Efficiency Levels in Sequential Auctions with Dynamic Arrivals Ron Lavi IE&M The Technion Ella Segev IE&M Ben-Gurion Univ. and

Efficiency Levels in Sequential Auctions with Dynamic Arrivals

Ron Lavi

IE&M

The Technion

Ella Segev

IE&M

Ben-Gurion Univ.

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.

1 2

vL

vH

v3

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• Strategies can be (“VAL”): all players

remain 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.

1 2

v1 = 3

v2 = 5v3 = 6

Do these strategies dominate all other?

• 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.

1 2

v1 = 3

v2 = 5v3 = 6

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. 1 2

v1 =

v2 = 1- v3 = 1

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 2

(1) Chooses number of players and arrival times:



1 2

(1) Chooses number of players and arrival times: r n-r



1 2

(2) Chooses a distribution F and draws values i.i.d.



1 2

(2) Chooses a distribution F and draws values i.i.d.



1 2

(3) Chooses an undominated strategy for each player (can depend on all actual values)



THM: For any such setting, E[V(A)] > 0.70711 E[V(OPT)]

1 2


Average-case analysis• In other words, for K=2, the adversary operates as follows:


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.

1 2


Average-case analysis• In other words, for K=2, the adversary operates as follows:


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.

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


Another Lemma: For any distribution F,

E[A] = 0 poly1(F(x))dx ; E[OPT] = 0

poly2(F(x))dx

0 [ j j,n,r·(F(x))j ] dx 0

[ j j,n,r·(F(x))j ] dx

= =

Proof of Lemma




poly2(F(x))dx

where the sum of coefficients of each polynomial is 0.

Proof of Lemma




poly2(F(x))dx


(1) EFp[A] = poly1(p):

x

Fp(x)

p

10

1

EFp[A] = 0 poly1(F(x))dx =

= 01 poly1(p)dx =

poly1(p)

Proof of Lemma




poly2(F(x))dx


(1) EFp[A] = poly1(p) ; EFp[OPT] = poly2(p)

=> poly1(p) - poly2(p) > 0 ( for any 0 < p < 1 )

Proof of Lemma




poly2(F(x))dx




(2) EF[A] - EF[OPT] > 0

< = > 0 [ poly1(F(x)) - poly2(F(x)) ] dx > 0

Proof of Lemma




poly2(F(x))dx




(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




poly2(F(x))dx




(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.

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Efficiency Levels in Sequential Auctions with Dynamic Arrivals Ron Lavi IE&M The Technion Ella Segev IE&M Ben-Gurion Univ. and