Bundling As An Optimal Selling Mechanism For A Multiple-Good

Monopolist

Alejandro M. Manelli ∗

Department of Economics

College of Business

Arizona State University

Tempe, AZ 85287-3806

and

Daniel R. Vincent ∗

Department of Economics

University of Maryland

College Park, MD 20742

Draft: March 6, 2002

(This version is preliminary and incomplete. Comments are welcomed.)

Abstract

Zero-one mechanisms are generalized take-it-or-leave-it mechanisms that consist of a pricefor each possible collection of goods. We illustrate that previously suggested conditions for whensuch mechanisms are optimal in the class of all incentive compatible and individually rationalmechanisms are not strong enough. We provide necessary conditions for a schedule of pricesto be optimal within the class of take-it-or-leave-it prices. We utilize the linear programmingtheorem along with these necessary conditions to determine sufficient conditions for when suchoptimal zero-one mechanisms are optimal mechanisms over the class of all incentive compatibleand individually rational mechanisms. We illustrate these conditions in applications to two-goodand three-good cases.

Keywords: auctions, multi-dimensional incentive compatibility, adverse selection.

∗We are grateful to Roberto Munoz for excellent research assistantship and to R. Preston McAfee for discussing

the relationship of this paper with McAfee and McMillan (1988). The work was funded, in part, by NSF Grant

01-5-23853.

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1 Introduction

A commonly used and simple method of price discrimination when selling multiple indivisible,distinct objects is to offer them in bundles. The most general form of bundling is to post a scheduleof prices, one for each possible collection of the objects, and allow the buyer the opportunity toselect which price/bundle pair he prefers. In this paper, we derive necessary conditions for theoptimality of such a schedule of prices for any number of objects the seller has to sell. We usethese conditions to determine when bundling in this manner is optimal over the class of all feasibletrading mechanisms. Thus, we offer a partial answer to the question: what is the best way to selln indivisible goods to a single buyer who has linear utility for the goods and money and whosevaluation for each of the objects is privately known?

In the case where n = 1, the answer to this question has long been known. An appropriatelyselected take-it-or-leave-it offer to the buyer generates the highest expected revenue among allincentive compatible and individually rational mechanisms.1 ‘Bundling’ is the natural extensionof this result for the case n > 1. In its most general form, the policy of bundling is a 2n − 1‘take-it-or-leave-it-offer’ mechanism — each element in the vector corresponds to a price for one ofthe 2n − 1 possible subsets of goods that the seller could offer. 2 Because these mechanisms implythat a buyer of a given type will buy a given subset with probability that is either zero or one,these mechanisms can be thought of as ‘zero-one’ mechanisms. These mechanisms are ‘simple’ inthat randomization in the assignment of goods is not used. We denote this practise by the morerecognizable term of bundling and show circumstances under which bundling is optimal over allpossible incentive compatible and individually rational (IC and IR) mechanisms.

McAfee and McMillan (1988)suggested a solution for the case with two goods and left openthe question of optimal mechanisms for more than two goods. We show that their solution fortwo goods is incorrect and the sufficient condition that they offer is not strong enough. 3 Thecounterexample we provide illustrates that, unlike the one good case, some environments requirerandomized assignments for optimality. The question then remains: When is bundling an optimalmechanism? To answer this question, we first characterize the bundling prices which are optimalonly within the class of bundling prices. Such mechanisms partition the set of buyer types intoconvex subsets. For a given subset of goods, each buyer type subset represents the set of buyerswho buy that subset of goods in the mechanism. The necessary condition for optimization yields acondition that must be satisfied for each of these subsets of buyer types. We then use these condi-tions to exploit the linear programming feature of the seller’s problem in the space of mechanisms.We provide sufficient conditions for when optimal zero-one mechanisms are optimal in the class ofall IC and IR mechanisms. To illustrate the power of this approach, the results are applied to yield

1See, for example, Myerson (1981) or Samuelson (1984).2It is known from the literature on bundling that mixed bundling – prices both for individual goods and for

collections of goods – weakly dominate a schedule of prices for the individual goods alone. Thus, a mechanism with

n single good prices will typically be suboptimal. See, for example, McAfee, McMillan and Whinston (1989).3A similar counterexample was discovered independently and simultaneously by Thannasoulis (2001).

2

sufficient conditions for when ‘bundling’ is optimal, first in a class of environments for the case inwhich two goods are to be sold and then in a specific example for when there are three goods.

This work is related to research on multidimensional mechanism design. McAfee and McMillan(1988) first examined the question of when zero-one mechanisms are optimal for multidimensionaluncertainty and offered a partial characterization of conditions required for these mechanisms tobe optimal. Armstrong (1998) showed that all optimal mechanisms must generate a set of positivemeasure of buyer types who never trade. Rochet (1995) and Rochet and Chone (1997) extendedboth of these papers and illustrated that optimal mechanisms typically require ‘bunching’ (even inthe case where goods are divisible). Bunching implies that buyer types virtually always pool intoa set of positive measure of other buyer types. Rochet (1995) also offers an example of discretelydistributed buyer types in which zero-one mechanisms are suboptimal. 4

Our approach extends techniques that we developed in Manelli and Vincent (1995). Considera specific IC and IR mechanism which has some simple features in terms of its probability oftrade functions. By the duality theorem of linear programming, this mechanism is optimal if andonly if there exists a feasible solution to the dual to the seller’s optimization problem (the ‘primalproblem’)which yields a dual value equal to the primal value given by the candidate mechanism.To show that a candidate mechanism (in our case, the optimal bundling prices) solves the overallmechanism design problem, we construct feasible dual variables that achieve the same value in thedual problem as the proposed mechanism achieves in the primal problem.

The sufficient conditions for bundling to be optimal even for the case n = 2 are quite strong.This is no accident. In a companion paper, Manelli and Vincent (2001), we offer an analysis of thefeasible set for the seller’s problem. This set is a closed convex set and because the seller’s objectivefunctional is linear, the solution set will always contain an extreme point of the feasible set. In thecase of n = 1, the extreme points of the set of IC and IR mechanisms is, in fact, the set of zero-onemechanisms. Thus, the well-known results follows almost immediately. The set of extreme pointsof IC and IR mechanisms in the case n > 1 is far richer and generally includes mechanisms withsignificant randomization in the allocation of objects.

This paper is a contribution to the literature on the use of bundling as a form of second degreeprice discrimination. However, the results may point to solutions to a broader, open question. If n =1, it is known that the optimal take-it-or-leave-it offer to a single buyer corresponds to the optimalreserve price in an independent private values auction environment with m buyers and, further,that such auctions are optimal over the class of all IC and IR mechanisms . 5 The use of the linearprogramming techniques that we make in this paper offers the opportunity to extend the results inan analogous way. Specifically, what is the optimal mechanism for selling n indivisible goods to m

buyers? Since randomized mechanisms are more complicated than the deterministic mechanismsgenerated by bundling, the understanding of when these simpler mechanisms are optimal with a

4This was not a counterexample to the McAfee and McMillan result since their model and their sufficient condition

was restricted to the case of types whose distributions could be represented by densities.5See Myerson, (1981) or Riley and Samuelson, (1984), for example.

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single buyer is important to extending the theory of optimal auctions to the case of more than oneobject. We are currently working on extending the analysis in this direction.

2 The Model and A CounterExample

A seller with n different objects attempts to trade with a single buyer. The buyer’s preferencesover consumption and money transfers are given by

U(x, q, t) = x · q − t,

where x is the vector of buyer’s valuations, q is the quantity consumed of each good, and t is themonetary transfer made to the seller. Since the buyer has demand for at most one unit of eachgood, the vector q is an element of {0, 1}n; for ease of notation we assume that x is an element ofIn where I = [0, 1], and t is in IR.

Valuations for each object are the buyer’s private information. The seller does not observethe buyer’s valuation but knows only that valuation x is distributed according to a prior densityfunction f(x). The density function is common knowledge and represents the seller’s beliefs aboutthe buyer’s private information. We require that the densities are such that f(x) > 0 for allx ∈ In and ∂f(x)

∂xi≡ fi(x) is continuous for all i, x ∈ In. In characterizing sufficient conditions

for a take-it-or-leave-it mechanism to be optimal, we will eventually need to impose some furtherrestrictions on f . For reference, we list, here, conditions that will be invoked at various points inthe analysis, however, unless otherwise stated, it is to be assumed only that f is strictly positiveand continuously differentiable over I2.

(Ind) f(x) = Πni=1f

i(xi).

(Mhr1) ∀i∀x, f(x) + xifi(x) ≥ 0.

(Mhr2) ∀i∀x, xifi(x)/f(x)increasing in xi.

(Mhr1) implies that (n + 1)f(x) +∑n

i=1 xifi(x) ≥ 0 which is the sufficient condition invoked byMcAfee and McMillan (1988). Mhr2 is less familiar. As far as we know, it is not a commonlyimposed restriction, however, the class of distributions, Πn

i=1xαii , αi ≥ 0 satisfies the condition as

well as Mhr1 and Ind as does the class of distributions, KΠni=1(e

αixi − 1), αi ≥ 0,K > 0.6

In searching for an optimal mechanism, one may restrict attention to direct revelation mech-anisms where buyers report their types truthfully. A direct revelation mechanism is a pair offunctions

p : In −→ In

t : In −→ IR,

6The family of beta distributions also satisfy Mhr2 for β ≥ 1, however, neither the normal nor the Gamma

distributions satisfy the condition. See DeGroot, 1970, pp. 37-40.

4

where pi(x), the ith component of p(x), is the probability that the buyer will obtain good i when hisvaluation is x, and t(x) is the transfer made by the buyer to the seller when valuations are x.7 Inaddition, the buyer must have adequate incentives to reveal his information truthfully—incentivecompatibility (IC)—and to participate in the mechanism voluntarily—individual rationality (IR).The buyer’s expected payoff π(x′|x) under the mechanism (p, t) when the buyer has valuation x

and reports x′ isπ(x′|x) = p(x′) · x− t(x′).

For ease of notation, π(x|x) is denoted π(x). Then, (p, t) must satisfy

(IC) ∀x, π(x) ≥ π(x′|x) ∀x′

(IR) ∀x, π(x) ≥ 0.

(As stated, the constraints hold everywhere; it suffices that they hold almost everywhere.)We summarize below some readily available properties of IC and IR mechanisms that have been

noted and used in the literature.8

1. If (p, t) is a mechanism that satisfies IC, then the buyer’s expected payoff π(x) is convex, andits gradient ∇π(x) =

[∂π(x)∂xi

]belongs to In for most x ∈ In. Indeed ∇π(x) = p(x) almost

everywhere.9

2. If ϕ(x) is a convex function, and its gradient ∇ϕ(x) =[

∂ϕ(x)∂xi

]∈ In for most x ∈ In, then

there exists a mechanism (p, t) satisfying IC. The mechanism is defined by p(x) = ∇ϕ(x)almost everywhere, and t(x) = p(x) · x− ϕ(x). Under these definitions, π(x) = ϕ(x).

Intuitively, a mechanism is IC if and only if the corresponding buyer’s payoffs are convex, withpartial derivatives between zero and one. Furthermore, the partial derivative of the payoff of abuyer with type x with respect to his valuation for good i represents the probability that the buyerof type x receives good i in equilibrium.

The preceding properties completely characterize IC mechanisms in terms of the buyer’s expected-payoff function π(x). Individual rationality requires that π be non-negative. Since the objectiveis to find an optimal policy for the seller, and since the buyer’s expected payoff is non-decreasing,there is no loss of generality in restricting attention to payoff function where π(0) = 0.

Let F be the set of continuous functions, π : In → IR with the dual space, F∗. Let C be theclosed, convex cone in F given by

C ={π ∈ F | π(x) is convex, ∇π(x) ∈ IRn

+ a.e., and π(0) = 0}

.

7In order to compute expected payoffs, the functions p and t must be integrable.8See Rochet (1985), Armstrong (1996), and Jehiel, Moldovanu, and Stacchetti (1998), Krishna and Maenner

(2000).9Since π(x) is convex, it is a continuous function, and almost everywhere differentiable so its gradient is defined

almost everywhere .

5

Thus, the set of IC, IR mechanisms is the set of buyer expected utility functions, π ∈ C,∇π(x) ≤1.10

Given any IC, IR mechanism π, a buyer with type x receives a payoff π(x) = ∇π(x) · x− t(x).Thus, the seller’s expected revenue when using the mechanism π(·) is

E[t(x)] =∫

In

(∇π(x) · x− π(x)) f(x)dx.

and the seller’s problem is to select a mechanism π ∈ C to maximize expected revenue,

maxπ∈C,∇π≤1

∫

In

(∇π(x) · x− π(x)) f(x)dx. (1)

This paper is concerned with a special class of IC and IR mechanisms. A zero-one mechanismwith monitoring can be characterized by a 2n vector of prices, each price PJ corresponds to theprice a consumer has to pay in order to purchase the subset of goods, J . Note that, since the sellercan observe the bundles selected by the buyer, she can prevent the buyer from constructing her ownlarger bundle from subsets of that bundle. The assumption that the seller can monitor purchasesimplies that, if the seller chooses, she can offer a price profile such that PJ∪I > PJ + PI and stillensure that if the bundle J ∪ I is purchased, a price of PJ∪I is received. 11

A zero-one mechanism is an IC and IR mechanism. To gain an understanding of these mech-anisms, consider the case n = 2. Here, a zero-one mechanism is generally a triple of prices,{P10, P01, P11} where {10, 01, 11} represents the set of the three (nontrivial) bundles and PJ is theprice of bundle J . The buyer expected utility function induced by this mechanism can be picturedin Figure 1. Buyer types are distributed over the unit square. The square is partitioned into fourregions. For example, A10 = {x : x1−P10 ≥ x2−P01, x1−P10 ≥ x1 +x2−P11, x1−P10 ≥ 0} is theset of buyer types who buy the bundle consisting of good one alone. Generic level sets of a buyerutility function, π that is generated by a zero-one mechanism take the same shape as the dottedline. Buyers in the set A00 gain zero utility. Within the class of zero-one mechanisms, altering anyof the prices typically changes the size and position of any of these sets, {AJ} but their shapesremain roughly similar.

McAfee and McMillan (1988) argue that for n = 2 and if 3f(x) +∇f(x) · x ≥ 0 then zero-onemechanisms are optimal within the class of all IC and IR mechanisms. Consider the followingcounterexample. In the unit square, let f(x) be uniform over the region above the line joiningthe points (0, 1) and (1, .5) and zero elsewhere. Since ∇f(x) = 0 a.e., the McAfee and McMillancondition is satisfied a.e. (The gradient is undefined on the measure zero line defining the lowerboundary of the support but notice that, since f(x) is everywhere non-decreasing, any continuousapproximation of f(x) would have to satisfy the condition as well.) For the case of buyer types withthis density, direct calculations show that the optimal zero-one mechanism is, in fact, a two pricemechanism, P01 = 0.85, P11 = 1.26.12. This mechanism generates expected revenue 0.291. Consider

10Functions such as .5(Pn

i=1 x2i ) illustrate that the constraint set has a non-empty interior.

11For the remainder of the paper, we refer zero-one mechanisms with monitoring simply as ‘zero-one mechanisms.’12The proof is provided in Appendix 6.1

6

-

6

x1

x2

@@@@@@@@

1P01P11 − P01

P11 − P10

1

P01

A00

A01 A11

A10

Figure 1: A Zero-One Mechanism

an alternative mechanism which offers the full bundle of goods for the same price, P11 = 1.26, butinstead of offering good two alone, it offers a randomized product consisting of good two withprobability one and good one with probability one-half for a price of exactly 1. Notice that thismechanism is more efficient than the optimal zero-one mechanism because when the latter does notsell the full bundle, it never sells good one. The randomized mechanism offers at least a chance atgood one. This increased efficiency also raises seller revenues which now are 0.298. 13

The intuition for why the McAfee and McMillan condition is insufficient lies partly in the failureof the condition to rule out negative correlation across valuations. A modified example helps toillustrate this. 14 Suppose that buyer types are uniformly distributed on the line joining the points(0, 1) and (1, .5) (rather than distributed over the set above the line.) In this example, buyer’s twovaluations are perfectly negatively correlated. A two price mechanism will be optimal within theclass of zero-one mechanisms and any such mechanism can be pictured by a horizontal line at priceP01 < 1 and a 45o line satisfying x1 + x2 = P11 ≥ 1. Refer to Figure 2. This mechanism fails to

13For computational ease, the counterexample offered in this section violates the continuous differentiability condi-

tion (on a manifold of zero-measure.) However, Appendix 6.1 shows that revenues are continuous in measures which

approximate the one in the counterexample and which are continuously differentiable so there exist densities which

satisfy continuous differentiability everywhere and yield similar results.14This is not a true counterexample because any continuous approximation to the density must be steeply non-

monotone over the unit square and will, therefore, violate the McAfee and McMillan condition.

7

sell to agents on the line between points B and C. The randomized mechanism described above,sells the full bundle to the same types as the zero-one mechanism but sells the random bundle toeveryone else at a price of 1 > P01 and therefore dominates the zero-one mechanism.

-

6

x1

x2

HHHHHHHHHHHHHHHHHHHHHHHHHH

@@@

1P11 − P01

1

P01

C

B

A01

A11

Figure 2: A Zero-One Mechanism Which Can Be Dominated

The next section offers a more detailed characterization of zero-mechanisms which can be usedto rule out examples of this type.

3 Features of Zero-one Mechanisms

3.1 Some Characteristics of the Partition of Buyer Types

Casual observation suggests that sellers often set prices for the different bundles leaving consumersonly with the choice of what bundle to purchase. This practice constitutes an institutional repre-sentation of an IC and IR zero-one mechanism, a mechanism where the buyer either gets a bundlewith probability one or with probability zero after truthfully reporting his/her valuation. In thissection we identify necessary conditions for a set of prices, one for each bundle, to maximize ex-pected revenue within the class of zero-one mechanisms. In the next section we use these conditionsto determine when zero-one mechanisms are optimal over all IC and IR mechanisms.

As already noted in the literature (McAfee, McMillan and Whinston (1988)), it is important

8

to specify whether or not the seller can keep the buyer from constructing his own bundles byaggregating sub-bundles into larger sets of goods. We consider the case where the seller has thispower. When the seller is facing a single buyer, such power seems natural. Later in this section,we consider solutions to the seller’s problem with the feature that the buyer has no incentive toengage in aggregation.

Index the n goods by i = 1, . . . , n. Let N represent the set of available goods. Any bundleJ ⊂ N can be represented by an n-dimensional vector of zeros and ones, aJ = (aJ

1 , aJ2 , ..., aJ

n) whereaJ

i takes the value 1 if i ∈ J and the value 0 otherwise. Both representations of a bundle will beuseful in the analysis. The utility of a buyer of type x ∈ In who acquires the bundle J ⊂ N at pricePJ is aJ · x− PJ . The notation (xi, x−i) denotes a vector of valuations with the i′th component asxi. Similarly, for any subset, J with cardinality j, (xJ , xN/J) denotes the valuation vector with thecomponents corresponding to valuations for the goods in J given by the j-vector, xJ . For i ∈ J ,(xi, xJ/i, xN/J) is the valuation vector with valuation xi for good i and the j−1 vector, xJ/i, yieldsvaluations for the remaining goods in J .

A pricing rule P stipulates a price PJ for each possible bundle of goods J ⊂ N . (We adopt theconvention that P∅ = 0.) Given a pricing rule P , let

AJ = {x ∈ In | aJ · x− PJ ≥ aK · x− PK ∀K ⊂ N}.

Thus, AJ represents the set of buyer types that will buy the bundle J given the pricing rule P .(Note that AJ or AaJ represent the same set.)

To determine some properties of AJ used in the paper, note that the valuation x ∈ In is in AJ

if and only if (aJ − aK) · x ≥ PJ − PK for all K ⊂ N . It follows that AJ is the intersection ofIn with finitely many half spaces in Rn, and it is therefore convex and compact. The collection ofsets, {AJ}J∈S forms a partition of In. 15

If given any two bundles J and K, AJ ∩ AK 6= ∅, then AJ ∩ AK is a subset of the hyperplane{x | (aJ − aK) ·x = PJ −PK}. Thus aK − aJ is the outer normal of the set AJ along the boundarygiven by AJ ∩ AK . For instance, let n = 3 and consider the set A(1,1,0) where aJ = (1, 1, 0)corresponds to the bundle J = {1, 2}. The outer normal of A(1,1,0) along its boundary with A(0,0,0)

is the vector, (−1,−1, 0).Fix a bundle J and its corresponding set of types AJ . For each i ∈ J ,

BiJ = {x−i ∈ In−1 | (1, x−i) ∈ AJ}

represents the intersection of AJ with the boundary of In along the coordinate i = 1.The definition of AJ implies that its indicator function is increasing in xi, i ∈ J .

Lemma 1 For i ∈ J, if (xi, x−i) ∈ AJ then (x′i, x−i) ∈ AJ for all x′i > xi.For i /∈ J, if (xi, x−i) ∈ AJ

then (x′i, x−i) ∈ AJ for all x′i < xi.

15This is an abuse of terminology since sets AJ and AK often share a boundary. However, such boundaries have

lebesgue measure zero in IRn so this informality is inconsequential. When we refer to partitions over subsets of IRn

in this paper, we typically allow for such deviations from the formal definition.

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Proof (xi, x−i) ∈ AJ implies that aJ · x − PJ ≥ aK · x − PK for all K. If i ∈ K, then raising xi

to x′i raises both the left and right side without affecting the inequality. If i /∈ K, then raising xi

to x′i increases the left side but the right remains fixed. The argument for the second statement issimilar. Q.E.D.

The lemma implies that for i ∈ J, if (xi, x−i) ∈ AJ then (1, x−i) ∈ AJ . Conversely, if x−i is suchthat x−i /∈ Bi

J , then there does not exist any xi such that (xi, x−i) ∈ AJ . Finally, for i ∈ J, BiJ is

empty if and only if AJ is empty.In general, the construction of the sets, AJ requires the comparison of utility obtained from

AJ with the utility obtained from purchasing any other set K. If the profile of prices has someadditional structure, though, the number of relevant comparisons is much smaller. Consider twopotential conditions that P might satisfy:

(WSM) ∀I, J ∈ S, PI∪J < PI + PJ .

(SSM) ∀I, J ∈ S, PJ∪I ≤ PJ + PI − PI∩J

and if I ∩ J 6= I, J, PJ∪I < PJ + PI − PI∩J .

The conditions may be thought of as weak and strong submodularity conditions. If WSM issatisfied, then there is no incentive for a buyer to attempt to purchase bundle I ∪ J by purchasingI and J separately even if I and J are non-overlapping. If SSM holds, then the incentives for adifferent sort of arbitrage vanish. Suppose that a buyer could buy or sell at the outstanding prices,P . If I and J overlap (that is, I ∩ J 6= ∅) and the condition is violated, a buyer could form bundleI ∪ J more cheaply by buying I and J separately and then selling back the duplicated goods inI ∩ J. For n = 2, WSM implies SSM but the implication is not true in general. It is important tonote that we do not impose this as a constraint on the seller, rather, the condition considers simplywhat is the nature of the sets, AJ , for any price profile such that these inequalities are satisfied.16

(Posing the conditions as a constraint corresponds to zero-one mechanisms without monitoring andwould require modifying the seller’s program in Theorem 1 to account for this limitation on herpricing ability.)

Lemma 2 Let P be a 2n − tuple of prices that satisfy WSM . Then x ∈ AJ if and only if i)aJ · x− PJ ≥ 0, and ii) for all M, M ∩ J 6= ∅, aJ · x− PJ ≥ aM · x− PM .

Proof (Only if) Condition i) is clearly necessary by individual rationality and if ii) fails, thenutility is higher buying M than I.

(If) The condition aJ · x− PJ ≥ 0 implies that some bundle will be purchased. Suppose that I

is contained in N/J so that aJ∪I = aJ + aI . We show that aJ · x − PJ ≥ aJ∪I · x − PJ∪I impliesaJ ·x−PJ > aI ·x−PI . By WSM , aJ∪I ·x−PJ∪I = aJ ·x+aI ·x−PJ∪I > aI ·x−PI +aJ ·x−PJ ≥aI · x− PI where the last inequality comes from condition i). The only other alternatives a buyer

16In particular, the strong condition does not seem to be true in optimal zero-one mechanisms in general, however,

we can show, computationally, that in the case of independent and identically distributed valuations with F i(xi) =

xαi , n = 3 for α ≤ 3, optimal zero-one mechanisms satisfy this condition.

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has to consider, are bundles which do not belong to N/J and condition ii) covers all those relevantcomparisons. Q.E.D.

This lemma indicates that, under weak submodularity, in constructing the sets, AJ , we need onlyconcern ourselves with the comparisons of J and any other bundles with non-null intersections withJ. This means that with n goods, for any subset of size m, we compare it with 2n−1−(2n−m−1) =2n−m(2m − 1) other subset prices.

The next lemma shows that if we restrict attention only to profiles of prices which are stronglysubmodular, then the comparisons are even fewer.

Lemma 3 Suppose P is such that SSM holds. Then AJ ∩AK 6= ∅ implies J ⊂ K or K ⊂ J .

Proof Let x ∈ AJ ∩AK . From Lemma 2, we have

aJ · x− PJ ≥ aJ∩K · x− PJ∩K

aK · x− PK ≥ aJ∪K · x− PK∪J

ThereforeaJ · x− PJ + aK · x− PK ≥ aJ∩K · x− PJ∩K + aJ∪K · x− PK∪J

But aJ + aK = aJ∩K + aK∪J . Thus, we have PK∪J ≥ PJ + PK − PJ∩K which violates SSM forK ∩ J 6= K, J . Q.E.D.

This result simplifies the comparisons further because, along with Lemma 2, it implies that forany J, we need only compare the purchase of J with any K such that either K ⊂ J or J ⊂ K. Theresult also yields a type of independence of the set of valuations for the goods outside of the set J

from the valuations for the goods in J as follows.

Lemma 4 Suppose SSM holds. Let x = (xJ , xN/J) ∈ AJ , x′ = (x′J , x′N/J) ∈ AJ . Then x =(x′J , xN/J) ∈ AJ .

Proof By Lemma 3, to determine if x ∈ AJ we need only compare bundle J with K,K ⊂ J orJ ⊂ K. First consider K ⊂ J. The inequality x ·aJ−PJ ≥ x ·aK−PK is independent of xN/J so thedominance of AJ over AK holds for x. Now consider J ⊂ K and suppose x · aJ −PJ < x · aK −PK .

Since J ⊂ K, x · aK − x′ · aJ + x · aJ = x · aK and x · aJ − x′ · aJ + x · aJ = x · aJ . Therefore,x · aJ −PJ < x · aK −PK implies x · aJ −PJ < x · aK −PK which contradicts the assumption thatx ∈ AJ . Q.E.D.

This result implies that if valuations for each good are independently distributed, holding anybuyer valuations for goods in bundle J, xJ , fixed, the probability that the buyer has valuations forgoods outside of J such that J is his preferred bundle is independent of xJ .

For any pricing rule, P , for any bundle, J , of cardinality, j, and for any i ∈ J define

AJJ ≡ {xJ ∈ Ij | ∃y ∈ In−j , (xJ , y) ∈ AJ},

AN/JJ ≡ {y ∈ In−j | (xJ , y) ∈ AJ , xJ ∈ AJ

J}Di

J ≡ {z ∈ Ij−1 | (1, z) ∈ AJJ}.

11

Observe that, by Lemma 4, if the pricing rule satisfies SSM , for any xJ ∈ AJJ and any y ∈ A

N/JJ ,

(xJ , y) ∈ AJ . Thus AN/JJ can be expressed independent of xJ . (This is not true in general.)

Then y ∈ AN/JJ , z ∈ Di

J implies (y, z) ∈ BiJ . Therefore, under SSM , we can decompose Bi

J intoindependent component sets, Bi

J = AN/JJ ×Di

J .By Lemma 1, we can define the functions xi : In−1 → IR by

xi(x−i) = min{xi | (xi, x−i) ∈ ∪J⊂N,i∈JAJ}. (2)

If the AJ ’s are generated by a pricing rule satisfying SSM , Lemma 4 implies that the integral ofany φ over AJ can be expressed, after selecting any i ∈ J , as

∫

AJ

φ(x)dx =∫

AN/JJ

∫

DiJ

∫ 1

xi(x−i)φ(x)dxidxJ/idxN/J . (3)

3.2 First Order Necessary Conditions

Theorem 1 identifies a necessary condition for the optimality of a pricing rule P , when the seller’sstrategies are restricted to the selection of a pricing rule. The condition uses the density f(x)governing the distribution of valuations to characterize a relationship between the interior of eachset, AJ , and its outer boundary, ∪i∈JBi

J .

Theorem 1 Let P be a pricing rule generating {AJ}J⊂N . If P is optimal among zero-one mech-anisms, then for all J 6= ∅

∫

AJ

[(n + 1)f(x) +n∑

i=1

xifi(x)]dx−∑

i∈J

∫

BiJ

f(1, x−i)dx−i = 0.

Proof Note that if AJ has zero measure, then the condition holds trivially, so we restrict theanalysis to subsets with strictly positive measure. Also, a P such that PJ = 0 for any J cannot beoptimal, since this would imply that the seller gains zero on buyers who purchase J and can alwaysdo better by charging a slightly higher price (See Armstrong (1996) for a fuller discussion.) For agiven profile of prices, P, the revenue function is given by

R(P ) =∑

J∈S

PJ

∫

AJ

fdx.

Let SJ be the set of bundles, K, such that {x | (aJ − aK) · x + PK − PJ = 0} ∩ AK has positivelebesgue measure in IRn−1. Since PJ > 0, a necessary condition for the optimal choice of P is, forall J,

∂R(P )∂PJ

= 0 =∫

AJ

fdx +∑

K∈SJ

PK∂

∂PJ

∫

AK

fdx.

Recall that the set AJ ∩AK is a subset of the hyperplane given by ∆ = {x ∈ IRn | (aK − aJ) · x =PK−PJ}. For each AK with AK∩AJ 6= ∅, for K 6= J , there is an i ∈ K∪J such that (aK

i −aJi ) 6= 0.

12

Then, xi = g(x−i) where g(·) is a linear function defined on proj−i(∆). Given two points x and y

∈ ∆, let d(x−i, y−i) = ‖((g(x−i), x−i)− (g(y−i), y−i)‖. Let λKJ be the (N-1)-dimensional Lebesguemeasure with support ∆ obtained from using the distance d(·, ·). The partial derivative can thenbe written as (see Appendix 6.2)

0 =∫

AJ

fdx +∑

K∈SJ

(PK − PJ)∫

AK∩AJ

fdλKJ . (4)

Note that x ∈ AK ∩AJ implies

x · aJ − PJ = x · aK − PK

This implies x · (aK − aJ) = PK − PJ . Therefore, we have

0 =∫

AJ

fdx +∑

K∈SJ

∫

AK∩AJ

fx · (aK − aJ)dλKJ .

Let H(x) ≡ (x1f, x2f, ..., xnf) so fx · (aK − aJ) = H · (aK − aJ) and recall that (aK − aJ) is theouter normal (away from the set AJ) of the set AJ ∩AK . Denote it by n(AK ∩AJ). Thus we have

0 =∫

AJ

fdx +∑

K∈SJ

∫

AK∩AJ

H · n(AK ∩AJ)dλKJ . (5)

The outer normal of the set 1×BiJ , n(Bi

J), is the vector with one in the i′th coordinate and zeroeseverywhere else. Thus,

f(1, x−i) = H(1, x−i) · n(BiJ).

The term ∑

K∈SJ

∫

AK∩AJ

H · n(AK ∩AJ)dλKJ +∑

i∈J

∫

BiJ

H · n(BiJ)dx−i

defines the integral of H · n around the boundary of AJ . (Note that, for all i, the outer normal tothe set {(0, x−i) ∈ In} is the vector n(0i) with −1 in the i′th coordinate and zeroes elsewhere soH(0, x−i) · n(0i) = 0). Since H is a C1 vector field, by the general form of the divergence theorem(Edwards, p. 380) we get 17

∑

K∈SJ

∫

AK∩AJ

H · n(AK ∩AJ)dλKJ =∫

AJ

divHdx−∑

i∈J

∫

BJi

H · n(BiJ)dx−i

=∫

AJ

n∑

i=1

(f + xifi)dx−∑

i∈J

∫

BiJ

f(1, x−i)dx−i.

Substituting this result into Equation 5 yields

0 =∫

AJ

{(n + 1)f +n∑

i=1

xifi}dx−∑

i∈J

∫

BiJ

f(1, x−i)dx−i

17The first use of the divergence theorem in these types of problems appears to be by McAfee and McMillan (1988)

(using Stoke’s Theorem) and, much more extensively, by Rochet (1995).

13

as the necessary condition for the optimal choice of PJ or, alternatively, the construction of AJ .

Q.E.D.In words, Theorem 1 states that for any set, AJ generated by the pricing scheme, P , the

integral of (n + 1)f +∇f(x) · x over the interior of AJ must equal the integral of f restricted tothe intersection of AJ with the outside ‘edge’ of the set In.

Note that for n = 1, using direct integration,

0 =∫ 1

P1

{2f(x) + xf ′(x)}dx− f(1)

= −(P1f(P1)− (1− F (P1))).

Thus, Theorem 1 generalizes the well-known necessary condition for the optimal take-it-or-leave-itprice in the case n = 1. In that case, the optimal price, P1 must be a zero of the buyer’s ”virtualutility” function, x− (1− F (x))/f(x). (Myerson, 1979 or 1981.)

In Section 4, we show how this condition can be used to characterize restrictions on f suchthat zero-one mechanisms are optimal within the class of all IR and IC mechanisms. Since thisclass of mechanisms is larger than the class of all zero-one mechanisms, under these restrictions, thenecessary conditions are also sufficient. Here, we provide two direct applications of the result. Thefirst offers a multi-dimensional generalization of the McAfee, McMillan and Whinston (1988) resultthat, within the class of zero-one mechanisms, mixed bundling generally dominates individual goodpricing. The second application shows that (Ind) and (Mhr1) are sufficient to ensure that, forn = 2, the optimal pricing rule satisfies SSM.

McAfee, McMillan and Whinston (1988) show that, for n = 2, mixed bundling often dominatesindividual pricing. In our environment, pure individual good pricing corresponds to the case wherePi < 1 for i = 1, 2, ..., n, PJ =

∑i∈J Pi for | J |> 1.Thus, we can characterize the sets, AJ = {x ∈

In | xi ≥ Pi, i ∈ J, xi < Pi, i /∈ J}.18

Theorem 2 Assume (Ind) and (Mhr1). A schedule of prices, P , such that ∀J, PJ =∑

i∈J Pi,that is, individual good pricing, is not optimal.

Proof For clarity, restrict attention to the symmetric case, f i(·) = f(·) for all i. Consider thebundle N consisting of all n goods. Observe that for any i, Bi

N = {x−i ∈ In−1 | xj ≥ Pj , j 6= i}.Thus, the necessary condition for AN (using the symmetry) can be written as

0 =n∑

i=1

∫

BiN

∏

j 6=i

f(xj){f(1)−∫ 1

P1

n + 1n

f(xi) + xif′(xi))dxi}dx−i.

This implies that at least one of the terms in the summation is non-positive. Wlog, supposethat it is for i = 1. Therefore,

0 ≥ f(1)−∫ 1

P1

n + 1n

f(x1) + x1f′(x1))dx1.

18Restricted to the case examined by McAfee, McMillan and Whinston, n = 2, conditions (Ind) and (Mhr1) are

stronger than the condition they invoke.

14

For the set consisting of good 1 alone, the necessary condition from Theorem 1 is given by

0 =∫

B11

∏

j 6=1

f(xj){f(1)−∫ 1

P1

[n + 1

nf(x1) + x1f

′(x1) +n∑

i=2

n + 1n

f(xi) + xif′(xi)]dx1}dx−1.

Condition Mhr1 and the assumption that f(·) > 0 implies that n+1n f(xi)+xif

′(xi) > xif′(xi)+

f(xi) ≥ 0. The necessary condition for AN implies that f(1)−∫ 1P1

(f(x1)/n+x1f′(x1)+f(x1))dx1 ≤

0 which yields a contradiction. Q.E.D.The final equation in the proof can be used to provide an insight into computing optimal pricing

rules when the joint distribution, f , is independent and uniform. In this case, fi ≡ 0. If the optimalpricing rule satisfies SSM , Lemma 4 implies that the necessary condition determining the priceof any single good, i, call it pi, can be expressed as

0 =∫

Bii

{1−∫ 1

pi

(n + 1)dxi}dx−i.

Solving this equation yields pi = nn+1 .

Much of the analysis of the sets of types who purchase a given bundle, J , is made simplerwhen the profile of prices satisfy the SSM condition. The next theorem illustrates some sufficientconditions for when the optimal zero-one mechanism satisfies WSM when n = 2 and, thus, satisfies,SSM .

Theorem 3 Assume (Ind) and (Mhr1) and n = 2. The optimal zero-one mechanism satisfiesWSM .

Proof See Appendix 6.3.

4 When Zero-One Mechanisms Are Optimal

4.1 Sufficient Conditions for Optimal Mechanisms

Following McAfee and McMillan (1988), since f is C1, we can employ integration by parts andrepresent the seller’s objective functional in ( 1) as

< π, T >≡∫

In

{n∑

i=1

π(1, x−i)f(1, x−i)− π(x)((n + 1)f(x) +n∑

i=1

xifi(x))}dx.

Let {PJ}J∈S denote the 2n vector of optimal prices in a zero-one mechanism with the set of buyertypes who buy bundle J given these prices as AJ . Let π(x) denote the expected utility of a buyerof type x in this mechanism.

The constraint set, {π ∈ C | 1 − ∇π ≥ 0}, is a composed mapping from C to the space offunctions, C : [0, 1]n → IRn which are piecewise-continuous. The dual space to this constraint is not

15

very convenient, algebraically. A more useful characterization of the constraint is the following. Fixa set of functions, xi(·) : In−1 → [0, 1], i = 1, 2, ..., n. Define I(x) ≡ ∑n

i=1 xi and the transformation,Gφ = I − φ so Gφ ∈ F .

Lemma 5 Suppose φ ∈ C. ∇φ(x) ≤ 1, ∀x ∈ In implies

Gφ(1, x−i)−Gφ(z, x−i) ≥ 0,∀z ∈ [xi(x−i), 1],∀x−i ∈ In−1, ∀i. (6)

Proof If xi(x−i) = 1 the inequality is trivial. Otherwise, for any i, x−i and any z ≥ xi(x−i),

φ(1, x−i) = φ(z, x−i) +∫ 1

z

∂φ(w, x−i)∂w

dw ≤ φ(z, x−i) + 1− z

where use is made of the fact that ∇φ ≤ 1. Rearranging gives us the desired expression. Q.E.D.Thus, if φ is feasible for the seller’s problem, Equation 6 holds. The set of all non-negative

functions in F , denoted by F+, is a convex cone. If φ is feasible, then Gφ(1, x−i) − Gφ(x) liesin F and a natural choice for the corresponding positive cone in the dual space might appear tobe F⊕, the set of all linear functionals, < ·, ω >, such that < φ, ω >≥ 0 for all φ ∈ F+. Forexample, if t(x) ≥ 0, then the linear functional, < φ, t >=

∫In φ(x)t(x)dx, is in the positive cone

of the dual space of all non-negative functions. However, the constraint set is somewhat morerestrictive. If φ is feasible then Gφ(1, x−i) is increasing in x−i. Since the set of all non-negativefunctions, Gφ(1, x−i) − Gφ(x) such that Gφ(1, x−i) is increasing in x−i is contained in the set ofall non-negative functions, the corresponding positive cone of the dual space to the seller’s actualproblem is larger than F⊕. This fact plays an important role in the proof of Theorems 5 and 6because the candidate dual variable we use does not lie in F⊕.

We draw from Lemma 5 to guess that the objective function in the dual to the seller’s problemcan be represented by a functional of the form < I, ω > and the constraint to the seller’s problemcan be represented by < I −φ, ω >≥ 0. That is, we guess < ·, ω >∈ F∗ is such that φ ∈ C,∇φ ≤ 1implies < I−φ, ω >≥ 0. The heart of our approach is to show that there exists a functional of thisform which indeed solves the dual to the seller’s problem. Let F∗+ represent the set of such linearfunctionals, F⊕ ⊂ F∗+ ⊂ F∗. The lagrangian corresponding to the seller’s problem can be writtenas

L(π, ω) = < π, T > + < I − π, ω >

= < I, ω > + < π, T − ω >

where the second representation helps to illustrate the dual to the seller’s problem. The comple-mentary slackness conditions for both problems can be combined to yield a sufficient condition fora candidate (feasible) π to be a solution to the primal problem. The proof is an adaptation ofLuenberger, (1969), p. 221, Theorem 2.

16

Theorem 4 Let π be feasible for the seller’s problem, ( 1). Suppose there exists < ·, ω >∈ F∗ suchthat

(CSD) < π, T − ω >= 0,

(CSP ) < I − π, ω >= 0,

(FD) < φ, T − ω >≤ 0, ∀φ ∈ C,

(PC) < I − φ, ω >≥ 0, ∀φ ∈ C,∇φ ≤ 1.

Then π is optimal in the class of all IC and IR mechanisms.

Proof Write the lagrangian evaluated at π as

L(π, ω) = < π, T > + < I − π, ω >

= < π, T >

= < I, ω > + < π, T − ω >

= < I, ω > .

The first and third equalities are by definition of L, the second equality comes from CSP and thefinal equality comes from CSD. Thus,

L(π, ω) = < I, ω >

≥ < I, ω > + < φ, T − ω >,∀φ ∈ C

= < φ, T > + < I − φ, ω >

≥ < φ, T >,∀φ ∈ C,∇φ ≤ 1.

The first inequality comes from FD, the next equality by rewriting the expression and the finalinequality comes from PC. Thus, we have, for all φ ∈ C,∇φ ≤ 1, < π, T >≥< φ, T > and, so, π

solves the constrained maximization problem. Q.E.D.A comment may help clarify the role of Theorem 4. The theorem is true, independent of whether

our guess for the form of the dual functional is correct. If our guess is incorrect, however, then we willnot be able to find such a functional which satisfies the four hypotheses of the theorem. Therefore,we are seeking to show the existence of a functional of this form which satisfies CSP, CSD,FD

and PC.If the density f satisfies (Ind),(Mhr1), and (Mhr2) and the pricing rule satisfies SSM there is

a natural candidate for the definition of < ·, ω > such that CSP,CSD, FD are fairly easily shownto be satisfied. This < ·, ω > is not typically in F⊕ so it needs to be determined if it lies in F∗+.We now proceed to define < ·, ω >. The assumptions, (Ind),(Mhr1), (Mhr2) and SSM will bemaintained for the remainder of this section, however, for some of the subsequent lemmas, they arenot always needed. The statement of the lemmas will be explicit about when they are used.

17

Define

t(x) ≡ {(n + 1) +n∑

i=1

xifi′(xi)

f i(xi)}f(x)

where f i′(x) ≡ df i(x)dxi

. Observe that (Ind) and (Mhr2) imply that the term in brace brackets is

increasing in all its arguments. Define KJ = E[∑

k/∈Jxkf ′(xk)

f(xk) | xN/J ∈ AN/JJ ] and the functions,

tJ : AJJ → IR by t∅ ≡ 0 and

tJ(xJ) = n + 1 +∑

i∈J

xifi′(xi)

f i(xi)+ KJ .

Note that if P satisfies SSM and xJ ∈ AJJ , by Lemma 6 and by construction,

∫

AN/JJ

tJ(xJ)f(xJ , xN/J)dxN/J =∫

AN/JJ

t(xJ , xN/J)dxN/J . (7)

Since t(·) ≥ 0, this implies tJ(·) ≥ 0. For x ∈ AJ , i ∈ J , define

T iJ(xi, x−i) = f(1, x−i)−

∫ 1

xi

tiJ(v, xJ/i)f(v, x−i)dv (8)

where tiJ : AJJ → IR, tiJ(xJ) ≥ 0,

∑

i∈J

tiJ(xJ) = tJ(xJ).

Setting tiJ ≡ tJ for some i ∈ J , and tkJ ≡ 0 for k 6= i, k ∈ J illustrates that there always existsa collection of tiJ ’s that satisfy the conditions in the definition. By construction, T i

J(xi, x−i) isincreasing in xi and T i

J(1, x−i) = f(1, x−i).The linear functional, < ·, ω > is defined as

< φ, ω >=n∑

i=1

∫

In−1

φ(1, x−i)f(1, x−i)dx−i −∑

J⊂N

∫

AJ

φ(x)tJ(xJ)f(x)dx.

Observe that < ·, ω >∈ F∗.The following lemma exploits the first order conditions derived in Theorem 1 and is used to

show that < ·, ω > satisfies CSP and PC. The statement employs the sequence of functions{xi(·)}n

i=1 defined in Equation 3 that characterize the ‘lower’ bounds of AJ .

Lemma 6 Assume (Ind), suppose P is an optimal zero-one mechanism satisfying SSM and let{T i

J} be any collection of functions satisfying 8. Then,

< φ, ω >=∑

J⊂N

∑

i∈J

∫

BiJ

{∫ 1

xi(x−i)[φ(1, x−i)− φ(x)]tiJ(xJ)f(x)dxi + φ(1, x−i)T i

J(xi(x−i), x−i)}dx−i

(9)and, for all J ⊂ N , xN/J ∈ A

N/JJ ,

0 =∑

i∈J

∫

DiJ

T iJ(xi(xJ/i, xN/J), xJ/i, xN/J)dxJ/i. (10)

18

Proof See Appendix 6.4.

Lemma 7 (CSD) Assume SSM and (Ind). Then < π, T >=< π, ω >.

Proof The definition of < ·, ω > implies

< π, ω > − < π, T >=∑

J⊂N

∫

AJ

π(x)(t(x)− tJ(xJ)f(x))dx.

Using the definition of tJ(xJ), since, x ∈ AJ implies that π(x) = x · aJ − PJ which does not varywith xi, i ∈ N/J we can write a generic term in the summation as

∫

AJJ

π(x){∫

AN/JJ

t(x)dxN/J − tJ(xJ)∫

AN/JJ

f(x)dxN/J}dxJ = 0.

The equality follows using Equation 7, SSM and (Ind).Therefore, < π, ω >=< π, T > since x ∈ A∅ implies π(x) = 0. Q.E.D.

Lemma 8 (FD) i) Assume (Ind) and (Mhr2). If AN/JJ is a sublattice for all J 6= ∅, then for all

φ ∈ C, < φ, T−ω >≤ 0. ii) Assume F (x) =∏n

i=1(xi)αi , αi > 0, then for all φ ∈ C, < φ, T−ω >≤ 0.

Proof By definition,

< φ, T − ω >= −∫

A∅φ(x)t(x)dx−

∑

J⊂N

∫

AJ

φ(x)[t(x)− tJ(xJ)f(x)]dx

= −∫

A∅φ(x)t(x)dx−

∑

J⊂N

∫

AJJ

∫

AN/JJ

φ(x)[t(x)− tJ(xJ)f(x)]dxN/JdxJ . (11)

Since t(x) ≥ 0,the first term is non-positive for positive φ. For all xJ ∈ AJJ , t(x) − tJ(xJ)f(x) =

{∑i6∈Jxif

i′(xi)f i(xi)

− E[∑

i6∈Jxif

i′(xi)f i(xi)

| xN/J ∈ AN/JJ ]}f(x). Under the hypothesis for part ii), the

bracketed term is identically zero and the lemma follows. Otherwise, conditions (Ind) and (Mhr2)imply that the function in the brackets is increasing in xi, i /∈ J. If f(x) is affiliated, for any twomonotone functions, φ, ψ, and any sublattice, A

N/JJ ,

∫

AN/JJ

φψfdxN/J ≥∫

AN/JJ

φfdxN/J

∫

AN/JJ

ψfdxN/J

(see Karlin and Rinott, Theorems 2.3, 4.2). Since independence satisfies affiliation, we have∫

AJJ

∫

AN/JJ

φ(x){t(x)− tJ(xJ)f(x)}dxN/JdxJ

=∫

AJJ

∫

AN/JJ

φ(x){∑

i6∈J

xifi′(xi)

f i(xi)− E[

∑

i 6∈J

xifi′(xi)

f i(xi)| xN/J ∈ A

N/JJ ]}f(x)dxN/JdxJ

≥∫

AJJ

∫

AN/JJ

φ(x)f(x)dxN/J

∫

AN/JJ

{t(x)− tJ(xJ)f(x)}dxN/JdxJ

= 0.

19

The final equality follows using Equation 7, SSM and (Ind). Thus the final terms in Equation11 are negative so (FD) is satisfied. Q.E.D.

Note that if n = 2 and P satisfies WSM then for, say, J = 10, AN/JJ = {x2 | x2 ≤ P11 − P10}

for all x1 ∈ AJJ which is a sublattice. Of course, if J = 11,AN/J

J = ∅ which is trivially a sublattice.Thus, the condition in Lemma 8i) is satisfied for the two good case .

Lemma 6 is used to prove (CSP ) as follows.

Lemma 9 (CSP )Assume (Ind) and that P is an optimal pricing rule satisfying SSM . Then< I − π, ω >= 0.

Proof Note that I − π ∈ F+. The buyer utility function can be represented as

π(x) =∑

J⊂N

1AJ(aJ · x− PJ).

Therefore,φ(x) ≡ I − π(x) =

∑

J⊂N

1AJ(aN/J · x + PJ).

For any x ∈ AJ , i ∈ J , φ(1, x−i) − φ(x) = 0. Therefore, applying Equations 9 and 10 in Lemma6 yields

< I − π, ω >=∑

J⊂N

∑

i∈J

∫

AN/JJ

(aN/J · x− PJ)∫

DiJ

T iJ(xi(x−i), x−i)dxJ/idxN/J = 0.

Q.E.D.To show that < ·, ω > is in the positive cone of the dual space of the constraint set, we use the

following condition applied to a function T : In−1 → IR and a subset of its domain, B.

(Covar) ∀ψ : In−1 → IR, ψ increasing ,

∫

Bψ(y)T (y)dy ≥ 0.

Lemma 10 (PC) Assume (Ind) and let P be an optimal pricing rule satisfying SSM and {T iJ}

be any collection of functions satisfying 8. If, for all J ⊂ N , for all i ∈ J , (T iJ(xi(x−i), x−i), Bi

J)satisfy (Covar), then φ ∈ C,∇φ ≤ 1 implies < I − φ, ω >≥ 0.

Proof By Lemma 5, for φ ∈ C,∇φ ≤ 1, I(1, x−i) − φ(1, x−i) − (I(x) − φ(x)) is non-negative.Therefore, since tiJ(xJ) is positive, applying Lemma 6, Equation 9,

< I − φ, ω >≥∑

J⊂N

∑

i∈J

∫

BiJ

(I(1, x−i)− φ(1, x−i))T iJ(xi(x−i), x−i)dx−i.

Since φ ∈ C,∇φ ≤ 1 implies that I(1, x−i)−φ(1, x−i) is increasing, the conclusion follows. Q.E.D.The hypothesis of Lemma 10 is a type of positive covariance condition. Collecting the conditions

in Lemmas 9 through 10, we have, either for f satisfying (Ind), (Mhr1), (Mhr2) and an optimalpricing rule, P , satisfying SSM and A

N/JJ a sublattice, or for f i(xi) = αix

αi−1i and an optimal

20

pricing rule, P , satisfying SSM if (Covar) holds, we can apply Theorem 4 to conclude that anoptimal pricing rule is also optimal over the class of all IC and IR mechanisms. The next subsectionillustrates that, if n = 2, (Ind), (Mhr1), (Mhr2) implies the remaining conditions and that if n = 3,f i = 1, i = 1, 2, 3 implies the remaining conditions and, thus, Theorem 4 can be applied to showthe optimality of bundling in these cases.

4.2 Applications

We show how to check the condition invoked in Lemma 10 to illustrate some optimal mechanisms,first, for the case n = 2 and, second, for the case n = 3 with F i(xi) = xi.

Theorem 5 Suppose f satisfies (Ind), (Mhr1), (Mhr2) and n = 2. The optimal zero-one mecha-nism is optimal in all IR and IC mechanisms.

Proof Theorem 3 shows that {PJ} satisfies WSM and therefore (since n = 2) SSM also. Thus,the conclusions of Lemmas 7, 8 and 9 follow. We show that Lemma 10 holds by defining thetiJ ’s that generate the {T i

J}i∈J ’s for each J ∈ {10, 01, 11}. The spirit of the definitions can be seenin Figure 3.

-

6

x1

x2

@@@@@@@@

1P01P11 − P01

1

P01

A00

A01t211 =

t11,

t111 = 0t101 = 0, t201 = t01

t00 = 0

t111 + t211 = t11

t111 = t11, t211 = 0

t110 = t10,

t210 = 0

x1(β)

x2(β)

P11 − x1(β)

P11 − P10

1

A1111

A10

A211

Figure 3: Proof of Theorem 5

Consider the one-good bundles, J = {i}, i = 1, 2. The adding up condition in Equation 8implies that tiJ ≡ tJ , i ∈ J . SSM implies that AJ = {x | xi ≥ PJ , x−i ≤ P11 − PJ}. This implies

21

that xi(x−i) = PJ , x−i ≤ P11 − PJ . Lemma 6, Equation 10 yields,

0 =∫ P11−PJ

0T i

J(xi(x−i), x−i)dx−i

= (PJf i(PJ)− (1− F i(PJ))(2 + KJ))∫ P11−PJ

0f−i(x−i)dx−i (12)

Equation 12 follows from the definition of T iJ and direct integration. It implies, for J = {i}, i =

1, 2, for all increasing ψ,

0 =∫ P11−PJ

0ψ(1, x−i)T i

J(PJ , x−i)dx−i. (13)

Now consider the full bundle. The possible asymmetries of the fs require a ‘rebalancing’ viathe parameter, β. For β ∈ IR and define xi(β) implicitly by

0 = xi(β)f i(xi(β))− (1− F i(xi(β)))(1/2 + (−1)iβ).

This is uniquely defined for each i by Condition Mhr1. Also, x1(·) is a decreasing function andx2(·) is an increasing function. Furthermore, x1(1/2) = x2(−1/2) = 0 so xi(·) > 0, i = 1, 2 impliesβ ∈ (−1/2, 1/2).

The set A11 is the union of three sets,

A111 = {x | x2 ∈ [P11 − P10, x2(β)], x1 ≥ P11 − x2}

A211 = {x | x1 ∈ [P11 − P01, x1(β)], x2 ≥ P11 − x1}

A1111 = {x | x1 ≥ x1(β), x2 ≥ x2(β), x2 ≥ P11 − x1}.

Define tiJ : AJ → IR as follows:

tiJ(x;β) = tJ(x), x ∈ Ai11,

=xif

i′(xi)f i(xi)

+ 3/2 + (−1)iβ, x−i ≥ x−i(β), x ∈ A1111,

= 0, x ∈ A−i11 .

We must show that the tiJs as defined satisfy Equation 8. By definition, Ai11 ∩ A11

11 has measurezero but it is not immediate that Ai

11 ∩A−i11 is zero measure. Lemma 11i)(below) implies this since

if x ∈ A111, then x1 ≤ x1(β) ≤ P11 − x2(β). But x ∈ A2

11 implies x1 ≥ P11 − x2 ≥ P11 − x2(β). Thedefinitions of tiJ yield that t1J(x) > 0, t2J(x) > 0 imply that x ∈ A11

11 and t1J(x)+ t2J(x) = tJ(x).DefineT i

J(x; β) analogously to Equation 8.

T iJ(xi, x−i; β) = f(1, x−i)−

∫ 1

xi

tiJ(v, xJ/i; β)f(v, x−i)dv.

Part i) of the next lemma shows that this definition satisfies Equation 8. Part ii) confirms a single-crossing property used to verify the hypothesis of Lemma 10 and part iii) addresses the potentialasymmetries of the fs. The proof is in Appendix 6.5

22

Lemma 11

(i) For all β ∈ (−1/2, 1/2), xi(β) ≤ P11 − x−i(β).

(ii) T i11(xi(x−i), x−i, β)dx−i ≤ 0, x−i ≤ x−i(β)i = 1, 2.

T i11(xi(x−i), x−i, β)dx−i ≥ 0, x−i ≥ x−i(β)i = 1, 2.

(iii) ∃β ∈ (−1/2, 1/2) such that∫ 1

P11−P{i}T i

11(xi(x−i), x−i, β)dx−i = 0, i = 1, 2.

Thus, selecting a β satisfying Lemma 11iii), for any increasing ψ,∫ 1

P11−P01

ψ(x1, 1)T 211(x1, x2(x1);β)dx1

=∫ x1(β)

P11−P01

ψ(x1, 1)T 211(x1, x2(x1);β)dx1 +

∫ 1

x1(β)ψ(x1, 1)T 2

11(x1, x2(x1);β)dx1

≥ ψ(x1(β), 1)∫ x1(β)

P11−P01

T 211(x1, x2(x1);β)dx1 + ψ(x1(β), 1)

∫ 1

x1(β)T 2

11(x1, x2(x1);β)dx1

= 0.

The inequality follows from the single-crossing property of T 211(x1, x2(x1);β) in x1 (Lemma 11ii)

and the monotonicity of ψ. The equality follows from Lemma 11iii). A similar argument followsfor T 1

11. Combined with Equation 13 and applying Lemma 10 and Theorem 4 the conclusionfollows. Q.E.D.

Note that the program addressed in Theorem 1, selecting an optimal set of bundle prices amongall bundle prices, is typically not concave. Thus, the necessary conditions yielded by the theoremare only partially conclusive. Since Theorem 5 solves an optimization problem over a larger feasibleset, it illustrates that, for n = 2, if f satisfies (Ind),(Mhr1), and (Mhr2), the necessary conditionsare also sufficient for the more limited problem as well.

Not surprisingly, checking that (Covar) is satisfied becomes progressively more challenging asn grows. However, if more structure is placed on the problem, it is possible to apply the result asthe following analysis shows.

Theorem 6 Suppose n = 3, f i = 1, i = 1, 2, 3. The optimal zero-one mechanism is optimal in allIR and IC mechanisms.

Proof Direct computation shows that the optimal zero-one mechanism is P{i} = 3/4, P{i,j} =1.14, j 6= i, P{1,2,3} = 1.22, which satisfies SSM and is symmetric. The uniform and independentdensity assumption implies that T i

J = 4xi − 3 for (xi, x−i) ∈ AJ , i ∈ J . The symmetry of pricesimplies that if (v, w, y) ∈ A100, then (w, v, y) ∈ A010 and so on. Thus we can restrict attention tothe argument for one good and bundles containing it, say, good 3.

The decomposition of the tJ ’s that we will use is the following. Define

SiJ = {x ∈ AJ | xi ≥ xk, i ∈ J, k 6= i}.

23

We show first that for each J , {SiJ}i∈J is a partition of J . Suppose (xj , x−j) ∈ AJ , j /∈ J, xj >

xk, k 6= j. Define K = J ∪ j. By Lemma 1, (xj , xJ , xN/K) ∈ AJ implies xj aJ ∈ AJ . Byhypothesis and the fact that x ∈ AJ implies aJ · x ≥ PJ , xj > maxk∈Jxk ≥ PJ

|J | . Inspection yieldsPJ|J |a

J ∈ AJ ∩ A∅ and PK|K|a

K ∈ AK ∩ A∅. Lemma 4 implies x′ ≡ xj a{j} + PJ|J |a

J ∈ AJ ∩ A∅. ButxJ > PJ

|J | ≥ PK|K| implies that aK · x′ − PK > 0 which contradicts the claim that x′ ∈ AJ ∩ A∅.

Therefore, x ∈ AJ implies x ∈ SiJ for some i ∈ J .

The definition of tiJ is as follows:

tiJ(xJ) = tJ(xJ)1xi≥xk,i,k∈J .

Definexi(x−i) = min{xi | (xi, x−i)Si

J , i ∈ J}.Recall from Lemma 4 that, for x ∈ J , both xi(·) and xi(·) are independent of components inN/J . Since tiJ(xJ) = 0 for xi < xi(xJ/i, z), z ∈ A

N/JJ , we have T i

J(xi(x−i), x−i) = T iJ(xi(x−i), x−i)

and the hypothesis of Lemma 10 follows if we show it for the latter. This motivates the need tocharacterize xi(·).

For J = 001, S3J = AJ . For xN/J ∈ A

N/JJ , x3(x1, x2) = 3/4, and T 3

J = 0 so the condition forLemma 10 is satisfied in this case.

For (say) J = 101, x2 ≤ P111 − P101, and

x3(x1, x2) = P101 − x1, x1 ∈ [P101 − P001, P101/2]

= x1, x1 ∈ [P101/2, 1]

Note that this implies that T 3J (x1, x2, x3(x1, x2)) ≤ 0 if and only if x1 ≤ 3/4 and, since T 3

J (x1, x2, x3(x1, x2))does not vary with x2 for (x1, x2, x3(x1, x2)) ∈ B3

J , Lemma 6 along with symmetry implies∫ 1

P101−P001

T 3J (x1, x2, x3(x1, x2))dx1 = 0.

for all x2 ≤ P111 − P101. This implies that for all increasing ψ,∫

B3J

ψ(x1, x2, 1)T 3J (x1, x2, x3(x1, x2))dx1dx2

=∫ P111−P101

0

∫ 1

P101−P001

ψ(x1, x2, 1)T 3J (x1, x2, x3(x1, x2))dx1dx2

≥∫ P111−P101

0ψ(3/4, x2, 1)

∫ 1

P101−P001

T 3J (x1, x2, x3(x1, x2))dx1dx2

= 0.

The symmetric argument shows the same inequality for J = 011.We use the following lemma, shown in Appendix 6.6. Define the function

Gi(xi) =∫ 1

max{xi,P111−P001−xi}T 3

111(x1, x2, x3(x1, x2))dx−i.

24

Lemma 12 Suppose x1 ≥ x2.

(i) T 1113 (x1, x2, x3(x1, x2)) > 0 ⇔ x1 > P001 = 3/4.

(ii)∫ 1

P111−P101

Gi(xi)dxi = 0.

(iii) ∃a < P001 such that Gi(xi)dxi > 0 ⇔ xi > a.

Let ψ be any increasing function. We can now apply the following inequalities:∫

B3111,{x1≥x2}

ψ(x1, x2, 1)T 3111(x1, x2, x3(x1, x2))dx1dx2

=∫ P001

P111−P101

∫ 1

max{x2,P111−P001−x2}ψ(x1, x2, 1)T 3

111(x1, x2, x3(x1, x2))dx1dx2

+∫ 1

P001

∫ 1

x2

ψ(x1, x2, 1)T 3111(x1, x2, x3(x1, x2))dx1dx2

≥∫ P001

P111−P101

∫ 1

max{x2,P111−P001−x2}ψ(3/4, x2, 1)T 3

111(x1, x2, x3(x1, x2))dx1dx2

+∫ 1

P001

∫ 1

x2

ψ(3/4, x2, 1)T 3111(x1, x2, x3(x1, x2))dx1dx2

=∫ 1

P111−P101

ψ(3/4, x2, 1)G(x2)dx2

=∫ a

P111−P101

ψ(3/4, x2, 1)G(x2)dx2 +∫ 1

aψ(3/4, x2, 1)G(x2)dx2

≥∫ a

P111−P101

ψ(3/4, a, 1)G(x2)dx2 +∫ 1

aψ(3/4, a, 1)G(x2)dx2

= 0.

The first inequality follows because ψ is increasing in x1 and applying Lemma 12(i). The nextinequality follows from Lemma 12(iii) and because ψ(3/4, x2, 1) is increasing in x2. The finalequality follows from Lemma 12(ii).

A symmetric argument shows∫

B3111,{x2≥x1}

ψ(x1, x2, 1)T 3111(x3(x−3), x−3))dx2dx1 ≥ 0.

Thus, ∫

B3111

ψ(y)T 3111(x3(y), y)dy ≥ 0.

Applying the same arguments to B2111, B

1111 and combining with the argument for J = 001, J =

101 and applying Lemma 10 and Theorem 4 as before, yields the conclusion.Q.E.D.

25

5 Conclusion

We conclude with a brief discussion of the possibilities for weakening some of the conditions invokedin Lemmas 7 through 10.

Some conditions arise as a consequence of the strategy of proof. Condition Mhr1 appearsin many single-dimensional applications. In the one good, one buyer case, it is known not tobe required, however, it is often invoked to simplify the analysis. It implies the monotonicity ofthe virtual utility function. In the context of our approach, it allows us to ignore the convexityconstraint on the utility functions that comes from incentive compatibility because the requirementthat utility be convex is not binding at a solution. Refer to the proof of Lemma 8 and note that(Mhr1) implies t ≥ 0 which yields the negativity of the first term of Equation 11. In the one goodcase, if (Mhr1) is violated, then FD is shown by employing the dual variable on the convexityconstraint. In the multiple good case, the form of this constraint is not known (to us at least)and so the tractability gain from invoking (Mhr1) is significant. The counterexample suggests thatnegative covariance of valuations pose problems,so, it may be possible to weaken (Ind). A potentialconjecture to explore is whether (Ind) can be weakened to the requirement that f satisfy affiliation.

(Mhr2) is the most unusual condition. It does not arise in the one good case. However, theproof of Lemma 8 relies critically on the assumption and its role is clearly tied to the multiplegood problem. We have constructed examples which satisfy (Ind) and (Mhr1) but not (Mhr2)and which appear to show the suboptimality of zero-one mechanisms, however, the comparisonsare in the order of the sixth digit and we do not have that much confidence in their results.

Finally, the extension to n > 2 places stringent restrictions on the family of distributions asdescribed in Lemma 8. The difficulty lies in a conflict between conditions invoked in Lemma 8 andLemma 9. Lemma 8 requires A

N/JJ be a sublattice while Lemma 9 requires P to satisfy SSM . If

n = 3, SSM implies that both (1, P111 − P110, P101 − P100) and (1, P110 − P100, P111 − P101) belongto A100. SSM also implies P111 − P101 ≤ P101 − P100. But (1, P110 − P100, P101 − P100) ∈ A111

so AN/100100 is not a sublattice. If we restrict attention to distributions of the form F i(xi) = xα

i ,however, then the sublattice condition used in Lemma 8 is no longer needed.

6 Appendix

6.1 Counterexample

This section provides the argument that an optimal zero-one mechanism can be dominated evenwhen the McAfee and McMillan condition is satisfied.

Buyer types are distributed uniformly over the region {(x1, x2) : .5x1 + x2 ≥ 1, 1 ≥ x1, 1 ≥ x2}.A (two- price) zero-one mechanism in this framework consists of the prices (Y, Z) where Y is theprice for good 2, and Z is the price for the bundle. A typical zero-one mechanism is represented inFigure 4. 19

19Note that a separate price for good 1 is never optimal if it is such that it intersects the line x1 + x2 = Z below

26

-

6

x1

x2

HHHHHHHHHHHHHHHHHHHHHHHHHH

@@@@@@

1Z − Y2(1− Y ) 2Z − 2

1

Y

a b

f

edg

h

c′c

Figure 4: The Optimal Zero-One Mechanism

The set A01 is given by region b, A11 is given by regions c + c′ + d + e + f . The areas of theseregions are:

a + b = (Z − Y )(1− Y )

a = 1/2(1− Y )(2(1− Y )) = (1− Y )2

b = (1− Y )(Z − Y − (1− Y ))

= (1− Y )(Z − 1)

c + c′ = (1− Y )(1− Z + Y )

e = (Y + Z − 2)(3− 2Z)

f = 1/2(3− 2Z)(3/2− Z) = (3− 2Z)2/4

d = (Y + Z − 2)2/2

Therefore E[Revenues(Y,Z)] = Y b + Z(c + c′ + e + f + d). To determine the optimal Y , first note

the line .5x1 +x2 = 1 since it must be strictly less than 1 and, in this case, will only draw buyers away from the more

profitable bundle priced at Z. If the intersection is above this line, it is conceivable that a price for good one below

2(Z−1) could add more sales but, intuitively, it would have to be significantly below to add much and the costs from

lost bundle sales seem very large. A formal analysis which shows that such X’s are not optimal is available from the

authors.

27

that ∂(c + c′ + e + f + d)/∂Y = −(1−Z + Y ) + 1− Y + 3− 2Z + Y + Z − 2 = 1− Y Using this inthe expected revenue function yields

0 = ∂E[Revenues(Y,Z)]/∂Y

= Y ∂b/∂Y + b + Z∂(c + c′ + e + f + d)/∂Y

= (1− 2Y )(Z − 1) + Z(1− Y )

= (2Z − 1)− (2(Z − 1) + Z)Y

Therefore Y = (2Z−1)/(3Z−2). Optimizing revenues with respect to Z does not yield an analyticsolution but plugging the optimized value for Y into the expected revenue function and maximizingover Z gives via Mathematica, Z = 1.26, Profit(Z)= 0.291, Y = 0.85.

Now, suppose that instead of the zero-one mechanism, a buyer is offered the bundle at priceZ = 1.26 or a stochastic bundle at price 1 which consists of good two with probability one andgood one with probability one-half . This mechanism can also be seen in Figure 4. Observe that ifa buyer type (x1, x2) is indifferent between the two choices, x2 + x1−Z = x2 + .5x1− 1 so a buyerof type (x1, x

′2), x

′2 > x2 is also indifferent. Therefore the set of buyers who accept the full bundle

is c′ + e + f . The buyers who accept the random bundle is b + c + d + h + g.If we compare this mechanism using the optimal Z computed from before, the trade-offs are as

follows. In g and h, the seller gains price 1 from types who did not trade in the other mechanismand in b the seller gains 1 instead of Y < 1. In regions c and d the seller loses Z− 1 as some buyerschoose the randomized bundle over the full bundle. The new areas are now given by

c′ + e + f = (3− 2Z)(Z − 1) + (3− 2Z)(3/2− Z)/2

= (3− 2Z)(Z − 1 + (3/2− Z)/2) = (3− 2Z)(Z − 1/2)/2

b + c + d + h + g = (Z − 1)2(Z − 1)/2

= (Z − 1)2.

Therefore, profits from this mechanism are (Z − 1)2 + (3 − 2Z)(Z/2 − 1/4)Z. Evaluated at theoptimal zero-one Z, this gives profits of .2975 which dominates the zero-one profits. Optimizingwithin this class of mechanisms improves slightly to .298 and yields a bundle price of 1.28.

In their proof, as in our proofs, McAfee and McMillan make liberal use of integration by partswhich, indirectly, requires the assumption of absolute continuity on the function, xπ(x)f(x). Thiscondition is clearly violated by the counterexample. However, the seller’s optimized expectedrevenue function is continuous in measures that are both absolutely continuous and approximatethe probability density function in the example. Thus, there is a class of absolutely continuousfunctions whose optimal zero-one solutions yield values arbitrarily close to the solution computedabove and which can be dominated by a random mechanism.

To see this, let gδ be a measure with support contained in [0, 1]2 that equals f when f > 0 andgoes smoothly to zero outside of the support of f such that∫

I2

(gδ − f) = δ.

28

(For example, draw a line parallel to and below the lower boundary of the support of f and let gδ

rise smoothly from 0 to 1 over that sleeve.) Define ∆ ≡ gδ − f ≥ 0. Let P δ be a zero-one solutionto the seller’s problem against the measure gδ and Rδ be the corresponding value. Let Aδ

J be thesets of types who buy bundle J when P δ is offered. Then

Rδ =∑

J∈S

P δJ

∫

AδJ

gδ

=∑

J∈S

P δJ

∫

AδJ

∆ + P δJ

∫

AδJ

f

≤ 2δ +∑

J∈S

P δJ

∫

AδJ

f

The inequality follows by definition of gδ and the fact that we can restrict attention to PJ ≤ 2 forany bundle. Now suppose there exists an ε > 0 such that Rδ ≥ R0 + ε for all δ > 0. Since P δ isfeasible for the problem with δ = 0, we have

R0 + ε ≥∑

J∈S

P δJ

∫

AδJ

f + ε

but combining with the inequality above yields 2δ ≥ ε. Setting δ < ε/2 gives a contradiction. Thusfor δ small enough but positive, we can still apply integration by parts but the expected revenuefrom the optimal zero-one mechanism is arbitrarily close to the expected revenue computed above.The expected revenue from the randomized mechanism is close to but slightly higher than the valuecomputed above since the set of types who buy the random bundle remains the same and the setof types who buy the full bundle under measure gδ strictly includes the types who bought it underf .

6.2 Derivation of Equation 4

Lemma 13 Let P be a pricing rule segmenting consumers into {AJ}J⊂N and generating a seller’srevenue of R(P ) =

∑J∈S PJ

∫AJ

f(x) dx. Then,

∂R(P )∂PJ

=∫

AJ

f(x) dx +∑

K⊂N

(PK − PJ)∫

AK∩AJ

f(x) dλKJ ,

where λKJ , the (N − 1)-dimensional Lebesgue measure defined earlier.

Proof We prove the lemma by using the definition of derivative. The market segments {AK}K⊂N

depend on the price PJ . We denote this dependence by writing AK(PJ) when convenient.If R(P ) is differentiable with respect to PJ then

∂R(P )∂PJ

=∫

AJ

f(x) dx +∑

K⊂N

PK

∂∫AK(PJ ) f(x) dx

∂PJ(14)

29

We calculate the partial derivative in the last term of the expression above by first studyingthe behavior of the Lebesgue measure and then integrating. To that effect suppose AK ∩ AJ 6= ∅.Then, there is (aK

i − aJi ) 6= 0. Let E1 = {x ∈ IRN | x = α(aK − aJ), α ∈ IR}. Similarly, let

EN−1 = {x ∈ IRN | (aK − aJ) · x = 0}. Thus EN−1 and E1 are complement spaces in IRN .Let λ1 and λN−1 be the Lebesgue measures on E1 and EN−1 respectively, both compatible withthe norm in IRN . Note that for any basic rectangle B′ × B′′ with B′ ⊂ E1 and B′′ ⊂ EN−1,λ(B′ ×B′′) = λ1(B′)× λN−1(B′′).

Let δn > 0 and define

∆(δn) = {x ∈ IRN | (PK − PJ) ≥ (aK − aJ) · x ≥ (PK − PJ − δn)}.

DefiningE1

δn = {x ∈ E1 | x = α(aK − aJ), α ∈ [0, δn]},we may write

∆(δn) = E1δn ×EN−1.

For any open rectangle (B′ ×B′′), define λn ∈ M(IRN ) by

λn(B′ ×B′′) =λ1(B′ ∩ E1

δn)× λN−1(B′′ ∩ EN−1)δn

,

For n sufficiently large, λ1(B′ ∩ E1δn) = δn. Thus, λn ⇒ λN−1.

A similar argument could be made for δn < 0.By definition, the partial derivative in the last term of the expression (14) is,

∂∫AK(PJ ) f(x) dx

∂PJ= lim

δn→0

∫AK(PJ+δn) f(x) dx− ∫

AK(PJ ) f(x) dx

|δn|

= limδn→0

∫AK(PJ+δn)\AK(PJ ) f(x) dx

|δn| .

= limδn→0

∫

Yf(x)dλn,

=∫

Yf(x)dλN−1.

where Y = {x ∈ IN | (aK−aL) ·x ≥ PK−PL, ∀L ⊂ N}. The last line follows because λn ⇒ λN−1.Setting λKJ = λN−1 completes the proof of the lemma.Q.E.D.

6.3 Proof of Theorem 3

Proof (The explicit asymmetry of the f i’s is suppressed but understood.) Suppose P11 > P10+P01.

Since for all J , PJ is part of an optimal zero-one mechanism, Theorem 1 implies

2∑

i=1

∫ 1

P11−Pi

{f(1)−∫ 1

P11−P−i

(xif′(xi) + 3/2f(xi))dxi}f(x−i)dx−i = 0.

30

Therefore, at least one element of the sum is non-positive. Wlog suppose it is the element i = 1.Integration then implies

(P11 − P01)f(P11 − P01)− (1− F (P11 − P01))/2 ≤ 0.

Since xf(x) is non-decreasing and f(·) > 0, this implies that xf(x) − (1 − F (x))/2 < 0 for allx < P11 − P01. Define l1(x2) = min{x1 | (x1, x2) ∈ A10}. Applying Theorem 1 to A10, then yields

∫ P11−P10

0{f(1)f(x2)−

∫ 1

l1(x2)(f(x2)(x1f

′(x1)+3/2f(x1))+f(x1)(x2f′(x2)+3/2f(x2)))dx1}dx2 = 0.

Mhr1 implies that the final term is strictly negative and the argument for A11 implies that thefirst terms are non-positive. A contradiction. Q.E.D.

6.4 Proof of Lemma 6

Proof By definition

< φ, ω > =n∑

i=1

∫

In−1

φ(1, x−i)f(1, x−i)dx−i −∑

J⊂N

∫

AJ

φ(x)tJ(xJ)f(x)dx

=∑

J⊂N

∑

i∈J

∫

BiJ

φ(1, x−i)f(1, x−i)dx−i −∑

J⊂N

∑

i∈J

∫

AJ

φ(x)tiJ(xJ)f(x)dx

=∑

J⊂N

∑

i∈J

∫

BiJ

φ(1, x−i)f(1, x−i)dx−i −∑

J⊂N

∑

i∈J

∫

BiJ

∫ 1

xi(x−i)φ(x)tiJ(xJ)f(x)dx (15)

=∑

J⊂N

n∑

i=1

{∫

BiJ

[φ(1, x−i)f(1, x−i)−∫ 1

xi(x−i)φ(x)tiJ(xJ)f(x)dxi]dx−i}

=∑

J⊂N

∑

i∈J

{∫

BiJ

[φ(1, x−i)(f(1, x−i)− T iJ(xi(x−i), x−i))−

∫ 1

xi(x−i)φ(x)tiJ(xJ)f(x)dxi

+ φ(1, x−i)T iJ(xi(x−i), x−i)]dx−i}

=∑

J⊂N

∑

i∈J

{∫

BiJ

∫ 1

xi(x−i)(φ(1, x−i)− φ(x))tiJ(xJ)f(x)dxi + φ(1, x−i)T i

J(xi(x−i), x−i)dx−i}.

The second equality follows since {BiJ}i∈J,J⊂N is a partition of In−1 and

∑i∈J tiJ = tJ from

Equation 8. The next equality follows from Equation 3. The fourth equality collects all termsin the summation in i and the next one adds and subtracts T i

J(xi(x−i), x−i). The final equalityfollows by Equation 8.

31

The first order condition from Theorem 1 implies for all J ,

0 =∑

i∈J

∫

BiJ

[f(1, x−i)dx−i −∫

AJ

t(x)dx

=∑

i∈J

∫

BiJ

[f(1, x−i)dx−i −∫

AJ

tJ(x)dx

=∑

i∈J

{∫

BiJ

[f(1, x−i)−∫ 1

xi(x−i)tiJ(xJ)dxi]dx−i}

=∑

i∈J

{∫

BiJ

∫ 1

xi(x−i)T i

J(xi(x−i), x−i)dx−i}

=∑

i∈J

{∫

AN/JJ

∫

DiJ

T iJ(xi(x−i), x−i)dxJ/idxN/J}.

The second equality follows from Equation 7. The next equality uses the same argument as in 15above. The fourth equality comes by definition of T i

J . The independence of the integrals is impliedby SSM and Lemma 4. Note that T i

J(xi(x−i), x−i) can be expressed as giJ(xJ)

∏k/∈J fk(xk)

so the result follows. (If | J |= 1, then the integration is over AN/JJ and x−i ∈ A

N/JJ implies

T Ji (xi(x−i), x−i) = 0). Q.E.D.

6.5 Proof of Lemma 11

The equation defining x(·) can be represented as

wi(z; β) = zf i(z)− (1− F i(z))(1/2 + (−1)iβ)

= f i(1)−∫ 1

zvf i′(v) + (1/2 + (−1)iβ)f i(v)dv (16)

(Mhr1) implies that wi is increasing in z and the definition of xi(β) implies wi(xi(β);β) = 0 Thefollowing inequality is used in the proof. For z ≤ Pi and x−i ≥ P11 − Pi, Equation 12 implies

0 = (Pifi(Pi)− (1− F i(Pi))(2 + E[

wf−i′(w)f−i(w)

| w ≤ P11 − Pi]))f−i(u)

≥ (Pifi(Pi)− (1− F i(Pi))(2 +

x−if−i′(x−i)

f−i(x−i)))f−i(x−i)

≥ (zf i(z)− (1− F i(z))(2 +x−if

−i′(x−i)f−i(x−i)

))f−i(x−i)

= f i(1)f−i(x−i)−∫ 1

zxif

i′(xi)f−i(x−i) + 3f i(xi)f−i(x−i) + f i(xi)x−if−i′(x−i)dxi

= f i(1)f−i(x−i)−∫ 1

zt(x)dxi. (17)

The first inequality follows by Mhr2 and x−i ≥ P11 − P{−i}. The second inequality follows byMhr1. The following equality follows from direct differentiation with respect to z. The last linecomes by definition of t.

32

Proof (Part i) Suppose that xi(β) > P11− x−i(β). Define li(x−i) = max{P11−x−i, x−i(β)} wherethe dependence on β is suppressed for brevity. The first order condition from Theorem 1 appliedto A11 can be written as

0 =2∑

i=1

∫ P11−xi(β)

P11−Pi

{f i(1)f−i(x−i)−∫ 1

P11−x−i

t(x)dxi}dx−i

+∫ 1

P11−xi(β){f i(1)f−i(x−i)−

∫ 1

li(x−i)(xif

i′(xi) + (3/2 + (−1)iβ)f i(xi))f−i(x−i)dxi}dx−i

≤2∑

i=1

∫ 1

P11−xi(β)w(li(x−i);β)f−i(x−i)dx−i

< 0

which is a contradiction. The equality follows because the limits of integration partition A11 intothree disjoint sections. Two sections are reflected in the limits of integration in the first term of thesummation. The third section is repeated in the second term of the summation but adding eachof the integrands yields t(x). The first inequality follows by Equation 17. The second inequalityfollows because wi(x;β) is strictly increasing in x (by Mhr1 and the fact that β ∈ (−1/2, 1/2))and is zero at x = xi(β) and li(x−i) < xi(β).Q.E.D.Proof (Part ii)Consider i = 2. Equation 17 implies for x1 ∈ [P11 − P01, x1(β)],

0 ≥ (f2(1)−∫ 1

P11−x1

t211(x1, x2;β)f(x2)dx2)f1(x1)

= T 211(x1, x2(x1);β).

The first inequality follows because in this region, t211 = t11.For x1 ∈ (x1(β), P11 − x2(β)],

T 211(x1, x2(x1);β) = (f2(1)−

∫ 1

P11−x1

t211(x1, x2;β)f(x2)dx2)f1(x1)

= w2((P11 − x1);β)f1(x1)

≥ w2(x2(β);β)f1(x1)

= 0.

The second equality applies Equation 16 and the defintion of t211. The inequality follows sincex1 ≤ x2(β) implies (P11 − x1) ≥ x2(β). The equality is by definition of x2(β).

Finally, for x1 ≥ P11 − x2(β),

T 211(x1, x2(x1);β) = (f2(1)−

∫ 1

P11−x1

t211(x1, x2;β)f(x2)dx2)f1(x1)

= (f2(1)−∫ 1

x2(β)t211(x1, x2;β)f(x2)dx2)f1(x1)

= 0.

33

The second equality follows because for x2 ≤ x2(β), t211(x) = 0. The final equality follows byEquation 16. The same argument follows for i = 1 Q.E.D.Proof (Part iii) Part i) implies that T i

J(·; β) satisfies Equation 8. Applying Lemma 6 to J = 11gives

0 =2∑

i=1

∫ 1

P11−P{i}T i

11(x1, x2;β)dxi

=2∑

i=1

{∫ x−i(β)

P11−P{i}T i

J(P11 − x−i, x−i)dx−i +∫ P11−xi(β)

x−i(β)T i

J(P11 − x−i, x−i)dx−i}

≡2∑

i=1

{Fi + Gi}

The second equality uses the fact that T iJ(x−i(β), x−i) = 0 for x−i ≥ P11 − xi(β) and T i

J(1, x−i) =f i(1)f−i(x−i). Suppose that for some i, Fi+Gi is strictly negative (and therefore F−i+G−i > 0) forall β. Equation 17 implies that Fi is non-positive and Gi is non-negative. Note that Fi +Gi variescontinuously with β. Let β increase or decrease as necessary so that xi(β) approaches P11 − Pi.Since the measure of (P11 − Pi, x−i(β)] goes to zero, Fi goes to zero. But Gi is non-negative byPart ii) and, by assumption, Fj + Gj is strictly positive. This yields a contradiction. Q.E.D.

6.6 Proof of Lemma 12

Proof We first characterize x3(x1, x2) on B = B3J∩{(x1, x2) | x1 ≥ x2}. The points (P101/2, P111−

P101, P101/2), (P111/3, P111/3, P111/3) are in A111 ∩ A∅ ∩ S3111. Since this set is convex, so is the

line connecting them given by {x | x3 = x1, x2 = P111 − 2x1, x3 ∈ [P111/3, P{1,3}/2]}. SinceAJ is increasing in xi, i = 1, 2, 3, all points {x | x3 ≥ x1, x1 ≥ P111/3, x2 ≥ P111 − 2x1} are inA111. This implies that for {(x1, x2) | x1 ≥ x2, x1 ≥ P111/3, x2 ≥ P111 − 2x1}, x3(x1, x2) = x1.Any point {x | x3 = x1, x1 ≤ P{1,3}/2, x2 < P111 − 2x1} yields strictly negative utility if N

was purchased, so it is not in A111. Thus, for {(x1, x2) | x1 ≥ x2, x1 ≤ P{1,3}/2, x2 ≤ P111 − 2x1},x3(x1, x2) = P111−x1−x2. Symmetry implies that x3(v, w) = x3(w, v) and, therefore G1(v) = G2(v)for all v ∈ [P111 − P101, 1].

Part (i) now follows since 4x1 − 3 < 0 if and only if x1 > 3/4 and 4(P111 − x1 − x2) − 3 ≤4P001 − 3 = 0 since (x1, x2) ∈ B3

J implies x1 + x2 ≥ P111 − P001.

Note thatT 3

111(x3(x1, x2), x1, x2) = T 3101(x3(x1, x2), x1, x2) (18)

at x2 = P111 − P101, x1 ≥ P101 − P001. Furthermore, given the definition of x3(x1, x2),

T 3111(x3(x1, x

′2), x1, x

′2) ≤ T 3

101(x3(x1, x2), x1, x2) (19)

for x1 ≥ x′2 ≥ x2, x1 ≥ P101 − P001.

34

For x2 ∈ [P111 − P101, P101 − P001],

G2(x2) =∫ P111−P101

max{x2,P111−P001−x2}T 3

111(x3(x1, x2), x1, x2)dx1 +∫ 1

P101−P001

T 3111(x3(x1, x2), x1, x2)dx1

≤∫ 1

P101−P001

T 3111(x3(x1, x2), x1, x2)dx1

≤∫ 1

P101−P001

T 3111(x3(x1, P111 − P101), x1, P111 − P101)dx1

= 0

The first inequality comes because T 3111 ≤ 0 for x1 ≤ P111 − P101 ≤ 3/4. The second comes from

Equation 19, the equality comes from Equation 18 and Lemma 6.Part (i) implies that G2(x2) > 0 for x2 > P001. By direct differentiation, G2(x2) is shown to be

increasing for x2 ∈ [P111/3, P001]. It is convex for x2 ∈ [P101−P001, P111/3] which implies there is av such that G2(·) is decreasing for x2 ∈ [P101−P001, v] and increasing thereafter. Thus, G2(x2) ≤ 0for x2 ≤ v, is increasing for x2 ∈ [v, P001] and strictly positive for x2 ≥ P001. Therefore, there existsa < P001 such that x2 ≤ a implies G(x2) ≤ 0 and x2 ≥ a implies G(x2) ≥ 0 which yields Part (ii).

Finally, Lemma 6 and symmetry implies

0 =∫ 1

P111−P101

G2(x2)dx2 +∫ 1

P111−P101

G1(x1)dx1.

G1 = G2 and P101 = P011 then implies Part (ii). Q.E.D.

35

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