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AN INTRODUCTION TO AUCTION THEORY

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An Introduction to

Auction Theory

FLAVIO M. MENEZES

University of Queensland

PAULO K. MONTEIRO

EPGE/FGV

1

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To Our Wives, Laura and Nair


Preface

Hugo Sonnenschein, in his 1983 inaugural Nancy Schwartz Memorial Lecture,1

argued that one of the most important contributions of economics has beento the understanding of how incentives work—in particular, of how to designinstitutional arrangements that might induce individuals to behave in a way sothat a certain outcome (e.g., an ex-post efficient outcome) prevails.

About twenty years later economists have been recognized for their contribu-tion to the design of several auction-like mechanisms such as the U.S. FederalCommunications Commission spectrum auctions, the 3G auctions in Europeand beyond, the auction markets for electricity in Australia and elsewhere inthe world, the allocation of the rights to land at airports, regulations governingaccess pricing in natural gas pipelines, and the sale of former government-ownedcompanies around the globe. Perhaps it is significant that “Market Architec-ture” was the title chosen by Robert Wilson for his 1999 Econometric SocietyPresidential Address (Wilson 2002). Similarly, Alvin Roth’s Fischer-SchultzLecture at 1999 European Econometric Society Meeting was entitled “TheEconomist as an Engineer” (Roth 2002).

The concept of market architecture or engineering relies on insights fromgame theory (in particular, games of incomplete information) and mechanismdesign. It also relies on our understanding of how to tackle informational issuesbut perhaps some of the most important insights come from auction theory. Thepractical and theoretical importance of auction theory is widely recognized.Indeed, some of the more celebrated results from the single-object auctiontheory (e.g., the revenue equivalence theorem or the characterization of theoptimal auction) are now usually taught in advanced undergraduate courses onthe economics of information.

However, a step-by-step self-contained treatment of the theory of auctionsdoes not exist to the best of our knowledge. Thus, our aim is to provide anintroductory textbook that will allow students and readers with a calculusbackground, and armed with some degree of persistence, to work through allthe basic results. For example, the reader will be able to derive by himself or

1 Frontiers of Research in Economic Theory; The Nancy L. Schwartz Memorial Lectures(1983–1997), eds. D. Jacobs, E. Kalai and M. Kamien, Cambridge University Press, 1998.

vii

viii Preface

herself the celebrated Revenue Equivalence Theorem and to evaluate the effectsof introducing affiliation into the standard auction theory model.

Graduate microeconomics textbooks (such as Mas-Colell et al. 1995) typi-cally approach auctions as applications of mechanism design techniques. Gametheory textbooks (such as Fudenberg and Tirole 1991) examine auctions as anapplication of Bayesian games. Thus, their focus is on techniques rather thanon results. On the other hand, there are several excellent surveys focusing onthe results of auction theory rather than on techniques. In contrast we willfocus both on results and techniques. The main idea is that although readingsome of the original papers can be quite a daunting task for an advanced under-graduate student or for a first-year graduate student, it is possible to presentthis material in a more friendly way. Paul Klemperer (1999), when referringto some of the earlier literature on auctions, argues that with the exception ofVickrey’s first 1961 paper, the other papers “are no longer for the beginner.”Our aim is to make this material available to the “beginner.”

Acknowledgements

This book was born from the lecture notes used to teach courses atthe Australian National University and at EPGE-FGV. Its completion wasfacilitated by an Australian Research Council Grant (no. A0000000055) thatallowed Flavio Menezes to visit EPGE-FGV and Paulo Monteiro to visitthe ANU. Paulo Monteiro acknowledges the support of CNPq-Brazil. Severalstudents (current and former) have contributed to the improvement of this text.In particular, we would like to thank Joisa Dutra, Craig Malam, GuilhermeNorman and Louise Sutherland for very detailed comments. Many colleagueshave been very supportive of this project but special thanks go to Simon Grantand Steve Dowrick for their advice and encouragement. Menezes acknowled-ges the financial support from the Australian Research Council (ARC GrantsDP 0557885 and 0663768) for the revisions and additions in this new edition.

ix


Contents

1 Introduction 1

2 Preliminaries 52.1 Notation 52.2 Bayesian Nash Equilibrium 62.3 Auctions as Games 9

2.3.1 What is an Auction? 92.3.2 Auction Types 102.3.3 Auction as Bayesian Games 11

3 Private Values 133.1 The Independent Private Values Model 13

3.1.1 First-price Auctions 143.1.2 Second-price Auctions 183.1.3 Revenue Equivalence 203.1.4 Reserve Prices and Entry Fees 22

3.2 The Correlated Private Values Model 253.2.1 Second-price Auction 263.2.2 First-price Auction 273.2.3 Comparison of Expected Payment 29

3.3 The Effect of Risk Aversion 323.3.1 Revenue Comparison 33

3.4 The Discrete Valuation Case 343.5 Exercises 35

4 Common Value 394.1 An Example with Independent Signals 40

4.1.1 First-price Auction 404.1.2 Second-price Auction Example 42

4.2 An Example with Correlated Types 424.2.1 First-price Auction 434.2.2 Second-price Auction 45

4.3 The Symmetric Model with Two Bidders 46

xi

xii Contents

4.3.1 Second-price Auctions 504.3.2 First-price Auctions 514.3.3 Revenue Comparison 54

4.4 Exercises 55

5 Affiliated Values 575.1 The General Model 585.2 Second-price Auctions 635.3 First-price Auctions 655.4 English Auctions 675.5 Expected Revenue Ranking 685.6 Exercises 70

6 Mechanism Design 716.1 The Revelation Principle 716.2 Direct Mechanisms 756.3 Revenue Equivalence and the

Optimal Auction 796.4 Some Extensions 85

6.4.1 Non-monotonic Marginal Valuation 856.4.2 Correlated Values 976.4.3 Several Objects 1116.4.4 Common Values Auction 113

6.5 Exercises 115

7 Multiple Objects 1177.1 Sequential Auctions 1187.2 Simultaneous Auctions 120

7.2.1 Discriminatory Auctions 1217.2.2 Uniform price Auctions 126

7.3 Optimal Auction 1337.4 Exercises 139

8 What is Next? 1418.1 Distribution Hypotheses in Auction Theory 143

A Probability 151A.1 Probability Spaces 151A.2 Uncountable Sample Space Case 153A.3 Random Variables 154A.4 Random Vectors and their Distribution 156A.5 Independence of Random Variables 158A.6 The Distribution of the Maximum of Independent Random

Variables 158

Contents xiii

A.7 The Distribution of the Second Highest Value 159A.8 Mean Value of Random Variables 160A.9 Conditional Probability 162

B Differential Equations 165B.1 The Simplest Differential Equation 165B.2 Integrating Factor 165

C Affiliation 167

D Convexity 171

References 175

Index of Notations 179

Index of Proper Names 181

Index 183


1

Introduction

The theory of auctions is one of the most successful modern economic theories.Its success is reflected in a coherent body of theory but also in its ability toprovide insights into many practical policy issues. Indeed, we claim that thelabel “auction theory” is somewhat misleading—although we will use it for theremainder of this book—as the economics behind auction theory are actuallycommon to many other applications. We will not elaborate here on the indirectconnection between auction theory and economics in general. Instead, we referthe reader to Klemperer (2004) for a detailed discussion. What we do belowis to provide three examples to illustrate the importance of auction theory tomodern economics.

First, consider the problem faced by a regulator who wants to regulatea monopolist with unknown costs: a regulator wants to choose instruments(a price or a quantity, a subsidy or a tax, issuing a license to operate) so thatthe regulated monopolist chooses the action (how much to produce or howmuch to charge) to promote efficiency (second-best efficiency as in the presenceof asymmetric information: first-best efficiency is not possible). It turns outthat this is analogous to the problem faced by a seller who wants to extractthe most expected revenue possible when selling an object without knowing thevaluation of the buyers. (Indeed Roger Myerson wrote both one of the seminalpapers on the optimal auction (Myerson 1981) and together with David Baron,the paper on how to regulate a monopolist with unknown costs (Baron andMyerson 1982).) When regulating a monopolist, the regulator fixes a pricemenu (and a subsidy to cover fixed costs) that rewards low marginal cost typesfor choosing the “right quantity or price”—this is the informational rent kept bythe monopolist. In an auction, you allocate the object to the individual with thehighest valuation at a price equal to the largest of the second highest valuationand the optimally chosen reserve price—the difference is the informational rentkept by the buyer.

Our second example also relates to regulation. When it is not possible tohave competition in the market, we have to design a mechanism that will

1

2 Introduction

establish competition for the market. For example, when allocating spectrum(for mobile telephony) or cable-TV licenses or the license to build a transmissionline, we want to design an auction that will allocate the licenses to those whovalue them the most highly. Some of these auctions raised several billion dollarsfor governments around the globe.

Finally, our third example is the design of the spot market for electricity:one wants to design a market (a rule that will determine who is going tosell and the price at which they will sell) that reflects the cost of the mar-ginal generator—the highest cost generator that has to supply to meet systemdemand. This will guarantee dynamic efficiency. Electricity restructuring invol-ving the establishment of a spot market is pervasive in developed and developingcountries alike.

Comparison with recent contributionsThere are several excellent surveys of the auction theory literature availa-ble, including Engelbrecht-Wiggans (1980), Maskin and Riley (1985), Milgrom(1985, 1987, 1989), McAfee and McMillan (1987), Riley (1989), Wilson (1992),Wolfstetter (1996), and Klemperer (1999). These surveys provide a guide to theauction theory literature that covers papers that are not realistically accessibleto a wider audience. Although these surveys are excellent and do provide somehelp in reading the original papers, our aim here is to provide considerablymore detail so that readers can work their way through the most importantresults. We also include exercises.

As was noted above, current treatment of auctions in existing graduatemicroeconomics textbooks is limited to applications either of Bayesian gamesor mechanism design. Two exceptions are Laffont and Tirole (1993) andWolfstetter (1999). Laffont and Tirole devote a chapter to auctions cover-ing the optimal auction and the revelation principle in auctions. Wolfstetteralso devotes a chapter (60 pages) to auction theory. The main topics coveredinclude some of the basic theory, auction rings, optimal auctions and commonvalue auctions. It is also worth mentioning Vijay Krishna’s Auction Theory.Krishna’s book is well-written and provides a very comprehensive overview ofauction theory. Our aim is different; our objective is to start, whenever pos-sible, from basic principles (calculus and introductory probability is the onlyassumed knowledge) and to equip students with the techniques that are neces-sary to master the theory of auctions. As a result, our coverage has to be lesscomprehensive than Krishna’s. Our hope is that by having worked through thebook, the reader will have the confidence and technical ability to derive allthe results in the book and to construct and solve simple and sensible auctionmodels. In addition, the reader will have a “working knowledge” of mechanismdesign—as applied to auctions.

Finally, there is a high degree of complementarity between this textbookand Klemperer (2004). Klemperer’s approach is to introduce the basic auc-tion theory in a non technical fashion, relegating to appendices some of the

Introduction 3

more technical material. Thus, the reader, for example, can read Klemperer’sexcellent book to obtain an overview of basic theory. However, to master thetechniques and to develop a working knowledge of the subject, the reader canthen work through this book. Once this is done, the reader can then return toKlemperer for applications of auction theory.

Intended audienceThis book is intended for first-year graduate students and advanced honoursundergraduates in economics and related disciplines. This text can also beused to teach a special topics course on auction theory or, more likely, it couldbe used as a supplementary textbook for an advanced microeconomics coursefocusing on the economics of information. In our experience, teaching auctiontheory prior to introducing mechanism design is helpful to students as they canrelate these more abstract techniques to the more concrete auction context.

In addition, this book could be used as a supplementary textbook forcourses on game theory or for a stand-alone graduate or advanced undergra-duate course on auction theory, perhaps in conjunction with Klemperer (2004).Of course, this book can also be used by independent readers who want tounderstand auction theory.

OrganizationChapter 2 introduces the equilibrium concept used throughout the book,namely, that of a Bayesian Nash equilibrium. It also introduces the idea ofstudying auctions as games and defines some notation. Chapters 3, 4, and 5cover the private, common and affiliated values models, respectively. Chapter 6relates the field of mechanism design to auction theory by deriving a generalversion of the Revenue Equivalence Theorem and by characterizing the optimalauction, that is, the auction that maximizes the seller’s expected revenue. InChapter 7, we opted for covering some existing multi-object auction models inorder to complement the analysis in Vijay Krishna’s book. Chapter 8 providessome guidance on how the reader can extend his or her knowledge of auctionsbeyond what is covered in this book. Finally, there are four appendices coveringprobability theory, differential equations and affiliation.

Using this book as a textbookA graduate or advanced undergraduate course in auction theory should coverChapters 2–6. Chapter 7 is optional and includes material that is substantiallymore difficult than the rest of the book. If used as a supplement to a graduate oradvanced undergraduate course on the economics of information, the instructorcan concentrate on the case of independent private values and cover Chapters 2,3, and 6. Chapter 6 can be read on its own as an introduction to mechanismdesign and Chapter 5 can be read on its own as an introduction to affiliation.


2

Preliminaries

In this chapter, we introduce the equilibrium notion to be used in the book,namely that of a Bayesian Nash equilibrium. We then formally define an auctionas a game of incomplete information. We first introduce some notation that willbe used throughout the book.

2.1 Notation

In this book, we try to follow as closely as possible the standard notationof auction theory papers. Some notation is standard in other fields, other ispeculiar to auction theory.

We denote by N = 1, 2, . . . the set of natural numbers. The set of realnumbers is denoted by R and the set of non-negative real numbers is denoted byR+ or R+. If Xi is a set for i = 1, 2, . . . , n, n ∈ N then X = X1 ×X2 ×· · ·×Xn

is the Cartesian product of X1, X2, . . . , Xn. If the sets Xi are the same, sayXi = C for all i then we write Cn for C × · · · × C (n times). Thus [0, 1]2 =[0, 1] × [0, 1]. And if Xi = high, low, i = 1, 2 then

high, low2 = (high, high), (high, low), (low, high), (low, low).The following convention will be used throughout the book. If x =

(x1, x2, . . . , xn) is a vector of n coordinates we denote by x−i the vector obtainedfrom x by the removal of the ith coordinate. Thus,

x−i = (x1, x2, . . . , xi−1, xi+1, . . . , xn)

is a vector with n − 1 coordinates. For example, if s = (s1, s2, s3) then s−1 =(s2, s3).

The maximum of a finite sequence of real numbers, x1, x2, . . . , xn is deno-ted either by maxx1, x2, . . . , xn or by x1 ∨ x2 ∨ · · · ∨ xn. The maximumbetween x and 0 is denoted by x+. Thus x+ = maxx, 0 = x∨0. If Z1, . . . , Zn

are random variables, we will frequently consider the maximum of Z1, . . . , Zn

5

6 Preliminaries

denoted by maxZ1, . . . , Zn or Z1 ∨ Z2 ∨ . . . ∨ Zn. Similarly we denote byZ1∧ Z2 ∧ . . . ∧ Zn the minimum of Z1, Z2, . . . , Zn. We will denote by Y therandom variable that represents the highest number amongst Z2, . . . , Zn. ThusY = maxZ2, . . . , Zn.

A function f : R → R is increasing if x < y implies that f(x) ≤ f(y). It isstrictly increasing if x < y implies f(x) < f(y). Thus, an increasing functionmay be flat for parts of the domain. A strictly increasing function will have noflat part in the domain.

We say that a function f : R → R is continuously differentiable if for everyx ∈ R it has a derivative f ′(x) := limy→x(f(y)−f(x))/(y−x) and x → f ′(x) isa continuous function. The inverse of a (injective and onto) function f : A → Bis denoted by f−1 : B → A. The composition of the function g : B → C andf : A → B is denoted g f . That is g f(a) = g(f(a)).

If a set X is finite we denote by #X the number of elements of X.Occasionally we use the notation ∀x which translates to “for all x”.

2.2 Bayesian Nash Equilibrium

Throughout the book, we will use the notion of a Bayesian Nash equilibrium asdefined by Harsanyi (1967). His approach is to transform a game of incompleteinformation into one of imperfect information; any buyer who has incompleteinformation about other buyers’ values is treated as if he were uncertain abouttheir types. It is like introducing an extra player—nature—that chooses thetype for each player.

We can think of games of incomplete information as a two-stage game. Priorto the beginning of the game, before players make a decision, nature choosesa type for each player. At this stage, each player knows his own type but notthe types of other players. In the second stage, each players chooses a strategyknowing his own type and the initial distribution of all types.

To introduce formally the equilibrium notion we will need some notation.This notation will also be used in the remaining chapters. The set of players willbe denoted by I = 1, 2, . . . , n. The set of possible types for each player i ∈ Iis denoted by Xi. This set will be an interval, [0, v], in most of the book. Wedenote by F (·) the probability distribution over X = X1×X2×· · ·×Xn, whichreflects the probabilities attached to each combination of types occurring.

We denote by Si the set of strategies for player i ∈ I and by si : Xi → Si

the decision function of player i. It is a mapping from the set of possible typesto the set of possible strategies. (In a particular case below we will set Si = R+

and Xi = Xj , for all i, j). We denote by Fi(x−i |xi) the probability distributionof types x−i of the players j = i given that i knows his type is xi. That is,player i updates his prior information about the distribution of the other typesusing Bayes rule upon learning that his type is xi.

2.2 Bayesian Nash Equilibrium 7

We let πi(si, s−i, xi, x−i) denote i’s profits given that his type is xi, thathe chooses si ∈ Si and that the other players follow strategies s−i(x−i) =(sj(xj))j =i (the function sj : Xj → Sj being j’s decision function) and theirtypes are x−i. For each vector (x1, x2, . . . , xn) chosen by nature, there areupdated beliefs given by F1(x−1 |x1), . . . , Fn(x−n |xn).

A Bayesian Game is defined as a five-tuple

G = [I, Sii∈I , πi(·)i∈I , X1 × · · · × Xn, F (·)].

That is, it is a set of players, a strategy set for each player, a payoff (or utility)function for each player, a set of possible types and a distribution over the setof types.

A Bayesian Nash equilibrium is a list of decision functions (s∗1(·), . . . , s∗n(·))such that ∀i ∈ I,∀xi ∈ Xi and ∀si ∈ Si:∫

x−i∈X−i

πi(s∗i , s∗−i, xi, x−i) dFi(x−i |xi)

≥∫

x−i∈X−i

πi(si, s∗−i, xi, x−i) dFi(x−i |xi).

In words, each player chooses a strategy contingent on his type—that is,he uses a Bayesian decision function. We can then apply the Nash equilibriumnotion to these decision functions: each player forms a best response strategyof choosing the best Bayesian decision functions, based on the best responsestrategies of other players (who are choosing their Bayesian decision functions).

In part of the book, the distribution of players types is independent. That is,

F (x) = F1(x1)F2(x2) · · ·Fn(xn).

In this case,

Fi(x−i) = F1(x1) · · ·Fi−1(xi−1)Fi+1(xi+1) · · ·Fn(xn).

Remark 1 A symmetric Bayesian Nash equilibrium is such that all playerschoose the same decision function.

In the next few chapters, the reader will have ample opportunity to checkhis or her understanding of the symmetric Bayesian Nash equilibrium notionin the context of auctions. For the remainder of this section we work throughan example to apply this equilibrium notion when the set of types is discreteand in a context that will be familiar to many readers.

Example 1 Consider a Cournot model where two firms, 1 and 2, produce ahomogeneous good and compete in quantities. The inverse market demand isgiven by p = 1−Q, where Q is the sum of quantities produced by each firm. Unit

8 Preliminaries

costs of both firms are constant. However, the unit cost may be either high, ch,or low, cl. We assume that 4 − 5ch + cl ≥ 0. The joint probability distributionis given by

F (ch, ch) = F (ch, cl) = F (cl, cl) = F (cl, ch) =14.

Let us compute the symmetric Bayesian Nash equilibrium.First, note that we can apply Bayes rule to compute F (ch | ch), F (cl | ch),F (cl | cl), and F (ch | cl). Since the distribution determining the types is the samefor both players, F (ch | ch), for example, denotes the probability that Player 1who is of type ch faces a Player 2 of type ch but also the probability that a typech Player 2 faces a type ch Player 1. Thus,

F (ch | ch) =F (ch, ch)

F (ch, ch) + F (cl, ch)=

12.

Similar calculations yield:

F (cl | ch) = F (cl | cl) = F (ch | cl) =12.

Note that the symmetric Bayesian Nash equilibrium is a pair of decision functi-ons (q∗(·), q∗(·)), one for each player, indicating that player’s action if his typeis ch and his action if his type is cl. We will proceed along the following line ofreasoning, which will be used in the entire book: We posit that Player 2 is fol-lowing a strategy q(·) = (q(cl), q(ch)), in this case one action for each possibletype, and compute 1’s best reply. In the symmetric equilibrium both players areusing the same strategy.

Accordingly, suppose Player 2 is choosing q(·) and Player 1 is of type cl

and has to choose a “number” sl determining how much he will produce. Theexpected profits of Player 1 are given by:

π1(sl, q(·), cl) = 12 (1 − q(cl) − sl − cl)sl + 1

2 (1 − q(ch) − sl − cl)sl.

Maximizing with respect to sl we obtain:12 (1 − q(cl) − 2sl − cl) + 1

2 (1 − q(ch) − 2sl − cl) = 0.

Given that we are looking at symmetric equilibrium, we can set sl = q(cl) toobtain

q(ch) = 2 − 5q(cl) − 2cl. (2.1)

Similarly we need to consider the case where Player 2 is choosing q(·) andPlayer 1 is of type ch and has to choose a “number” sh determining how muchhe will produce. His expected profits are given by:

π1(sh, q(·), ch) = 12 (1 − q(cl) − sh − ch)sh + 1

2 (1 − q(ch) − sh − ch)sh.

We then obtain

q(cl) = 2 − 5q(ch) − 2ch. (2.2)

2.3 Auctions as Games 9

By solving the “best-response functions” (2.1) and (2.2) simultaneously, weobtain the symmetric Bayesian Nash equilibrium:

q∗(ch) =4 − 5ch + cl

12

and

q∗(cl) =4 − 5cl + ch

12.

Note that a high cost producer in the symmetric equilibrium produces less thana low cost producer.

2.3 Auctions as Games

In this section we provide a brief introduction to auctions, explain how we viewauctions as games of incomplete information and define some notation.

2.3.1 What is an Auction?

Cassady (1967) provides a very nice guide to the various practical uses ofauctions. If such a guide were to be revised today, it would be many timesthicker than the original version. The reason is that auctions have becomean effective tool to implement public policy. Their use now ranges from theallocation of radio spectrum necessary for mobile communication, to spot mar-kets trading electricity and pollution permits, as well as being widely used ingovernment procurement.

We can define an auction by one of its central properties: as a marketclearing mechanism, to equate demand and supply. Other market mechanismsinclude fixed price sales (as in a supermarket) or bargaining (as in the nego-tiated sale of a house or a used car). Within the class of market mechanismswhich allocate scarce resources, one particular characteristic of the auction isthat the price formation process is explicit. That is, the rules that determinethe final price are usually well-understood by all parties involved.

Auctions are often used in the sale of goods for which there is no establishedmarket. Auctions were instrumental in the mass privatization in Eastern Europegiven the absence of a price system that could guide the valuation process forfirms being privatized. Rare or unique objects are typically sold in auctionsas the markets for these objects are likely to be very thin. However, auctionsare also used to sell Treasury bills and the markets for these assets are verythick. The reason is that only governments can legally produce such bonds andtherefore the sale in an auction is an exercise in revenue maximization.

Auctions are more flexible than a fixed price sale and perhaps less time-consuming than negotiating a price. Auctions are used to sell hundreds of

10 Preliminaries

goods, such as bales of wool or used cars, in a few hours. One can imaginehow many hours it would take to sell 100 used cars through negotiated sales.The reader should then ask the question why then car dealers do not switch toauctions as a sales mechanism. Although a complete answer to this question isbeyond the scope of this book, we could expect that under certain conditionsa negotiated sale or even a fixed price might result in higher expected revenuefor the seller. We will touch on this in Chapter 6.

2.3.2 Auction Types

Auctions can be classified according to several distinct criteria. For example,we distinguish between open auctions and sealed-bid auctions. In the formertype of auction, all bids are publicly observable whereas in the latter they arenot. We can also differentiate between ascending and descending price auctions.In both types of auctions bids are public, but the ascending auction starts ata low price and bids have to be increasing, whereas in the latter bidding startsat a high price that continuously declines until one of the bidders stops theprocess by acquiring the object.

Auctions for single objects are also distinct from auctions for multipleobjects. There are several possible designs available when selling multipleobjects which are not available when selling a single indivisible good. Forexample, a multiple object auction format might allow for bids on combina-tions of items (combinatorial auctions) or objects might be sold sequentially.Chapter 7 describes some multiple object auction types (simultaneous versussequential, discriminatory versus uniform price). However, most of this book(Chapters 3–6) deals with single-object auctions. We will examine four basicformats: English, Dutch, First-Price, and Vickrey auctions. Although most rea-ders are familiar with at least some of these auction formats, for completenesswe describe them below.

The English or ascending-price auction is the best-known format. It is anopen auction where an auctioneer (there are also electronic implementations)starts requesting bids at a low price and bidders bid by meeting the incrementsproposed by the auctioneer. The auction stops when no bidder is willing toincrease his bid above the highest standing bid. The bidder with the higheststanding bid wins the auction and pays the highest bid. This auction is com-monly used in the sales of rare paintings, used cars, houses and many otherobjects. There can be several aspects such as secret reserve prices, dummybids (bids made by the seller or the auctioneer, perhaps without the know-ledge of the bidders) and sometimes even the possibility of negotiation betweenthe winner of the auction and the seller. While auction theory can be used toaccommodate these possibilities, we ignore such issues in the theory expoundedin this book. We will model English auctions in the tradition of Milgrom and

2.3 Auctions as Games 11

Weber (1982) and model English auctions as “button auctions”.1 Later wediscuss the implications of such a modeling assumption.

In both first- and second-price (or Vickrey) sealed-bid auctions, each biddersubmits his or her bid without the knowledge of the bids made by others. Thewinner in both cases is the bidder with the highest bid. He or she will pay hisor her bid in a first-price auction and the second highest bid in a second-priceauction. Whereas first-price auctions are typically used in the procurement ofgoods and services,2 second-price auctions have remained relatively rare untilmore recently when they have been adopted by business to business platforms.

Finally, the Dutch auction is an open descending price auction. It is widelyknown for its use in selling flowers in the Netherlands. Bidding starts at a highprice that continuously decreases on an automated clock. The auction endswhen one of the participants stops the clock. This bidder wins the object andpays the price at which the clock stopped.

2.3.3 Auction as Bayesian Games

To define an auction as a Bayesian game G, we will keep the notation definedabove for the set of potential bidders I = 1, 2, . . . , n, Xi = [0, v] will denotethe set of possible types of player i, i = 1, . . . , n, and vi the type received byplayer i. F (·) : [0, v]n → [0, 1] is the joint distribution of types and the associateddensity is denoted by f(·) : [0, v]n → R+. The set of possible bids or strategiesfor player i, i = 1, . . . , n, is Si = R+.

It should be noted here that for simplicity, we assume that the seller’svaluation is zero (and the seller will therefore not accept any negative bids).Moreover, it is assumed throughout the book that there is no secondary market,and no other resale possibility. This is because our objective is to provide athorough exposition of standard auction theory for the beginner, rather thancovering all existing research in auctions.3

Finally, the payoff to player i will depend on his or her attitude towardsrisk, on a valuation or utility function ui(v1, . . . , vn) and on the rules of theauction. The precise nature of this relationship will be made explicit in the

1 Each bidder presses a button while the price increases continuously. A participant dropsout when she takes her hand off the button. The auction ends when there is only one bidderleft pressing the button. This bidder wins the auction and pays the price at which thenext-to-last player stopped pressing the button.

2 In a procurement auction, several sellers are competing to sell a good to the buyer. Ina first-price procurement auction, the winner is the seller with the lowest bid and the buyerpays the equivalent to this bidder’s bid. The analysis is completely analogous to that ofa standard first-price auction. In this book we concentrate on the latter. Note that bothgovernments and large private buyers are increasingly using alternative auction formats suchas electronically descending auctions to buy goods and services.

3 While this research currently includes numerous interesting and relevant examples forpractitioners (see, e.g., Grant et al. 2006), our objective is instead to provide the buildingblocks that are necessary to understand this research.

12 Preliminaries

chapters below. Here, we will offer a general view on the existing auctionmodels. Auction models typically fall into three categories. In a private valuesmodel, each potential buyer knows his or her own value for the object, whichis not influenced by how other potential buyers value it (see Chapter 3). If indi-viduals’ types are independent from each other—for example, one may think oftypes being determined by independent draws from a fixed distribution—thenwe have the independent private value (IPV) model. If valuations are depen-dent on one another, then we have the correlated private value model. Moregenerally, a private values model might be more appropriate for non-durablegoods with no resale value.

In the common value model (see Chapter 4), the object is worth the sameto every potential bidder, but this value is unknown at the time of bidding.Typically, individuals have some information about the (unknown) true valueof the object. If information is correlated across individuals, then we have adependent common value model. If information is independent across indivi-duals, then we have an independent common value model. The common valuemodel is often more appropriate for analyzing the sale of mineral rights andoffshore oil drilling leases.

Finally, Milgrom and Weber (1982) introduce the notion of affiliated values(see Chapter 5), which includes both private and common values as specialcases. Roughly speaking, affiliated values capture the idea that individuals’valuations for an object have a private component but are influenced by howother people value it. In most sales we can imagine, a bidder’s valuation forthe object being sold does have a private component, but that valuation isalso influenced by other individuals’ valuations. For instance, when bidding fora house, one takes into account both the personal value of the house as wellas how easily it would be to resell it in the future.4 Affiliation, however, is anotion of global positive correlation and this has particular implications for theranking of auction formats according to the expected revenue they generate, aswill be discussed in Chapter 5.

4 This relates to how individuals value objects. Of course, even in the IPV model, bidders’bidding behavior will depend on how they think others will bid.

3

Private Values

In this chapter, we examine the case where bidders’ values for the object beingauctioned off is a function only of their own types. As seen in Chapter 2,individuals’ types can be either independent or correlated. In the case of inde-pendent types we have the IPV model. This is the benchmark model for auctiontheory and it provides several useful insights. This model will be covered in thenext section. The correlated private values model will be covered in the subse-quent section. The last section examines the effects of risk aversion on biddingbehavior and on the seller’s expected revenue for the private values model.

3.1 The Independent Private Values Model

A single object will be sold to one of n bidders. Each bidder i, i = 1, . . . , n,receives a type vi and his valuation is equal to ui(vi) = vi. The implicit assump-tion here is that buyers are risk-neutral, that is, they are indifferent between alottery that yields an expected value of x and receiving x for certain.

Each bidder knows his own valuation vi and that his opponents’ valuationsare drawn independently from the distribution F (·) with density f(·) > 0 inthe interval [0, v]. (Appendix A contains an introduction to probability theory.)That is, F (x) denotes the probability that the random variable v is less thanor equal to a certain number x.

This is the IPV model where the value of the object to a bidder dependsonly on his own type. Bidding behavior, however, depends on one’s expectationabout other bidders’ valuations and about how they bid. Although the inde-pendent private value model is only appropriate to describe the case where theobject does not have a resale value (or it is too costly to resell), it allows us toderive several important insights. For simplicity, we assume that the seller setsthe reserve price at zero and that there are no entry fees.

13

14 Private Values

In this chapter, we will compute the equilibrium bidding strategies and theseller’s expected revenue in four distinct types of auctions: first- and second-price sealed-bid, English, and Dutch auctions. As we have seen in Chapter 2,each bidder submits his bid without observing the bids made by other playersin a sealed-bid auction. In a first-price auction, the winner is the bidder withthe highest bid and he pays his bid. In a second-price auction, the winner isstill the bidder with the highest bid but he pays the second highest bid.

A naive commentator would argue that a first-price auction should gene-rate more revenue than the second-price auction as the winner pays his bid inthe former and the second highest bid in the latter. However, this argumentfails because bidders behave strategically. We will show below that bidders bidless than their valuations in the unique symmetric equilibrium of a first-priceauction and bid their valuations in the unique symmetric equilibrium of thesecond-price auction.

3.1.1 First-price Auctions

We start our search for a symmetric Bayesian Nash equilibrium by analyzingthe game from the point of view of one of the players, say Player 1. Suppose thisplayer has a valuation v = v1 and believes that other players follow a biddingstrategy b(·). Knowing only his value and the distribution of the valuations ofplayers 2, . . . , n, Player 1 has to figure out what is his best reply. Suppose bidderi = 2, . . . , n has valuation vi. Thus bidder i ≥ 2 bids bi = b(vi). Then if Player 1bids b1 the object is won if b1 > bi for i ≥ 2. That is if b1 > maxb2, . . . , bn.If b1 < maxb2, . . . , bn Player 1 does not win the object. Let us suppose thatin case of a draw, that is if b1 = maxb2, . . . , bn, the object is not sold. ThusPlayer 1’s payoff is

v − b1 if b1 > maxb(v2), . . . , b(vn)0 if b1 ≤ maxb(v2), . . . , b(vn).

The expected profits from bidding b1 are given by

π(b1) = π(v, b1, b(·)) = (v − b1) Pr(b1 > maxb(v2), . . . , b(vn)).We can rewrite the expression above as

π(b1) = (v − b1) Pr(b1 > b(v2), . . . , b1 > b(vn)).

For the moment assume that the function b(·) is strictly increasing and diffe-rentiable. (We will later verify that our equilibrium strategy is indeed increasingand differentiable in the domain and thus our analysis is justified.) Thus, therange of b(·) is an interval: b([0, v]) = [b, b]. The bidder will never bid hig-her than b since the payment will be higher and the object will be won aswell. Any bid lower than b is a losing bid. Thus we may suppose, withoutloss of generality that b1 ∈ [b, b]. Therefore there exists x ∈ [0, v] such that

3.1 The Independent Private Values Model 15

b1 = b(x). The problem of bidder 1 is therefore equivalent to choose x ∈ [0, v]to maximize expected utility

π(x) = π(b(x)) = (v − b(x)) Pr(b(x) > b(v2), . . . , b(x) > b(vn))

= (v − b(x)) Pr(x > v2, . . . , x > vn). (3.1)

In the second line of the above equation, we used the fact that b(·) is strictlyincreasing and that all players follow the same strategy in equilibrium sinceall of them are faced with the same maximization problem. Since the vjs areindependent and identically distributed random variables, we can rewrite (3.1)as follows:

π(x) = (v − b(x)) Pr(x > v2) · · ·Pr(x > vn) = (v − b(x))F (x)n−1. (3.2)

The derivative of π is now easy to calculate:

π′(x) = (v − b(x))(n − 1)f(x)F (x)n−2 − b′(x)F (x)n−1. (3.3)

In a symmetric equilibrium, the expected profit is maximized at x = v.1 Thusthe first-order condition is π′(v) = 0. Using (3.3) we obtain

b′(v)F (v)n−1 = (v − b(v))(n − 1)f(v)F (v)n−2. (3.4)

This differential equation can be easily solved. Note that using (3.4), wecan write

(b(v)F (v)n−1)′ = b′(v)F (v)n−1 + b(v)(n − 1)f(v)F (v)n−2

= v(n − 1)f(v)F (v)n−2. (3.5)

The Fundamental Theorem of Calculus yields:

b(v)F (v)n−1 =∫ v

0

x(n − 1)f(x)F (x)n−2 dx + k,

where k is the constant of integration. If v → 0 the left-hand side tends to zerosince b(·) is bounded. Thus, we conclude that k = 0. That is, the candidateequilibrium bidding strategy is given by

b∗(v) =

(n − 1)∫ v

0

xf(x)F (x)n−2 dx

F (v)n−1 if 0 < v ≤ v;

0 if v = 0.

(3.6)

1 It is convenient to pause and think through the approach taken: we posited the existenceof an increasing, symmetric equilibrium function b(·). We then consider a “direct revelation”game where bidders are asked to announce a signal and their bids are then defined using thefunction b(·). Further, we assume that bidders 2, . . . , n announce their true signals and askwhat is Bidder 1’s best response. This approach is pursued throughout the remainder of thebook and it is formally expounded in Chapter 6.

16 Private Values

We need to check that b(v) is continuous. It suffices to show this only for v = 0.Note that when v > 0,

b∗(v) =(n − 1)

∫ v

0xf(x)F (x)n−2 dx

F (v)n−1(3.7)

<(n − 1)

∫ v

0vf(x)F (x)n−2 dx

F (v)n−1= v. (3.8)

Thus b(v) is continuous at zero and hence everywhere. Now let us check thatb∗ is indeed an equilibrium. From (3.3) and (3.4) we see that

π′(x) = (v − b(x))(n − 1)f(x)F (x)n−2 − b′(x)F (x)n−1

= (v − x)(n − 1)f(x)F (x)n−2.

Therefore if x < v, π′(x) > 0. And if x > v, π′(x) < 0. It is clear then that x = vmaximizes the expected utility. Note that (3.6) has a revealing interpretation.The equilibrium bid of a player with value v is equal to the expected valueof the individual with the second highest valuation conditional on v being thehighest valuation (see Appendix (A.20)). If my value v is the highest amongall players, then in a symmetric equilibrium where strategies are increasing,it suffices for me to bid just to outbid the opponent with the second highestvaluation.

It is a simple task to check that the equilibrium bidding strategy in (3.6) isstrictly increasing in v (simply differentiate 3.6).

From (3.7) we conclude that b∗(v) < v. Now from (3.4) we get that(b∗)′(v) > 0. Thus, the amount v − b∗(v) indicates by how much a biddershades his bid in equilibrium. In particular it says how much the bidder redu-ces his bid compared to his valuation. To calculate the shading we integrateexpression (3.6) by parts. The rule for integration by parts is as follows:

∫ b

a

u dz = uz |ba −∫ b

a

z du.

Letting z = F (x)n−1 implies that dz = (n − 1)F (x)n−2f(x) dx. Similarly,letting du = dx implies (by integration) that u = x. Therefore,

(n − 1)∫ v

0

xf(x)F (x)n−2 dx =∫ v

0

u dz

= xF (x)n−1 |v0 −∫ v

0

F (x)n−1 dx

= vF (v)n−1 −∫ v

0

F (x)n−1 dx. (3.9)


Replacing (3.9) into (3.6), we obtain

b∗(v) = v −∫ v

0F (x)n−1 dx

F (v)n−1. (3.10)

The amount of shading is therefore∫ v

0(F (x)/F (v))n−1 dx. It decreases with

the number of bidders. The larger is the number of my opponents, the closerto my valuation I will bid.

Now that we have a prediction for how bidders will behave in a first-priceauction, it is possible to ask what is the expected revenue for the seller from afirst-price auction, denoted by R1. The expected revenue is simply the expectedvalue of the highest bid, that is,

R1 = E[maxb∗(v1), . . . , b∗(vn)] = E[b∗(maxv1, . . . , vn)]

From the viewpoint of the seller, buyers are ex-ante identical. Thus, the pro-bability that all valuations are below a given value v is simply F (v)n and itsdensity is nF (v)n−1f(v) (see Appendix A). As a result, the expected revenuecan be written as:

R1 =∫ v

0

nb∗(v)F (v)n−1f(v) dv. (3.11)

In the remainder of this section we investigate individual behavior and com-pute the seller’s expected revenue from a Dutch auction. We will need thefollowing definition.

Definition 1 Two games with the same set of players and the same strategyspace are said to be strategically equivalent if each player’s expected profitsunder one of the games are identical to his expected profits in the other game.

We show that the Dutch auction is strategically equivalent to the first-priceauction. A bidding strategy in a Dutch auction is a function b(·) : [0, v] → R+.For example, consider the strategy profile (b∗1, . . . , b

∗n). Suppose b∗1 is the highest

bid. In a first-price auction, player 1 wins the object and his profits are v1 − b∗1,while the profits of all other players are equal to zero. In a Dutch auction, ifplayer 1 is the one stopping the clock at price b∗1, his profits are equal to v1−b∗1,while the profits of all other players are equal to zero. Player 1, however, waschosen arbitrarily. The conclusion is that for any player with the highest bid,if the same profile of strategies is used in both auctions, this profile yieldsthe same profits for all players. That is, the first-price auction and the Dutchauction are strategically equivalent. Thus, these two auction formats yield thesame expected revenue given by (3.11).

18 Private Values

3.1.2 Second-price Auctions

In a second-price sealed-bid auction, players submit their bids simultaneouslywithout observing the bids made by other players. We now explain Vickrey’s(1961) original insight that in such auctions it is in a bidder’s best interest toalways bid his own valuation. We will need the following definitions.

Definition 2 A strategy bi ∈ [0, v] is a dominant strategy for player i if

πi(vi, bi, b−i) ≥ πi(vi, bi, b−i)

for all bi ∈ [0, v] and for all b−i ∈ [0, v]n−1.

In words, bi is a dominant strategy for player i if it maximizes i’s expectedprofits for any strategies of the other players. An equilibrium in dominantstrategies is one where every bidder plays his dominant strategy. Formally,

Definition 3 An outcome (b∗1, . . . , b∗n) is said to be an equilibrium in

dominant strategies if b∗i is a dominant strategy for each player i, i = 1, . . . , n.

The reader can immediately show an equilibrium in dominant strategies isa Bayesian Nash equilibrium. The converse is not always true. Also it is easy tosee that bidding one’s true valuation is a dominant strategy in a second-priceauction. This is a remarkable property of the Vickrey auction. We explainintuitively why truth telling is a dominant strategy in a second-price auction.

Let us look at bidder 1 who has valuation equal to v1. Denote by b thehighest bid among players 2, . . . , n. Assume first that bidder 1 bids b1 < v1. Ifb1 > b then bidder 1 wins the object as he would have won with a bid equalto v1. However, if b1 < b < v1 then bidder 1 loses the auction. By bidding hisvaluation he would have won the auction and earned expected profits equal tov1− b. Therefore, bidder 1 does not gain by bidding less than his valuation andcould possibly lose. That is, his expected profits decrease with a bid b1 < v1.

Now suppose that bidder 1 bids b1 > v1. If b1 < b, then bidder 1 losesthe auction as he would have lost if he had bid his valuation. However, ifv1 < b < b1, then Player 1 wins the object and pays more than his valuation.That is, he loses b−v1. Therefore, bidder 1 does not gain by bidding more thanhis valuation but could possibly lose. Thus, his expected profits decrease witha bid b1 > v1.

We now show formally that telling the truth is a Bayesian Nash equilibriumbidding strategy. We examine the auction from the viewpoint of bidder 1, whohas a value equal to v1, and chooses a bid b1 to maximize his expected profitsgiven that players 2, . . . , n follow some strategy b(·). Bidder 1’s expected profitscan be written as

π1(v1, b1, b(·)) = E[(v1 − Y ) Ib1>Y ], (3.12)


where Ib1>Y denotes an indicator variable that is equal to 1 when b1 > Y andtakes the value 0 otherwise. Moreover, we suppose that bidder 1 assumes that hereceives the object in case of a draw2 and we let Y denote the highest valuationamong players 2, . . . , n. That is, bidder 1’s expected profits are equal to theexpected value of the difference between 1’s valuation and the second highestbid for the case when 1’s bid is greater than Y . The distribution function of thehighest among n − 1 samples is simply F (x)n−1 (see Appendix A). Therefore,we can take the expected value in (3.12) to obtain

π1(v1, b1, b(·)) =∫ b1

0

(n − 1)(v1 − x) f(x) F (x)n−2 dx. (3.13)

Bidder 1’s problem is to choose a b1 to maximize (3.13). Suppose first thatb1 < v1. Then if b1 is increased to v1 the integral in (3.13) increases by theamount ∫ v1

b1

(n − 1)(v1 − x)f(x) F (x)n−2 dx.

This is true since if b1 < x < v1, we have that v1 − x > 0. The reverse happensif b1 > v1 since in the region b1 > x > v1 and the integrand is negative. Thus,the expected profit maximizing bid is b1 = v1.

What is the expected revenue generated by the second-price auction? Giventhat each bidder bids his true valuation, the expected revenue is the expectedvalue of the second highest valuation. From Appendix A, the probability thatthe second largest of n draws from a fixed distribution is less than a certainvalue v is equal to F (v)n + nF (v)n−1[1 − F (v)]. The first term of the sumdenotes the probability that all draws are less than v and the second term ofthe sum presents the probability that v is the second highest value. (Consideringthat there are n ways to choose the highest valuation, F (v)n−1 represents theprobability of n − 1 valuations being smaller than v, and 1 − F (v) denotesthe probability of exactly one valuation being higher than v.) Therefore, thedensity of the second highest is n(n− 1)F (v)n−2[1−F (v)]f(v) and the seller’sexpected revenue is given by

R2 =∫ v

0

n(n − 1)vF (v)n−2[1 − F (v)]f(v) dv. (3.14)

As noted in the previous chapter, oral or English auctions are perhaps themost popular amongst auction mechanisms. Is it possible to analyze biddingbehavior in such complex auction format? Of course, these auctions are verydifficult to formalize. What should be the strategy space? For example, it isnot uncommon for bidders to make their bids by raising a hand or nodding tothe auctioneer instead of calling out their bids. It is also common for bidders

2 It is left as an exercise for the reader to prove that the following reasoning holds whateveris the tie-breaking method used.

20 Private Values

to wait until the very last minute to make a bid after being silent for most ofthe auction. We will ignore many of these complications and will refer to thefollowing version (sometimes referred to as Japanese auctions and introducedby Milgrom and Weber (1982)): each bidder presses a button while the priceincreases continuously. A participant drops out when he takes his hand off thebutton. The auction ends when there is only one bidder left pressing the button.This bidder wins the auction and pays the price at which the next-to-last playerstopped pressing the button. A strategy in this auction is a function from [0, v]into the non-negative real numbers. The strategy says that price at which thebidder releases the button.

Consider a strategy profile (b(·), . . . , b(·)) = (v1, . . . , vn). Suppose that b(v1)is the highest bid and that b(v2) is the second highest bid. In a second-priceauction, bidder 1 wins the auction and has profits equal to v1 − v2. Player2, . . . , n receive zero profits. In the oral auction—represented by the buttonauction—bidder 1 is the last pressing the button, while bidder 2 takes his handoff the button when the price reaches v2. Bidder 1’s profits are equal to v1−v2,while bidders 2, . . . , n earn zero profits. Note that the choice of players 1 and 2was completely arbitrary. Thus, the same profile of strategies in both auctionsyields the same profits for all players. That is, oral auctions and second-priceauctions are strategically equivalent. The expected revenue generated by bothtypes of auction is given by (3.14).

3.1.3 Revenue Equivalence

Among the four types of auctions considered above, first- and second-price,Dutch, and English auctions, which one generates the highest expected revenuefor the seller? It turns out that with independent private values, these fourauction formats generate the same expected revenue! This result is actuallyquite general as we will see in Chapter 6 and it is a by-product of the EnvelopeTheorem. A direct proof of the result below can be provided by just comparingexpressions (3.11) and (3.14).

Theorem 1 (revenue equivalence) With private independent values, the fourauction formats analyzed, first- and second-price, Dutch and Oral, yield thesame expected revenue.

Proof:

R1 =∫ v

0

b∗(x)nFn−1(x)f(x) dx =∫ v

0

nf(x)[b∗(x)Fn−1(x)]dx.

We can then use integration by parts to rewrite the above expression as:

= nF (x)b∗(x)Fn−1(x)|v0 −∫ v

0

nf(x)[b∗(x)Fn−1(x)]′dx.

= nb∗(v) −∫ v

0

n(n − 1)f(x)xFn−1(x)dx


We can use (3.6) to rewrite this expression as:

= n

∫ v

0

x(n − 1)Fn−2(x)f(x)dx −∫ v

0

n(n − 1)xf(x)Fn−1(x)]dx

=∫ v

0

n(n − 1)xFn−2(x)f(x)[1 − F (x)]dx = R2.

What happens to the seller’s expected revenue if the number of participantsincreases? Since the addition of one bidder valuation does not decrease thesecond highest valuation and might increase it, expected profits should increasewith the number of bidders. The proof is not difficult. Rewrite R2 = R2(n),

R2 =∫ v

0

n(n − 1)v[F (v)n−2 − F (v)n−1]f(v) dv

=∫ v

0

v[nF (v)n−1 − (n − 1)F (v)n]′ dv

= v[nF (v)n−1 − (n − 1)F (v)n]|v0 −∫ v

0

[nF (v)n−1 − (n − 1)F (v)n] dv

= v −∫ v

0

[nF (v)n−1 − (n − 1)F (v)n] dv.

Now if h(n) = −[nF (v)n−1 − (n − 1)F (v)n] we have that h(n + 1) − h(n)

= − ((n + 1)F (v)n − nF (v)n+1) + (nF (v)n−1 − (n − 1)F (v)n)

= nF (v)n+1 − 2nF (v)n + nF (v)n−1

= nF (v)n−1[F (v)2 − 2F (v) + 1]

= nF (v)n−1(F (v) − 1)2 ≥ 0.

Thus it follows that R2 =∫

h(n) dv is strictly increasing with n. Thus we haveproved the following.

Corollary 1 The seller’s expected revenue in any of the four auction formatsincreases with the number of participants.

The Revenue Equivalence Theorem is really quite remarkable. In its gene-ral form it establishes that any auction that allocates the object to the bidderwith the highest valuation (and satisfies a technical condition on assigningzero expected profits to the player with the lowest possible valuation) yieldsthe same expected revenue. The astute reader, however, will point out thatin the introduction we gave several examples of objects that are sold exclu-sively by oral auctions (e.g., houses, paintings, wool, etc.), objects that aresold by first-price auctions (e.g., government purchases), objects that are sold

22 Private Values

exclusively by Dutch auctions (e.g., flowers) and that second-price auctionsare rare. The Revenue Equivalence Theorem would predict that the auctionmechanism does not matter so we would expect to see flowers, for example,being sold by different auction formats.

One could argue that tradition plays an important role in the establishmentof the auction format. This argument is difficult to justify, however, as in somecases these are new markets (such as auctions of used cars). Although we doobserve changes in auction formats (e.g., wool in Australia is now sold byelectronic auctions whereas it used to be sold through oral English auctions) insome markets, there are several examples of little experimentation with otherauction formats. This leads us to conclude that there may be other factors atwork that are not captured by the IPV model.

Indeed later we will examine several extensions of the Independent Pri-vate Values model where revenue equivalence breaks down; for example, whenbidders are risk averse or when their valuations are correlated.

Although the revenue equivalence result is not robust, some of the insightsdeveloped above are robust and have been applied successfully to the designof several new markets. In a later chapter we will pursue a more abstractapproach and analyze the private independent values model under the realm ofthe revelation principle. In the remainder of this section we examine the effectof the seller setting a reserve price (above his valuation, which is equal to zeroby assumption) or an entry fee or both to try to raise his expected revenue.

3.1.4 Reserve Prices and Entry Fees

We denote the reserve price by r and the entry fee by δ. The reserve price isassumed to be known to all bidders and the seller is assumed to have committedto not selling below the reserve. In essence, the reserve price is the minimum bid.

Note that these two instruments, the reserve price and the entry fee, gene-rate two opposing effects: they reduce bidder’s incentives to participate in theauction but they might increase revenue as the seller collects extra revenue eit-her via the entry fee—those bidders who enter have to pay the entry fee to theseller—or via the effect of a reserve price on bidding behavior—those bidderswho enter bid more aggressively. We allow the seller to set both a reserve priceand an entry fee concurrently.

We will only examine the effects of a reserve price and entry fees on equi-librium behavior and on the seller’s expected revenue in a first-price auction.The analysis of their effects in second-price auctions is left as an exercise forthe reader. The objective here is to illustrate that these two instruments canbe used to increase the seller’s expected revenue. This theme will be discussedagain in Section 6.3 when we use the mechanism design approach to identifythe optimal auction—the auction that maximizes the seller’s expected revenue.


Below we will find a cut-off value ρ for player i such that if vi < ρ, i doesnot participate in the auction. If vi ≥ ρ then i does participate. We will assumethat players 2, . . . , n follow this participation rule and bid according to a strictlyincreasing differentiable function b(·). Then we compute 1’s best reply. We findan equilibrium such that b(ρ) = r. Suppose Player 1’s valuation is v1 = v. Hisproblem is to choose a participation rule and a bid b1 so as to maximize hisexpected profits:

π1 = E[(v − b1)Ib1 ≥ maxb(Z), r] − δ,

where Z = maxvj ; vj ≥ ρ, j = 2, . . . , n if the set is non-empty and Z = 0otherwise. In order to bid, bidder 1’s bid must be greater than or equal to r.Thus, 1’s expected profits can be rewritten as

π1 = −δ + (v − b1) Pr[b1 ≥ maxb(Z), r]

= −δ + (v − b1) Pr[b1 ≥ maxb(Z), b(ρ)].

If b1 = r then π1 = −δ + (v − r)Fn−1(ρ). If b1 = b(s) > r then

π1 = −δ + (v − b(s)) Pr[s > Z] = −δ + (v − b(s))Fn−1(s). (3.15)

Note that s > ρ. We can now compare (3.15) with (3.2) and conclude thatthe first-order condition is the same as in the case where both the reserveprice and the entry fee were equal to zero. The only distinction is that theboundary condition has to reflect the fact that a bidder with value equal to ρhas to be indifferent between entering or not and therefore b(ρ) = r. Thus, theequilibrium bidding function is given by:

b∗(v) =

v −

δ +∫ v

ρF (x)n−1dx

F (v)n−1, if v ≥ ρ

not bid, otherwise. (3.16)

We leave as an exercise to the reader to analyze the case b(ρ) > r and showthat this cannot arise as a symmetric equilibrium.

We can now compare (3.16) with the equilibrium bidding strategy (3.10)under zero entry fees and zero reserve price. Note that while entry fees will onlyaffect a bidder’s decision whether or not to enter, a non-zero reserve price willaffect both the decision to enter and the bidding strategies—those bidders whodo enter bid more aggressively in the symmetric equilibrium. Nevertheless,entry fees and reserve prices affect the seller’s expected revenue in a similarfashion as we will explain below.

To fully characterize equilibrium behavior we still need to compute the cut-off value ρ. Recall that a player with value ρ is indifferent between participatingor not:

−δ + (ρ − r)Fn−1(ρ) = 0.

24 Private Values

That is, when the indifferent player participates, he pays the entry fee δ. Sincehe only wins if he is the only participant, the price paid in the auction is thereserve price r. Fn−1(ρ) denotes the probability that the indifferent bidder isthe only participating bidder. We can rewrite this expression as:

(p − r)Fn−1(ρ) = δ. (3.17)

Clearly we have ρ > r and δ < v − r. This last inequality follows from the factthat the maximum value of ρ− r is equal to v− r and that Fn−1(ρ) < 1. Thus,we can conclude that δ + r < v. If this inequality were not true, then no bidderwould ever participate in this auction.

From (3.16) and (3.17) we note that there are two effects on equilibriumbehavior from imposing an entry fee and a reserve price. Firstly, lower valuationbidders will not participate in the auction. Secondly, those who do participatebid more aggressively—the reason is that to win the auction now a player hasto bid the expected value of the highest among his opponents with values bet-ween ρ and v, and not between 0 and v as before. Lower participation reducesthe seller’s expected revenue but more aggressive bidding increases it. Thus, wecan now ask what is the combination of reserve price and entry fee that maxi-mizes the seller’s expected revenue. Firstly, let us write the seller’s expectedrevenue:

R1 =∫ v

ρ

b∗(v)nFn−1(v)f(v)dv + nδ(1 − F (ρ)),

where the second term in the RHS (right hand side) represents the expectedrevenue from the entry fee when players follow b∗(·), that is

δ

n∑k=0

(n

k

)(n − k)F (ρ)k(1 − F (ρ))n−k = nδ(1 − F (ρ)).

Note that the entry fee δ and the reserve price r are linked via equation (3.17).For example, we can set δ = 0 and r = ρ. Alternatively, we can set r = 0and δ = ρFn−1(ρ) or any pair satisfying (3.17). Let us assume that δ = 0 andr = ρ and so we can rewrite R1 as

R1 =∫ v

ρ

(v −

∫ v

ρF (x)n−1 dx

F (v)n−1

)nFn−1(v)f(v) dv

=∫ v

ρ

vnF (v)n−1f(v) dv −∫ v

ρ

(∫ v

ρ

F (x)n−1 dx

)nf(v) dv.

Changing the order of integration in the double integral yields (as ρ < x < vand ρ < v < v, we have that when integrating over v, first we obtain

3.2 The Correlated Private Values Model 25

∫ v

xf(v) dv = 1 − F (x)):

R1 =∫ v

ρ

vnF (v)n−1f(v) dv −∫ v

ρ

n(1 − F (v))F (v)n−1 dv.

Since we are trying to find the value of ρ that maximizes R1, we use Leibnitz’srule to differentiate the above expression and obtain:

∂R1

∂ρ= −ρnF (ρ)n−1f(ρ) + n(1 − F (ρ))F (ρ)n−1

= nF (ρ)n−1−ρf(ρ) + 1 − F (ρ).

At an interior maximum:

nF (ρ)n−1f(ρ)(−ρ +

1 − F (ρ)f(ρ)

)= 0 (3.18)

or

ρ =1 − F (ρ)

f(ρ). (3.19)

That is, condition (3.19) tells us the level of the reserve price that maximizesthe expected revenue of a seller using a first-price auction when the seller doesnot charge any entry fees. It turns out that setting a positive reserve price (i.e.,a reserve price above the seller’s valuation) maximizes the seller’s expectedrevenue. We will discuss this in more detail in Chapter 6 but the economicsbehind it is very simple. It relates to the standard monopoly pricing: justas a standard monopolist charges a price higher than the marginal cost toextract surplus from higher valuation buyers at the sacrifice of lower valuationsbuyers who do not consume the good, a seller sets a reserve price to extractmore expected surplus from the highest valuation bidder but it excludes theparticipation of lower valuation buyers.

Of course, this is not (ex-post) efficient as the seller will not sell the objectin some events whereas efficiency dictates that it should be sold given that theseller’s value is assumed to be zero. This is analogous to the deadweight lossresulting from a standard monopolist not serving lower valuation costumers.

3.2 The Correlated Private Values Model

In this section, we relax one of the main hypotheses of the IPV model of auc-tions. More specifically, we assume that individuals’ types are correlated. Theset of types is [0, v]n. Bidder i knows his type but does not know the otherbidders’ types. The distribution of types is common knowledge but it is no lon-ger independent. When a bidder receives his type, he updates his beliefs aboutother bidders’ types. More specifically, we assume that f : [0, v]n → [0,∞) is

26 Private Values

the density of the vector of types (X1, X2, . . . , Xn). As mentioned in Chapter 2,our viewpoint when comparing auctions is that of the seller or from an analystthat does not have information about individuals’ types. Moreover, we focus onsettings where the identity of an individual bidder is not particularly relevant.That is, we assume that bidders are ex-ante symmetric. In order to introducethe notion of symmetry we need the following definition.

Definition 4 A permutation of the set S is a bijection σ : S → S.

We can now define the notion of a symmetric function.

Definition 5 A function u : [0, v]n → R is symmetric if for every permu-tation σ of the set 1, . . . , n and for every x ∈ [0, v]n, u(x1, x2, . . . , xn) =u(xσ(1), xσ(2), . . . , xσ(n)).

For example, if n = 3, u(a, b, c) is symmetric if

u(a, b, c) = u(a, c, b) = u(b, a, c) = u(b, c, a)

= u(c, a, b) = u(c, b, a)

for every (a, b, c) ∈ [0, v]3.We will apply this concept to the distribution of types. For example,

exchanging the types of bidders 1 and 2 in a vector (x1, x2, x3, . . . , xn) isof no consequence, that is, f(x1, x2, x3, . . . , xn) = f(x2, x1, x3, . . . , xn). Thisassumption is stated below along with some regularity restrictions.

Symmetry The density f is symmetric, strictly positive and continuous.

We now investigate equilibrium bidding behavior in second- and first-priceauctions. We should note that with private values, the English button auctionand the second-price auction are strategically equivalent. There is no informa-tion being revealed in an English (button) auction that changes an individual’svaluation. As we mentioned in the previous section, Dutch and first-priceauctions are always strategically equivalent no matter what the structure ofvalues is.

3.2.1 Second-price Auction

Let us show that in the second-price auction the equilibrium is still to bid one’svalue which is given by one’s type in our model. Recall that Y = maxj≥2 Xj .Thus, suppose bidder j bids xj , j = 2, . . . , n and let us find the best reply of


bidder 1. If he bids t ≥ 0, then his expected utility is given by

h(t) = E[(x − Y )It>Y |X1 = x].

That is, when bidders 2, . . . , n bid their true values, then bidder 1 wins thedifference between his value and the second highest value if he bids more than Y .Note that the expectation is taken with respect to the highest value amongst1’s opponents conditional on 1’s type being x. Otherwise, his profits are zero.

Suppose that t > x. Then we have that x − Y < 0 whenever t > Y > x.Thus, bidder 1’s expected profits can be written as:

h(t) = E[(x − Y )It>Y ≥x + (x − Y )It>x>Y |X1 = x]

≤ E[(x − Y )Ix>Y |X1 = x] = h(x).

That is, bidder 1 can increase his expected profits by decreasing his bid t whent > x. Now suppose that t < x. Then x− Y > 0 whenever x > t > Y . Thus bya similar reasoning we can conclude that

h(t) ≤ E[(x − Y )Ix>Y |X1 = x] = h(x).

That is, bidder 1 can increase his expected profits by increasing his bid t whent < x. Therefore, the best reply of bidder 1 is to bid t = x, his type. If wedenote by fX the marginal density of X1 the expected payment of bidder 1when his type is x is given by

Ps(x) := E[Y Ix≥Y |X1 = x]

=∫

xj≤x,j≥2

max2≤j≤n

xjf(x, x2, . . . , xn)

fX(x)dx2 · · · dxn.

That is, bidder 1’s expected payment in a second-price auction is simply equalto the expected value of the highest type amongst his opponents, Y , conditionalon bidder 1’s having received a type x > Y . Note that the seller’s revenue isequal to expected payment of the highest valuation bidder.

3.2.2 First-price Auction

To find the equilibrium equation for the first-price sealed-bid auction withcorrelated types is a harder task. Denote by fY |X the conditional density of Ygiven X and by FY |X the conditional distribution. That is,

fY |X(y |x) =f(x, y)∫f(x, y) dy

and FY |X(y |x) =∫ y

0

fY |X(z |x) dz.

28 Private Values

Suppose bidders j = 2, . . . , n bid accordingly to b(·). Suppose bidder 1 bidst = b(s). Then his expected utility is

h(s) = (x − b(s))E[Is>Y |X = x] = (x − b(s)) FY |X (s |x).

Differentiating, we obtain

h′(s) = −b′(s) FY |X (s |x) + (x − b(s)) fY |X (s |x). (3.20)

If s = x is to be optimum then

−b′(x) FY |X(x |x) + (x − b(x)) fY |X (x |x) = 0.

Solving for b′ we obtain the differential equation

b′(x) = (x − b(x))fY |X (x |x)FY |X (x |x)

. (3.21)

To solve this equation we use the integrating factor method which is explainedin Appendix B. Define γ(x) = fY |X(x |x)/FY |X(x |x). We may rewrite theequation as

b′(x) + γ(x)b(x) = xγ(x).

It is now clear that an integrating factor must solve P ′ = Pγ. Therefore P (x) =exp[−

∫ v

xγ(u)du] is an integrating factor. Thus,

(Pb)′(x) = P (x)b′(x) + P ′(x)b(x)

= P (x)(b′(x) + γ(x)b(x))

= P (x)xγ(x).

Now since b(0) = 0 and P (0) ≤ 1 we obtain integrating between 0 and x that:

P (x)b(x) =∫ x

0

P (u)γ(u)u du =∫ x

0

P ′(u)u du. (3.22)

Thus, we get from the first equality

b(x) =∫ x

0

exp[−∫ x

u

γ(v) dv

]γ(u)u du.

If we integrate by parts the last term of equation (3.22) we obtain that

b(x) =

∫ x

0P ′(u)u du

P (x)= x −

∫ x

0P (u) du

P (x).

To show that this is an equilibrium, we go back to (3.20):

h′(s) = −b′(s)FY |X(s |x) + (x − b(s))fY |X(s |x)

= −b′(s)FY |X(s |x) + (x − s)fY |X(s |x) + (s − b(s))fY |X(s |x).


Using (3.21) we may write

h′(s) = −b′(s)FY |X(s |x) + (x − s)fY |X(s |x) + b′(s)fY |X(s |x)

γ(s)

= fY |X(s |x)(

x − s + b′(s)(

1γ(s)

− FY |X(s |x)fY |X(s |x)

)).

Suppose that x → fY |X(s |x)/FY |X(s |x) is increasing. Then h′(s) > 0 if andonly if x > s. Thus we proved the following:

Theorem 2 If the ratio fY |X(s|x)/FY |X(s|x) is increasing in x then a sym-metric equilibrium of the first-price sealed-bid auction is given by the functionbf(·) defined by

bf(x) =∫ x

0

exp[−∫ x

u

γ(v) dv

]γ(u)u du

= x −∫ x

0

exp[−∫ x

u

γ(v) dv

]du. (3.23)

where γ(u) = fY |X(u |u)/FY |X(u |u). Moreover, bf(x) solves the differentialequation

b′(x) = (x − b(x)) γ(x)

with initial condition b(0) = 0.

Remark 1 The condition that fY |X(s |x)/FY |X(s |x) is increasing in x issatisfied if f(X,Y ) has the monotone likelihood ratio property (see Definition 6on page 47). This property will be studied in detail in Chapter 4.

3.2.3 Comparison of Expected Payment

We may easily compare the expected payment of the bidders in the first- andsecond-price auctions. The expected payment of a bidder with valuation x inthe first-price auction is given by bf(x)FY |X(x |x), that is, his bid times theconditional probability of winning. Thus

Pf(x) := bf(x)FY |X(x |x)

= FY |X(x |x)∫ x

0

exp[−∫ x

u

γ(v) dv

]γ(u)u du.

30 Private Values

We can also write the expected payment of a bidder with valuation x in asecond-price auction as:

Ps(x) = E[Y |X = x] =∫ x

0

yfY |X(y |x) dy

=∫ x

0

(y − bf(y)) fY |X(y |x) dy +∫ x

0

bf(y) fY |X(y |x) dy

=∫ x

0

b′f (y)fY |X(y |x)

γ(y)dy +

∫ x

0

bf(y)fY |X(y |x) dy.

Now note that since y < x, 1/γ(y) = FY |X(y | y)/fY |X(y | y) ≥ FY |X(y |x)/fY |X(y |x). Therefore, the inequality fY |X(y |x)/γ(y) ≥ FY |X(y |x) istrue. Thus

Ps(x) ≥∫ x

0

b′f(y)FY |X(y |x) dy +∫ x

0

bf(y) fY |X(y |x) dy

=∫ x

0

[b′f(y)FY |X(y |x) + bf(y)fY |X(y |x)] dy

=∫ x

0

[bf(y)FY |X(y |x)]′ dy = bf(x)FY |X(x |x) = Pf(x).

That is, with correlated types, the second-price auction yields more expectedrevenue than a first-price auction.

Example 2 Suppose f(x, y) = (1 + f(x)f(y))/2 where f(x) is an increasingdensity on [0, 1]. Then

fY |X(s |x)FY |X(s |x)

=1 + f(x)f(s)s + f(x)F (s)

is increasing. To see this note that the marginal density is

fX(x) =∫

f(x, y) dy =1 + f(x)

2= fY (x).

The conditional density

fY |X(y |x) =f(x, y)fX(x)

=1 + f(x)f(y)

1 + f(x)

and the conditional distribution is

FY |X(y |x) =y + f(x)F (y)

1 + f(x).

Thus, the likelihood ratio is given by:

γ(y, x) =fY |X(y |x)FY |X(y |x)

=1 + f(x)f(y)y + f(x)F (y)

. (3.24)


Then

∂γ(y, x)∂x

=∂

∂x

(1 + f(x)f(y)y + f(x)F (y)

)

=f ′(x)f(y)

y + f(x)F (y)− (1 + f(x)f(y))f ′(x)F (y)

(y + f(x)F (y))2

= f ′(x)f(y)(y + f(x)F (y)) − (1 + f(x)f(y))F (y)

(y + f(x)F (y))2

= f ′(x)yf(y) − F (y)

(y + f(x)F (y))2≥ 0,

since F (y) =∫ y

0f(z) dz ≤

∫ y

0f(y) dz = f(y)y.

Let us particularize further with an example to explicitly compute theequilibrium strategies.

Example 3 Suppose F (x) = x2. Then

γ(x) =1 + 4x2

x(1 + 2x2)and P (x) = x

√1 + 2x2

3.

The first-price auction equilibrium bidding function is given by:

b(x) = x −∫ x

0u√

(1 + 2u2)/3 du

x√

(1 + 2x2)/3.

This function is plotted below, for comparison, together with the functionx → x/2.

0.5

0.375

0.25

0.125

0 0.25 0.5 0.75 1x

y

32 Private Values

3.3 The Effect of Risk Aversion

We start with the second-price auction. We are now assuming that biddersare risk averse and calculate their utility with the concave von Neumann-Morgenstern utility u(·). We suppose u′ > 0 ≥ u′′. We normalize u(0) = 0.We want to prove that to bid this type is still an equilibrium bidding function.Thus suppose bidders j = 2, . . . , n bid xj . If bidder 1 bids t ≥ 0 his expectedutility is

h(t) = E[u(x − Y )It>Y |X = x] =∫ t

0

u(x − y)fY |X(y |x) dy.

Differentiating we have that

h′(t) = u(x − t)fY |X(t |x).

Thus, h′(t) > 0 if and only if u(x − t) > 0. That is, h′(t) > 0 if and only ift < x. Hence, t = x maximizes the expected utility.

We now consider the first-price auction. Suppose b(·) is a strictly increasingcontinuous bidding strategy played by bidders i = 1. Let us find bidder 1’s bestreply. If he bids t = b(s) the expected utility is

h(s) = u(x − b(s)) Pr(s > Y |X = x) = u(x − b(s))FY |X(s |x).


h′(s) = −b′(s)u′(x − b(s))FY |X(s |x) + u(x − b(s))fY |X(s |x).

If s = x is to be the optimal then h′(x) = 0 and therefore,

b′(x) =u(x − b(x))u′(x − b(x))

fY |X(x |x)FY |X(x |x)

. (3.25)

Suppose now that b(·) solves this differential equation. Then

h′(s)FY |X(s |x)

= −b′(s)u′(x − b(s)) + u(x − b(s))fY |X(s |x)FY |X(s |x)

= − u(s − b(s))u′(s − b(s))

fY |X(s | s)FY |X(s | s)u′(x − b(s)) + u(x − b(s))


.

3.3 The Effect of Risk Aversion 33

If s > x then fY |X(s |x)/FY |X(s |x) ≤ fY |X(s | s)/FY |X(s | s) and x − b(s) <s − b(s). Thus,

− u(s − b(s))u′(s − b(s))

fY |X(s | s)FY |X(s | s)u′(x − b(s)) + u(x − b(s))


≤(− u(s − b(s))

u′(s − b(s))u′(x − b(s)) + u(x − b(s))

)fY |X(s | s)FY |X(s | s)

=(− u(s − b(s))

u′(s − b(s))+

u(x − b(s))u′(x − b(s))

)fY |X(s | s)u′(x − b(s))

FY |X(s | s)

≤(− u(s − b(s))

u′(s − b(s))+

u(s − b(s))u′(s − b(s))

)fY |X(s | s)u′(x − b(s))

FY |X(s | s) = 0.

The first inequality follows fY |X(s |x)/FY |X(s |x) ≤ fY |X(s | s)/FY |X(s | s) andthe second inequality follows from u(x)/u′(x) being increased. Analogously wecan show that if s < x, h′(s) > 0. Thus b(·) defined by the differential equationabove is the equilibrium.

For example suppose u(x) = xt, t ∈ (0, 1). Then

u(x − b(x))u′(x − b(x))

=(x − b(x))t

t(x − b(x))t−1=

x − b(x)t

.

In this case, the differential equation becomes:

b′(x) =u(x − b(x))u′(x − b(x))

γ(x) = (x − b(x))γ(x)

t.

3.3.1 Revenue Comparison

Define b1(·) as the equilibrium bidding function in a first-price auction whenthere is indifference to risk. Suppose b2(·) is the equilibrium bidding functionin a first-price auction when risk aversion is captured by the utility function u.Since u is concave, u(z) ≥ u′(z)z. Thus,

b′2(x) =u(x − b2(x))u′(x − b2(x))

γ(x) ≥ (x − b2(x))γ(x).

The following inequality is immediate from this:

(b2(x) − b1(x))′ = b′2(x) − b′1(x)

≥ (x − b2(x))γ(x) − (x − b1(x))γ(x)

= (b1(x) − b2(x))γ(x).

34 Private Values

Thus if P (x) = e−∫ vx

γ(s)ds then P ′ = Pγ(x) and therefore (we omit theargument x for conciseness)

ddx

(P (b2 − b1)) = Pγ(b2 − b1) + P (b2 − b1)′

≥ Pγ(b2 − b1) + P (b1 − b2)γ = 0.

Finally, we have that P (x)(b2(x) − b1(x)) ≥ 0. Thus, a risk-averse bidder bidsuniformly more aggressively than a risk-neutral bidder. Hence, we have justproved the following result.

Theorem 3 If there are risk aversion and private values the first-price auc-tion generates more revenue than the second-price auction or the Englishauction.

3.4 The Discrete Valuation Case

In this section, we relax another assumption from the standard IPV model.We allow valuations to be drawn from a discrete distribution. We will usean example to illustrate that with discrete types we will have to look for aBayesian Nash equilibrium in mixed strategies as a pure strategy equilibriumwill not exist.

Example 4 Suppose we have two bidders with valuations x ∈ 0, 1 with equalprobabilities. Then, there is no equilibrium of the first-price auction in purestrategies. To see this, suppose bidder 2 bids 0 if his valuation is 0 (a bidderwith zero valuation will never bid higher than zero). And if his valuation is 1his bid is b ≥ 0. Similarly, bidder 1 bids zero if his valuation is zero. What ishis bid if his valuation is 1? The expected utility of bidder 1 is

π1 = 12 (1 − 0) + 1

2 (1 − b1) if b1 > b.

Thus if b ≥ 1, bidder 1 bids b1 = 0. If b < 1 and b1 > b, π1 = 12 + 1

2 (1 − b1).Thus, if b ≥ 1 the best reply of bidder 1 is to bid 0 and if b < 1 there is no bestreply since the expected utility increases as b1 > b decreases. That is, there isno pure strategy Bayesian Nash equilibrium.

Let us find a mixed strategy equilibrium. First, note that if bidder 2 has azero valuation, then he bids zero for sure. If bidder 2 has valuation 1, supposethat his bid belongs to [0, x] with probability G(x). For G to define an equilibriumstrategy, we need that Pr(x) = 0 whenever x > 0—given that if Pr(x) > 0whenever bidder 1 intends to bid x, he may slightly increase his bid increasinghis probability of winning by Pr(x). What is the best reply for bidder 1? Hisexpected utility is

π1 = 12 (1 − b) + 1

2 (1 − b)G(b).

3.5 Exercises 35

Thus, bidder 1 will bid b such that b maximizes (1− b)(1 + G(b)). Thus if F isthe distribution of bidder’s 1 bids we have that

F (b; b does not maximize (1 − b)(1 + G(b))) = 0;

G(b; b does not maximize (1 − b)(1 + F (b))) = 0.

Suppose k = maxb≥0(1−b)(1+G(b)). Then if x, x′ are such that F (x)F (x′) > 0,

(1 − x)(1 + G(x)) = k = (1 − x′)(1 + G(x′)).

The infimum of such x must satisfy (1−x) = k or x = 1−k. The supremum ofx′ : (1−x′)2 = k. Thus the support of the distribution is [1−k, 1− k

2 ]. If k = 1,

F (x) = G(x) =1

1 − x− 1 =

x

1 − x;

x ∈[0, 1

2

].

1

0.8

0.6

0.2

0.4

0 0.1 0.2 0.3 0.4 0.5x

y

These probability distributions represent the mixed strategy equilibrium forthis example. They are depicted in the diagram above. The interpretation isthat a bidder who receives a type 1 will submit a bid in [0, α] with probabilityα/(1 − α) if α ≤ 1

2 and will never submit a bid higher than 12 .

3.5 Exercises

1. Compute the equilibrium bidding strategy in both first- and second-price auctions when the seller sets a reserve price equal to v0. That is,the seller only accepts bids that are greater or equal to v0. What is theseller’s expected revenue in both auctions? Does revenue equivalence stillhold?

36 Private Values

2. (Riley and Samuelson 1981): Consider an auction with two buyers withvaluations drawn independently from the uniform [0, 1] distribution. Theseller sets a reserve price equal to 1

2 and she employs the following auctionrules:(a) There is a single round of bidding. Buyer 1 is given the opportunity

to make a bid a price b1 ≥ 12 .

(b) If buyer 1 bids b1 ≥ 12 , buyer 2 can match b1 and win the object. If

buyer 1 makes no bid, buyer 2 can obtain the good at price 12 if he

so chooses.(i) Does this auction resemble any selling mechanism that you

know of?(ii) Can you compute the buyers’ equilibrium bidding strategies

and the seller’s expected revenue?(iii) Is the object in equilibrium always allocated to the individual

with the highest valuation?(iv) Compare the expected revenue generated by this auction with

the expected revenue generated by a second-price auction withreserve price equal to 1

2 .3. The deduction leading to equation (3.1) is incomplete as b1 may not be

in the range of b(·). Reduce the general case to the case b1 ∈ b([0, 1]).4. Compute the seller’s expected revenue if the number of bidders goes to

infinity.5. In the text we assumed that in case of a tie the object is not deli-

vered to any bidder. Show that the equilibrium strategies obtainedabove (for both first- and second-price auctions) still hold under anytie-breaking rule.

6. Show formally that to bid one’s own value is a dominant strategy in theVickrey auction.

7. Suppose we have an auction in which every bidder pays his bid whetheror not he wins the object. Find the equilibrium strategies in this case ifthere are n bidders and the distribution is uniform. (This auction formatis called an all-pay auction for obvious reasons.)

8. Suppose there are two bidders and the distribution is uniform. Supposeif you win or not you pay the second highest bid. Find the equilibriumstrategies. What is peculiar about them? (This auction format is referredto as the war of attrition.)

9. Characterize equilibrium behavior and the seller’s expected revenue in asecond-price auction with independent private values, a reserve price rand an entry fee δ. Show that the expected revenue is the same as theone generated by a first-price auction with reserve price r and entry feeδ as derived in Section 3.1.4.

10. Suppose there are two bidders. Calculate the expected revenue from afirst-price auction as a function of the reserve price r for the following

3.5 Exercises 37

distributions defined for x ∈ [0, 1]:(a) F (x) = x(b) F (x) = x2

(c) F (x) =√

x.11. Find the optimal reserve price for each distribution of the previous

exercise.12. Now suppose the two bidders of exercise (3.5) collude to bid together

and that the auctioneer knows this. Find the new optimal reserve pricesfor each of the distributions above. If the numbers are different fromthose calculated in (a), (b) and (c) above, explain why this does notcontradict the fact that the optimal reserve price is independent of thenumber of bidders.

13. In Example 2, suppose that F (X) = xθ and therefore f(x) = θxθ−1.Show that

P (x) = exp[−∫ 1

x

γ(x) dx

]=

x√

a + θx2θ−2

√a + θ

.

And that

bf(x) = x − 1x√

a + θx2θ−2

∫ x

0

u√

a + θu2θ−2 du.

14. Show that k = 1 is necessary for an equilibrium in the reasoningunderlying Example 4.

15. Find the equilibrium distribution if the set of valuations is 0, 1, 2 andthere are two bidders.

16. Is the equilibrium in the discrete case efficient?17. Show that setting the reserve price ρ = 0 is never optimal for the seller.18. Find the equilibrium bidding functions of the first-price auction for

distribution F (x) = xθ, θ > 0 and n > 1 bidders.19. Show that γ(y, x) in (3.24) is increasing in x if f(x) is decreasing.20. Show that (3.10) is increasing in n and that the limit of the seller’s

expected revenue when n → ∞ is equal to v.


4

Common Value

In this chapter, we consider a situation where the value of the object being soldis the same for all players but this value is not known at the time that biddingtakes place. A typical example includes bidding for the rights of exploration ofsome natural resource. For given international prices, the amount of oil availablein a particular area is fixed but it is only really known after the auction andwhen the successful bidder starts drilling.

Most auctions will have a common value component associated withthe object being sold. For example, in a house auction the common valuecomponent might be the expected market value of the house.

Common value auctions are usually associated with a phenomenon knownas the “winner’s curse”. Capen et al. (1971) claim that “the winner tends tobe the player who most overestimates the true tract value” and that resultedin low profits earned by oil companies on offshore tracts in the 1960s. Ourreader will have to ask himself or herself how a fully rational bidder, indeedlike the bidders we consider in this book, can overestimate the true value ofthe object. Given that equilibrium behavior requires maximization of ex-antepayoffs, there cannot be a winner’s curse in our setting. That is, bidders willtake into account the possibility that they might overestimate the value and willbid more cautiously. How cautiously they bid will vary with the auction formatand this will have implications for auction design and the seller’s expectedrevenue. A full account of this relationship is given in the next chapter. Herewe work our way through some examples to illustrate the effects of the commonvalue assumption in bidding behavior and on the seller’s expected revenue. Inthis process, we will touch on several topics that will be dealt with in greaterdetail in the next chapter.

In common value auction models we usually refer to the bidders’ types asbidders’ signals. We do this because we prefer to think of types as completelydetermining the bidders preferences whereas in common values models the bid-ders’ preferences might be a function of the other bidders types/signals as well.In our first example we assume that the individual signals about the true value

39

40 Common Value

of the object are independently determined. In the second example, we allowfor types to be correlated.

4.1 An Example with Independent Signals


Let us consider the case where the true value of the object is V = v1+v2, wherethe vi’s are independent and uniformly distributed on [0, 1]. For example, itmight be the case that the true value of the object has two components, bidder 1receives a signal about the first component, bidder 2 about the second, and thetrue value of the object is the sum of both components. For example if V isthe amount of oil on a particular tract, v1 might indicate the amount of oil onpart of the tract and v2 the amount of oil in the complementary area.

We will now look at bidding behavior in a first-price auction and com-pare with that developed in Chapter 3. Suppose bidder 2 follows some strictlyincreasing strategy b(·). Bidder 1, who has received signal v1 = v, will chooseb1 = b(ω) to maximize expected profits:

π1 =∫ ω

0

(v + y − b(ω)) dy = vω +ω2

2− b(ω)ω,

where we are integrating over all cases where Player 1 wins the auction, that is,when b(ω) > b(y) or ω > y. Thus differentiating with respect to ω and settingthe derivative equal to zero yields

v + ω − b(ω) − ωb′(ω) = 0.

We can now set ω = v as we are looking for the symmetric equilibrium:

2v − b(v) − vb′(v) = 0.

Thus,

2v = (vb(v))′.

If a bidder with signal v = 0 bids b0 ≥ 0 (why we cannot dismiss b0 > 0?),integrating both sides yields:

vb(v) = vb(v) − 0 · b0 =∫ v

0

2y dy = v2.

Thus b(v) = v if v > 0. Since we are supposing b(·) continuous, b0 = b(0) =limv→0 v = 0. Therefore,

bF(v) = v.

This is the equilibrium bidding strategy in this first-price auction. Note thatin a symmetric increasing equilibrium it suffices to outbid one’s opponent.

4.1 An Example with Independent Signals 41

By bidding one’s signal, in a symmetric equilibrium, one is guaranteed to winthe auction if one has the highest signal. Note that by following such biddingstrategy a bidder will never pay more than the true value of the object! Is thisthe rule in common values models? To analyze further this possibility let usconsider a more general distribution. Suppose each bidder valuation v ∈ [0, 1]is drawn from the distribution F (·). Suppose also that it has a density f = F ′

which is continuous and strictly positive. We look again for a symmetric equi-librium b(·). Suppose bidder 1 has a signal v. If bidder 2 bids according to b(·),bidder 1’s expected profit by bidding b = b(ω) is given by

π =∫ ω

0

(v + y − b(ω))f(y) dy =∫ ω

0

(v + y)f(y) dy − b(ω)F (ω). (4.1)

Thus, the first-order condition is

π′ = (v + ω)f(ω) − (b(ω)F (ω))′ = 0. (4.2)

In an equilibrium ω = v and thus we obtain the differential equation

(b(v)F (v))′ = 2vf(v). (4.3)

If b(0) = b0, the solution of this differential equation is given by:

b(v)F (v) = b(v)F (v) − b0F (0) =∫ v

0

2yf(y) dy.

Therefore,

b(v) =

∫ v

02yf(y) dy

F (v). (4.4)

The reader can verify that the continuity of b(·) implies that b0 = 0. It isa simple matter to check that this is really a symmetric equilibrium biddingstrategy. Just use (4.3) to substitute (b(ω)F (ω))′ by 2ωf(ω) in (4.2). Let usparticularize further and assume that F (v) = vθ, θ > 0. Then by following thesteps above we obtain

b(v) =2θ

θ + 1v.

Thus, a bidder will bid higher than his signal v if θ > 1, and he will bid lessthan his signal if θ < 1. If θ > 1 it might happen that after the auction thebidder discovers (i.e., if the losing bid is announced as well) that his bid ishigher than the true value of the object. That is the bidder regrets bidding ashe bid. However, this does not mean that the bidder has suffered the winner’scurse—this bidder has fully taken into account that he might pay more thanthe true value of the object. In comparison, if θ < 1 winning the auction isquite a pleasurable experience.

42 Common Value

4.1.2 Second-price Auction Example

Now let us examine the bidding behavior in a second-price auction. Supposefirst the distribution is uniform on [0, 1]. Assume again that bidder 2 followssome strictly increasing strategy b(·). Player 1, who has received signal v1 = v,will choose b1 = b(ω) to maximize expected profits:

π1 =∫ ω

0

(v + y − b(y)) dy.

Differentiating with respect to ω and setting the derivative equal to zero yields:

v + ω − b(ω) = 0.

Therefore, the equilibrium bidding strategy in this second-price auction is

bS(v) = 2v.

That is, a bidder bids more than the expected value of the object conditionalon having the highest signal. (This expected value is 3v/2.) Does this meanthat this player is suffering from a winner’s curse? The answer again is no.The winning bidder does not pay his bid but rather the second highest bid.Conditional on having the highest signal, this bidder expected to pay pS =2(v/2) = v—this is the bid submitted by his opponent. Note that the seller’sexpected revenue is again the expected payment of the highest bidder: 2

3 . Butthis is the same expected revenue from the first-price auction above! Is this acoincidence?

The fact that the two auctions formats generate the same expected revenuein this example is no coincidence. In Chapter 6, we will prove a general versionof the Revenue Equivalence Theorem that will cover this example.

4.2 An Example with Correlated Types

We will now consider a simple example where signals are correlated. We willmaintain the assumption that the true value of the object is V = v1 + v2.However, the joint distribution of the vi’s is now given by:

f(v1, v2) = 45 (1 + v1v2), for v1, v2 ∈ [0, 1].

To be able to obtain the density f(v1) we integrate the above expression withrespect to v2:

f(v1) =∫ 1

0

45 (1 + v1v2) dv2 = 4

5

(1 +

v1

2

).

4.2 An Example with Correlated Types 43

Similarly,

f(v2) =∫ 1

0

45 (1 + v1v2) dv1 = 4

5

(1 +

v2

2

).

We also need the conditional densities. For example,

f(v2 | v1) =f(v1, v2)

f(v1)=

2(1 + v1v2)2 + v1

.

Similarly,

f(v1 | v2) =f(v1, v2)

f(v2)=

2(1 + v1v2)2 + v2

.


Now we are ready to compute equilibrium bidding behavior in a first-price auc-tion. We consider that both bidders are using the same strategy b(·), bidder 2bids b(v2) when he receives signal v2. Bidder 1, who receives signal v1, willchoose to “announce” a value s so as to maximize his expected profits:

π1(s) = E[(v − b(s))Is>v2 | v1]

= E[(v1 + v2 − b(s))Is>v2 | v1].

Taking the expected value yields:

π1(s) =∫ s

0

(v1 + v2 − b(s))2(1 + v1v2)

2 + v1dv2

=∫ s

0

2(v1 + v2)(1 + v1v2)2 + v1

dv2 − b(s)∫ s

0

2(1 + v1v2)2 + v1

dv2.

Differentiating with respect to s yields:

(2 + v1)π′1(s)

= 2(v1 + s)(1 + v1s) − b′(s)∫ s

0

2(1 + v1v2) dv2 − b(s)2(1 + v1s).

This derivative is equal to zero when s = v1. That is, in equilibrium our bidderchooses b(s) = b(v1). Thus,

4v1(1 + v21) = b′(v1)

∫ v1

0

2(1 + v1v2) dv2 + b(v1)2(1 + v21).

Integrating this expression we obtain the following differential equation:

b′(v) + 2b(v)(1 + v2)2v + v3

=4(1 + v2)2 + v2

.

44 Common Value

We solve this differential equation by using the integrating factor method (seeAppendix B). Define P = v

√2 + v2. Then

(v√

2 + v2 b(v))′ = v√

2 + v2 b′(v) + 21 + v2√(2 + v2)

b(v)

= v√

2 + v2

(b′(v) + 2

1 + v2

v(2 + v2)b(v))

=4(1 + v2)2 + v2

v√

2 + v2 =4v(1 + v2)√

2 + v2.

Integrating both sides of our differential equation and using the boundarycondition b(0) = 0 to obtain the value of the constant, we obtain our symmetricequilibrium

bF(v) =43

√2 + v2(v2 − 1) +

√2

v√

2 + v2.

1

0.8

0.4

0.6

0 0.2 0.4 0.6 0.8 1v

Bidding function bF (v )

b(v

)

0.2

The dotted line is the graph of x, it is included just for comparison. Tocompute the seller’s expected revenue we have to consider the distribution ofthe maximum between v1 and v2:

Pr[maxv1, v2 ≤ λ] = Pr[v1 ≤ λ, v2 ≤ λ]

=∫ λ

0

∫ λ

0

45 (1 + v1v2) dv1 dv2 =

45

(λ2 +

λ4

4

).

Thus, the density of the maximum V1 ∨ V2 is

fV1∨V2(x) =45(2x + x3).

4.2 An Example with Correlated Types 45

The seller’s expected revenue then can be written as:

RF =∫ 1

0

bF (x) 45 (2x + x3) dx.

The reader can check that RF = 0.735.

4.2.2 Second-price Auction

We can now compute equilibrium behavior in a second-price auction. Let usfollow the same approach and write bidder 1’s expected profits given that bothplayers are using strategy b(·), bidder 1 has received signal v1 but chooses toannounce a value s to maximize his expected value:

E[(v1 + v2 − b(v2))Is>v2 | v1].

Taking the expected value yields:

π1(s) =∫ s

0

(v1 + v2 − b(v2))2(1 + v1v2)

2 + v1dv2.

Differentiating with respect to s yields:

π′1(s) = (v1 + s − b(s))

2(1 + v1s)2 + v1

.

In equilibrium the derivative equals to zero when s = v1. Thus,

bS(v1) = 2v1.

That is, in the symmetric equilibrium of the second-price auction bidders bidmore than their expected value for the object conditional on having the highestsignal. However, as in the example with independent signals, this does not meanthat they expect to pay more than the true value of the object. Recall that inthis auction format they pay the second highest bid!

We can now compute the seller’s expected revenue. First, we need todetermine the distribution of the minimum of the two signals:

Pr[minv1, v2 ≤ λ] = 1 − Pr(minv1, v2 > λ)

= 1 −∫ 1

λ

∫ 1

λ

45 (1 + v1v2) dv1 dv2

= 1 − 45

[(1 − λ)2 +

(1 − λ2

2

)2]

.

Thus, the density of the minimum of v1 and v2 (V1 ∧ V2) is given by

fV1∧V2(x) =85(1 − x) +

45x(1 − x2).

46 Common Value

Therefore, the seller’s expected revenue is equal to:

RS =∫ 1

0

2x(

85 (1 − x) + 4

5x(1 − x2))

dx = 0.746.

Unlike the example with independent signals, the expected revenue genera-ted by the second-price auction is higher than that generated by the first-priceauction. The reason is related to the so-called linkage principle. In a second-price auction, the price paid by the winner is based on the signal received bythe runner up whereas in a first-price auction it is based on his own signalonly and signals are positively correlated. This means that rational bidders bidmore cautiously in first-price auctions than in second-price auctions resultingin lower expected revenue for the seller. This will be explained in detail in thenext chapter in the context of affiliation.

4.3 The Symmetric Model with Two Bidders

To prepare the terrain for the next chapter, we consider a model which is ageneralization of both the private values model and the common values model.However, in this section we limit ourselves to the simple case of two symmetricbidders as defined below.

Valuation symmetry The valuation of bidder i is given byui : [0, v]2 → R satisfying u2(x1, x2) = u1(x2, x1).

Example 5 Suppose u1(x1, x2) = x1+x2/2 and u2(x1, x2) = x2+x1/2. Thenthe bidders’ valuations are symmetric.

Thus, if bidder 1 were in the same situation as bidder 2, bidder 1 will havethe same valuation as bidder 2. Thus this example is neither a pure commonvalues model (since in general u1(x1, x2) = u2(x1, x2)) nor a private valuesmodel (since u1(x1, x2) = x1).

Define u(a, b) = u1(a, b). Then u2(x1, x2) = u(x2, x1). Define also D =[0, v]2. We now introduce a regularity assumption:

Regularity u : D → R, is non-negative, continuously differentiable andstrictly increasing in the first variable and increasing in the second variable.

Regularity is a purely technical assumption. Its role will be clearer below.We suppose the distribution of the random vector (X,Y ) has a density

f(·, ·). Although not strictly necessary we impose the following additionalassumption:

Symmetric density The density f : D → R is continuous, strictly positiveand symmetric. That is, for every (x, y) ∈ D, f(x, y) = f(y, x).


Example 6 The density f(v1, v2) = (1 + v1v2)/2 defined for v1, v2 ∈ [0, 1] isa symmetric density. If g(v1, v2) is a continuous strictly positive density thenf(v1, v2) = (g(v1, v2) + (g(v2, v1))/2 is a symmetric density.

The next property plays an important part in the definition of affiliationfor two random variables.

Definition 6 The positive function f satisfies the monotone likelihood ratioproperty if for every (a, c), (b, d) ∈ D, whenever a > b and c > d, then

f(a, c)f(a, d)

≥ f(b, c)f(b, d)

.

Alternatively, we could say that the ratio f(x, c)/f(x, d) increases in xif c > d.

Example 7 If X and Y are independent then the joint density satisfiesthe monotone likelihood ratio property. This is not difficult to see. Supposef(x, y) = f1(x)f2(y). Then

f(a, c)f(a, d)

=f1(a)f2(c)f1(a)f2(d)

=f2(c)f2(d)

=f(b, c)f(b, d)

.

Example 8 Suppose f(ω, x) = β(ω)exωg(x), (ω, x) ∈ [0, 1]2 where g(x) > 0and β(ω) > 0. Then f satisfies the monotone likelihood ratio property. To provethis consider a > b and c > d. Then

f(a, c)f(a, d)

=β(a)eacg(c)β(a)eadg(d)

= ea(c−d) g(c)g(d)

≥ eb(c−d) g(c)g(d)

=β(b)ebcg(c)β(b)ebdg(d)

=f(b, c)f(b, d)

.

An immediate consequence of the definition is as follows.

Lemma 1 If the density of (X,Y ) satisfies the monotone likelihood ratioproperty then for every y ∈ [0, v], FY |X(y |x)/fY |X(y |x) decreases with x.

Proof: The conditional density of Y given X = x is fY |X(y |x) =f(x, y)/

∫ v

0f(x, z) dz. The conditional distribution is therefore given by:

FY |X(y |x) =∫ y

0

fY |X(z |x) dz =

∫ y

0f(x, z′) dz′∫ v

0f(x, z) dz

.

48 Common Value

Thus,

FY |X(y |x)fY |X(y |x)

=

∫ y

0f(x, z′) dz′∫ v

0f(x, z) dz

/f(x, y)∫ v

0f(x, z) dz

=

∫ y

0f(x, z′) dz′

f(x, y).

Suppose x′ > x. Then for every z′ < y, the monotone likelihood ratio pro-perty implies that f(x′, y)/f(x′, z′) ≥ f(x, y)/f(x, z′). Taking the reciprocal:f(x, z′)/f(x, y) ≥ f(x′, z′)/f(x′, y) and integrating we obtain

FY |X(y |x)fY |X(y |x)

=

∫ y

0f(x, z′) dz′

f(x, y)≥∫ y

0f(x′, z′) dz′

f(x′, y)=

FY |X(y |x′)fY |X(y |x′)

.

Definition 7 The random variables (X,Y ) are affiliated if the joint densitysatisfies the monotone likelihood ratio property.

When f is twice differentiable it turns out that we can characterize affiliationin the following way.

Proposition 1 If the density f : D → R++ is twice differentiable,then it satisfies the monotone likelihood ratio property if and only if∂2/∂x∂y(log f(x, y)) ≥ 0.

Proof: First note that the following equivalences are true:

f(x, c)f(x, d)

is increasing in x

(4.5)

log(

f(x, c)f(x, d)

)is increasing in x

For every x,∂

∂x(log f(x, c) − log f(x, d)) =

∂

∂xlog(

f(x, c)f(x, d)

)≥ 0.

Suppose first that f satisfies the monotone likelihood ratio property. Thenwhenever c > d, f(x, c)/f(x, d) is increasing and therefore for every x,∂∂x log f(x, c) − ∂

∂x log f(x, d) ≥ 0. Hence dividing by c − d and making c con-verges to d we conclude that ∂2

∂x∂y (log f(x, y)) ≥ 0. Reciprocally, if for every x,∂2

∂x∂y (log f(x, y)) ≥ 0 then if c > d the last line of (4.5) is true. Hence, the firstline is true and the monotone likelihood property as well.

Remark 3 The proof above is from Karlin (1957). This proposi-tion is easy to apply. For example, if f(x, ω) = β(ω) exωg(x) then


log f(x, ω) = log(β(ω)) + xω + log g(x). Therefore, ∂∂x (log f(x, ω)) = ω +

g′(x)/g(x) and hence ∂2

∂ω∂x log f(x, ω) = 1 > 0.

The next theorem will not be used in this chapter but illustrates quitenicely a property of the densities with the monotone likelihood ratio property.This property is important in establishing the results in Chapter 5, specifi-cally (5.10).

Theorem 4 If u(x, y) is increasing in both variables then E[u(X,Y ) |Y = y]is increasing in y.

Proof: Suppose y′ < y′′. We suppose first that u does not depend on y. Thensince E[u(X)|Y = y] =

∫u(x)f(x, y) dx/

∫f(a, y) da we have to prove that

∫u(x)f(x, y′) dx∫

f(a, y′) da≤∫

u(x)f(x, y′′) dx∫f(a, y′′) da

.

This is equivalent to prove that

∫f(a, y′′)f(b, y′)u(b) da db =

∫f(a, y′′) da ·

∫u(x)f(x, y′) dx

≤∫

u(x)f(x, y′′) dx ·∫

f(a, y′) da

=∫

f(a, y′)f(b, y′′)u(b) dadb.

We now rewrite the first integral as

∫f(a, y′′)u(b)f(b, y′) dadb =∫a<b

f(a, y′′)u(b)f(b, y′) dadb +∫

b≤a

f(a, y′′)u(b)f(b, y′) dadb =

∫a<b


a≤b

f(b, y′′)u(a)f(a, y′) dadb =

∫a<b


a<b

f(b, y′′)u(a)f(a, y′) dadb =

∫a<b

(f(a, y′′)f(b, y′)u(b) + f(b, y′′)f(a, y′)u(a)) dadb.

50 Common Value

Analogously we write∫f(a, y′)u(b)f(b, y′′) dadb

=∫

a<b

(f(a, y′)f(b, y′′)u(b) + f(b, y′)f(a, y′′)u(a) dadb.

Finally, since a < b we have that

f(a, y′)f(b, y′′)u(b) + f(b, y′)f(a, y′′)u(a) − f(a, y′′)f(b, y′)u(b)

− f(b, y′′)f(a, y′)u(a)

= f(a, y′)f(b, y′′)(u(b) − u(a)) + f(b, y′)f(a, y′′)(u(a) − u(b))

= (f(a, y′)f(b, y′′) − f(b, y′)f(a, y′′))(u(b) − u(a)) ≥ 0.

This ends the proof if u does not depend on y. Let us consider now the generalcase. Suppose again that y′ < y′′. Define u(x) = u(x, y′). Then by the previouscase we have that

E[u(X,Y ) |Y = y′] = E[u(X, y′) |Y = y′] = E[u(X) |Y = y′]

≤ E[u(X) |Y = y′′] = E[u(X, y′) |Y = y′′]

≤ E[u(X, y′′) |Y = y′′] = E[u(X,Y ) |Y = y′′].

4.3.1 Second-price Auctions

We now find equilibrium strategies for the second-price auction. We look fora symmetric equilibrium. We will first find a candidate equilibrium. Then weshall prove that our candidate is indeed an equilibrium. Thus suppose bidder 2plays b(y), a strictly increasing continuous bidding strategy. Let us find thebest reply of bidder 1. If he bids t ≥ 0 then he wins the object if t > b(y) andthe payment is b(y). The expected utility is

E[(u(x, y) − b(y))It>b(y) |X = x].

However, since b(·) is continuous, the range of b(·) is a closed interval, say[α, β]. If t < α the bidder never wins the object. If t > β the bidder wins theobject but also wins with t = β. Thus, to maximize the expected utility we mayassume that t ∈ [α, β]. That is, we assume that t = b(s), s ∈ [0, v]. Bidder 1’sexpected profits, as a function of the announced type s (that implicitly definesthe bid t), can be written as:

π(s) := E[(u(x, y) − b(y))Is>y |X = x]

=∫ s

0

(u(x, y) − b(y))fY |X(y |x) dy.


Differentiating with respect to s we obtain

π′(s) = (u(x, s) − b(s))fY |X(s |x).

Thus if s = x is to be optimal we need that u(x, x) = b(x). We found thecandidate equilibrium. We proceed to show that it is an equilibrium. First,note that b(x) = u(x, x) is strictly increasing since u(·, ·) is non-decreasing inthe second variable and strictly increasing in the first variable. Moreover, b(x)is continuous as u(·, ·) is continuous by assumption. Thus,

π′(s) = (u(x, s) − u(s, s))fY |X(s |x).

If s > x thenu(s, s) > u(x, s) and thenπ′(s) < 0. Analogously if s < x, π′(s) > 0.Therefore it is optimum to set s = x. We have proved the following result.

Theorem 5 (second-price auction equilibrium) A symmetric equilibrium bid-ding strategy for the second-price auction with two bidders with affiliated valuesis bs(x) = u(x, x).

Our reader will ask himself or herself whether affiliation was used at all inthe above reasoning. It was not! This result is quite general. Note, however,that the interpretation of the symmetric equilibrium here is quite differentfrom the interpretation under the IPV model. In the latter, a bidder bids histrue valuation, whereas under affiliation a bidder bids:

b(x) = u(x, x) = E[u(x, y) |X = x, Y = x],

that is, the expected value of the object to this player conditional on the otherplayer having received the same type. Recall that Theorem 4 implies that whenu(x, y) is increasing in both variables, E[u(x, y) |Y = y] is also increasing in y.

4.3.2 First-price Auctions

In this section, we find a symmetric equilibrium of the first-price auction withaffiliation. To be more precise, we want to prove the following.

Theorem 6 The first-price auction has a symmetric equilibrium underaffiliation described by a bid function bf : [0, v] → R such that

bf(x) = u(x, x) −∫ x

0

exp[−∫ x

t

fY |X(u |u)FY |X(u|u)

du

](ddt

(u(t, t)))

dt. (4.6)

Moreover, bf solves

b′f(x) = (u(x, x) − bf(x))fY |X(x |x)FY |X(x |x)

; (4.7)

bf(0) = u(0, 0).

52 Common Value

Before proving the theorem, we first show where the differentialequation (4.7) comes from. Suppose b(·) is continuous and strictly increasing.We assume that bidder 2 bids according to the function b(·) and we will find thebest reply of bidder 1. The expected utility of bidder 1 when he bids t = b(s)is given by:

π(s) = E[(u(x, y) − t)It>b(y) |X = x]

= E[(u(x, y) − b(s))Is>y |X = x]

=∫ s

0

(u(x, y) − b(s))fY |X(y |x) dy

=∫ s

0

u(x, y)fY |X(y |x)dy − b(s)FY |X(s |x).

Note that once more we transformed bidder 1’s problem from one where he waschoosing a bid t, to one where he is choosing to announce a type s and thenthe bid is set at b(s).

Differentiating π(s) with respect to s we obtain

π′(s) = −b′(s)FY |X(s |x) + (u(x, s) − b(s))fY |X(s |x).

Thus if s = x is to be optimal:

π′(x) = −b′(x)FY |X(x |x) + (u(x, x) − b(x))fY |X(x |x) = 0.

This yields the differential equation

b′(x) = (u(x, x) − b(x))fY |X(x |x)FY |X(x |x)

. (4.8)

Note that if we set u(x, x) = x in the expression above and the random variablesX and Y are independent, we get

b′(x) = (x − b(x))fY (x)FY (x)

.

This is the same differential equation as in the IPV case.We now solve differential equation (4.8). We will use the integrating factor

method. Define

γ(s, x) =fY |X(s |x)FY |X(s |x)

and γ(s) =fY |X(s | s)FY |X(s | s) .

We can rewrite (4.8) as

b′(x) + b(x)γ(x) = u(x, x)γ(x).


Define our integrating factor as P (x) = exp[−∫ v

xγ(u) du

]. Thus

(Pb)′(x) = P (x)b′(x) + P ′(x)b(x)

= P (x)b′(x) + γ(x)P (x)b(x)

= P (x)u(x, x)γ(x).

If b(0) = u(0, 0) = 0 then integrating between 0 and x:

P (x)b(x) =∫ x

0

P (t)u(t, t)γ(t) dt.

Thus

b(x) = (P (x))−1

∫ x

0

P (t)u(t, t)γ(t) dt

= (P (x))−1

∫ x

0

u(t, t)P ′(t) dt.

Integrating by parts the last integral we obtain:

b(x) = u(x, x) − (P (x))−1

∫ x

0

P (t)(

ddt

(u(t, t)))

dt.

Substituting P (x) in the expression we obtain (4.6). Note that so far we havenot used the affiliation assumption. Thus the solution above is potentially anequilibrium for any distribution function. However, we still need to check thatb(·) is indeed an equilibrium. To do this will require affiliation.

From the reasoning above, since b(·) is strictly increasing and differentiablethen

π′(s) = −b′(s)FY |X(s |x) + (u(x, s) − b(s))fY |X(s |x)

= FY |X(s |x)[−b′(s) + (u(x, s) − b(s))γ(s, x)]

= FY |X(s |x)[−b′(s) + (u(x, s) − u(s, s))γ(s, x) + (u(s, s) − b(s))γ(s, x)]

= FY |X(s |x)[−b′(s) + (u(x, s) − u(s, s))γ(s, x) + b′(s)

γ(s, x)γ(s)

]

= FY |X(s |x)[(u(x, s) − u(s, s))γ(s, x) + b′(s)

(γ(s, x)γ(s)

− 1)]

.

Thus if s > x,

γ(s, x) − γ(s) =FY |X(s |x)fY |X(s |x)

− FY |X(s | s)fY |X(s | s) ≥ 0

and π′(s) < 0. Also if s < x then π′(s) > 0. Thus s = x maximizes π(s) endingthe proof of the theorem.

54 Common Value

4.3.3 Revenue Comparison

First note that bidder 1’s expected payment in the second-price auctionconditional on having received x is given by

Psp(x) =∫ x

0

u(y, y)fY |X(y |x) dy.

Thus, from the seller’s point of view, her expected revenue is simply twice theexpected value of Psp(x)—the seller does not know bidder 1’s (and bidder 2’s)type and therefore she has to compute this expected value; her expectedrevenue.

Bidder 1’s expected payment in the first-price auction is simply his bid timesthe probability that he wins with this bid:

Pfp(x) = bf(x)FY |X(x |x).

Note that we can rewrite Psp(x) as:

Psp =∫ x

0

(u(y, y) − bf(y))fY |X(y |x) dy +∫ x

0

bf(y)fY |X(y |x) dy.

By using (4.8) we can write:

Psp =∫ x

0

b′f(y)FY |X(y | y)fY |X(y | y)

fY |X(y |x) dy +∫ x

0

bf(y)fY |X(y |x) dy

≥∫ x

0

b′f(y)FY |X(y |x) dy +∫ x

0

bf(y)fY |X(y |x) dy

=∫ x

0

(bf(y)FY |X(y |x))′ dy = bf(x)FY |X(x |x) = Pfp.

The last inequality follows from Lemma 1 which implies that:

FY |X(y | y)fY |X(y | y)

≥ FY |X(y |x)fY |X(y |x)

for y ≤ x.

Therefore, the expected revenue of the second-price sealed-bid auction is notless than the expected revenue of the first-price sealed-bid auction: 2E[Psp] ≥2E[Pfp].

Example 9 Suppose the joint density of (X,Y ) is

f(x, y) =4(1 + xy)

5, (x, y) ∈ [0, 1]2.

It is easy to check that f has a monotone likelihood ratio. The marginaldensity of X is fX(x) =

∫ 1

0f(x, y) dy = (4 + 2x)/5 the conditional density

is fY |X(y |x) = f(x, y)/fX(x) = 2(1 + xy)/2 + x and the conditional distribu-tion is FY |X(y|x) =

∫ y

0(2(1+xz)/(2 + x)) dz = (2y+xy2)/(2+x). We suppose

4.4 Exercises 55

that u(x, y) = x + by, b ≥ 0. Thus the second-price auction equilibrium biddingfunction is bs(x) = u(x, x) = (1+b)x. To find the first-price auction equilibriumbidding function define γ(x) = fY |X(x |x)/FY |X(x |x) = 2(1 + x2)/x(2 + x2).Thus, from equation (4.6), the first-price auction equilibrium bidding function is

bf(x) = (1 + b)x − (1 + b)∫ x

0

exp[−∫ x

t

21 + u2

u(2 + u2)du

]dt

= (1 + b)(

x − (2 + x2)3/2 − 23x

√2 + x2

).

Bidder 1’s expected payment in a second-price auction is

2(1 + b)∫ x

0

y1 + xy

2 + xdy =

1 + b

3x2 2x2 + 3

2 + x.

Bidder 1’s expected payment in a first-price auction is

bf(x)FY |X(x |x) = (1 + b)(

x2 − (2 + x2)3/2 − 23√

2 + x2

)2 + x2

2 + x.

Thus, we can write:

Psp − Pfp

1 + b=

x2(2x2 + 3)3(2 + x)

−(

x2 − (2 + x2)3/2 − 23√

2 + x2

)2 + x2

2 + x.

See the plot of (Psp − Pfp)/(1 + b) below. Note that Psp − Pfp is bounded awayfrom zero.

x

y0.195

0.19

0.185

0.175

0.17

0.18

0.1650 0.2 0.4 0.6 0.8 1

4.4 Exercises

1. Suppose there are two bidders with types distributed uniformly on theinterval [0, 1] and that the common value is given by u(x1, x2) = 2x1−x2.

56 Common Value

Find the equilibrium bidding strategies for the first- and second-priceauctions. Compute the expected revenue associated with each auctionformat.

2. Suppose we have two bidders and that the common value is given byu(x1, x2). Assume that types are independent. Give general conditi-ons on u(·, ·) so that b(x) = u(x, x) is a symmetric equilibrium biddingstrategy of the second-price auction.

3. Check if the functions below satisfies the monotone likelihood ratioproperty:(a) p(x, y) = nxn−1

yn , 0 < x < y, and n a natural number;

(b) p(x, y) = n(n − 1)xn−2(y−x)yn , 0 < x < y and n ≥ 2 an integer;

(c) p(x, y) = 1π

11+(x−y)2 , x, y ∈ R.

4. Calculate the seller’s expected revenue for the second- and first-priceauctions if the joint density is f(x, y) = 4(1+xy)

5 , (x, y) ∈ [0, 1]2 and thecommon utility is x + y.

5. Discuss if the regret mentioned in (4.1.1) is an important concept?6. Show that in the derivation of the second-price auction equilibrium it

suffices that u(x, y) is strictly increasing in the first variable and thatu(x, x) is increasing. Thus u(x, y) = x − y/2 is allowed?

7. Show that

F (x) = 1 − k

(√ax2 + bx + c

1 − x

)−1/a+b+c

exp[−γ arctan

(2xa + b√4ac − b2

)]

defined on x ∈ [0, 1] is a distribution function if c > 0, 4ac > b2, γ =(2a + b)/(b + a + c)

√4ac − b2 and k is chosen so that F (0) = 0. Show

that you may chose a, b, c so that

ρ =1 − F (ρ)

f(ρ)

has three roots. Find the optimal reserve price for your choice ofparameters.

8. Show that the bidding function (4.4) is continuous at zero if b0 = 0.9. Suppose F (v) = v2 in the first-price auction example. Calculate the

expected profit of the bidders and determine the “regret” region, that is,

(v, v2); bidder 1 regrets winning the auction.10. Generalize Theorem 4 to prove that E[u(X,Y ) |Y = y] is strictly

increasing if u(x, y) is strictly increasing in y.

5

Affiliated Values

As in the previous chapters, a single object will be sold to one of n bidders.Throughout this chapter, we assume once more that the seller’s valuation forthe object is equal to zero and she is risk neutral. Each bidder j, j = 1, . . . , n,receives a type xj and if x = (x1, x2, . . . , xn) bidder i valuation is equal to Vi =ui(x)—note that the value of the object for player i does depend on variablesnot observable by him, x−i = (x1, . . . , xi−1, xi+1, . . . , xn). Buyers’ valuationsare expressed in dollars and bidders are risk-neutral. The main distinction nowvis-a-vis the previous chapters is that we will consider an auction model thatwill have the private and common value models seen before as special cases. Inthis chapter, we will explain in detail the notion of affiliation and analyze itsimplications in terms of equilibrium bidding functions and revenue.

The notion of affiliation, to be formally defined below, captures a globalpositive correlation between individuals’ types, and allows the generalization ofthe theory of auctions beyond the private independent values case. The natureof correlation, however, will imply that the main result from the IPV case,namely, the Revenue Equivalence Theorem, will no longer hold. In particular,second-price sealed-bid auctions will generate (weakly) more expected revenuethan first-price auctions. The English auctions, modeled as button auctions—and this modeling assumption will be extremely important for the result as wewill argue below—will dominate in expected revenue second-price sealed-bidauctions. The strategic equivalence between the English button auction andthe second-price auction will vanish under affiliation.

The main reference for this chapter is Milgrom and Weber (1982). However,we concentrate on distributions with density and provide more details. Thechapter being on the whole more demanding and longer than the other chapterswe deliberately leave out the results concerning the effects on the seller’s revenueof credibly revealing information. The reader who wants to pursue these topicswill do well to read the Milgrom and Weber paper. We expect that her or histask will be much facilitated by the study of this chapter. We also omit Milgromand Weber’s results regarding reserve prices and entry fees.

57

58 Affiliated Values

This chapter is organized as follows. In Section 5.1, we introduce the notionof affiliation in a general multivariate context and demonstrate a few theoremsthat are essential to understanding how affiliation works. Sections 5.2, 5.3,and 5.4, respectively, characterize equilibrium behavior in second-, first-price,and English button auctions. Section 5.5 contains the expected revenue com-parisons for these three auction formats and Section 5.6 examines the effects ofrisk aversion.

5.1 The General Model

We now consider the case of n ≥ 2 bidders. The set of possible types isD = [0, v]n. We will also impose that bidders are ex ante identical. To be ableto define what that means with n bidders, we need to cover some preliminaryground work. Recall (Definition 4) that a permutation of S is a bijection on S.The symmetry definition to follow is very similar to Definition 5.

Definition 8 A function u : D → R is symmetric in the last n − 1 varia-bles if for every permutation σ of the set 2, . . . , n and for every x ∈ D,u(x1, x2, . . . , xn) = u(x1, xσ(2), . . . , xσ(n)).

For example, when n = 3, u(a, b, c) is symmetric in the last two variables ifu(a, b, c) = u(a, c, b) for every (a, b, c) ∈ [0, v]3. With four variables, u(a, b, c, d)is symmetric in the last three variables if

u(a, b, c, d) = u(a, b, d, c) = u(a, c, b, d)

= u(a, c, d, b) = u(a, d, b, c) = u(a, d, c, b).

For example, the reader can check directly that the function

u(a, b, c, d) = a2 +b + c + d

2+ (bcd)3

is symmetric in the last three variables.

Valuation symmetry There exists a function u : D → R such thatui(x) = u(xi, x−i), i = 1, . . . , n. The function u is symmetric in the lastn− 1 variables, strictly increasing in the first variable and non-decreasing inthe remaining variables.

Regularity u : D → R is non-negative, continuous and strictly increasing.

Example 10 The valuation function ui(x) = axi + b∑

j =i xj where a> 0and b ≥ 0 satisfies valuation symmetry. If b = 0 we have private values as inChapter 3. If a = b we have pure common values as in Chapter 4.

5.1 The General Model 59

We denote by f(x) the joint probability density of the random variablesX1, . . . , Xn. The density f(·) satisfies the following assumption:

Symmetry f(·) is symmetric in its n arguments.

For example, if f is symmetric then f(x1, x2, x3, . . . , xn) =f(x2, x1, x3, . . . , xn). Thus f(x1, x2) = x1 + x2 is symmetric but g(x1, x2) =2x1 + 6x2

2 is not.We will also assume that f(·) is strictly positive and continuous in its

domain. To define affiliation we need to specify some notation.For every x, y ∈ Rn we denote the coordinate-wise supremum by x ∨ y and

the coordinate-wise minimum by x ∧ y. Thus

x ∨ y = (maxx1, y1, . . . ,maxxn, yn)

and

x ∧ y = (minx1, y1, . . . ,minxn, yn).

For example take two vectors x, y ∈ R4, x = (1, 2, 3, 4) and y = (0, 1, 5, 2). Thenx ∨ y = (1, 2, 5, 4) and x ∧ y = (0, 1, 3, 2).

A non-negative function f : D → R has the multivariate monotone likelihoodratio property if for every x, y ∈ D,

f(x ∨ y)f(x ∧ y) ≥ f(x) f(y).

We now define affiliation.

Definition 9 The random variables (X1, . . . , Xn) are affiliated if the jointdensity f : D → R has the multivariate monotone likelihood ratio property.

For example, let us check that independent random variables areaffiliated. Independence implies that the joint density can be written asf(x) = f1(x1)f2(x2) · · · fn(xn). Then

f(x ∨ y)f(x ∧ y) =n∏

i=1

fi(maxxi, yi) fi(minxi, yi)

=n∏

i=1

fi(xi) fi(yi) = f(x) f(y).

Above we used that maxxi, yi,minxi, yi = xi, yi. We now prove aseries of theorems that will be needed to allow us to characterize equilibriumbidding strategies for different auction formats and to compare the expectedrevenue they generate. The next result allows us to check directly whethera symmetric density is affiliated.


Theorem 7 A symmetric density function f : D → R is affiliated if for everyz ∈ [0, v]n−2, the function

(x1, x2) → f(x1, x2, z)

satisfies the monotone likelihood ratio property.

Proof: From Proposition (10) in Appendix C it is enough to show that thedensity has a monotone likelihood ratio for every pair of variables (xi, xj).From symmetry, however, it is enough to show the monotone likelihood ratioproperty for (x1, x2).

Corollary 2 The multivariate monotone likelihood ratio property holds for atwice differentiable function f if and only if ∂2

∂xi∂xjlog(f(x)) ≥ 0 for any i = j.

For example if the random variables X1, X2, . . . , Xn are independent thenf(x1, x2, . . . , xn) = f1(x1)f2(x2) · · · fn(xn) has the monotone likelihood ratioproperty since ∂2

∂xi∂xjlog f(x) = 0.

As in the private and common value cases, the properties of the distributionof the highest types play an important role in deriving equilibrium behavior.Following the notation of previous chapters, we consider the different auctiongames from bidder 1’s perspective and we denote by y1 the largest type amongx2, . . . , xn, by y2 the second largest and so on until yn−1 the n − 1 largest(that is, the smallest) types among x2, . . . , xn. Our symmetry assumptionallows us to write

u1(x) = u(x1, x−1) = u(x1, y), with y = (y1, y2, . . . , yn−1).

If (X1, X2, . . . , Xn) is a vector of random variables the (X1, Y1, Y 2, . . . , Y n−1)

is the vector of random variables where Y 1(ω) is the largest of the numbersX2(ω), . . . , Xn(ω). Then Y 2(ω) is the second largest and so on. Let us findthe density of (X1, Y

1, Y 2, . . . , Y n−1), knowing the density f(x1, x2, . . . , xn) of(X1, X2, . . . , Xn).

Lemma 2 The density of (X1, Y1, Y 2, . . . , Y n−1) is

f(x1, y) =

(n − 1)!f(x1, y1, . . . , yn−1) if y1 ≥ · · · ≥ yn−1,

0 otherwise.(5.1)

Proof: We first find the distribution of (X1, Y1, Y 2, . . . , Y n−1). This is, for a

given (x1, a1, . . . , an−1) defined by

FX1,Y (x1, a1, . . . , an−1) = Pr[(X1, Y1, Y 2, . . . , Y n−1) ≤ (x1, a1, a2, . . . , an−1)]

= Pr[X1 ≤ x1, Y1 ≤ a1, . . . , Y

n−1 ≤ an−1].

5.1 The General Model 61

Now define Pn−1 as the set of all permutations of 2, . . . , n. That is, σ :2, . . . , n → 2, . . . , n is a permutation if it is a bijection. If σ ∈ Pn−1 wedefine the set

Cσ = (x2, . . . , xn) : xσ(2) > · · · > xσ(n).

The sets Cσ, σ ∈ Pn−1 are pairwise disjoint. The union of Cσ is Rn−1 exceptfor the subset ∪i=jx ∈ Rn−1;xi = xj. Since Pr[Xi = Xj ] = 0 (see Lemma 11on page 151.) Hence

Pr[X1 ≤ x1, Y1 ≤ a1, . . . , Y

n−1 ≤ an−1]

=∑

σ∈Pn−1

Pr[X1 ≤ x1, (X2, . . . , Xn) ∈ Cσ, Xσ(2) ≤ a1, . . . , Xσ(n) ≤ an−1]

=∑

σ∈Pn−1

Pr[X1 ≤ x1, Xσ(2) > · · · > Xσ(n), Xσ(2) ≤ a1, . . . , Xσ(n) ≤ an−1]

= (n − 1)! Pr[X1 ≤ x1, X2 > · · · > Xn, X2 ≤ a1, . . . , Xn ≤ an−1]

= (n − 1)!∫ x1

0

∫ a1

x2=0

· · ·∫ an−1

xn=0

Ix2>x3>...>xnf(x) dx1dx2 . . . dxn.

In the third line to the fourth line above we used the symmetry of the densityf(x) and that there are (n− 1)! = (n− 1)(n− 2) · · · 3 · 2 permutations of n− 1elements. Therefore the distribution has a density (n−1)!Ix2>x3>...>xn

f(x).

Theorem 8 If f is affiliated in X and symmetric then the random variablesX1, Y1, . . . , Yn−1 are affiliated.

Proof: First note that joint density of X1, Y1, . . . , Yn−1 is f(x, y)We now show that f has the multivariate monotone likelihood ratio

property. We must show that

f(x1 ∨ x′1, y ∨ y′)f(x1 ∧ x′

1, y ∧ y′) ≥ f(x1, y)f(x′1, y

′).

The right-hand side is different from zero only when y1 ≥ · · · ≥ yn−1 andy′1 ≥ · · · ≥ y′

n−1. In this case f(x1, y)f(x′1, y

′) = ((n − 1)!)2f(x1, y)f(x′1, y).

Since yj ∨ y′j ≥ yj+1 and yj ∨ y′

j ≥ y′j+1 we have that yj ∨ y′

j ≥ yj+1 ∨ y′j+1 and

so on: y1 ∨ y′1 ≥ y2 ∨ y′

2 ≥ · · · ≥ yn ∨ y′n. Analogously for the infimum. Thus,

f(x1 ∨ x′1, y ∨ y′)f(x1 ∧ x′

1, y ∧ y′)

= ((n − 1)!)2f(x1 ∨ x′1, y ∨ y′)f(x1 ∧ x′

1, y ∧ y′)

≥ ((n − 1)!)2f(x1, y)f(x′1, y

′) = f(x1, y)f(x′1, y

′).

We will now show that the function E[V1 | X1 = x, Y1 = y, . . . , Ym−1 =ym−1] is strictly increasing in x. Of course, there is no significance attached to


the subscript 1 in V1, since by symmetry if this function is increasing, it willstill be increasing when we exchange subscripts.

Consider now the conditional expectation

h(xm+1, . . . , xn) = E[u(x) | xm+1, . . . , xn]

=∫

u(x)f(x)

f(x1, . . . , xm)dx1 · · · dxm,

where f(x1, . . . , xm) =∫

f(x1, . . . , xn) dxm+1 · · · dxn. If we prove that h isstrictly increasing we will obtain our result for the conditional expectation. Todo this, the following theorem is necessary.

Theorem 9 Let f1, f2, f3, and f4 be non-negative functions on Rn such thatfor all x, y ∈ Rn, f1(x)f2(y) ≤ f3(x ∨ y)f4(x ∧ y). Then∫

f1(x) dx

∫f2(x) dx ≤

∫f3(x) dx

∫f4(x) dx.

This theorem is proved in Appendix C. We can now prove that theconditional expectation is increasing.

Theorem 10 Suppose (X1, . . . , Xn) is a random vector with affiliated density.Then for any increasing function u : X → R+ we have that

E[u(X) | Xk+1 = xk+1, . . . , Xn = xn]

is increasing in (xk+1, . . . , xn).

Proof: Define z = (xk+1, . . . , xn), z′ = (x∗k+1, . . . , x

∗n) and suppose xk+1 ≤

x∗k+1, . . . , xn ≤ x∗

n. Define

g(z) =∫

f(x1, . . . , xk, z) dx1 · · · dxk

and

g(z′) =∫

f(x1, . . . , xk, z′) dx1 · · · dxk.

We now choose

f1(x) = u(x)f(x | z), f2(x) = f(x | z′), f3(x) = u(x)f(x | z′)

and finally f4(x) = f(x | z). Since

E[u(X) | Xk+1 = xk+1, . . . ,n = xn]

=∫

u(x)f(x | z) dx1 · · · dxk =∫

f1(x) dx1 · · · dxk ·∫

f2(x) dx1 · · · dxk

5.2 Second-price Auctions 63

and

E[u(X) | Xk+1 = x∗k+1, . . . , Xn = x∗

n]

=∫

u(x)f(x | z′) dx1 · · · dxk =∫

f3(x) dx1 · · · dxk ·∫

f4(x) dx1 · · · dxk

it suffices to show that f1(x)f2(y) ≤ f3(x ∨ y)f4(x ∧ y) to finish the proof:

f1(x)f2(y)f3(x ∨ y)f4(x ∧ y)

=u(x)f(x | z)f(y | z′)

u(x ∨ y)f(x ∨ y | z′)f(x ∧ y | z)

=u(x)f(x, z)f(y, z′)

g(z)g(z′)u(x ∨ y)f(x ∨ y | z′)f(x ∧ y | z)

=u(x)f(x, z)f(y, z′)

u(x ∨ y)f(x ∨ y, z′)f(x ∧ y, z)≤ 1.

Corollary 3 If u(x) is strictly increasing in xi then

E[u(X) | Xk+1 = xk+1, . . . , Xn = xn], k < i

is strictly increasing in xi too.

Proof: Suppose i = k + 1 and x′k+1 > xk+1. Define v(x) = u(x′

i, x−i). Sincev is increasing the last theorem implies that

E[v(X) | Xi = xi, . . . , Xn = xn] ≤ E[v(X) | Xi = x′i, . . . , Xn = xn].

On the other hand, u(x) < u(x−i, x′i) = v(x) for every x−i. Thus,

E[u(X) | Xi = xi, . . . , Xn = xn] < E[v(X) | Xi = xi, . . . , Xn = xn]

≤ E[v(X) | Xi = x′i, . . . , Xn = xn].

5.2 Second-price Auctions

The theoretical results established in the previous section will be useful incharacterizing bidders’ equilibrium behavior and comparing the expected reve-nue generated by different auction formats. We start by computing equilibriumbidding strategies in a second-price sealed-bid auction. As before we willconsider the auction game from the perspective of bidder 1. Thus, it isconvenient to define

w = maxj =1

b(xj).


Define the following function

v(x, y) = E[V1 | X1 = x, Y 1 = y]

=∫

u(x, y, y2, . . . , yn−1)f(x, y, y2, . . . , yn−1)

fX,Y (x, y)dy2 · · · dyn−1.

Here

fX,Y (x, y) =∫

f(x, y, y2, . . . , yn−1) dy2 · · · dyn−1,

is the joint density of (X,Y 1). By the Corollary 3, v(x, y) is strictly increasingin x and increasing in y. We will show that the equilibrium bidding function isgiven by:

bs(x) = v(x, x) = E[V1 | X1 = x, Y 1 = x]. (5.2)

Suppose bidders i = 2, . . . , n bid accordingly to b(·) which is continuous andstrictly increasing. Bidder 1’s expected profits conditional on having receiveda type x, chosen a bid t and given that other players are following biddingstrategy b(·), are given by:

π(x, t) = E[(V1 − w)It>w | X1 = x], (5.3)

where I indicates a characteristic function determining the cases where player 1wins the auction. Thus, bidder 1’s problem is to choose a bid t ≥ 0 to maximizehis or her expected utility. We are after the symmetric equilibrium, so we willassume that bidders 2, . . . , n are following the equilibrium bidding strategy b(·)given in (5.2) and we will find bidder 1’s best response. From Theorem 5, wecan conclude that b(·) is increasing in x and therefore we can write

w = max b(xj) = b(Y1).

Thus we can rewrite (5.3) as:

π(x, t) = E[(V1 − b(Y1))It>b(Y1) | X1 = x]. (5.4)

Since b(·) is continuous, the range of b(·) is an interval [u, v]. If t < u, bidder 1will never win the object. If t > v bidder 1 wins the object but he wins witht = v as well. Thus we may suppose t ∈ [u, v] and, therefore there existss ∈ [0, v] such that b(s) = t. Hence, we can write bidder 1’s expected profits asa function of s alone:

h(s) = π(x, t) = E[(V1 − b(Y1))Is>Y1 | X1 = x].

Let fY |X denote the conditional density of Y1 given X1 and fX,Y the jointdensity of (X1, Y

1). By using the theorem of iterated conditional expectations

5.3 First-price Auctions 65

(see Lemma 12), we can write:

h(s) = E[E[(V1 − b(Y1)Is>Y1 | X1, Y1)] | X1 = x]

= E[(E[V1 | X1, Y1] − b(Y1))Is>Y1 | X1 = x]

=∫ s

0

(v(x, y) − b(y))fY |X(y | x) dy.

Thus, differentiating we obtain

h′(s) = (v(x, s) − b(s))fY |X(s | x).

If s = x is the optimal solution, b(x) = v(x, x).Hence, we have shown that bidder 1, who has received a type x, will bid

according to (5.2). At this stage we can compute bidder 1’s expected paymentconditional on receiving a type X1 = x:

P SP = E[b(Y1) | X1 = x, x > Y1]

= E[v(Y1, Y1) | X1 = x, x > Y1] =∫ x

0

v(ω, ω)fY |X(ω | x) dω. (5.5)

The seller’s expected revenue from a second-price auction in a symmetricenvironment is simply n times this expected payment.

5.3 First-price Auctions

Our main task in this section is to characterize equilibrium behavior in a first-price auction game. We will use the same approach as above and show directlythat the symmetric equilibrium bidding function is given by (see equation 4.6):

bf(x) = v(x, x) −∫ x

0

e∫ tx

γ(s) ds ddt

(v(t, t)) dt. (5.6)

Recall that γ(s) = fY |X(s | s)/FY |X(s | s). To find an equilibrium biddingstrategy we first find a candidate equilibrium. Then we will show that ourcandidate is indeed an equilibrium. So suppose b(·) is a strictly increasing bid-ding strategy played by bidders i ≥ 2. Let us find bidder 1’s best response.The range of b(·) is an interval [u, v]. If bidder 1 bids t < u, he will neverwin the object. If t > v, then the bidder always wins but pays more thannecessary. Thus we may suppose t ∈ [u, v]. We may therefore look for s ∈ [0, v]to maximize

h(s) = E[(V1 − b(s))Ib(s)>b(Y1) | X1 = x]

= E[(V1 − b(s))Is>Y1 | X1 = x].


If v(x, y) = E[V1 | X1 = x, Y1 = y] then

h(s) = E[(v(x, Y1) − b(s))Is>Y1 | X1 = x]

=∫ s

0

(v(x, y) − b(s))fY |X(y | x) dy

=∫ s

0

v(x, y)fY |X(y | x) dy − b(s)FY |X(s | x).


h′(s) = v(x, s)fY1|X1(s | x) − b′(s)FY1|X1(s | x) − b(s)fY1|X1(s | x).

If s = x is to be optimal then

v(x, x)fY |X(x | x) − b′(x)FY |X(x | x) − b(x)fY |X(x | x) = 0.

Or

b′(x) = (v(x, x) − b(x))fY |X(x | x)FY |X(x | x)

. (5.7)

The initial condition is b(0) = v(0, 0) = 0. This differential equation is an oldcompanion (see equation (4.7, page 51)) and its solution is:

bf(x) = u(x, x) −∫ x

0

exp[−∫ x

t

γ(u) du

](ddt

(u(t, t)))

dt. (5.8)

We now show that bf(·) as defined in (5.8) is a symmetric equilibrium biddingfunction.

It is clear that bf is differentiable. From the differential equation we seethat bf(·) is strictly increasing. Thus, if bidders play bf the expected utility ofbidder 1 when he bids t = bf(s) is

h(s) =∫ s

0

v(x, y)fY |X(y | x) dy − bf(s)FY |X(s | x).

Thus

h′(s) = (v(x, s) − bf(s))fY |X(s | x) − b′f(s)FY |X(s | x).

We now use equation (5.7) in the first summand above:

(v(x, s) − bf(s))fY |X(s | x)

= (v(x, s) − v(s, s))fY |X(s | x) + (v(s, s) − bf(s))fY |X(s | x)

= (v(x, s) − v(s, s))fY |X(s | x) + b′f(s)fY |X(s | x)

γ(s).

5.4 English Auctions 67

Thus

h′(s) = (v(x, s) − v(s, s))fY |X(s | x) + b′f(s)(

fY |X(s | x)γ(s)

− FY |X(s | x))

= (v(x, s) − v(s, s))fY |X(s | x) + b′f(s)fY |X(s | x)(

1γ(s)

− 1γ(s, x)

).

If s > x then v(x, s) − v(s, s) < 0 and fY |X(s | x)/FY |X(s | x) − fY |X(s | s)/FY |X(s | s) ≤ 0. And reciprocally if s < x, h′(s) > 0. Thus s = x maximizesh(s). That is, the strategy described in (5.8) is a symmetric equilibrium for thefirst-price auction.

We can now compute bidder 1’s expected payment in a first-price auction.It is simply his bid multiplied by the probability that he has the highest bidand therefore wins the auction. Note that the probability of winning is equal tothe probability of having the highest type, given that the equilibrium biddingstrategies are strictly increasing. Thus, bidder 1’s expected payment is:

PFP = bf(x)FY |X(x | x)

=(

v(x, x) −∫ x

0

e−∫ xt

γ(u) du

(ddt

(v(t, t)))

dt

)FY |X(x | x). (5.9)

The seller’s expected revenue from a first-price auction is simply n timesthe expected payment from bidder 1. Recall that first-price auctions are stillstrategically equivalent to Dutch auctions under affiliation and therefore theygenerate the same expected payment.

5.4 English Auctions

We will model the English auction as in Milgrom and Weber (1982); biddersbid by pressing a button. The price rises continuously on an automatic clock.By releasing the button a bidder drops out of the auction. The auction endswhen the next-to-last person releases the button. The winner is the bidder whois still pressing the button and he wins at a price equal to the amount at whichthe next-to-last person has dropped out.

Of course, actual ascending auctions might be quite different in manyrespects. For example, you might not know when a bidder has dropped out.In some sense, this particular formulation of the English auction as a buttonauction probably reveals the most amount of information; remaining biddersknow how many bidders have dropped out and at what price. In particular, inthis auction, it is a weakly dominant strategy to bid up to one’s estimate of thevalue of the object. Instead of characterizing equilibrium behavior in general,we chose to consider the case of three bidders who have received actual privatetypes x1 = a, x2 = b, x3 = c, respectively, with a < b < c.


The implicit assumption is that the price starts at zero and risescontinuously. The first player to drop out is player 1 at price p1 such that:

E[V1 | x1 = a = X2 = X3] = p1.

Prior to bidder 1 dropping out, the expected values of bidders 2 and 3 weregiven, respectively, by:

E[V2 | x2 = b = X1 = X3]

and

E[V3 | x3 = c = X1 = X2].

After bidder 1 dropping out, these expected values are revised in thefollowing way:

E[V2 | x2 = b, E[V2 | X1 = X2 = X3 = a] = p1]

and

E[V3 | x3 = c, E[V3 | X1 = X2 = X3 = a] = p1].

The price keeps rising until the final price p2 and the winner is determined asfollows:

E[V2 | x2 = b = X3, p1] = p2.

The value of the object to player 3 prior to knowing p2 is

E[V3 | x3 = c = X2, p1].

Bidder 3 wins the auction at price p2. The generalization to n players shouldbe clear to the reader and the expected price in the English button auction(i.e., the expected payment from say bidder 1) is then

PE = E[E[V1 | X1 = y1 = Y2, Y3, · · · , Yn] | X1 = x > Y1].

Note that the button English auction is not strategically equivalent tosecond-price auctions as in the case of independent private values. The rea-son is that now there is information being revealed that is used by playersto update their valuations for the good. The implications of this informationrevelation for the ranking of auction formats according to the expected revenuethey generate will be explored in the next section.

5.5 Expected Revenue Ranking

We will start by showing that the second-price auction generates (weakly) morerevenue than the first-price auction. This generalizes the revenue comparison ofSection 4.3.3. Recall that for the special case when types are independently andidentically distributed, these different auctions formats yield the same expected

5.5 Expected Revenue Ranking 69

revenue (see Chapter 3). Thus, given that IPV is a special case of affiliation,the ranking of auction formats according to the expected revenue they generatecannot be strict.

We will compare the expected payment by the winner in a second-priceauction (5.5) with that in a first-price auction (5.9). Recall that

P SP =∫ x

0

v(y, y)fY |X(y | x) dy.

Thus

P SP =∫ x

0

(v(y, y) − bf(y))fY |X(y | x) dy +∫ x

0

bf(y)fY |X(y | x) dy

=∫ x

0

b′f(y)FY |X(y | y)fY |X(y | y)

fY |X(y | x) dy +∫ x

0


≥∫ x

0

b′f(y)FY |X(y | x) dy +∫ x

0


= bf(x)FY |X(x | x) = PFP.

We now show that expected payment in an English button auction is(weakly) higher than that in a second-price auction. Recall that

v(y, y) = E[V | X1 = y, Y1 = y].

The theorem of iterated expectations allows us to write

v(y, y) = E[E[V | X1 = y, Y1 = y, Y−1] | X1 = y, Y1 = y]

≤ E[E[V | X1 = y, Y1 = y, Y−1] | X1 = x, Y1 = y]. (5.10)

This last inequality follows from Theorem 7, which is implied by affiliation.The expected payment from bidder 1 in a second-price auction with x > y isgiven by

P SP = E[E[v(y, y) | X1 = x, x > Y1 = y]].

By applying the conditional expectation to both sides of (5.10) we obtain:

E[E[v(y, y) | X1 = x, x > Y1 = y]]

≤ E[E[E[V | X1 = Y1, Y−1] | X1 = x > Y1 | X1 = x > Y1]

= E[E[V | X1 = Y1, Y−1] | X1 = x > Y1] = PE.

We will now refer to the linkage principle (Milgrom and Weber: 1110) toexplain the ranking of auction formats when types are affiliated. Let’s thinkof the auction games as direct revelation games—that is, instead of computingb∗(·) each bidder will take the equilibrium bidding function as given and insteadwill report a type t.


In these direct revelation games, prices (at equilibrium) depend on thewinner’s type and on other information (e.g., on other bidders’ types). In afirst-price auction, the equilibrium price is a function only of the winner’s type.In a second-price auction, the expected price is a function of the highest andthe second highest types. In an English button auction, the expected price is afunction of all types. That is, in a first-price auction there is no linkage to otherbidders’ types whereas in the other auction formats the linkage is stronger in theEnglish button auction. The linkage works to increase the expected value of theobject to a bidder conditional on having received the highest type. The strongerthe linkage the stronger is the increase (due to affiliation) and therefore thehigher is the bid (at equilibrium) and, thus, the higher is the expected price.

5.6 Exercises

1. Prove that the product of functions with the monotone likelihood ratioproperty also has the multivariate monotone likelihood ratio property.

2. Justify the reasoning that lead to (4.1).3. Suppose g(x1, . . . , xn) has the monotone likelihood ratio property.

Suppose also that φi(xi) is increasing for every i. Show thatg(φ1(x1), . . . , φn(xn)) has the monotone likelihood ratio property. Applythis result to

g(x, y) = 1 + xy

to conclude that

f(x, y) =1 + f(x)f(y)

1 +(∫ 1

0f(x) dx

)2

has a monotone likelihood ratio if f(x) is increasing in [0, 1]. Is the sameconclusion true if f(x) is decreasing?

6

Mechanism Design

In this chapter, we will approach auctions from a more abstract perspective.We will use the approach of Myerson (1981) to show that much of auctiontheory under independent types can be cast as a mechanism design problem.In particular, we can use this approach to show that the revenue equivalencebetween first- and second-price, Dutch, and English auctions is in fact moregeneral: under some hypotheses (that hold in the analysis of Chapter 3), anytwo mechanisms that allocate the object in the same way and that yield thesame expected surplus to the individual with the lowest type will generate thesame expected revenue for the seller. We will also use these tools to characterizethe optimal auction—the one that maximizes the seller’s expected revenue.

In Section 6.1, we set up the basic notation, explain the revelation principleand why we can restrict ourselves to direct mechanisms. In Section 6.2, weexamine direct mechanisms. In Section 6.3, we demonstrate a general versionof the Revenue Equivalence Theorem and characterize the optimal auction.Section 6.4 extends the analysis of the optimal auction to non-monotonic mar-ginal valuations. In this section, we also discuss the optimal auction when typesare correlated, revenue equivalence for common value auctions and when thereare several identical objects for sale but demand is unitary.

6.1 The Revelation Principle

Although this chapter is more abstract than previous ones, the reader will beamply compensated by the generality of the results.

There are n bidders. Each player i = 1, . . . , n receives a type xi ∈ Xi = [0, v]that is distributed accordingly to the distribution Fi(x) =

∫ x

0fi(z) dz, x ∈ Xi.

The density fi : Xi → R is continuous and strictly positive. The set of individualtypes is denoted by X = Πn

i=1Xi = [0, v]n.

71

72 Mechanism Design

Each bidder knows his type and knows the distribution of the types of otherbidders. That is, i knows that the other bidders’ types, x−i ∈ X−i = [0, v]n−1,are distributed according to the density f−i(x−i) = Πj =ifj(xj). We assumethat the types are independent random variables. Thus, the joint density is

f(x1, . . . , xn) = f1(x1) × · · · × fn(xn) for x ∈ X. (6.1)

If W (x) is a random variable we denote by E−i[W (x)] its expectation withrespect to the variables x−i. That is,

E−i[W (x)] =∫

W (xi, x−i)f−i(x−i) dx−i.

We also assume, as in previous chapters, that the seller values the object atzero and that bidder i’s utility is Vi = ui(x) = xi. That is, we consider theIndependent Private Values model, bidders are risk neutral and their utilitiesare expressed in dollars.

We now must consider the mechanisms the seller may use when designing therevenue maximizing auction. The simplest procedure is to allow every bidder toannounce a number (the bid) and by means of some auction rule decide who getsthe object and how much each participating bidder pays. This setup would coverthe first- and second-price auctions as particular cases. It would not, however,cover the English auction as it does not allow bidders to rebid. Therefore, theseller has to be allowed to use a more flexible mechanism. Thus, suppose thatthe seller designs, for each bidder i, a set Si of the possible choices allowed forbidder i. Define S = Πn

i=1Si as the set of joint choices of all bidders. The sellerhas to choose, for each s ∈ S, an allocation rule q(s)—that is, a probabilityvector q(s) = (q1(s), . . . , qn(s)), where qi(s) is the probability that bidder i winsthe object for sale—and a payment vector, P (s) = (P1(s), . . . , Pn(s)), wherePi(s) ∈ R is the expected payment for bidder i when he participates in theauction.

Since participation in the auction is voluntary, we assume that for every ithere exists np ∈ Si, which is the non-participation strategy. Naturally, if s ∈ Sand si = np, then Pi(s) = 0 and qi(s) = 0. Formally,

Definition 10 An auction rule (or mechanism) is a triple (S, q, P ) suchthat:

1. S = Πni=1Si;

2. q = (q1, . . . , qn) : S → [0, 1]n,∑n

i=1 qi(s) ≤ 1;3. P = (P1, . . . , Pn) : S → Rn;4. If s ∈ S and si = np then qi(s) = 0 and Pi(s) = 0.

Example 12 (first-price auction) We can write the first-price auction as anauction rule. Define Si = R+ ∪ np, i = 1, . . . , n. If s ∈ S \ (np, . . . , np)

6.1 The Revelation Principle 73

define

b∗(s) = maxsi; si = np, 1 ≤ i ≤ n and

J(s) = i ≤ n; si = b∗(s).

That is, b∗(s) is the highest bid when s is chosen and at least one bidderparticipates and J(s) denotes the set of bidders who have the highest bid. Then

qi(s) =

1#J(s)

if i ∈ J(s)

0 if i /∈ J(s)and Pi(s) = qi(s)b∗(s).

define the auction rule where the object is allocated to the highest bidder andif there is more than one bidder with the highest bid, these bidders have anidentical probability of winning.

The second-price auction is formalized in a similar fashion. The Englishauction, however, is more difficult to formalize.

Example 13 (English button auction) First, recall that in the English buttonauction bidders keep pressing a button while the price rises continuously. Forsimplicity, we assume that pt = t. If a bidder releases the button then he leavesthe auction and cannot bid again. The object goes to the bidder who is last torelease the button and the price paid is the price at which the last but one bidderleft.

To formalize the auction we define the strategy set of a bidder. It has aninitial decision to participate (p) or not (np) in the auction. Then it has adeparture rule for each time t > 0 as a function of:

(i) the set of bidders who are still active at time t;(ii) the set of bidders who left the auction and the price at which they left

the auction.

We assume that the auction ends at time T > 0 if there is more than oneactive bidder at this time. Define for each bidder i the sets

∆i = p, npn−1 ∪ (t, x−i); t ∈ (0, T ), xj ∈ p, t and

Γi = φ;φ : ∆i → [0, T ], φ(t, x−i) ≥ t,∀t ∈ (0, T ).

The interpretation is that if xj = p bidder j is still bidding. If xj = p thenbidder j left at time xj. For φi ∈ Γi and z ∈ p, npn−1 we have that φi(z) isthe time at which bidder i leaves the auction if the set of active bidders (i.e.,j; zj = p) do not change. If t > 0, φi(t, x−i) is the time at which i quits ifall the bidders who are active at time t (i.e., j;xj = p) remain active untiltime T . Note that i may quit immediately at time t. Finally, we specify the

74 Mechanism Design

strategy set:

Si = Γi ∪ np.

Thus, S = S1 × · · · × Sn.We must now define, for each φ ∈ S, the allocation rule q(φ). That is, we

need to specify who receives the object.At time t = 0 we define z0

j = np if φj = np and z0j = p if φj = np. If

φi(0, z0−i) = 0 for every participating bidder i, then the auctioneer keeps the

object and charges nothing. If there is only one i such that φi(0, z0−i) > 0, this

bidder i wins the object at price 0. Otherwise, define

τ1 = minφi(0, z0−i);φ

i = np, φi(0, z0−i) > 0 and

H1 = i;φi(0, z0−i) = τ1, φi = np.

If there is no bidder i such that φi(0, zo−i) > τ1, then the object is allocated to

one of the members of H1 with probability 1/#H1 at price τ1. Otherwise, theauction proceeds.

Define z1i = p for every participating bidder such that φi(0, zo

−i) > τ1. Definealso z1

i = τ1 if i ∈ H1. Finally z1i = np if φi = np. We proceed inductively

defining

τ2 = minφi(τ1, z1−i);φ

i(τ1, z1−i) > τ1,

z2j =

np if φj = np

τ1 if φj(τ1, z1−j) = τ1

p if φj(τ1, z1−j) > τ1

and so on. The process stops eventually since at every step at least one bidderleaves. The object is allocated to the last bidder to leave and he pays the priceat which the next to last bidder dropped out.

For the following definition, we consider a given auction mechanism(S, q, P ).

Definition 11 A Bayesian Nash equilibrium of the auction mechanism(S, q, P ) is a vector of strategies, (b∗i (·))n

i=1, where b∗i : Xi → Si is such that forevery i,

E−i[qi(b∗i (xi), b∗−i(x−i))xi − Pi(b∗i (xi), b∗−i(x−i))]

= maxy∈Si

E−i[qi(y, b∗−i(x−i))xi − Pi(y, b∗−i(x−i))].

Thus, in a Bayesian Nash equilibrium, if bidders j = i are using strategiesb∗j (xj), the best bidder i with signal xi can do is to choose strategy b∗i (xi). Wealso need the following definitions.

6.2 Direct Mechanisms 75

Definition 12 An auction mechanism (S, q, P ) is direct if for every i,Si = Xi. The direct mechanism (X, q, P ) is incentive compatible if

E−i[(qi(x)xi − Pi(x))] = maxy∈Xi

E−i[qi(y, x−i)xi − Pi(y, x−i)].

Finally, the direct mechanism (X, q, P ) is individually rational if

E−i[(qi(x)xi − Pi(x))] ≥ 0.

Thus, we will examine strategies of the type b∗i (xi) = xi as candidate for aBayesian Nash equilibrium. The revelation principle says that we only need toconsider direct mechanisms.

Theorem 12 (Revelation principle) For any Bayesian Nash equilibrium(b1(·), . . . , bn(·)) of an auction mechanism (S, q, P ), there exists an incentivecompatible, individually rational, direct mechanism that yields the seller andbidders the same expected utilities as in the original auction mechanism.

Proof: Define for every x ∈ X the mechanisms

q(x) = q(b1(x1), . . . , bn(xn)) and P (x) = P (b1(x1), . . . , bn(xn)).

Now from

E−i[(qi(x)xi − Pi(x))] = E−i[qi(b(x))xi − Pi(b(x))]

and

n∑i=1

Pi(x) =n∑

i=1

Pi(b(x)),

it is immediate that the direct mechanism (q, P ) gives the same expected utili-ties to seller and bidders as the original mechanism. The mechanism is incentivecompatible since b(x) is a Bayesian Nash equilibrium. Moreover, it is indivi-dually rational since bidders always have the non-participation option in theoriginal mechanism.

6.2 Direct Mechanisms

As a result of the revelation principle, we can without loss of generality concen-trate our efforts into analyzing direct mechanisms. In what follows we make theassumption that the direct mechanisms are Riemann integrable. This assump-tion is necessary to make sense of the integrals and expected values that will

76 Mechanism Design

be calculated. Thus, the direct mechanism (q, P ) is individually rational if (itis Riemann integrable and)

E−i[qi(x)xi − Pi(x)] ≥ 0. (6.2)

Condition (6.2) says that player i’s expected profits when his type is xi andwhen all players announce their true types is non-negative.

Now we can turn our attention to the incentive compatibility constraintthat for each player i, i = 1, . . . , n, and for every y ∈ Xi, requires:

E−i[qi(x)xi − Pi(x)] ≥ E−i[qi(x)xi − Pi(x)], (6.3)

where x = (y, x−i). That is, the incentive compatibility condition (6.3) gua-rantees that when all players other than player i are revealing their true types,it is optimal for player i to reveal his true type. Now we can write player i’sexpected profits when his type is xi and when he reveals xi as

πi(xi) = Qi(xi)xi − E−i[Pi(x)], (6.4)

where Qi(xi) = E−i[qi(x)]. Condition (6.3) implies that

πi(xi) ≥ Qi(y)xi − E−i[Pi(y, x−i)]

= Qi(y)(xi − y) + Qi(y)yi − E−i[Pi(y, x−i)]

= Qi(y)(xi − y) + πi(y).

Thus, we can conclude that

πi(xi) − πi(y) ≥ (xi − y)Qi(y) for any xi, y ∈ Xi. (6.5)

It is immediate from (6.5) that πi(xi) is non-decreasing. We now prove that Qi

is non-decreasing as well.

Proposition 3 Qi(·) is a non-decreasing function.

Proof: Let us exchange xi and y in (6.5) above to obtain

πi(y) − πi(xi) ≥ (y − xi)Qi(xi). (6.6)

Adding equations (6.5) and (6.6) yields

0 ≥ (xi − y)[Qi(y) − Qi(xi)] = −(xi − y)[Qi(xi) − Qi(y)].

Our next task is to show that condition (6.3) also requires that πi(·) takesa particular form. We need the following intermediate result.

Lemma 3 Let xi > y. Then

Qi(xi) ≥πi(xi) − πi(y)

xi − y≥ Qi(y).

6.2 Direct Mechanisms 77

The lemma follows in a direct fashion from equations (6.5) and (6.6). Wenow show that πi is completely defined by πi(0) and qi.

Theorem 13 Player i’s expected profits in any direct incentive compatiblemechanism is given by

πi(xi) = πi(0) +∫ xi

0

Qi(t) dt

= πi(0) +∫ xi

0

∫qi(t, x−i)f−i(x−i) dx−i dt. (6.7)

Proof: Partition the interval [0, xi] as follows:[0,

xi

n,2xi

n, . . . ,

nxi

n

].

We can write

πi(xi) − πi(0) =n∑

j=1

[πi

(jxi

n

)− πi

((j − 1)xi

n

)]

= πi

(xi

n

)− πi(0) + πi

(2xi

n

)− πi

(xi

n

)+ · · ·

+ πi

(nxi

n

)− πi

((n − 1)xi

n

),

where the first equality comes from “telescoping” (that is, cancelling the termswith opposite signs). Lemma 3 then allows us to write

πi(xi) − πi(0) ≥n∑

j=1

Qi

((j − 1)xi

n

)xi

n.

We have shown that Qi(·) is monotone (therefore it has at most a countablenumber of discontinuities). Thus, Qi(·) is Riemann integrable (in the expressionabove we have the lower sums) and as a result as we can take n to infinity (i.e.,as our partition becomes finer and finer) to obtain

πi(xi) − πi(0) ≥∫ xi

0

Qi(t) dt.

By following a similar process for the upper sums we can show that

πi(xi) − πi(0) ≤∫ xi

0

Qi(t) dt.

An immediate consequence of this theorem is the determination of thepayment mechanism.

78 Mechanism Design

Corollary 4 If the direct mechanism (q, P ) is incentive compatible andindividually rational, then there exists an ai ≥ 0 such that

E−i[Pi(x)] = Qi(xi)xi − ai −∫ xi

0

Qi(t) dt.

Proof: From the definition of expected profits (6.4) we have that

E−i[Pi(x)] = Qi(xi)xi − πi(x).

It follows from (6.7) that

E−i[Pi(x)] = Qi(xi)xi − πi(0) −∫ xi

0

Qi(t) dt.

The individual rationality constraint implies that πi(0) ≥ 0 and, thus, bydefining ai = πi(0), we are done.

The following definition will be useful below.

Definition 13 An allocation rule (qi(x))ni=1 is feasible if

Qi(xi) = E−i[qi(x)] =∫

qi(x)f−i(x−i) dx−i

is increasing and there are payments (Pi(x))ni=1 such that

E−i[Pi(x)] = Qi(xi)xi − ai −∫ xi

0

Qi(t) dt, ai ≥ 0. (6.8)

Thus, if (q, P ) is incentive compatible and individually rational, the alloca-tion rule q is feasible. Reciprocally, if q is feasible and (Pi(x))n

i=1 are associatedpayments, then (q, P ) is incentive compatible and individually rational.

Remark 5 Note that there are many ways to choose payments such that (6.8)is true. One simple way is to choose

Pi(xi) = Qi(xi) − ai −∫ xi

0

Qi(t) dt.

This corresponds to an auction in which every bidder pays whether winning theobject or not. Another possible payment rule is given by

Pi(x) = qi(x)xi −∫ xi

0

qi(s, x−i) ds.

In this case the bidder only pays as long as qi(x) > 0.

6.3 Revenue Equivalence and the Optimal Auction 79

6.3 Revenue Equivalence and theOptimal Auction

We start this section by reminding the reader that we have just shown that anyincentive compatible direct mechanism is such that condition (6.7) holds. Theseller’s expected revenue in any such mechanism is given by

R =∫ v

0

n∑i=1

Pi(t)f(t) dt.

That is, the seller’s expected revenue is simply the sum of the expectedpayments from players i = 1, . . . , n. Recall that from (6.4) we can write:

E−i[Pi(x)] = Qi(xi)xi − πi(xi) = Qi(xi)xi − πi(0) −∫ xi

0

Qi(t) dt, (6.9)

where the last equality follows from (6.7). From E[Pi(x)] = Ei[E−i[Pi(x)]] wehave that

E[Pi(x)] = Ei

[Qi(xi)xi − πi(0) −

∫ xi

0

Qi(t) dt

]

= −πi(0) + Ei

[Qi(xi)xi −

∫ xi

0

Qi(t) dt

].

Thus we have proved the following general form of the revenue equivalencetheorem:

Theorem 14 (General Revenue Equivalence Theorem) If (q, P ) and (q, P )are direct mechanisms with the same allocation rule, i.e. q = q, and such that ifa bidder has the lowest type his expected utility is the same in both mechanisms,then the seller’s expected revenue is the same.

This theorem generalizes the revenue equivalence theorem obtained in Chap-ter 3. To see this, suppose we are in the symmetric IPV case. Assume furtherthat the bidder with the highest type receives the object. That is, qi(x) = 1 ifxi > maxj =i xj . This implies that

Qi(xi) =∫

qi(xi, x−i)f−i(x−i) dx−i

=∫

xi>maxj =i xk

∏j =i

fj(xj) dx−i = F (xi)n−1.

bidder i’s expected payment is therefore,

E−i[Pi(x)] = Qi(xi)xi − πi(xi) = F (xi)n−1xi − πi(0) −∫ xi

0

F (t)n−1 dt.

(6.10)

80 Mechanism Design

In the first-price, second-price, Dutch, and English auctions, the bidder withlowest type (i.e., xi = 0) receives expected utility equal to zero. Thus πi(0) = 0.Hence, it follows that they all have the same expected revenue. This is summarizedby the following result.

Theorem 15 Consider any two auction mechanisms that allocate the objectto the bidder with the highest type. Suppose further that in these two mecha-nisms the expected profits from the bidder with the lowest possible type is zero.Then these two mechanisms yield the same expected payment for each biddergiven his type and the same expected revenue for the seller.

Theorem 15 and equation (6.10) can be quite useful in determining equili-brium bidding strategies for any auction format. For example, consider afirst-price sealed-bid auction. If bidder i wins the auction, he pays his bid.Therefore, his expected payment in equilibrium is equal to his bid multipliedby the probability of winning:

F (xi)n−1bF (xi) = F (xi)n−1xi −∫ xi

0

F (t)n−1(t) dt.

Thus, his equilibrium bid is equal to

bF (xi) = xi −∫ xi

0F (t)n−1(t) dt

F (xi)n−1,

which coincides with the expression obtained in Chapter 3.Consider now an all-pay auction, where the winner is the bidder with the

highest type but all bidders pay their bids. In this case, i’s expected paymentcoincides with his equilibrium bid as he will pay regardless of whether he winsor not:

bA(xi) = F (xi)n−1xi −∫ xi

0

F (t)n−1(t) dt.

We can now turn our attention to the optimal auction—the auction that maxi-mizes the seller’s expected revenue. We can use expression (6.9) determiningindividual i’s expected payment in any direct mechanism to determine theseller’s revenue. An individual’s expected payment can be written as:

Pi =∫

Pi(x)f−i(x−i)fi(xi) dx−i dxi

=∫

xi

[∫x−i

Pi(x)f−i(x−i) dx−i

]fi(xi) dxi

=∫

xi

[Qi(xi)xi − πi(0) −

∫ xi

0

Qi(t) dt

]fi(xi) dxi.


The last equality follows from (6.9). Since πi(0) ≥ 0 we set πi(0) = 0, withoutloss of generality. Recall that

Qi(xi) =∫

x−i

qi(x)f−i(x−i) dx−i.

Thus we have

P =∫

xiqi(x)f(x) dx −∫ (∫ xi

0

Qi(t) dt

)fi(xi) dxi.

We can now change the order of integration in the last integral:∫ (∫ xi

0

Qi(t) dt

)fi(xi) dxi =

∫ ∫ v

t

Qi(t)fi(xi) dxi dt

=∫

Qi(t)(1 − Fi(t)) dt

=∫

qi(x)(1 − Fi(t))f−i(x) dx−i dt

=∫

qi(x)1 − Fi(xi)

fi(xi)f(x) dx.

In the last line we replaced t by xi as the label of the variable of integration.Substituting this in the expression for Pi :

Pi =∫

xiqi(x)f(x) dx −∫

qi(x)1 − Fi(xi)

fi(xi)f(x) dx

=∫ [

xi −1 − Fi(xi)

fi(xi)

]qi(x)f(x) dx.

This is i’s expected payment in any incentive compatible direct mechanism. Tofind the seller’s expected revenue we simply sum over all bidders:

R =∫ n∑

i=1

[xi −

1 − Fi(xi)fi(xi)

]qi(x)f(x) dx. (6.11)

Note that

R ≤∫

maxi=1,...,n

[xi −

1 − Fi(xi)fi(xi)

]+f(x) dx,

where [r]+ = maxr, 0, stands for the non-negative part of [·]. We can nowcharacterize the optimal auction when the expression inside the brackets isincreasing.

82 Mechanism Design

Theorem 16 (optimal auction) Suppose that for every bidder i,

Ji(x) = x − 1 − Fi(x)fi(x)

(6.12)

is increasing. Then the optimal auction allocates the object to the individualwith the highest Ji(xi) if Ji(xi) ≥ 0. Otherwise the auctioneer keeps the object.Specifically, the allocation rule, (q∗i (x))n

i=1, is given by

q∗j (x) =

0 if j /∈ W (x) or Jj(xj) < 0,1

#W (x) if j ∈ W (x), and Jj(xj) ≥ 0,

where W (x) = j;Jj(xj) = max1≤i≤n Ji(xi).

Remark 6 The function Ji(xi) in (6.12) is referred to as the marginalvaluation. This is due to an analogy with the monopolist’s marginal revenue.See, for example, Bulow and Roberts (1989).

The payments by the bidders associated with the allocation rule (q∗j )nj=1

can be specified in several ways. See remark 5 above. The following corollarygives the optimal auction in the symmetric case:

Corollary 5 If the distribution of types is the same for every bidder, thatis, Fi = F for every i, and if x − (1 − F (x))/f(x) is strictly increasing, theoptimal auction allocates the object with probability one to the individual withthe highest type if his marginal valuation is non-negative at the highest type.

Proof of Theorem 16 Define the set

W (x) =

1 ≤ j ≤ n;Jj(xj) = max1≤i≤n

Ji(xi)

.

For every j we define

q∗j (x) =

0 if j /∈ W (x) or Jj(xj) < 0.

1#W (x) if j ∈ W (x), and Jj(xj) ≥ 0.

First note that q∗j (x) is increasing in the variable xj and therefore Q∗j (xj) =∫

q∗j (x) dx−j is also increasing. Thus if we define

P ∗j (xi) = Q∗

i (xi)xi −∫ xi

0

Q∗i (t) dt,

the resulting mechanism (q∗j , P ∗j )j≤n satisfies the individual rationality and

incentive compatibility constraints. It remains to check that it maximizes theseller’s expected revenue. So suppose the allocation rule (qi)n

i=1 is feasible.1

1 See Definition 13.


From equation (6.11) we see that

R =∫ n∑

i=1

[xi −

1 − Fi(xi)fi(xi)

]qi(x)f(x) dx

≤∫

maxi

[xi −

1 − Fi(xi)fi(xi)

] n∑i=1

qi(x)f(x) dx

=∫

max1≤i≤n

Ji(xi)

(n∑

i=1

qi(x)

)f(x) dx

≤∫ (

max1≤i≤n

Ji(xi))+

f(x) dx

=∫ n∑

i=1

[xi −

1 − Fi(xi)fi(xi)

]q∗i (x)f(x) dx.

We now give a direct proof of the corollary.

Proof: Define J(x) = x − (1 − F (x))/f(x). We define q∗i (x) = 0 if xi <maxj =i xj . Therefore Qi(xi) = Fn−1(xi) is increasing. Thus the associatedmechanisms satisfy (6.2) and (6.3) and for every feasible q:

R =∫ n∑

i=1

[xi −

1 − F (xi)f(xi)

]qi(x)f(x) dx

≤∫

maxi

J(xi)n∑

i=1

qi(x)f(x) dx

=∫ (

J

(max

1≤i≤nxi

))+

f(x) dx =∫ n∑

i=1

Ji(xi)q∗i (x)f(x) dx.

The next example illustrates how an optimally chosen reserve price can beused to increase the seller’s expected revenue.

Example 14 Suppose fi(x) = 110 , x ∈ [0, 10]. Then J(x) = x − 1−(x/10)

1/10 =2x − 10. If maxi xi < 5 the auctioneer keeps the object. Otherwise, the objectis allocated to the bidder with the highest xi. The seller’s expected revenue ina symmetric equilibrium of any auction that allocates the object to the highestbidder and with a reserve price of 5 is equal to:

R =∫

maxi xi≥5

(2 max

1≤i≤nxi − 10

)1

10ndx1 · · · dxn

=∫ 10

5

(2y − 10)nyn−1

10ndy =

101 + n

(n − 1 + 2−n).

84 Mechanism Design

The seller’s expected revenue for the Vickrey auction without a reserve price is

R2 =∫ 10

0

n(n − 1)( y

10

)n−1 (1 − y

10

)dy = 10

(n − 1n + 1

).

The difference is R − R2 = 10/2n(1 + n). Thus if we have three bidders thedifference is 10

32 = 0.31, which is about 6% of the maximum R. The full surplusis∫

max xi dx =∫ 10

0nyn

10n dy = 10nn+1 . Thus the consumer’s surplus is

10n

n + 1− 10

1 + n(n − 1 + 2−n) =

10(1 − 2−n)n + 1

≤ 10n + 1

.

We now show that the optimal auction, in the symmetric bidders case, canbe seen as a second-price sealed-bid auction with a single reserve price.

In the symmetric case we have that maxi J(xi) = J(maxi xi). Definex∗ = maxi xi. If there is only one bidder j such that xj = x∗ and J(xj) ≥ 0, thisbidder wins the object. If there are m bidders in this situation each one wins theobject with probability 1/m. Now we define the following payment (see Remark 5above):

Pj(x) = qj(x)xj −∫ xj

0

qj(s, x−j) ds

=

maxj =i xj if xj = x∗ > maxi=j xi

1#l;xl = x∗x∗ if xj = x∗.

Above we used that if xj > maxi=j xi then

qj(x)xj −∫ xj

0

qj(s, x−j) ds = xj −(

xj − maxi=j

xi

)= max

i=jxi.

That is, the winner’s payment is equal to the largest of : (1) the maximum typeamongst all other players or, (2) the optimal reserve price.

Example 15 There are many probability distributions that satisfy the incre-asing marginal valuations assumption. For example, the uniform distributionF (x) = x in [0, 1] satisfies it since

J(x) = x − 1 − x

1= 2x − 1

is strictly increasing. More generally if F (x) = xθ then

J(x) = x − 1 − xθ

θxθ−1=

x(θ + 1) − x−θ+1

θ

is increasing if θ ≥ 1 since

J ′(x) =θ + 1 − (1 − θ)x−θ

θ≥ 0

6.4 Some Extensions 85

in this case. If θ < 1 then J is not increasing. For example F (x) =√

x thenJ(x) = 3x − 2

√x is plotted below:

–0.2

00.2 0.6 0.8 1

0.2

0.4

0.6

0.8

1

x

y

Marginal valuation of F(x) = √x

0.4

Remark 7 If the density is increasing then the marginal valuation isincreasing. To see this, differentiate J(x):

ddx

J(x) = 2 +(1 − F (x))f ′(x)

f2(x)

and, therefore, f ′ ≥ 0 implies J ′ > 0. Thus, convex distributions haveincreasing marginal valuations.

Remark 8 A simple but useful generalization of the optimal auction theoremcan be obtained if instead of increasing marginal valuations we assume thatx, J(x) ≥ 0 is an interval and J(·) is increasing in this interval. Thedistributions F (x) = xθ, 0 < θ < 1 are covered by this assumption.

6.4 Some Extensions

6.4.1 Non-monotonic Marginal Valuation

In this section, we allow for marginal valuations that are not everywhereincreasing. We begin by showing that there are many distributions with thisproperty.

Proposition 4 Suppose h : [0, 1] → R is a continuous function. Assumealso that

(i) h(x) < x for every x ∈ [0, 1);(ii) h(1) = 1;(iii)

∫ 1

0dy

y−h(y) = ∞.

86 Mechanism Design

Then there exists a continuously differentiable distribution F (r) such thatfor every r ∈ [0, 1),

r − 1 − F (r)F ′(r)

= h(r).

Proof: We first remark that the continuity of h and (i) imply that for everyx < 1, miny≤x(h(y) − y) > 0. Thus, the integral

∫ x

0dy

y−h(y) > 0 is finite.Define F : [0, 1] → R by

F (x) =

1 − exp

[−∫ x

0

dy

y − h(y)

], if 0 ≤ x < 1,

1, if x = 1.

It is immediate that 0 ≤ F (x) < 1 for every x < 1. From (iii) we have thatlimx→1 F (x) = 1. The density of F is given by

f(x) = F ′(x) = exp[−∫ x

0

dy

y − h(y)

]1

x − h(x)=

1 − F (x)x − h(x)

,

0 ≤ x < 1. Thus,

x − 1 − F (x)f(x)

= x − (x − h(x)) = h(x), 0 ≤ x < 1.

Remark 9 A simple condition for the validity of (iii) above is h′(1) > 1.

We see from the proposition above that practically any behavior on (0, 1)is possible for a marginal valuation J(·). And since there are no theoreticalgrounds to assume that the distribution of bidders’ types has a monotonic mar-ginal valuation, it is important to consider the general case of non-monotonicvaluations.

We now return to the general development. Our aim is to compute theoptimal auction when the marginal valuation is not monotonic. It is instructiveto start with the simplest case.

One-bidder caseThe seller allocates the object if and only if t ≥ t0 where t0, is such that∫ v

t0

J(s)f(s) ds = maxt∈[0,v]

∫ v

t

J(s)f(s) ds. (6.13)

To see this, suppose q(t) ∈ [0, 1] is increasing. The seller maximizes∫ v

0

q(t)J(t)f(t) dt.


Suppose first that q(·) is a step function. That is, q(t) = ai ∈ [0, 1] if t ∈(ti, ti+1]. Then, defining bi =

∫ ti

0J(t)f(t) dt, we obtain

∫ v

0

q(t)J(t)f(t) dt =n∑

i=0

ai

∫ ti+1

ti

J(t)f(t) dt

=n∑

i=0

ai(bi+1 − bi) =n∑

i=0

aibi+1 −n∑

i=0

aibi

=n+1∑j=1

aj−1bj −n∑

j=1

ajbj = an

∫ v

0

J(t)f(t) dt −n∑

j=1

(aj − aj−1)bj

≤ an

∫ v

0

J(t)f(t) dt − (an − a0)minj≥1

bj

= an maxj≤n

∫ v

tj

J(t)f(t) dt + a0 minj≥1

bj

since

n∑j=1

(aj − aj−1)bj ≥n∑

j=1

(aj − aj−1)min bj = (an − a0)minj≥1

bj .

The maximum in this case is attained for a0 = 0 and aj = 1 if tj ≥ t. Thus,our solution maximizes expected revenue amongst the family of increasing stepfunctions.

Now consider an increasing function q(x) ∈ [0, 1]. Suppose ε > 0 is given.Divide the interval [0, 1] in N parts where 1/N < ε. Define Ij = ( j

N , j+1N ],

j = 0, . . . , N−1. Then the inverse image q−1(Ij) is an interval (possibly empty).Also ∪jq

−1(Ij) = [0, v]. Thus, if we define q(x) = inf q(Ij) for x ∈ q−1(Ij) thisfunction, q, is increasing, constant in the intervals q−1(Ij) and differs from qby no more than ε. Now note that

∫|J(t)|f(t) dt < ∞. Therefore if ε > 0 is

given and q(x) ∈ [0, 1] is increasing and q is a step function distant from q byno more than ε then

maxt∈[0,v]

∫ v

t

J(s)f(s) ds ≥∫

J(s)q(s)f(s) ds

≥∫

J(s)f(s)(q(s) − ε) ds =∫

J(s)f(s)q(s) ds − ε

∫|J(s)|f(s) ds.

Since ε is arbitrary we proved the general case.

88 Mechanism Design

Several-bidders caseIn this section we need the concept of the Stjeltjes integral. Most readers maywant to skip the proof and go directly to the examples below. First, we define

hi(q) = Ji(F−1i (q)) = F−1

i (q) − 1 − q

fi(F−1i (q))

.

Second, define Hi(q) =∫ q

0hi(r)dr. Now let Gi(q) be the convex hull2 of Hi(q).

By this we mean the largest convex function Gi such that Gi ≤ Hi. Finally,define3 gi(x) = (Gi)′+(x) the right-hand derivative of Gi. The function gi isincreasing. Define Ji(ti) = gi(Fi(ti)). The optimal auction gives the object tothe bidder with the greatest Ji(ti) as will be seen in the proof of the followingtheorem.

Theorem 17 (optimal auction—general case) The optimal auction allocatesthe object to the bidder with the highest Ji(ti) if this number is nonnega-tive. Otherwise, the auctioneer keeps the object. That is, the allocation ruleis given by

qi(t) =

1#W (t)

if i ∈ W (t) and Ji(ti) ≥ 0,

0 if i /∈ W (t).

where W (t) = j; Jj(tj) = maxi Ji(ti).

First note that Qi(ti) =∫

qi(t)f−i(t−i) dt−i is increasing and therefore themechanism defined by (qi(t))n

i=1 is feasible. We now divide the proof into twolemmas.

Lemma 4 The following inequality is true for every i:

∫Ji(ti)qi(t)f(t) dt ≤

∫Ji(ti)qi(t)f(t) dt. (6.14)

Lemma 5 Moreover, the equality is true if and only if

∫ v

0

(Hi(Fi(ti)) − Gi(Fi(ti))) dQi(ti) = 0. (6.15)

2 See equation (D.1).3 See Appendix D.3.


Proof: First we have for every i that∫Ji(ti)qi(t)f(t) dt

=∫

hi(Fi(ti))qi(t)f(t) dt

=∫

(hi(Fi(ti)) − gi(Fi(ti)))qi(t)f(t) dt +∫

Ji(ti)qi(t)f(t) dt

=∫

(hi(Fi(ti)) − gi(Fi(ti)))Qi(ti)fi(ti) dti +∫

Ji(ti)qi(t)f(t) dt

=∫ (

ddti

Hi(Fi(ti)) −d

dtiGi(Fi(ti))

)Qi(ti) dti +

∫Ji(ti)qi(t)f(t) dt.

In the first line above we just used that hi(Fi(ti)) = Ji(ti). In the second linewe used that Ji(ti) = gi(Fi(ti)). We now integrate by parts the first integral ofthe last line. Thus, writing s = ti and omitting the subscript i for conciseness:∫ v

0

(dds

H(F (s)) − dds

G(F (s)))

Q(s) ds

= (H(F (s)) − G(F (s)))Q(s)|v0 −∫ v

0

(H(F (s)) − G(F (s))) dQ(s)

= −∫ v

0

(H(F (s)) − G(F (s))) dQ(s) ≤ 0. (6.16)

Above we used that that Hi(1) = Gi(1) and Hi(0) = Gi(0). Therefore, wehave that ∫

Ji(ti)qi(t)f(t) dt ≤∫

Ji(ti)qi(t)f(t) dt.

It is also immediate from (6.16) that the equality is true if and only if∫ v

0

(Hi(Fi(ti)) − Gi(Fi(ti))) dQi(ti) = 0.

Lemma 6 For the mechanism (qi(t))ni=1 it is true that

Bi =∫ v

0

(Hi(Fi(ti)) − Gi(Fi(ti))) dQi(ti) = 0. (6.17)

Proof: Consider ti ∈ (0, v) such that

Hi(Fi(ti)) > Gi(Fi(ti)).

In this case Gi is linear in a neighborhood (see Lemma 14 in Appendix D) ofFi(ti). So Ji(s) = gi(Fi(s)) is constant in a neighborhood of ti and finally we

90 Mechanism Design

have that Qi(s) = Qi(ti) in the same neighborhood. Therefore we concludethat Bi = 0.

We now prove the theorem.

Proof: It follows from (6.14) that for every feasible mechanism (qi(t))ni=1,∫ n∑

i=1

Ji(ti)qi(t)f(t) dt ≤∫ n∑

i=1

Ji(ti)qi(t)f(t) dt ≤∫ n∑

i=1

Ji(ti)qi(t)f(t) dt.

The last inequality above follows directly from the definition of (qi(t))ni=1. We

now show that∫ n∑i=1

Ji(ti)qi(t)f(t) dt =∫ n∑

i=1

Ji(ti)qi(t)f(t) dt

and we are done with the proof. This follows since to have equality in (6.14) itsuffices that (6.15) is true. But (6.17) says exactly this.

We now apply the optimal auction theorem when the virtual valuation is notmonotonic. We remark first that if W : [0, 1] → [0, 1] is strictly increasing andW (0) = 0, W (1) = 1 then its inverse W−1 is also strictly increasing. Thus, theinverse of a strictly increasing distribution W : [0, 1] → [0, 1] is also a (strictlyincreasing) distribution.

Example 16 The function

W (x) =3x − 8x2 + 8x3

3, 0 ≤ x ≤ 1

1

0.8

0.6

0.2

0.4

0 0.2 0.4Distribution W(x)

0.6 0.8 1x

y


is a strictly increasing distribution function. Thus F (x) = W−1(x) is adistribution. We can even write F (x) explicitly as:

F (x) =z1/3 − 2z−1/3 + 4

12(6.18)

z = z(x) = −44 + 324x + 18√

6 − 88x + 324x2.

1

0.8

0.6

0.2

0.4

0 0.2 0.4 0.6 0.8 1x

y

Distribution F(x)

If J(x) = x − (1 − F (x))/F ′(x) is the marginal valuation then

h(y) = J(F−1(y)) = W (y) − 1 − y

F ′(W (y))= W (y) − (1 − y)W ′(y)

=32y3 − 48y2 + 22y − 3

3=

323

(y − 1

4

)(y − 1

2

)(y − 3

4

).

This function is plotted below:

–1

–0.5

0

0.5

1

0.4 0.6 0.8 1y

h(y) = J(F –1 (y))

0.2

92 Mechanism Design

The derivative of h is given by

h′(y) =13(96y2 − 96y + 22) = 32

(y − 1

2−

√3

12

)(y − 1

2+

√3

12

).

Thus, h has a local maximum at 12 − 1

12

√3 0.35 and a local minimum

at 12 + 1

12

√3 = 0.64. The function h is increasing in [0, 1

2 − 112

√3] and in

[ 12 + 112

√3, 1]. It is decreasing in [12 − 1

12

√3, 1

2 + 112

√3]. Now we define H(q) =∫ q

0h(y) dy = 8

3q4 − 163 q3 + 11

3 q2 − q.

0

–0.02

–0.04

–0.06

–0.08

0.2 0.4

H(q) = q

0h(y ) dy

0.6 0.8 1

The function H is convex in the intervals [0, 12 − 1

12

√3] and [12 + 1

12

√3, 1]. We

now have to find the greatest convex function that is pointwise less or equalto H. See the plot above. The vertical dotted lines indicate x = 1

2 − 112

√3

and x = 12 + 1

12

√3. The horizontal dotted line indicates the value of H(1

4 ) =H( 3

4 ) = − 332 . From the graph it shall be clear that

G(x) =

H(x) if 0 ≤ x ≤ 14

− 332

if14

< x ≤ 34

H(x) if34

< x ≤ 1


is the convex hull of H. This is plotted below:

0

–0.02

–0.04

–0.06

–0.08

0.250 0.5 0.75 1

y

x

G the convex hull of H

We are almost done with the example. Now define g(x) = (G′)+(x) the right-hand derivative of G. This is given by

g(x) =

13 (32x3 − 48x2 + 22x − 3) if 0 ≤ x ≤ 1

4

0 if 14 < x ≤ 3

413 (32x3 − 48x2 + 22x − 3) if 3

4 < x ≤ 1.

Finally, J(x) = g(F (x)) is plotted below. For comparison we plot also thegraph of J.

–1

–0.5

0

0.5

1

0.2 0.4 0.6 0.8 1x

The marginal valuation J

J (x )

We see from the figure that no bidder with type xi < 14 receives the object. If

the highest type is higher than 34 then this bidder receives the object. If the

highest type is between 14 and 3

4 and (say) there are m bidders with types inthis range then each bidder receives the object with probability 1/m.

94 Mechanism Design

Example 17 We now consider another example where it will be moredifficult to find the convex hull. Define the inverse distribution F−1(x) =13 (4y − 10y2 + 9y3).

1

0.8

0.6

0.2

0.4

0 0.2 0.4 0.6 0.8 1F –1 (x )

1

0.8

0.6

0.2

0.4

0 0.2 0.4 0.6 0.8 1F (x )

The distribution F (x) is plotted above. It has an analytical expression similarto (6.18) and the reader will certainly forgive us for not explicitly writing it. IfJ is the marginal valuation for F then

h(x) = J(F−1(x)) = F−1(x) − (1 − x)(F−1)′(x)

= −43

+28x

3− 19x2 + 12x3

= 12(

x − 14

)(x − 2

3

)2

.


This is plotted below.

h (x )

1/4 2/3

Then, the function H(x) =∫ x

0h(y) dy = − 4

3x+ 143 x2− 19

3 x3 +3x4 has the graph:

0

–0.02

–0.04

–0.06

–0.08

–0.1

–0.12

0.2 0.4 0.6 0.8 1

H (x )

To find the convex hull of H, we first find the regions for which H′′ ≥ 0, that

is, the regions for which h′ ≥ 0. The derivative of h is given by

h′(x) =283

− 38x + 36x2 = 36(

x − 718

)(x − 2

3

).

The vertical dotted lines in the graph above connects(

718 , 0

)to(

718 ,H

(718

))and(

23 , 0)

to(

23 ,H

(23

)). Thus, h is increasing in

[0, 7

18

]and in

[23 , 1]. So suppose

G is the convex hull of H. To the left of the point x = 7/18 0.39 it mustcoincide with H. This follows since the function:

G(x) =

H(x) if x ≤ 1

4

H(1/4) if 14 ≤ x ≤ 1

is convex and less than or equal to H. No point of the graph of H to the rightof x = 7

18 and to the left of 23 belongs to the graph of G since H is concave in

96 Mechanism Design

this region. So what we need to find are points a ∈(

14 , 7

18

)and b ∈

(23 , 1)

suchthat the line connecting (a,H(a)) and (b,H(b)) satisfies the conditions:

h(a) = H ′(a)

H ′(b) = h(b);

H(b) − H(a)b − a

= H ′(a).

This gives a system:

12b2 + (12a − 19)b + 12a2 − 19a +283

= 0,

3b3 +9a − 19

3b2 +

14 − 19a + 9a2

3b +

−14a + 38a2 − 27a3

3= 0.

In the first equation above we solve for b and take the highest root (why?). Thisyields

b∗ =19 − 12a +

√−87 + 456a − 432a2

24.

Substituting this in the second equation yields an equation in variable a:

−2041288

a − 9a3 +574

a2 +3781 − 25

√−87 + 456a − 432a2

3456= 0.

Solving this equation we obtain:

a∗ =19 − 5

√3

36 0.29,

b∗ =19 + 5

√3

36 0.77.

The graph of the convex hull G is plotted below:

0

–0.02

–0.04

–0.06

–0.08

–0.1

–0.12

0.2 0.4 0.6 0.8 1

G (x )


Analytically, if ω(x) = − 43x + 14

3 x2 − 193 x3 + 3x4 we have

G(x) =

ω(x) if x ≤ 19 − 5√

336

− 20449139968

+1251944

x if19 − 5

√3

36< x ≤ 19 + 5

√3

36

ω(x) if19 + 5

√3

36< x ≤ 1.

We then find g(x) = G′(x) :

g(x) =

28x − 43

− 19x2 + 12x3 if x ≤ 19 − 5√

336

1251944

if19 − 5

√3

36< x ≤ 19 + 5

√3

3628x − 4

3− 19x2 + 12x3 if

19 + 5√

336

< x ≤ 1

and the associated marginal valuation J(x) = g(F (x)). Thus, if there aretwo bidders with types between W ((19 − 5

√3)/36) = F−1((19 − 5

√3)/36) and

W ((19 + 5√

3)/36) and all others have smaller types, those two bidders receivethe object with 50 percent probability.

6.4.2 Correlated Values

The optimal auction result we presented above is very general when typesare independent. If the marginal valuations are increasing and the bidders areex-ante symmetric the optimal auction is a second-price sealed bid auction withan optimally chosen reserve price. The optimal auction can also be implementedas a first-price auction with a reserve price or through any other efficient auctionformat with an appropriately chosen reserve price.

However, the existence of correlation between types is also an importantpossibility. For example how much you appreciate a painting, a rare book or afine wine is certainly a private matter but it is also naturally influenced by howmuch you think others appreciate them. What is the optimal auction if typesare correlated? It turns out that, in general, it is not the Vickrey auction witha reserve price. It includes a Vickrey auction and also a lottery. This lotterywill include auctioneer payments to the bidders! Before going into these detailslet us see where the approach used in the IPV case gets into difficulties. Themain reason is that bidder’s expected utility is not uniquely given as a functionof the allocation rule. We saw this on the affiliated values chapter. First- andsecond-price auctions have the same allocation rule but give different expectedrevenue.

As before, we have n bidders with valuation Vi = xi. Bidder i type varies inxi ∈ Xi = [0, v]. The set of joint types is X = [0, v]n and the joint distribution of

98 Mechanism Design

types has a continuously differentiable density f : X → R++. The distributionof types of bidder i has density fi(xi) =

∫f(x) dx−i. The equality (6.1) is not

valid since we do not have independence. If bidder i has type xi, the distributionof x−i, that is, the distribution of the other bidders’ types has the conditionaldensity given by

f(x−i |xi) =f(xi, x−i)∫f(x)dx−i

.

The incentive compatibility constraints (6.3) must be rewritten as

E[qi(x)xi − Pi(x) |xi]

=∫

(qi(x)xi − Pi(x))f(x−i |xi) dx−i

≥ E[qi(x)xi − Pi(x) |xi].

Here x = (y, x−i) as before. Now defining,

Qi(xi) = E[qi(x) |xi] and πi(xi) = Qi(xi)xi − E[Pi(x) |xi]

the analogous equation to (6.5) is

πi(xi) − πi(y) ≥ Qi(y)(xi − y) + E[Pi(y, x−i) | y] − E[Pi(y, x−i) |xi]

= Qi(y)(xi − y) +∫

Pi(y, x−i)(f(x−i | y) − f(x−i |xi)) dx−i.

If xi > y we have that

πi(xi) − πi(y)xi − y

≥ Qi(y) −∫

Pi(y, x−i)f(x−i |xi) − f(x−i | y)

xi − ydx−i.

Suppose now that xi < y. Then

πi(xi) − πi(y)xi − y

≤ Qi(y) −∫

Pi(y, x−i)f(x−i |xi) − f(x−i | y)

xi − ydx−i.

Thus, if πi is differentiable at y we have that

π′i(y) = Qi(y) −

∫Pi(y, x−i)

∂

∂yf(x−i | y) dx−i. (6.19)

We see from this equation that πi is not a function only of qi but of Pi as well.The next section shows how to characterize the optimal auction in general.

Full surplus extractionFor this part we rely on Cremer and MacLean (1988) and McAfee and Reny(1992). By surplus we mean the sum of the bidder’s surplus and the seller’ssurplus. A bidder’s surplus is his utility. The seller’s surplus is the paymentshe receives. Thus, the total surplus is equal to maxi xi if the bidder withthe highest type receives the object. By full surplus extraction we mean that


the seller receives on average maxi xi. Finally, by approximate full surplusextraction we mean that for every ε > 0 there is an auction mechanism in whichthe seller surplus is at least maxi xi − ε in expectation. It is quite remarkablethat, for a reasonably general class of density functions approximate full surplusextraction is possible. The key condition for approximate full surplus extractionis sufficient variability of the conditional density.

Discrete caseWe begin our analysis with a simple discrete model. There are two bidders andtwo types. Say bidder i = 1, 2 can be of type ti = 1 or ti = 2. Let us supposethe following probability distribution of types:(

Pr(1, 1) Pr(1, 2)Pr(2, 1) Pr(2, 2)

)=(

a bb d

),

where a, b, d > 0 and a + 2b + d = 1 and ad − b2 = 0. That is Pr(t1, t2)is the probability that bidder 1 is of type t1 ∈ 1, 2 and bidder 2 is oftype t2 ∈ 1, 2. The model is symmetric since Pr(t1, t2) = Pr(t2, t1). Theconditional probabilities are given in matrix form below:

(Pr(1 | 1) Pr(2 | 1)Pr(1 | 2) Pr(2 | 2)

)=

a

a + b

b

a + b

b

b + d

d

b + d

.

For example, Pr(1 | 1), the conditional probability that bidder two is of type 1given that bidder 1 is of type 1, is equal to

Pr(1, 1)Pr(1, 1) + Pr(1, 2)

=a

a + b.

Let us specify the mechanism that extracts the full surplus. There is nothingspecial about the allocation rule, if both bidders announce 1, the auctioneerassigns the object to one of the bidders. If both announces 2 the auctioneerallocates the object to either one of the bidders. In this case, each bidderreceives the object with probability 50 percent. If only one bidder announces2, this bidder receives the object. The novelty is in the payment rule:

(x1(1, 1) x1(1, 2)x1(2, 1) x1(2, 2)

)=

(12 01 1

)+

− b2

ad − b2

ab

ad − b2

− b2

ad − b2

ab

ad − b2

.

The first matrix after the equality corresponds to the payments in the second-price sealed bid auction. The second matrix is a lottery that depends on theother bidder’s announcement. For example, if bidder 1 is of type 1, then withprobability Pr(1 | 1) = a/(a + b) he receives a payment b2/(ad − b2) and with

100 Mechanism Design

probability Pr(2 | 1) = b/(a + b) he pays ab/(ad − b2). The expected value ofthis lottery is

− a

a + b

b2

ad − b2+

b

a + b

ab

ad − b2= 0.

If bidder 1 is of type 2, his lottery has an expected value given by

− b

b + d

b2

ad − b2+

d

b + d

ab

ad − b2=

b

b + d.

Hence, a bidder with type 1 has an expected utility of 0. A bidder with type 2has an expected utility equal to

Pr(1 | 2)(2 − 1) + Pr(2 | 2)(1 − 1) − b

b + d= 0.

We leave as an exercise to check the incentive compatibility constraints. Thatis, the reader needs to check that given the allocation and payment rules, abidder will find it in his interest to report his true valuation when his opponentdoes the same. We now calculate the auctioneer’s expected revenue.

R =∑

i∈1,2

∑j∈1,2

Pr(i, j)(x1(i, j) + x2(i, j))

= a

(1 − 2b2

ad − b2

)+ 2b

(1 +

ab − b2

ad − b2

)+ d

(2 +

2ab

ad − b2

)

= 4b + a + 2d = E[maxt1, t2].

Remark 10 One curious aspect of the full surplus extracting auction is thatin one case (t1 = t2 = 1) the auctioneer makes a net payment to bidders. Forexample if 2b2/(ad − b2) − 1 > 2 (i.e., if 5b2 > 3ad), the seller makes a netpayment to bidders higher than the object’s valuation. This may favor bidders’collusion.

Before we proceed to the continuum of types case, let us see what happensif the seller’s information about the distribution is imprecise while the biddersknow the correct distribution. Thus, we suppose that due to estimation errors,imperfect knowledge, or any other related reason, the seller assumes that thedistribution of types is given by

(Pr′(1, 1) Pr(1, 2)Pr(2, 1) Pr′(2, 2)

)=(

a + ε bb d − ε

), ε = 0,


which implies that the conditional distribution is as follows:

(Pr′(1 | 1) Pr′(2 | 1)Pr′(1 | 2) Pr′(2 | 2)

)=

a + ε

a + b + ε

b

a + b + ε

b

b + d − ε

d − ε

b + d − ε

.

Define δ(ε) = (a + ε)(d− ε)− b2. The seller then announces the allocation ruleas before and the payment rule is given by:

(x1(1, 1) x1(1, 2)x1(2, 1) x1(2, 2)

)=

(12 01 1

)+

− b2

δ(ε)(a + ε)b

δ(ε)

− b2

δ(ε)(a + ε)b

δ(ε)

.

This mechanism is incentive compatible as before. Let us check the participationconstraints. Suppose bidder 1 has a type t1 = 1. His expected utility is

Pr(1 | 1)(

12− 1

2+

b2

δ(ε)

)+ Pr(2 | 1)

(− (a + ε)b

δ(ε)

)= −ε

b2

(a + b)δ(ε).

Now if t1 = 2, his expected utility is

Pr(1 | 2)(

2 − 1 +b2

δ(ε)

)+ Pr(2 | 2)

(1 − 1 − (a + ε)b

δ(ε)

)= −ε

b(a + ε)(b + d)δ(ε)

.

Hence, if δ(ε) > 0 and ε > 0, the participation constraints are not satisfiedfor any type of bidder! That is, by using a full-extracting surplus auction, amisinformed seller can end up receiving zero-revenue.

Continuum caseFor simplicity we analyze only the two-bidder case. For every b ∈ [0, v], definethe function fb : [0, v] → R by fb(x) = f(b |x) = f(b, x)/f2(x). Analogously,for every a ∈ [0, v], we define fa(y) = f(a | y) = f(a, y)/f1(y). Here f1(a) =∫

f(a, y)dy and f2(b) =∫

f(x, b)dx. The following lemma provides necessaryand sufficient conditions for full surplus extraction. These conditions determinethe allocation and payment rules.

Lemma 7 Suppose (q, P ) is an incentive compatible, voluntary participationdirect mechanism. Then (q, P ) extracts the full surplus if and only if for all aand b except on a measure zero set:

1. Q1(a)a − E[P1(a, b) | a] = 0 and Q2(b)b − E[P2(a, b) | b] = 0;2. If a > b then q1(a, b) = 1 and if b > a, q2(a, b) = 1.

Proof: We prove first the “only if” part. Voluntary participation implies that

Q1(a)a − E[P1(a, b) | a] ≥ 0 and Q2(b)b − E[P2(a, b) | b] ≥ 0. (6.20)


That is, the expected profits from participation must be non-negative. Takingthe expectations and adding the two expressions above we obtain:

E[P1 + P2] =∫

E[P1(a, b)| a]f1(a) da +∫

E[P2(a, b)| b]f2(b) db

≤∫

Q1(a)af1(a) da +∫

Q2(b)bf2(b) db

=∫

q1(a, b)af(a, b) dadb +∫

q2(a, b)bf(a, b) dadb

=∫

(q1(a, b)a + q2(a, b)b)f(a, b) dadb

≤∫

maxa, bf(a, b) dadb.

On the other hand since the seller is extracting all the surplus, E[P1 + P2] =∫maxa, bf(a, b) dadb. Hence we must have equality in (6.20). Moreover, we

must have also that maxa, b = q1(a, b)a+q2(a, b)b except on a set of types thatoccurs with probability zero. It is an immediate consequence that q1(a, b) = 1if a > b and q2(a, b) = 1 if b > a. The “if” part is immediate since we haveequalities in place of inequalities above.

Example 18 Suppose the density function is f(a, b) = (1 + 4ab)/2. Then itis not possible to extract all the surplus. To prove this by contradiction, assumethat the incentive compatible mechanism (q, P ) extracts all surplus. Then thevoluntary participation and incentive compatibility constraints can be writtenrespectively, as:

a

∫ a

0

f(b | a) db −∫

P1(a, b)f(b | a) db

= a

∫q1(a, b)f(b | a) db −

∫P1(a, b)f(b | a) db

= aQ1(a) − E[P1(a, b)| a] = 0.

and

0 ≥ a

∫ y

0

f(b | a) db −∫

P1(y, b)f(b | a) db

= aQ1(y) − E[P1(y, b)| a], y ∈ [0, v].

Since f(b | a) = f(a, b)/f1(a) we may multiply both equations by f1(a) andrewrite them as: ∫

P1(a, b)(

1 + 4ab

2

)db = a

∫ a

0

1 + 4ab

2db


and ∫P1(y, b)

(1 + 4ab

2

)db ≥ a

∫ y

0

1 + 4ab

2db.

Define φ1(y) =∫

P1(y, b) db and φ2(y) =∫

P1(y, b)bdb. Then the last equalityand inequality can be rewritten as:

φ1(a)2

+ 2aφ2(a) = a

∫ a

0

1 + 4ab

2db =

a2

2+ a4

and

φ1(y) + 2aφ2(y) ≥ a

∫ y

0

1 + 4ab

2db =

ay

2+ a2y2.

Subtracting φ1(y)+ 2yφ2(y) = (y2/2)+ y4 from both sides of the last inequalitywe obtain:

2(a − y)φ2(y) ≥ ay

2+ a2y2 − y2

2− y4

=(a − y)y

2+ (a − y)(a + y)y2.

Hence, we have that (a−y)[2φ2(y)−(y/2)−(a+y)y2] ≥ 0 for every y and a. If wemake a decrease to y we have that 2φ2(y)−(y/2)−2y3 ≥ 0. Now make a increaseto y to obtain 2φ2(y) − (y/2) − 2y3 ≤ 0 and, therefore, φ2(y) = (y/4) + y3.This yields φ1(y) = (y2/2)+ y4 − 2y((y/4)+ y3) = −y4. Now we go back to theincentive compatibility constraints

−y4 + 2a(y

4+ y3

)≥ ay

2+ a2y2 (6.21)

for every y and a. However, a = 0 implies y = 0: a contradiction.

Remark 11 It shall be clear that if f(a, b) =∑n

l=1 gl(a)hl(b) is a density, itwill not be possible, in general, to extract the full surplus.

To present a positive result we need some definitions.

Definition 14 Let C([0, v]) denote the set of continuous functionsg : [0, v] → R. The set G ⊂ C([0, v]) is dense if for every continuoush : [0, v] → R and every ε > 0 there exists g ∈ G such that

supx∈[0,v]

|h(x) − g(x)| < ε.


Thus, the vector space generated by G ⊂ C([0, v]), namely4

〈G〉 :=

n∑

l=1

λigi;n ∈ N, λi ∈ R, gi ∈ G, 1 ≤ i ≤ n

is dense if for every continuous h : [0, v] → R and every ε > 0 there exist n ∈ N,g1, . . . , gn ∈ G and real numbers λ1, . . . , λn such that

supx∈[0,v]

∣∣∣∣∣h(x) −n∑

l=1

λlgl(x)

∣∣∣∣∣ < ε.

Example 19 A classical example of a dense set is the set of polynomialfunctions. That is, the vector space generated by set G = xn;n ∈ N∪0 ⊂C([0, v]). This is a consequence of the Stone–Weierstrass theorem. Anotherclassical example is the set of trigonometric functions: sin kx, k ∈ N ∪cos kx; k ∈ N∪0. In contrast, if G = x2n;n ∈ N∪0 the vector spacegenerated by G is not dense since it contains only even functions.

The following is true.

Theorem 18 Suppose the density f(a, b) is symmetric and that the vectorspace generated by the functions

a →

∫ v

0

z(b)f(b | a) db; z(·) ∈ C([0, v])

(6.22)

is dense in the space of the continuous functions on [0, v]. Then there isapproximate full surplus extraction.

Proof: We consider bidder 1 first. Suppose ε > 0 is given. We will show thatit is possible to extract bidder 1 surplus with an error of at most ε. Define theallocation rule as in Lemma 7(ii). Define the payment function as

P1(a, b) =

b + z(b) if a ≥ b,

z(b) if a < b.

4 That is, the vector space generated by a set G is the smallest set of continuous functionsthat is closed for addition and multiplication by scalars.


The function z(·) will be defined below. The expected utility of bidder 1 isgiven by

aQ1(a) − E[P1(a, b) | a]

= a

∫ a

0

f(b | a) db −∫ a

0

(b + z(b))f(b | a) db −∫ v

a

z(b)f(b | a) db

=∫ a

0

(a − b)f(b | a) db −∫ v

0

z(b)f(b | a) db.

The incentive compatibility constraint is satisfied for any z(·) since for every y,∫ a

0(a − b)f(b | a) db ≥

∫ y

0(a − b)f(b | a) db and therefore:∫ a

0

(a − b)f(b | a) db −∫ v

0

z(b)f(b | a) db

≥ aQ1(y) − E[P1(y, b) | a]

=∫ y

0

(a − b)f(b | a) db −∫ v

0

z(b)f(b | a) db.

The function h(a) =∫ a

0(a − b)f(b | a) db is continuous. Hence, by the density

assumption, there exists a z(·) ∈ C([0, 1]) such that

supa

∣∣∣∣h(a) −∫ v

0

z(b)f(b | a) db

∣∣∣∣ < ε

2.

Now define z(b) = z(b) − ε/2 and since

h(a)−∫ v

0

z(b)f(b | a) db =h(a) −∫ v

0

z(b)f(b | a) db +ε

2∈(− ε

2,ε

2

)+

ε

2= (0, ε),

the voluntary participation constraint is satisfied and the surplus is extractedexcept at most for ε.

Remark 12 Note that the mechanism constructed above has two parts. Thepart constituted by z(·) corresponds to a lottery—it is analogous to the discretecase. This lottery can be implemented in the following way. The auctioneerreceives bidders’ types (a, b) and charges bidder 1, 0 or b according to whethera < b. This corresponds to a Vickrey auction. Additionally, the auction chargesbidder 1, an amount equal to z(b). This is the lottery part. The lottery is drawnfrom the density f(b | a).

Remark 13 In the independent case the range of (6.22) is one-dimensionalsince

∫z(b)f(b | a) db is constant with a. Thus, the density assumption in (6.22)

is as far as possible from the independence assumption.

Let us now see that the condition (6.22) is not empty. To this purpose weprovide an example of a density that has a dense range and therefore almost


full surplus extraction is possible. This is accomplished by choosing a set oforthogonal polynomials. If we define for f, g ∈ C([0, 1]) the function 〈f, g〉 =∫ 1

0f(x)g(x) dx is an internal product in C([0, 1]). Thus two functions f, g are

orthogonal if 〈f, g〉 =∫

f(x)g(x) dx = 0. If f and g are not orthogonal we maychoose r ∈ R such that f and g − rf are orthogonal by choosing r such that

∫ 1

0

f(x)(g(x) − rf(x)) dx = 〈f, g〉 − r〈f, f〉 = 0.

That is r = 〈f, g〉/〈f, f〉. In general the Gram–Schmidt orthogonalizationmethod applied to a linearly independent set of functions (fn)n results in alinearly independent set (gn)n such that if n = m, 〈gn, gm〉 = 0. If we apply thisorthogonalization method to 1, x, x2, . . . , xn, . . . we obtain a set of orthogonalpolynomials,

P = P0(x), P1(x), . . . , Pn(x), . . ..

For example,

P0(x) = 1, P1(x) = 2x − 1, P2(x) = 6x2 − x + 1.

Example 20 We take v = 1. For every P ∈ P define |P |∞ = max|P (x)|; 0 ≤x ≤ 1. For conciseness define rn = 1/2n(n + 1)|Pn|2∞. We may now defineour density function:

f(a, b) = 1 +∞∑

n=1

rnPn(a)Pn(b).

We have that f(a, b) ≥ 12 , since∣∣∣∣∣

∞∑n=1

rnPn(a)Pn(b)

∣∣∣∣∣ ≤∞∑

n=1

rn|Pn(a)Pn(b)|

=∞∑

n=1

12n(n + 1)

|Pn(a)Pn(b)||Pn|2∞

≤∞∑

n=1

12n(n + 1)

=12

∞∑n=1

(1n− 1

n + 1

)=

12.

Now from∫

Pn(b) db = 〈Pn, P0〉 = 0, we have that

∫ 1

0

f(a, b) db = 1 +∞∑

n=1

rnPn(a)∫ 1

0

Pn(b) db = 1.


Therefore f(b | a) = f(a, b)/∫ 1

0f(a, b) db = f(a, b). To show approximate full

surplus extraction we use Theorem 18. For any integer m ≥ 1,∫ 1

0

Pm(b)f(b | a) db

=∫ 1

0

Pm(b)

(1 +

∞∑n=1

rnPn(a)Pn(b)

)db

=∫ 1

0

Pm(b) db +∞∑

n=1

rnPn(a)∫ 1

0

Pm(b)Pn(b) db

=(

rm

∫ 1

0

P 2m(b) db

)Pm(a).

Thus for every m ≥ 1, by choosing z(b) = Pm(b)/(rm

∫ 1

0P 2

m(b) db) we obtainthat

∫ 1

0z(b)f(b | a) db = Pm(a). Thus the set

R =

a →∫

z(b)f(b | a); z ∈ C([0, 1])

⊃ Pm(x);m ≥ 1.

But R also contains P0 (for z ≡ 1) and therefore R is dense.

One question that arises is whether we can say anything about the optimalauction in the intermediate case when the range in (6.22) is not dense and is notunidimensional. We consider only dominant strategy optimal auctions (definedbelow). Note that the Vickrey auction and the auction in Theorem 18 aredominant strategy auctions. We will maintain the restriction to a two-biddersymmetric model.

Definition 15 Suppose (q, P ) is a direct mechanism. It is a dominant strategymechanism if for every type a of bidder 1 and every type b of bidder 2 it istrue that

qi(a, b)a − Pi(a, b) ≥ qi(a′, b)a − Pi(a′, b), ∀a′. (6.23)

Thus, regardless of bidder 2’s bid, bidder 1 is better off (weakly better offat least) telling the truth than to pretend to be of type a′. We say that (q, P )is incentive compatible if∫

(qi(a, b)a − Pi(a, b))f(b | a) db ≥ 0.

That is, the incentive compatibility constraint is the same as before. DefineT (a, b) = qi(a, b)a − Pi(a, b). The following proposition is analogous toProposition 3.


Proposition 5 For every b, a → qi(a, b) is increasing.

Proof: The incentive compatibility constraint implies for every a′,

T (a, b) ≥ qi(a′, b)a − Pi(a′, b) = qi(a′, b)(a − a′) + T (a′, b).

Therefore, we have that

T (a, b) − T (a′, b) ≥ (a − a′)qi(a′, b).

Analogously, we have that

T (a′, b) − T (a, b) ≥ (a′ − a)qi(a, b).

Adding both inequalities we obtain

0 ≥ (a′ − a)(qi(a, b) − qi(a′, b))

and therefore if a′ > a, then qi(a′, b) ≥ qi(a, b).

In a similar fashion to the proof of (6.7) we may have the following.

Proposition 6 For every b, it is true that

T (a, b) = T (0, b) +∫ a

0

qi(s, b) ds.

From this proposition we conclude that the payment satisfies the following:

Pi(a, b) = qi(a, b)a −(

T (0, b) +∫ a

0

qi(s, b) ds

)

= qi(a, b)a −∫ a

0

qi(s, b) ds − T (0, b). (6.24)

Let z(b) be a function and define T (0, b) = z(b). Hence (q, P ), with Pi definedby (6.24) and q is increasing in each variable. Then (q, P ) is a dominant strategymechanism. It is also incentive compatible if and only if∫ (

z(b) +∫ a

0

qi(s, b) ds

)f(b | a) db ≥ 0, ∀a.

The auctioneer’s expected revenue is given by:

R = R1 + R2,

R1 =∫

P1(a, b)f(a, b) dadb

=∫ (

q1(a, b)a −∫ a

0

q1(s, b) ds − z1(b))

f(a, b) dadb,

the expression for R2 being analogous.


We now make an additional restriction. We restrict our mechanisms toefficient mechanisms. That is, mechanisms that satisfy (2) of Lemma 7. Thus∫ a

0q1(s, b) ds = (a−b)+. Also

∫ b

0q2(a, s) ds = (b−a)+. The revenue from bidder

1 can be rewritten as

R1 =∫

a

(∫ a

0

f(a, b) db

)da

−∫

(a − b)+f(a, b) dadb −∫

z1(b)f(a, b) dadb.

And z1(b) satisfies the restriction:∫(z1(b) + (a − b)+)f(a, b) db ≥ 0, ∀a.

Note that we used the fact that f(b | a) = f(a, b)/∫

f(a, b) db. Define φ1(a) =∫(a− b)+f(a, b) db. Therefore to maximize revenue the auctioneer must choose

z1(b) such that∫z1(b)f(a, b) db + φ1(a) ≥ 0, ∀a, and

∫∫z1(b)f(a, b) dbda

is minimized.

Example 21 We now fulfill our promise and find the optimal auction in acase in which the range is a two-dimensional vector space. Suppose the densityis f(a, b) = (1 + 4ab)/2. The function

φ1(a) =∫

(a − b)+f(a, b) db =∫ a

0

(a − b)1 + 4ab

2db =

a4

3+

a2

4.

The integral∫z1(b)f(a, b) db =

∫z1(b)

1 + 4ab

2db =

∫z1(b) db

2+ 2a

∫z1(b)bdb.

Thus the seta →

∫z(b)f(b | a) db

=

a → µ1

2+ 2µ2a;µi ∈ R and exists z(·),

∫z(b)bl db = µl, l = 0, 1

Define λ1 =∫

z1(b) db and λ2 =∫

z1(b)bdb. Then we want to find λ1 and λ2

such that

λ1

2+ 2aλ2 +

a4

3+

a2

4≥ 0, ∀a,


and

γ :=λ1

2+ λ2 =

∫ (λ1

2+ 2aλ2

)da

is a minimized. Choosing a = 0 we see that necessarily λ1 ≥ 0. Since λ1 =λ2 = 0 is a possibility we see that γ ≤ 0 and we may suppose without loss ofgenerality that λ2 < 0. The function

v(a) = 2aλ2 +a4

3+

a2

4

is convex in [0, 1] and decreases in a neighborhood of a = 0. If v′(1) ≤ 0 then

v′(1) = 2λ2 +43

+12≤ 0, and

λ1

2+ 2λ2 +

13

+14≥ 0.

The first inequality implies λ2 ≤ − 1112 . The second implies γ ≥ −λ2 − 7

12 ≥ 412 > 0.

Suppose now that v′(1) > 0. Then there exists x ∈ (0, 1) such that v′(x) = 0.From this we have that 2λ2 = −( 4x3

3 + x2 ) and hence

λ1

2− x

(4x3

3+

x

2

)x4

3+

x2

4≥ 0.

This inequality is equivalent to λ1 ≥ 89x8 + 1

3x6 − 12x2 and therefore,

γ =λ1

2+ λ2 ≥ 4

9x8 +

16x6 − 1

4x2 − 2

3x3 − 1

4x.

The minimum of the right-hand side is at x 0.9023. If now we choose

λ2 = −12

(4(0.9023)3

3+

0.90232

)= −0.715,

λ1 =89(0.9023)8 +

13(0.9023)6 − 1

2(0.9023)2 = 0.163

γ = 0.1632 − 0.715 = −0.6335 is minimum. To finish note that z1(b) = 4.942 −

9.558b is such that∫z1(b) db = λ1 and

∫z1(b)bdb = λ2.

What conclusion can we drawn from the full surplus extraction results? Alt-hough the independent private values model plays a central role in the develop-ment of theory, it is reasonable to expect that bidders’ types will exhibit somedegree of correlation. However, auctions with random payments, if they exist atall, will be rare. This is true both in traditional auction settings, such as thoseused to sell houses, paintings, and wine, but also of the new engineered auctionssuch as the famous airwaves auctions. Below are some possible explanations forthe nonexistence of these complex auctions involving lotteries, none of which


has been formalized in the existing literature:

1. Too much information about the primitives (distribution of values,number of bidders, etc.) is needed to design the lottery;

2. The lottery might involve large payments from the auctioneer/bidders toprovide bidders with the right incentives. In practice, bidders and theseller might be cash constrained;

3. Bidders have better things to do than participate in unprofitable auctions;4. A risk averse seller might not be willing to risk making net payments to

bidders; and5. A full surplus extraction auction might encourage bidders’ collusion.

Nevertheless, this is clearly an area that deserves more research.

6.4.3 Several Objects

We now return to the IPV set up. We will obtain a revenue equivalence resultfor the case where there are several identical objects for sale.

We suppose that there are K > 1 objects to sell and n bidders. Bidders onlywant one object each. If bidder i has type ti ∈ [0, 1] and receives object k hisutility is tiαik where αik > 0 is given. Define the vector αi = (αi1, . . . , αiK).Thus, if qik is the probability that bidder i receives object k then his utility is

ti · (qi · αi) = ti

K∑k=1

qikαik.

We denote by A the set of possible allocations of the objects amongst thebidders. Thus,

A =(a1, . . . , an); ai ∈ 0, 1, . . . ,K,#i ≤ n; ai = k ≤ 1 or k = 0.

That is, each bidder receives at most one of the K existing objects. A directmechanism, (q, P ) = (q, P1, . . . , Pn), is therefore given by

q(t) = (qa(t))a∈A, qa(t) ≥ 0,∑a∈A

qa(t) ≤ 1;

and P i(t) ∈ R is the payment of bidder i. The probability that i receivesobject k is therefore, qik(t) =

∑a∈A,ai=k qa(t). The expected utility of i under

the mechanism is

πi(ti) = tiE−i

[K∑

k=1

qik(t)αik

]− E−i[P i(t)]

= tiE−i[qi(t) · αi] − E−i[P i(t)].


If we define Qi(ti) = (Qi1(ti), . . . , Q

iK(ti)), Qi

k(ti) = E−i[qik(t)], we may writethe incentive compatibility constraints as

πi(ti) ≥ tiQi(t′) · αi − E−i[P i(t′)], ∀t′.

From this we obtain the inequality

πi(ti) ≥ (ti − t′)Qi(t′) · αi + πi(t′).

Thus, following the same method as in the one-object case, we may prove that

πi(ti) = πi(0) +∫ ti

0

Qi(s) · αi ds.

Hence, the bidder’s expected payment is

E−i[P i(t)] = tiE−i[qi(t) · αi] − πi(ti)

= tiE−i[qi(t) · αi] −∫ ti

0

Qi(s) · αi ds − πi(0).

Accordingly the following result holds:

Theorem 19 Assume bidders demand only one object in a multi-objectauction with IPV. Assume that (q, P ) and (q, P ) are two individually rationalincentive compatible mechanisms such that:

1. The allocation rule is the same, that is, q = q2. The lowest type has the same expected utility in both mechanism, that is,

E−i[P i(0, t−i)] = E−i[P i(0, t−i)].

Then, the bidders’ expected payments and the seller’s expected revenue arethe same in both auctions respectively. That is,

E−i[P i(t)] = E−i[P i(t)] and

E

[n∑

i=1

P i(t)

]= E

[n∑

i=1

P i(t)

].

Proof: The proof is easy. Note that E−i[P i(0, t−i)] = E−i[P i(0, t−i)] impliesthat E−i[P i(t)] = E−i[P i(t)] which is the expected payment of a type ti bidder.The seller expected revenue is

E

[∑i

P i(t)

]=∑

i

E[P i(t)] =∑

i

Ei[E−i[P i(t)]]

=∑

i

Ei[E−i[P i(t)]] =∑

i

E[P i(t)] = E

[∑i

P i(t)

].


6.4.4 Common Values Auction

In this section, we show that the Revenue Equivalence Theorem is also truein the common value independent signals5 case. We will be even more general.We consider interdependent valuations possibly distinct for each bidder. Soinstead of supposing that a bidder with signal xi values the object as xi wesuppose his valuation is V i = ui(x1, . . . , xn) where xj is bidder j’s signal.Naturally bidder i does not know the other bidders’ signals. Let us rewrite theincentive compatibility and individual participation constraints. If (q, P ) is adirect mechanism then it is incentive compatible if

E−i[ui(x)qi(x) − Pi(x)] ≥ E−i[ui(x, x−i)qi(x) − Pi(x)], ∀x′i,

where x = (x′i, x−i). The mechanism is individually rational if

E−i[ui(x)qi(x) − Pi(x)] ≥ 0.

We begin with a lemma.

Lemma 8 If we define Ti(xi) = E−i[ui(x)qi(x) − Pi(x)] then for every x′i:

Ti(xi) − Ti(x′i) ≥ E−i[(ui(x) − ui(x′

i, x−i))qi(x′i, x−i)].

Proof: Rewrite the incentive compatibility constraint as

Ti(xi) ≥ E−i[ui(x)qi(x′i, x−i) − Pi(x′

i, x−i)]

= E−i[(ui(x) − ui(x′i, x−i))qi(x′

i, x−i)] + Ti(x′i).

Therefore, we conclude that

Ti(xi) − Ti(x′i) ≥ E−i[(ui(x) − ui(x′

i, x−i))qi(x′i, x−i)].

We want to prove the following lemma:

Lemma 9 Suppose ui is continuously differentiable. Then,

Ti(xi) = Ti(0) +∫ xi

0

E−i

[∂ui

∂xi(s, x−i)qi(s, x−i)

]ds.

Proof: For a given n ∈ N, define the partition 0, xi

n , 2xi

n , . . . , xi. For each

j = 1, 2, . . . , n there exists a ξj ∈(

(j−1)xi

n , jxi

n

)such that

ui

(jxi

n, x−i

)− ui

((j − 1)xi

n, x−i

)=

∂ui

∂xi(ξj , x−i)

xi

n.

5 Recall that in the common value model we prefer to use “signal” for “types”.


Therefore,

Ti

(jxi

n

)− Ti

((j − 1)xi

n

)≥ E−i

[∂ui

∂xi(ξj , x−i)

xi

nqi

((j − 1)xi

nx−i

)],

j = 1, . . . , n. Adding all terms we obtain,

Ti(xi) − Ti(0) ≥n∑

j=1

xi

nE−i

[∂ui

∂xi(ξj , x−i)qi

((j − 1)xi

nx−i

)].

As n approaches ∞, the right-hand side converges to∫ xi

0

E−i

[∂ui


]ds.

Therefore

Ti(xi) − Ti(0) ≥∫ xi

0

E−i

[∂ui

∂xi(s, x−i)qi(sx−i)

]ds.

The reverse inequality is proved in an analogous way. Hence, we haveestablished the following.

Theorem 20 (general revenue equivalence) If (q, P ) and (q, P ) are incentivecompatible and individually rational mechanisms such that q = q and a bidderwith the lowest signal has the same expected utility in both mechanisms thenthe expected payments and the sellers’ expected revenue are the same for bothmechanisms. That is

E−i[Pi(t)] = E−i[Pi(t)] and E

[n∑

i=1

Pi(t)

]= E

[n∑

i=1

Pi(t)

].

Proof: If ai is the expected payment of a bidder type xi = 0 for (q, P ) andai is the expected payment for (q, P ) then if ai = ai it is true that

E−i[Pi(x)] = E−i[ui(x)qi(x)] − Ti(xi)

= E−i[ui(x)qi(x)] − ai −∫ xi

0

E−i

[∂ui


]ds

= E−i[ui(x)qi(x)] − ai −∫ xi

0

E−i

[∂ui


]ds

= E−i[Pi(x)].

Thus bidders’ expected payments are the same for each bidder type. Moreo-ver, the auctioneer’ expected revenue is the same as well since E[Pi(x)] =Ei[E−i[Pi(x)]].

6.5 Exercises 115

6.5 Exercises

1. (a) Show that if f : R → R and g : R → R are convex functions thenmaxf, g is also a convex function.

(b) Generalize the above result to show that if fi; i ∈ I is a family ofconvex functions and supi∈I fi(x) < ∞ for every x then supi∈I fi isalso a convex function. Conclude that if H : R → R is a function thenthere exists the largest convex function G such that G(x) ≤ H(x)for every x.

2. Prove the assertions on Remark 8.3. Show that t0 that solves (6.13) is such that J(t0) = 0. Moreover show

also that the optimum exists and is interior.4. Find the optimal auction revenue if the individual distribution is uniform

in [0, 1] and there are n bidders. Compare it with the revenue of first-price sealed bid auctions (without reserve prices). How many biddersare necessary so that the auction without reserve prices generates morerevenue than the optimal auction?

5. Repeat the last exercise if the individual distribution, instead of uniform,is any continuously differentiable distribution.

6. Prove the assertion on remark 9. Show also that if we want the dis-tribution to have a continuous density on [0, 1] it is necessary thath′(1) = 2.

7. An all-pay auction is an auction in which every bidder pays his bidwhether or not receiving the object. In terms of mechanisms this meansthat Pi(x) depends only on xi. Suppose we have private values but typesare correlated. Find the optimal auction amongst the class of all-payauctions.

8. Suppose the utility of bidder i with type xi is ui(xi) = x2i . Reduce

this case to the case ui(xi) = xi. Suggestion: if x2i has density f(xi)

find first the density of xi. If x2i is uniformly distributed what is the

distribution of xi. If xi is uniformly distributed what is the distributionof x2

i ?9. Do the same calculation of Example 20 if

f(a, b) = 1 +∞∑

n=1

anPn(a)Pn(b)

|Pn|2∞,

∞∑n=1

|an| < 1.

10. The formalization of the English auction we presented is designedto cover the correlated values case. If types are independent, amuch simpler strategy set can be used. Find one such simplifiedset up.


11. Suppose we have two bidders and the distribution is uniform. Showthat if the auction is efficient the following payment rule is a possibleone:

P (v, y) =

v

2+

1√v− 1

2√

yif v > y

0 otherwise.

Show that the seller’s payments are unbounded.12. Find the war of attrition equilibrium for two bidders using the revenue

equivalence theorem.

7

Multiple Objects

In this chapter we consider the sale of several objects. The objective of thischapter is not to generalize the theory developed in previous chapters to coverthe sale of several objects. Instead, the purpose of this chapter is to illustrateseveral difficulties that arise when developing a general theory of auctions ofmultiple objects.

The reason for these difficulties does not come from the nature of the valua-tions (independent private values versus affiliated values) but rather from theadditional strategic considerations when bidding for multiple objects—an indi-vidual’s bid for one object might influence the allocation (price and likelihoodof winning) not only of this particular object but also of other objects. Thus,unlike in the single-object case where a bidder’s only concern was the bid ofthe opponent with the highest type, in a multi-object setting, bidders have totake into account several bids.

Suppose there are K = 2 objects for sale. These objects can be sold eithersequentially or simultaneously. The objects are identical in the sense that bid-ders only care about how many objects they obtain. Bidders may demand oneor more units. We consider the symmetric, Independent Private Values model.Thus every bidder i receives a type x ∈ [0, 1] distributed with density f(x). Ifthe bidder with type x receives one object his utility is U1(x) = x. If he receives2 objects his utility is U2(x) = α2x. We assume that α2 ≥ 1. Thus if the bidderdemands at most one object, α2 = 1. If α2 > 2 we say that there is positivesynergy since two objects together generate more utility than twice the utilitygenerated from a single object. If 1 ≤ α2 < 2, we say that there is negativesynergy.

This chapter is organized as follows. Section 7.1 examines sequential second-price auctions. Section 7.2 studies discriminatory and uniform auctions. Finally,we characterize the optimal auction and compare its expected revenue with thatof standard auction formats.

117

118 Multiple Objects

7.1 Sequential Auctions

For convenience, let us define r = α2−1. Thus, if r > 1 there is positive synergyand if r < 1 negative synergy.

Sequential auctions present a complication that does not arise in simulta-neous auctions; the first auction generates information about bidders’ typesthat can be used in the second auction. For example, if strategies are strictlyincreasing in the first auction and the winning bid is announced (this is a mini-mum requirement) then bidders may infer the highest type. It may not be inthe interest of the winner that this inference be made. Note that if all bidsare announced then there is a complete revelation of all bidders’ types. Thispossibility of information revelation makes the analysis more interesting butmuch harder. Thus, in this section we look only at Second-price sequentialauctions. The reason that it is easier to examine second-price auctions thanfirst-price auctions is the fact that second-price auctions in this context areVickrey auctions—the revelation of types is a dominant strategy. Thus, theinformation that is revealed in the first round of a sequence of second-priceauctions does not interact strategically with the information to be used inthe second round.

Let us find equilibrium bidding strategies for a sequence of two second-price auctions. The equilibrium bidding strategy in the second auction is easyto calculate. If bidder i does not win one object in the first auction he bidsb2i (x) = x in the second auction. If he does win one object in the first auction

he bids b2i (xi) = α2xi−xi = (α2−1)xi = rxi− his value for an additional unit,

in the second auction.Now let us find equilibrium bidding strategies in the first auction. Assume

that bidder i, i = 2, . . . , n bids according to the strictly increasing strategy b(x)in the first auction. Define Y = maxxj ; j ≥ 2. We also need to define Y 2, thesecond highest among xj ; j ≥ 2. Thus, if bidder 1, with type x, bids b(z) hisexpected utility is:

φ(z) = E[(x − b(Y ) + (rx − Y )+)Iz>Y ] + E[(x − maxrY, Y 2)+Iz<Y ].(7.1)

If player 1’s type is higher than Y , then he wins the first object and bids rxin the second auction. This is the first part of (7.1). The term maxrY, Y 2in the second part appears for the following reason. If bidder 1 does not winthe first object his type is smaller than Y. The bidder with type Y wins thefirst object and therefore bids rY in the second auction. (Ties occur with zeroprobability). The non-winning bidders bid their types, the highest of which isY 2, in the second auction. If we have positive synergy then rY ≥ Y ≥ Y 2 andmaxrY, Y 2 = rY . If we have negative synergy, then this maximum can beequal to Y 2. (For example, consider the limiting case where α2 = 1.) We willconsider both cases together. To write (7.1 ) in terms of fY , the density of Y ,

7.1 Sequential Auctions 119

consider

ψ(y) = E[(x − maxrY, Y 2)+ |Y = y],

the conditional expectation of (x− maxrY, Y 2)+ given Y . The expectedutility (7.1) can be rewritten as

φ(z) =∫ z

0

(x − b(y) + (rx − y)+)fY (y) dy +∫ 1

z

ψ(y)fY (y) dy.

In the last summand we used iterated expectations. That is, E[ψ(Y )Iz<Y ] =E[(x − maxrY, Y 2)+Iz<Y ]. Differentiating we obtain,

φ′(z) = x − b(z) + (rx − z)+ − ψ(z)fY (z). (7.2)

Thus if z = x maximizes φ, then φ′(x) = 0 :

x + (r − 1)+x − ψ(x) = b(x).

Substituting this b(·) into (7.2) yields

φ′(z)fY (z)

= x − z − (r − 1)+z + (rx − z)+

=

x − z + (rx − z)+ if α2 ≤ 2;x − rz + (rx − z)+ if α2 > 2.

We examine the different possible cases.

(i) r ≤ 1 In this case, if z < x then φ′(z) > 0. Suppose now that z > x.Then (rx − z)+ = 0 and φ′(z) < 0. Therefore z = x maximizes φ.

(ii) r > 1 If z < x then rx > x > z and

φ′(z)fY (z)

= x − rz + rx − z = α2(x − z) > 0.

Finally if z > x then φ′(z) < 0 when (rx − z)+ = 0 and if rx − z ≥ 0then

φ′(z)fY (z)

= x − rz + rx − z = α2(x − z) < 0.

Hence, in all cases z = x is the optimum. We summarize our conclusions asfollows.

Proposition 7 The equilibrium bidding strategy for the first of the twosequential second-price auction of identical goods is given by:

b(x) = x + (r − 1)+x − E[(x − maxrY, Y 2)+ |Y = x].


In particular,

1. If there is positive synergy, then

b(x) = rx.

2. If there is negative synergy, then

b(x) = E[maxrx, Y 2 |Y = x].

Note that if bidders demand only one object, then the bid in the first auctionis b(x) = E[Y 2 |Y = x]. In what follows we present a numerical examplewhere we can explicitly compute equilibrium bidding strategies and the seller’sexpected revenue.

Example 22 Suppose there are two bidders, the types are distributed uni-formly on [0, 1] and the synergy is negative, say α2 = 3

2 . Define X1 =maxx1, x2 and X2 = minx1, x2. With two bidders, Y 2 = 0 and thusb(x) = x/2. The seller’s expected revenue is

E

[b(X1) +

X1

2IX1>2X2 + X2IX1<2X2

]

= 2∫

x1>x2

(b(x1) +

x1

2Ix1>2x2 + x2Ix1<2x2

)dx2 dx1

= 2∫ 1

0

x21

2dx1 + 2

∫ 1

0

(x1

2

)2

dx1 + 2∫

x1>x2

x2Ix1<2x2 dx1 dx2

=13

+16

+ 2∫ 1

0

∫ x1

x1/2

x2 dx2 dx1 =34.

Example 23 If there is positive synergy, then the bidder with the highest typewins both objects. Thus, the sequential auction is efficient and the RevenueEquivalence Theorem holds. If there is negative synergy and when X2 > rX1,where X1 is the highest type and X2 is the second highest, then the bidder withthe highest type may not win both objects. In this case, the revenue equivalencetheorem breaks down.

7.2 Simultaneous Auctions

We now consider the simultaneous sale of two objects. Simultaneous auctionshave attracted a lot of attention as governments usually sell Treasury billsin this way. There are two common types: discriminatory and uniform-priceauctions. In the former, all winning bidders pay their bids. In the latter, allthe winning bidders pay the same amount—either the highest losing bid or thelowest winning bid. Below we analyze these two types of auction formats.

7.2 Simultaneous Auctions 121

7.2.1 Discriminatory Auctions

In the discriminatory auction each bidder i bids a vector of K bids,(bi1, . . . , b

iK

), bi

1 ≥ bi2 ≥ · · · ≥ bi

K .

The K ′s higher bids amongbik; 1 ≤ i ≤ n, 1 ≤ k ≤ K

are the winning bids. If

bidder i has one bid amongst the winning bids his payment is bi. If he has k bidsamong the winning bids his payment is bi

1 + · · ·+ bik and he receives k objects.

(The uniform-price auction is similar except in the payment; every winningbidder pays the same unitary price, which is equal either to the K +1th highestbid or to the Kth highest bid.) While the discriminatory auction is similar to thethe first-price auction (you pay your bid), the uniform-price auction is similarto the second-price auction. This analogy, however, is imperfect as biddersin general will not bid their true valuations in a simultaneous uniform-priceauction for strategic reasons (see next section).

We now consider a simple example with two bidders and two objects. If thereis positive synergy, then the discriminatory auction has a simple equilibrium.The bidders bid as they were bidding in a first-price auction for a object withutility α2x. The reason is that in an increasing, symmetric equilibrium, theindividual with the highest type wins both objects. This individual pays thesum of his bids. This is equivalent to bidding for a single object that is worthα2x. Below we formalize this intuition. We denote by b(·) the symmetric biddingfunction for the first object and by c(·) the symmetric bidding for the secondobject.

Theorem 21 If α2 ≥ 2, then a symmetric equilibrium strategy in the dis-criminatory auction of two objects is to bid the same amount for each object,b(x) = c(x), where

b(x) =α2

2

∫ x

0yfY (y) dy

FY (x). (7.3)

We first note that b(x) defined in (7.3) satisfies the differential equation

b′(x)FY (x) + b(x)fY (x) =α2

2xfY (x). (7.4)

If bidder 1 bids b ≥ c for the first and second object his expected utility isgiven by

φ(b, c) = (α2x − b − c) Pr(c > b(Y )) + (x − b) Pr(b > b(Y ) > c).

Note that any bid c > b(1) can be profitably reduced to c = b(1). Any bidb > b(1) can be profitably reduced as well. Thus, we may suppose that b = b(α)and c = b(β), where 1 ≥ α ≥ β ≥ 0. As a result, we can rewrite the expected


utility as

ψ(α, β) = (α2x − b(α) − b(β)) Pr(β > Y ) + (x − b(α)) Pr(α > Y > β)

= (α2x − b(α) − b(β))FY (β) + (x − b(α))(FY (α) − FY (β))

= ((α2 − 1)x − b(β))FY (β) + (x − b(α))FY (α).

The constrained maximization problem

maxα≥β

ψ(α, β)

may be solved using Kuhn–Tucker. If λ is the Kuhn–Tucker multiplier then

∂ψ

∂α= −b′(α)FY (α) + (x − b(α))fY (α) = λ;

∂ψ

∂β= −b′(β)FY (β) + ((α2 − 1)x − b(β))fY (β) = −λ;

λ · (α − β) = 0.

The differential equation (7.4) allows us to rewrite ∂ψ/∂α and ∂ψ/∂β to obtain(x − α2α

2

)fY (α) = λ; (7.5)(

(α2 − 1)x − α2β

2

)fY (β) = −λ; (7.6)

λ · (α − β) = 0.

At the optimum we must have α = β. To see why, suppose not; then λ = 0.Thus, x − (α2α/2) = 0 and

0 = (α2 − 1)x − α2α

2= (α2 − 2)x > 0.

Hence, α = β. Adding (7.5) and (7.6) we obtain(x − α2α

2+ (α2 − 1)x − α2α

2

)fY (α) = 0

and this yields α = x = β, ending the proof.The task of finding equilibrium strategies under negative synergy is harder.

We will obtain the solution as a pair of differential equations. Suppose bidder 2bids according to the strictly increasing continuous strategies (b(·), c(·)), b(·) ≥c(·). If bidder 1 bids (b, c), b ≥ c his expected utility is

ψ(b, c) = (α2x − b − c) Pr(c > b(Y )) + (x − b) Pr(b > c(Y ), b(Y ) > c).

The first summand corresponds to winning two objects and the second one towinning one of the objects only. Now note that a bid b > c(1) can be reducedto c(1), increasing utility. Therefore, we suppose without loss of generalitythat b ≤ c(1). An analogous reasoning allows us to suppose that c ≤ b(1).


If equilibrium exists, c = c(1) and b = b(1) will be the choices when x = 1.Therefore, necessarily c(1) ≤ b(1) and b(1) ≤ c(1) implying b(1) = c(1). Hence,a necessary condition for equilibrium is that b(1) = c(1). We may rewrite

Pr(c > b(Y )) = Pr(b−1(c) > Y ) = F (b−1(c)).

We have also that

Pr(b > c(Y ), b(Y ) > c) = Pr(c−1(b) > Y > b−1(c))

= F (c−1(b)) − F (b−1(c)).

The maximization problem can be reduced to the problem of

maxb,c

(α2x − b − c)F (b−1(c)) + (x − b)F (c−1(b)) − F (b−1(c))

subject to b ≥ c.To see that it is always true that c−1(b) ≥ b−1(c), write b = b(α) and c = c(β).Then

c−1(b) = c−1(b(α)) ≥ c−1(c(α)) = α and c−1(b) ≥ c−1(c) = β.

Thus c−1(b) ≥ maxα, β. Analogously minα, β ≥ b−1(c). Simplifying thelast equation we have

ψ(b, c) = ((α2 − 1)x − c)F (b−1(c)) + (x − b)F (c−1(b)). (7.7)

The first-order conditions for optimum at b = b(x), and c = c(x) are:

∂ψ

∂b|b=b(x) = −F (c−1(b(x))) + (x − b(x))f(c−1(b(x)))(c−1)′(b(x)) = 0; (7.8)

∂ψ

∂c|c=c(x) = −F (b−1(c(x))) + ((α2 − 1)x − c(x))f(b−1(c(x)))(b−1)′(c(x)) = 0.

If we make a change of variables y = b(x) in the first equation and y = c(x) inthe second equation, we have the system of differential equations:

−F (c−1(y)) + (b−1(y) − y)f(c−1(y))(c−1)′(y) = 0;

−F (b−1(y)) + ((α2 − 1)c−1(y) − y)f(b−1(y))(b−1)′(y) = 0.

We have just proved the necessity part of the following theorem.

Theorem 22 The discriminatory price auction has a unique strictly increa-sing differentiable equilibrium. The strategies (b(·), c(·)) are equilibrium bidding


strategies of the discriminatory price auction if and only if b−1(·) and c−1(·)satisfy the system of differential equations:

(c−1)′(y) =F (c−1(y))

(b−1(y) − y)f(c−1(y));

(b−1)′(y) =F (b−1(y))

((α2 − 1)c−1(y) − y)f(b−1(y)),

(7.9)

and the initial conditions

b(1) = c(1);

b(0) = 0 = c(0).

Let us now prove the sufficiency part. Thus, suppose we have a solution(b−1, c−1) of (7.9) such that b(1) = c(1) and b(0) = 0 = c(0). From (7.8) wesee that

∂ψ

∂b= −F (c−1(b)) + (x − b)f(c−1b)(c−1)′(b)

= −F (c−1(b)) + (x − b)f(c−1(b))F (c−1(b))

(b−1(b) − b)f(c−1(b))

= F (c−1(b))−b−1(b) + x

(b−1(b) − b).

Thus, ψ is decreasing if b−1(b) > x and is increasing if b−1(b) < x. Thereforeb = b(x) is the optimum. Analogously, we prove that c = c(x) is optimum.

We will not prove, however, that the differential equations system has asolution. For this we refer the reader to Lebrun and Tremblay (2003).

Remark 14 A system of two differential equations depends on two constantswhich are the initial conditions. For example if x0 ∈ (0, 1) and C1 and C2 areconstants then the initial conditions have the form

c−1(x0) = C1; (ic)

b−1(x0) = C2.

The condition b(1) = c(1) is not of this form. To write as (ic) define y = b(1) =c(1). Then b−1(y) = 1 and c−1(y) = 1.Thus, instead of having two arbitraryconstants we have C1 = C2 = 1 but y is also to be found. The additionalcondition is to choose y so that c−1(0) = 0 = b−1(0). All this is needed sincewe cannot simply choose x0 = 0 as (7.9) is not defined for y = 0.

Remark 15 There are methods to numerically approximate solutions of dif-ferential equations. These methods, however, are not applied directly to systems


of the form (ic). Instead, we have to look in the parametrized family of solu-tions (b−1

k (·), c−1k (·)) with initial condition b−1

k (k) = 1 = c−1k (1), and one that

satisfies also b−1k (0) = 0 = c−1

k (0).

Below we solve this system for an example with the uniform-[0, 1]distribution.

Example 24 Suppose there are two bidders. Suppose α2 = 3/2 and F (x) = x.Write φ(y) = c−1(y) and ψ(y) = b−1(y). Let us solve the system

φ′(y) =φ(y)

ψ(y) − y,

ψ′(y) =ψ(y)

φ(y)/2 − y.

Eliminating the denominator in both equations we obtain,

φ′(y)ψ(y) = yφ′(y) + φ(y) = (yφ(y))′and ψ′(y)φ(y) = 2(yψ(y))′.

Adding both equations yields

(φ(y)ψ(y))′ = (yφ(y))′ + 2(yψ(y))′.

Therefore, by the Fundamental Calculus Theorem, φ(y)ψ(y)= yφ(y)+2yψ(y)+ C. Considering y = 0, we see that C = 0. Thus, we have that

φ(y) =2yψ(y)

ψ(y) − y. (7.10)

Substituting this result into the equation for ψ we obtain:

ψ′(y) =ψ(y)(ψ(y) − y)

y2.

Define z(y) by yz(y) = ψ(y) and substituting this into the last equation weobtain an equation for z′:

yz′(y) = z(y)(z(y) − 2).

Using that 1/z(z − 2) = −1/2z + 1/2(z − 2) we rewrite the last equation as

− z′(y)2z(y)

+z′(y)

2(z(y) − 2)=

1y

(7.11)

which is easily integrated to yield

z(y) − 2z(y)

= Ky2.


Hence z(y) = 2/(1 − Ky2) and ψ(y) = 2y/(1 − Ky2). Using (7.10) we obtainthat φ(y) = 4y/(1 + Ky2). We are almost done now. To determine K we haveto solve (why? Hint: y gives us b(1)),

φ(y) = ψ(y) = 1.

The equality φ(y) = ψ(y) gives y = 1/√

3K. Substituting this in the last equalitywe find K = 3. This gives b−1(y) = 2y/(1 − 3y2). One can now easily checkthat the equilibrium bidding strategies are:

b(x) =x

1 +√

1 + 3x2; c(x) =

x

2 +√

4 − 3x2.

7.2.2 Uniform price Auctions

Uniform-price auctions have been suggested as an alternative format for thesale of Treasury bills and are currently used in electricity spot markets aroundthe globe. Given that the debate on which auction format is superior (discrimi-natory versus uniform), our aim in this section is modest. Our objective here isto illustrate that, in the context of the sale of multiple objects, auction formatsmatter quite considerably in terms of bidding behavior and the seller’s expectedrevenue.

As above, we concentrate on the Independent Private Values model andon the sale of K objects. Each bidder i bids a vector of bids

(bi1, . . . , b

iK

),

bi1 ≥ · · · ≥ bi

K . The winners are the Kth highest bidders. The payment for eachobject will be the same for the winning bidders. This payment is either equalto the K + 1th highest bid (that is, the highest losing bid) or the Kth highestbid (that is, the lowest winning bid). Both types occur in practice. We considerboth cases below.

Winning bidders pay the highest losing bid.It is tempting to conclude that this auction format is similar to the Vickreyauction and that bidding one’s true valuation for each object is an equilibrium.Such a conclusion would be flawed. This auction format has properties that aredistinct from those of the Vickrey auction as illustrated below for the case ofnegative synergies.

We consider again K = 2 and first we examine the positive synergy case.The next result should come as no surprise. With positive synergies, in anyincreasing, symmetric equilibrium, the individual with the highest type receivesboth objects. Thus, it does not matter if the two objects are sold together orseparately. In the former setting, the equilibrium bidding strategy is such wherea player with value x bids α2x. In the latter, this individual’s bids for bothobjects are identical and equal to (α2/2)x as we show below. In this case, theanalogy with the Vickrey auction is precise.


Theorem 23 The uniform price auction of two objects with positivesynergy(α2 ≥ 2) has as equilibrium strategies b(x) = c(x) = (α2/2)x.

Proof: Suppose bidders i = 2, . . . , n bid according to (b(·), c(·)), b(x) =c(x) = rx, r = α2/2. Let us find bidder 1’s best response. If he bids (b, c), b ≥c ≥ 0 his expected utility is given by

φ(b, c) = E[(2r(x − Y ))Ic>rY + (x − rY )Ib>rY >c]

= 2r

∫ c/r

0

(x − y)fY (y) dy +∫ b/r

c/r

(x − ry)fY (y) dy.

Note that the payment in case b > rY > c is rY since rY is the third highestbid. The constrained maximization problem

maxb≥c

φ(b, c) (7.12)

has a Kuhn–Tucker multiplier λ ∈ R so that at the optimum (b, c).

∂φ

∂b= λ

∂φ

∂a= −λ

λ(b − c) = 0.

Thus

(x − b)fY (b/r)1r

= λ

((2r − 1)x − c)fY

( c

r

) 1r

= −λ

λ(b − c) = 0.

If b > c then λ = 0 and x = b. But, then, (2r − 1)x − c ≥ x − c > x − b = 0, acontradiction. Therefore, at the optimum b = c and from

((2r − 1)x − c + x − c)fY

( c

r

) 1r

= 0,

we conclude that c = rx.

The following limiting case is of interest. Suppose that there is no synergy,that is, α2 = 2. Then b(x) = c(x) = x is an equilibrium. If bidders want onlyone object (α2 = 1) then (x, 0) is an equilibrium. We consider now the negativesynergy case. That is, consider 1 < α2 < 2 and that bidder i, i = 2, . . . , n bidsaccording to strictly increasing, continuous bidding strategies (b(·), c(·)). Wewill find bidder 1’s best response. We begin with a lemma.


Lemma 10 To maximize his expected utility, bidder 1 can restrict his bids tob = x and c ≤ x.

Proof: If bidder 1 bids b ≥ c we may write the expected utility as

ψ(b, c) = E[(α2x − 2b(Y ))Ic>b(Y ) + (x − maxc, c(Y ))Ib>c(Y ),b(Y )>c].

If c > x, bidder 1’s expected utility is negative when only one object is won:x−maxc, c(Y ) ≤ x− c < 0. This is always true when two objects are won ifc > b(Y ) ≥ x. Thus, reducing to c = x increases or at least does not decreasehis expected utility. If b > x ≥ c, reducing b to x does not alter the firstsummand but it increases the second summand since it eliminates the cases inwhich b > c(Y ) > x. Now if x > b ≥ c, increasing b to x does not reduce andmay increase the second summand by including the cases where x > c(Y ) ≥ b.

The conclusion to draw from the lemma above is that we shall look for anequilibrium with b(x) = x ≥ c(x). We take this into account in ψ(b, c) andrewrite it as

φ(c) = ψ(x, c)

= E[(α2x − 2Y )Ic>Y + (x − maxc, c(Y ))Ix>c(Y ),Y >c]. (7.13)

Define k = 1 if x > c(1) and k = c−1(x) otherwise. To obtain first-orderconditions we now restrict ourselves to the case where c = c(β) ≤ x. This isequivalent to β ≤ k. Thus, the expected utility may be written as

φ(β) =∫ c(β)

0

(α2x − 2y)fY (y) dy +∫ k

c(β)

(x − maxc(β), c(y))fY (y) dy

=∫ c(β)

0

(α2x − 2y)fY (y) dy +∫ k

c(β)

(x − c(maxβ, y))fY (y) dy

=∫ c(β)

0

(α2x − 2y)fY (y) dy +∫ β

c(β)

(x − c(β))fY (y) dy

+∫ k

β

(x − c(y))fY (y) dy

=∫ c(β)

0

(α2x − 2y)fY (y) dy + (x − c(β))(FY (β) − FY (c(β)))

+∫ k

β

(x − c(y))fY (y) dy. (7.14)


Differentiating we obtain

φ′(β) = (α2x − 2c(β))fY (c(β))c′(β) − c′(β)(FY (β) − FY (c(β)))

+ (x − c(β))(fY (β) − fY (c(β))c′(β)) − (x − c(β))fY (β)

= fY (c(β))c′(β)rx − c(β) − c′(β)(FY (β) − FY (c(β)))

= c′(β)[fY (c(β))rx − c(β) − (FY (β) − FY (c(β)))].

Thus if β = x is to be the optimum:

c′(x) [fY (c(x))rx − c(x) − (FY (x) − FY (c(x)))] = 0.

Therefore, whenever c′(x) > 0,

(rx − c(x))fY (c(x)) + FY (c(x)) = FY (x). (7.15)

For example, if F (x) = x this equation implies that α2 = 2. If F (x) =√

x,there is a solution:

c(x) = (3 − α2 − 2√

2 − α2)x.

The first-order condition above suggests that perhaps we should look at a con-stant equilibrium, that is, c′ = 0. So, to this effect, suppose (x, 0) is the bidsubmitted by bidders i = 2, . . . , n. If bidder 1 bids (x, c), x ≥ c his expectedutility is (see equation (7.13)):

φ(c) = E[(α2x − 2Y ) Ic>Y + (x − c) IY >c]

=∫ c

0

(α2x − 2y)fY (y) dy + (x − c)(1 − FY (c)).

Thus,

φ′(c) = (α2x − 2c)fY (c) − (1 − FY (c)) − fY (c)(x − c)

= ((α2 − 1)x − c)fY (c) − (1 − FY (c)).

The following theorem is now easy to prove. Recall that r = α2 − 1.

Theorem 24 Suppose the synergy is negative. Suppose also that fY (0) < ∞,and f ′

Y (y) ≥ 0 for every y < r. Then, (x, 0) is a symmetric equilibrium biddingstrategy.

Proof: If 0 < c < rx then φ′′(c) = (rx − c)f ′Y (c) ≥ 0. Therefore φ′ is

increasing. Thus,

φ′(c) ≤ φ′(rx) = −(1 − F (rx)) ≤ 0.

Therefore, φ(c) ≤ φ(0). This ends the proof.

Note that with negative synergies bidders do not bid their true valuations!Indeed, this auction format can perform poorly in terms of the seller’s expected


revenue. The seller’s expected revenue in the symmetric equilibrium is equal tozero as shown in the next example.

Example 25 We can apply the theorem to FY (x) = xt if t ≥ 1. If FY (x) =Fn(x), n ≥ 2 then fY (0) = nFn−1(0)f(0) = 0 if f(0) < ∞.

Bidders pay the lowest winning bidAn alternative formulation is for winning bidders to pay the lowest winningbid as the constant per unit price. The reader might conjecture that such asmall change will not impact much on the analysis. It turns out that sucha conjecture would be wrong as we show next.

Suppose (b(·), c(·)) is a symmetric, strictly increasing equilibrium. If bid-ders 2, . . . , n bid according to (b(·), c(·)) and bidder 1 bids (b, c), b ≥ c hisexpected utility is

ψ(b, c) = E[(α2x − 2c)Ic>b(Y ) + (x − minb, b(Y ))Ib>c(Y ),b(Y )>c].

We can assume without loss of generality that c ≤ α2/2. We can also assumethat b ≤ x. Note that b can be strictly less than x. Moreover, if b > c(1),bidder 1 can reduce b without reducing his expected utility to b = c(1). Sinceb = b(1) is the choice if x = 1, we have in equilibrium b(1) = c(1). As before,we may rewrite ψ as

ψ(b, c) = (α2x − 2c)F (b−1(c)) +∫ c−1(b)

b−1(c)

(x − minb, b(y))f(y) dy

= (α2x − 2c)F (b−1(c)) +∫ b−1(b)

b−1(c)

(x − b(y))f(y) dy

+∫ c−1(b)

b−1(b)

(x − b)f(y) dy

= (α2x − 2c)F (b−1(c)) +∫ b−1(b)

b−1(c)

(x − b(y))f(y) dy

+ (x − b)∫ c−1(b)

b−1(b)

f(y) dy.

If b > c , the first-order conditions for the optimum (i.e., ∂ψ/∂b = 0 and∂ψ/∂c = 0, respectively) are (collecting terms in (x − b) and simplifying):

(x − b)(F c−1)′(b) − (F (c−1(b)) − F (b−1(b))) = 0;

− 2F (b−1(c)) + (α2x − 2c)(F b−1)′(c) − (x − c)f(b−1(c))(b−1)′(c) = 0.


Substituting b = b(x) and c = c(x) we obtain:

(x − b(x))(F c−1)′(b(x)) − (F (c−1(b(x))) − F (x)) = 0;

− 2F (b−1(c(x))) + (α2x − 2c(x))(F b−1)′(c(x))

− (x − c(x))f(b−1(c(x)))(b−1)′(c(x)) = 0.

Changing variables (y = b(x) in the first line and y = c(x) in the second line)yields:

(b−1(y) − y)(F c−1)′(y) − (F (c−1(y)) − F (b−1(y))) = 0;

− 2F (b−1(y)) + (α2c−1(y) − 2y)(F b−1)′(y)

− (c−1(y) − y)f(b−1(y))(b−1)′(y) = 0.

We now use that (F c−1)′ = (f c−1)(c−1)′ and (F b−1)′ = (f b−1)′(b−1)′

to get

(c−1)′(y) =F (c−1(y)) − F (b−1(y))(b−1(y) − y)f(c−1(y))

;

(b−1)′(y) =2F (b−1(y))

((α2 − 1)c−1(y) − y)f(b−1(y));

b(1) = c(1).

(7.16)

Example 26 Let us find the equilibrium for the uniform distribution withoutsynergy (α2 = 2). In this case we have to solve,

(c−1)′(y) =c−1(y) − b−1(y)

b−1(y) − y;

(b−1)′(y) =2b−1(y)

c−1(y) − y;

b(1) = c(1).

To simplify the notation, denote by φ = φ(y) = c−1(y) and ψ = ψ(y) = b−1(y):

φ′ =φ − ψ

ψ − y;

ψ′ =2ψ

φ − y.

Eliminating the denominator in both equations, we get:

φ′ψ = yφ′ + φ − ψ = (yφ)′ − ψ,

ψ′φ = yψ′ + 2ψ = (yψ)′+ ψ.


Adding both equations we obtain that (φψ)′ = (yφ)′ + (yψ)′ and thus

φ =yψ

ψ − y.

Substituting this in the equation for ψ′ we get

ψ′ =2ψ

yψ/(ψ − y) − y=

2ψ(ψ − y)y2

.

Now define yz = ψ to obtain the following equation for z:

z′ =z(2z − 3)

y.

The decomposition in partial fractions 1z(2z−3) = − 1

3z + 23(2z−3) yields

− z′

3z+

2z′

3(2z − 3)=

1y.

There is, therefore, a constant T > 0 such that (2z − 3)/z = Ty3, which gives

z(y) =3

2 − Ty3, ψ(y) =

3y

2 − Ty3and φ(y) =

3y

1 + Ty3.

To find T we solve the system φ(y) = ψ(y) = 1 which gives T = 4. Finallythe equilibrium strategies are obtained by inverting φ and ψ, c(x) = φ−1(x)and b(x) = ψ−1(x). The expressions are cumbersome. For example the solutionfor b(x):

b(x) =12

(2x + 4x5/2/(

√1 + 4x3 + 1)

)((2x3/2 +

√1 + 4x3

)1/3+ 1)

(2x3/2 +

√1 + 4x3

)1/3+(2x3/2 +

√1 + 4x3

)2/3+2x3/2 +

√1 + 4x3

.

0.5

0.4

0.3

0.2

0.2 0.4 0.6 0.8 1x

y

b (x )

0.1

0

7.3 Optimal Auction 133

7.3 Optimal Auction

When there are multiple objects for sale but bidders demand only one object,the Revenue Equivalence Theorem holds; both the uniform-price auctionsand the discriminatory auction are efficient and yield the same expectedrevenue. The reader is invited to find the equilibrium bidding functions forthese auction formats in this case.

In this section, we look for the optimal auction in the presence of synergy.For simplicity, we consider only two objects and symmetric bidders. Supposea bidder has type x ∈ [0, 1].Then if qik is the probability that bidder i receivesk objects, k = 0, 1, 2, and his payment is P i, then his utility is

x(qi1 + qi2α2) − P i.

There are several possible ways of allocating the objects amongst the bidders.This set is given by

A =

(k1, . . . , kn); ki ∈ 0, 1, 2,

n∑i=1

ki ≤ 2

.

If bidder 1 receives two objects, then the allocation is (2, 0, . . . , 0). If bidder 1receives one object and bidder 3 the other the allocation is (1, 0, 1, . . . , 0) andso on.

A direct mechanism to sell the two objects is(q, P ) = (q(t), P 1(t), . . . , Pn(t))t∈[0,1]n such that

q(t) = (qa(t))a∈A, qa(t) ∈ R+,∑a∈A

qa(t) ≤ 1;

P i(t) ∈ R is the expected payment of bidder i.

That is, qa(t) represents the probability of occurrence of the allocation athat determines how many objects each bidder will win. The probability thatbidder i wins one object is therefore qi1 = qi1(t) =

∑a∈A,ai=1 qa(t). The

probability that i wins two objects is qi2 = qi2(t) =∑

a∈A,ai=2 qa(t).Define the expected utility of i under this mechanism as:

πi(ti) = tiE−i[qi1(t) + qi2(t)α2] − E−i[P i(t)]. (7.17)

If we define Qi(ti) = (Qi1(ti), Q

i2(ti)), Qi

k(ti) = E−i [qik(t)] we may write theincentive compatibility constraints as

πi(ti) ≥ ti(Qi1(t

′) + Qi

2(t′)α2) − E−i[P i(t′, t−i)], t′ ∈ [0, 1].

From this we obtain the inequality

πi(ti) ≥ (ti − t′i)(Qi

1(t′) + Qi

2(t′)α2

)+ πi (t′i) .


Thus, following the same method as in the one object case, we can prove that

πi(ti) = πi(0) +∫ ti

0

(Qi1(s) + Qi

2(s)α2) ds. (7.18)

Combining (7.17) and (7.18) we have that

E−i[P i(t)] = tiE−i[qi1(t) + qi2(t)α2] − πi(ti)

= tiE−i [qi1(t) + qi2(t)α2] − πi(0) −∫ ti

0

(Qi1(s) + Qi

2(s)α2) ds.

Hence the seller’s expected revenue is

R =n∑

i=1

∫P i(t)f(t) dt =

n∑i=1

∫E−i[P i(t)]f(ti) dti

= −n∑

i=1

πi(0) +n∑

i=1

∫ (tiE−i[qi1(t) + qi2(t)α2]

−∫ ti

0

(Qi1(s) + Qi

2(s)α2)ds

)f(ti) dti.

Therefore, the Revenue Equivalence Theorem generalizes to multiple objects.That is, if q(t) is the same in two auction formats and the lowest type has azero expected utility (i.e., πi(0) = 0) then the seller’s expected revenue is thesame in the two auctions. In particular, all efficient auction formats yieldingthe lowest type zero expected utility have the same expected revenue.

Let us now find the optimal auction. Voluntary participation means thatπi(ti) ≥ 0 for every type. In particular πi(0) ≥ 0. Thus, we choose πi(0) = 0for all i. As before we can write∫ (∫ ti

0

(Qi

1(s) + Qi2(s)α2

)ds

)f(ti) dti

=∫

1 − F (ti)f(ti)

(qi1(t) + qi2(t)α2)f(t) dt.

Thus, to maximize revenue we have to choose mechanisms (q, P ) that areincentive compatible and maximize

n∑i=1

∫ (ti [qi1(t) + qi2(t)α2] −

1 − F (ti)f(ti)

(qi1(t) + qi2(t)α2))

f(t) dt

=∫ n∑

i=1

(ti −

1 − F (ti)f(ti)

)(qi1(t) + qi2(t)α2)f(t) dt.


Define

lia =

1 if ai = 10 otherwise

and mia =

1 if ai = 20 otherwise.

Define also J(ti) = ti − (1−F (ti))/f(ti). In this notation we have that qi1(t) =∑a∈A

liaqa(t) and qi2(t) =∑

a∈Ami

aqa(t). Then, the revenue can be written as

∫ n∑i=1

(ti −

1 − F (ti)f(ti)

)(qi1(t) + qi2(t)α2)f(t) dt

=∫ n∑

i=1

J(ti)∑a∈A

qa(t)(lia + α2mia)f(t) dt

=∫ ∑

a∈A

qa(t)n∑

i=1

J(ti)(lia + α2m

ia

)f(t) dt (mr)

≤∫

maxa∈A

n∑i=1

J(ti)(lia + α2m

ia

)f(t) dt. (7.19)

Finally, define for each t ∈ [0, 1]n the set

S(t) =

a∗ ∈ A;

n∑i=1

J(ti)(lia∗ + α2m

ia∗)

= maxa∈A

n∑i=1

J(ti)(lia + α2m

ia

).

Define

q∗a(t) =

1

#S(t) if a ∈ S(t),

0 otherwise.

We want to establish the following result that characterizes the optimal auction.

Theorem 25 Suppose J(ti) is increasing. Then the optimal auction allocatesthe objects accordingly to (q∗a(t))a∈A

and charges

P i(t) = ti[qi1(t) + qi2(t)α2] −∫ ti

0

qi1(s, t−i) ds −∫ ti

0

qi2(s, t−i)α2 ds.

Proof: We have to check the incentive compatibility and voluntary partici-pation constraints. First note that

E−i[P i(t)] = ti(Qi

1(ti) + α2Qi2(ti)

)−∫ ti

0

(Qi

1(s) + α2Qi2(s))ds.


A bidder’s expected utility is given by

πi(ti) = tiE−i[qi1(t) + qi2(t)α2] − E−i[P i(t)]

= ti(Qi

1(ti) + α2Qi2(ti)

)− ti

(Qi

1(ti) + α2Qi2(ti)

)+∫ ti

0

(Qi

1(s) + α2Qi2(s))ds

=∫ ti

0

(Qi

1(s) + α2Qi2(s))ds ≥ 0.

Thus voluntary participation is assured. Let us check incentive compatibility.If bidder i being of type ti declares type u under the mechanism his expectedutility is

ti(Qi

1(u) + α2Qi2(u)

)− E−i[P i(u, t−i)]

= ti(Qi

1(u) + α2Qi2(u)

)−(

u(Qi

1(u) + α2Qi2(u)

)

−∫ u

0

(Qi

1(s) + α2Qi2(s))ds

)

= (ti − u)(Qi

1(u) + α2Qi2(u)

)+∫ u

0

(Qi

1(s) + α2Qi2(s))ds. (7.20)

We need to compare the last number with πi(ti). Subtracting πi(ti) from (7.20)we obtain∫ ti

0

(Qi

1(s) + α2Qi2(s))ds − (ti − u)

(Qi

1(u) + α2Qi2(u)

)−∫ u

0

(Qi

1(s) + α2Qi2(s))ds

=∫ ti

u

(Qi

1(s) + α2Qi2(s))ds − (ti − u)

(Qi

1(u) + α2Qi2(u)

).

We only have to show that qi1(s, t−i)+α2qi2(s, t−i) is increasing in s. This willimply that Qi

1(s) + α2Qi2(s) is increasing as well and this will end the proof.

Recall how q∗ is defined. It allocates the objects according to the maximum of∑ni=1 J(ti)

(lia + α2m

ia

). Now note that

n∑i=1

J(ti)(lia + α2m

ia

); a ∈ A

= J1 ∪ J2 ∪ J3,

where

J1 = J(ti); 1 ≤ i ≤ n,J2 = J(ti) + J(tj); i, j ≤ n, i = j


and

J3 = α2J(ti); 1 ≤ i ≤ n.

If for some s, qi1(s, t−i) = 0 and qi2(s, t−i) = 0 then any other s will notyield less than 0. Suppose bidder i receives one object. Suppose also withoutloss of generality that J(s) > 0. There must be another bidder receiving anobject too (why? This is left as an exercise). If J(s) + J(tj) is the maximum,then either J(s′) + J(tj) is the maximum when i’s type increases to s′ orα2J(s′) is the new maximum. In either case, bidder i receives two objects andqi1(s′, t−i) + qi2(s′, t−i)α2 = α2. If bidder i is already receiving two objects,increasing s only increases his expected utility.

Remark 16 The specification of payments, P i(t) is of course not unique.Any specification that yields the same E−i[P i(t)] is correct.

Example 27 As an example let us consider K = 2 and the uniform distri-bution. And suppose α2 = 3

2 . We calculate the optimal revenue and allocation.Note that J(t) = t− (1− t) = 2t−1. If t1, t2, . . . , tn is the realization of typesand Z = max ti and Z2 the second highest, then the objects are allocated to thebidder with the highest marginal valuation amongst

0, 2ti − 1, 2ti − 1 + 2tj − 1, 32 (2ti − 1), i, j ≤ n, i = j

,

or equivalently, amongst0, 2Z − 1, 2Z + 2Z2 − 2, 3

2 (2Z − 1)

.

However, note that if 2t−1 ≥ 0 then 32 (2t−1) ≥ 2t−1. Therefore, the optimal

auction is such that it never happens that only one object is allocated. It mayhappen that no objects are allocated. This is so if Z ≤ 1

2 . If Z > 12 , the highest

type bidder receives two objects if

32 (2Z − 1) > 2Z + 2Z2 − 2.

That is if Z > 2Z2 − 12 . If Z > 1

2 and Z ≤ 2Z2 − 12 , the bidder with the highest

and second highest signals receives one object each. We now suppose n = 2 andcalculate the optimum revenue. From (mr), the optimal revenue is

∫maxa∈A

n∑i=1

(2ti − 1)(lia + α2m

ia

)dt1 dt2

=∫

max0, 2maxt1, t2 − 1, 2(t1 + t2 − 1), 3maxt1, t2 − 3

2

dt1 dt2


= 2∫

u>v

max0, 2u − 1, 2(u + v − 1), 3u − 3

2

du dv

= 2∫

u>12

∫ v=1

v= u2 + 1

4

(2(u + v − 1)) du dv + 2∫

u>1/2

∫ v= u2 + 1

4

v=0

(3u − 3

2

)du dv

= 2∫ 1

1/2

(94u − 5

4u2 − 9

16

)du + 2

∫ 1

1/2

(u

2+

14

)(3u − 3

2

)du

=4348

= 0.89.

For the sake of comparison, consider the optimal auction with the requirementthat the object is always delivered. In this case, the revenue is given by (why?)

R =∫

max2t1 − 1 + 2t2 − 1, 3

2 (2t2 − 1), 32 (2t1 − 1)

dt1 dt2.

We decompose R in four parts accordingly to whether ti ≥ 12 or not, i = 1, 2.

If t1 ≤ 12 and t2 ≤ 1

2 then the integrand is 2t1 − 1 + 2t2 − 1. Thus

I1 =∫ 1/2

0

∫ 1/2

0

(2u − 1 + 2v − 1) du dv = −14.

If u ≥ 12 and v ≥ 1

2 then since the cases u ≥ v and v ≥ u are symmetrical,

I2 = 2∫ 1

12

∫ u

12

max2u − 1 + 2v − 1, 3

2 (2u − 1)

du dv

= 2∫ 1

12

∫ 12+ 2u−1

4

0

32 (2u − 1) du dv

+ 2∫ 1

12

∫ u

12+ 2u−1

4

(2u − 1 + 2v − 1) du dv =3148

.

It remains to consider the cases u ≥ 12 ≥ v and v ≥ 1

2 ≥ u. From the symmetryboth integrals will be equal. Thus

I3 = I4 =∫ 1

1/2

∫ 1/2

0

max2u − 1 + 2v − 1, 3

2 (2u − 1)

du dv

=∫ 1

1/2

∫ 1/2

0

32 (2u − 1) dv du =

316

.

The total is R = I1 + I2 + I3 + I4 = − 14 + 31

48 + 616 = 37

48 = 0.770.That is, to guarantee efficiency the seller has to give up some of its expected

revenue. This is the same trade-off (expected revenue versus efficiency) thatone faces in the one object auction model.

7.4 Exercises 139

7.4 Exercises

1. Suppose there are three objects to sell. How much synergy is necessaryso that in a symmetric equilibrium each bidder bids the same amount foreach object?

2. Solve equation (7.11).3. Suppose 0 < r < 1 in Example 24. Check that the strategies

b(x) =rx

r +√

r2 + 1 − r2x2;

c(x) =rx

1 +√

1 − x2(1 − r2),

satisfy the differential equations (7.9) in the uniform [0, 1] distributioncase and that they are equilibrium bidding strategies.

4. In Theorem 23 we only checked the first-order conditions. Show thatb(x) = c(x) = rx are indeed optimal solutions of the expected utilitymaximization problem (7.12).

5. Check that (7.15) has no solution if FY (x) = x2. More generally, showthat it has no solution if 0 < f ′

Y (0) < ∞.6. Consider the two-bidder two-object uniform-price auction. Suppose the

distribution is uniform on [0, 1] and that c′(x) > 0 for every x. Show that(x, c(x)) is a symmetric equilibrium when α2 = 2.

7. Calculate the seller’s expected revenue in Example 24. Compare yourresult with the expected revenue of the sequential second-price auction.


8

What is Next?

The reader who has persevered with the book will now be fully versed inthe mechanics of single-object auction theory and will also have developedan appreciation of the difficulties involved in analyzing the sale of two or moreobjects.

However, this reader will only have a limited understanding of the ongoingand potential research questions. This was our explicit choice. We chose insteadto focus on the mechanics of basic theory rather than offer a broader (but lessdetailed) coverage of topics. To fill this gap, we suggest two complementaryapproaches.

The first approach we suggest is a conventional one. The reader can con-sult the excellent books by Krishna (2002), Klemperer (2004), and Milgrom(2004). These books, although perhaps offering a less detailed exposition of themechanisms of the theory, are more comprehensive in their coverage of some ofthe recent research topics. This includes the auction of multiple objects, such asthe analysis of recent designs including ascending clock auctions and hybrid—combining sealed-bid and ascending/descending stages—designs, as well as theanalysis of topics such as entry, collusion, and the effects of the existence ofbudget-constrained bidders.

The second approach that can be followed by readers who want to beintroduced to ongoing (and potential) research questions consists of review-ing the literature that confronts auction theory with practice. There are threestreams of literature that are relevant for this approach. First, there is anincreasing body of literature that empirically tests the various auction theorymodels. See, for example, the survey of Athey and Haile (2005). There is alsoan empirical literature that is more exploratory in nature and aims at unco-vering auction phenomena that might not have been considered yet by thetheoretical literature. Examples include the survey of empirical studies of artauctions undertaken by Ashenfelter and Graddy (2003) and the work of Jones,Menezes, and Vella (2004) on wool auctions in Australia. For example, thelatter work documents violations of the law of one price—identical objects

141

142 What is Next?

being auctioned for different prices—and has resulted in a stream of theoreticalpapers attempting to identify the reasons for the violation.

The experimental economics literature offers an alternative approach totesting auction theory. The survey by Kagel (1995) although somewhat datedprovides a comprehensive review of the literature that tests auction theory ina laboratory. This is an important body of literature covering both classicalresults demonstrating more aggressive bidding (i.e., bids that are closer to thebidder’s value) than suggested by theory, as well as tests of specific featuresand new auction designs.

The final stream of literature aims to reconcile theory with the actual prac-tice of designing and running auctions. A very good reference in this stream isKlemperer (2004). Although we are not aware of a comprehensive survey of thedifficulties encountered in the practice of auction design, many useful referencescan be found which are usually associated with policy-making initiatives. Thisis the case for the design of spectrum auctions in telecommunications and morerecently the design of auctions to allocate CO2 allowances under various initia-tives to reduce greenhouse gas emissions. See, for example, Holt et al. (2007)and Evans and Peck (2007).

Reading this type of literature seems to us a necessity for any serious studentof auction theory. In doing so one appreciates both the limitations as well asthe value of the insights provided by the theory developed in this book. Itbecomes apparent that in many applications fundamental assumptions do nothold.

For example, the value that potential bidders assign to an object hasboth private and common components. An individual bidding for a residen-tial property will have both an idiosyncratic value for the property and somecommon component related to the expected resale value and interest rates. Agenerator bidding to supply electricity will have considerations including itsown costs (not necessarily known by other bidders), its expectations aboutthe behavior of other generators and, importantly, common factors such asweather forecasts. The difficulty this poses is that valuations might not beaffiliated—for example, the private and common components might move inopposite directions across individuals. Many results (e.g., ranking auctionsaccording to the expected revenue they generate) do not necessarily hold in thiscase.

Similarly, there is no reason to expect that the distribution of types orvaluations of a bidder is an interval of the real line as typically assumed.However, as is demonstrated in the next section, in this case one can still adaptthe techniques and approaches developed above to obtain results that are moregeneral in nature and to show that under some circumstances standard resultsdo not hold for distributions without density.

A second observation that emerges from the practice of running and partici-pating in auctions is that it is rare for an auction to involve the sale of only oneobject. Typically, similar (or identical) objects are sold either simultaneously

8.1 Distribution Hypotheses in Auction Theory 143

or sequentially. Importantly, often bidders demand more than one object andare asymmetric in their demands. For example, the asymmetry might be alongthe lines of incumbent versus entrants or related to size or choice of techno-logy. Auction practitioners are also often concerned with entry. Ensuring theparticipation of enough competitors can mitigate the exercise of market powerby large bidders (who, for example, might bid less than their demand) or ofcollusion (tacit or otherwise).

Finally, literature concerning the practical design of auctions highlights thatoften there is enough flexibility for the auction designer to change some para-meters generally considered fixed in theory. For example, consider the auctionof the rights to access an essential input produced by a vertically integratedincumbent. Suppose further that even an efficient entrant requires two to threeyears of operation to recover the fixed costs of entry. The auction designer canpropose to auction one-year contracts for each of the following three years—making the objects complementary—or to auction, for example, one-year andthree-year supply contracts. In the latter case, objects are substitutes, whichavoids the complications that emerge under complementarity.1 The auctiondesigner can also affect participation by choosing a platform making it easierfor agents to bid or by setting appropriate penalties for non-compliance (e.g.,in the case of auctions of licences). In addition, auction designers can influencethe definition of the property right—for example, by including transferabilityor alienability of the right—so that a secondary market can help mitigate anyinefficiencies that might arise from the auction allocation.

In summary, we highly recommend that the reader venture into the prac-tice of designing, running, and participating in auctions; it is a sure wayto understand the usefulness and limitations of the theory contained in thisbook and a quick path to the discovery of some very exciting researchquestions.

8.1 Distribution Hypotheses in Auction Theory2

In this book and in most of the economics literature the set of possible typesor valuations of a bidder i is an interval T := Ti = [a, b], 0 ≤ a < b. LetF : R → [0, 1] be the distribution of types t ∈ T . That is: F is an increasingright-continuous function such that F (a−) = 0 and F (b) = 1. The distributionusually satisfies condition (M):

1 Recall that in Chapter 7 we illustrated the difficulties involved when auctioning multipleobjects with synergies. More generally, see Milgrom (2007).

2 This section is based on Monteiro (2007) and Monteiro and fux Svaiter (2007).

144 What is Next?

Definition 16 A distribution F : R → R satisfies condition (M) if there is acontinuous function f : [a, b] → R++ such that

F (x) =∫ x

a

f (u) du, a ≤ x ≤ b.

That is, a distribution satisfies condition (M) if it has a continuous strictlypositive density on its support [a, b]. There are, however, several economicallymeaningful examples that do not satisfy condition (M):

1. Suppose that in a population of potential bidders there are two typesof bidders. Half of the bidders have possible types that are uniformlydistributed in [0, 1] and the other half are uniformly distributed in [2, 3].If two bidders are drawn randomly from this distribution and it is notpossible to distinguish from which part of the population they come, thedistribution of types is given by

F (x) =

x2 if 0 ≤ x ≤ 1;12 if 1 ≤ x ≤ 2;

x−12 if 2 ≤ x ≤ 3.

This distribution has density:

f (x) =

12 if 0 ≤ x ≤ 1;0 if 1 < x < 2;12 if 2 ≤ x ≤ 3.

Thus the density is not continuous in 1, 2 and f (x) = 0, 1 < x < 2.2. More generally, if the set of types is a closed set such that a, b ⊂ T

[a, b], then the distribution of types F has a density in [a, b] \ T butf (x) = 0 if x ∈ [a, b] \ T .

3. Suppose that types are drawn from a population such that half of thepopulation has types uniformly distributed in [0, 1]. And the other half ofthe population is identical and has type t = 1

2 . The distribution of typesin this case is

F (x) =

x2 if 0 ≤ x < 1

2 ;34 if x = 1

2 ;x+12 if 1

2 ≤ x ≤ 1.

Thus the distribution has a size 1/2 jump at x = 1/2.

What can we say about independent private value auctions, for a generaldistribution? We consider three topics:

a) Is the revenue equivalence theorem valid?b) Can we characterize the optimal auction in this case?c) What is the equilibrium of the first-price auction?


We begin addressing item (a). If we review section 6.2 we see that nothingchanges if we define Qi (x) :=

∫qi (x, x−i) dF−i (x−i) where F−i = Πj =iFj .

Thus, if (q, P) is an individually-rational, incentive-compatible mechanismthen

E−i [Pi (x)] :=∫

Pi (x) dF−i (x−i) = xiQi (xi) − ai −∫ xi

0

Qi (t) dt. (8.1)

Thus, any such mechanism yields the same expected revenue. Let us examinethis revenue equivalence from another perspective and, thus, consider thefollowing distribution of types:

F (x) =

0 if x < 0;12 if 0 ≤ x < 1;1 if 1 ≤ x.

(*)

That is, bidders are of type t = 0 or type t = 1 with equal probability. Typest ∈ (0, 1) have zero probability. If we only allow bidders to declare a typet ∈ T = 0, 1 we have in principle more leeway to define mechanisms since wedo not have to worry about incentive compatibility for types 0 < t < 1. Let ussee an example.

Example 28 Consider a second-price auction with two bidders of typest1, t2 ∈ T = 0, 1. Types are independent with distribution (*). In a secondprice auction we have the following allocation rule for bidder 1,(

q1 (0, 0) q1 (0, 1)q1 (1, 0) q1 (1, 1)

)=(

12 01 1

2

).

That is, in the case of bidder 1 and bidder 2 announcing the same type, there isa tie and the object is allocated to bidder 1 with 50% probability. The allocationrule for bidder 2 is q2 (i, j) = 1 − q1 (i, j). Payment is given by:(

P1 (0, 0) P1 (0, 1)P1 (1, 0) P1 (1, 1)

)=(

0 00 1

)(P1)

if the bidder receives the object. Thus expected payment is P1 (0) = 0 andP1 (1) = 1

4 .Consider now the following expected payment rule:(

P ∗2 (0) = P ∗

1 (0)P ∗

2 (1) = P ∗1 (1)

)=(

01/2

). (P*)

Let us show that the allocation (q1, q2) above and payment P ∗ = (P ∗1 , P ∗

2 ) satis-fies both the incentive compatibility and voluntary participation constraints. Ifbidder 1 is of type 0 his expected utility is 0. If bidder 1 is of type 1 his expectedutility is

34

− 24

=14

> 0.

146 What is Next?

Thus, voluntary participation is satisfied. A type-0 bidder will not announce histype as type 1 since this would yield expected utility equal to −1

2 < 0. A type-1bidder also has no incentive to announce his type as 0 since his expected utilityfrom truthful revelation would be equal to 1

4 .

This example shows that revenue equivalence is not valid if T = 0, 1.However if T = [0, 1] and the distribution is the same, revenue equivalenceis valid. The following theorem summarizes the relationship between the twosetups.

Theorem 26 Suppose Ti ⊂ [0, 1] is the set of types of bidder i, 1 ≤ i ≤ n.Let Fi be the distribution of types ti ∈ Ti and suppose we have independentprivate values. If (q, P ) : Πn

i=1Ti → [0, 1]n × Rn is a voluntary participation,incentive compatible mechanism then it is possible to extend (q, P ) to a mecha-nism (q′, P ′) : [0, 1]n → [0, 1]n × Rn that is incentive-compatible and satisfiesvoluntary participation. Moreover the expected payment implied by P ′ is com-pletely specified as a function of q′ and the utility of the lowest type as in(8.1).

The reasoning above allows us to assume, without loss of generality, thatthe set of types is an interval. And as a bonus of extending the set of types toan interval, we restore revenue equivalence.

We now address question (b). The optimal auction studied in Chapter 6relies on condition (M) and is not useful for an arbitrary distribution. Notethat the allocation rule stated in Theorem 17 (Chapter 6) is not even definedif the distribution does not have a density or if the density exists but is nullsometimes. In this case, it is possible to show that the following generalizationholds:

Theorem 27 Suppose the distribution of types of bidder i is Fi : [0, 1] → [0, 1],1 ≤ i ≤ n. Then there are l0 ≡ 0, l1 (·) , . . . , ln (·) increasing functions such thatthe optimal mechanism is defined by (q, p) where pi is defined by (8.3) andq(s) = (q1 (s) , . . . , qI (s)) by (8.2).

Definition 17 For each s ∈ S let H(s) = i ≥ 1; li(si) = maxj≥0 lj(sj).Define

qi(s) = 1

#H(s) if i ∈ H(s);0 if i ∈ H(s).

(8.2)

Definition 18 The payment of bidder i is for s ∈ S given by

pi (s) = siqi (s) −∫ si

ai

qi (y, s−i) dy. (8.3)


It would take a great deal of effort to go through a formal and didacticallyappealing proof of this result. Instead, we illustrate the application of thistheorem through a simple example:

Example 29 Let the distribution of types of bidders i = 1, 2 be given by

F (u) =

u3 if 0 ≤ u < 1;

u+13 if 1 ≤ u ≤ 2.

This distribution has a jump of size 1/3 at u = 1. The function l1 = l2 is givenby

l1 (x) =

2x − 3 if 0 ≤ x < 1;2(√

2 − 1)

if 1 ≤ x ≤√

2;2 (x − 1) if

√2 ≤ x ≤ 2.

We now calculate the revenue of the second-price auction (with anoptimal reserve price) and the optimal auction revenue to illustrate how theoptimal auction differs from the standard case where the set of types is aninterval. It should be clear that it is optimal to set r = 1 for a second-price auction. The expected revenue of such an auction can be calculated asfollows:

2∫

x1≥r>x2

rdF (x1) dF (x2) +∫

minx1,x2≥r

min x1, x2 dF (x1) dF (x2)

= 2F (1−) (1 − F (1−)) +∫

x1≥x2≥1

x2dF (x1) dF (x2)

+∫

x2>x1≥1

x1dF (x1) dF (x2) =49

+∫

x2≥1

(1 − F (x2−)) x2dF (x2)

+∫

x1≥1

(1 − F (x1)) x1dF (x1) =79

+∫

x2>1

(1 − x2 + 1

3

)x2

3dx2

+∫

x1>1

(1 − x1 + 1

3

)x1

3dx1 =

79

+23

∫ 2

1

(z − z2 + z

3

)dz

=79

+23

∫ 2

1

2z − z2

3dz =

79

+29

(z2 − z3

3

)|21 =

2527

.

148 What is Next?

In contrast, the revenue in the optimal auction can be calculated as follows:∫max l (x) , l (y) , 0 dF (x) dF (y)

=∫

maxx,y≥1

l (max x, y) dF (x) dF (y) =∫ 2

1

l (z) dF 2 (z)

= l (1)13

+∫ √

2

1

2(√

2 − 1) 2 (z + 1)

9dz +

∫ 2

√2

2 (z − 1)2 (z + 1)

9dz

= 2(√

2 − 1) 1

3+ 2(√

2 − 1)(z2

9+ z

)|√

21 +

49

(z3

3− z

)|2√

2

= 2(√

2 − 1) 1

3+ 2(√

2 − 1)(1

9+√

2 − 1)

+49

(83− 2

√2

3− 2 +

√2

)

= 2(√

2 − 1)(√

2 − 59

)+

49

(23

+√

213

)=

32862700

>25002700

.

We now address question (c). Let us first consider a continuous distributionF : [0, 1] → [0, 1]. Recall expression (3.10):

b∗ (v) = v −∫ v

0F (x)n−1

dx

F (v)n−1 .

This is the equilibrium bid of the first-price auction when there are n bidderseach with valuation distributed as F . The expression (3.10) was obtained underthe assumption that F has a density. The density however does not appear inthe definition of b∗ (·). We will now prove that this expression works also ifthe distribution does not have a density. Suppose bidders i = 2, . . . , n bidsaccording to b∗ (·) and let us find the best response of bidder 1 with valuation0 < v ≤ 1. We note first without proof that b∗ (·) is strictly increasing. If bidder1 bids b∗ (ω) his expected utility is

(v − b∗ (ω)) Pr (b∗ (ω) > max b∗ (v2) , . . . , b∗ (vn))

= (v − b∗ (ω)) Fn−1 (ω) = (v − ω) Fn−1 (ω) +∫ ω

0

F (x)n−1dx

=∫ v

ω

Fn−1 (ω) dx +∫ ω

0

F (x)n−1dx.

If ω < v then∫ v

ω

Fn−1 (ω) dx +∫ ω

0

F (x)n−1dx =

∫ v

0

minFn−1 (ω) , Fn−1 (x)

dx

<

∫ v

0

minFn−1 (v) , Fn−1 (x)

dx =

∫ v

0

Fn−1 (x) dx.


And if ω > v,∫ v

ω

Fn−1 (ω) dx +∫ ω

0

F (x)n−1dx =

∫ v

0

Fn−1 (x) dx

+∫ ω

v

[Fn−1 (x) − Fn−1 (ω)

]dx <

∫ v

0

Fn−1 (x) dx.

This expression however is not valid if the distribution is not continuous. Apure strategy equilibrium does not exist if the distribution has jumps. For-tunately it is possible to combine the above expression and a generalizationof the mixed equilibrium of section 3.4. The following theorem gives thesymmetric equilibrium of the first-price auction for an arbitrary distributionF : [0, 1] → [0, 1].

Theorem 28 The symmetric first-price auction equilibrium is a mixed-strategy equilibrium. It is a mixed strategy Gv with support [b∗ (v−) , b∗ (v)]if F is discontinuous at v and is the pure strategy b∗ (v) if F is continuousat v.

The mixed strategy has support [b (v−) , b (v)] and is defined by

Gv (x) =F (v−)

F (v) − F (v−)

(−1 +

(v − b∗ (v−)

v − x

) 1n−1)

. (8.4)


Appendix A

Probability

The requirements on probability theory for this book are modest. In this appen-dix, we collect the definitions and results needed in the book. Probability theoryis an important area of mathematics with several and diverse applications ineconomics. A thorough understanding of probability theory requires measuretheory and Lebesgue’s integration theory. For this book we assume howeverthat the reader is familiar with the Riemann integral that is taught in Calculuscourses.

A.1 Probability Spaces

A probability space is defined by three objects: the sample space, the set ofevents and the probability measure. The sample space is the set formed of allpossible results of an experiment. For example, if the throwing of a coin isthe experiment, the sample space is the set H,T. If the experiment is thethrowing of a dice the sample space is 1, 2, 3, 4, 5, 6.

The sample space is not uniquely defined. For example, the throwing of acoin could have the sample space 1, 2. We denote by Ω the sample space. Letus denote by P (Ω) the set of subsets of Ω. That is,

P (Ω) = A;A ⊂ Ω .

The next step is to define the set of events. A subset of Ω, say B, is an eventif the probability of B is defined. Thus the set of events is the domain of theprobability measure (to be defined below). Suppose E ⊂ P (Ω) is the set ofevents. If A and B are events, that is, if A,B ∈ E , the union of the eventsA ∪ B is an event as well:

A ∪ B = ω ∈ Ω;ω ∈ A or ω ∈ B ∈ E .

151

152 Probability

If A is an event then the complement of A is an event:

If A ∈ E then Ac = ω ∈ Ω;ω /∈ A ∈ E .

If the set of events is closed for unions and complements, then it is closed forintersections as well since (A ∩ B)c = Ac∪Bc implies that A∩B = (Ac ∪ Bc)c.In general, if the sample space is infinite we need to consider the generalcountable union of events.

Definition 19 (σ-algebra) A set E ⊂ P (Ω) is a sigma-algebra if

1. ∅ ∈ E;2. for any A ∈ E then Ac ∈ E;3. for any family An ∈ E , n ≥ 1 then ∪∞

n=1An ∈ E.

We now define probability measure. We say that events A and B are disjoint(or mutually exclusive) if A ∩ B = ∅. If An, n ≥ 1 is a family of events, we saythat they are pairwise disjoint if for every n = m then An ∩ Am = ∅.

Definition 20 (probability measure) A probability measure, P , is a functionP : E → [0, 1], E ⊂ P (Ω) a σ-algebra of events, such that:

p-1 P (Ω) = 1,p-2 If A,B ∈ E are mutually exclusive (i.e. A ∩ B = ∅) then P (A ∪ B) =

P (A) + P (B),p-3 If An ∈ E for every n ≥ 1 are pairwise disjoint then P (∪nAn) =∑∞

n=1 P (An).

We can now formally define a probability space.

Definition 21 (probability space) A probability space is a triple (Ω, E , P )such that E ⊂ P (Ω) is a σ-algebra and P : E → [0, 1] is a probability measure.

If Ω is finite it is usually possible to assign a probability to every subset of Ω.The same is true if Ω is countable. However if Ω is uncountable it is not possible,in general, to assign a probability to every subset of Ω. The reason for this istechnical and need not concern us here.

Example 30 The finite and countably infinite sample space case.Suppose Ω = ωi; i ∈ N is a countable set. If the sequence of non-negative realnumbers, ri ≥ 0 has a sum

∑∞i=1 ri = 1 we may define a probability space

(Ω,P (Ω) , P ) by

P (A) =∑

i : ωi∈A

ri if A ⊂ Ω. (A.1)

A.2 Uncountable Sample Space Case 153

A.2 Uncountable Sample Space Case

The case of uncountable sample spaces is very important. Uniform distributi-ons, normal distributions, Brownian motions, etc. can only exist on uncountablesample spaces. On uncountable sample spaces it is not possible in general todefine probability for every event. One approach is to try to have a probabilitydefined for a large enough class of events. If C ⊂ P (Ω) is a class of events, thereexists the smallest σ-algebra of events that contains C. Suppose Ω = R. We saythat I ⊂ R is a bounded interval if there exist a and b, with a < b such that Iis one of the sets

[a, b] , (a, b], [a, b), (a, b).

If C1 = I ⊂ R; I is a bounded interval, the smallest σ-algebra that containsC1 is denoted by B1 and is called the Borelean σ-algebra on R. Thus, a countableunion of open intervals belongs to the Borelean σ-algebra. Since every opensubset of R can be written as a countable union of open intervals, it is thereforealso a Borelean set. The closed subsets are Borelean since a closed set is thecomplement of an open set.

Example 31 (uniform distribution) Suppose Ω = [0, 1]. We want to formalizethe experiment of taking a point x ∈ [0, 1] at random. A natural beginning isto consider that the probability of an interval [c, d] ⊂ [0, 1] is the length of theinterval. Thus P ([c, d]) = d − c. Hence P (Ω) = 1 − 0 = 1.

For instance, the probability of taking a point in the interval[0, 1

2

]is 1

2 . Theprobability of taking a point in

[13 , 2

3

]is 1

3 . We can also evaluate the probabilityof more complicated sets. Thus if I and J are disjoint intervals, I ∪ J ⊂ [0, 1]then P (I ∪ J) = P (I) + P (J). If we have a countable set of intervals In ⊂[0, 1] and the family In, n ≥ 1 is pairwise disjoint then we define P (∪nIn) =∑∞

n=1 P (In).A natural question is then when should we stop? Another important question

is: Do we need to be restricted to sets that are obtained from intervals after afinite number of operations of union/complementation?

The answer to this last question is given in measure theory by Caratheodory’sextension theorem. The application of Caratheodory’s theorem to our exampleyields the existence of a unique probability measure λ : B1 → [0, 1] such thatλ ([a, b]) = b − a whenever 0 ≤ a ≤ b ≤ 1. The Borelean σ -algebra contains allsets obtained from intervals after a countable number of operations of unionsand complements. It is, however, not exhausted by these sets.

Instead of considering only intervals, we can consider the Cartesian productof intervals. Define

Cn =n∏

j=1

C1 =

n∏j=1

Ij ; Ij is an interval

.

154 Probability

The smallest σ-algebra of Rn that contains Cn is denoted Bn and is called theBorel σ-algebra on Rn. The more important examples of probability spacesare the probability spaces with density functions. The uniform and normaldistributions have densities. We restrict ourselves in this book to continuousdensity functions. Thus suppose K =

∏nj=1 [ai, bi] is a product of intervals.

A continuous density function is a continuous function f : K → R such that

1. f (k) ≥ 0 for every k ∈ K;2.∫

Kf (x) dx =

∫K

f (x1, x2, . . . , xn) dx1 dx2 · · · dxn = 1.

The following theorem—a particular case of Caratheodory’s extensiontheorem—is proved in measure theory courses.

Theorem 28 For any density function f : K → R, there is a uniqueprobability measure P : Bn → [0, 1] such that:

Ifn∏

j=1

Ij ⊂ K then P

n∏

j=1

Ij

=

∫∏n

j=1 Ij

f (x) dx.

For example if f (k) = 1/Πnj=1 (bi − ai), the probability that emerges from

applying the theorem is called the uniform probability measure on K. It is there-fore the unique probability measure defined on the Borelean σ-algebra such thatP(Πn

j=1 [cj , dj ])

= Πnj=1(dj −cj)/(bj −aj) whenever aj ≤ cj ≤ dj ≤ bj . Usually

when we mention the uniform distribution we are referring to the uniform pro-bability measure on [0, 1]. If another type of uniform distribution is consideredthis will be explicitly mentioned.

A.3 Random Variables

Suppose (Ω,A) is given, where Ω is the sample space and A is a sigma-algebraof events in Ω. Usually if ω ∈ Ω is drawn, we obtain as a measurement somereal number X (ω). For example, if Ω is the set of all humans, X (ω) may bethe height of human ω. In most experiments we only obtain functions of thesample space as a result. The following definition is important.

Definition 22 (random variable) A function X : Ω → R is a random variableif for every interval (a, b) of real numbers, the set ω ∈ Ω; a < X (ω) < b is anevent. That is, for a random variable, the probability of its value being in agiven interval is well defined.

Remark 17 It is possible to show that if X is a random variable then the setω ∈ Ω;X (ω) ∈ B is an event for every B in the Borelean σ-algebra. That is,for every B in the σ-algebra generated by the intervals.

A.3 Random Variables 155

We need to introduce the following notation: If X is a random variable wedenote the set ω ∈ Ω;X (ω) ∈ B by [X ∈ B]. That is we omit the ω in thenotation.

If X is a random variable we may define its distribution function. This isone of the fundamental concepts in probability theory.

Definition 23 The distribution function of a random variable X : Ω → R isthe function FX : R → [0, 1] defined by

FX (x) = P ([X ≤ x]) . (A.2)

Thus, FX (x) is the probability that the random variable is less than orequal to x. The distribution function of a random variable is a particular caseof the general concept of distribution functions.

Definition 24 (distribution) A function F : R → [0, 1] is a distributionfunction if:

1. F is non-decreasing, that is, F (x) ≤ F (y) if x ≤ y.2. F is right-continuous: I.e. F (x+) := limy↓x F (y) = F (x).3. F (−∞) := limx→−∞ F (x) = 0, F (∞) := limx→∞ F (x) = 1.

Remark 18 It is possible to show that for any distribution function F thereis a probability space (Ω,A, P ) and a random variable X : Ω → R such thatFX = F . Thus in this sense distribution of random variables and distributionfunctions concepts coincide.

Example 32 (uniform distribution) A random variable U : Ω → R such thatits distribution is given by FU (x) = x for every x ∈ [0, 1] is said to havea uniform [0, 1] distribution. It shall be clear that F (x) = 0 if x < 0 andF (x) = 1 if x ≥ 1.

Definition 25 (density of a distribution) The distribution function F has adensity if there is a (Riemann integrable) function f : R → R+ such that forevery x we have that F (x) =

∫ x

−∞ f (u) du.

We remind the reader that a continuous function is Riemann integrable.If a function has a finite or infinite countable number of discontinuities it isalso Riemann integrable. This is enough for our purposes. Reciprocally if the(Riemann integrable) function f : R → R is non-negative and

∫∞−∞ f (x) dx = 1

then defining F (x) =∫ x

−∞ f (u) du we have that F is a distribution function.

156 Probability

Remark 19 The Fundamental Calculus Theorem implies that at every pointof continuity of f (x) that F ′ (x) = f (x). In this way we recover the densityfunction f from the distribution F .

The uniform distribution is a distribution with a density. Consider

f (x) =

1 if x ∈ [0, 1] ,0 if x /∈ [0, 1] .

Since F (x) =∫ x

−∞ f (y) dy = x if x ∈ [0, 1], f (·) is the density of the uniformdistribution.

Definition 26 The distribution has a bounded support if there is an interval[a, b] such that F (a−) = supx<a F (x) = 0 and F (b) = 1. Moreover, F (x) isstrictly increasing then the interval [a, b] is the support of the distribution F .

A.4 Random Vectors and their Distribution

A random variable is a function of the result of an experiment. Recall theexample of the height of human ω. We could however obtain more information.We could measure the height and weight of human ω. In this case, we wouldhave X : Ω → R2. We are, however, not limited to two measures. The generalconcept is that of a random vector.

Definition 27 A function X : Ω → Rn is a random vector ifω ∈ Ω;X (ω) ∈ Πn

i=1 (ai, bi) is an event for every Cartesian product ofintervals Πn

i=1 (ai, bi) ⊂ Rn.

It is possible to show that if X is a random vector then [X ∈ B] is an eventfor every B ⊂ Rn Borelean. If we write X = (X1, . . . , Xn), then X is a randomvector if and only if X1, . . . , Xn are random variables. It is convenient to definerandom vector distributions.

Definition 28 The distribution of a random vector X is the functionFX : Rn → [0, 1] given by

FX (x1, . . . , xn) = P ([X1 ≤ x1, . . . , Xn ≤ xn]) .

The distribution FX has a density if there exists a Riemann integrable functionf : Rn → R+ such that for every x ∈ Rn,

FX (x) =∫ x1

−∞

∫ x2

−∞. . .

∫ xn

−∞f (y1, y2, . . . , yn) dy1 dy2 · · · dyn.

A.4 Random Vectors and their Distribution 157

From the joint distribution of X we may obtain the distribution of each Xj

(the marginal distribution). Simply note that

FX1 (x1) = P (X1 ≤ x1) = P (X1 ≤ x1, X2 < ∞, . . . , Xm < ∞)

= FX (x1,∞, . . . ,∞) = limx2→∞,...,xm→∞

FX (x1, x2, . . . , xm) .

If X has a density then X1 has a density as well. This density is easily obtained:Since

FX1 (x1) = P (X1 ≤ x1, X2 < ∞, . . . , Xm < ∞)

=∫ x1

−∞

(∫ ∞

−∞. . .

∫ ∞

−∞f (y1, y2, . . . , yn) dy2 · · · dyn

)dy1

Thus f1 (x1) :=∫∞−∞ . . .

∫∞−∞ f (x1, y2, . . . , yn) dy2 · · · dyn is a density function

for FX1 . The following lemma gives an important property of distributions withdensity.

Lemma 11 If the random vector (X1, X2, . . . , Xn) has a density then forevery pair of indices i = j we have that

Pr[Xi = Xj ] = 0.

Proof: Let M be an upper bound of f (x) ;x ∈ Rn. Let N,m > 0 be inte-gers. For each n ∈ Z we divide the interval [n, n + 1] in N pairwise disjointintervals,

Ikn, k = 1, 2, . . . , N

. Thus

Pr[Xi ∈ Ik

n, Xj ∈ Ikn, Xl ∈ [−m,m] , l /∈ i, j

]=∫

xi∈Ikn

∫xj∈Ik

n

∫xl∈[−m,m]

f (xi, xj , x−l) dxi dxj dx−l

≤ M (2m)n−2 1N2

.

Hence

Pr [Xi = Xj , Xi ∈ [n, n + 1] , |Xl| ≤ m, l = i, j]

≤N∑

k=1

Pr[Xi ∈ Ik

n, Xj ∈ Ikn, Xl ∈ [−m,m] , l = i, j

]

≤ M (2m)n−2 N

N2=

M (2m)n−2

N.

Since this is true for every N we have that

Pr[Xi = Xj , Xi ∈ [n, n + 1] , |Xl| ≤ m, l = i, j] = 0.

158 Probability

The countable union of sets with probability 0 has probability 0 thereforePr[Xi = Xj ] = 0.

A.5 Independence of Random Variables

Suppose Xi : Ω → R, 1 ≤ i ≤ m are random variables.

Definition 29 We say that X1, X2, . . . , Xm are independent if for allB1, B2, . . . , Bm, Borel subsets of R, it is true that

P ([Xi ∈ Bi, for all i, 1 ≤ i ≤ m]) = P (X1 ∈ B1) · · ·P (Xm ∈ Bm) . (A.3)

To see that if X1, . . . , Xm are independent then every sub-family is alsoindependent, note that by choosing Bj = R the random variable Xj can beomitted from (A.3). If X1, . . . , Xm are independent then

FX (x) = P ([X1 ≤ x1, . . . , Xm ≤ xm])

= P ([X1 ≤ x1]) · · ·P ([Xm ≤ xm])

= FX1 (x1) . . . FXm(xm) .

Reciprocally if FX (x) = FX1 (x1) . . . FXm(xm) then the random variables

X1, . . . , Xm are independent. As a corollary we have that if FX has a densitythen the coordinates are independent if and only if f (x) = f1 (x1) , . . . ,fm (xm).Thus we have the following proposition.

Proposition 8 1. If X has a density f (x) then Xi has a density fi givenby fi (xi) =

∫∞−∞ . . .

∫∞−∞ f (y1, . . . , yi−1, xi, yi+1, . . . , yn) dy−i.

2. If the random variables X1, . . . , Xn have density then X1, X2, . . . , Xn

are independent if and only if the random vector X = (X1, . . . , Xn) hasdensity f (x) = f1 (x1) , . . . ,fn (xn).

A.6 The Distribution of the Maximum ofIndependent Random Variables

We begin recalling the definition of the maximum. If a and b are real numbersthe maximum of a and b, denoted maxa, b is a if a ≥ b. And it is b ifb ≥ a. More generally the maximum of the n real numbers, a1, a2, . . . , an is aj

such that aj ≥ al, l = 1, 2, . . . , n. We write max a1, a2, . . . , an = aj . SupposeX1, X2, . . . , Xn are independent random variables, identically distributed. Thatis, FX1 = FX2 = · · · = FXn

. To each ω ∈ Ω define maxX1, . . . , Xn (ω) =max X1 (ω) , . . . , Xn (ω). We want to find the distribution of the maximum

A.7 The Distribution of the Second Highest Value 159

of the n functions, max X1, . . . , Xn. Call F the common distribution FX1

and call G the distribution of max X1, . . . , Xn. Thus,

G (r) = P (ω ∈ Ω; max X1 (ω) , . . . , Xn (ω) ≤ r) (A.4)

= P ([Xi ≤ r, 1 ≤ i ≤ n]) = P (∩ni=1 [Xi ≤ r])

=n∏

i=1

P ([Xi ≤ r]) =n∏

i=1

F (r) = Fn (r) . (A.5)

If F has a density, G has a density too (recall remark (19)):

g (r) = G′ (r) = nFn−1 (r) f (r) . (A.6)

Example 33 The maximum of n independent uniform random variables hasa distribution with density g (r) = nrn−1, r ∈ [0, 1].

A.7 The Distribution of the Second HighestValue

Let us consider a finite group of numbers, x1, x2, . . . , xm. Let us re-orderthese numbers from the largest to the smallest, y1, y2, . . . , ym. Thus y1 ≥y2 ≥ · · · ≥ ym. If there is repetition among the x′

js there will be also repetitionamongst the y

′js. The number y1 = max x1, x2, . . . , xm is the maximum. The

number y2 is the second largest and so on. The number ym is the mth largest.For example, suppose our group is 1, 2, 2. The y1 = 2 = y2, y3 = 1.

We want to find the distribution of the second highest value of the inde-pendent, identically distributed random variables, X1, X2, . . . , Xm. Call Y 2 thesecond highest and Y 1 the highest. We need to evaluate the probability thatY 2 ≤ r for a given real number r. If Y 1 ≤ r then a fortiori Y 2 ≤ r. Supposethe maximum is X1 (ω) > r. Then Y 2 ≤ r if and only if Xj (ω) ≤ r, j = 1. Alsoif the maximum is Xk (ω) > r then Y 2 ≤ r if and only if Xj (ω) ≤ r, for allj = r. Thus,

[Y 2 ≤ r

]=[Y 1 ≤ r

]∪ ∪m

k=1 Xk > r ≥ Xj ,∀j = k . (A.7)

The sets in (A.7) are pairwise disjoint. Therefore,

FY 2 (r) = P([

Y 1 ≤ r])

+m∑

k=1

P ([Xk > r ≥ Xj ,∀j = k]) . (A.8)

160 Probability

We know already that P([

Y 1 ≤ r])

= Fm (r). To evaluate the summandsabove we use independence:

P ([Xk > r ≥ Xj ,∀j = k]) (A.9)

= P ([Xk > r]) · P ([r ≥ Xj ,∀j = k]) (A.10)

= P ([Xk > r]) ·∏j =k

P ([r ≥ Xj ]) = (1 − F (r)) Fm−1 (r) (A.11)

Finally we have that FY 2 (r) = Fm (r)+∑m

k=1 (1 − F (r)) Fm−1 (r) = Fm (r)+m (1 − F (r)) Fm−1 (r). If F has density f then the density of Y 2 is given by

fY 2 (r) = m (m − 1) (1 − F (r)) Fm−2 (r) f (r) . (A.12)

Let us find the joint distribution of the highest and second highest types. Wehave n bidders. Thus if a ≥ b then

Pr(Y ≤ a, Y 2 ≤ b

)= Pr (Y ≤ b) + Pr

(b < Y ≤ a, Y 2 ≤ b

)= Fn (b) + n(F (a) − F (b))Fn−1 (b) .

From this we get the density f(Y,Y 2) easily:

f(Y,Y 2) (a, b) = nf (a) (n − 1) Fn−2 (b) f (b) , b ≤ a.

A.8 Mean Value of Random Variables

Suppose we have n real numbers, x1, x2, . . . , xn. The mean of x1, x2, . . . , xn isby definition

∑ni=1 xi/n. It is possible that some of the xi are equal. Suppose

that by omitting repetitions we obtain the set y1, y2, . . . , yl of real numbers.And define k1 ∈ N the number of times y1 is repeated in x1, x2, . . . , xn, k2 thenumber of times y2 is repeated and so on until kl the number of times yl isrepeated. By construction

∑li=1 ki = n. Thus, the average of a collection of

numbers with repetition is∑li=1 kiyi∑li=1 ki

=l∑

i=1

(ki∑li=1 ki

)yi. (A.13)

Let us consider now a discrete random variable. That is, a random variableX : Ω → R with a finite range, X (Ω) = y1, y2, . . . , yl. The average of X is bydefinition

E [X] =l∑

i=1

yiP ([X = yi]) . (A.14)

Comparing (A.13) with (A.14) we see that the discrete random variable meansdefinition is natural. If the random variable is not discrete the definition of

A.8 Mean Value of Random Variables 161

the mean is more complicated. A particularly important case is when therandom variable has a distribution with density. If the random variable Xhas a distribution FX with density fX its mean is defined by

E [X] :=∫ ∞

−∞xfX (x) dx. (A.15)

This definition is also natural. Rewriting (A.14) as

E [X] =l∑

i=1

yi (P ([X ≤ yi]) − P ([X ≤ yi−1]))

=l∑

i=1

yi (FX (yi) − FX (yi−1)) ,

we may consider (A.14) as a Riemann sum of the integral∫∞−∞ xfX (x) dx. Thus,

if we use finer and finer partitions, the discrete mean will converge to (A.15).The general definition is given in terms of the Lebesgue integral

∫X (ω) dP (ω)

and will not be used in this book. We summarize the discussion above in thefollowing definition:

Definition 30 The mean of the distribution F with density f is defined by∫∞−∞ xf (x) dx. If the random variable X has distribution FX with density fX

then the mean of X is defined by E [X] :=∫∞−∞ xfX (x) dx.

If X is a random variable and h : R → R is a continuous function the meanof hX can be proved to be equal to

∫∞−∞ h (x) fX (x) dx. This result is true in

much more generality for any h (·) which is measurable. That is, for any h (·)such that h−1 ((−∞, a)) is a Borelean set for any real number a. If h (x) = xp

the mean E [Xp] is the moment of order p. The variance of X is given byE[(X − E [X])2].

Example 34 Let us calculate the mean of some random variables.

1. Given the uniform [a, b] distribution. It has a density function f (x) =1/(b − a) if a ≤ x ≤ b and f (x) = 0 otherwise. Thus if the randomvariable X has a uniform [a, b] distribution its mean is

E [X] =∫ b

a

x1

b − adx =

b2 − a2

2 (b − a)=

b + a

2.

The uniform [0, 1] distribution has mean 12 .

2. Consider now the power distribution: F (x) = xγ , γ > 0. It is defined forx ∈ [0, 1]. If x > 1, F (x) = 1. If x < 0, F (x) = 0. Its density is easilycalculated by f (x) = F ′ (x) = γxγ−1, x ∈ (0, 1). Since f (·) is continuous

162 Probability

on (0, 1) it is indeed the density of F . Hence, the mean of the powerdistribution xγ is

∫ 1

0x(γxγ−1

)dx = γ

∫ 1

0xγ dx = γ/(γ + 1).

3. Another common distribution is the exponential distribution. It is givenby F (x) = 1 − e−rx, x ≥ 0, with parameter r > 0. The exponentialdistribution has a continuous density f (x) = re−rx. Thus, the mean is∫ ∞

0

xre−rx dx =∫ ∞

0

y

re−y dy =

∫∞0

ye−y dy

r=

1r.

The last integral is calculated using integration by parts.

Example 35 Let us calculate the mean of the maximum of independent uni-formly distributed random variables. Suppose X1, X2, X3 are independentlyuniformly distributed in [0, 1]. From (A.6) we obtain the mean of the maximum:

E [max X1, X2, X3] =∫

x(3F 2 (x) f (x)

)dx (A.16)

=∫ 1

0

x(3x2 · 1

)dx =

34. (A.17)

The generalization to any number of independently uniformly distributedrandom variables is immediate:

E [max X1, X2, . . . , Xm] =∫ 1

0

xmxm−1 dx =m

m + 1. (A.18)

The mean of the second highest value can also be easily calculated using ( A.12).We obtain

E[Y 2]

=∫ 1

0

m (m − 1) (1 − x)xm−1 dx =m − 1m + 1

.

The following proposition has an immediate proof using (A.6) and (A.12).

Proposition 9 Suppose X1, . . . , Xm is a sequence of independent identicallydistributed random variables. Suppose also that the common distribution F (·)has a density f (·) which is zero outside the interval [a, b]. Then the meanvalue of the maximum is

∫ b

armFm−1 (r) f (r) dr. The mean value of the second

highest value is∫ b

arm (m − 1) (1 − F (r)) Fm−2 (r) f (r) dr.

A.9 Conditional Probability

Consider a probability space (Ω,A, P ) and suppose that the event A ∈ A,P (A) > 0 is known to occur. We often want to know the probability of anevent B given that A occurs. For example, when throwing a dice we may want

A.9 Conditional Probability 163

to know the probability of number 2 occurring given that an even numberedside has occurred. The conditional probability of event B given the event A isby definition

P (B |A) =P (A ∩ B)

P (B). (A.19)

Thus P (A |A) = 1. If B and C are disjoint events then P (B ∪ C |A) =P (B |A) + P (C |A). If we define A = B ∩ A;B ∈ A we have that(A, A, P (· |A)) is a probability space. The conditional probability is very sim-ple if A and B are independent. In this case P (B |A) = P (B). So knowingthat A occurs doesn’t affect the probability of the occurrence of B.

Let us now consider the case of conditioning for events with probabilityzero. This is a much harder case and the general answer needs the Radon–Nikodym theorem which is not covered here (see Chung’s book, e.g.) If werestrict ourselves to distributions with density a definition can be made withoutmuch difficulty.

Suppose we have a random vector (X,Y ) with distribution F and densityf (·, ·). A natural definition of the conditional density fX|Y =y is fX|Y =y(x) =f(x, y)/fY (y). The conditional distribution FX|Y =y is

FX|Y =y (x) =∫ x

−∞fX|Y =y(u) du =

∫ x

−∞

f (u, y)∫∞−∞ f (a, y) da

du.

Example 36 Suppose X1, . . . , Xm are independent, uniformly distributed in[0, 1]. Let us find the conditional distribution of Y 2 given Y 1. We first find thedistribution of

(Y 2, Y 1

):

F (y, z) = P(Y 2 ≤ y, Y 1 ≤ z

).

If z ≤ y then F (y, z) = P(Y 1 ≤ z

)=∫ z

0fY 1 (u) du. If y < z. Then F (y, z) =

P(Y 1 ≤ y

)+P

(Y 2 ≤ y < Y 1 ≤ z

). The last summand can be calculated using

the reasoning in (A.7) and (A.8). It is P(Y 2 ≤ y < Y 1 ≤ z

)= m (z − y) ym−1.

Thus F (y, z) = ym + m (z − y) ym−1. The density of F is f (y, z) = ∂2F∂y∂z =

m (m − 1) ym−2 if 0 ≤ y ≤ z ≤ 1 and 0 otherwise. The conditional density istherefore

fY 2|Y 1=z =m (m − 1) ym−2

mzm−1=

(m − 1) ym−2

zm−1if y ≤ z.

The mean of Y 2 conditional on Y 1 = z is∫ z

0

(m − 1) ym−1

zm−1dy = z

m − 1m

.

164 Probability

Lemma 12 Suppose X = (X1, . . . , Xm) is a random vector with joint densityfX (x) . Suppose also that u (x) is a Riemann integrable function. Then

E [u (X) |X1 = x1] = E [E [u (X) |X1, X2] |X1 = x1] .

Proof: If fX1 (x1) =∫

fX (x) dx2 . . . dxm is the marginal density of X1 then

E [u (X) |X = x1] =∫

u (x1, x2, . . . , xm)fX (x)

fX1 (x1)dx2 · · · dxm.

Now if fX1X2 is the joint density of (X1, X2) we have that

E [u (X) |X1 = x1, X2 = x2] =∫

u (x)f (x1, x2, x3, . . . , xm)

fX1X2 (x1, x2)dx3 · · · dxm.

Thus∫E [u (X) |X1 = x1, X2 = x2]

fX1X2 (x1, x2)fX1 (x1)

dx2

=∫ (∫

u (x)f (x1, x2, x3, . . . , xm)

fX1X2 (x1, x2)dx3 · · ·dxm

)fX1X2 (x1, x2)

fX1 (x1)dx2

=∫

u (x)f (x1, x2, x3, . . . , xm)

fX1 (x1)dx2 dx3 · · · dxm = E [u (X) |X = x1] .

In equation (3.6), we mentioned that the equilibrium bidding function ofthe first-price auction is the expected value of the second highest type giventhe highest type. The conditional density of Y 2 given Y = a is

fY 2|Y =a (b) =(n − 1) Fn−2 (b) f (b)

Fn−1 (a), b ≤ a.

Thus

E[Y 2 |Y = v

]=∫

yfY 2|Y =v dy

=∫ v

0

y(n − 1) Fn−2 (y) f (y)

Fn−1 (v)= b∗ (v) . (A.20)

Appendix B

Differential Equations

We collect here a few results on ordinary differential equations that are used inthe text.

B.1 The Simplest Differential Equation

Suppose f : R → R is a continuous function. Let x0 and A be given realnumbers. If we want to find a differentiable function y : R → R such that

y′(x) = f(x), ∀x ∈ R;

y(x0) = A,(B.1)

we say that we want to solve the differential equation y′ = f with initialcondition y(x0) = A. This equation is solved using the Fundamental Theoremof Calculus.

Theorem 29 Fundamental Theorem of Calculus If two differentiable functi-ons g(x) and h(x) have the same derivative then they differ by a constant.

To solve (B.1) consider ψ(x) = A +∫ x

x0f(y) dy. Then note that ψ(x0) = A

and ψ′(x) = f(x) for every x since f(·) is continuous. Suppose now that y(·)solves (B.1). Then since y′(x) = f(x) = ψ′(x) we have by the fundamentaltheorem of calculus that ψ(x) = y(x) + C for some constant C. Choosingx = x0 we see that C = 0.

B.2 Integrating Factor

The integrating factor is used to solve linear differential equations. It is usedseveral times in this book. It can be applied to equations of the following form:

b′(x) + b(x)Q(x) = R(x).

165

166 Differential Equations

To understand how it works note that the left-hand side of the above equationhas a slight similarity to the derivative of a product: (bp)′ = b′p + bp′. Supposewe multiply the left-hand side by a function P (x). It becomes

b′(x)P (x) + b(x)P (x)Q(x).

If we can choose P (x) such that P ′(x) = P (x)Q(x) (this P is called an integra-ting factor) we transform the left-hand side into the derivative of the productb(x)P (x):

(b(x)P (x))′ = b′(x)P (x) + b(x)P ′(x)

= (b′(x) + b(x)Q(x))P (x)

= R(x)P (x).

Now to find P such that P ′ = PQ is easy. Note that

(log P )′ =P ′

P= Q.

Thus P (x) = e∫

Q(x) dx is an integrating factor. And our differential equationhas a solution:

b(x) =∫

R(x)P (x)P (x)

.

The reader may wonder why we did not mention the initial condition. This will,however, be taken care of in the text each time we use the integrating factor.

Appendix C

Affiliation

Proposition 10 Suppose that for every i, j, p(·) satisfies the monotone likeli-hood property for the variables (xi, xj). Then the density p(·) satisfies themultivariate monotone likelihood property.

Proof: We follow the proof in Karlin and Rinott (1981) Proposition 2.1. Weprove by induction on dimension n. Suppose it is true for n. Suppose p is defi-ned on X = Πn+1

i=1 [ai, bi] and satisfies the monotone likelihood property for anypair of variables. By the induction hypotheses it also satisfies the multivariatemonotone likelihood for any set of n variables. Consider x, y ∈ X. Withoutloss of generality, we may suppose that x = (x∗

1, . . . , x∗k, xk+1, . . . , xn+1),

y = (y1, . . . , yk, y∗k+1, . . . , y

∗n+1) where x∗

l ≥ yl, l = 1, . . . , k and y∗l ≥ xl for the

other coordinates. Then

p(x ∨ y)p(x ∧ y)p(x)p(y)

=p(x∗

1, . . . , x∗k, y∗

k+1, . . . , y∗n+1) p(y1, . . . , yk, xk+1, . . . , xn+1)

p(x∗1, . . . , x

∗k, xk+1, . . . , xn+1) p(y1, . . . , yk, y∗

k+1, . . . , y∗n+1)

=p(x∗

1, . . . , x∗k, y∗

k+1, . . . , y∗n+1) p(x∗

1, y2, . . . , yk, xk+1, . . . , xn+1)p(x∗

1, . . . , x∗k, xk+1, . . . , xn+1) p(x∗

1, y2, . . . , yk, y∗k+1, . . . , y

∗n+1)

×p(x∗

1, y2, . . . , yk, y∗k+1, . . . , y

∗n+1) p(y1, . . . , yk, xk+1, . . . , xn+1)

p(x∗1, y2, . . . , yk, xk+1, . . . , xn+1) p(y1, . . . , yk, y∗

k+1, . . . , y∗n+1)

≥ 1.

The term in the second line above is not less than one by the inductionhypothesis—x∗

1 is the variable that is fixed. In the third line above the samereasoning applies—(y2, . . . , yk) is fixed.

167

168 Affiliation

Lemma 13 If f1(x)f2(y) ≤ f3(x ∨ y)f4(x ∧ y) and φi(x) =∫

fi(x) dxn,i = 1, 2, 3, 4 then

φ1(x)φ2(y) ≤ φ3(x ∨ y)φ4(x ∧ y). (C.1)

Proof: Write y = (x1, . . . , xn−1). To prove (C.1) is equivalent to prove that∫f(y, a)f(y′, b) dadb ≤

∫f(y ∨ y′, a)f(y ∧ y′, b) dadb.

Or rather that∫a<b

[f(y, a)f(y′, b) + f(y, b)f(y′, a)] dadb

≤∫

a<b

[f(y ∨ y′, a)f(y ∧ y′, b) + f(y ∨ y′, b)f(y ∧ y′, a)] dadb.

Define

t1 = f(y, a)f(y′, b), t2 = f(y, b)f(y′, a),

s1 = f(y ∨ y′, a)f(y ∧ y′, b), s2 = f(y ∨ y′, b)f(y ∧ y′, a).

It is immediate that t1 ≤ s2, t2 ≤ s2. Moreover, t1t2 ≤ s1s2. If s2 = 0 thent1 = t2 = 0. If

s2 > 0, s2[s2 + s1 − t1 − t2] = s22 + s2s1 − s2t1 − s2t2

= (s2 − t2)(s2 − t1) + (s2s1 − t1t2) ≥ 0

imply that s2 + s1 ≥ t1 + t2.

We now prove Theorem 9 repeated here for convenience:

Theorem 9 Let f1, f2, f3, and f4 be non-negative functions on Rn such thatfor all x, y ∈ Rn, f1(x)f2(y) ≤ f3(x ∨ y)f4(x ∧ y). Then∫

f1(x) dx

∫f2(x) dx ≤

∫f3(x) dx

∫f4(x) dx.

Proof: The proof is by induction on the dimension n. For n = 1 the result istrue if ∫

x<y

(f1(x)f2(y) + f1(y)f2(x)) dxdy

≤∫

x<y

(f3(x)f4(y) + f3(y)f4(x)) dxdy.

Define

a = f1(x)f2(y); b = f1(y), d = f3(y)f4(x).

Affiliation 169

It is straight forward to verify that the following inequalities hold a ≤ d, b ≤ dand ab ≤ cd. The inequality c + d ≥ a + b follows from the inequality d(c +d − a − b) = (d − a)(d − b) + cd − ab ≥ 0. Suppose now the result is true fordimension n − 1. If f1(x)f2(y) ≤ f3(x ∨ y)f4(x ∧ y) then integrating in xn theresulting function

φi(x−n) =∫

fi(x) dxn

also satisfies the same inequality. See the Lemma 13. Thus,∫φ1(x) dx−n ·

∫φ2(x) dx−n ≤

∫φ3(x) dx−n ·

∫φ4(x) dx−n.

And now the result is immediate since∫

φi(x) dx−n =∫

fi(x) dx.


Appendix D

Convexity

In this section, we consider a few results on convex functions that are used inChapter 6. We begin with the convex hull.

Definition 31 The convex hull of a function f(·) is the greatest convex func-tion that is pointwise smaller than f . That is, if f : [0, 1] → R then g : [0, 1] → Ris the convex hull of f if:

1. g is convex;2. g ≤ f , that is g(x) ≤ f(x), for all x ∈ [0, 1];3. if h is a convex function and h ≤ f then h ≤ g.

The convex hull of a function exists and has an explicit expression (seeRockafellar: 36). We do not consider here the more general case.

Proposition 11 If h : [0, 1] → R is continuous then the function g definedbelow is the convex hull of h:

g(q) = min

n∑

l=1

λlh(rl);λl, rl ∈ [0, 1],n∑

l=1

λl = 1,n∑

l=1

λlrl = q

(D.1)

Proof: The minimum is attained since h is continuous. It is immediate thatg(q) ≤ h(q) for every q. Now if f ≤ h is a convex function then for any convexcombination q =

∑nl=1 λlrl we have that

f(q) = f

(n∑

l=1

λlrl

)≤∑

l

λlf(rl) ≤∑

l

λlh(rl).

171

172 Convexity

Hence f(q) ≤ g(q). It remains to prove that g is convex. Suppose λ ∈ (0, 1) andq1, q2 ∈ [0, 1]. Then there exists ωl, rl, ω

′l, r

′l such that

q1 =n∑

l=1

ωlrl;ωl ≥ 0, rl ∈ [0, 1],n∑

l=1

ωl = 1;

q2 =n∑

l=1

ω′lr

′l;ω

′l ≥ 0, r′l ∈ [0, 1],

n∑l=1

ω′l = 1;

g(q1) =n∑

l=1

ωlg(rl), g(q2) =n∑

l=1

ω′lh(r′l).

We have that

λg(q1) + (1 − λ)g(q2) = λ

n∑l=1

ωlh(rl) + (1 − λ)n∑

l=1

ω′lh(r′l)

=n∑

l=1

λωlh(rl) +n∑

l=1

(1 − λ)ω′lh(r′l)

=2n∑i=1

θih(si);

θi =

λωi if i ≤ n

(1 − λ)ω′l if i = l + n, l ≤ n

and

si =

ri if i ≤ n

r′l if i = l + n, l ≤ n.

Thus,

g(λq1 + (1 − λ)q2) ≤2n∑i=1

θih(si) = λg(q1) + (1 − λ)g(q2).

Lemma 14 If g(x) < f(x) then g′(y) = g′(x) in a neighborhood of x.

Proof: Suppose g(x) < f(x). There are λl ≥ 0,∑

l λl = 1, xl ∈ [0, 1],x =

∑l λlxl such that g(x) =

∑l λlf(xl). Since

g(x) ≤∑

l

λlg(xl) ≤∑

l

λlf(xl) = g(x),

we conclude that g(xl) = f(xl) for every l. We first show that there is a θ ∈ [0, 1]such that g(x) = θf(ak) + (1 − θ)f(al) and θak + (1 − θ)al = x. To see this

Convexity 173

note that if λ∗ is a solution of

minn∑

l=1

λlbl,

λl > 0,

n∑l=1

λl < 1,

n∑l=1

λlal = x

then for some µ real

bl = µal, 1 ≤ l ≤ n

and in this case∑

l λlbl = µx for every λ and therefore for some λ with λl = 0as well. Now consider the problem

minn∑

l=1

λlbl,

λl ≥ 0,n∑

l=1

λl = 1,n∑

l=1

λlal = x,

and consider the solution λ∗ with the greatest number of zero coefficients λ∗l .

From the previous case we see that n = 2. Define α = ak and β = al, we maywrite

g(θα + (1 − θ)β) = θf(α) + (1 − θ)f(β),

θα + (1 − θ)β = x.

Without loss of generality α < x < β. Now consider y, y′ such that α < y < x <y′ < β. There are a, b ∈ (0, 1) such that aα+(1−a)x = y and bx+(1−b)β = y′.Namely a = (x − y)/(x − α) and b = (β − y′)/(β − x). Then

g(y) ≤ ag(α) + (1 − a)g(x),

g(y′) ≤ bg(x) + (1 − b)g(β).

Moreover, r := y′−xy′−y ∈ (0, 1) is such that ry + (1 − r)y′ = x. Then

g(x) ≤ rg(y) + (1 − r)g(y′)

≤ rag(α) + (r(1 − a) + (1 − r)b)g(x) + (1 − r)(1 − b)g(β), (D.2)

and thus,

g(x) ≤ ra

ra + (1 − r)(1 − b)f(α) +

(1 − r)(1 − b)ra + (1 − r)(1 − b)

f(β)

=−β + x

−β + αf(α) + − x − α

−β + αf(β)

= θf(α) + (1 − θ)f(β) = g(x).

Thus the inequalities in (D.2) are equalities and this ends the proof.

174 Convexity

We now consider the one-sided derivative of a convex function.

Proposition 12 If g : [0, 1) → R is convex then for every x ∈ [0, 1) thereexists

g′+(x) = limr↓0

g(x + r) − g(x)r

. (D.3)

Moreover, the function x → g′+(x) is increasing.

Proof: First note that if a < b and c < d, a ≤ c, b ≤ d then

g(b) − g(a)b − a

≤ g(d) − g(c)d − c

.

To see this consider first a = c. That is a = c < b ≤ d. Then if r = (d−b)/(d−a)we have that

g(b) = g(ra + (1 − r)d) ≤ d − b

d − ag(a) +

b − a

d − ag(d).

The case b = d is analogous. Thus we have that

g(b) − g(a)b − a

≤ g(d) − g(a)d − a

, a < b < d.

Therefore, the ratio (g(b) − g(a))/(b − a) decreases with b > a for every a.The same is therefore true of (g(d) − g(c))/(d − c). We see from (D.3) that(g(d) − g(c))/(d − c) has a lower bound (namely (g(b)−g(a))/(b−a)). Thereforethere exists

g′+(c) = limd↓c

g(d) − g(c)d − c

.

We see also from (D.3) that g′+(a) ≤ g′+(c) whenever a < c.

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Index of Notations

I, I = 1, 2, . . . , n, the set of bidders 6F (·), a distribution function 143–4G, a Bayesian game 7g o f , composition of g and f 6N, the set of natural numbers 5R, the set of real numbers 5R+, the set of non-negative real

numbers 5x−i 5

x1 ∨ x2 ∨ . . . ∨ xn 5

x ∧ y, minimum of x and y 59

x+, the positive part of x 5

X1 × X2 × . . . × Xn, Cartesian productof Xi, . . . , Xn 5

[0, v], set of types 11

Y, Y = Z2 ∨ . . . ∨ Zn 6

#X, number of elements of X 6

179


Index of Proper Names

Ashenfelter, O. 141, 175Athey, S. 141, 175

Baron, D. 1, 175Bulow, J. 82, 175Burtraw, D. 176

Campbell, W. 175Capen, E. 39, 175Cassady, R. 9, 175Chung, K. 163, 175Clapp, R. 175Cremer, J. 98, 175

Dowrick, S. ixDutra, J. ix

Engelbrecht-Wiggans 2, 175Evans and Peck 142, 175

Fudenberg, D. viii, 175fux Svaiter, B. 143, 177

Goeree, J. 176Graddy, K. 141, 175Grant, S. ix, 11, 175Green, J. 176

Haile, P. 141, 175Harsanyi, J. 6, 176Holt, C. 142, 176

Jones, C. 141, 176

Kagel, J. 142Kaiji, A. 175Karlin, S. 48, 167, 176Klemperer, P. viii, 1–3, 141, 142, 176Krishna, V. 2, 3, 141, 170, 176

Laffont, J. 2, 176Lebrun, B. 124, 176

MacLean, R. 98, 175Malam, C. ixMas-Colell, A. viii, 176Maskin, E. 2, 176McAfee, R. 2, 98, 176, 177McMillan, J. 2, 176Menezes, F. 141, 175, 176Milgrom, P. 2, 10, 12, 20, 57, 67, 69,

141, 143, 177Monteiro, P. K. 143, 177Myerson, R. 1, 71, 175, 177

Norman, G. ix

Palmer, K. 176

Reny, P. 98, 177Riley, J. 2, 36, 176, 177Rinott, Y. 167, 176Roberts, J. 82, 175Rockafellar, T. 171, 177Roth, A. vii, 176, 177Rubin, H. 176Ryan, M J 175

Samuelson, W. 36, 177Shobe, W. 176Sonnenschein, H. vii, 177Sutherland, L. ix

Tirole, J. viii, 2, 175, 176Tremblay, M. 124, 176

Vella, F. 141, 176Vickrey, W. viii, 18, 177

Weber, R. 11, 12, 20, 57, 67, 69, 177Whinston, M. 176Wilson, R. vii, 2, 177, 178Wolfstetter, E. 2, 178

181


Index

ANU, Australian National University ixauction:

all-pay 36ascending 10button 11, 20common values 12, 142descending 10discriminatory 121distribution hypotheses 143–9dominant strategy 18Dutch 11English 10, 67equilibrium, first-price 15, 148–9first-price 11

revenue 17hybrid 141Japanese 20literature on 141–3open 10optimal 82, 88, 146–8practical design 142–3private values 12, 142sealed-bid 10second-price 11

expected revenue 19strategically equivalent 17uniform price 121Vickrey 11war of attrition 36

Bayesian Game 7Bayesian Nash, see equilibrium

CNPq, Conselho Nacional de Pesquisasix

common value, see auctionconvex hull 88, 171

dense set of functions 103

density of a distribution 155distribution hypotheses 143–9dominant strategy, see auction

entry fee 22EPGE, Escola de Pos-Graduacao em

Economia ixequilibrium:

Bayesian Nash 7discriminatory auction:

negative synergy 123positive synergy 121

dominant strategies 18first-price 148–9first-price with affiliation 51, 65first-price with common value 41first-price with risk aversion 33first-price without independence 29second-price with affiliation 51second-price with common value 42sequential auction 119symmetric 7uniform price, positive synergy 127

feasible allocation rule 78FGV, Fundacao Getulio Vargas ixfunction:

continuously differentiable 6increasing 6strictly increasing 6symmetric 26, 58

Gram-Schmidt 106

increasing function, see functionintegrating factor 28, 44, 52, 165IPV, independent private values 12iterated expectations 164

183

184 Index

linkage principle 69

marginal valuation 82, 85mechanism:

allocation rule 72auction rule 72direct 75English auction 73first-price 72incentive compatible 75individually rational 75

monotone likelihood ratio 47multivariate monotone likelihood 59

negative synergy 118

optimal auction, see auctionorthogonalization, see Gram-Schmidt

permutation 26positive synergy 118private values, see auction

random variables 154

affiliated 48, 59distribution 155independent 158

random vector 156distribution 156

regret 41reserve prices 22revelation principle 75revenue equivalence 20, 79, 80, 114,

144–6

second highest value:density 160distribution 160

shading 17Stone-Weierstrass 104strategically equivalent, see auctionstrictly increasing function, see function

uniform distribution 153, 155uniform price auction 126

winner’s curse 39

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