efficient allocation of heterogenousasen/papers/mitrasen.pdf · commodities with balanced...

EFFICIENT ALLOCATION OF HETEROGENOUSCOMMODITIES WITH BALANCED TRANSFERS

MANIPUSHPAK MITRA AND ARUNAVA SEN

Abstract. In this paper we characterize heterogenous commodi-ties allocation problems where an efficient rule can be implementedin dominant strategies with balanced transfers. We show that theclass of such problems is non-trivial and that the associated differ-ence domain is one-dimensional. If the domain satisfies convexitythere exists a permutation of the objects for which the associateddifference domain is a straight line passing through the origin withpositive slope. An interesting feature of the domains is that atmost two orderings of the commodities are permissible and, inparticular, one ordering must be the reverse of the other.

Keywords: Allocation Problem, Efficiency, Budget Balance, Domi-

nant Strategy Implementation.

JEL Classification Numbers: C72, C78, D82.

1. Introduction

In this paper we provide a complete answer to the following ques-

tion: what is the class of convex domains of valuations over which effi-

cient outcomes in heterogenous commodities allocation problem can be

achieved in dominant strategies and balanced transfers? The problem

Date: March 10, 2008.This is a substantially revised version our earlier paper “Efficient allocation of

commodities with balanced transfers”. The current version contains significantlyimproved results. We thank Sushil Bikhchandani, Herve Moulin, Georg Noldekeand William Thomson and various seminar participants for helpful comments. Mi-tra gratefully acknowledges financial support from the Deutsche Forschungsgemein-schaft Graduiertenkolleg (DFG) 629 at the University of Bonn.

1

2 MANIPUSHPAK MITRA AND ARUNAVA SEN

is to assign n heterogenous commodities or objects (such as jobs or

houses) to n agents where each agent requires only one object. The

setting for the problem is standard. Preferences of the agents are as-

sumed to be quasi-linear and their valuations (of relevant commodities)

assumed to be private information. The goal of the planner is to design

a mechanism that attains the following objectives: (i) each agent has

dominant strategy incentives to reveal her private information truth-

fully and (ii) the outcome in every state of the world is efficient. The

latter requires the allocation to maximize the sum of utilities it gener-

ates and also for aggregate transfers to be balanced.

1.1. Motivation. One of the most important results in mechanism

design theory is that in the quasi-linear environment, the class of

Vickrey-Clarke-Groves (or VCG) mechanisms (Vickrey [32], Clarke [5]

and Groves [11]) induces truth telling in dominant strategies and guar-

antees an efficient allocation in every state. Moreover under certain

mild richness assumptions on the domain, the VCG mechanisms are

the only ones that have these properties (Holmstrom [14], Laffont and

Maskin [20]).

The main difficulty with VCG mechanisms is that in typical do-

mains, they are not budget balancing (Groves and Ledyard [12], Green

and Laffont [10], Hurwicz [15], Hurwicz and Walker [16] and Walker

[33]). Thus equilibrium outcomes in certain states will involve wastage

of money and a consequent departure from “full” efficiency. There is

a growing literature, both in economics and computer science, where

the broad issue is to identify VCG mechanisms that lead to “minimum

efficiency loss” (Cavallo [3], Guo and Conitzer [8], [9] and Moulin [25],

[26] and [27]). However, Zhou [35] provides examples of both public

and private good allocation problems for which (a) no budget balanc-

ing VCG mechanisms exists and (b) all VCG mechanisms are infe-

rior to other “reasonable” non-VCG mechanisms. Zhou [35] concludes

therefore that unless one can find budget balancing VCG mechanism,

one should limit its use. We note also that budget balance is a stan-

dard assumption in a variety of mechanism design problems such as in

EFFICIENT ALLOCATION OF HETEROGENOUS COMMODITIES 3

d’Aspremont and Gerard-Varet [7], Cramton, Gibbons and Klemperer

[6], Makowski and Mezzetti [22] and Myerson and Satterthwaite [28]).

The budget imbalance problem with VCG mechanisms has led to

the advocacy of a weaker solution concept in mechanism design, viz.

Bayesian incentive compatibility. Here it has been shown that full

efficiency and budget balance are typically compatible, for instance,

d’Aspremont Gerard-Varet [7], Makowski and Mezzetti [22]) and Kr-

ishna and Perry [18]). However, mechanism design using Bayesian in-

centive compatibility is subject to well-known criticisms (Wilson [34]).

The validity of outcomes of a Bayesian mechanism depends on strong

informational and behavioural assumptions. In particular, such mecha-

nisms depend critically on prior beliefs held by agents about the “types”

of other agents; mild requirements on robustness to these beliefs are

usually equivalent to dominant strategy implementation (Ledyard [21]).

Several recent papers on mechanism design express skepticism regard-

ing Bayesian mechanisms (Bikchandani et al [2], Chung and Ely [4],

Bergemann and Morris [1] and Jehiel, Meyer-ter-Vehn, Moldovanu and

Zame [17]).

Summarizing, we feel that in view of (i) the lack of a good justifica-

tion for using VCG mecahnisms when budget balance is not attained

and (ii) the unrealistic assumptions that are required to justify the use

of Bayesian mechanisms, we feel that there are compelling reasons to

investigate the exact conditions when VCG mechanisms do work, i.e.

are budget balancing. We analyze this general question in the specific

context of a natural multi-dimensional problem, that of allocating n

distinct objects to n agents.

1.2. Discussion of the main result. Our characterization result is

in terms of the difference domain of feasible valuations. A difference

vector is an n− 1 dimensional vector of differences of valuations for a

particular arrangement of the commodities. For instance, if there are

three commodities and the valuation vector (5, 12, 6) is feasible, the

associated difference vector is (7,−6). Note that differences depend on


the way the commodities are arranged. Our characterization result can

be verbally stated as follows.

Suppose that the domain of feasible valuations satisfies convexity. Then,

there exists an arrangement of commodities for which the difference do-

main is a straight line (in <n−1). Moreover the slope of this line has

very particular properties.

An important consequence of this result is that the set of admissible

types is one-dimensional, i.e. the planner needs to elicit only the value

of a single parameter from each agent. An interesting feature of the

domains is that at most two orderings of the commodities are permis-

sible in the domain. For instance if the domain includes the ordering

where object 1 is preferred to 2, preferred to 3 . . . preferred to n, the

only other ordering which can be present is the dual or reverse of this

ordering, i.e n is preferred to n− 1 . . . preferred to 1.

It is no surprise, perhaps, that the class of domains over which full ef-

ficiency can be attained in dominant strategies, is restricted (although

their exact structure is of interest). However, we wish to emphasize

that these domains and their associated incentive problems are not

trivial. Consider again the case of three commodities and three indi-

viduals. Consider the domain Θ = {(s, 2s, 3s) | s ∈ (−∞,∞)}, that

is the valuations of the first, second and third commodities are s, 2s

and 3s respectively.1 Consider the following mechanism. Each agent

announces their value of the parameter s. The agent with the highest

announced s get object 3, the second highest gets 2 and the lowest gets

1.2 Moreover the agent with the highest announcement pays the agent

with the lowest announcement an amount equal to the second highest

announcement. It is not difficult to verify that this mechanism induces

truth-telling in dominant strategies and is fully efficient.

1There is a natural interpretation of this model. Imagine that the three objectsare “tickets” to use an ATM machine in periods one, two and three. The machineneeds to be used by a customer for exactly one unit of time but can be used by onlyone customer at a time. Interpret −s as the per period waiting cost of a customer.

2Ties are broken arbitrarily.


We note that the domain and mechanism described in the paragraph

above is closely related to incentive problems in sequencing and queue-

ing models studied in Suijs [30], Mitra [23] and [24]. Interestingly, a

consequence of our result is that domains of this kind are necessary if

truthful implementation in dominant strategies with balanced transfers

is to be attained in the allocation problem we study.

1.3. Related Literature. There are a number of papers such as Groves

and Loeb [13], Tian [31] and Liu and Tian [19] which have identified

some pure public commodities problems where full efficiency can be

attained with dominant strategies. The results in these papers pertain

to problems where types are one-dimensional. In our problem they are

possibly multi-dimensional although we show that domains over which

full efficiency with dominant strategies can be achieved must be, in

effect, one-dimensional. Also unlike the papers cited above, we have

a complete characterization result. Finally, another paper which is re-

lated to ours is Hurwicz and Walker [16]. There the authors show that

if the environment is “indecomposable”, then every continuous mecha-

nism where truth-telling is a dominant strategy, full efficiency will fail

“generically”. We make no continuity assumptions; moreover in the

problems we consider, we identify exactly the domains over which full

efficiency and dominant strategy implementation are consistent.

2. Notation and Definitions

We consider the problem of allocating exactly n different commodi-

ties amongst n agents. We shall let N = {1, 2, . . . , n} denote the (finite)

set of both agents and commodities. We assume that each agent has

use for exactly one object; the problem is therefore of assigning each

object to a single agent. We let Σ(N) denote the set of all permutations

of the set N . An allocation is an element of Σ(N). A typical allocation

is denoted by x and xj denotes the object received by the jth agent.

The utility derived by the jth agent from the kth object is denoted by

θj(k) ≥ 0. The n dimensional vector θj = (θj(1), . . . , θj(n)) which spec-

ifies agent j’s valuation for each object is referred to as the type of agent


j. It is assumed that the set of types for each agent is the same and

this set is denoted by Θ. A state is an n- tuple θ = (θ1, . . . , θn) ∈ Θn

which is a collection of types, one for each agent. Agents are assumed

to have preferences that are quasi-linear in money. Thus, for any allo-

cation x ∈ Σ(N), the utility of an agent j with type θj ∈ Θ is given by

Uj(x, tj; θj) = θj(xj) + tj where tj ∈ < is the monetary transfer that

she receives. An allocation problem Γ is a pair 〈N, Θ〉.

Definition 1. An allocation x∗ ∈ Σ(N) is efficient in state θ ∈ Θn if

x∗ ∈ arg maxx∈Σ(N)

∑j∈N θj(xj).

Abusing notation slightly we shall say that an efficient rule x∗ is a

mapping x∗ : Θn → Σ(N) which assigns an efficient allocation x∗(θ)

for every state θ ∈ Θn. There may exist certain states where more

than one allocation is efficient. In these states an efficient rule selects

an allocation from those that are efficient.

The goal of the planner is to ensure an efficient allocation in every

state. The difficulty however is that agents have private information

about their valuations. The planner, therefore has to design a mecha-

nism to extract this private information. Applying the Revelation Prin-

ciple we can concentrate on direct revelation mechanism where agents

report their types and, based on their reports, the planner decides (a)

an allocation of the n commodities and (b) a transfer for each agent.

Formally, a (direct) mechanism M is a pair 〈x, t〉, where x ∈ Σ(N) and

t ≡ (t1, . . . , tn) : Θn → <n. If M = 〈x, t〉 is the mechanism in oper-

ation, then an announcement θ = (θ1, . . . , θn) ∈ Θn, results in agent

j of type θj getting utility Uj(xj(θ), tj(θ); θj) = θj(xj(θ)) + tj(θ). We

assume that the mechanism must be designed so as to provide agents

dominant strategy incentives to reveal their private information truth-

fully.

Definition 2. An efficient rule x∗ : Θn → Σ(N) for Γ = 〈N, Θ〉 is

implementable, if there exists a mechanism M = 〈x∗, t〉 such that, for

all j ∈ N , all θj, θ′j ∈ Θ and all θ−j ∈ Θn−1, we have

Uj(x∗(θj, θ−j), tj(θj, θ−j); θj) ≥ Uj(x

∗(θ′j, θ−j), tj(θ′j, θ−j); θj).


In other words the mechanism induces each agent to reveal their type

truthfully irrespective of what they believe about the announcements

of the other agents. Of course when agents are truthful, an efficient

allocation is achieved. In addition to the requirements above, we are

going to assume that the transfers are balanced.

Definition 3. An efficient rule x∗ in Γ = 〈N, Θ〉 is implementable

with balanced transfers if there exists a mechanism M = 〈x∗, t〉 that

implements it and furthermore∑

j∈N tj(θ) = 0 for all θ ∈ Θn.

The primary restriction that domains under consideration satisfy is

convexity.

Definition 4. The domain Θ is convex if ∀ x, y ∈ Θ, λx+(1−λ)y ∈ Θ

for all λ ∈ (0, 1).

Convexity is a standard and natural “richness” assumption on the

domain. It is used widely in the mechanism design literature (see for

example, Green and Laffont [10]). One of the consequences of convexity

is that it implies that implementability and allocation efficiency implies

that the search for mechanisms can be restricted to the VCG class.

3. The result

Our main result is a characterization of allocation problems where

an efficient rule can be implemented with balanced transfers. Before

describing the result, we need to introduce the concept of a difference

domain.

Let µ ∈ Σ(N). For all θj ∈ Θ, let ∆µθj ≡ (∆µθj(1), . . . , ∆µθj(n −1)) be an n − 1 dimensional vector whose kth component ∆µθj(k) =

θj(µ(k+1))−θj(µ(k)). Thus ∆µθj is the vector of differences generated

by θj under the permutation µ. For instance, suppose N = {1, 2, 3}and µ is given by µ(1) = 2, µ(2) = 3 and µ(3) = 1. Let θj = (5, 12, 21).

Then ∆µθj is the vector (9,−16). For any µ, we shall let ∆µΘ denote

the set of difference vectors which can be obtained from vectors in Θ

and refer to it as its associated difference domain.


Theorem 1. Let Γ = 〈N, Θ〉 be an allocation problem where Θ is con-

vex. Then an efficient rule in Γ is implementable by balanced transfers

if and only if there exists µ ∈ Σ(N) such that the associated differ-

ence domain is of the form ∆µΘ = {(1 − s).δ + s.δ′ | s ∈ I} where

I ⊂ < is an interval and δ, δ′ ∈ <n−1 are such that (i) δ′ ≥ δ and (ii)∑n−1k=1(−1)k−1

(n−2k−1

)δk =

∑n−1k=1(−1)k−1

(n−2k−1

)δ′k.

The Theorem says that an efficient rule in an allocation problem with

a convex domain, is implementable with balanced transfers if and only

if there exists a permutation of the objects such that the associated

difference domain is a segment of a straight line in <n−1. Moreover

this line must have a non-negative slope and satisfy another restriction

(part (ii) in the statement of the Theorem) that we shall clarify later

by means of examples.

The proof of Theorem 1 is provided in the Appendix. The arguments

are largely combinatorial and geometric. It is difficult to provide “intu-

ition” for the necessity part of the result and we can do no better than

to provide an informal account of the steps involved in the proof. We

first show that for any two vectors α, β ∈ Θ, there exists a permutation

of the objects µ ∈ Σ(N) such that the associated difference vectors are

ordered, i.e. ∆µα ≤ ∆µβ. We assume without loss of generality that

this µ is in fact, the identity permutation. The convexity of the domain

implies that the only mechanisms that need to be considered are VCG

mechanisms (Holmstrom [14]). Using the general necessary condition

for budget balance VCG mechanisms (Walker [33]) we show that, for

the selected α, β ∈ Θ, the associated difference vectors ∆α, ∆β ∈ ∆Θ

must be such that∑n−1

k=1(−1)k−1(

n−2k−1

)∆αk =

∑n−1k=1(−1)k−1

(n−2k−1

)∆βk.

We then show that in an appropriate neighbourhood of the vectors

(∆α, ∆β) (this is actually the cube constructed with ∆α and ∆β be-

ing the ends of the main diagonal), the feasible difference domain is

the line joining (∆α and ∆β). The final step in the proof is to show

that all points in the difference domain must lie on this line.

The sufficiency part of the result states that an efficient rule in al-

location problems defined over domains of the type specified in the


Theorem, can be implemented by balanced transfers. If the difference

domain is a straight line with non-negative slope, then in every state,

all agents are ordered according to their difference vectors. Efficiency

then implies that agents with higher differences will get commodities

with “higher” indices. The following important independence property

also holds. Consider an efficient allocation and suppose an agent is

“removed” by assigning the object with lowest index. In the problem

of assigning the n − 1 commodities to the remaining n − 1 agents, ef-

ficiency requires that the object assignment of the agents with higher

difference vectors than that of the removed agent remains unchanged.

On the other hand, those with lower difference vectors moves “up one

place”. This independence property together with condition (ii) allows

for an efficient rule to be implemented with balanced transfers. This

is similar to the independence property and the combinatorial prop-

erty obtained in the queueing problems (Mitra [23], [24]), though the

constructions required in the proof are much more general.

A sharper version of Theorem 1 can be obtained if an additional

(weak) assumption is made on domains.

Definition 5. The domain Θ satisfies Property I (Indifference) if there

exists α ∈ Θ such that α1 = α2 = ... = αN .

A domain satisfies Property I if it includes a type vector in which all

objects are valued identically. In what follows, 0 is the origin in <n−1.

Corollary 1. Let Γ = 〈N, Θ〉 be an allocation problem where Θ is

convex and satisfies property I . Then an efficient rule in Γ is imple-

mentable by balanced transfers if and only if there exists µ ∈ Σ(N)

such that ∆µΘ = {s.δ | s ∈ I} where I ⊂ < is an interval which

includes 0 and δ ∈ <n−1 is such that (i) δ ≥ 0 or 0 ≥ δ and (ii)∑n−1k=1(−1)k−1

(n−2k−1

)δk = 0.

Corollary 1 is an easy consequence of Theorem 1. Since Property I

is satisfied, 0 ∈ ∆µΘ for all µ ∈ Σ(N). Therefore, 0 can be taken to be

one of the vectors δ or δ′ in Theorem 1.

Finally, we note an interesting implication of Corollary 1.


Remark 1. Suppose and efficient rule in Γ = 〈N, Θ〉 can be imple-

mented with balanced transfers. Suppose Θ is convex and satisfies Prop-

erty I. Then Θ can admit at most two orderings of the n objects. If µ

is the permutation with respect to which Corollary 1 is satisfied, then

all agents in any state of the world either prefer object µ(1) to µ(2) and

so on till object µ(n) or its exact inverse, i.e. object µ(n) is preferred

to object µ(n− 1) till object µ(1).

4. Examples

We give examples illustrating our results. Throughout we assume

that the domains under consideration are convex. We begin with a

trivial case.

Example 1. Let Θ = {{c, c, ..., c}, c ∈ <}. Then ∆µΘ = 0 for all µ ∈Σ(N). It is obvious that the domain satisfies our sufficient condition.

This is not surprising because all allocations are efficient in this case.

Full efficiency can be attained by allocating the objects arbitrarily - no

information revelation or transfers are required.

Example 2. Let n = 2. In this case all difference vectors have a sin-

gle component. The necessity requirement that δ1 = 0 for all feasible

difference vectors implies ∆µΘ = 0 for some µ ∈ Σ(N). This implies

∆µΘ = 0 for all µ and we conclude that an efficient rule can be im-

plemented with balanced transfers only in the trivial case described in

Example 1.

Example 3. Let n = 3 and suppose that the domain Θ satisfies Prop-

erty I. From Corollary 1 we know that there exists µ (which we assume

w.l.o.g to be the identity map) such that∑2

k=1(−1)k−1(

1k−1

)δk = 0 for

all δ ∈ ∆µΘ which implies δ1 = δ2. Therefore the difference domain in

two-dimensional space is an interval of the 45o line through the origin

containing the origin. Note that the horizontal and vertical axis in this

space represent respectively the difference in valuation of object 2 over

object 1 and the difference in valuation of object 3 over object 2. There-

fore ∆µΘ = {(s, s) | s ∈ I} where I is a non-trivial interval of the real


line containing 0. Thus we could have Θ = {(s+b, 2s+b, 3s+b) | s ∈ I}where b is a constant and I is an appropriate interval. Note that if I is

of the form [0, s] (with 0 < s), then the only ordering represented in Θ

is the one where object 3 is strictly preferred to object 2 which in turn

is strictly preferred to object 1;3 if I is of the form [s, 0] (with s < 0),

then the only ordering represented in Θ is the reverse ordering of the

previous one where object 1 is strictly preferred to object 2 which in

turn is strictly preferred to object 3; if I is of the form [s, s] with s < 0

and s > 0, both orderings are represented.

We omit discussion of the sufficiency part of Corollary 1 because we

have already done so in the Introduction.

Example 4. Let n = 4 and suppose that the domain Θ satisfies Prop-

erty I. From Corollary 1 we know that there exists µ (which we assume

w.l.o.g to be the identity map) such that∑2

k=1(−1)k−1(

1k−1

)δk = 0 for

all δ ∈ ∆µΘ which implies that δ1 − 2δ2 + δ3 = 0. Thus, the difference

domain must be of the form ∆µΘ = {(2as, (a+ b)s, 2bs) | s ∈ I} where

I is a non-trivial interval of the real line containing 0 and a, b > 0

are constants. The interval I determines which of the two orderings,

object 4 preferred to 3 preferred to 2 preferred to 1 or its reverse, is

represented in Θ.

Consider the domain Θ = {(s, 2s, 3s, 4s) | s ∈ (−∞, +∞)}. Let x∗

be the allocation rule where, in each state s, the agent with the highest,

second highest, third highest and lowest values of s get commodities

4, 3, 2 and 1 respectively with ties are broken in favour of agents with

the lower index. Clearly x∗ is an efficient rule. Consider the follow-

ing mechanism. Each agent announces a real number which can be

interpreted as their value of the parameter s. If s is the announcement

3It could also be of the form [0, s).


vector, then x∗(s) is implemented and transfers are as follows:

ti(s) =

−sj − 1

2sl if x∗i (s) = 4, x∗j(s) = 3 and x∗l (s) = 2

−12sl if x∗i (s) = 3 and x∗l (s) = 2

12sj if x∗i (s) = 2 and x∗j(s) = 3

12sj + sl if x∗i (s) = 1, x∗j(s) = 3 and x∗l (s) = 2

In other words, the agent with the highest announcement pays an

amount equal to the announcement of the second highest plus half

the amount of the third highest; the agent with the second highest

announcement pays an amount equal to half the announcement of the

third highest; the agent with the third highest announcement receives

a subsidy equal to half the announcement of the second highest and

the agent with the lowest announcement receives a subsidy equal to

the third highest plus half the announcement of the second highest.

Suppose si > sj > sk > sl. Then the transfers of agents’ i, j, k and l

are −sj − 12sk, −1

2sk,

12sj and sk + 1

2sj. Aggregate transfers are clearly

zero. Observe that for agent i truth-telling payoff is 4si − sj − 12sk

while the payoffs he can get by lying are 3si − 12sk, 2si + 1

2sk and

si + sl + 12sk. Simple calculations confirm that truth-telling is better

than lying. Moreover, it is easy to verify that a similar claim holds

true for agents j, k and l.

5. Conclusion

In this paper we have characterized domains in a heterogenous ob-

jects model where an efficient rule can be implemented in dominant

strategies with balanced transfers. We have shown that these domains

are one-dimensional but non-trivial. Several questions remain open.

For instance our results do not cover the case where the number of

agents and objects are distinct. Are there any “nice” rules (other than

the efficient ones) that can be implemented with budget balanced dom-

inant strategy mechanisms? We hope to address some of these issues

in future research.


6. Appendix

Proof of Theorem 1: Necessity: We show that if an allocation prob-

lem defined over a convex domain is implementable by balanced trans-

fers, then the associated difference domain must be of the form de-

scribed in the statement of the Theorem. We start with two useful

lemmas.

Lemma 1. Let Γ = 〈N, Θ〉 be an allocation problem. Let µ ∈ Σ(N)

and θ ∈ Θn be such that for all i, j ∈ N , either ∆µθi ≥ ∆µθj or

∆µθj ≥ ∆µθi holds. Let x be an allocation such that for all i, j ∈ N ,

∆µθi ≥ ∆µθj ⇒ µ(xi) > µ(xj). Then x is efficient in state θ.

Proof: Suppose not i.e. let x′ be an allocation such that∑

i∈N θi(x′i) >∑

i∈N θi(xi). Clearly x′ can be obtained from x by a sequence of pair-

wise switches of objects between agents. Consider the allocation x ob-

tained by the switch of objects between agents j and k in the allocation

x, i.e. xi = xi for all i 6= j, k, xj = xk = µ(xk) and xk = xj = µ(xj).

Suppose that ∆µθj ≥ ∆µθk, so that µ(xj) > µ(xk). In particular, let

µ(xk) = µ(xj) + L where L > 0. Then∑i∈N θi(xi)−

∑i∈N θi(xi)

= [θj(µ(xj)) + θk(µ(xk))]− [θj(µ(xk)) + θk(µ(xj))]

= [θj(µ(xj))− θj(µ(xk))]− [θk(µ(xj))− θk(µ(xk))]

=∑L−1

s=0 [∆µθj(µ(xk) + s)−∆µθk(µ(xk) + s)]

≥ 0

The last inequality follows from the assumption that ∆µθj ≥ ∆µθk. If

∆µθj = ∆µθk, an analogous argument yields∑

i∈N θi(xi) =∑

i∈N θl(xi).

Hence,∑

i∈N θi(xi) ≥∑

i∈N θl(xi) in all cases. Moreover applying the

argument repeatedly, we obtain,∑

i∈N θi(xi) ≥∑

i∈N θl(x′i) which con-

tradicts our initial supposition. �

Lemma 1 says the following. Consider a state where the difference

vectors of all agents with respect to permutation µ are all ordered. Now

construct equivalence classes of agents according to their difference


vector, i.e. agent j belongs to a higher equivalence than agent k if

∆µθj ≥ ∆µθk. Two agents belong to the same equivalence class only

if their difference vectors are identical. Construct an allocation x as

follows: if agent j is in an equivalence class higher than k’s then j

gets an object with a higher “index” than k, i.e. µ(xj) > µ(xk). Ties

within equivalence classes can be broken arbitrarily. Then x is efficient

in state θ.

Lemma 2. Let Γ = 〈N, Θ〉 be an allocation problem and let µ ∈ Σ(N).

If Θ is convex, then ∆µΘ is also convex.

The proof of Lemma 2 is straightforward and therefore omitted.

We now turn to the proof of the Necessity part of the Theorem. The

proof involves five steps. In what follows we assume that Γ = 〈N, Θ〉is an arbitrary allocation problem that is implementable with balanced

transfers and that Θ is convex.

Step 1: Pick arbitrary α, β ∈ Θ. Then there exists a permutation

µ ∈ Σ(N) such that ∆µα ≤ ∆µβ.

Proof: Consider any pair α, β ∈ Θ and the difference vector β −α = (β1 − α1, . . . , βn − αn). Consider the permutation µ ∈ Σ(N) of

the elements of β − α with the property that βµ(1) − αµ(1) ≤ . . . ≤βµ(n) − αµ(n). Note that since for each k ∈ {1, . . . , n}, βk − αk ∈ <we can always arrange the elements of β − α in such a manner by

taking appropriate tie breaking rules. Moreover, for all k ∈ {1, . . . , n−1},βµ(k) − αµ(k) ≤ βµ(k+1) − αµ(k+1) ⇒ ∆αµ(k) = αµ(k+1) − αµ(k) ≤βµ(k+1)−βµ(k) = ∆βµ(k) for all k ∈ {1, . . . , n−1} and the result follows.

�

Step 1 says that if we pick any pair α, β from Θ, we can always per-

mute their elements in way such that the associated difference vectors


of α and β are comparable. In the next four steps, we refer to an ar-

bitrary but fixed pair of vectors α, β ∈ Θ. We assume two properties

about α and β without loss of generality

(i) that the permutation µ which makes the difference vectors ∆µα

and ∆µβ comparable (whose existence is guaranteed by Step 1) is the

identity permutation. Furthermore in order to economize on notation,

we will simply write ∆γ, for any γ ∈ Θ as the difference vector of γ

associated with the identity permutation.

(ii) ∆α 6= ∆β. If this condition were violated, then ∆α = c for some

constant c and all α ∈ Θ. This would make the Theorem trivially true.

We introduce some additional notation. Let α, β ∈ Θ be the vectors

that we have identified previously with ∆α ≤ ∆β.

Let Box(∆α, ∆β) =∏n−1

k=1 [∆αk, ∆βk]. Geometrically, it is the cube

obtained with ∆α and ∆β being the opposite ends of the main diagonal.

Let Q(∆α, ∆β) = {k | ∆αk = ∆βk}, i.e. the set of components of

∆α which do not differ from corresponding components of ∆β.

Let L(∆α, ∆β) denote the line joining ∆α and ∆β, i.e. L(∆α, ∆β) =

{s∆β + (1− s)∆α for some s ∈ [0, 1]}.Let L(∆α, ∆β) denote the line on which ∆α and ∆β lie, i.e.

L(∆α, ∆β) = {s∆β + (1− s)∆α for some s ∈ <}.

We now make some critical observations. Since Θ is convex, we can

apply the result of Holmstrom [14] to claim that the mechanism which

implements Γ must be a VCG mechanism, i.e. for all j and for all

θ ∈ Θn, tj(θ) =∑

i6=j θi(x∗i (θ)) − gj(θ−j) where x∗ is an efficient rule

in Γ. Furthermore, since transfers are balanced, we can also apply the

results of Walker [33] to conclude that for all pairs of profiles θ, θ′ ∈ Θn,

we must have 4

(1)∑S⊆N

(−1)|S|∑i∈N

θi(x∗i (θ(S))) = 0

4This is the so-called Cubical Array Lemma.


where θ(S) = (θ1(S), . . . , θn(S)) ∈ Θn is a state such that θj(S) = θj

if j 6∈ S and θj(S) = θ′j if j ∈ S.

Step 2 says the following. All feasible difference vector allocations in

Box(∆α, ∆β) must satisfy a special combinatorial property.

Step 2: There exists c ∈ < such that for all ∆γ ∈ Box(∆α, ∆β)∩∆Θ,

we have ρ.∆γ = c where ρ = (ρ1, . . . , ρn−1) and ρk = (−1)k−1(

n−2k−1

)for

all k = 1, . . . , n− 1.

Proof: Pick an arbitrary type γ ∈ Box(∆α, ∆β) ∩ ∆Θ. Hence for

γ, ∆α ≤ ∆γ ≤ ∆β. The existence of such a type γ is guaranteed by

convexity of ∆Θ domain and the fact that α 6= β. Consider two states

θ = (θ1, . . . , θn) and θ′ = (θ′1, . . . , θ′n) such that θ1 = γ, θj = β for all

j ∈ N−{1} and θ′i = α for all i ∈ N . We claim that applying equation

(1) to the pair θ, θ′ and simplifying, we get

(2)n∑

k=1

ρkγ(k)−n∑

k=1

ρkα(k) = 0

where ρk = (−1)k−1(

n−1k−1

)for all k = 1, . . . , n.

To see this, we evaluate the LHS of equation (1) for the profiles

θ and θ′ described above. Note that for all S ⊂ N , the expression∑i∈N θi(x

∗i (θ(S))) only involves the type vectors α, β and γ. Since

∆α ≤ ∆γ ≤ ∆β, Lemma 1 implies that in any state θ(S), we can

obtain efficient allocations by giving all agents with types β commodi-

ties with greater indices than those with types γ who in turn are given

commodities of indices greater than agents of type α.

We begin by evaluating all the terms involving β in the LHS of

equation (1). Let S ⊂ N be a set which does not contain agent 1.

Observe that the number of agents in S who have type β in the profile

θ(S) is exactly the same as the number of agents in S ∪ {1} who have

type β and this number is n−|S|−1. It follows therefore (from Lemma 1

and the observation above) that in the efficient allocation in state θ(S),

the n − |S| − 1 agents of type β get the commodities n, . . . , |S| + 2.


By the same reasoning, all efficient allocations in the state θ(S ∪ {1})have the property that the n−|S|−1 agents of type β get commodities

n, . . . , |S|+2. Observe that in the LHS of equation (1), the β’s from the

set S appear in the form (−1)|S|(β(n)+. . .+β(|S|+2)) while those from

the set S ∪ {1} appear in the form (−1)|S|+1(β(n) + . . . + β(|S| + 2)).

Since, (−1)|S| + (−1)|S|+1 = 0 these two terms cancel out. We can

conclude therefore that no term involving β appears in the LHS of

equation (1).

Now let us evaluate the terms involving γ. Note that θ(S) involves

γ only when S does not include agent 1. Observe that in θ(S), there

are |S| agents of type α and n − |S| − 1 agents of type β. The effi-

cient allocation in state θ(S) assigns the (|S|+ 1)th object, i.e |S| + 1

to agent 1 (who has type γ). The total number of ways to select

the set S is(

n−1|S|

)where |S| ranges from 0 to n − 1. Collecting all

the terms where γ appears in the LHS of equation (1) we obtain

the expression,∑n−1

|S|=0(−1)k−1(

n−1|S|

)γ(|S| + 1) which is equivalent to∑n

k=1(−1)k−1(

n−1k−1

)γ(k) which is also the first term on the LHS of equa-

tion (2).

Finally we evaluate terms involving α. Note that θ(S) will not in-

volve α only when S = φ. Therefore consider any S ⊆ N with S 6= φ.

There are exactly |S| agents with type α in θ(S) and Lemma 1 implies

that these agents get the commodities 1, . . . , |S| in any efficient alloca-

tion in state θ(S). The sum of valuations of all agents in S in this state

is∑|S|

k=1 α(k). Since the total number of ways in which we can choose

a set of size |S| is(

n|S|

), the expression for all terms involving α in the

LHS of equation (1) is∑

S⊆N,S 6=φ(−1)|S|(

n|S|

) [∑|S|k=1 α(k)

]. Simplifica-

tion of this sum yields the expression −∑n

k=1(−1)k−1(

n−1k−1

)α(k). This

completes the proof our claim that equation (1) reduces to equation

(2) when θ and θ′ are chosen as described.

Next we consider two other states θ and θ′ such that θ1 = γ, θj = α

for all j ∈ N − {1} and θ′i = β for all i ∈ N . Applying arguments

similar to the previous arguments, we can infer that


(3)n∑

k=1

ρkγ(n + 1− k)−n∑

k=1

ρkβ(n + 1− k) = 0

Using the relation ρn+1−k = (−1)n−1ρk in (3) and simplifying, we get

(4)n∑

k=1

ρkγ(k)−n∑

k=1

ρkβ(k) = 0

Therefore, from equations (2) and (4), we have

(5)n∑

k=1

ρkα(k) =n∑

k=1

ρkγ(k) =n∑

k=1

ρkβ(k)

Substituting(

n−10

)=

(n−2

0

)and

(n−1k−1

)=

(n−2k−2

)+

(n−2k−1

)for all k ∈

{2, . . . , n} in (5) it follows that for all γ ∈ Box(∆α, ∆β) ∩∆Θ,

(6)n−1∑k=1

ρk∆α(k) =n−1∑k=1

ρk∆γ(k) =n−1∑k=1

ρk∆β(k) = c

From (6), the result follows. �

According to Step 3, all feasible difference vectors in Box(∆α, ∆β)

must be ordered.

STEP 3: For all ∆γ, ∆γ′ ∈ Box(∆α, ∆β) ∩∆Θ, either ∆γ ≥ ∆γ′ or

∆γ ≤ ∆γ′.

Proof: Suppose that this is false. Then there exists γ, γ′ ∈ Θ for which

we get two mutually exclusive, non-empty and exhaustive subsets T1

and T2 of the set {1, . . . , n− 1} such that

(1) ∆γ(k) ≤ ∆γ′(k) for all k ∈ T1 (with at least one k ∈ T1 for

which the inequality is strict) and

(2) ∆γ(k) > ∆γ′(k) for all k ∈ T2.

Consider a pair of states θ, θ′ ∈ Θn such that θ1 = γ, θ2 = γ′, θj = β

for all j ∈ N − {1, 2} and θ′i = α for all i ∈ N . We evaluate the LHS

of condition (1) for θ and θ′ described above in terms of α, β, γ and γ′.

Let us first evaluate terms involving β and α. Let S ⊂ N − {1}.Observe that the number of agents with types β in the profile θ(S)


is exactly the same as the number of agents with type β in θ(S ∪{1}). Using arguments in Step 2, we conclude that there are no terms

involving β in the LHS of equation (1). Furthermore, we can apply

identical arguments to those in Step 2 with respect to α and infer that

the term involving α is∑n

k=1 ρkγ(qk). Since α ∈ Θ, we know from Step

2 that this sum is the constant c.

We now evaluate terms involving terms involving γ and γ′. There

are three distinct cases to consider.

Case A. S ⊂ N − {1, 2}Case B. S ⊂ N − {1} and 2 ∈ S

Case C. S ⊂ N − {2} and 1 ∈ S

We begin with Case A. Observe that Lemma 1 implies that all ef-

ficient allocations in state θ(S) involve agents 1 and 2 (those with

types γ and γ′ respectively) getting commodities |S| + 1 and |S| + 2.

In fact, 1 and 2 gets |S| + 1 and |S| + 2 respectively if and only if

γ(|S| + 1) + γ′(|S| + 2) ≤ γ(|S| + 2) + γ′(|S| + 1) i.e. if and only if

∆γ(|S| + 1) ≤ ∆γ′(|S| + 1). Observe that |S| ranges from 0 to n − 2

and that there are(

n−2|S|

)ways to pick such a set S. It follows therefore

that the term involving γ and γ′ in the LHS of equation (1) is given by

(7)n−2∑|S|=0

(−1)|S|(

n− 2

|S|

)max[γ(|S|+1)+γ′(|S|+2), γ(|S|+2)+γ′(|S|+1)]

which in turn can be written as

(8)n−1∑k=1

(−1)k−1

(n− 2

k − 1

)max[γ(k) + γ′(k + 1), γ(k + 1) + γ′(k)]

Note again that the expression above can be broken into the terms

(9)∑k∈T1

(−1)k−1

(n− 2

k − 1

)(γ(k)+γ′(k+1))+

∑k∈T2

(−1)k−1

(n− 2

k − 1

)(γ(k+1)+γ′(k))


Furthermore, the first term of the expression above can be further

rewritten as

(10)n−1∑k=1

(−1)k−1

(n− 2

k − 1

)(γ(k)+γ′(k+1))−

∑k∈T2

(−1)k−1

(n− 2

k − 1

)(γ(k+1)+γ′(k))

Therefore expression (8) reduces to

(11)n−1∑k=1

(−1)k−1

(n− 2

k − 1

)(γ(k)+γ′(k+1))+

∑k∈T2

(−1)k−1

(n− 2

k − 1

)(∆γ(k)−∆γ′(k))

We now turn to Case B. Observe that in state θ(S), agent 1 has type

γ, |S| agents have type α and the remaining agents have type β. It

follows from Lemma 1 that agent 1 gets the (|S|+ 1)th object. The set

S can be chosen in(

n−2|S|−1

)ways where |S| ranges from 1 to n−1. Thus,

the term contributed by these sets to the LHS of equation (1) is given

by

(12)n−1∑k=1

(−1)k−1

(n− 2

k − 1

)γ(k + 1)

Case C is the symmetric counterpart of Case B with γ replaced by

γ′. Therefore the term contributed by these sets to the LHS of equation

(1) is given by

(13)n−1∑k=1

(−1)k−1

(n− 2

k − 1

)γ′(k + 1)

The overall term involving γ and γ′ is the sum of expressions (11),

(12) and (13). Observe that the term involving γ′ in the first term of

(11) cancels with (13) because (−1)k + (−1)k−1 = 0. Also note that

the sum of the term involving γ in the first term of (11) and (12) can

be written as∑n

k=1(−1)k−1(

n−1k−1

)γ(k) since

(n−2k−1

)+

(n−2k−2

)=

(n−1k−1

). But

according to Step 2, this sum is c. Therefore, putting together the

terms involving β, α and Cases A, B and C, in equation (1), we obtain


(14)∑k∈T2

(−1)k−1

(n− 2

k − 1

)(∆γ(k)−∆γ′(k)) = −2c

If equation (14) does not hold for the chosen γ, γ′, then the proof

is complete. If however (14) holds, then we argue that we can find

another pair of states for which it does not.

Let θ(λ1) = λ1γ + (1 − λ1)α and θ′(λ2) = λ2γ′ + (1 − λ2)β where

λ1, λ2 ∈ [0, 1]. Convexity of Θ ensures that θ(λ1), θ′(λ2) ∈ Θ. Note that

θ(1) = γ, θ′(1) = γ′, θ(0) = α and θ′(0) = β. We can define two mutually

exclusive and exhaustive subsets T1(λ1, λ2) and T2(λ1, λ2) of the set

{1, . . . , n− 1} such that

(1) ∆θ(λ1)(k) ≤ ∆θ′(λ2)(k) for all k ∈ T1(λ1, λ2) and

(2) ∆θ(λ1)(k) > ∆θ′(λ2)(k) for all k ∈ T2(λ1, λ2)

We construct two states θ, θ′ ∈ Θn where θ1 = θ(λ1), θ2 = θ′(λ2),

θj = β for all j ∈ N − {1, 2} and θ′i = α for all i ∈ N . Applying

equation (1) for the pair θ, θ′ and then simplifying, we obtain(15)Z(λ1, λ2) =

∑k∈T2(λ1,λ2)

ρk {[λ1∆γ(k)− λ2∆γ′(k)]− [(1− λ2)∆β(k)− (1− λ1)∆α(k)]}

where ρk = (−1)k−1(

n−2k−1

)for k = 1, . . . , n− 1 and Z(λ1, λ2) = −2c.

The details behind the derivation of equation (15) are very similar to

those in the derivation of equation (14). Observe first, T2(0, 0) = φ

since ∆θ(1)(k) = ∆α(k), ∆θ′(1)(k) = ∆β(k) and ∆β > ∆α. Also

T2(1, 1) = T2 6= φ by hypothesis. Now consider a path in the unit

square from λ1 = λ2 = 0 to λ1 = λ2 = 1. Along this path we

have observed that T2(λ1, λ2) changes value at least once. However,

it follows from the definition of the function T2(λ1, λ2) that it can

change value only a finite number of times along the path. Since

Z(λ1, λ2) is continuous and non-constant in λ1 and λ2, there must

exist a pair (λ1, λ2) 6= (λ1, λ2) such that T2(λ1, λ2) = T2(λ1, λ2) but

Z(λ1, λ2) 6= Z(λ1, λ2). However this would contradict equation (15)

which requires Z(λ1, λ2) to be a constant for all λ1, λ2 ∈ [0, 1]. This

establishes Step 3. �


The next step establishes that feasible difference vectors in Box(∆α, ∆β)

must, in fact, lie on the line joining ∆α and ∆β.

Step 4: Box(∆α, ∆β) ∩∆Θ = L(∆α, ∆β).

Proof: Suppose that this is false. That is, we can find a vector ∆γ ∈Box(∆α, ∆β) ∩∆Θ such that ∆γ 6∈ L(∆α, ∆β). Since ∆Θ is convex

and α, β ∈ Θ by assumption, the triangle formed by the non co-linear

points (∆α, ∆β, ∆γ) also belongs to ∆Θ. But then we can find two

“non-comparable” vectors in such a triangle, i.e. we can find ∆γ′ and

∆γ′′ such that ∆γ′(k) > ∆γ′′(k) and ∆γ′(k) < ∆γ′′(k) for some k 6= k

and k, k ∈ {1, . . . , n − 1} which would contradict Step 3. Hence, the

result follows. �

The final step shows that all points in the difference domain ∆Θ lie

on the line on which ∆α and ∆β lie.

Step 5: L(∆α, ∆β) ⊆ ∆Θ ⊂ L(∆α, ∆β).

Proof: Pick an arbitrary ∆γ ∈ ∆Θ. Suppose that ∆γ /∈ L(∆α, ∆β),

i.e. ∆α, ∆β and ∆γ are not co-linear. Let D denote the convex hull

of these points. Formally, D = {λ1∆α + λ2∆β + λ3∆γ; λ1, λ2, λ3 ≥0, and λ1 + λ2 + λ3 = 1}. Since the difference domain ∆Θ is convex

(Lemma 2) and ∆α, ∆β, ∆γ ∈ ∆Θ, we have D ⊂ ∆Θ. Let ∆u =13∆α + 1

3∆β + 1

3∆γ. Let ε ≡ (ε1, ε2, ε3) be such that ε1 + ε2 + ε3 = 0

and let ∆v(ε) = (13

+ ε1)∆α + (13

+ ε2)∆β + (13

+ ε3)∆γ. Clearly ∆u ∈∆Θ (Lemma 2). In addition since ∆α, ∆β and ∆γ are not co-linear,

there exists a neighbourhood around (0, 0, 0) such that for all ε in this

neighbourhood, ∆v(ε) ∈ ∆Θ (also Lemma 2).

Now fix an arbitrary ε in this neighbourhood and consider the pair

u, v(ε) ∈ Θ. Applying Steps 1 through 4 to this pair, it follows that

there exists a permutation µ ∈ Σ(N) such that

(16) Box(∆µu, ∆µv(ε)) ∩∆µΘ = L(∆µu, ∆µv(ε))


Of course, ∆µu = 13∆µα+ 1

3∆µβ+ 1

3∆µγ and ∆µv(ε) = (1

3+ε1)∆

µα+

(13

+ ε2)∆µβ + (1

3+ ε3)∆

µγ. Let ∆µw be the mid-point in the line

L(∆µu, ∆µv(ε)), i.e. ∆µw = (13+ ε1

2)∆µα +(1

3+ ε2

2)∆µβ +(1

3+ ε3

2)∆µγ.

Now pick h ≡ (h1, h2, h3) 6= 0 with h1+h2+h3 = 0 such that ∆µw′ =

(13+ ε1

2+h1)∆

µα+(13+ ε2

2+h2)∆

µβ+(13+ ε3

2+h3)∆

µγ. Clearly, ∆µw′ ∈∆µΘ. Since Box(∆µu, ∆µv(ε)) =

∏k[

13∆µαk + 1

3∆µβk + 1

3∆µγk, (

13

+

ε1)∆µαk + (1

3+ ε2)∆

µβk + (13

+ ε3)∆µγk], we can ensure that ∆µw′ ∈

Box(∆µu, ∆µv(ε)) by choosing any h such that |hr| < | εr

2| for all r =

1, 2, 3. By choosing such an h, we have ∆µw′ ∈ Box(∆µu, ∆µv(ε)) ∩∆µΘ.

Equation (16) implies that ∆µw′ ∈ L(∆µu, ∆µv(ε)). This, in turn,

implies the existence of τ(h), 0 ≤ τ(h) ≤ 1 such that ( τ(h)3

+ (1 −τ(h))(1

3+ε1))∆

µα+( τ(h)3

+(1−τ(h))(13+ε2))∆

µβ+( τ(h)3

+(1−τ(h))(13+

ε3))∆µγ = (1

3+ ε

2+ h1)∆

µα + (13

+ ε2

+ h2)∆µβ + (1

3+ ε3

2+ h3)∆

µγ.

On simplification, this yields, a1∆µα + a2∆

µβ = (a1 + a2)∆µγ where

ar = [(1 − t(h))εr − εr

2− hr] for r = 1, 2. If a1 + a2 6= 0, we have

∆µγ =(

a1

a1+a2

)∆µα +

(a2

a1+a2

)∆µβ. Hence the points ∆µα, ∆µβ and

∆µγ are co-linear. If a1 + a2 = 0 then we obtain ∆µα = ∆µβ. The

latter implies αµ = βµ + K for some constant K where αµ and βµ are

the vectors α and β whose components have been permuted according

to µ . Therefore ∆α = ∆β which contradicts our initial assumption

that ∆α 6= ∆β. We conclude therefore that ∆µα, ∆µβ and ∆µγ are

co-linear. We complete the proof of Step 5 by showing that this cannot

be the case.

Since ∆µγ = λ∆µα+(1−λ)∆µβ, we have ∆µγ = ∆µ(λα+(1−λ)β).

Hence γµ = λαµ+(1−λ)βµ+K for some constant K and γ = λα+(1−λ)(β) + K. Therefore, ∆γ = ∆(λα + (1 − λ)β) = λ∆α + (1 − λ)∆β.

Consequently, ∆α, ∆β and ∆γ are co-linear which contradicts our

earlier assumption. This completes Step 5.

�

We can now complete the proof of the necessity part of the Theo-

rem. We have shown that there exists a permutation of the objects


such that the all points of the associated difference domain lie on a

straight line which satisfies parts (i) and (ii) of the statement of the

Theorem. Convexity of the difference domain (Lemma 2) imples that

the difference domain must in fact be an interval.

Sufficiency: We assume without loss of generality that the difference

domain associated with the identity permutation, is a straight line sat-

isfying the properties specified in the statement of Theorem 1. We will

show that an efficient rule in this allocation problem is implementable

with balanced transfers.

For all θ ∈ Θn, we define a permutation σ(θ) : N → N as follows:

for all i, j ∈ N , σi(θ) > σj(θ)5 if and only if either [∆θi > ∆θj] or

[∆θi = ∆θj and i > j] hold. In other words, all agents are ranked

according to their difference vectors with ties broken in favour of the

agent with the higher index. Observe that our assumption regarding Θ

implies that for all θ ∈ Θn and i, j ∈ N one of the following must hold:

∆θi > ∆θj, ∆θj > ∆θi or ∆θi = ∆θj. Therefore σ(θ) is well-defined.

Now define a function x : Θn → Σ(N) as follows. For all θ ∈ Θn,

and i ∈ N , xi(θ) = σi(θ). Since σ(θ) is an permutation of N , it follows

that x is an allocation. Moreover it is easy to verify using Lemma 1

that x(θ) is efficient in state θ.

For an arbitrary agent i ∈ N and θ−i ∈ Θn−1 we define a permu-

tation, σ(θ−i) : {1, . . . , n − 1} → {1, . . . , n − 1} as follows: for all

j, k ∈ N − {i}, σj(θ−i) > σk(θ−i) if and only if either [∆θj > ∆θk] or

[∆θj = ∆θk and j > k] hold.

It follows from the construction of σ(θ) and σ(θ−i) that the following

property holds: for all j 6= i,

σj(θ−i) =

{σj(θ) if σj(θ) < σi(θ)

σj(θ)− 1 if σj(θ) > σi(θ)

5We are abusing notation slightly here; σi(θ) is the value of the function σ(θ) ati etc


The last condition follows from the fact that the domain Θ contains

types that are all well ordered in terms of first differences. In any state

θ if an agent i ∈ N with type θi is removed, the relative ranking of first

differences of types of the remaining set of individuals (that is, ∆θj for

all j ∈ N − {i}) remains unchanged.

For any distinct i, j ∈ N and θj ∈ Θq, hj(σj(θ−i), θj) is defined as

follows:

(17)

hj(σj(θ−i), θj) =

σj(θ−i)∑r=1

(−1)(σj(θ−i)−r) (σj(θ−i)− 1)!(n− σj(θ−i)− 1)!

(r − 1)!(n− r)!θj(r)

where θj(r) = θj(r)− rn−1c(−1)n−1(n−1)!

for all r = 1, . . . , n.

We are now going to consider the sum∑

i:i6=j hj(σj(θ−i); θj) for agent

j i.e. we are going to “hold” j and sum over all i where i 6= j. Using the

relationship between σj(θ−i) and σj(θ) stated above and noting that

|{i|σj(θ) < σi(θ)}| = n− σj(θ) and |{i|σj(θ) > σi(θ)}| = σj(θ)− 1, we

obtain∑i:i6=j hj(σj(θ−i), θj)

= (n− σj(θ))hj(σj(θ), θj) + (σj(θ)− 1)hj(σj(θ)− 1, θj)

= θj(σj(θ))− [σj(θ)]n−1c

(−1)n−1(n−1)!

= θj(xj(θ))− [σj(θ)]n−1c

(−1)n−1(n−1)!

For all j ∈ N and θ−j ∈ Θn−1, define gj(θ−j) = (n−1)∑

i:i6=j hi(σi(θ−j), θi)+

A(c) where A(c) = c(−1)n−1(n−2)!n

∑nk=1 kn−1. Observe that∑

j∈N

gj(θ−j)

= (n− 1)∑j∈N

∑i6=j

hi(σi(θ−j), θi) + nA(c)

= (n− 1)∑j∈N

{∑i6=j

hj(σj(θ−i), θj)

}+ nA(c)

= (n− 1)∑j∈N

θj(xj(θ))− (n− 1) c(−1)n−1(n−1)!

∑j∈N

[rj(θ)]n−1 + nA(c)

= (n−1)∑j∈N

θj(xj(θ))−nA(c)+nA(c) (since∑j∈N

[σj(θ)]n−1 =

n∑k=1

kn−1)


= (n− 1)∑j∈N

θj(xj(θ)).

Now consider the VCG mechanism M = 〈x, t〉 where for all θ ∈ Θn,

tj(θ) =∑

i6=j θi(xi(θ))−gj(θ−j). Then,∑

j∈N tj(θ) = (n−1)∑

j∈N θj(xj(θ))−∑j∈N gj(θ−j) = 0. Therefore transfers are balanced and the proof is

complete. �

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Economic Research Unit, Indian Statistical Institute, Kolkata, In-

dia.

E-mail address: [email protected]

Planning Unit, Indian Statistical Institute, New Delhi, India.

E-mail address: [email protected]

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