Presentation on theme: "Bayesian games and their use in auctions Adapted from notes by Vincent Conitzer."— Presentation transcript:
Bayesian games and their use in auctions Adapted from notes by Vincent Conitzer
What is mechanism design? In mechanism design, we get to design the game (or mechanism) –e.g. the rules of the auction, marketplace, election, … Goal is to obtain good outcomes when agents behave strategically (game-theoretically) Mechanism design is often considered part of game theory 2007 Nobel Prize in Economics! Before we get to mechanism design, first we need to know how to evaluate mechanisms
Which auction generates more revenue? Each bid depends on –bidders true valuation for the item (utility = valuation - payment), –bidders beliefs over what others will bid ( game theory), –and... the auction mechanism used In a first-price auction, it does not make sense to bid your true valuation –Even if you win, your utility will be 0… In a second-price auction, it always makes sense to bid your true valuation 0 bid 1: $10 bid 2: $5 bid 3: $1 Are there other auctions that perform better? How do we know when we have found the best one? 0 bid 1: $5 bid 2: $4 bid 3: $1 a likely outcome for the first-price mechanism a likely outcome for the second- price mechanism
Collusion in the Vickrey auction 0 b = highest bid among other bidders Example: two colluding bidders price colluder 1 would pay when colluders bid truthfully v 2 = second colluders true valuation v 1 = first colluders true valuation price colluder 1 would pay if colluder 2 does not bid gains to be distributed among colluders
The term Bayesian "Bayesian" refers to the mathematician and theologian Thomas Bayes (1702–1761), who provided the first mathematical treatment of a non-trivial problem of Bayesian inference. Bayesian probability is one of the different interpretations of the concept of probability and belongs to the category of evidential probabilities
The probability of H and D (joint probability) can be thought of in two ways: P(H,D) = P(H|D) * P(D) = P(D|H)*P(H)
Now, driving is not independent from gender. Given someone is a girl, what is the probability she doesnt drive? Given someone doesnt drive, what is the probability she is female?
Bayes theorem: The posterior probability is proportional to the likelihood of the observed data, multiplied by the prior probability. Thuslikelihoodprior probability P(H|D) = P(D|H)*P(H)/P(D) P(H) is the prior probability of H: the probability that H is correct before the data D are seen.prior probability P(D|H) is the conditional probability of seeing the data D given that the hypothesis H is true. This conditional probability is called the likelihood.conditional probabilitylikelihood P(D) is the probability of D. P(H|D) is the posterior probability: the probability that H is true, given the data and previous state of beliefs about H.posterior probability
Example Marie is getting married tomorrow. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 5% of the time. What is the probability that it will rain on the day of Marie's wedding? P(rain) = 5/365 P(not raining) = 360/365 P(weatherman predicting rain|rain) =.9 P(weatherman predicting rain|does NOT rain) =.05
What is the probability of rain, given the weatherman predicts rain? P( rain | predict rain) = P(rain) P( predict rain| rain ) / P(predict rain) P(predict rain) = [P( rain)* P( predict rain| rain ) + P( no rain)* P( predict rain| no rain ) ] P( rain | predict rain) = (0.014)(0.9) / [ (0.014)(0.9) + (0.986)(0.05) P( rain | predict rain)] = 0.2035
Bayesian games What if we didnt even know what game is being played? (Note that in imperfect information games, we didnt know prior moves, but we at least knew the payoffs) On the exam, the gift question – both players knew whether or not a gift had been given, but suppose they didnt have that knowledge? We say the type of agent determines the game we are playing
Expected Utility Ex post: utility based on all agents actual types Ex interim: agent knows his own type, but not that of others Ex ante: the agent does not know anybodys type
Bayesian games In a Bayesian game a players utility depends on that players type as well as the actions taken in the game –Notation: θ i is player is type, drawn according to some distribution from set of types Θ i –In this example, each player knows/learns its own type, not those of the others, before choosing action –Pure strategy s i is a mapping from Θ i to A i (where A i is is set of actions) –In general players can also receive signals about other players utilities. Assume all know the probabilities of being in each game. 2,00,2 2,0 U D LR row player type 1 row player type 2 2,20,0 1,1 U D LR 2,20,3 3,01,1 2,10,0 1,2 column player (.3)(.1) (.2)(.4)
Induced normal form (based on probabilities – no types are known) LLLRRLRR UU2, 11,.71, 1.20,.9 UD.8,.21, 1.1.4, 1.6, 1.9 DU1.5, 1.4.5, 1.11.7,.4.7,.1 DD.3,.6.5, 1.51.1, 0.21.3, 1.1 If we had a NE in this induced normal form, it would be called a Bayes NE.
Induced normal form (based on probabilities given player 1 knows his type is type 1) LLLRRLRR UU2, 0.51.5,.75.5, 20, 2.25 UD2, 0.51.5,.75.5, 20, 2.25 DU.75. 1.5.25, 1.752.25, 01.75,.25 DD.75. 1.5.25, 1.752.25, 01.75,.25 Since player 1 knows he is in top games, in left game ¾ time, right ¼.
Could rewrite with only two rows LLLRRLRR U2, 0.51.5,.75.5, 20, 2.25 D.75. 1.5.25, 1.752.25, 01.75,.25 Since player 1 knows he is in top games, in left game ¾ time, right ¼. Meaningless to find Nash equilibrium, as payoffs arent common knowledge
Bayes-Nash equilibrium A profile of strategies is a Bayes-Nash equilibrium if it is a Nash equilibrium for the normal form of the game –Minor caveat: each type should have >0 probability Alternative definition: for every i, for every type θ i, for every alternative action a i, we must have: Σ θ -i P(θ -i ) u i (θ i, σ i (θ i ), σ -i (θ -i )) Σ θ -i P(θ -i ) u i (θ i, a i, σ -i (θ -i ))
First-price sealed-bid auction BNE Suppose every bidder (independently) draws a valuation from [0, 1] What is a Bayes-Nash equilibrium for this? Say a bidder with value v i bids v i (n-1)/n Claim: this is an equilibrium! Proof of equilibrium: suppose all others use strategy For a bid b < (n-1)/n, the probability of winning is when the other n-1 bidders bid less than b. For a bidder to bid b, his valuation would be v i = bn/(n-1) Since he actually bids v i (n-1)/n = bn/(n-1)(n-1)/n = b Because valuations are uniformly distributed (and the range is 0-1) the probability of anyone bidding b or less, is (bn/(n-1)) And the probability of all n-1 others bidding b or less is (bn/(n-1)) n-1
First-price sealed-bid auction BNE For a bid b < (n-1)/n, the probability of winning is (bn/(n-1)) n-1, so the expected value is (v i -b)(bn/(n-1)) n-1 Derivative w.r.t. b is (using product rule) - (bn/(n-1)) n-1 + (v i -b)(n-1)b n-2 (n/(n-1)) n-1 which should equal zero - (bn/(n-1)) n-1 + (v i -b)(n-1)b n-2 (n/(n-1)) n-1 = 0 Dividing both sizes by b n-2 (n/(n-1)) n-1 Implies -b + (v i -b)(n-1) = 0, which solves to b = v i (n- 1)/n
Analyzing the expected revenue of the first-price auction First-price auction: probability of there not being a bid higher than b is (bn/(n-1)) n (for b < (n-1)/n) –This is the cumulative density function of the highest bid Probability density function is the derivative, that is, it is nb n-1 (n/(n-1)) n Expected value of highest bid is n(n/(n-1)) n (n-1)/n b n db = (n-1)/(n+1) Upper bound of integral chosen as no bid will be higher
Revenue equivalence theorem Suppose valuations for the single item are drawn i.i.d. from a continuous distribution over [L, H] (with no gaps), and agents are risk-neutral Then, any two auction mechanisms that –in equilibrium always allocate the item to the bidder with the highest valuation, and –give an agent with valuation L an expected utility of 0, will lead to the same expected revenue for the auctioneer
What if bidders are not risk-neutral? Behavior in second-price/English/Japanese does not change, but behavior in first-price/Dutch does Risk averse: first price/Dutch will get higher expected revenue than second price/Japanese/English Risk seeking: second price/Japanese/English will get higher expected revenue than first price/Dutch
With common valuations E.g. bidding on drilling rights for an oil field Each bidder i has its own geologists who do tests, based on which the bidder assesses an expected value v i of the field If you win, it is probably because the other bidders geologists tests turned out worse, and the oil field is not actually worth as much as you thought –The so-called winners curse –So if you have common valuation, what would you bid in a second price auction?
With common valuations E.g. bidding on drilling rights for an oil field Each bidder i has its own geologists who do tests, based on which the bidder assesses an expected value v i of the field If you win, it is probably because the other bidders geologists tests turned out worse, and the oil field is not actually worth as much as you thought –The so-called winners curse Hence, bidding v i is no longer a dominant strategy in the second-price auction In English and Japanese auctions, you can update your valuation based on other agents bids, so no longer equivalent to second-price In these settings, English (or Japanese) > second-price > first-price/Dutch in terms of revenue
Problem You and your siblings have been left with an estate consisting of many items The value of each item is correlated (partially common and partially private) How can you divide the property so that the division is fair?
Auctions without a seller faculty from a certain department in a university may want to decide who get to use a limited number of seminar rooms on a certain day, so that the department as a whole benefits the most; a group of housemates who collectively own a car may want to decide who gets to use it on a particular weekend, so that the one who needs it the most gets to use it. Agents are self interested and may lie about their valuation.
Caution We are programmed to think that auctions are about earning money. It may be that we are only trying to distribute goods fairly. In the case of dividing the estate, if there is no cost, people could be greedy or overstate their utility (given an inefficient allocation, right?). Childrens book: Instead of charging people for items, everything was given away. The logic was that people who made baseball bats loved to do it, so they were happy to give them away to someone who would enjoy it. Then, they could make more. I remember thinking this was a great idea.
no deficit: it is reasonable to assume that a mechanism shall not subsidize any allocation it comes up with, i.e. the total payments made by the agents to the center must always be non- negative a stronger version of the no deficit property called strong budget balance, which says the total payments made by the agents to the center must be zero (useful property if jointly owned) strategy proofness means `truthful in dominant strategy'.
VCG mechanism fails to have the strong budget balance property (as the auctioneer makes money), it is in fact impossible to find a mechanism that has this property and at the same time is truthful and efficient in dominant strategy Solutions? Redistribute as much of the VCG revenue as possible back to the agents Sacrifice truthfulness Sacrifice efficiency (social optimality)
Vickrey auction without a seller v( ) = 2 v() = 4 v( ) = 3 pays 3 (money wasted!) But if there is no cost, people wont be honest about their valuation
Can we redistribute the payment? v( ) = 2 v( ) = 4 v( ) = 3 pays 3 receives 1 Idea: give everyone 1/n of the payment not strategy-proof: A loser (bidding higher) increases his redistribution payment He gets 1/n of the value as determined by the auction- so he benefits by increasing the selling price
Incentive compatible redistribution [Bailey 97, Porter et al. 04, Cavallo 06] v() = 2 v( ) = 4 v() = 3 receives 1 Idea: give everyone 1/n of second-highest other bids Strategy-proof Your redistribution does not depend on your bid; incentives are the same as in Vickrey Who pays what?
Incentive compatible redistribution [Bailey 97, Porter et al. 04, Cavallo 06] v() = 2 v( ) = 4 v() = 3 pays 3 receives 1 receives 2/3 Idea: give everyone 1/n of second-highest other bids Strategy-proof Your redistribution does not depend on your bid; incentives are the same as in Vickrey 2/3 wasted (22%)
Bailey-Cavallo mechanism… Bids: V 1 V 2 V 3... V n 0 First run Vickrey auction Payment is V 2 by bidder 1 First two bidders receive V 3 /n Remaining bidders receive V 2 /n Total redistributed: 2V 3 /n+(n- 2)V 2 /n So why cant we give it all to the remaining bidders? R 1 = V 3 /n R 2 = V 3 /n R 3 = V 2 /n R 4 = V 2 /n... R n-1 = V 2 /n R n = V 2 /n
Another redistribution mechanism Bids: V 1 V 2 V 3 V 4... V n 0 First run Vickrey Redistribution: Receive 1/(n-2) * second- highest other bid, - 2/[(n-2)(n-3)] third-highest other bid Total redistributed: V 2 -6V 4 /[(n-2)(n-3)] R 1 = V 3 /(n-2) - 2/[(n-2)(n-3)]V 4 R 2 = V 3 /(n-2) - 2/[(n-2)(n-3)]V 4 R 3 = V 2 /(n-2) - 2/[(n-2)(n-3)]V 4 R 4 = V 2 /(n-2) - 2/[(n-2)(n-3)]V 3... R n-1 = V 2 /(n-2) - 2/[(n-2)(n-3)]V 3 R n = V 2 /(n-2) - 2/[(n-2)(n-3)]V 3 Idea pursued further in Guo & Conitzer 07 / Moulin 07 Key point is their return is never a function of their bid