Bayes-Nash Price of Anarchy for GSP Renato Paes Leme va Tardos - PowerPoint PPT Presentation

Bayes-Nash Price of Anarchy for GSP Renato Paes Leme Éva Tardos Cornell University

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Auction Model $$$ b 1 $$$ b 3 b 2 $$ b 4 b 5 $$ $ b 6 $

Auction Model b b ?

Auction Model Idea: Optimize against a distribution. b b b b b b b b b b b b b

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

Model • n advertisers and n slots • v i ~ V i (valuations distribution) • player i knows v i and V j for j ≠ i • Strategy: bidding function b i (v i ) • Assumption: b i (v i ) ≤ v i

Model v 1 ~ V 1 b 1 (v 1 ) α 1 v 2 ~ V 2 b 2 (v 2 ) α 2 α 3 v 3 ~ V 3 b 3 (v 3 )

Model v 2 ~ V 2 b 2 (v 2 ) α 3

Model σ = π -1 α j v i ~ V i b i (v i ) i = π (j) j = σ (i) Utility of player i : u i (b) = α σ (i) ( v i - b π ( σ (i) + 1) )

Model α j v i ~ V i b i (v i ) i = π (j) j = σ (i) Utility of player i : u i (b) = α σ (i) ( v i - b π ( σ (i) + 1) ) next highest bid

Model α j v i ~ V i b i (v i ) i = π (j) j = σ (i) Bayes-Nash equilibrium: E[u i (b i ,b -i )|v i ] ≥ E[ u i ( b’ i ,b -i )|v i ]

Model α j v i ~ V i b i (v i ) i = π (j) j = σ (i) Bayes-Nash equilibrium: E[u i (b i ,b -i )|v i ] ≥ E[u i ( b’ i ,b -i )|v i ] Expectation over v -i

Bayes-Nash Equilibrium v i are random variables μ( i) = slot that player i occupies in Opt (also a random variable) E[ ∑ i v i α μ (i) ] Bayes-Nash PoA = E[ ∑ i v i α σ (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? v π (i) α j v π (j) α i + ≥ 1 v π (j) α i v π (i) α j • We need a structural characterization

How was pure PoA proved? v π (j) α i α j v π (j) + α i v π (i) ≥ α i v π (j) v π (i) α j • We need a structural characterization

New Structural Characterization Lemma: 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 ]

New Structural Characterization Lemma: 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 ] E[ ∑ i v i α π (i) ] ≥ (1/8) E[ ∑ i v i α μ (i) ]

Main Theorem Lemma: 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 ] Proof of main theorem: SW = (1/2) E[∑ i α i v π (i) + α σ (i) v i ] = = (1/2) E[∑ i α μ( i) v π ( μ( i)) + α σ (i) v i ] = = (1/2) E[∑ i E[ α μ( i) v π ( μ( i)) |v i ]+ v i E[ α σ (i) |v i ] ] ≥ (1/8) E[∑ i v i α μ (i) ]

New Structural Characterization Lemma: 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 ] 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: 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 ] How to prove it ? • 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 (k) |v i , μ (i) = k] i • 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[b π (k) |v i , μ (i) = k] decreases with k i (here we use independence) μ (i) | • Write Nash inequalities for those deviations: v i E[ α σ (i) |v i ] ≥ Σ j ≥k ½ P(μ (i)=k|v i ) α j (v i - B k ), k

New Structural Characterization How to prove it ? • Smart Dual averaging the expression: v i E[ α σ (i) |v i ] ≥ Σ j ≥k ½ P(μ (i)=k|v i ) α j (v i - B k ) , k • Maintain payments small and value large 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 ]

New Structural Characterization How to prove it ? • Dual averaging the expression: v i E[ α σ (i) |v i ] ≥ Σ j ≥k ½ P(μ (i)=k|v i ) α j (v i - B k ) • 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