Envy-free and budget-balanced assignment

of identical objects

Yan Long

⇤

November 15, 2017

Abstract

Strategy-proof, budget-balanced, and envy-free rank mechanisms

assign q identical objects to n agents. The efficiency loss is the largest

ratio of surplus loss to efficient surplus, over all profiles of non-zero

valuations. The smallest efficiency loss

n�qn2�n is uniquely achieved by

the following simple allocation rule: assign one object to each of the

q � 1 agents with the highest valuations, a large probability to the

agent with the qth highest valuation, and the remaining probability to

the agent with the (q + 1)th highest valuation.

Keywords

Multiple-object assignment; Budget balance; Envy-freeness; Worst

case analysis

JEL Classification D70; D82; D44

⇤New York University Abu Dhabi. Email: ylong@nyu.edu. I am grateful to HervéMoulin for very inspiring discussions and to referees for very helpful suggestions. Seminarparticipants at the University of Glasgow are acknowledged. This paper is part of thesecond chapter of my PhD thesis. The other part of that chapter is merged with a similarand independent work by Debasis Mishra and Tridib Sharma, and appears in a relatedpaper (Long et al. (2017)).

1

1 Introduction

Uniform randomization (“Lottery”) is common practice in assignment. Wheneveryone has equal claim, Lottery is appealing in that it is simple and fair.However, it is notably inefficient, since people usually appreciate the samething to different extent. If quasi-linear preference is assumed and moneytransfer is allowed, the following is an easy way to improve on efficiency whilekeeping the favorable aspects of Lottery. Say, one object is to be assigned ton agents and each agent i has a private valuation vi 2 R+. Denoted by v⇤k

the kth highest valuation. Let agent with v⇤1 get the object with probabilityn�1n and pay n�2

n v⇤2, agent with v⇤2 get it with probability 1n and pay nothing,

and all other agents receive 1nv⇤2. It is easy to check that this mechanism is

strategy-proof, budget-balanced, anonymous, and envy-free: nice propertiesshared by Lottery. As far as welfare is concerned, it loses 1

n of the efficientsurplus v⇤1 at most; Lottery, however, loses n�1

n v⇤1 in the worst case.For any budget-balanced mechanism, we define its worst efficiency loss

or simply efficiency loss to be the largest ratio of surplus loss to the ef-ficient surplus, over all profiles of non-zero valuations.1 It is well knowthat no strategy-proof and budget-balanced mechanisms can achieve zeroefficiency loss (Green and Laffont (1979)). We show that for single-objectallocation, the above mechanism is the unique one to achieve the smallestefficiency loss among all strategy-proof, budget-balanced, and envy-free rankmechanisms. By rank mechanisms, we mean anonymous mechanisms whoseallocation rules can be represented by a vector (a1, a2, · · · , an), where ak

is the probability that agent with the kth highest valuation gets an object.Both Lottery and VCG mechanisms (with uniform tie-breaking rule) are rankmechanisms. We focus on rank mechanisms for their simplicity and tractabil-ity. Not only are the allocation rules fairly simple, the efficiency measure andthe budget-balanced condition will also be simplified dramatically under rank

1Worst case analysis has been applied to various settings, see the introduction of Moulin(2009) for a brief survey.

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mechanisms.In general, when q identical objects are to be assigned to n agents (q < n)

and each agents demands at most one object, the allocation rule that archivesthe smallest efficiency loss is the following: give one object to each of the q�1

agents with the highest evaluations, a large probability to the agent withthe qth highest valuation, and the remaining probability to the agent withthe (q + 1)th highest valuation. Unlike Lottery, which randomize uniformlyamong all agents, this rule only randomize between the two agents at themargin. That is, the two agents with the qth and (q+1)th highest valuations.For q = 1, this mechanism is the same as “randomly selecting one agent tobe the residual claimer of the Vickery mechanism run among all the otheragents” (proposed first in Green and Laffont 1979); for q > 1, these two aredifferent.

To solve for the smallest efficiency loss, we first characterize the class ofall strategy-proof and budget-balanced rank mechanisms. This class was firststudied and characterized by Long et al. (2017). We give an alternative char-acterization here, which helps us to identify the envy-free sub-class. We thenobtain in the sub-class the mechanism that achieves the smallest efficiencyloss. This mechanism shares the good fairness properties owned by Lottery,and behaves much better in terms of efficiency. It is hoped to be a readyalternative.

Literature review

When allocating objects among a group of agents, allocation efficiency re-quires to assign the objects to the agents with the highest valuations; howeverno truth-telling mechanism can be efficient and budget-balanced at the sametime (Green and Laffont (1979)). To tackle the impossibility, different ap-proaches exist in literature.

Several papers enforce allocation efficiency and explore the idea of redis-tributing VCG payments while respecting incentives (Cavallo (2006), Moulin

3

(2009), Guo and Conitzer (2009), etc.). Moulin (2009) and Guo and Conitzer(2009) find independently the VCG mechanisms that minimize (relative) bud-get surplus.

In Guo and Conitzer (2014), Drexl and Kleiner (2015), etc., welfare lossis defined to be the sum of allocation efficiency loss and budget surplus. Guoand Conitzer (2014) give an algorithm to compute the smallest (relative)welfare loss in the restricted class of rank mechanisms. Drexl and Kleiner(2015) address a similar problem as in Guo and Conitzer (2014), in a Bayesiansetting, for two agents. Interestingly, they find out that under a mild distri-bution assumption, an optimal deterministic mechanism is budget-balanced.

Note that in all the papers mentioned above (except Drexl and Kleiner(2015)), efficiency loss is defined relative to the efficient surplus. Even if theworst efficiency loss is close to zero, it is still possible that a huge amount ofmoney will be burned if agents have high enough valuations. While lotteriesare wildly used, one rarely observes in reality allocations accompanied byburning money: mis-allocation seems to be more easily tolerated for whateverreason. A recent paper (Long et al. (2017)) enforces budget balance as ahard constraint and searches for mechanisms that behave well in terms ofefficiency. Efficiency loss is hence caused solely by misallocation. Buildingon the rank mechanism proposed in Guo and Conitzer (2014) (they use theterm “linear mechanisms”), Long et al. (2017) first characterize the class ofall strategy-proof and budget balanced rank mechanisms, and then find outmechanisms that achieve the smallest worst ratio of the efficiency loss to theoptimal surplus within that class. They show that the smallest efficiencyloss converges to zero at an exponential rate, i.e., the optimal mechanismconverges to efficiency at an exponential rate.

This paper works in the same framework as Long et al. (2017), but addsan additional dimension: fairness. Fair allocation is one of the main concernsparticularly in assignment settings where everyone has equal claim. We henceimpose envy-freeness, one of the most compelling fairness criteria, to the

4

mechanisms. The optimal mechanism still converges to efficiency, but ata slower rate. Our paper is also related to Moulin (2010) and Sprumont(2013). Both of them work in a single-object assignment setting, considermechanisms that does not necessarily satisfy allocation efficiency, and invokefairness properties. Moulin (2010) explores the trade-off between “k-fairness”(first proposed by Porter et al. (2004)) and the efficiency loss in the generalclass of strategy-proof (deterministic) mechanism. He finds the mechanismguaranteeing to each agent a fair share ( 1

n) of the qth highest valuation, forany integer q, 3 q n, while minimizing the worst ratio of the welfareloss to the efficient surplus. Sprumont (2013) offers an explicit description ofthe individually rational mechanisms which are Pareto-optimal in the classof strategy-proof, anonymous, and envy-free (deterministic) mechanisms.

Our paper is also related to a slightly different line of work, which studiesenvy-free VCG mechanisms in various quasilinear assignment models. Recentpapers are Svensson (2004), Pápai (2003), and Ohseto (2006). All these pa-pers restrict attention to deterministic mechanisms that never leave an objectunallocated. Since under these restrictions envy-freeness implies allocationefficiency, the mechanisms they look at are VCG mechanisms. For exam-ple, in a multiple-object assignment model similar to ours, Ohseto (2006)characterizes the set of envy-free VCG mechanisms and identifies the Paretoun-dominated subset.

2 Setting

We consider the following problem: q identical objects are to be assigned ton agents; and 1 q < n. Each agent i 2 N = {1, · · · , n} demands at mostone object and has a private valuation vi 2 R+ for the object. If an agentis given an object with probability �i 2 [0, 1] and she pays ti for it, then hernet utility is �ivi � ti.

For any profile v 2 RN+ of valuations and any i 2 N , v�i stands for the

5

vector obtained from v by deleting vi, and (v0i, v�i) stands for the valuationprofile obtained from v by replacing vi with v

0i.

An allocation rule is a mapping � : RN+ ! [0, 1]n, where we denote by

�i(v) the probability of agent i getting an object at valuation profile v. Weassume that for any v 2 RN

+ ,P

i2N �i(v) q.A payment function is a mapping t : RN

+ ! Rn, where we denote byti(v) agent i’s payment at valuation profile v. A mechanism is a pair (�, t).A mechanism (�, t) is

• strategy-proof if for any i 2 N , for any v�i 2 RN\{i}+ , and for any vi,

v

0i 2 R+

�i(vi, v�i) · vi � ti(vi, v�i) � �i(v0i, v�i) · vi � ti(v

0i, v�i);

• budget-balanced if for any v 2 RN+ ,

X

i

ti(v) = 0;

• anonymous if for any v 2 RN+ , any i 2 N , and any permutation ⇡ on

N ,�⇡(i)(⇡v) = �i(v) and t⇡(i)(⇡v) = ti(v),

where the profile ⇡v is defined by (⇡v)⇡(i) = vi for all i 2 N ;

• envy-free if for any v 2 RN+ and any i, j 2 N ,

�i(v) · vi � ti(v) � �j(v) · vi � tj(v).

Let M be the class of mechanisms that are strategy-proof, budget-balanced,anonymous and envy-free. Note that Lottery is a member of the class. Ourgoal is to find mechanisms in M that behaves well in terms of efficiency. Toso that, we adopt a prior-free notion to measure efficiency.

6

Note that for any budget-balanced mechanism (�, t), efficiency loss iscaused solely by mis-allocation. Let V ⌘ {v 2 RN

+ : v1 � · · · � vn and v1 >

0}. At any valuation profile v 2 V , the maximum possible social welfare isP

1iq vi, and the social welfare achieved by (�, t) isP

i2N �i(v) · vi. Theratio of the surplus loss to efficiency surplus

P1iq vi �

Pi2N �i(v) · viP

1iq vi

is a (relative) measure of efficiency loss at the valuation profile v. The worst

efficiency loss, or simply the efficiency loss of (�, t) is the largest ratio ofsurplus loss to efficient surplus, over all profiles of non-zero valuations:

supv2V

"1�

Pi2N �i(v) · viP

1iq vi

#.

3 Rank mechanisms

Ideally we would like to find mechanisms that achieve the smallest efficiencyloss in M, which we do not know how to do currently. Instead, we restrictour attention to “rank mechanisms”, a class of anonymous mechanisms inwhich the probability of an agent getting an object depends only on theranking of her valuation. This feature makes the allocation rule simple. Andas will be shown later, it also simplifies the efficiency measure, and makesthe budget-balanced condition tractable.

Note that anonymity requires agents with the same valuations get anobject with the same probability. We hence define the “set of rankings” foreach agent, which will play a role in the uniform tie-break. For any v 2 RN

+

and any i 2 N , let

ri(v) ⌘ {l 2 N : |{j 2 N : vj > vi}| < l n� |{j 2 N : vj < vi}|}.

7

That is, ri(v) is agent i’s set of rankings. Note that ri(v) is a singleton if andonly if vi 6= vj for any j 6= i.

Definition 1. A mechanism (�, t) is a rank mechanism if (1) it is anonymousand (2) there exists {al}nl=1 with each al 2 [0, 1] and

Pnl=1 al q such that

for any v 2 RN+ and any i 2 N , �i(v) =

Pl2ri(v)

al

|ri(v)| .

We hence use ({al}; t) to refer to a rank mechanism, and denoted by Mr

the class of rank mechanisms in M. Note that Lottery is a rank mechanism,where al =

1n for any l. Any VCG mechanism with uniform tie-breaking rule

is also a rank mechanism, where al = 1 for any l q and al = 0 otherwise.

3.1 Efficiency loss

Fix a mechanism ({al}; t) in Mr. The efficiency loss of the mechanism thentakes the following form:

supv2V

"1�

Pnl=1 al · vlP1lq vl

#,

and has a surprisingly simple representation.

Proposition 1. For any a1 � · · · � an � 0,

supv2V

"1�

Pnl=1 al · vlP1lq vl

#= 1�

Pql=1 al

q

. (1)

Proof. For any v 2 V and any i, j 2 N ,

(ai � aj)(vi � vj) � 0.

That isaivi + ajvj � aivj + ajvi.

8

Adding up the inequalities for all pairs of (i, j) such that 1 i, j q, we get

q(a1v1 + a2v2 + · · ·+ aqvq) � (a1 + a2 + · · ·+ aq)(v1 + v2 + · · ·+ vq). (2)

Therefore for any v 2 V ,Pn

l=1 alvlPlq vl

�P

lq alvlPlq vl

�P

lq al

q

. (3)

The first inequality is obvious; the second one is due to (2).If v1 = v2 = · · · = vq > 0 and vq+1 = vq+2 = · · · = vn = 0, then both

inequalities in (3) hold with equality. Hence

infv2V

"Pnl=1 alvlPlq vl

#=

Plq al

q

,

as desired.

Therefore, minimizing the efficiency loss of mechanisms in Mr is equiva-lent to maximizing

Pql=1 al of mechanisms in Mr.

Now we head towards a characterization of the set Mr.

3.2 Strategy-proofness

Lemma 1. A rank mechanism ({al}; t) is strategy-proof if and only if

an an�1 · · · a1;

and for any v 2 RN+ and any i 2 N ,

ti(v) = �i(v)vi �Z vi

0

�i(xi, v�i)dxi � h(v�i)

where �i(v) =P

l2ri(v)al

|ri(v)| and h(·) is a symmetric function from Rn�1+ to R.

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Proof. The proof is similar to the proof of Lemma 2 in Myerson (1981). Seealso Long et al. (2017) for the proof of the q = 1 case. The proof for themulti-unit case is omitted.

Remark 1. Note that for any v 2 V and any i 2 N ,R vi0 �i(xi, v�i)dxi =

Pnj=i(vj � vj+1)aj, where vn+1 = 0. Therefore ui(v) = �i(v)vi � ti(v) =

Pnj=i(vj � vj+1)aj + h(v�i). Let l 2 ri(v) and t

(l)i (v) = alvi �

Pnj=l(vj �

vj+1)aj � h(v�i). Then ui(v) = alvi � t

(l)i (v) and ti(v) =

Pl2ri(v)

t(l)i (v)

|ri(v)| .

We use ({al};h) to refer to a strategy-proof rank mechanism.

3.3 Budget balance

We call {�k}n�1k=1 2 Rn�1 a consistent coefficient vector of a mechanism

({al};h) if it satisfies the following system of linear equations: for all j,0 j n� 1,

j · �j + (n� j � 1) · �j+1 = j(aj � aj+1), (4)

where �0 = �n = a0 = 0.For any (x1, · · · , xn�1) 2 Rn�1

+ , let x⇤1, · · · , x⇤(n�1) be a reordering ofx1, · · · , xn�1 such that x⇤1 � · · · � x⇤(n�1).

Proposition 2. A strategy-proof rank mechanism ({al};h) is budget-balancedif and only if there exists a consistent coefficient vector {�k}n�1

k=1 2 Rn�1 suchthat for any x 2 Rn�1

+ , h(x) =Pn�1

k=1 �kx⇤k .

Proof. By Lemma 1, for any v 2 V ,

nX

i=1

ti(v) =nX

i=1

aivi �nX

i=1

nX

j=i

(vj � vj+1)aj �nX

i=1

h(v�i)

=n�1X

i=1

i · (ai � ai+1)vi+1 �nX

i=1

h(v�i)

, (5)

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where vn+1 = 0.We show in the appendix (Lemma 2) that

Pi ti(v) = 0 for any v 2 V

implies that h(·) takes a “linear” form. That is, there exists {�k}n�1k=1 2 Rn�1

+

such that for any x 2 Rn�1+ ,

h(x) = �1x⇤1 + �2x⇤2 + · · ·+ �n�1x⇤(n�1).

Therefore for any v 2 V ,

nX

i=1

h(v�i) =nX

i=1

((i� 1)�i�1 + (n� i)�i)vi,

where �0 = �n = 0. It is then easy to see thatP

i ti(v) = 0 for any v 2 V ifand only if the system of linear equations (4) holds.

For any n > 0 and any m, 0 m n, let Cnm be the binomial coefficient

“take m among n”. For convenience, let C00 ⌘ 1, and Cn

m ⌘ 0 for any m < 0

and any pair m, n such that m > n.

Corollary 1. A strategy-proof rank mechanism ({al}, h) is budget-balancedif and only if its allocation vector {al} satisfies the following equation:

nX

l=1

(�1)l�1 · �l · al = 0, (6)

where �l ⌘ Cn�2l�2 + Cn�2

l�1 for 1 l n.

Proof. Fix an allocation vector {al}nl=1. The system of linear equations (4)have n � 1 unknowns , i.e., {�k}n�1

k=1 , and n equations. By straightforwardcalculation (see Lemma 3 in the appendix for details), the linear system (4)has a unique solution if and only if Equation (6) of {al}nl=1 holds.

Therefore a strategy-proof rank mechanism ({al}, h) is budget-balancedif and only if its allocation vector {al} satisfies Equation (6). This is exactlythe characterization provided in Long et al. (2017).

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We hence use ({al}; {�k}) as well as ({al}) to refer to a strategy-proofand budget-balanced rank mechanism.

3.4 Envy-freeness

Proposition 3. A strategy-proof and budget-balanced rank mechanism ({al}; {�k})is envy-free if and only if �k � 0 for all k, 1 k n� 1.

Proof. First we show the “only if” part. Let v 2 V and l 2 {1, · · · , n � 1}.Assume that vk 6= vl for any k 6= l, and vk 6= vl+1 for any k 6= l+ 1. Then byLemma 1 and Proposition 2,

tl(v)� tl+1(v) = (al�al+1)vl+1��l(vl+1�vl) = �lvl+(al�al+1��l)vl+1. (7)

If �l < 0, then for some vl�1, vl and vl+1 such that vl�1 > vl > vl+1 andvl >

(al�al+1��l)vl+1

��l,

tl(v)� tl+1(v) = �lvl + (al � al+1 � �l)vl+1 < 0.

On the other hand, envy-freeness requires that al+1vl+1 � tl+1(v) � alvl+1 �tl(v). Therefore

tl(v)� tl+1(v) � (al � al+1)vl+1 � 0.

Contradiction!Now we show the “if” part. Suppose �k � 0 for all k, 1 k n�1. Then

according to the linear system (4),

ak � ak+1 � �k =n� k � 1

k

�k+1 � 0.

Without loss of generality assume v 2 V . Suppose vi = vj, then by anonymityof the rank mechanisms, ti = tj and ui = uj, as desired. Suppose vi > vj

12

(hence i < j). Note that i 2 ri(v) and j 2 rj(v). Note that for each l 2 N ,

t

(l)l (v) = alvl �

nX

j=l

(vj � vj+1)aj � (l�1X

k=1

�kvk +n�1X

k=l

�kvk+1).

Then

t

(i)i (v)� t

(j)j (v) =

j�1X

l=i

(al � al+1)vl+1 + �ivi �j�2X

l=i

(�l � �l+1)vl+1 � �j�1vj

= �ivi +j�2X

l=i

(al � al+1 � �l + �l+1)vl+1 + (aj�1 � aj � �j�1)vj

.

Since al � al+1 � �l + �l+1 � al � al+1 � �l � 0 for any i l j � 2 andaj�1 � aj � �j�1 � 0 for any j, we have

t

(i)i (v)�t

(j)j (v) �ivi+

j�2X

l=i

(al�al+1��l+�l+1)vi+(aj�1�aj��j�1)vi = (ai�aj)vi,

and

t

(i)i (v)�t

(j)j (v) � �ivj+

j�2X

l=i

(al�al+1��l+�l+1)vj+(aj�1�aj��j�1)vj = (ai�aj)vj.

Note that ui(v) = aivi � t

(i)i (v) (see Remark 1). Therefore for each l 2

rj(v), ui(v) � alvi � t

(l)l (v). Therefore ui(v) �

Pl2rj(v)

al

|rj(v)| vi �P

l2rj(v)t(l)l (v)

|rj(v)| =

�j(v)vi � tj(v). Agent i does not envy agent j. Similarly, agent j does notenvy agent i.

Corollary 2. A strategy-proof and budget-balanced rank mechanism ({al})is envy-free if and only if for each k, 2 k n� 1,

kX

j=1

(�1)k+1�j[n�jY

i=n�k

�i +n�1�jY

i=n�k

�i]aj � 0, (8)

13

where �n�1 = 0, and for each i, 1 i n� 2, �i =n�i�1

i .

Proof. We show in Lemma 3 in the appendix that �1 = 0, and for each k,2 k n� 1, �k equals the left hand side of the inequality (8).

4 Optimal rank mechanisms

We define below the class of mechanisms that achieve the smallest efficiencyloss in Mr, which we name “Randomize-between-Two” (RT) mechanisms. InRT mechanisms, only the two agents with the qth and (q+1)th valuations areinvolved in randomization of assignment. Each of the remaining agents withhigher valuations is assigned one object, and each of the remaining agentswith lower valuations is assigned zero probability of getting an object (andis subsidized).

Definition 2. For any n and q, 1 q < n, the Randomize-between-Two(RT) mechanism ({a?l }; {�?k}) is the following2:

a

?l =

8>>>>>><

>>>>>>:

1 l < q

(n�1)n�(n�q)qn2�n l = q

(n�q)qn2�n l = q + 1

0 l > q + 1

;

and

�

?k =

8>>>>>><

>>>>>>:

0 k < q

(q�1)qn2�n k = q

(n�q)qn2�n k = q + 1

0 k > q + 1

.

2We do not give the explicit form of the payment function, which is a routine calculationfollowing Lemma 1 and Proposition 2.

14

Remark 2. Note that (i)Pn

l=1 a?l = q, and for any l and l

0 such that l < l

0,1 � a

?l � a

?l0 � 0; (ii) given that for each l, al = a

?l , {�?k} is the unique solution

of the linear system (4); (iii) for any k, �?k � 0. Therefore ({a?l }; {�?k}) 2 Mr.Since

Pq+1l=1 a

?l = q, the efficiency loss is a?q+1

q = n�qn2�n . Note that if q = 1,

then a

?1 =

n�1n , a?2 = 1

n and a

?l = 0 for any l > 2. This is the allocation rule

pointed out at the beginning of the Introduction.

Example 1. The RT mechanism for n = 4, q = 2.The allocation rule is: a

?1 = 1, a?2 = 2

3 , a?3 =

13 , a

?4 = 0.

Since �

?1 = 0, �?2 = 1

6 , �?3 = 1

3 , the payment function is (under the assump-tion that v1 > v2 > v3 > v4):

t1 =(1

3v2 +

1

3v3 +

1

3v4)� (

1

6v3 +

1

3v4) =

1

3v2 +

1

6v3

t2 =1

3v3 +

1

3v4 � (

1

6v3 +

1

3v4) =

1

6v3

t3 =1

3v4 � (

1

6v2 +

1

3v4) = �1

6v2

t4 =� (1

6v2 +

1

3v3) = �1

6v2 �

1

3v3

.

One can check thatP4

i=1 ti = 0. And the efficiency loss is 16 .

Theorem 1. For any n and q, 1 q < n, RT mechanism is the uniquemechanism that achieves the smallest efficiency loss in Mr.

Proof. Suppose for the sake of contradiction that there exists another mech-anism ({al}; {�k}) 2 Mr that achieves the same or smaller efficiency loss.Then by Proposition 1, aq � a

?q, aq+1 a

?q+1, and aq � aq+1 > a

?q � a

?q+1.

Suppose q = n � 1. Since, by (4), �n�1 = an�1 � an, and an�1 � an >

a

?n�1 � a

?n, we have �n�1 > �

?n�1. However, since (n � 2)�n�2 + �n�1 = (n �

2)(an�2 � an�1), and �n�2 � 0 = �

?n�2, an�2 1 = a

?n�2, an�1 � a

?n�1, we

have �n�1 �

?n�1. Contradiction.

Suppose q n�2. Since (q+1)�q+1+(n�q�2)�q+2 = (q+1)(aq+1�aq+2),and �q+2 � 0 = �

?q+2, aq+1 a

?q+1, aq+2 � 0 = a

?q+2, we have �q+1 �

?q+1.

15

Since (q � 1)�q�1 + (n � q)�q = (q � 1)(aq�1 � aq), and �q�1 � 0 = �

?q�1,

aq�1 1 = a

?q�1, aq � a

?q, we have �q �

?q . However since q�q + (n � q �

1)�q+1 = q(aq�aq�1), and �q �

?q , aq�aq+1 > a

?q�a

?q+1, we have �q+1 > �

?q+1.

Contradiction.

Therefore the smallest efficiency loss achieved by mechanisms in Mr

is n�qn2�n . It is achieved uniquely by the RT mechanisms. Note that RT mech-

anism converges to efficiency as n increases. To understand the restrictionimposed by envy-freeness, we mention here some results in Long et al. (2017).Envy-freeness is costly in most cases. That is, for most n and q, smaller ef-ficiency loss can be achieved by rank mechanisms that are strategy-proof,budget balanced but not envy-free. In fact, fixed q, the smallest efficiencyloss generated by such mechanisms goes to zero exponentially in n. However,for n and q such that q < n 8, RT mechanism remains optimal among allrank mechanisms that are strategy-proof and budget balanced; for n > 8,as long as q is close to n

2 , the statement still holds. More surprisingly, inrare cases (e.g. n = 8, q = 7), RT mechanism remains optimal among allrank mechanisms that are strategy-proof (not necessarily budget balanced),as shown in Example 6 of Guo and Conitzer (2014) 3.

Finally we show that RT mechanisms also guarantee a “fair share” foreach agent.4 That is, no agent will get an outcome that is worse than whatshe can get in the Lottery.

Proposition 4. For any n and q, 1 q < n, and any v 2 RN+ , RT mecha-

nism guarantees a utility qn · vi for all agent i 2 N .

Proof. Without loss of generality assume v 2 V . Note that according to (4),a

?q � a

?q+1 � �

?q = n�q�1

q �

?q+1 � 0. Therefore by Lemma 1, the utility u

?i each

3The author thanks the referee who pointed out this interesting result.4This property, sometimes called “unanimity lower bound”, is another widely used fair-

ness criterion in the literature. (e.g. Moulin (1992), Cramton et al. (1987).)

16

agent i gets is:

u

?l = vl � (1� a

?q)vq � (a?q � a

?q+1 � �

?q )vq+1 � [1� (1� a

?q)� (a?q � a

?q+1 � �

?q )]vl

=q

n

· vl, 8l < q,

u

?q = a

?qvq � (a?q � a

?q+1 � �

?q )vq+1 � [a?q � (a?q � a

?q+1 � �

?q )]vq =

q

n

· vq,

u

?q+1 = a

?q+1vq+1 + �

?qvq � (a?q+1 + �

?q )vq+1 =

q

n

· vq+1,

u

?l = �

?qvq + �

?q+1vq+1 � (�?q + �

?q+1)vl =

q

n

· vl, 8l > q + 1,

as desired.

5 Appendix

5.1 Lemma 2 and its proof

Lemma 2. Suppose a symmetric function h : Rn�1+ ! R satisfies the follow-

ing functional equation: for any x = (x1, x2, · · · , xn) 2 V,

X

i

h(x�i) =nX

i=2

�ixi,

where �i � 0 for all i. Then there exist (�1, · · · , �n�1) 2 Rn�1 such that forany z 2 Rn�1

+ , h(z) = �1z⇤1 + �2z⇤2 + · · ·+ �n�1z⇤(n�1).

Proof. Note that the lemma states only a necessary condition of the solutions.We show it by induction. Since

�2x2 + �3x3 + · · ·+ �nxn = h(x2, x3, · · · , xn)

+ h(x1, x3, · · · , xn)

+ · · ·

+ h(x1, x2, · · · , xn�1)

,

17

taking x1 = · · · = xn = 0, we get h(0, 0, · · · , 0) = 0. Taking x1 � x2 = · · · =xn = 0, we get h(x1, 0, · · · , 0) = 0. Now suppose h(x1, x2, · · · , xk, 0, · · · , 0) =�

01x1 + �

02x2 + · · · + �

0kxk for any x1 � x2 � · · · � xk � 0. We show that

h(x1, x2, · · · , xk, xk+10, · · · , 0) = �

001x1 + �

002x2 + · · ·+ �

00kxk + �

00k+1xk+1 for any

x1 � x2 � · · · � xk � xk+1 � 0.Note that

�2x2 + �3x3 + · · ·+ �kxk + �k+1xk+1 = h(x2, x3, · · · , xk+1, 0, · · · , 0)

+ h(x1, x3, · · · , xk+1, 0, · · · , 0)

+ · · ·

+ h(x1, x2, · · · , xk, 0, · · · , 0)

+ (n� k � 1)h(x1, x2, · · · , xk, xk+1, 0, · · · , 0)

=k+1X

l=1

[(l � 1)�0l�1 + (k � l + 1)�0l]xl

+ (n� k � 1)h(x1, x2, · · · , xk, xk+1, 0, · · · , 0)

,

where �

00 = 0.

Therefore h(x1, x2, · · · , xk, xk+1, 0, ..., 0) = �

001x1+�

002x2+· · · �00kxk+�

00k+1xk+1,

where�

00l =

�l � [(l � 1)�0l�1 + (k � l + 1)�0l]

n� k + 1,

and �1 = 0.

5.2 Lemma 3 and its proof

Lemma 3. The system of linear equations (4) has a unique solution {�k}n�1k=1

if and only if Equation (6) holds.

Proof. Note that the system of linear equations (4) have n�1 unknowns andn equations. We show the “only if” part. We write out the equations:

18

(n� 1) · �1 = 0

�1 + (n� 2) · �2 = (a1 � a2)

2�2 + (n� 3) · �3 = 2(a2 � a3)

......

(n� 2) · �n�2 + �n�1 = (n� 2) · (an�2 � an�1)

(n� 1) · �n�1 = (n� 1) · (an�1 � an)

Note that according to the first three equations, �1 = 0, �2 = a1�a2n�2 , and

�3 = � 2

(n� 2)(n� 3)a1 + [

2

(n� 2)(n� 3)+

2

n� 3]a2 �

2

n� 3a3.

We show by induction that for each k, 2 k n� 1,

�k = (�1)k ·n�2Y

i=n�k

�i · a1 + (�1)k�1 · (n�2Y

i=n�k

�i +n�3Y

i=n�k

�i) · a2

+ · · ·+ (�1)2 · (n�k+1Y

i=n�k

�i + �n�k) · ak�1 + (�1) · �n�k · ak

, (9)

where �n�1 = 0, and for each i, 1 i n� 2, �i =n�i�1

i .Let k < n� 1. Suppose for each l, 2 l k, Equation (9) holds. Then

according to the (k + 1)th equation of the linear system (4),

�k+1 =k

n� k � 1(ak � ak+1)�

k

n� k � 1�k

= �n�k�1(ak � ak+1)� �n�k�1�k

.

Now it is easy to calculate �k+1. And Equation (9) holds for k+1, as desired.

19

Hence according to the first n� 1 equations of the system,

�n�1 = (�1)n�1 ·n�2Y

i=1

�i · a1 + (�1)n�2 · (n�2Y

i=1

�i +n�3Y

i=1

�i) · a2

+ · · ·+ (�1)2 · (2Y

i=1

�i + �1) · an�2 + (�1) · �1 · an�1

. (10)

Note thatQm

i=1 �i = Cn�2m = Cn�2

n�2�m =Qn�2�m

i=1 �i.According to the last equation of the system, we have

�n�1 = an�1 � an. (11)

Combining (10) and (11), we have (6).Now the “if” part is easy to see.

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