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# Auctions

Auctions. Strategic Situation. You are bidding for an object in an auction. The object has a value to you of \$20. How much should you bid? Depends on auction rules presumably. Review: Second Price Auctions. Suppose that the auction is a second-price auction High bidder wins

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## Auctions

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1. Auctions

2. Strategic Situation • You are bidding for an object in an auction. • The object has a value to you of \$20. • How much should you bid? • Depends on auction rules presumably

3. Review: Second Price Auctions • Suppose that the auction is a second-price auction • High bidder wins • Pays second highest bid • Sealed bids • We showed (using dominance) that the best strategy was to bid your value. • So bid \$20 in this auction.

4. Review: English Auctions • An English (or open outcry) auction is one where bidders shout bids publicly. • Auction ends when there are no higher bids. • Implemented as a “button auction” in Japan • Implemented on eBay through proxy bidding.

5. What to Bid • Again, suppose you value the object at \$20. • Dominance says to drop out when bid = value. • The fact that bidding strategies are the same in the two auction forms means that they are strategically equivalent.

6. Revenues • How much does the seller earn on the auction? • Depends on the distribution of values. • Suppose that there are 2 bidders and values are equally likely to be from \$0 to \$100. • The seller earns an amount equal to the expected losing bid.

7. Order Statistics • The seller is interested in the expected value of the lower of two draws from 0-100. • This is called the second order statistic of the distribution. • We will sometimes write this as E[Vk(n)] where the k denotes the order (highest, 2nd highest, etc.) of the draw and (n) denotes the number of draws. • So we’re interested in E[V2(2)]

8. Order Statistics of Uniform Distributions • There order statistics have simple regularity properties • The mean of a uniform draw from 0-100 is 50. • Note the mean could be written as E[V1(1)]. 100 0 50

9. Two Draws • Now suppose there are two draws. • What are the first and second order statistics? 100 0 66 33

10. Key Observation • With uniform distributions, the order statistics evenly divide the number line into n + 1 equal segments. • Let’s try 3 draws: 1st 3rd 2nd 50 0 75 100 25

11. Generalizing • So in general, • E[Vk(n)] = 100* (n – k + 1)/(n + 1) • So revenues in a second price or English auction in this setting are: • E[V2(n)] = 100 * (n – 1)/(n + 1) • As the number of bidders grows large, the seller’s revenues increase • As the number of bidders grows unbounded, the seller earns all the surplus, i.e. 100!

12. First Price Auctions • Now suppose you have a value of \$20 and are competing with one other bidder in a first-price auction • You don’t know the exact valuation of the other bidder. • But you do know that it is randomly drawn from 0 to 100. • How should you bid?

13. Setting Up the Problem • As usual, you want to bid to maximize your expected payoff • But now you need to make a projection about the strategy of the other bidder • Presumably this strategy depends on the particular valuation the bidder has. • Let b(v) be your projection for the bid of the other bidder when his valuation is v.

14. Bidder’s Problem • Choose a bid, B, to maximize expected profits. • E[Profit] = (20 – B) x Pr(B is the highest bid) • What is Pr(B is the highest bid)? • It is Pr(B > b(v))

15. What is Pr(B > b(v))? b(v) B v I lose I win b-1(B)

16. Conjectures about b(v) • Suppose that I believe that my rival’s strategy is to bid a constant fraction of his value • Then b(v) = av • Where a is some fraction • I win whenever • B >= av • Or, equivalently • v <= B/a • So Pr(B > b(v)) becomes: • Pr( v <= B/a) = B/100a

17. Bidder’s Problem Revisited • So now I need to choose B to maximize • E[Profit] = (20 – B)(B/100a) • Optimize in the usual way: • (1/100a) x (20 – 2B) = 0 • Or B = 10 • So I should bid 10 when my value is 20.

18. Other Values • Suppose my value is V? • E[Profit] = (V – B)(B/100a) • Optimize in the usual way: • (1/100a) x (V – 2B) = 0 • Or B = V/2 • So I should always bid half my value.

19. Equilibrium • My rival is doing the same calculation as me. • If he conjectures that I’m bidding ½ my value • He should bid ½ his value (for the same reasons) • Therefore, an equilibrium is where we each bid half our value.

20. Uncertainty about my Rival • This equilibrium we calculated is a slight variation on our usual equilibrium notion • Since I did not exactly know my rival’s payoffs in this game • I best responded to my expectation of his strategy • He did likewise

21. Bayes-Nash Equilibrium • Mutual best responses in this setting are called Bayes-Nash Equilibrium. • The Bayes part comes from the fact that I’m using Bayes rule to figure out my expectation of his strategy.

22. Comments • In this setting, dominant strategies were not enough • What to bid in a first-price auction depends on conjectures about how many rivals I have and how much they bid. • Rationality requirements are correspondingly stronger.

23. Revenues • How much does the seller make in this auction? • Since the high bidder wins, the relevant order statistic is E[V1(2)] = 66. • But since each bidder only bids half his value, my revenues are • ½ x E[V1(2)] = 33 • Notice that these revenues are exactly the same as in the second price or English auctions.

24. Revenue Equivalence • Two auction forms which yield the same expected revenues to the seller are said to be revenue equivalent • Operationally, this means that the seller’s choice of auction forms was irrelevant.

25. More Rivals • Suppose that I am bidding against n – 1 others, all of whom have valuations equally likely to be 0 to 100. • Now what should I bid? • Should I shade my bid more or less or the same? • In the case of second-price and English auctions, it didn’t matter how many rivals I had, I always bid my value • What about in the first-price auction?

26. Optimal Bidding • Again, I conjecture that the others are bidding a fraction a of their value. • E[Profit] = (V – B) x Pr(B is the high bid) • To be the high bid means that I have to beat bidder 2. • Pr( B >= b(v2)) = B/100a • But I also have to now beat bidders 3 through n.

27. Probability of Winning • So now my chance of winning is • B/100a x B/100a x …B/100a • For n – 1 times. • Or equivalently • Pr(B is the highest) = [B/100a]n-1

28. Bidder 1’s optimization • Choose B to maximize expected profits • E[Profit] = (V – B) x Pr(B is highest) • E[Profit] = (V – B) x [B/100a]n-1 • E[Profit] = (1/100a)n-1 x (V – B) x [B]n-1 • Optimizing in the usual way: • (1/100a)n-1 x ((n-1)V – nB) [B]n-2 = 0 • So the optimal bid is • B = V x (n-1)/n

29. Equilibrium • I bid a proportion of my value • But that proportion is (n-1)/n • As I’m competing against more rivals, I shade my bid less. • Since all my rivals are making the same calculation, in equilibrium everyone bids a fraction (n-1)/n of their value.

30. Revenues • How much does the seller make in this auction? • The relevant order statistic is E[V1(n)] = 100* n/(n + 1) • But eveyone shades by (n-1)/n so • Revenues = (n-1)/n x E[V1(n)] • Revenues = 100 x (n-1)/(n+1)

31. Comments • Revenues are increasing in the number of bidders • As that number grows arbitrarily large, the seller gets all the surplus, i.e. 100! • How does this compare to the English or Second-Price auction?

32. Comparing Revenues • First-price: • R = (n-1)/n x E[V1(n)] • R = 100 x (n-1)/(n+1) • Second-price: • R = E[V2(n)] • R = 100 x (n-1)/(n+1) • The auctions still yield the same expected revenues.

33. Revenue Equivalence Theorem • In fact, revenue equivalence holds quite generally • Consider any auction which: • Allocates the object to the highest bidder • Gives any bidder the option of paying zero • Then if bidders know their values • Values are uncorrelated • Values are drawn from the same distribution • Then all such auctions are revenue equivalent!

34. Implications • This means that we can determine the revenues quickly and easily for all sorts of auctions • Consider an all-pay auction • Bidders submit cash payments to the seller (bribes) • The bidder submitting the highest bribe gets the object • The seller keeps all the bribe money • This auction auction yields the same revenues as an English auction.

35. Other Strange Auction forms • Suppose that all bidders submit bribes to the auctioneer • The object is awarded to the person paying the highest bribe • And the seller gives back the bribe of the winner, but keeps all the others • This is also revenue equivalent.

36. Optimal Auctions • Revenue equivalence says that the form of the auction does not affect how much money the seller makes. • But there are other tools the seller has to make money.

37. One Bidder Auctions • Suppose that the seller is running an auction that attracts only one bidder. • What should he do? • If he goes with the usual auction forms, he’ll make nothing since the second highest valuation for the object is zero.

38. Monopoly • Since the seller is a monopoly provider of the good, maybe some tricks from monopoly theory might help. • Suppose a monopolist faced a linear demand curve and could only charge a single price • What price should he charge?

39. Monopoly Problem P Demand curve Q

40. Monopoly Problem • The monopolist should choose p to maximize profits • Profits = P x Q(P) – C(Q(P)) • Or equivalently, the monopolist could choose Q to maximize profits • Profits = P(Q) x Q – C(Q) • P(Q) is the inverse demand function • Optimizing in the usual way, we have: • MR = MC

41. Monopoly Problem P Marginal Revenue P* MC Q Q*

42. Back to Auctions • What is the demand curve faced by a seller in a one bidder auction? • One can think of the “quantity” as the probability of making a sale at a given price. • So if the seller asks for \$100, he will make no sales. • If he asks for \$0, he will sell with probability = 1 • If he asks \$50, he will sell with probability .5

43. Auction/Monopoly Problem P 100 50 0 1 Q = Pr of sale 1/2

44. Auction/Monopoly Problem P Q = 1 – F(p) 100 50 0 1 Q = Pr of sale 1/2

45. Demand Curve • So the demand curve is just the probability of making a sale • Pr(V > P) • If we denote by F(p) the probability that V <=p, then • Q = 1 – F(p) • But we need the inverse demand curve to do the monopoly problem the usual way. • P = F-1(1 – Q)

46. Auction/Monopoly Problem • Now we’re in a position to do the optimization. • The seller should choose a reserve price to maximize his expected profits • E[Profits] = p x (1 – F) • Equivalently, the auctioneer chooses a quantity to maximize • E[Profits] = F-1(1 – Q) x Q

47. Optimization • As usual the optimal quantity is where MR = MC • But MC is zero in this case • So the optimal quantity is where MR = 0

48. Auction/Monopoly Problem P Marginal Revenue 100 P* 0 1 Q = Pr of sale Q*

49. So what is Marginal Revenue? • Revenue = F-1(1 – Q) x Q • Marginal Revenue = F-1(1 – Q) – Q/f(F-1(1 – Q)) • where f(p) is (approximately) the probability that v = p • Now substitute back: • P – (1 – F(p))/f(p) = 0

50. Uniform Case • In the case where valuations are evenly distributed from 0 to 100 • F(p) = p/100 • f(p) = 1/100 • So • P – (1 – P) = 0 • Or • P = 50!

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