Axiomatic and strategic justifications for the
constrained equal benefits rule in the airport
problem∗
Cheng-Cheng Hu†
Min-Hung Tsay‡
Chun-Hsien Yeh§
March 2011
Abstract
We consider the “airport problem”, which is concerned with sharing the cost of an airstrip among agents who need airstrips of different lengths. We investigate the implications of two properties, Leftendpoint Subtraction (LS) bilateral consistency and LS converse consistency, in the airport problem. First, on the basis of the two properties, we characterize the constrained equal benefits rule, which equalizes agents’ benefits subject to no one receiving a subsidy. Second, we
introduce a 3-stage extensive form game that exploits LS bilateral consistency and LS converse consistency. We show that there is a unique
subgame perfect equilibrium outcome of the game and moreover, it is
the allocation chosen by the constrained equal benefits rule. Journal
of Economic Literature Classification Numbers: C71; C72; D63; D70.
Keywords: Bilateral consistency; converse consistency; constrained
equal benefits rule; subgame perfect equilibrium; airport problem
∗
We would like to thank William Thomson and Youngsub Chun for their comments.
Yeh greatly acknowledges financial support from 2010 Career Development Award of
Academia Sinica, Taiwan.
†
Department of Economics, National Cheng Kung University, Tainan 701, Taiwan.
E-mail: [email protected]
‡
Department of Finance, National Central University, Chungli 320, Taiwan. E-mail:
[email protected]
§
Corresponding author: Institute of Economics, Academia Sinica, Taipei 115, Taiwan.
E-mail: [email protected]
1
1
Introduction
We consider the following class of “airport problems”, which exemplifies the
problem of sharing the cost of a facility among agents who have different
needs for it, but serving a given agent allows serving all agents with smaller
needs than hers at no extra cost. An airstrip has to be built to serve a group
of airlines. Different airlines need airstrips of different lengths. The larger
a plane, the longer the airstrip it needs. An airstrip that accommodates a
given plane can accommodate all smaller planes than its at no extra cost.
To accommodate all planes, the airstrip must be long enough for the largest
plane. How should the cost of the airstrip for the largest plane be shared
among the airlines with different costs of the airstrips?1 A “rule” is a function
that associates with each airport problem, an allocation of the cost of the
airstrip. We call this allocation a “contributions vector”.
The objectives of our paper are to base axiomatic justifications for a
rule on two properties, and to introduce a game to implement the rule that
exploits the properties. A rule is bilaterally consistent if an alternative is
chosen by the rule for a problem, then for the associated “two-agent reduced
problem” obtained by imagining the departure of all other agents with their
“components of the alternative”, and reassessing the options open to the
two-agent subgroup, it chooses the restriction of the alternative to that subgroup. A rule is conversely consistent if an alternative is such that for each
problem and each proper two-agent subgroup, the rule chooses the restriction
of the alternative to this subgroup for the reduced problem it faces, then the
alternative should be chosen for the initial problem by the rule.2
We apply the two principles to the airport problem, calling the resulting properties “Left-endpoint Subtraction (LS) bilateral consistency” and
“LS converse consistency”, and study their implications. We show that the
“Constrained Equal Benefits (CEB) rule” (Potters and Sudhölter, 1999),3
1
For a survey of the literature, initiated by Littlechild and Owen (1973), see Thomson (2004).
2
The principles of bilateral consistency and converse consistency have been studied
extensively in a great variety of models such as taxation, bargaining, social choice etc.
For references, see Lensberg (1985), Peleg (1986), Young (1988), Chun (2002), and Yeh
(2006). For a survey of the literature on the consistency principle and its converse, see
Thomson (2010).
3
Potters and Sudhölter (1999) refer to the rule as the “modified nucleolus”. The ter-
2
which equalizes agents’ benefits (at a contributions vector, the benefit of an
agent is the difference between the cost of satisfying her need and her contribution) subject to no one receiving a subsidy, is LS bilaterally consistent
and LS conversely consistent. Moreover, it is the only LS bilaterally consistent (or LS conversely consistent) rule satisfying either “order preservation
for benefits” and “cost monotonicity”, or “equal treatment of equals” and
“last-agent cost additivity”.
We next introduce a 3-stage extensive form game to implement the CEB
rule that exploits its LS bilateral consistency and LS converse consistency.4
Stage 1 : Each agent, except for one of the agents with the largest need
called the responder,5 announces a number interpreted as a contribution to the total cost (the cost of satisfying an agent with the largest need).
Stage 2 : The responder takes one of the following two actions: (1) she
makes all other agents contribute the amounts they specified, whereas
she contributes the difference between the total cost and the sum of all
other agents’ contributions; (2) she takes one agent as her partner to
the next stage. All other agents contribute the amounts they specified.
Stage 3 : Nature chooses one of the two agents with equal probability. The
agent chosen by Nature is given priority to use the sum of those contributions already made to cover her cost (an agent’s cost is the cost
of satisfying her need). The other agent uses the remainder (the difference between the sum of the contributions already made and the
chosen agent’s cost if positive; zero, otherwise) to cover the difference
between the total cost and the chosen agent’s cost.
We show that for each airport problem, there is a unique Subgame Perfect
Equilibrium (SPE) outcome and moreover, it is the allocation chosen by the
CEB rule.
minology we adopt is due to Thomson (2004).
4
Krishna and Serrano (1996) suggest to use the properties of a rule as guides in designing a game that implements the rule. For this line of research, see also Serrano (1997),
Dagan et al. (1997), Serrano and Vohra (2002), Chang and Hu (2008), Chun et al. (2011b),
and others. For a survey on the Nash program, see Serrano (2005).
5
This particular selection of the responder is due to the adoption of LS bilateral consistency and LS converse consistency in designing the game. For a detailed discussion,
see Section 2.2.
3
Several features of our game are worth noticing. First, since the airport
problem deals with the provision of a public good, it is natural to require
each agent to announce an amount she wishes to contribute. Second, we
do not specify agents’ contributions based on a particular rule. Finally, at
equilibrium, the game ends in one round consisting of three stages, and the
equilibrium strategies played by the agents are straightforward.
The paper is organized as follows. Section 2 introduces the model, and
central rules and properties. Section 3 develops axiomatic justifications. Section 4 establishes a strategic justification. Section 5 is concluding remarks.
2
2.1
Notation and definitions
The model
Let U j N be a universe of agents with at least two elements, where N is
the set of natural numbers. An airport problem, or simply a problem, is a
pair (N, c) where N j U is a finite nonvoid agent set and c ≡ (ci )i∈N ∈ RN
+
is the profile of agents’ costs. Let A be the class of all problems on U . A
contributions vector of (N, c) ∈ A is a vector x ∈ RN satisfying two conditions. First is “efficiency”: the sum of all contributions should be equal to
P
the total cost. Formally, i∈N xi = maxi∈N ci . Second is “reasonableness”:
each agent should not receive a subsidy and should not contribute more than
her cost. Formally, for each i ∈ N , 0 ≤ xi ≤ ci . Let X(N, c) be the set of all
contributions vectors for (N, c) ∈ A. A rule is a function defined on A that
associates with each problem (N, c) ∈ A a vector x ∈ X(N, c). Let n ≡ |N |,
and for simplicity, assume that N ≡ {1, . . . , n} and c1 ≤ · · · ≤ cn . Thus, the
agents are ordered in terms of their costs. We refer to agent 1 (the agent
with the smallest cost and the lowest index) as the first agent and to agent n
(the agent with the largest cost and the highest index) as the last agent. Our
generic notation for a rule is ϕ. For each N 0 ⊂ N , we denote (ci )i∈N 0 by cN 0 ,
(ϕi (N, c))i∈N 0 by ϕN 0 (N, c), and so on.
4
2.2
Central rules and properties
We now introduce our central rules. First is defined only for two-agent problems. It says that the agent with the smaller cost contributes half of her cost,
and the other contributes the remainder. Second is the CEB rule (Potters
and Sudhölter, 1999).
Standard rule, S: For each ({i, j}, (ci , cj )) ∈ A with ci ≤ cj ,
Si ({i, j}, (ci , cj )) ≡ c2i
Sj ({i, j}, (ci , cj )) ≡ cj − Si ({i, j}, (ci , cj )) .
Constrained Equal Benefits rule, CEB: For each (N, c) ∈ A and each
i ∈ N,
CEBi (N, c) ≡ max {ci − β, 0} ,
P
where β ∈ R+ is chosen such that i∈N CEBi (N, c) = cn .
Remark 1: A number of rules (e.g., the CEB rule and the “nucleolus”)6
coincide with the standard rule for two-agent problems.
In contrast to other models of fair allocation, for which a unique formulation of a reduced problem usually stands out as the most natural, there are
at least two formulations of a reduced problem in airport problems. Informally, this is because what has to be divided is not a homogeneous whole,
but it is composed of segments used differently by different agents. Potters
and Sudhölter (1999) propose two (left-endpoint and right-endpoint) formulations.7 Our central properties are based on the left-endpoint formulation.
6
The nucleolus is a direct application of the “nucleolus” (Schmeidler, 1969), which
lexicographically maximizes the “welfare” of the worst-off coalitions, a central solution
in Transferable Utility (TU) games. Any solution defined on the class of TU games can
be used to provide recommendations for airport problems. There are several ways of
transforming an airport problem into an associated TU game. One way to do it is to
define the worth of a coalition as the largest cost of any member in that coalition. We
then apply a solution in TU games to solve the associated TU game. This yields a payoff
vector. We take this payoff vector as the contributions vector for the airport problem.
7
For detailed discussions on these formulations, see Thomson (2004).
5
Figure 1: Three-agent problem
To introduce it, consider the three-agent problem in Figure 1 and a contributions vector x chosen by a rule for it. Imagine that one agent, say agent 2,
pays her contribution x2 and “leaves”, and reassess the situation from the
viewpoint of the remaining agents. The remaining agents are now left with
x2 that can help cover part of the total cost c3 . Instead of thinking of x2
as covering an abstract part of the airstrip, it is natural to regard x2 as a
contribution to the costs of segments agent 2 uses (namely, the costs of segments 1 and 2). The question is which segmental cost should be covered first.
The left-endpoint formulation suggests to first cover the cost of segment 1
(the segment all agents use); unless x2 > c1 , in which case, c1 is completely
covered and the remainder (x2 − c1 ) would be used to help cover the cost
of segment 2 (the segment agents 2 and 3 use but agent 1 does not).8 This
8
The right-endpoint formulation suggests to first cover the cost of segment 2 (the
segment agents 2 and 3 use but agent 1 does not); unless x2 > c2 − c1 , in which case,
c2 − c1 is completely covered and the remainder (x2 − (c2 − c1 )) would be used to help
cover the cost of segment 1 (the segment all agents use). This implies that if c2 − c1 ≥ x2 ,
since x2 is too little to cover the cost of segment 1, agent 1’s cost is revised down by 0;
otherwise, c2 − c1 is completely covered and the remainder (x2 − (c2 − c1 )) would be used
to help cover the cost of segment 1. Thus, her cost is revised down by x2 − (c2 − c1 )
6
implies that for each agent, say agent j, her cost is revised down by x2 . Since
each agent’s cost is not less than 0, agent j’s revised cost is the maximum of
cj − x2 and zero.
A rule satisfies “Left-endpoint Subtraction (LS) consistency” if a contributions vector x is chosen by the rule for a problem, then for the reduced
problem just defined, the components of x pertaining to the remaining agents
should still be chosen by the rule.9
We adopt the above idea to define the two-agent reduced problem. Formally, let (N, c) ∈ A with n ≥ 2, i ∈ N \{n}, and x ∈ RN . The reduced
x
problem of (N, c) with respect to N 0 ≡ {i, n} and x, (N 0 , rN
0 ), is
defined by setting
n
o
P
x
(rN
ci − k6=i,n xk , 0 , and
0 )i ≡ max
x
(rN
0 )n ≡ max{cn −
P
k6=i,n
xk , 0}.
A rule is “LS bilaterally consistent” if a contributions vector x is chosen by
the rule for a problem, then for the two-agent reduced problem just defined,
the components of x pertaining to the remaining agents should still be chosen
by the rule.
Left-endpoint Subtraction (LS) bilateral consistency:For each (N,c) ∈
x
∈
A with |N | ≥ 2 and each i ∈ N \{n}, if x = ϕ (N, c), then {i, n} , r{i,n}
x
A and x{i,n} = ϕ {i, n} , r{i,n}
.
A rule is “LS conversely consistent” if a contributions vector x is such
that for each two-agent reduced problem involving agent n, the restriction
and her revised cost is c2 − x2 . Altogether, agent 1’s revised cost is the minimum of c1
and c2 − x2 . Since contributing to the segments agent 2 uses implies contributing to the
segments agent 3 uses, agent 3’s cost is then revised down by x2 . A rule satisfies “Rightendpoint Subtraction (RS) consistency” if a contributions vector x is chosen by the rule
for a problem, then for the two-agent reduced problem just defined, the components of x
pertaining to the remaining agents should still be chosen by the rule. For an additional
reference on RS consistency, see Hwang and Yeh (2010).
9
Potters and Sudhölter (1999) refer to LS consistency and RS consistency as ψconsistency and ν-consistency, respectively.
7
of x to this subgroup is chosen by the rule, then x should be chosen by the
rule for the original problem.10
LS converse consistency: For each (N, c) ∈ A with n > 2 and each
x
x ∈ X(N, c), if for each N 0 ⊂ N with |N 0 | = 2 and n ∈ N 0 , xN 0 = ϕ (N 0 , rN
0 ),
then x = ϕ (N, c).
In the formulations of LS consistency, LS bilateral consistency, and LS
converse consistency, the possibility of the last agent leaving is excluded since
the last-agent’s cost cn determines the total cost and after the departure of
the last-agent, the new total cost would have no reason to be related to the
sum of the contributions required of the remaining agents.
3
Axiomatic characterizations
Since the CEB rule is LS consistent (Potters and Sudhölter, 1999), it is also
LS bilaterally consistent. This fact and the following lemma play important
roles in providing our axiomatic characterizations of the CEB rule.
Lemma 1: The CEB rule is LS conversely consistent.
Proof. Let (N, c) ∈ A. Without loss of generality, assume that N ≡
{1, . . . , n} and c1 ≤ · · · ≤ cn . Let x ∈ X (N, c) be such that for each
x
. We show that x = CEB (N, c).
i ∈ N \{n}, x{i,n} = CEB {i, n} , r{i,n}
P
x
, by
xk = cn and x{i,n} = CEB {i, n} , r{i,n}
Let i ∈ N \ {n}. Since
k∈N
x
x
Remark 1, r{i,n}
= xi + xn and r{i,n}
= 2xi . Let β ≡ cn − xn . Since
n
i
P
cn = k∈N xk , we have
!
X
ci − β = ci − cn + x n = ci −
xk − xi .
k6=i,n
If ci −
P
k6=i,n
xk ≥ 0, then 2xi =
x
r{i,n}
i
= ci −
10
P
xk . Thus, max{ci − β, 0} =
k6=i,n
It can be shown that LS bilateral consistency and LS converse consistency are logically
independent.
8
xi . If ci −
P
k6=i,n
x
= 0. Since xi = 0 and ci − β < 0, we
xk < 0, then r{i,n}
i
have max{ci − β, 0} = 0 = xi . Altogether, x = CEB (N, c).
Q.E .D.
We consider the following additional properties of rules. First, two agents
with equal costs should contribute equal amounts.
Equal treatment of equals: For each (N, c) ∈ A and each pair {i, j} ⊆ N ,
if ci = cj , then ϕi (N, c) = ϕj (N, c).
The second requirement is that of two agents, the benefit of the agent (at
a contributions vector, the benefit of an agent is the difference between her
cost and her contribution) with the larger cost should be at least as much as
that of the other (Littlechild and Thompson, 1977). Clearly, this property
implies equal treatment of equals.
Order preservation for benefits: For each (N, c) ∈ A and each pair
{i, j} ⊆ N , if ci ≤ cj , then ci − ϕi (N, c) ≤ cj − ϕj (N, c).
Third, if an agent’s cost increases, all other agents should contribute at
most as much as they did initially.11
Cost monotonicity: For each pair {(N, c) , (N, c0 )} of elements of A and
each i ∈ N , if c0i ≥ ci and for each j ∈ N \{i}, c0j = cj , then for each
j ∈ N \{i}, ϕj (N, c0 ) ≤ ϕj (N, c).
Under the hypothesis of cost monotonicity, the agent whose cost increases
will contribute more. Since the airport problem deals with the provision of a
public good, if an agent contributes more due to her cost increase, all other
agents would enjoy a positive externality.
Our final property says that if the last agent’s cost increases by δ, then
this agent’s contribution should increase by δ. Since this addition to the
11
This property is introduced by Thomson (2004) under the name of “others-oriented
cost monotonicity”. It is a complement of “individual cost monotonicity” (Potters and
Sudhölter, 1999), which says that under the same hypothesis, agent i should contribute at
least as much as she did initially. For an additional reference, see Chun et al. (2011a).
9
airstrip is only used by the last agent, it is natural to ask her to cover all of
this incremental cost.
Last-agent cost additivity: For each pair {(N, c) , (N, c0 )} of elements of
A and each δ ∈ R+ , if c0n = cn + δ and for each j ∈ N \{n}, c0j = cj , then
ϕn (N, c0 ) = ϕn (N, c) + δ.
Potters and Sudhölter (1999) propose the property “stronger last-agent
cost additivity”, which says that under the same hypothesis, not only the
last agent’s contribution should increase by δ but also all other agents’ contributions should remain unchanged.
We first present two axiomatic characterizations of the standard rule.
Proposition 1: For |N | = 2. The standard rule is the only rule satisfying
order preservation for benefits and cost monotonicity.
Proof. It is clear that the standard rule satisfies the two properties of the
proposition. Conversely, let (N, c) ∈ A with N ≡ {i, j} and ci ≤ cj . Let ϕ
be a rule satisfying the two properties. Let x ≡ ϕ(N, c) and y ≡ S(N, c).
We first show that xi = yi , and then by efficiency, we conclude that x = y.
The proof is divided into two cases.
Case 1: ci = cj . Since order preservation for benefits implies equal treatment
of equals, xi = xj = c2i = yi .
Case 2: ci < cj . Let (N, c0 ) ∈ A with c0i = ci and c0j = ci . Let x0 ≡ ϕ(N, c0 ).
By Case 1, x0i = c2i . Since c0i = ci and c0j < cj , by cost monotonicity, xi ≤ x0i .
Thus, xi ≤ c2i . Since for the two-agent case, order preservation for benefits
implies that xi ≥ c2i , then xi = c2i = yi .
Q.E .D.
Proposition 2: For |N | = 2. The standard rule is the only rule satisfying
equal treatment of equals and last-agent cost additivity.
Proof. It is clear that the standard rule satisfies the two properties of the
proposition. Conversely, let (N, c) ∈ A with N ≡ {i, j} and ci ≤ cj . Let ϕ
be a rule satisfying the two properties. Let x ≡ ϕ(N, c) and y ≡ S(N, c).
We show that x = y. The proof is divided into two cases.
10
Case 1: ci = cj . By equal treatment of equals, xi = xj . By efficiency,
xi = xj = c2i = yi = yj .
Case 2: ci < cj . Let (N, c0 ) ∈ A with c0i = ci and c0j = ci . Let x0 ≡ ϕ(N, c0 ).
By Case 1, x0j = c2i . By last-agent cost additivity, xj = x0j +δ, where δ ≡ cj −ci .
Thus, xj = cj − c2i = yj . By efficiency, xi = yi .
Q.E .D.
In the formulations of LS bilateral consistency and LS converse consistency, we do not consider all two-agent subgroups. The following lemma is
an application of the Elevator lemma (Thomson, 2010).
Elevator Lemma If a rule ϕ is LS bilaterally consistent and coincides with
a LS conversely consistent rule ϕ0 in the two-agent case, then ϕ coincides
with ϕ0 in general.
It is easy to verify that the CEB rule satisfies order preservation for benefits and cost monotonicity. The rule also satisfies equal treatment of equals,
last-agent cost additivity, and LS consistency (Potters and Sudhölter, 1999).
Moreover, it is also LS bilaterally consistent and LS conversely consistent
(Lemma 1). With the help of the Elevator Lemma and Propositions 1 and 2,
the following results are immediate.12
Theorem 1: The CEB rule is the only LS bilaterally consistent rule satisfying order preservation for benefits and cost monotonicity.
Theorem 2: The CEB rule is the only LS conversely consistent rule satisfying order preservation for benefits and cost monotonicity.
Theorem 3: The CEB rule is the only LS bilaterally consistent rule satisfying equal treatment of equals and last-agent cost additivity.
Theorem 4: The CEB rule is the only LS conversely consistent rule satisfying equal treatment of equals and last-agent cost additivity.
12
Our theorems are special applications of the lemma. To illustrate how the lemma
can be applied to here, we give a formal proof of Theorem 1. Let (N, c) ∈ A with
N ≡ {1, . . . , n} and c1 ≤ · · · ≤ cn . Of course, the CEB rule satisfies the properties of the
theorem. Conversely, let ϕ be a LS bilaterally consistent rule satisfying the two properties.
Let x ≡ ϕ(N, c) and y ≡ CEB(N, c). We show that x = y. By LS bilateral consistency
x
of ϕ, for each i ∈ N \{n}, ϕ({i, n}, r{i,n}
) = x{i,n} . By Proposition 1 and Remark 1,
x
x{i,n} = CEB({i, n}, r{i,n} ). Since ϕ is LS bilaterally consistent and the CEB rule is LS
conversely consistent, by the Elevator Lemma, x = y.
11
Potters and Sudhölter (1999) show that the CEB rule is the only LS consistent rule satisfying equal treatment of equals, strong last-agent cost additivity, and “homogeneity” (if all costs are multiplied by the same positive number, so should the contributions).13 Their characterization and Theorem 3
provide axiomatic justifications for the CEB rule. In that sense, the two results are closely related. To compare them, note that we include efficiency in
the definition of a rule, while on the contrary, Potters and Sudhölter (1999) do
not. Therefore, Theorem 3 says that a characterization of the CEB rule can
be obtained from Potters and Sudhölter (1999)’s characterization by replacing homogeneity with efficiency, and weakening LS consistency and strong
last-agent cost additivity to LS bilateral consistency and last-agent cost additivity, respectively. Since homogeneity, strong last-agent cost additivity, and
LS consistency altogether imply efficiency (Potters and Sudhölter, 1999),
Theorem 3 implies their characterization of the CEB rule.
4
A strategic justification
We now establish a strategic justification for the CEB rule that exploits its
LS bilateral consistency and LS converse consistency. If one agent’s cost is
equal to 0, it implies that this agent has no need for the airstrip. To make
our analysis more interesting, we require that each agent’s cost should be
positive. Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . We introduce the following
3-stage extensive form game Γ(N, c). The game tree is depicted in Figure 2.
Stage 1 : Each agent k ∈ N \{n} independently announces a number xk ∈ R
P
such that 0 ≤ xk ≤ ck . Let xn ≡ cn − k6=n xk . We refer to x ≡ (xk )k∈N
as the proposal.
Stage 2 : Agent n adopts one of the following two strategies:
1. Action A (Accept xn ). The game ends with x as the outcome.
2. Action R (Reject xn ). Agent n takes one agent from N \ {n}, say
agent i, as her partner to the next stage. Each agent in N \ {i, n},
13
Potters and Sudhölter (1999) combine strong last-agent cost additivity and homogeneity as a property, referred to as covariance.
12
Stage 1:
Each agent k ∈ N \ {n} announces
a number xk . Let x n = c n − ∑ xi .
i≠n
Stage 2:
Agent n decides to take A
(accept xk ) or R (reject xk ).
If agent n chooses R, she takes one
agent from N \ {n} , say agent i, as R
her partner to the next stage. Each
k ∈ N \ {i, n} contributes xk and leaves.
A
x ≡ ( xk ) k ∈N
Stage 3:
Nature chooses either agent i or
agent n with probability 1 .
2
Agent i
Agent n
⎛
⎞
⎜⎜ (ci − ∑ xk ) + , xi + xn − (ci − ∑ xk ) + , x N \{i ,n} ⎟⎟
k ≠i ,n
k ≠i ,n
⎝
⎠
⎞
⎛
⎜⎜ ( xi + xn − (cn − ∑ xk ) + , (cn − ∑ xk ) + , x N \{i,n} ⎟⎟
k ≠i ,n
k ≠i ,n
⎠
⎝
+
Note that ( x) ≡ max{0, x} , where x is an arbitrary real number.
Figure 2: The game tree of Γ(N, c)
13
say agent k, contributes xk and leaves with the understanding that the
airstrip will be built.
Stage 3 : Nature chooses either agent i or agent n with equal probability.
n
o
P
1. Agent i is chosen. Agents i and n contribute zi ≡ max ci − k6=i,n xk , 0
and zn ≡ xi + xn − zi , respectively.
n
o
P
2. Agent n is chosen. Agents n and i contribute z̄n ≡ max cn − k6=i,n xk , 0
and z̄i ≡ xi + xn − z̄n , respectively.
We elaborate Γ(N, c) and explain how it exploits the LS bilateral consistency and LS converse consistency.
1. Each agent k ∈ N \{n} in Stage 1 announces a number xk interpreted
as a contribution to the total cost. Due to the exclusion of agent n’s
leaving in the definitions of LS bilateral consistency and LS converse
consistency, agent n plays no role in Stage 1. We refer to her as the
responder. In Stage 2, the responder either accepts or rejects xn interpreted as the residual cost.
2. In Stage 3, the contributions of agent n and her partner, say agent i, are
specified based on the reduced problem underlying LS bilateral consistency and LS converse consistency. We can think of the contributions
as made by imaging that the agent chosen by Nature is given priority
P
to use k6=i,n xk to cover her cost, and the other agent uses the reP
mainder (the difference between k6=i,n xk and the chosen agent’s cost
if positive; 0, otherwise) to cover the difference between the total cost
and the chosen agent’s cost.
3. The game exploits LS bilateral consistency and LS converse consistency since (1) partial agreements are allowed (all agents, except for
the responder and her partner, contribute the amounts they specified
and leave) and (2) bilateral renegotiation takes place (the responder
bargains with her partner on their contributions at the final stage). It
is also worth noticing that agents’ contributions are not specified based
on any particular rule.
14
We use the following notation to construct a SPE of Γ (N, c). Let x ∈ RN
be a proposal. Suppose that agent n chooses agent i ∈ N \ {n} in Stage 2.
Let the expected contributions of agents i and n in Stage 3 be denoted by
(
)
X
X
1
τ ii (c, x) ≡
max{0, ci −
xk } + xi + xn − max{0, cn −
xk }
2
k6=i,n
k6=i,n
and
τ in (c, x) ≡ xi + xn − τ ii (c, x) ,
respectively. Let τ i (c, x) ≡ (τ ii (c, x) , τ in (c, x)).
As shown next, if the proposal x is a contributions vector and agent n
chooses agent i ∈ N \{n} in Stage 2, then the expected contributions of
agents i and n are prescribed by the CEB rule for the reduced problem with
respect to {i, n} and x.
Lemma 2: Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . If x ∈ X (N, c)
is the proposal, and agent n chooses agent i ∈ N \{n} in Stage 2, then
x
.
τ i (c, x) = CEB {i, n} , r{i,n}
Proof. Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . Let x ∈ X (N, c) be
the proposal.
thatagentn chooses agent i ∈ N \{n} in Stage 2.
Suppose
P
1
x
x
= max{0, cn − k6=i,n xk } = xi + xn .
Let β ≡ 2 r{i,n} . Thus, r{i,n}
n
i
Since τ ii (c, x) + τ ni(c, x) = xi + xn , it suffices to show that τ ii (c, x) =
x
CEBi {i, n} , r{i,n}
, which is derived by noticing that
τ ii (c, x) ≡
1
2
(
max{0, ci −
)
X
xk } + xi + xn − max{0, cn −
k6=i,n
X
xk }
k6=i,n
o
1n x x
=
r{i,n} i + xi + xn − r{i,n} n
2
1 x =
r
i
2 {i,n}
x
= r{i,n} i − β
x
= CEBi {i, n} , r{i,n}
.
Q.E .D.
15
The next two results establish a strategic justification for the CEB rule.
First says that the allocation chosen by the CEB rule can be supported by
a SPE of Γ(N, c). Second says that each SPE outcome of Γ(N, c) is the
allocation chosen by the CEB rule. As usual, we solve Γ(N, c) by backward induction.
Theorem 5: Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . There is a SPE of
Γ (N, c) with outcome CEB (N, c).
Proof. Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . The proof is given by
construction of a strategy profile that is a SPE of Γ (N, c) and generates
CEB (N, c) as the outcome. Let f be the strategy profile defined as follows:
Stage 1 : Each agent k ∈ N \{n} announces CEBk (N, c).
Stage 2 : Suppose that x is the proposal. Let µ ≡
min τ kn (c, x). If
k∈N \{n}
xn ≤ µ, agent n takes action A; otherwise, she takes action R and
chooses one agent from N \ {n}, say agent i, such that τ in (c, x) = µ.
Step 1: Following f, the outcome is CEB (N, c). Following fk , each
agent k ∈ N \{n} announces CEBk (N, c). Let x̄ ≡ CEB (N, c). Then
x̄ ∈ X(N, c) is the proposal. By LS bilateral consistency of the CEB rule
and Lemma 2, for each k ∈ N \{n},
x̄
x̄n = CEBn {k, n} , r{k,n}
= τ kn (c, x̄) .
Thus, x̄n = min τ kn (c, x̄) ≡ µ. Following fn , agent n takes action A. The
k∈N \{n}
game ends with outcome CEB (N, c).
Step 2: f is a SPE. It is easy to verify that following fn is a best response
for agent n in Stage 2 provided each agent k ∈ N \{n} follows fk . Let
i ∈ N \ {n}. Suppose that each agent k ∈ N \{i} follows fk . We show that
following fi is a best response for agent i. We consider that agent i could
deviate by announcing a real number x0i , which is either x0i < x̄i or x0i > x̄i .
Case 1: x0i < x̄i . There is ε ∈ R such that ε > 0 and x0i = x̄i − ε. We show
that agent i is not better off deviating. Let x0 be the proposal after agent i’s
16
deviation. Note that x̄i + x̄n ≥ 0, x0n = x̄n + ε, and for each k ∈ N \{i, n},
x0k = x̄k . Since x̄ ∈ X(N, c), by Lemma 2, τ ii (c, x̄) = x̄i . Since
X
X
1
τ ii (c, x0 ) ≡ {max{0, ci −
x0l } + x0i + x0n − max{0, cn −
x0l }}
2
l6=i,n
l6=i,n
X
X
1
x̄l } + x̄i + x̄n − max{0, cn −
x̄l }}
= {max{0, ci −
2
l6=i,n
l6=i,n
≡ τ ii (c, x̄)
= x̄i ,
τ in (c, x0 ) ≡ x0i + x0n − τ ii (c, x0 ) = x̄n < x0n . Agent n takes action R. Let
P
k ∈ N \ {i, n}. Since x0k + x0n ≥ 0 and cn = l∈N x0l ,
X
X
1
τ kk (c, x0 ) ≡ {max{0, ck −
x0l }}
x0l } + x0k + x0n − max{0, cn −
2
l6=k,n
l6=k,n
X
1
= max{0, ck −
x̄l + ε}.
2
l6=k,n
It follows that
τ kn (c, x0 ) ≡ x0k + x0n − τ kk (c, x0 )
X
1
= x̄k + x̄n + ε − max{0, ck −
x̄l + ε}.
2
l6=k,n
P
P
If ck − l6=k,n x̄l +ε ≤ 0, then τ kn (c, x0 ) > x̄n = τ in (c, x0 ). If ck − l6=k,n x̄l +ε >
0, by order preservation for benefits of the CEB rule,
X
1
τ kn (c, x0 ) = x̄k + x̄n + ε − max{0, ck −
x̄l + ε}
2
l6=k,n
(
)
X
ε 1
x̄k +
x̄l − ck
= x̄n + +
2 2
l6=n
ε 1
+ {cn − x̄n − (ck − x̄k )}
2 2
ε
≥ x̄n +
2
> x̄n
= x̄n +
= τ in (c, x0 ) .
17
In either case, following fn , agent n takes action R and chooses agent i. Since
agent i ends up with contribution τ ii (c, x0 ) = x̄i , she is not better off deviating.
Case 2: x0i > x̄i . There is ε ∈ R such that ε > 0 and x0i = x̄i + ε. We
show that agent i is not better off deviating. If agent n takes either action A,
or takes action R and chooses agent k ∈ N \ {i, n}, then agent i ends up
with contribution x̄i + ε, which implies that she is not better off deviating.
If agent n takes action R and chooses agent i, let x0 be the proposal after
agent i’s deviation. Since x0i = x̄i + ε, x0n = x̄n − ε, and for each j ∈ N \{i, n},
P
x0j = x̄j , we have x0i + x0n = x̄i + x̄n ≥ 0 and cn = j∈N x0j . Since by LS
x̄
bilateral consistency of the CEB rule, x̄i = CEBi {i, n}, r{i,n} , then
1
τ ii (c, x0 ) ≡
2
)
(
max{0, ci −
X
x0j } + x0i + x0n − max{0, cn −
X
x0j }
j6=i,n
j6=i,n
X
1
= max{0, ci −
x̄j }
2
j6=i,n
x̄
= CEBi {i, n}, r{i,n}
= x̄i ,
which implies that agent i is not better off deviating. Thus, f is a SPE. Q.E .D.
Theorem 6: Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . Each SPE outcome of
Γ (N, c) is CEB (N, c).
Proof. Let (N, c) ∈ A with c ≡ (ci )i∈N ∈ RN
++ . Let g be a SPE. Suppose
that following gk , each k ∈ N \{n} announces xk ∈ R such that 0 ≤ xk ≤ ck
P
in Stage 1. Let xn ≡ cn − k6=n xk and x ≡ (xk )k∈N be the proposal. In
Stage 2, following gn , agent n takes either action A or action R.
Case 1: Agent n takes action A. The game ends with outcome x. We first
show that x ∈ X (N, c), and then use Lemma 2 to show that x = CEB(N, c).
Suppose, by contradiction, that x ∈
/ X(N, c). Since for each k ∈ N \{n},
P
following gk , 0 ≤ xk ≤ ck , and i∈N xi = cn , then either xn < 0 or xn > cn .
P
If xn > cn , since k∈N xk = cn , there is j ∈ N \{n} such that xj < 0, which
18
P
gives a contradiction. Thus xn < 0. Since k∈N xk = cn , there is i ∈ N \ {n}
P
such that xi > 0 and cn − k6=i,n xk < xi . We show that agent i is better
off deviating, which violates the assumption that g is a SPE. Suppose that
agent i announces 0. Let x0 be the proposal after agent i’s deviation. If
agent n takes either action A, or action R and chooses agent k ∈ N \{i, n},
then since agent i ends up with contribution 0 < xi , she is better off deviating.
If agent n takes action R and chooses agent i, the proof is divided into two
subcases.
P
Subcase 1.1: cn − k6=i,n xk < 0. Note that x0i = 0, x0n = xi + xn , and
P
for each k ∈ N \{i, n}, x0k = xk . Since ci ≤ cn , ci − k6=i,n xk < 0. Since
xn < 0 < xi ,
(
)
X
X
1
τ ii (c, x0 ) ≡
max{0, ci −
x0k } + (x0i + x0n ) − max{0, cn −
x0k }
2
k6=i,n
k6=i,n
1
(xi + xn )
2
< xi .
=
Since agent i is better off deviating, the assumption that g is a SPE is violated.
P
P
Subcase 1.2: cn − k6=i,n xk ≥ 0. Since xn < 0 and cn =
j∈N xj ,
P
P
cn − k6=i,n xk < xi . Since ci ≤ cn , ci − k6=i,n xk < xi . Since xi > 0,
)
(
X
X
1
0
0
0
0
i
0
xk }
xk } + (xi + xn ) − max{0, cn −
τ i (c, x ) ≡
max{0, ci −
2
k6=i,n
k6=i,n
(
)
X
X
1
=
max{0, ci −
xk } + (xi + xn ) − max{0, cn −
xk }
2
k6=i,n
k6=i,n
)
(
X
1
<
xi + (xi + xn ) − cn +
xk
2
k6=i,n
< xi .
Since agent i is better off deviating, the assumption that g is a SPE is violated. By Subcases 1.1 and 1.2, x ∈ X (N, c).
19
We now show that x = CEB (N, c). Suppose, by contradiction, that
x 6= CEB (N, c). By LS converse consistency
of the CEB rule, there is
x
i ∈ N \{n} such that x{i,n} 6= CEB {i, n} , r{i,n} . Since x ∈ X (N, c),
n
o
P
x
≡ max 0, cn − k6=i,n xk = xi + xn , then
xi + xn ≥ 0. Since r{i,n}
n x
x
xi + xn = CEBi {i, n} , r{i,n} + CEBn {i, n} , r{i,n} . Since x{i,n} 6=
x
x
< xn , or
, it follows that either CEBn {i, n} , r{i,n}
CEB {i, n} , r{i,n}
x
x
< xn , we show that
< xi . If CEBn {i, n} , r{i,n}
CEBi {i, n} , r{i,n}
agent n is better off taking action R and choosing agent i. To see this,
note that
x ∈ X(N,c). By Lemma 2, agent n ends up with contribution
x
CEBn {i, n} , r{i,n}
< xn . Since she is better off deviating, the assump
x
tion that g is a SPE is violated. If CEBi {i, n} , r{i,n}
< xi , we show
x
rather than xi in
that agent i is better off announcing CEBi {i, n} , r{i,n}
Stage 1. If agent n takes either action A, or takes action R and
chooses agent
x
<
k ∈ N \{i, n}, then agent i ends up with contribution CEBi {i, n} , r{i,n}
xi . If agent n takes action R and chooses agent
i, let x0 be the proposal after
x
0
agent i’s deviation. Note that xi = CEBi {i, n} , r{i,n} , x0n = xi + xn − x0i ,
and for each k ∈ N \{i, n}, x0k = xk . Since x ∈ X(N, c), by Lemma 2,
)
(
X
X
1
x0k }
x0k } + (x0i + x0n ) − max{0, cn −
max{0, ci −
τ ii (c, x0 ) ≡
2
k6=i,n
k6=i,n
(
)
X
X
1
=
max{0, ci −
xk } + (xi + xn ) − max{0, cn −
xk }
2
k6=i,n
k6=i,n
≡ τ ii (c, x)
x
= CEBi {i, n} , r{i,n}
.
x
Since in any case, agent i ends up with contribution CEBi {i, n} , r{i,n}
<
xi , she is better off deviating, which violates the assumption that g is a SPE.
We conclude that x = CEB(N, c).
Case 2: Agent n takes action R. Suppose that agent n takes action R
and chooses one agent from N \{n}, say agent i, in Stage 2. Let y ≡
20
τ i (c, x) , xN \{i,n} be the outcome. We show that y = CEB (N, c). If |N | =
2, then y = τ i (c, x) = ( 21 ci , cn − 21 ci ) = CEB (N, c). Let |N | > 2. We first
show that x ∈ X(N, c) and then use Lemma 2 to show that y = CEB (N, c).
P
Suppose, by contradiction, that x ∈
/ X(N, c). Since j∈N xj = cn and
for each j 6= n, 0 ≤ xj ≤ cj , then either xn < 0 or xn > cn . If xn > cn ,
P
then since j∈N xj = cn , there is l ∈ N \{n} such that xl < 0, which give a
P
contradiction. Thus, xn < 0. Since j∈N xj = cn , there is k ∈ N \{n} such
that xk > 0. We consider two possibilities: either k ∈ N \{i, n} or k = i.
• k ∈ N \{i, n}. We show that agent k is better off proposing 0 in Stage 1,
which violates the assumption that g is a SPE. Let x0 be the proposal after
agent k’s deviation. If agent n either takes action A, or takes action R and
chooses agent j ∈ N \{k, n}, then since agent k ends up with contribution
0 < xk , she is better off deviating. If agent n takes action R and chooses
agent k, we consider two possibilities.
P
(1) cn − j6=k,n xj < 0. Note that x0k = 0, x0n = xk + xn , and for each
P
j ∈ N \{k, n}, x0j = xj . Since ck ≤ cn , ck − j6=k,n xj < 0. Since xn < 0 < xk ,
τ kk (c, x0 ) ≡
1
2
)
(
max{0, ck −
X
x0j } + (x0k + x0n ) − max{0, cn −
X
j6=k,n
j6=k,n
1
(xk + xn )
2
< xk ,
=
which implies that agent k is better off deviating.
21
x0j }
P
P
(2) cn − j6=k,n xj ≥ 0. Since xn < 0, ck ≤ cn , and cn = j∈N xj , then
P
P
ck − j6=k,n xj ≤ cn − j6=k,n xj < xk . Since xk > 0,
1
2
(
1
=
2
(
1
<
2
(
τ kk (c, x0 ) ≡
max{0, ck −
max{0, ck −
)
X
x0j } + (x0k + x0n ) − max{0, cn −
X
j6=k,n
j6=k,n
X
X
x0j }
)
xj } + (xk + xn ) − max{0, cn −
j6=k,n
xj }
j6=k,n
)
X
xk + (xk + xn ) − cn +
xj
j6=k,n
< xk ,
which implies that agent k is better off deviating.
• k = i. We show that agent n is better off deviating, which violates the
assumption that g is a SPE. The proof is divided into two possibilities.
P
(1) cn − j6=k,n xj < 0. Then τ kk (c, x) = 21 (xk + xn ) and yn = τ kn (c, x) =
xk + xn − τ kk (c, x) = 21 (xk + xn ). Since xn < 0 < xk , yn > xn . If agent n
takes action A rather than action R, she ends up with contribution xn < yn ,
which implies that she is better off deviating.
P
P
(2) cn − j6=k,n xj ≥ 0. Since xn < 0 < xk , ck ≤ cn and cn = j∈N xj ,
P
P
then ck − j6=k,n xj ≤ cn − j6=k,n xj < xk . It can be shown that τ kk (c, x) < xk
and yn = τ kn (c, x) = xk + xn − τ kk (c, x) > xn . If agent n takes action A rather
than action R, she ends up with contribution xn < yn , which implies that
she is better off deviating.
Thus, we conclude that x ∈ X(N, c).
We now show that y = CEB(N, c). The proof is divided into two subcases.
Subcase 2.1: For each j ∈ N \ {i,
n}, xj = 0. Then, yN \{i,n} = 0N \{i,n}
1
1
i
and y{i,n} = τ (c, x) = 2 ci , cn − 2 ci . We consider two possibilities.
22
(1) For each j ∈ N \ {i, n}, cj ≤ c2i . To solve
β ∈ R+ in the definition
of the CEB rule must be
1
1
c , c − 2 ci , 0N \{i,n} , y = CEB(N, c).
2 i n
P
k∈N CEBk (N, c) = cn ,
ci
. Since CEB (N, c) =
2
(2) For some j ∈ N \ {i, n}, cj > c2i . We show that yN \{i,n} 6= 0N \{i,n} ,
which gives a contradiction. We then conclude that at equilibrium, this
possibility never happens. Suppose, by contradiction, that yN \{i,n} = 0N \{i,n} .
Note that ci > 0. We show that agent i is better off deviating. Let ε ∈ R
be such that 0 < ε < min{cj − c2i , c2i }. Suppose that agent i announces
1
c − ε. Let x0 be the proposal after agent i’s deviation. Thus, x0i = c2i − ε,
2 i
x0n = cn − c2i + ε, and for each j ∈ N \{i, n}, x0j = xj = 0. If agent n
chooses one agent from N \{i, n}, say agent j, then since yn = cn − c2i and
τ jj (c, x0 ) = 12 (cj − 21 ci + ),
τ jn (c, x0 ) ≡ x0j + x0n − τ jj (c, x0 )
ci
= cn − + ε − τ jj (c, x0 )
2
1
1
= yn + ε −
cj − ci + ε
2
2
1
1
= yn −
cj − ci − ε
2
2
< yn ,
which implies that agent n does not choose agent i. Since agent i is better
off ending up with contribution 12 ci − ε < c2i , the assumption that g is a SPE
is violated.
Subcase 2.2: For some j ∈ N \ {i, n}, xj > 0. Let j ∈ N \{i, n} be
such that xj > 0. Since y is the outcome, by subgame perfection, for each
k ∈ N \ {i, n}, yn ≤ τ kn (c, x). We consider two possibilities.
(1) For each k ∈ N \ {i, n}, yn < τ kn (c, x). We show that agent j is
better off deviating. Let ε ∈ R be such that 0 < ε < min {τ jn (c, x) − yn , xj }.
Suppose that agent j announces xj − ε. Let x0 be the proposal after agent j’s
23
P
deviation. Since x ∈ X(N, c) and max{0, cn − k6=i,n xk } = xi + xn ≥ 0,
(
)
X
X
1
max{0, ci −
τ ii (c, x) ≡
xk } + xi + xn − max{0, cn −
xk }
2
k6=i,n
k6=i,n
X
1
xk }.
= max{0, ci −
2
k6=i,n
P
Since yn = τ in (c, x) = xi +xn −τ ii (c, x), yn = xi +xn − 12 max{0, ci − k6=i,n xk }.
Note that x ∈ X(N, c), x0j = xj − ε, x0n = xn + ε, and for each k ∈ N \{j, n},
x0k = xk . We have x0i + x0n = xi + xn + ε ≥ 0. Since τ ii (c, x0 ) = 12 max{0, ci −
P
P
1
i
0
k6=i,n xk + ε} ≥ τ i (c, x),
k6=i,n xk } = 2 max{0, ci −
τ in (c, x0 ) ≡ x0i + x0n − τ ii (c, x0 )
= xi + xn + ε − τ ii (c, x0 )
≤ xi + xn + ε − τ ii (c, x)
= yn + ε
< τ jn (c, x)
= τ jn (c, x0 ) ,
which implies that agent n does not choose agent j. Since agent j is better
off ending up with contribution xj − ε < xj , the assumption that g is a SPE
is violated.
(2) For some k ∈ N \ {i, n}, yn = τ kn (c, x). Let k ∈ N \ {i, n} be such
that yn = τ kn (c, x). The proof is divided into two possibilities.
• y{i,n} = x{i,n} . Then y = x. It suffices to show that x = CEB(N, c). Suppose, by contradiction, that x 6= CEB(N, c). By LS converse
consistency
x
of the CEB rule, there is l ∈ N \{n} such that x{l,n} 6= CEB {l, n}, r{l,n} .
x
x
Since x ∈ X(N, c) and (r{l,n}
)n = xl +xn , then either xn > CEBn {l, n}, r{l,n}
x
x
or xl > CEBl {l, n}, r{l,n}
. If xn > CEBn {l, n}, r{l,n}
, by Lemma 2, it
can be shown that agent
n is better
off choosing agent l and ending up with
x
contribution CEBn {l, n}, r{l,n} < xn , which violates the assumption that
x
g is a SPE. If xl > CEBl {l, n}, r{l,n}
, we show that agent l is better off
24
x
. If agent n either takes action A, or takes
announcing CEBl {l, n}, r{l,n}
action R and chooses one agent
from N \{l,n}, say agent p, then agent l ends
x
up with contribution CEBl {l, n} , r{l,n}
. If agent n takes action R and
0
chooses agent
l, then let x be the proposal after agent l’s deviation. Since
x
x0l = CEBl {l, n} , r{l,n}
, x0n = xl + xn − x0l , and for each j ∈ N \{l, n},
x0j = xj , we have
1
τ ll (c, x0 ) ≡
2
(
1
=
2
(
=
τ ll
max{0, cl −
max{0, cl −
)
X
x0j } + (x0l + x0n ) − max{0, cn −
X
j6=l,n
j6=l,n
X
X
x0j }
)
xj } + (xl + xn ) − max{0, cn −
j6=l,n
xj }
j6=l,n
(c, x) .
x
Since x ∈ X(N, c), by Lemma 2, τ ll (c, x) = CEBl {l, n} , r{l,n}
. Thus,
x
. Since in any case, agent l is better off
τ ll (c, x0 ) = CEBl {l, n} , r{l,n}
x
< xl , the assumption that
ending up with contribution CEBl {l, n} , r{l,n}
g is a SPE is violated. We conclude that x = CEB(N, c).
P
P
• y{i,n} 6= x{i,n} . Since yN \{i,n} = xN \{i,n} and p∈N yp = p∈N xp , then
either yn > xn or yn < xn . If yn > xn , agent n takes action A rather
than action R, and ends up with contribution xn < yn . Since she is better off
deviating, the assumption that g is a SPE is violated. Thus, yn< xn and xi <
x
yi . Since x ∈ X(N, c), by Lemma 2, y{i,n} = τ i (c, x) = CEB {i, n}, r{i,n}
.
By “order preservation for contributions”14 of the CEB rule, 0 < yi ≤ yn .
We show that agent i is better off deviating. Let ε ∈ R be such that 0 <
ε < min{yi − xi , xn }. Suppose that agent i announces xi + ε. Let x0 be the
proposal after agent i’s deviation. Note that x0i = xi + ε, x0n = xn − ε, and
for each l ∈ N \{i, n}, x0l = xl . Since k ∈ N \{i, n} and yn = τ kn (c, x), then
P
x0k + x0n = xk + xn − ε ≥ 0. Note that τ kk (c, x0 ) = 21 max{0, ck − l6=k,n x0l }.
14
This property says that of two agents, the contribution of the agent with the larger
cost should be at least as much as that of the other (Potters and Sudhölter, 1999). Clearly,
the CEB rule satisfies the property.
25
Since
(
)
X
1
τ kn (c, x0 ) = xk + xn − ε − max 0, ck −
xl − ε
2
l6=k,n
(
)
X
1
= xk + xn − max 2ε, ck −
xl + ε
2
l6=k,n
(
)
X
1
< xk + xn − max 0, ck −
xl
2
l6=k,n
= τ kn (c, x)
= yn
= τ in (c, x)
= τ in (c, x0 ) ,
agent n does not choose agent i. Since agent i is better off ending up with
contribution xi + ε < yi , the assumption that g is a SPE is violated.
By Subcases 2.1 and 2.2, we conclude that y = CEB (N, c).
Q.E .D.
5
Concluding remarks
In transferable utility games, it is well-known that different formulations of
consistency axiom provide justifications for different rules: the equal allocation of nonseparable cost value (Moulin, 1985), the Shapley value (Hart and
Mas-Colell, 1989), and the prenucleolus (Orshan, 1993).
In our companion paper (Hu et al., 2011), we address whether the same
phenomenon occurs in the airport problem. We adopt the right-endpoint formulation (Potters and Sudhölter, 1999) to define “Right-endpoint Subtraction (RS) bilateral consistency” and “RS converse consistency”. We show
that the nucleolus (Schmeidler, 1969) is the only RS bilaterally consistent
(or RS conversely consistent) rule satisfying equal treatment of equals and
last-agent cost additivity. In addition, we modify Γ(N, c) by revising agents’
payoffs at the final stage based on the reduced problem underlying RS bilateral consistency and RS converse consistency. We show that there is a
unique SPE outcome of the new game, which is the allocation chosen by the
nucleolus.
26
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