j h t(h t. Therefore, for each j, x L j = ˆx L j. Snce the aucton does not end at L n the hstory w L +1, Ĉ L = M j ˆx L j = M j x L j. On the other hand, snce the aucton ends at L n the hstory z L+1, C L M j x L j. Therefore, C L ĈL. Hence, Q (p L C L ĈL. Furthermore, for a s L + 1, Q (p s Ĉs, because Q (p s Q (p L and ĈL Ĉs. Thus, π ((σ j h t j N π (σ, (σ j h t j. Case 3. L = L = t. For each j N and each s t 1, obvousy x s j = ˆx s j. Furthermore, for each j, x L j = σj h t(h t = ˆx L j. Snce for each s L 1, C s = Ĉs, we cacuate C L and ĈL. Case 3-1. C L = x L. By the defnton of sncere bddng, x L = mn{x L 1, max{c L 1, Q (p L }}. 10
Snce x L 1 C L 1 and x L 1 > Q (p L 1 Q (p L, x L = max{c L 1, Q (p L }. If x L = Q (p L, then C L = Q (p L and we have Suppose that x L and C L 1 Ĉ L 1 π ((σ j h t j N π (σ, (σ j h t j. = C L 1. Then, C L 1 Q (p L and C L = C L 1. Snce ĈL = ĈL 1, Ĉ L C L. Therefore, Ĉ L C L Q (p L. Hence π ((σ j h t j N π (σ, (σ j h t j. Case 3-2. C L > x L. Frst, we show that for each j {1,..., 1}, Cj L = x L 1 j. Suppose that there exsts j {1,..., 1} such that Cj L x L 1 j. By the defnton of our ratonng rue, C L j = mn{x L 1 j, x L j + M = x L j + M j 1 x L k Ck L } k=j j 1 x L k Ck L. k=j k=1 k=1 Therefore, j M = x L k Ck L. k=j+1 k=1 Snce for each k N, x L k CL k, and k N CL k = M, j M = x L k Ck L = M. k=j+1 k=1 C L k k N Therefore, for each k j + 1, x L k = CL k. Because j + 1, ths contradcts to C L > x L. 11
Next, we show that C L ĈL. By the defnton of our ratonng rue, C L Ĉ L = mn{x L 1, x L + M 1 x L j Cj L } j= = mn{x L 1, ˆx L + M ˆx L j j=+1 j=1 1 x L j Ĉj L } j=1 = mn{x L 1, M = mn{x L 1, M j=+1 j=+1 1 x L j Cj L }, j=1 1 x L j Ĉj L }. j=1 For each j 1, snce x L 1 j = C L j and x L 1 j ĈL j, Therefore, mn{x L 1, M Hence, C L ĈL. C L j=+1 j=1 C L j ĈL j. 1 x L j Cj L } mn{x L 1, M j=+1 1 x L j Ĉj L }. Smary to case 2, we can show that Q (p L C L. Therefore, Q (p L ĈL. Thus, π ((σ j h t j N π (σ, (σ j h t j. j=1 12