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Course administration

Course administration. Tirgul : will be held every only after exercise submission. Teaching Assistant : Liri Finkelstein, lirifink@gmail.com Exercises : four exercises. Each one is 8% of the final grade.

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Course administration

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  1. Course administration • Tirgul: will be held every only after exercise submission. • Teaching Assistant: Liri Finkelstein, lirifink@gmail.com • Exercises: four exercises. Each one is 8% of the final grade. • Rest of grade: a final homework. Four hard exercises. To be submitted by the end of the semester. • Office hours: Flexible, email me (ronlavi@ie.technion.ac.il) • Course website:http://ie.technion.ac.il/~ronlavi/auction-theory/ • I will put all lecture slides there, plus different announcements and supporting material.

  2. What is Auction Theory? • Auctions are becoming more and more popular • The Internet only helps. • We see different auction formats, different rules. • Can we say anything about the difference, what will be the result of one format vs. another. • Examples: Ebay has proxies, different sites has different end rules, etc. • How to analyze this? We use Game Theory, that models strategic decisions of players. • A mathematical analysis, we will see one or two examples, but the main focus is on the mathematical modeling and techniques.

  3. Course Overview • Basic game-theoretic notions, The basic auction model, description and analysis of classic one-item auctions • Revenue-maximizing auctions • Auctions with interdependent values • Auctions for multiple identical goods • Combinatorial Auctions (if time permits) No special knowledge is required, I only assume basic math. Skills So don’t hesitate to stop me and ask any question whatsoever.

  4. First topic: analysis of classic one-item auction types • Basic game-theory notions: normal-form games, dominant strategies, games in incomplete information, Nash equilibrium. Bayesian-Nash equilibrium. • Analysis and comparison of four auction types: • First-price, second-price, English, Dutch • Efficiency and revenue.

  5. A game (in complete information) Player II’s actions Action “L” Action “R” Here’re the utilities if player I plays T and II plays L Action “T” Player I’s actions 5 , -3 Action “B” If player I plays B and II plays R then player I gets 5 and II gets -3

  6. Formalism (a “normal-form” game) • A game is composed of: • An action space Ai for each player i. • A utility function ui : A1 x …x An -> R • This modeling implies that players choose actions simultaneously.

  7. Example: Prisoner’s dilemma • The story: Two partners wish to terminate the partnership. Each one needs to hire a lawyer, to reach a settlement on how to split their property. Total property is worth 10. • Two types of lawyers: cheap costs 1, expensive costs 5. • If both choose same type, property is split equally. • Otherwise the one with the expensive lawyer gets everything. • So, the game is: “C” “E” 4 , 4 -1 , 5 “C” “E” 5 , -1 0 , 0

  8. What will happen in this game? DFN: Action A dominates action B (for player i) if for any combination of actions of the other players, a-i, ui(A,a-i) > ui(B,a-i).Action A is a dominant action if it dominates all other actions. • Observation:“E” is a dominant strategy. • Assumption: Players indeedplay a dominant strategy ifthey have one. “C” “E” 4 , 4 -1 , 5 “C” “E” 5 , -1 0 , 0

  9. A one-item auction The story: • A seller wishes to sell a good to n players. Player i will obtain a value of vi > 0 from having the good. vi is known only to i. • The seller can charge a payment pi from player i.In this case player i’s utility is: vi – pi A popular auction format (“English auction”): • The seller starts from a very low price. All players that are interested in buying the item in this price raise their hands. • The seller gradually increases the price, and uninterested buyers lower their hands. • When only one hand remains up, the price stops. The remaining player gets the item, and pays the last price.

  10. How should we analyze this? • Problem: this is not a normal-form game. • So, let’s assume for a moment that each player drops exactly when the price reaches his value: • Then the player with the highest value wins, and pays the second highest price. • Therefore we get the following “sealed-bid” 2nd price auction: • Each player submits a bid (integer number). • The player with the highest bid gets the item, and pays the second highest bid. • Now, how should we analyze this?

  11. Drawing the game (for two players) • The strategy space of each player is composed of all integers. • For fixedv1, v2, if player 1 plays b1 and player 2 plays b2,and b2 > b1, then the utility of player 1 is 0, and of 2 is v2 – b1. b2 . . . b1 0, v2 – b1

  12. Analysis for n=2 Claim: For fixedv1, v2, a dominant action for player i=1,2 is to play bi=vi Proof: Fix any b2 • If v1 > b2 then winning is better than losing for player 1. Declaring b1=v1willcause player 1 to win. • If v1 < b2 then losing is better than winning for player 1. Declaring b1=v1willcause player 1 to lose.

  13. Analysis for n>2 • For n > 2 players the game and the claim are similar. Claim: For fixedv1, …,vn the dominant action of player i is to play bi=vi Proof: Fix any b-i , and let x = max j  i bj • If vi > x then winning is better than losing for player i. Declaring bi=viwillcause player i to win. • If vi < x then losing is better than winning for player i. Declaring bi=viwillcause player i to lose.

  14. A game in incomplete information • For each tuple of vi’s we get a new game: the strategy spaces are all the same, but the utility functions may be different. • A player knows his own type (the vi), but not the other types. • Thus a player do not fully know the matrix of the game. • A strategy si : vi -> Ai determines the player’s action, as a function of his type. • A strategy si is dominant if the action si(vi) is dominant for every viand v-i THM: si(vi) = viis a dominant strategy in the 2nd price auction. Proof: Immediate from the previous claim.

  15. Few more definitions and remarks • A mechanism is essentially a game of incomplete information • A mechanism is direct-revelation if the bidders are supposed to declare their type (like the 2nd price auction). • A direct-revelation mechanism is truthful if, in equilibrium, players declare their true types (like the 2nd price auction).

  16. Checkpoint By now we know: • Normal-form games and dominant actions. • Mechanisms with incomplete information and dominant strategies. • The 2nd price auction (description, its truthfulness)

  17. 1st price auction • What about the natural 1st price rule: The highest bidder wins, and pays his bid? • Observation: There is no dominant strategy in this game (if a player wins with a bid b he would have preferred to say slightly less than b).

  18. Reminder: continuous probability • Suppose X is a random variable that takes values in theinterval [a,b]. F() is its cumulative distribution functionif Pr(X < ) = F() for all  in [a,b]. • f(x)=F’(x) is called a probability distribution function. • Remark: In contrast to discrete probability, here there is no positive probability for a specific point, only to an interval. • Example: The uniform distribution over [a,b] is F(x)=(x-a)/(b-a) • F(a)=0, F(b)=1, F((a+b)/2)=1/2 • The expectation of X is E(X) = ab f(x)·x·dx • The interpretation (by the central limit theorem): the average of many samples from f() will approach the expectation. • If X is uniformly distributed over [a,b] then E(X)=(a+b)/2

  19. Nash Equilibrium (in complete information) “A” “B” • In the “coordination game”, players gain a positive utility if and only if they play the same action. DFN • Action ai is best response to a-i iffor any other action a’i of player i,ui(ai, a-i) > ui(a’i, a-i) • The actions (a1,…,an) are in Nash equilibriumai is best response to a-i for all players i. • What are the Nash equilibria in the coordination game? “A” 2 , 2 0 , 0 0 , 0 “B” 1 , 1

  20. Bayesian-Nash Equilibrium(in incomplete information) • Since players do not know the matrix of the game, we extend the notion of Nash to Incomplete settings in several ways. • Assumption: vi is drawn from a probability distribution fi, and these distributions are known to all (“common priors”). DFN: The strategies s1,…, snare in Bayesian-Nash equilibrium if for any i, vi, ai : Ev-i[ui(si(vi),s-i(v-i)] > Ev-i[ui(ai,s-i(v-i)] Remark: One could think of two other definitions: (I) where the expectation is on vi as well (this is called ex-ante equilibrium), and (II) the “for any” quantifier is also on v-i, and no expectation is involved (this is called ex-post equilibrium).

  21. Analysis of the 1st price auction • Clearly, in the 1st price auction a bidder needs to “shade down” his value. The question is by how much? • We will find a Bayesian-Nash equilibrium under the following simplifying assumptions: • vi is independently drawn from the uniform distribution on [0,1] (Pr[vi< y] = y). • si(vi) = xi·vi (the bidder shades down his value by a constant fraction) • We want to find strategies si(vi) that form a Bayesian-Nash equilibrium.

  22. We will “guess” that si(vi) = x · vi and we will find the appropriate x that indeed satisfies the equilibrium properties. • Given that all players j  i play sj(vj), player i chooses a bid b to maximize her expected utility:ui(b) = (vi – b) · Pr[maxji (x· vj) < b] • We have: • Pr[maxji (x· vj) < b] = jiPr[x· vj< b] • Pr[x·vj <b] = Pr[vj<b / x] = min (b/x , 1) • If b/x > 1 then player i wins with probability 1, but this b clearly does not maximize the utility since by taking b’ < b such that b’/x > 1 we increase the utility. Thus b/x< 1. • We get ui(b) = (vi – b) · bn-1 / xn-1

  23. We want to find the b that maximizes the utility, so we take the derivative: u’(b) = [(n-1) · vi· bn-2 – n · bn-1]/xn-1 = 0  b=[(n-1)/n] · vi THM: A Bayesian-Nash equilibrium of the 1st price auction is when every player i bids si(vi)=[(n-1)/n] · vi

  24. Remarks • 1st price auction is equivalent to a descending (“Dutch”) auction: the auctioneer gradually lowers the price, the first to accept wins, for this price. • A comparison to second price: • No dominant strategies, so less obvious how to play. • In both auctions, in equilibrium, the bidder with the highest value gets the item (the “efficient” outcome). • 2nd price may look bad - the winner’s price may be much lower than his bid. An extreme once happened in a New-Zealand government-auction: One firm bid NZ$100,000 for a license, and paid the second-highest price of only NZ$6 (http://www.economicprincipals.com/issues/06.05.21.html). • But which auction indeed raises more revenue?

  25. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)].

  26. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. What are its statistical properties?

  27. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. • F1(n)(x)=[F(x)]n

  28. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. • F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1

  29. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. • F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1 • E[Y1(n)] = 01 f(x)xdx = 01nxndx = [n/(n+1)]xn+1|01

  30. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. • F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1 • E[Y1(n)] = 01 f(x)xdx = 01nxndx = [n/(n+1)]xn+1|01= n/(n+1)

  31. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. • F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1 • E[Y1(n)] = 01 f(x)xdx = 01nxndx = [n/(n+1)]xn+1|01= n/(n+1) • E[Revenue of 1st price] = E[s(Y1(n))] = E[ ((n-1)/n) Y1(n) ]

  32. 1st price • We have n random variables X1,…,Xnwith • The probability density function is f(x)=1 (for 0<x<1) • The cumulative distribution function is F(x)=x • Let Y1(n) = maxiXi . Y1(n)is a random variable [with c.d.f. F1(n)(x) and p.d.f. f1(n)(x)]. • F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1 • E[Y1(n)] = 01 f(x)xdx = 01nxndx = [n/(n+1)]xn+1|01= n/(n+1) • E[Revenue of 1st price] = E[s(Y1(n))] = E[ ((n-1)/n) Y1(n) ] = = [(n-1)/n]E[Y1(n)] = (n-1)/(n+1)

  33. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x))

  34. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = ??

  35. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x))

  36. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x)) = F(x)n + nF(x)n-1-nF(x)n = = nF(x)n-1 - (n-1) F(x)n

  37. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x)) = F(x)n + nF(x)n-1-nF(x)n = = nF(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x)

  38. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x)) = F(x)n + nF(x)n-1-nF(x)n = = nF(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x) => E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)]

  39. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x)) = F(x)n + nF(x)n-1-nF(x)n = = nF(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x) => E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)] = n(n-1)/n – (n-1)n/(n+1) = (n-1)(1 – n/(n+1))= = (n-1)/(n+1)

  40. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x)) = F(x)n + nF(x)n-1-nF(x)n = = nF(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x) => E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)] = n(n-1)/n – (n-1)n/(n+1) = (n-1)(1 – n/(n+1))= = (n-1)/(n+1) • E[Revenue of 2nd price] = E[Y2(n)] = (n-1)/(n+1)

  41. 2nd price • Let Y2(n) be the second highest of X1,…,Xn(with c.d.f. F2(n)(x)) • F2(n)(x) = F(x)n + nF(x)n-1(1-F(x)) = F(x)n + nF(x)n-1-nF(x)n = = nF(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x) => E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)] = n(n-1)/n – (n-1)n/(n+1) = (n-1)(1 – n/(n+1))= = (n-1)/(n+1) • E[Revenue of 2nd price] = E[Y2(n)] = (n-1)/(n+1) • Thus E[Revenue of 1st price] = E[Revenue of 2nd price] !!

  42. Questions • Is this by accident? • Can we design an auction with a higher revenue?

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