Bayes-Nash Price of Anarchy for GSP Renato Paes Leme

Éva Tardos

Cornell University

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Auction Model b1

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b2

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b5

b6

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Auction Model b

b?

Auction Model Idea: Optimize against a distribution. b

b

b

b

b

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b

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b

b

b

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b

Bayes-Nash solution concept • Bayes-Nash models the uncertainty of other players about valuations • Values vi are independent random vars • Optimize against a distribution Goal: bound the Bayes-Nash Price of Anarchy

Bayes-Nash solution concept • Bayes-Nash models the uncertainty of other players about valuations • Values vi are independent random vars • Optimize against a distribution Thm: Bayes-Nash PoA ≤ 8

Model • n advertisers and n slots • vi ~ Vi (valuations distribution) • player i knows vi and Vj for j ≠ i • Strategy: bidding function bi(vi)

• Assumption: bi(vi) ≤ vi

Model v1 ~ V 1

b1(v1)

α1 v2 ~ V 2 v3 ~ V 3

b2(v2)

b3(v3)

α2 α3

Model

v2 ~ V 2

b2(v2)

α3

Model vi ~ V i

bi(vi)

σ = π-1

i = π(j)

Utility of player i :

αj j = σ(i)

ui(b) = ασ(i) ( vi - bπ(σ(i) + 1))

Model vi ~ V i

bi(vi)

αj

i = π(j)

j = σ(i)

Utility of player i :

ui(b) = ασ(i) ( vi - bπ(σ(i) + 1)) next highest bid

Model vi ~ V i

bi(vi)

αj

i = π(j)

j = σ(i)

Bayes-Nash equilibrium:

E[ui(bi,b-i)|vi] ≥ E[ui(b’i,b-i)|vi]

Model vi ~ V i

bi(vi)

αj

i = π(j)

j = σ(i)

Bayes-Nash equilibrium:

E[ui(bi,b-i)|vi] ≥ E[ui(b’i,b-i)|vi] Expectation over v-i

Bayes-Nash Equilibrium vi are random variables μ(i) = slot that player i occupies in Opt (also a random variable) Bayes-Nash PoA =

E[∑i vi αμ(i)] E[∑i vi ασ(i)]

Related results • [PL-Tardos 09] prove a bound of 1.618 for (full information) PoA of GSP. • [EOS] [Varian] analyze full information setting • [Gomes-Sweeney 09] study Bayes-Nash equilibria of GSP and characterize symmetric equilibria.

How was pure PoA proved? αj vπ(i) ≥ 1 + αi vπ(j)

v π(j)

αi

v π(i)

αj

• We need a structural characterization

How was pure PoA proved?

αjv π(j)+αivπ(i) ≥ αivπ(j)

v π(j)

αi

v π(i)

αj

• We need a structural characterization

New Structural Characterization Lemma: viE[ασ(i)|vi] + E[αμ(i) vπμ (i)|vi] ≥ ¼ viE[αμ(i)|vi]

New Structural Characterization Lemma: viE[ασ(i)|vi] + E[αμ(i) vπμ (i)|vi] ≥ ¼ viE[αμ(i)|vi]

E[∑i vi απ(i)] ≥ (1/8) E[∑i vi αμ(i)]

Main Theorem Lemma: viE[ασ(i)|vi] + E[αμ(i) vπμ (i)|vi] ≥ ¼ viE[αμ(i)|vi] Proof of main theorem: SW = (1/2) E[∑i αi vπ(i) + ασ(i) vi] = = (1/2) E[∑i αμ(i) vπ(μ(i)) + ασ(i) vi] = = (1/2) E[∑i E[αμ(i) vπ(μ(i)) |vi]+ vi E[ασ(i)|vi] ] ≥ (1/8) E[∑i vi αμ(i)]

New Structural Characterization Lemma: viE[ασ(i)|vi] + E[αμ(i) vπμ (i)|vi] ≥ ¼ viE[αμ(i)|vi] How to prove it ? • Find the right deviation. • But player i doesn’t know his true slot • Solution: try all slots

New Structural Characterization Lemma: viE[ασ(i)|vi] + E[αμ(i) vπμ (i)|vi] ≥ ¼ viE[αμ(i)|vi] How to prove it ? • • • •

Player i gets k or better if he bids > bπi(k) But this is a random variable … Deviation bid: 2 E[bπi(k) |vi, μ(i) = k] Gets slot k with ½ probability (Markov)

New Structural Characterization How to prove it ? • also gets slot j ≤ k whenever μ(i) = j : 2 E[bπi(k) |vi, μ(i) = k] decreases with k | (here we use independence) • Write Nash inequalities for those deviations: μ (i)

viE[ασ(i)|vi] ≥ Σj≥k ½ P(μ(i)=k|vi) αj (vi - Bk),

k

New Structural Characterization How to prove it ? • Smart Dual averaging the expression: viE[ασ(i)|vi] ≥ Σj≥k ½ P(μ(i)=k|vi) αj (vi - Bk) ,

k

• Maintain payments small and value large Structural characterization: viE[ασ(i)|vi] + E[αμ(i) vπμ (i)|vi] ≥ ¼ viE[αμ(i)|vi]

New Structural Characterization How to prove it ? • Dual averaging the expression:

viE[ασ(i)|vi] ≥ Σj≥k ½ P(μ(i)=k|vi) αj (vi - Bk) • Maintain payments small and value large

Not a smoothness proof.

Conclusion • Constant bound for Bayes-Nash PoA • Uniform bounds across all distributions

• Future directions: • Improve the constant • Get rid of independence

Sponsored Search Equilibria for Conservative ... - Cornell University

New Structural Characterization v i. E[α σ(i). |v i. ] + E[α μ(i) v πμ (i). |v i. ] ≥ ¼ v i. E[α μ(i). |v i. ] Lemma: • Player i gets k or better if he bids > b π i. (k). • But this is a random variable … • Deviation bid: 2 E[b π i. (k). |v i. , μ(i) = k]. • Gets slot k with ½ probability (Markov). How to prove it ?

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