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Efficiency Levels in Sequential Auctions with Dynamic Arrivals. Ella Segev IE&M Ben-Gurion Univ. Ron Lavi IE&M The Technion. and. Sequential Auctions. K identical items. Each one is sold in a separate English auction. Auctions are conducted one after the other.
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Efficiency Levels in Sequential Auctions with Dynamic Arrivals Ella Segev IE&M Ben-Gurion Univ. Ron Lavi IE&M The Technion and
Sequential Auctions • K identical items. • Each one is sold in a separate English auction. • Auctions are conducted one after the other. • Buyers have private-values and unit-demand. • (Buyers do not discount time) • Milgrom and Weber (1983/2000) first studied this model, describing Bayesian-Nash equilibrium strategies, if each auction is either a first-price or a second-price auction. • In equilibrium, full efficiency is obtained, i.e. the bidders with the K highest values are the winners.
Dynamic Arrivals • In early models: All players are present from the beginning. • More natural: New players (may) join in every auction. • General models of “dynamic mechanism design” were studied recently by Athey and Segal (2007); Bergemann and Valimaki (2007); Cavallo, Parkes, and Singh (2007). • Said (2008) explicitly demonstrates the connection of Bergemann and Valimaki to sequential ascending auctions.
Example • Two items, two players present at the first auction, a third player with a high value might join the second auction. • If the higher player at auction 1under-estimates the probabilitythat player 3 will join in auction 2she will lose the first auction, andthe result will not be efficient. • Previous works recovered full efficiency by either assuminga common prior or by using other techniques that are not natural in the setting of sequential auctions. v3 vH vL 1 2
Motivation and Goals • We argue that some loss of efficiency is inherent in this dynamic setting, a real phenomena that we wish to highlight and analyze. • Our goal: to quantify how much efficiency (social welfare) is lost in the standard sequential auction mechanism. • Instead of trying to obtain equilibrium strategies, we analyze the set of undominated strategies. Thus, we do not assume: • Common priors • Risk neutrality • Complex rationality assumption needed to reason about the strategic foundations of equilibrium behavior. • This is in the spirit of Wilson’s critique (1987) and the recent suggestions for “robust” analysis by Bergemann & Morris (2005).
Our results (1)Analyze the set of undominated strategies. • We need to include a simple “activity rule” to obtain this. (2)Any tuple of undominated strategies results in a social welfare of at least half of the optimal social welfare. • Worst-case bound: adversary sets players’ values, arrival times, and their (undominated) strategies. (3)For K=2, expected social welfare is at least 70% of expected optimal welfare. • Adversary draws players’ values independently from a fixed distribution. Other parameters are set adversarially. • Bound for uniform distribution is at least 80%
Related work in CS • Such models are termed “online auction” in CS. (Lavi & Nisan 2004). • Usually: design dominant-strategy mechanisms with good approximations, for example: • Hajiaghayi et al. (2005): prices are charged only after the last auction, dominant-strategies with 2-approx. • Cole et al. (2008): designer can direct bidders to one specific auction, dominant-strategies with 2-approx. • Here, in contrast, we analyze an existing (popular) auction format, that has no dominant strategies, using similar reasoning. • Close in spirit to Lavi and Nisan (2005): Study a more general model, give weaker results: • Weaker game theoretic analysis. • Weaker bounds on the efficiency loss.
Outline for rest of the talk • Technical details: • Model of the auction, activity rule, undominated strategies • worst-case analysis • average-case analysis • Summary
Description of auction • In each period an ascending auction: • price clock continuously ascends • players drop one at a time • last to remain wins • pays last price. • Extensive form game, a strategy is a function from the history to drop/stay decision. • “Stopping the clock” assumption like in Ausubel (2004) to allow cascade of drops at the same price.
Example v3 = 6 v2 = 5 • Strategies can be (“VAL”): all playersremain until value in both auctions.Result: highest player wins first auctionand pays 5, second highest wins secondauction and pays 3. • Strategies can be (“ED”): in first auction, all players remain until value or until exactly one other player remains (the earliest event). In second auction they remain until value.Result: two highest players win, both pay 3. • Which strategy should the highest player choose? • VAL if she believes price for 2nd auction > 5. • ED if she believes price for 2nd auction < 5. v1 = 3 1 2
Do these strategies dominate all other? v3 = 6 v2 = 5 • No. They do not dominate the strategy“BAD”: in first auction drop at zero,in second auction remain until value. • A counter-example forthe ED strategy of player 3: • Strategies of the others: • Player 1: remain until value in both auctions. • Player 2: In first auction, if player 3 drops atzero, remain until value, otherwise drop immediatelyafter zero. In second auction remain until value. • If player 3 plays ED: she wins second auction, pays 5. • If player 3 plays BAD: she wins second auction, pays 3. v1 = 3 1 2
Solution • An activity rule: • Only the K-t+1 highest bidders of auction t are qualified to continue to auction t+1 (plus the new arrivals in auction t+1). • The price at auction t+1 starts from the level where there remained K-t+1 bidders in auction t. • For example, K=2, t=1 (K-t+1=2): only the two highest bidders of auction 1 can continue to auction 2. Proposition: In any undominated strategy, while more than K-t+1 players remain in auction t, a player drops if and only if price is equal to value. Corollary: Under undominated strategies, in every auction t, the values of the qualified bidders are the K-t+1highest values among all non-winners that arrived at or before auction t.
Remarks • Activity rules are becoming popular both in theory and in practice. Although a brute-force solution to the technical problem, it demonstrates the usefulness of such rules. • A very similar activity rule that is used in practice is “indicative bidding” (Ye 2007; GEB): players place bids and the few highest bids continue to a qualifying round. • this was recently used e.g. to auction assets in the electricity market in the US (Central Maine Power, Pacific Gas and Electric, Portland General Electric…)
Worst-case efficiency loss for K=2 • Easy implication of undominated strategies: the highest value player must win. • Simple analysis: Let V(A) denote the resulting social welfare of the two auctions, V(OPT) denote the maximal possible social welfare, VH denote the maximal value (in both auctions).We get: V(A) / V(OPT) > VH / (2 VH) = 1/2
Worst-case efficiency loss for K=2 • Easy implication of undominated strategies: the highest value player must win. • Simple analysis: Let V(A) denote the resulting social welfare of the two auctions, V(OPT) denote the maximal possible social welfare, VH denote the maximal value (in both auctions).We get: V(A) / V(OPT) > VH / (2 VH) = 1/2 • This is tight:Possible result ofundominated strategies:Players 1 and 3 winplayer 2 loses. v3 = 1 v2 = 1- v1 = 1 2
Analysis of worst-case efficiency loss • Let V1(OPT) > V2(OPT) > … > VK(OPT) be the values of the winners in the efficient allocation.Define in a similar way V1(A) > V2(A) > … > VK(A). Lemma: For any index 0 < L <K/2, VL+1(A) > V2L+1(OPT) In other words: V1(A) > V1(OPT) > V2(OPT) (L=0) V2(A) > V3(OPT) > V4(OPT) (L=1) … and so on … => V(OPT) < 2 [V1(A) + … + VK/2(A)] < 2V(A) => V(A) / V(OPT) > 1/2
Suggestion for an average-case analysis • Common sense suggests: if worst-case bound is 50%, most cases are much better. How to make this more rigorous? • Implicit in the worst-case analysis: a powerful adversary sets • Number of players • Arrival times • Players’ values • Players’ strategies (restricted to be undominated strategies).
Suggestion for an average-case analysis • Common sense suggests: if worst-case bound is 50%, most cases are much better. How to make this more rigorous? • Implicit in the worst-case analysis: a powerful adversary sets: • Number of players • Arrival times • Players’ values • Players’ strategies (restricted to be undominated strategies). Fixes any distribution, draws players’ value i.i.d. from it (a more restrictive adversary).
Average-case analysis • In other words, for K=2, the adversary operates as follows: (1) Chooses number of players and arrival times: 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: (1) Chooses number of players and arrival times: r n-r 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: (2) Chooses a distribution F and draws values i.i.d. 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: (2) Chooses a distribution F and draws values i.i.d. 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: (3) Chooses an undominated strategy for each player (can depend on all actual values) 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: THM: For any such setting, E[V(A)] > 0.70711 E[V(OPT)] (3) Chooses an undominated strategy for each player (can depend on all actual values) 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: THM: For any such setting, E[V(A)] > 0.70711 E[V(OPT)] Worst setup: • Distribution: Pr(v = 0) = 2-2 ; Pr(v = 1) = 1 - Pr(v = 0) • Two players at auction 1, infinite number of players at auction 2. (3) Chooses an undominated strategy for each player (can depend on all actual values) 1 2
Average-case analysis • In other words, for K=2, the adversary operates as follows: THM: For any such setting, E[V(A)] > 0.70711 E[V(OPT)] Remarks: • For other distributions the bound is higher, e.g. at least 80% for a uniform distribution on any interval. • Obtaining the other parameters in a distributional way will most likely increase the ratio even more. (3) Chooses an undominated strategy for each player (can depend on all actual values) 1 2
Remarks • Recall that the worst-case scenario also required only two values, and 1. • However the worst-case scenario required only three players, while here we need many players. The reason: • The gap between OPT and A results only from the event A=1 and OPT=2. • This happens if exactly one player at auction 1 has value 1, and at auction 2 there is at least one other player with value 1. • The probability for this increases as the number of players in auction 2 increases, and the number of players is auction 1 decreases.
Proof of theorem • Two random variables: • OPT = max value at time 1 + max remaining value • A = second-highest value at time 1 + max remaining value • Bernoulli distribution Fp (Pr(v=0)=p, Pr(v=1)=1-p): • OPT, A {0,1,2} => easy to compute explicit formula for EFp[A], EFp[OPT]. We then analytically find worst n,r,p. Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >.
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx 0 [ jj,n,r·(F(x))j ] dx 0 [ jj,n,r·(F(x))j ] dx = =
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx where the sum of coefficients of each polynomial is 0.
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx where the sum of coefficients of each polynomial is 0. (1) EFp[A] = poly1(p): EFp[A] = 0 poly1(F(x))dx = =01 poly1(p)dx = poly1(p) Fp(x) 1 p x 0 1
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx where the sum of coefficients of each polynomial is 0. (1) EFp[A] = poly1(p) ; EFp[OPT] = poly2(p) => poly1(p) - poly2(p) >0 ( for any 0 < p < 1 )
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx where the sum of coefficients of each polynomial is 0. (1) EFp[A] = poly1(p) ; EFp[OPT] = poly2(p) => poly1(p) - poly2(p) >0 ( for any 0 < p < 1 ) (2) EF[A] - EF[OPT] >0 < = > 0 [ poly1(F(x)) - poly2(F(x)) ] dx >0
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx where the sum of coefficients of each polynomial is 0. (1) EFp[A] = poly1(p) ; EFp[OPT] = poly2(p) => poly1(p) - poly2(p) >0 ( for any 0 < p < 1 ) (2) EF[A] - EF[OPT] >0 < = > 0 [ poly1(F(x)) - poly2(F(x)) ] dx >0 < = x, poly1(F(x)) - poly2(F(x)) >0
Proof of Lemma Lemma: Fix some s.t. for any n,r,p, EFp[A] / EFp[OPT] >. Then for any n,r,F, EF[A] / EF[OPT] >. Another Lemma: For any distribution F, E[A] = 0 poly1(F(x))dx ; E[OPT] = 0 poly2(F(x))dx where the sum of coefficients of each polynomial is 0. (1) EFp[A] = poly1(p) ; EFp[OPT] = poly2(p) => poly1(p) - poly2(p) >0 ( for any 0 < p < 1 ) (2) EF[A] - EF[OPT] >0 < = > 0 [ poly1(F(x)) - poly2(F(x)) ] dx >0 < = x, poly1(F(x)) - poly2(F(x)) >0 which follows from (1) (by taking p = F(x)).
Summary • Analyzed the popular sequential auction mechanism, under a no-common-priors assumption. • Characterized the set of undominated strategies • Worst-case efficiency loss is 50% • Average-case loss for two items > 70% • This is different than most of the auction theory in two aspects: • No equilibrium analysis -- following Battigalli and Siniscalchi (2003), Dekel and Wolinsky (2003), Cho (2005). • A quantitative rather than a dichotomous judgment of efficiency.