Strategic Divide and Choose∗
Antonio Nicolò†
and Yan Yu‡
January 7, 2008
∗ We
thank Giovanni Andreatta, Steven Brams, Marco Li Calzi, Ehud Kalai,
Jordi Massò, Hervè Moulin, James Schummer, William Thomson and the participants to the Second World Congress of the Game Theory Society 2004 in Marseille,
for useful comments.
† Department
of Economics, University of Padua, via del Santo
33, 35123 Padova, Italy.
Email: [email protected]; home page:
http://www.decon.unipd.it/~nicolo.
‡ Department of Economics, Hong Kong University of Science & Technology,
Clear Water Bay, Kowloon,Hong Kong. Email: [email protected].
Abstract
We consider the classic cake-divison problem when the cake is a heterogeneous good represented by an interval in the real line. We provide a mechanism to implement, in an anonymous way, an envy-free and efficient allocation
when agents have private information on their preferences. The mechanism
is a multi-step sequential game form in which each agent at each step receives
a morsel of the cake that is the intersection of what she asks for herself and
what the other agent concedes to her.
JEL Classification: D63, C78
1
Introduction
Thomson (1996) pointed out that an allocation rule is conceptually different from its selections and the normative properties of its outcomes do not
coincide with those of the rule itself. This observation generates two basic
questions about fairness: should we focus exclusively on the set of allocations
in order to determine criteria for fairness or should we also look at the rule
through which the final outcome is obtained? Should we take the view of
procedural fairness or the view of "ex-post" fairness, or both?
The classic problem of dividing a heterogeneous good (a cake) between
two agents offers a great opportunity to analyze these questions in a simple
framework. As already noted by Crawford (1977), and previously by Kolm
(1972), the classic divide and choose rule provides an efficient and envy free
outcome, but it is hardly considered “fair” when there is complete information on agents’ preferences. The divide and choose rule leads to a no-envy
outcome but the rule itself is not envy free: the chooser envies the role of
the divider. Agents are not treated symmetrically in the divide and choose
rule. Fairness can be translated in requirements like anonymity, which is
directed to guarantee an ex-ante symmetric treatment of the agents, or like
no-envy, which demands an ex-post symmetry among the actual allocations
of the agents. It is quite obvious that an allocation rule can satisfy some of
these requirements while violating others. Keeping with our simple problem,
a rule which assigns the entire cake to an agent flipping a (fair!) coin, satisfies anonymity (or procedural no-envy) but clearly violates the (“ex-post”)
no-envy criterion, while the divide and choose rule when the divider is fixed,
satisfies no-envy but violates anonymity.
Reconciling efficiency, procedural fairness, and “ex-post” fairness is not
so simple as it could appear at first glance. For instance, one could think
that the divide and choose rule when the divider is randomly chosen is the
(simplest) way to make the rule “fair”. Nevertheless, introducing a random
element in the mechanism has many consequences. The set of alternatives
over which agents’ preferences are defined is now a set of lotteries. The
random mechanism which assigns with equal probability to both agents the
role of the divider is equivalent to the lottery which assigns to each agent with
equal probability one the following two envy-free allocations: the allocation
such that agent 1 is indifferent over the two portions and the allocation
such that agent 2 is indifferent over the two portions. Therefore, we need
to make assumptions on how agents evaluate lotteries and the normative
1
content of any proposed mechanism will in general be sensitive to the different
assumptions. Moreover, even if we assume standard preferences over lotteries,
representable by von Neumann-Morgenstern utility functions, the simple rule
which randomly selects the divider may open the door to inefficiency when
agents differ in the degree of risk aversion.
In this paper, we focus on the fair division problem when the good to be
divided is representable by a linear segment of length one and agents’ utility
profiles are such that single cut divisions are efficient. Many problems, such
as time sharing problems, belong to this class. Consider, for instance, two
security guards deciding their shifts during the night: if their preferences
depend not only on the number of working hours, but also on their schedule,
then the good to be divided (the night hours) is heterogeneous and it can
be fair to have shifts of different length; nevertheless it turns to be efficient
to divide the night in no more than two shifts, one for each guard. Other
examples are related to classic Hotelling models: two ice-cream pedlars have
to decide how to partition a beach in two selling regions which can be of
different length, since density of bathers may vary along the beach. Again, in
order to minimize the pedlars’ effort in commuting, it is efficient to partition
the beach in two intervals, one for each pedlar.
We propose a normative property which identifies one allocation among
those which are envy-free, and provide an anonymous deterministic mechanism which implements it. Our mechanism is a sequential multi-step version
of the divide and choose rule. The assumption that single cut partitions
of the cake are efficient, allows to compare our mechanism with the classic
divide and choose rule only on the ground of fairness.
Let’s consider the following example. Two kids, Anna and Bob, have
to divide a rectangular cake which can be represented as the interval [0, 1].
The cake is 1/3 white chocolate, the interval [0, 1/3), 1/3 dark chocolate, the
interval (2/3, 1] , and in the middle, the interval [1/3, 2/3], an even mixture
of white and dark chocolate. Suppose that Anna prefers white chocolate and
Bob the dark one, but they are both greedy and to any portion prefer a bigger
portion that contains it. Note that any single cut partition of the cake which
assigns the left portion to Anna and the right portion to Bob is efficient.
The problem is where to put the knife in order to be fair. There exists an
interval of single cut points, each of them generating an envy-free and efficient
allocation with different utility levels to the greedy kids. The divide and
choose rule where either Anna or Bob is the divider implements among the
efficient and no-envy allocations the one preferred by the divider. Suppose
2
that Anna enjoys white chocolate twice as much as she enjoys dark chocolate
and she is indifferent between [0, 0.39] and [0.39, 1], while Bob enjoys dark
chocolate twice as much as he enjoys white chocolate and he is indifferent
between [0, 0.61] and [0.61, 1]. For any s in [0.39, 0.61] , the allocation such
that Anna gets [0, s], while Bob getting (s, 1] is efficient and envy-free. The
divide and choose rule with Anna as the divider implements s = 0.61, i.e.,
Anna gets [0, 0.61], while Bob gets (0.61, 1]. The divide and choose rule with
Bob as the divider implements s = 0.39. It’s clear that although the divide
and choose mechanism implements an efficient and envy-free allocation, it
does not treat agents symmetrically. In this example, the single-cut point
s = 0.5 is the most reasonable partition of the cake. Had the mother known
their preferences, she would have given the left half of the cake to Anna and
the right half to Bob.
In order to avoid noisy discussions on who is the divider, their mother
would help the kids by providing them a way to select an envy-free and efficient allocation in an anonymous way. Uniqueness is relevant in our problem
because we cannot leave the kids to choose one allocation in a set of possible solutions, if we really want to avoid noisy discussions. To describe the
mechanism in a very intuitive way suppose that the mother knows that Anna
prefers white chocolate and Bob prefers the dark. She knows that, once she
decides where to put the knife, it is efficient to give the left portion to Anna
and the right portion to Bob. Unfortunately, she does not know how strong
the kids’ preferences are over the two types of chocolate (while the kids know
each other’s preferences). Therefore, she let them choose how to cut the cake.
In fact, she proposes the following cake-cutting mechanism to the kids.
(1) Anna proposes to Bob to cut the cake at x1 ∈ [0, 1] . (By proposing a
single cut point at x1 , Anna implicitly asks for herself the portion [0, x1 ] and
concedes to Bob the portion [x1 , 1] .)
(2) Bob may take either the portion [0, x1 ] or the portion [x1 , 1] , or make
a counter proposal.
(2a) If Bob takes one of the two portions, then Anna gets the other portion
and the game ends (all the cake has been assigned).
(2b) If Bob does not take any portion, then he has to propose a different
cut at y 1 such that y 1 < x1 . (By proposing a single cut point at y 1 , Bob
implicitly asks for himself the portion [y 1 , 1] and concedes to Anna the portion
[0, y 1 ] .)
(3) Anna can now choose to take either the portion [0, y 1 ] or the portion
1
[y , 1] , or she can choose to continue.
3
(3a) If Anna takes one of the two portions, then Bob gets the other portion
and the game ends.
(3b) If Anna chooses to continue, then she receives the portion [0, y 1 ] and
Bob receives the portion [x1 , 1] . That is, each kid receives the morsel which
is the intersection between the portion she wants for herself and what the
other kid concedes to her. The interval [y 1 , x1 ] has still to be assigned and
the mechanism is iterated following the same rules until one of the two kids
takes one of the portion proposed by the other kid.
The mechanism we propose can be interpreted as a step-by-step negotiation process in which agents reach partial agreements. Whenever both agents
agree that some part (subset) of the cake should be consumed by one agent,
then they accept to assign this part to her. In this way they "reduce" the
object over which they dispute and therefore they can more easily reach a
definitive agreement.
From a normative point of view, should any kid complain with her mother?
Should Anna and Bob compete to be the first one to make proposal? The
answer is “no”. No matter who moves first, the mechanism implements the
same equilibrium allocation: in our symmetric example, [0, 0.5] to Anna and
(0.5, 1] to Bob. First let’s see that the mechanism won’t end in one step in
the example∗ . If Anna proposes x1 = 0.5 right away, then Bob can propose
y 1 = 0.4 and then Anna’s best reply will be choosing to continue and receiving [0, 0.4] in the first stage. In this case, Bob will get more than [0.5, 1].
Therefore, Anna should not start with such a large concession.
The rule anonymously selects a no-envy and efficient allocation which
has the following characteristics. Consider a subgame starting at any stage
t of the dividing game and let [a, b] denote the cake still to be divided. In
equilibrium, each agent receives at the current stage t a morsel which has the
same value (to her/him) as the overall portion that the other agent receives.
Let ([a, S] , (S, b]) be the (efficient) subgame perfect equilibrium allocation,
where [a, S] is Anna’s portion. Let xt and y t be respectively Anna and Bob’s
proposals at the current stage t according to the subgame perfect equilibrium.
Then, Anna is indifferent between the morsel [a, y t ] , the morsel she receives
at stage t, and the portion (S, b], i.e. the portion that Bob receives in all the
remaining stages. Similarly, Bob is indifferent between the morsel [xt , b] , the
∗
If Anna and Bob’s preferences are identical and both are indifferent between white
and dark chocolate, then the mechanism will end in one step: Anna proposes 0.5 and Bob
accepts either portion.
4
morsel he receives at stage t, and the portion [a, S] . Therefore in each stage
t of the game, both agents receive the minimal efficient and no-envy morsel,
that is the efficient morsel which has the same value as the overall portion
that the other agent receives in the SPNE allocation of the subgame starting
at stage t.
The mechanism only ends when there exists a single-cut point t such that
both agents are indifferent between the two parts of the remaining cake: [a, t]
and (t, b]. Otherwise, the mechanism will take infinite steps. Note that within
each stage, the order of agents is inconsequential: Anna will make the same
proposal regardless whether she is the first or the second to propose, so will
Bob. Therefore, this rule is anonymous.
The mechanism described above is slightly more complicated in the general case when the arbitrator does not know how to efficiently divide the
cake. But the logic of the mechanism is the same. At each stage agents
sequentially propose a partition of the cake and, in case no agent takes one
of the portions of the allocation proposed by the counterpart, each of them
receives the intersection between what she asks for herself and what the other
agent concedes her to consume.
Fair division of an heterogeneous good has been widely analyzed both
in the mathematical literature and, more recently, in the economic one; see,
respectively, Brams and Taylor (1996) and Robertson and Webb (1998) for
two recent books on cake-cutting. Mathematicians have devoted great efforts
in order to find the minimal number of cuts needed to fairly cut a "cake"
according to the number of eaters (Lester and Spanier (1961), Stromquist
(1980), Barbanel and Brams (2001)), while economists have been more interested in generalizing the problem, allowing for larger domains of preferences
(Berliant, Dunz and Thomson (1992)), analyzing the case of indivisible goods
(Crawford and Heller (1979), Demko and Hill (1988), Alkan, Demange and
Gale (1991), Brams and Fishburn (2000), Edelman and Fishburn (2001)), or
imposing additional properties, such as consistency or monotonicity requirements (Thomson (1994a), (1994b) Maniquet and Sprumont (2000), among
others). In most of these contributions great attention has been devoted to
the existence and the axiomatic characterization of a normative solution to
this fair division problem, but much less attention to a strategic approach.
A relevant exception is the recent paper by Thomson (2005), who showed a
simple game form called "divide and permute" to fully implement in Nash
equilibrium the no-envy solution in n-person fair division problem. In this
5
paper we follow a similar approach by proposing a game form of a two-agent
fair division problem to implement in subgame perfect equilibrium an envyfree and efficient solution in an anonymous way by means of a deterministic
mechanism.
2
The Iterated Divide and Choose Rule
Our model is a simple version of the classical two-agent cake division problem.
There is a measurable space (Ω, F ), where Ω ≡ [0, 1] (a cake) is the object
to be divided between two agents that can be represented by an interval in
the real line, and F is a σ-algebra that contains all Lebesgue measurable
subsets of [0,1]. We say that an element of F is a portion. Agents have
preferences over portions of Ω. Each agent i = 1, 2 is endowed with a utility
function ui : F → R+ that is a nonatomic probability measure on F.† (Since
preferences are invariant up to a positive rescaling of the utility function,
ui (Ω) = 1 is only a normalization). In particular, we assume the following.
For both i = 1, 2, let vi > 0 be acontinuous function on [0, 1] . Agent i’s
utility of the portion B is ui (B) = B vi (s)ds. When a portion is an interval,
we identify this portion by means of the two extremes
of the interval, that is
b
if B ≡ [a, b] ⊆ [0, 1], we write ui (B) = ui (a, b) = a vi (s)ds. Let Ui be the set
of agent i’s utility functions and U = (U1 , U2 ) be the set of all utility profiles.
An (ordered) partition P = (P1 , . . . , Pk ) of Ω is called a portioned kpartition. An allocation P = (P1 , P2 ) is a portioned two-partition, where
Pi is the portion assigned to agent i = 1, 2. An allocation which consists
in the partition of the cake in two intervals is called a single cut allocation.
An allocation P = (P1 , P2 ) is efficient at u ∈ U if there exists no other
allocation P = (P1 , P2 ) such that ui (Pi ) ≥ ui (Pi ) for all i, with the strict
inequality holding for some i. An allocation P is envy-free at u ∈ U (or
satisfies no-envy), if ui (Pi ) ≥ ui (Pj ) for i = 1, 2. Let P denote the set of
allocations.
An allocation rule is a function f : U → P. Let fi (u) be the portion
assigned by the allocation rule f to agent i = 1, 2 at u ∈ U . An allocation
rule f is envy-free if f (u) is envy-free at every u ∈ U. An allocation rule f is
efficient if f(u) is efficient at every u ∈ U. An allocation rule is anonymous
†
A measure ui is nonatomic if, for each portion A and each x in (0, u(A)), there exists
another portion B ⊆ A such that ui (B) = x.
6
if interchanging the preferences means interchanging the assigned portions,
that is for any (u1 , u2 ) ∈ U, fi (u1 , u2 ) = fj (u1 , u2 ) for both i, j = 1, 2, i = j.
In the paper we concentrate on multi-stage sequential mechanisms. Let
Z+ be the set of positive integers. Let Ωt be the amount of the heterogeneous
good still to be divided at stage t = {1, ..., T } with T ∈ Z+ . By assumption
Ω1 ≡ Ω.For sake of clarity, from now on we call an agent’s final allocation
of the cake a portion and an agent’s share of cake allocated in one stage a
morsel.
The Iterated Divide and Choose Mechanism
The mechanism consists of multiple stages t = {1, ..., T } with T ∈ Z+ ,
and each stage t is formed by three steps. At stage t = 1 step 1, agent 1
proposes a single cut allocation X 1 = (X11 , X21 ). At step 2, agent 2 can either
take one of the two portions, X11 or X21 , or he‡ can propose a different single
cut allocation Y 1 = (Y11 , Y21 ), such that at least for some i = 1, 2, Xi1 ∩Yi1 has
positive Lebesgue measure. In case agent 2 takes one of the two portions,
X11 or X21 , agent 1 gets the other portion and the game ends. In case agent
2 proposes a different allocation Y 1 , at step 3 agent 1 decides either to take
one of the two portions, Y11 or Y21 , or to continue. If agent 1 takes one of the
two portions, Y11 or Y21 , agent 2 gets the other portion and the game ends. In
case agent 1 chooses to continue, two cases have to be considered. If Xi1 ∩ Yi1
has zero Lebesgue measure for some i = 1, 2, then the entire cake Ω1 is given
to the agent j = i and the game ends. If Xi1 ∩ Yi1 has positive measure for
both i = 1, 2, then each agent i receives the morsel Pi1 = Xi1 ∩ Yi1 and stage
1 ends.
Note that, first, either the game ends at stage 1, or each agent receives
a morsel of the cake with positive size. Second, in case the game does not
end at stage 1, the amount of cake that has still to be assigned, denoted by
Ω2 ⊂ Ω1 , is an interval. The mechanism can therefore be iterated following
the same rules.
Formally, consider any stage t = {1, ..., T } with T ∈ Z+ and denote by
Ω ⊆ [0, 1] the cake still to be assigned at stage t.
Stage t
Step 1 Agent 1 proposes a single cut allocation X t of the cake Ωt .
Step 2. Agent 2 can either
t
‡
We refer to agent 1 as a female agent and to agent 2 as a male agent
7
• take one of the two portions, X1t or X2t . Agent 1 gets the other portion
and the game ends;
• propose a different single cut allocation Y t , such that at least for some
j = 1, 2, Xjt ∩ Yjt has positive Lebesgue measure.
Step 3 Agent 1 can either
• take one of the two portions, Y1t or Y2t . Agent 2 gets the other portion
and the game ends;
• choose to continue; in this case
— if Xit ∩ Yit has zero Lebesgue measure for some agent i = 1, 2, then
the entire cake is given to the agent j = i and the game ends.
— if Xit ∩ Yit has positive Lebesgue measure for both agent i = 1, 2,
then each agent i ∈ N receives the morsel Pit = Xit ∩ Yit in Stage
t. The amount Ωt+1 = Ωt − (P1t ∪ P2t ) must still be divided, and
the mechanism proceeds to Stage t + 1.
The mechanism ends at stage T when either one of the agents takes a
portion proposed by the counterpart or the entire cake ΩT is assigned to
some agent at stage T.
2.1
An Example
We first illustrate features of this procedure with the simple example in the
introduction. Anna’s utility density function is
 4
 3 if s ∈ [0, 13 ]
1 if s ∈ ( 13 , 23 ] ;
vA (s) =
 2
if s ∈ ( 23 , 1]
3
while Bob’s utility density function is
 2
 3 if s ∈ [0, 13 ]
vB (s) =
1 if s ∈ ( 13 , 23 ] .
 4
if s ∈ ( 23 , 1]
3
The complete proof of the existence of a unique subgame perfect Nash
equilibrium and of the anonymity of the mechanism is provided later. Here
8
we illustrate agents’ equilibrium proposals (concessions). In this example,
the outcome of the iterated divide and choose mechanism is the single cut
allocation ([0, 0.5], (0.5, 1]) where the first interval is Anna’s portion and the
second Bob’s portion.
In stage 1 step 1, Anna would like to propose a single cut allocation
1
X = ([0, x1 ], (x1 , 1]) with the largest single cut point x1 (hence conceding as
little to Bob as possible in stage 1) so long as Bob would still weakly prefer
continuing and getting (s, 1] (his subgame perfect Nash equilibrium payoff
contingent on Anna’s first proposal X 1 ) to taking [0, x1 ] and ending the game
now. In fact, in equilibrium, Anna proposes the allocation ([0, x1 ], (x1 , 1])
such that Bob is indifferent between getting (s, 1] and [0, x1 ]. In equilibrium,
s = 0.5. The point x1 making Bob indifferent between (0.5, 1] and [0, x1 ] is
0.7083. If Anna proposes a single cut allocation X̂ 1 = ([0, x̂1 ], (x̂1 , 1]) with
x̂1 > 0.7083, and hence concedes to Bob less in stage 1, Bob will take the
portion [0, x̂1 ] (if Bob chooses to propose Y 1 , in any subgame perfect Nash
equilibrium contingent on X̂ 1 and any possible Y 1 , Bob gets a utility level
lower than the utility level obtained by consuming the portion (0.5, 1]) and
Anna gets (x̂1 , 1] and is worse off than when proposing X 1 . If Anna proposes
a single cut allocation X̂ 1 = ([0, x̂1 ], (x̂1 , 1]) with x̂1 < 0.7083, and hence
concedes to Bob more in stage 1, Bob will choose to continue, but the final
outcome ([0, ŝ], (ŝ1 , 1]) in the subgame perfect Nash equilibrium will have
ŝ < 0.5 and Anna is worse off in this case.
In stage 1 step 2, Bob would like to propose the allocation ([0, y 1 ], (y 1 , 1])
with the smallest single cut y 1 so long as Anna would still weakly prefer
continuing and getting [0, s) to taking (y 1 , 1] and ending the game now. In
7
the equilibrium, y 1 = 24
≈ 0.2917.
In stage 1 step 3, Anna will choose to continue. Anna gets [0, 0.2917) and
Bob gets (0.7083, 1]. There are [0.2917, 0.7083] remaining cake to be divided
in stage 2.
Note that Bob is indifferent between [0, 0.7083] and (0.5, 1]. Equivalently,
Bob is indifferent between (0.7083, 1] and [0, 0.5], i.e., Bob is indifferent between getting his current morsel of cake in Stage 1 and getting Anna’s whole
portion in the subgame perfect Nash equilibrium. Similarly, Anna is indifferent between getting her current morsel of the cake in stage 1, [0, 0.2917),
and getting Bob’s whole portion in the subgame perfect Nash equilibrium
(0.5, 1].
Note also that if Bob is the first to propose and Anna the second, in
equilibrium Bob still proposes to cut the cake at 0.2917 and Anna proposes
9
to cut the cake at 0.7083. The ordering of agents has no impact on equilibrium
payoff in this procedure.
In the equilibrium, in stage 2, Anna proposes the allocation ([0.2917, x2 ], (x2 , 0.7083])
such that Bob is indifferent between getting (0.5, 0.7083] and taking [0.2917, x2 ]
( ending the game in stage 2 step 2). Anna’s equilibrium proposal is such that
x2 = 19
≈ 0.5278. Bob proposes the allocation ([0.2917, y 2 ], (y 2 , 0.7083]) with
36
17
y 2 = 36 ≈ 0.4722. There are [0.4722, 0.5278] remaining cake to be divided in
Stage 3.
Anna and Bob’s preferences are the same over the remaining cake [0.4722, 0.5278].
In the equilibrium in stage 3, Anna proposes the allocation ([0.4722, 0.5], (0.5, 0.5278])
and Bob takes (0.5, 0.5278] (in fact, Bob is indifferent between the two portions).
In this example, the iterated divide and choose procedure implements the
symmetric outcome in three stages.
2.2
Residual-Equivalent Envy-Free allocation
The iterated divide and choose procedure implements an efficient, envy-free,
and anonymous allocation in the above example. We first define the equilibrium allocation implemented by the iterated divide and choose mechanism
and then provide the equilibrium analysis of the game form.
Let Pit denote the morsel of the cake that agent i = 1, 2 receives at
stage t and Pjt the morsel that the other agent receives at the same stage
t. Hence, P1t ∪ P2t ∪ Ωt+1 = Ωt for all t = {1, ..., T − 1} . We call Pit agent
i’s current morsel at stage t. For any t ≤ T, let Pti = ∪Tk=t Pik denote the
overall portion that agent i’s will receive playing the mechanism from stage
t onwards; therefore P1i = Pi . We call Pti agent i’s residual portion at stage
t.
Now we are ready to introduce a more demanding property than no-envy.
A multi-stage mechanism is residual-equivalent envy-free if, at each stage,
each agent is indifferent between getting her current morsel and getting the
other agent’s residual portion. Any residual-equivalent envy-free mechanism
not only is envy-free in each stage, but also it equalizes the extent to which
an agent prefers her own portion to the other agent’s portion. Note that
although we introduce the concept of residual-equivalent envy-free in the
context of multistage mechanisms, it is well defined for any allocation P, as
10
in the formal definition below. Let (Pi1 , ..., PiT ) be a partition of agent i’s
portion Pi in T morsels.
Definition 1 An allocation P = (P1 , P2 ) is residual-equivalent envy-free
(REEF) at u ∈ U if for both agents i = 1, 2 there exists a partition (Pi1 , ..., PiT )
of Pi such that ui (Pit ) = ui (Ptj ) with j = 1, 2, j = i and for all t = {1, 2, ...T }
with T ∈ Z+ . An allocation rule f is residual-equivalent envy-free if f(u) is
residual-equivalent envy-free at every u ∈ U .
We do not provide an axiomatic foundation for this requirement. It is a
useful tool to prove that the mechanism we present implements a no-envy
and efficient allocation in an anonymous way. In the next section, in fact, we
characterize a utility domain in which for each utility profile there exists a
unique allocation that satisfies the above condition. For all utility profiles u
belonging to this domain, the REEF allocation at u turns out to be the SPNE
outcome of the iterated divide and choose game when agents are endowed
with the utility profile u.
2.3
Existence and Uniqueness of REEF allocations
The classic divide and choose mechanism generates a partition of the cake
in two intervals. We propose our mechanism as a way of ameliorating it by
guaranteeing an anonymous selection of an envy-free and efficient allocation.
Hence, we focus on the utility profile domain in which partitioning the cake
in two intervals is efficient.
(A1) For all x ∈ [0, 1] , either the allocation ([0, x], (x, 1]) or ([x, 1], [0, x))
is efficient.
Let U sc denote the domain of utility profiles for which condition A1 holds.
Hence, for all utility profiles in U sc and for all x ∈ [0, 1] , there always exists
an allocation generated by the single-cut x which is efficient. The following
Lemma characterizes the set U sc .
Lemma 1 A sufficient and necessary condition for (A1) is that v1 (x)−v2 (x)
is (weakly) monotonic in x.
Proof. Sufficiency. Without loss of generality, suppose v1 (x) − v2 (x) is
(weakly) decreasing. Suppose that there is a single cut allocation ([0, a), [a, 1])
which is not efficient, then there exists another allocation (P1 , P2 ) such that
11
u1 (P1 ) ≥ u1 (0, a) and u2 (P2 ) ≥ u2 (a, 1) with at least one strict inequality. Because u2 (P2 ) ≥ u2 (a, 1) and v2 (x) > 0 by definition, it is impossible that [0, a) ⊂ P1 such that the (Lebesgue) measure of P1 is larger
than a. Similarly [a, 1] ⊂ P2 is not possible. Let A = [0, a) ∩ P2 and
B = [a, 1] ∩ P1 . We know that A and B have positive Lebesgue measure.
There are three possible scenarios. (1) If u1 (A) > u1 (B), then there exists a
set A , A ⊂ A, such that u1 (A ) = u1 (B). Note that P1 = ({[0, a) − A}∪B) ⊂
({[0, a) − A } ∪ B). Hence, u1 (P1 ) < u1 ({[0, a) − A } ∪ B) = u1 ([0, a)). Contradictory to the claim that (P1 , P2 ) is Pareto superior to ([0, a), [a, 1]). (2)
If u1 (A) < u1 (B), because v1 (x) − v2 (x) is (weakly) decreasing and A is
to the left of B, u2 (A) < u2 (B). There exists a set B , B ⊂ B, such that
u2 (B ) = u2 (A). Note that P2 = ({[a, 1] − B} ∪ A) ⊂ ({[a, 1] − B } ∪ A).
Hence, u2 (P2 ) < u2 ({[a, 1] − B } ∪ A) = u2 ([a, 1]). Contradictory to the
claim that (P1 , P2 ) is Pareto superior to ([0, a), [a, 1]). (3) If u1 (A) = u1 (B),
because v1 (x) − v2 (x) is (weakly) decreasing and A is to the left of B,
u2 (A) ≤ u2 (B). If u2 (A) < u2 (B), apply the same argument as in (2) and
find a contradiction. If u2 (A) = u2 (B), this is contradictory to the claim
that (P1 , P2 ) is Pareto superior to ([0, a), [a, 1]).
Necessity. Suppose that v1 (x) − v2 (x) is not monotonic in x. Without
loss of generality, suppose that there exist three points: a, b, c, with 0 < a <
b < c < 1, such that v1 (b) − v2 (b) < v1 (c) − v2 (c) < v1 (a) − v2 (a)§ . Let y
be any point between b and c, i.e., a < b < y < c. We now show that both
([0, y), [y, 1]) and ([y, 1], [0, y)) are not efficient allocations. Let’s first look
at ([0, y), [y, 1]). Intuitively, agent 1 can exchange a tiny slice of the cake
centered around b with agent 2 for a tiny slice of the cake centered around c,
to make both agents better off. Let !b , !c be sufficiently small such that for
any x ∈ [b, b + !b ], v1 (x) − v2 (x) < v1 (c) − v2 (c), where !b < y − b, and for any
x ∈ [c, c + !c ], v1 (x) − v2 (x) > v1 (b) − v2 (b); moreover, the following equation
holds:
b+b
c+c
−
v1 (x)dx +
v1 (x)dx = 0.
b
c
We can find such !b , !c due to the continuity of agents’ utility density functions. By construction, agent 1 is indifferent between [0, y) and [0, b) ∪ [b +
!b , y) ∪ (c, c + !c ]; while agent 2 strictly prefers [y, c] ∪ (c + !c , 1] ∪ [b, b + !b ) to
§
If v1 (x) − v2 (x) is not monotonic in x, then either v1 (x) − v2 (x) is U-shaped over
certain interval or it is ∩-shaped over certain interval. We can find three points: a, b, c,
with 0 < a < b < c < 1, such that either v1 (b) − v2 (b) < v1 (c) − v2 (c) < v1 (a) − v2 (a) or
v1 (a) − v2 (a) < v1 (c) − v2 (c) < v1 (b) − v2 (b). The proofs of the two cases are symmetric.
12
b+
c+
[y, 1], because b b v2 (x)dx − c c v2 (x)dx > 0. Hence, ([0, b) ∪ [b + !b , y) ∪
(c, c + !c ], [y, c] ∪ (c + !c , 1] ∪ [b, b + !b )) Pareto dominates ([0, y), [y, 1]). Now
let’s check ([y, 1], [0, y)). Similarly, agent 1 can exchange a tiny slice of the
cake centered around c with agent 2 for a tiny slice of the cake centered
around a, to make both agents better off. The formal proof is omitted.
Note that the U sc domain contains the domain of single-peaked (or singleplateaued) utility functions where agents’ peaks are on the opposite extremes
[a,b]
of the segment
[0,1]. Let Fi
be the point of the interval [a, b] such that
[a,b]
[a,b]
[a,b]
1
ui a, Fi
agent i’s indifference
= ui Fi , b = 2 ui (a, b). We call Fi
[a,b]
point over [a, b]. With a little abuse in notation we write ui Fi
to denote
[a,b]
the utility of agent i in taking one of the two portions, either a, Fi
[a,b]
[a,b]
or Fi , b , respectively. We call ui Fi
agent i’s half-cake-equivalent
utility of [a, b].
Lemma 2 For any preference profile u ∈ U sc , if F1[0,1] ≤ F2[0,1] F1[0,1] ≥ F2[0,1]
[a,b]
[a,b]
[a,b]
[a,b]
then F1 ≤ F2
F1 ≥ F2
for all [a, b] ⊆ [0, 1].
Proof. By Lemma 1, v1 (x) − v2 (x) is either weakly increasing or weakly
[0,1]
[0,1]
decreasing. If F1 ≤ F2 , then v1 (x) − v2 (x) is weakly decreasing. Since
[a,b]
[a,b]
v1 (x)−v2 (x) is weakly decreasing over [0, 1], F1 ≤ F2 for all [a, b] ⊆ [0, 1].
Proposition 1 For any preference profile u ∈ U sc , there exists a unique
efficient residual-equivalent envy-free allocation.
We provide the intuition of the proof here. The formal proof is in the
[0,1]
[0,1]
appendix. For any u ∈ U sc , if F1
= F2
= c, then ([0, c), [c, 1]) is the
unique (in terms of utility) allocation which satisfies REEF property. With[0,1]
[0,1]
out loss of generality, assume F1 < F2 , then in any efficient single cut
allocation, agent 1 gets the left portion and agent 2 gets the right portion of
the cake. For any c ∈ [0, F1[0,1] ), ([0, c), [c, 1]) is not equivalent to an efficient
REEF allocation because agent 1 envies agent 2. Similarly, ([0, c), [c, 1]) is not
[0,1]
equivalent to an efficient REEF for any c ∈ (F2 , 1]. Hence, if ([0, c), [c, 1])
[0,1]
[0,1]
is equivalent to an efficient REEF allocation, then c ∈ [F1 , F2 ].
13
[0,1]
[0,1]
For any point c ∈ [F1 , F2 ], define yc1 such that u1 ([0, yc1 )) = u1 ([c, 1]),
and x1c such that u2 ([x1c , 1]) = u2 ([0, c]). If ([0, c), [c, 1]) is equivalent to an
efficient REEF, [0, yc1 ) is equivalent to agent 1’s morsel in stage 1, P11 , and
(x1c , 1] is equivalent to agent 2’s morsel in stage 1, P21 . Note that x1c and yc1
are decreasing and continuous in c. Now we look at agents’ division of the
remaining cake [yc1 , x1c ].
[y1 ,x1 ]
[y1 ,x1 ]
Let F11 ≡ F1 c c (F21 ≡ F2 c c ) denote agent 1’s (agent 2’s ) indifferent
point over [yc1 , x1c ], i.e., u1 ([yc1 , F11 ]) = u1 ([F11 , x1c ]). By Lemma 2, F11 ≤ F21 .
Similar to the above reasoning, if ([0, c), [c, 1]) is equivalent to an efficient
REEF allocation, then c must be in the interval [F11 , F21 ]. We can find an
[0,1]
[0,1]
interval [c1 , c1 ], which is a strict subset of [F1 , F2 ], such that if c ∈
/
1 1
1
1
[c , c ], then c ∈
/ [F1 , F2 ]. Hence, if ([0, c), [c, 1]) is equivalent to an efficient
REEF allocation, then c ∈ [c1 , c1 ]. If c1 = c1 = c∗ , we can show that F21 =
F12 = c∗ and ([0, c∗ ), [c∗ , 1]) is the unique (in terms of utility) efficient REEF
allocation.
Suppose that c1 < c1 .
For any c ∈ [c1 , c1 ], define x2c , yc2 such that if ([0, c), [c, 1]) is equivalent
to an efficient REEF, (yc1 , yc2 ] is equivalent to agent 1’s morsel in stage 2
and [x2c , x1c ) is equivalent to agent 2’s morsel in stage 2, i.e., (P11 , P21 ) =
((yc1 , yc2 ], [x2c , x1c )). We then look at agents’ division of the remaining cake
[yc2 , x2c ].
Define xtc , yct ; F1t , F2t , and ct , ct similarly. There are two possible scenarios.
(1) At some stage t, t ∈ Z+ , ct = ct = c∗ , and then ([0, c∗ ), [c∗ , 1]) is the
unique (in terms of utility) efficient REEF allocation. (2) At any stage t,
t ∈ Z+ , ct < ct . Since limt→∞ (xtc − yct ) = 0, then limt→∞(ct − ct ) = 0;
let c∗ = lim ct = lim ct , then ([0, c∗ ), [c∗ , 1]) is the unique efficient REEF
allocation.
2.4
Equilibrium Analysis
Proposition 2 The efficient residual equivalent envy-free allocation at u ∈
U sc is the unique SPNE outcome of the iterated divide and choose game when
agents are endowed with utility profile u.
Proof. See the appendix.
Since we proved that the residual equivalent envy-free allocation is unique,
it follows that the SPNE outcome of the game does not depend on the order
in which agents play, as its symmetric structure suggests.
14
2.5
K-truncated divide and choose mechanism
The mechanism might be infinite, and therefore it is interesting to know if a
finite version still has any nice property. Consider a K-truncated version of
the mechanism when we exogenously fix the maximum number of iterations
and whenever agents reach the last stage K,then they play the classic divide
and choose mechanism. That is, at stage K step 1 agent 1 proposes a single
cut allocation and at step 2 agent 2 chooses the portion he prefers.
For a given utility profile u ∈ U sc , if the iterated divide and choose game
ends in a number of stages lower than K, the SPNE outcome of the K −
truncated game coincides with the SPNE of the iterated divide and choose
game. Hence, we focus on the interesting case in which K < T ∗ (u),where
T ∗ (u) ∈ Z+ is the number of stages played by the agents in the SPNE of the
(non-truncated) iterated divide and choose game when agents are endowed
with utility profile u ∈ U sc .
Before stating a general result, we illustrate the features of the truncated
procedure with our leading example, described in Section 2.1. Let K = 2
and let s, as usual, be the single cut which characterizes the allocation,
SPNE outcome of this game. In stage 1 step 1, Anna (agent 1) proposes a
single cut allocation X 1 = ([0, x1 ], (x1 , 1]) with the largest single cut point
x1 so long as Bob weakly prefers continuing and getting (s, 1] to taking
[0, x1 ] and ending the game. At step 2 Bob proposes a single cut allocation
Y 1 = ([0, y 1 ], (y 1 , 1]) with the smallest single cut point y 1 so long as Anna
weakly prefers continuing and getting [0, s] to taking [y 1 , 1]. At stage 2 step 1,
[y1 ,x1 ]
[y1 ,x1 ] 1
), [F2
, x ]) and at step 2 agent
agent 1 proposes the allocation ([y 1 , F2
1 ,x1
y
[
] 1
[y1 ,x1 ]
2 takes the portion [F2
, x ]. Hence, in equilibrium s = F2
, Anna
1 ,x1
y
]
[
proposes the single cut allocation X 1 such that u2 (0, x1 ) = u2 (F2
, 1), and
1 ,x1
y
[
]
Bob proposes the single cut allocation Y 1 such that u1 (0, F2
) = u1 (y 1 , 1).
Simple computation shows that x1 = 59
, y 1 = 27 and s = 32
. Hence in this
84
63
example, the 2−truncated iterated divide and choose mechanism implements
the outcome [0, 32
), [ 32
, 1]. It is straightforward to note that agent 1 obtains a
63
63
portion larger than the portion she obtains in the iterated divide and choose
mechanism; the opposite result holds in case agent 2 moves first in each stage.
Nevertheless, agent 2’s utility is higher than he obtains in the classical divide
and choose rule where agent 1 is the divider.
The following Lemma generalizes the above observations.
15
Lemma 3 Let u ∈ U sc be the utility profile of the agents. The SPNE outcome of the K-truncated iterated divide and choose game is an efficient and
envy-free allocation, and
(i) for all 1 < K ≤ T ∗ (u) the utility of both agents is higher than the utility
level that the chooser achieves if agents play the divide and choose mechanism;
(ii) for all K < T ∗ (u) both agents prefer to be agent 1 of the iterated divide and choose game, but agent 2’s utility is increasing in the number of the
iterations K.
Proof. See the appendix.
Hence, even if the iterated divide and choose mechanism might be infinite,
an arbitrator can properly approximate its SPNE outcome by choosing a
relatively large number of iterations. Intuitively, the SPNE outcome of the
K-truncated divide and choose mechanism approaches the REEF allocation
very quickly. Because each agent is indifferent between his current morsel
and the other agent’s total share in the remaining game (including current
stage). Each agent’s morsel in stage t decreases significantly as t increases.
In our leading example, outcome of the 2-truncated version is very close to
the REEF allocation.
A statement similar to Lemma 3 holds for the iterated divide and choose
infinite mechanism with discounting where at each round there exists a positive probability of breakdown at the end of the round. In case of breakdown
the cake still to be divided is disposed. Let α be the probability of breakdown
at each round. The case with α = 1 is equivalent to the divide and choose
mechanism. It is easy to prove that the SPNE outcome of the iterated divide
and choose game with discounting is an efficient and envy-free allocation and
for all α ∈ [0, 1) the utility of both agents is higher than the utility level that
the chooser achieves if agents play the divide and choose mechanism; moreover both agents prefer to be agent 1 of the iterated divide and choose game
with discounting, and agent 2’s utility is decreasing in α.
3
Conclusion
The divide and choose rule implements an envy-free outcome, but the rule
itself is not envy-free: the chooser envies the divider. We propose the iterated
16
divide and choose rule which is both procedural envy-free and outcome envyfree: the two individuals are treated symmetrically. The proposed mechanism
implements the Residual-Equivalent Envy-Free allocation: the outcome is
not only envy-free, but also envy-free to the same extent. In each stage,
each agent is indifferent between her/his own morsel in the current stage and
the other agent’s residual portion in all the remaining stages (including the
current stage).
The iterated divide and choose rule gives rise to a gradual process in
which each agent takes turns to make concessions to the other agents. This
resembles the gradual bargaining process. Compte and Jehiel (2004) provided
a novel explanation for gradualism in bargaining. Two important features of
their model are (1) making a concession increases the other party’s outside
option pay-off and (2) each party can opt out at any time and outside option
incurs an efficiency loss in comparison with a negotiated agreement. If an
agent makes a large concession, it may increase the other agent’s outside
option value so much that the other agent prefers to opt out. Therefore,
agents only make small concessions resulting the gradualism in bargaining.
The opt-out efficiency loss is crucial in their model: it gives agents incentives
to make concessions to reach agreement and it is one of the most important
factor in determining the size of concession: the smaller the opt-out efficiency
loss, the more like the other agent will opt out, and hence the smaller should
the concession be. Our mechanism exploits their intuitions in the specific
context of divide and choose games. According to our mechanism, making a
concession increases the other agent’s payoff (which is the same as in Compte
and Jehiel (2004)) and therefore it induces an upper bound on the size of the
concession. Moreover, in our mechanism agents can stop the mechanism by
taking any of the two portions of the allocation proposed by the counterpart.
This possibility not only may lead (again as in Compte and Jehiel (2004))
to an opt-out efficiency loss, but also it induces a lower bound in the size of
the concession. Actually, this is the main difference with respect to Compte
and Jehiel (2004) model: in our mechanism a "too small" concession leads to
an opt-out inefficiency loss, while a "too large" concession does not induces
the counterpart to opt out inefficiently, but it directly decreases the payoff
of the proposer in the negotiated agreement.
17
References
[1] Alkan, A., G. Demange and D. Gale (1991). Fair Allocation of Indivisible
Goods and Criteria of Justice. Econometrica 59, 1023-1039.
[2] Barbanel, J. and S. Brams (2001). Cake Division with Minimal Cuts:
Envy-Free Procedures for 3 Person, 4 Persons, and Beyond. Mathematical Social Sciences 48, 251-270.
[3] Berliant, M., K. Dunz, and W. Thomson (1992). On the Fair Division
of a Heterogeneous Commodity. J. Math. Econ. 21, 201-16.
[4] Brams, S. and A. Taylor (1996). Fair Division: From cake-cutting to
Dispute Resolution. Cambridge UK, Cambridge University Press.
[5] Brams, S. and P. Fishburn (2000). Fair Division of Indivisible Items
between Two People with Identical Preferences: Envy-Freeness, ParetoOptimality and Equity. Soc. Choice Welfare 17, 247-267.
[6] Brams, S., M. Jones and C. Klamler (2004). Perfect Cake-Cutting Procedures with Money. Mimeo.
[7] Brams, S. (2004). Fair Division. In Barry R. Weingast and Donald
Wittman (eds.), Oxford Handbook of Political Economy, Oxford UK,
Oxford University Press.
[8] Compte O., and P. Jehiel (2004). Gradualism in Bargaining and Contribution Games. Rev. Econ. Stud. 71, 975-1000.
[9] Crawford, V. (1977). A Game of Fair Division. Rev. Econ. Stud. 44,
235-47.
[10] Crawford, V. and W. Heller (1979). Fair Division with Indivisible Commodities. J. Econ. Theory 21, 10-27.
[11] Demko, S. and T. Hill (1988). Equitable Distribution of Indivisible Objects. Math. Soc. Sciences 16, 145-158.
[12] Edelman, P. and P. Fishburn (2001). Fair Division of Indivisible Items
among People with Similar Preferences. Math. Soc. Sciences 41, 327-347.
18
[13] Kolm, S. C. (1972). Justice et Equité. Paris, Centre National de la
Recherche Scientifique.
[14] Lester, E. and E. Spanier (1961). How to Cut a Cake Fairly. American
Mathematical Monthly 68, 1-17.
[15] Maniquet, F. and Y. Sprumont (2000). On Resource Monotonicity in
the Fair Division Problem. Econ. Letters 68, 299-302.
[16] Robertson, J. and W. Webb (1998). Cake-Cutting Algorithms: Be Fair
If You Can. Natick MA, A. K. Peters Ltd.
[17] Stromquist, W. (1980). How to Cut a Cake Fairly. American Mathematical Monthly 106, 922-934.
[18] Thomson W. (1994a). Resource-Monotonic Solutions to the Problem of
Fair Division When Preferences Are Single-Peaked. Soc. Choice Welfare,
11, 205-223.
[19] Thomson W. (1994b), Consistent Solutions to the Problem of Fair Division When Preferences Are Single-Peaked, J. Econ. Theory, 63, 219-45.
[20] Thomson W. (1996). Concepts of Implementation. Japanese Econ. Rev.,
47(2), 133-143.
[21] Thomson W. (2005). Divide-and-Permute. Games Econ. Behav 52, 186200.
4
Appendix
[0,1]
[0,1]
Proof of Proposition 1 For any u ∈ U sc , if F1
= F2
= c, then
([0, c), [c, 1]) is the unique (in terms of utility) allocation which satisfies REEF
[0,1]
[0,1]
property. Without loss of generality, assume F1
< F2 , then in any
efficient single cut allocation, agent 1 gets the left part and agent 2 gets the
[0,1]
right part of the cake. For any c ∈ [0, F1 ), ([0, c), [c, 1]) is not equivalent
to an efficient residual-equivalent envy-free allocation because agent 1 envies
[0,1]
agent 2. Similarly, for any c ∈ (F2 , 1], ([0, c), [c, 1]) is not equivalent to an
efficient REEF because agent 2 envies agent 1.
[0,1]
[0,1]
For any c ∈ [F1 , F2 ], define yc1 such that u1 ([0, yc1 )) = u1 ([c, 1]), and
x1c such that u2 ([x1c , 1]) = u2 ([0, c]). If ([0, c), [c, 1]) is equivalent to an efficient
19
REEF, allocation, [0, yc1 ) is equivalent to agent 1’s portion in stage 1, P11 , and
(x1c , 1] is equivalent to agent 2’s portion in stage 1, P21 . Note that x1c and yc1 are
[y 1 ,x1 ]
[y1 ,x1 ]
decreasing and continuous in c. Let F1 c c (F2 c c ) denote agent 1’s (agent
[y1 ,x1 ]
[y1 ,x1 ]
2’s ) indifferent point over [yc1 , x1c ], i.e., u1 ([yc1 , F1 c c ]) = u1 ([F1 c c , x1c ]).
[y1 ,x1 ]
[y1 ,x1 ]
[0,1]
[0,1]
C 1.1 F1 c c and F2 c c are continuous functions of c from [F1 , F2 ]
[0,1]
[0,1]
into [F1 , F2 ].
[0,1]
[0,1]
[y1 ,x1 ]
P
Because u ∈ U sc and F1
< F2
, by Lemma 2, F1 c c ≤
[y1 ,x1 ]
[y 1 ,x1 ]
F2 c c . Since x1c and yc1 are decreasing and continuous in c, F1 c c and
[y1 ,x1 ]
F2 c c are decreasing and continuous in c. (See figure 1 for illustration.)
[y1 ,x1 ]
[0,1]
[0,1]
Let F 11 denote the lower bound of F1 c c for c ∈ [F1 , F2 ]. When c =
[0,1]
[y1 ,x1 ]
F2 , F1 c c
[0,1]
[y1 [0,1] ,F2 ]
F
achieves its minimum on
[0,1]
[0,1]
[F1 , F2 ].
F 11
[y1 [0,1] ,x1 [0,1] ]
= F1
F2
F2
=
[0,1]
F1 2
= F1 . (See figure 2 for illustration). The last equality fol[0,1]
[0,1]
lows from the fact that u1 ([F2 , 1]) = u1 ([0, y 1 [0,1] ]) and u1 ([0, F1 ]) =
F2
1
[0,1]
[y1 ,x1 ]
[0,1]
[0,1]
u1 ([F1 , 1]). Let F 2 denote the upper bound of F2 c c for c ∈ [F1 , F2 ].
1 1
When c = F1[0,1] , F2[yc ,xc ] achieves its maximum on [F1[0,1] , F2[0,1] ]. Similarly, we
1
[0,1]
[y1 ,x1 ]
[y 1 ,x1 ]
[0,1]
[y 1 ,x1 ]
[y1 ,x1 ]
have F 2 = F2 . Since F1 c c ≤ F2 c c , we have F1 ≤ F1 c c ≤ F2 c c ≤
[0,1]
[0,1]
[0,1]
[y1 ,x1 ]
[y 1 ,x1 ]
F2 for any c ∈ [F1 , F2 ]. Therefore, F1 c c and F2 c c are continuous
functions of c from [F1[0,1] , F2[0,1] ] into itself.
(Insert figure 1 and 2 here)
[y1 ,x1 ]
By Brouwer’s fixed point theorem, there exists c such that F1 c c = c.
[y1 ,x1 ]
Let c1 denote the smallest fixed point such that F1 c c = c. Note that
[0,1]
[0,1]
[y1 ,x1 ]
[0,1]
[0,1]
when c = F1 , yc1 = F1 and F1 c c > F1 , so c = F1 is not a fixed
1
1
1
1
[y ,x ]
[0,1]
[y ,x ]
point of F1 c c . Therefore c1 > F1 . Since F1 c c is decreasing in c, for
1
1
1
1
[0,1]
[y ,x ]
[y ,x ]
any c ∈ [F1 , c1 ), c < c1 ≤ F1 c c ≤ F2 c c ; hence, ([0, c), [c, 1]) is not an
[0,1]
efficient REEF allocation for any c ∈ [F1 , c1 ), because agent 1 envies agent
2 over the division of [yc1 , x1c ]. Similarly, by Brouwer’s fixed point theorem,
[y 1 ,x1 ]
there exists a c such that F2 c c = c. Let c1 denote the largest fixed point such
[y1 ,x1 ]
[0,1]
[y1 ,x1 ]
that F2 c c = c. Note that c = F2 is not a fixed point of F2 c c , therefore
[0,1]
[0,1]
[y1 ,x1 ]
[y1 ,x1 ]
c1 < F2 . Similarly, for any c ∈ (c1 , F2 ], c > F2 c c ≥ F1 c c , therefore,
20
[0,1]
([0, c), [c, 1]) is not an efficient REEF allocation for any c ∈ (c1 , F2 ]. Since
[y1 ,x1 ]
[y1 ,x1 ]
[0,1]
[0,1]
F1 c c ≤ F2 c c for all c ∈ [F1 , F2 ] and both are decreasing in c, c1 ≤ c1 .
We just established that if ([0, c), [c, 1]) is equivalent to an efficient REEF
allocation, then c ∈ [c1 , c1 ].
[y1 ,x1 ]
[y1 ,x1 ]
If c1 = c1 = c∗ , then by the definition of fixed points, F2 c c = F1 c c =
c∗ and the proposition is proved: the allocation ([0, c∗ ), [c∗ , 1]) is the unique
(in terms of utility) efficient REEF allocation.
If c1 < c1 , for any c ∈ [c1 , c1 ], define x2c such that u2 ([x2c , x1c ]) = u2 ([yc1 , c])
and yc2 such that u1 ([yc1 , yc2 ]) = u1 ([c, x1c ]). If ([0, c), [c, 1]) is equivalent to an
efficient REEF allocation, (yc1 , yc2 ] is equivalent to agent 1’s morsel in stage
2, P12 , and [x2c , x1c ) is equivalent to agent 2’s morsel in stage 2, P22 . Let
[y2 ,x2 ]
[y2 ,x2 ]
F1 c c (F2 c c ) denote agent 1’s (agent 2’s) indifference point over [yc2 , x2c ],
[y2 ,x2 ]
[y2 ,x2 ]
i.e., u1 ([yc2 , F1 c c ]) = u1 ([F1 c c , x2c ]).
2 2
Similar to the proof of above claim, we can establish that F1[yc ,xc ] and
[y2 ,x2 ]
F2 c c are continuous functions of c from [c1 , c1 ] into [c1 , c1 ]. Therefore, by
[y2 ,x2 ]
[y2 ,x2 ]
Brouwer’s fixed point theorem, F1 c c and F2 c c have fixed points. Let
[y2 ,x2 ]
c2 denote the smallest fixed point such that F1 c c = c and let c2 denote
[y 2 ,x2 ]
[y 1 ,x1 ]
the largest fixed point such that F2 c c = c. When c = c1 , F1 c c = c (by
[y2 ,x2 ]
definition of the fixed point) and yc2 = c, therefore F1 c c > c. So c = c1 is
[y2 ,x2 ]
not a fixed point of F1 c c . Moreover, c2 ∈ [ c1 , c1 ], so c1 < c2 . Similarly we
establish c1 < c2 ≤ c2 < c1 .
If c2 = c2 = c∗ , then the proposition is proved and the allocation ([0, c∗ ), [c∗ , 1])
is the unique (in terms of utility) efficient REEF allocation. If c2 < c2 , for
[y3 ,x3 ]
[y3 ,x3 ]
any c ∈ [c2 , c2 ], define yc3 , x3c , F1 c c , F2 c c , and c3 , c3 similarly. Hence, for
[yct+1 ,xt+1
]
[y t+1 ,xt+1 ]
c
any c ∈ [ct , ct ], define yct+1 , xt+1
, F2 c c , and ct+1 , ct+1 simic ; F1
larly. By definition ct < ct+1 ≤ ct+1 < ct for all t. Also by definition, for
any c ∈ [ct , ct ], yct < ct < ct < xtc . If for any t < ∞, ct = ct = c∗ , then
the proposition is proved and the allocation ([0, c∗ ), [c∗ , 1]) is the unique (in
terms of utility) efficient REEF allocation. If ct < ct for all t < ∞, and
lim ct < lim ct , then for any c ∈ (lim ct , lim ct ), lim yct < c < lim xtc , which implies that u2 ([lim yct , c]) = 0, contradicting v2 > 0. Therefore lim ct = lim ct .
Let c∗ = lim ct = lim ct . It is straightforward that ([0, c∗ ), [c∗ , 1]) is the unique
(in terms of utility) efficient REEF allocation when F1[0,1] < F2[0,1] .
Proof of Proposition 2 To prove this proposition we proceed by proving
some easy lemmata. Let at = min{xt−1 , y t−1 } and bt = max{xt−1 , y t−1 } for
21
all t > 1 and a1 = 0 and b1 = 1. We assume that agents only use stationary
strategies in the sense that at each stage t agents’ strategies only depend
upon the cake still to be divided, Ωt = [at , bt ], and on the proposals made
[0,1]
[0,1]
at this stage. We suppose, without loss of generality, that F1 ≤ F2 . In
t
t
[a ,b ]
order to simplify notation we write Fit instead of Fi
, when it does not
generate confusion in the reader.
Lemma 4 Consider any subgame starting at step 1 of some stage t. Let Pit
denote agent i’s portion in the subgame perfect equilibrium allocation of the
subgame Ωt . Then, ui (Pit ) ≥ ui (Fit ) = 12 ui (Ωt ), for both i = 1, 2, and for all
t = {1, 2, ..., T } .
Proof. We actually prove a stronger claim, that is for all t = {1, 2, ..., T }
each agent has a strategy that guarantees her to obtain ui (Pti ) ≥ ui (Fit )
(not only in equilibrium). Agent 2 takes his preferred portion of the allocation proposed by agent 1. Agent 1 proposes X t with X1t = [at , F1t ] and, in
case agent 2 makes a proposal, takes her preferred portion of the allocation
proposed by agent 2.
Lemma 5 In all subgame perfect equilibria of the game the mechanism ends
only if at stage T ∈ Z+ , agent 2 chooses one portion of the allocation proposed
by agent 1 and both portions of the allocation are indifferent for agent 2.
Proof. Consider any stage t of the game. By design, the mechanism ends
either if one of the agent takes one portion of the allocation proposed by the
other agent, or if for some j ∈ N Xjt ∩Yjt = ∅. In this last case there exists one
agent who receives a portion of zero Lebesgue measure contradicting Lemma
4. Now we prove the following two claims.
C 2.1 Agent 1 never chooses to end the game at step 3.
P
. Suppose that X1t = [at , xt ]. In equilibrium Y1t = [at , y t ] (otherwise there exists at least one agent j for which Xjt ∩ Yjt = ∅). There are two
cases: (i) y t > xt and agent 1 takes the portion [at , y t ]. But then agent 2
would obtain a higher payoff taking [xt , bt ] at step 2; (ii) y t < xt . If agent 1
takes the portion [y t , bt ], then agent 2 would obtain a higher payoff taking
[at , xt ] at step 2. If agent 1 takes the portion [at , y t ] , she could obtain a
greater utility by not taking any portion, in which case agent 1 receives the
morsel [at , y t ] at this stage and some morsel with positive Lebesgue measure
22
in the ensuing stages (by Lemma 4). The proof of the case X1t = [xt , bt ]
follows the same argument.
C 2.2 Agent 2 chooses to end the game only if agent 1 partitions the
cake in two portions which are indifferent for agent 2.
P
. Suppose that agent 2 takes a portion which is strictly preferred
[at ,bt ]
by him to the other one. Then, either xt > F2
and agent 2 takes the
t
t
t
t
portion [a , x ] or x < F2 and he takes the portion [xt , bt ] . The first case
contradicts Lemma 4 since agent 1 receives less than her half-cake-equivalent
utility. Hence, it must be that F1t ≤ xt < F2t . To take the portion [xt , bt ]
is a best response only if F1t = xt , otherwise proposing Y1t = [at , y t ] with
F1t < y t < xt agent 2 obtains a higher payoff. However, to propose X1t =
[at , F1t ] is not a best response for agent 1, since proposing X̃1t = [at , x̃t ] with
F1t < x̃t < F2t agent 1 obtains a payoff strictly higher than her half-cakeequivalent utility.
Now we can easily prove the following.
Lemma 6 In any subgame perfect equilibrium path, the mechanism ends at
stage T ∈ Z+ if and only if there exists a unique envy free allocation of the
cake ΩT .
Proof. Necessity. Suppose that there are more than one envy-free allocation: therefore F1t < F2t . We have proved above that the mechanism ends
at stage t only if xt = F2t . Suppose that agent 1 proposes X1t = [at , F2t ] . To
take a portion is not a best response for agent 2, since he obtains a larger
payoff by proposing Y1t = [at , y t ] with F1t ≤ y t < F2t . Suppose that agent
1 proposes X1t = [F2t , bt ]. Then agent 2 will announce Y1t = [at , F1t ] forcing
agent 1 to take the portion [at , F1t ] at stage 3 (otherwise agent 2 gets all the
cake), but then to propose X1t = [F2t , bt ] is not a best response for agent 1.
Sufficiency. Suppose now that at some T ∈ Z+ there exists a unique envy
free allocation, characterized by the point z = F1t = F2t . The result directly
follows from Lemma 4.
Lemma 7 In all subgame perfect equilibria X1t = [at , xt ] and Y1t = [at , y t ]
with xt > y t for all t = T.
Proof. Consider any stage t < T. Suppose X1t = [xt , bt ] . If xt ≤ F1t then
agent 2 will pick the portion [xt , bt ] , contradicting Lemma 6. If xt > F1t ,
then agent 2 can propose Y1t = [at , F1t ] forcing agent 1 to take the portion
23
[at , F1t ] at step 3 (note that ∩i∈N P1i (Ωt ) = ∅). But by proposing X1t = [at , F2t ]
at step 1 agent 1 can obtain a payoff strictly higher than the half-cake utility
level. Hence X1t = [at , xt ] with xt > F1t . By Lemma 6 agent 2 makes a
proposal that induces agent 1 to not take any portion and by Lemma 4 both
agents receive a morsel with positive Lebesgue measure. Then, it must be
that Y1t = [at , y t ] . Finally note that if y t > xt agent 2 is not playing a best
response since he would obtain a higher payoff by picking up the portion
[xt , bt ] . Hence P12 (Ωt ) = [at , y t ] with y t < xt .
Corollary 1 If there exists a subgame perfect equilibrium outcome of the
game, then it is efficient.
It follows that, if it exists, the SPNE outcome of the game is a single
cut partition that we denote by (P1 , P2 ) . Let S ∈ [0, 1] the point which
characterizes the SPNE outcome. By Corollary 1 P1 = [0, S] and P2 = (S, 1] .
By Lemma 4, S ∈ [F1t , F2t ] for all t = 1, 2, ..., T.
We now state a strategy profile (T1,T2) which is a subgame perfect equilibrium of the game.
Let R[at ,bt ] denote the single-cut point of the unique efficient REEF allocation of the cake [at , bt ]. As shown in the proof of Proposition 1, F1t <
R[at ,bt ] < F2t .
T1: In any t = 1, 2, ..., T,
Step 1: agent 1 proposes X1t = [at , xt ] with xt ≥ F2t such that u2 (X2t ) =
u2 (at , R[at ,bt ] );
Step 3:
1. if X1t ∩ Y1t = ∅ then agent 1 takes her preferred portion in Y t ;
2. if X2t ∩ Y2t = ∅ then agent 1 does not take anything;
3. if Xjt ∩ Yjt = ∅ for both j = 1, 2, and
a. Y1t = [at , y t ] ⊂ X1t = [at , xt ] , then agent 1 does not take anything if u1 (Y1t ) + u1 (y t , R[yt ,xt ] ) ≥ u1 (Y2t ), takes the portion Y2t if u1 (Y1t ) +
u1 (y t , R[yt ,xt ] ) < u1 (Y2t );
b. Y1t = [at , y t ] ⊃ X1t = [at , xt ] , then agent 1 takes her preferred
portion in Y t ;
c. Y1t = [y t , bt ] ⊂ X1t = [xt , bt ] , then agent 1 does not take anything
if u1 (Y1t ) + u1 (xt , R[xt ,yt ] ) ≥ u1 (Y2t ), takes the portion Y2t otherwise;
d. Y1t = [y t , bt ] ⊃ X1t = [xt , bt ] , then agent 1 takes her preferred
portion in Y t ;
24
T2: In any t = 1, 2, ..., T,
Step 2: agent 2
1. takes [xt , bt ] if xt ≤ F1t ;
2. proposes Y1t = [at , F1t ] if xt > F1t , X1t = [xt , bt ] and u2 (F1t , bt ) ≥
u2 (at , xt );
3. proposes Y1t = [at , y t ] such that u
1 (Y2t ) = u1 (at , R[yt ,xt ] ), if X1t =
[at , xt ] , xt > F1t and u2 (X1t ) ≤ u2 (X2t ) + u2 R[yt ,xt ] , xt ;
4. takes X1t = [at , xt ], otherwise.
The mechanism we propose results in games that are either with finite
horizon or continuous at infinity. Therefore, we can apply the one-stagedeviation principle¶ , i.e., a strategy profile, s, is subgame perfect if and only
if it satisfies the one-stage-condition that no player i can gain by deviating
from s in a single stage and conforming to s thereafter.
The following Lemma is the last ingredient we need before proving that
the strategy profile (T1,T2) is a SPNE of the game (Proof of this Lemma is
available for the authors’ website.)
Lemma 8 For any u ∈ U sc , let R[a,b] denote the single cut point of the
REEF allocation for the interval [a, b]. For any b̃ < b, R[a,b̃] < R[a,b] and for
any ã > a, R[ã,b] > R[a,b] .
We first note that if the strategy profile (T1,T2)
t is a SPNE
of the tgame,
then the equilibrium outcome is the allocation [a , R[at ,bt ] ), R[at ,bt ] , b . It
is easy to check that the strategy T1 is a best response to strategy T2 at
step 3 of any stage t < T. To check that it is also a best response at step 1
of any stage t note that if agent 1 deviates and proposes an allocation with
single-cut point xt ≤ F1t , then agent 2 takes the portion [xt , bt ] and therefore
agent 1 obtains a payoff equal to u1 (at , xt ) < u1 (at , R[at ,bt ] ). If she proposes
X1t = [xt , bt ] with xt > F1t , then agent 2 at step 2 proposes Y1t = [at , F1t ] and
agent 1 obtains a payoff equal to u1 (at , F1t ) < u1 (at , R[at ,bt ] ). Finally, if she
proposes X1t = [at , x̃t ] with x̃t < xt , by Lemma 8 she cannot obtain a higher
payoff, while if she offers x̃t > xt ≥ F2t then agent 2 will take the portion
[at , xt ] and therefore agent 1 obtains a payoff u1 (xt , bt ) < u1 (at , R[at ,bt ] ).
To check that the strategy T2 is a best response we consider the three
following cases.
¶
See Theorem 4.1 and Theorem 4.2 in Fudenberg and Tirole "Game Theory" page
109-110.
25
(i) X1t = [xt , bt ] and xt ≤ F1t . Agent 2 takes the portion [xt , bt ] . Note
that the allocation ([at , xt ] (xt , bt ]) is efficient and that agent 1’s utility level
is lower than 12 u1 (Ωt ). If agent 2 deviates, he can either take the portion
[at , xt ] ,which is obviously a less valuable portion than (xt , bt ] , or make a different proposal. But agent 1 at step 3 will choose an action which guarantees
her a utility equal or higher than 12 u1 (Ωt ).
(ii) X1t = [xt , bt ] and xt > F1t . Agent 2 proposes Y1t = [at , F1t ] if u2 (F1t , bt ) ≥
u2 (at , xt ), takes the portion [at , xt ] otherwise. If agent 2 proposes Y1t = [at , F1t ]
agent 1 takes the portion (at , F1t ) and agent 2 receives the portion (F1t , bt ].
If agent 2 deviates, then either he takes the portion [xt , bt ] ⊂ (F1t , bt ] or he
makes a different proposal. But then agent 1’s utility is strictly higher than
her half-cake utility level and therefore agent 2’s utility will be lower than
u2 (F1t , bt ]).
(iii) X1t = [at , xt ] . Agent 2 proposes Y1t = [at , y t ] such that u1 (Y2t ) =
u1 (at , R[yt ,xt ] ) if u2 (at , xt ) ≤ u2 (xt , bt ) + u2 R[yt ,xt ] , xt , takes the portion
[at , xt ] otherwise. Note that Y1t = [at , y t ] is the proposal, given strategy T1
and Lemma 8, that maximizes agent 2 payoff’s when he decides to continue
the game. Agent 2 decides to end the game if and only if the portion [at , xt ]
gives him a higher payoff than the utility of the final payoff induced by his
best proposal, and therefore strategy T2 is a best response.
Finally, to prove uniqueness, suppose that there exists a SPNE outcome
different than [0, R[a,b] ), [R[a,b] , 1]. By Corollary 1 all SPNE outcomes are efficient. Let vi be the SPNE payoff to player i in any subgame starting at
stage t beginning with player i s offer. Let x̂t and ŷ t be the proposal along
the equilibrium path generated by the strategy profile T1 and T2 at stage t.
C 2.3 Let [a, b] be the cake still to be divided at stage t. In any
subgame
v1 ≥ u1 (a, R[a,b] ) and v2 ≥ u2 (R[a,b] , b).
P
. Suppose first that v2 < u2 (R[a,b] , b). Let xt agent 1’s offer at
step 1 of stage t. If xt > x̂t , then by taking [a, xt ] agent 2 gets u2 (a, xt ) >
u2 (a, x̂t ) = u2 (R[a,b] , b) > v2 contradicting the assumption that v2 is a SPNE
payoff. Suppose that R[a,b] < xt < x̂t (if xt < R[a,b] agent 2 can take [xt , b]
obtaining more than v2 ). Then, there exists a proposal y t ≤ ŷ t such that
u1 (y t , b) < u1 (a, R[yt ,xt ] ), and
u2 (R[yt ,xt ] , b) > u2 (R[a,b] , b)
26
If agent 2 plays y t at stage t step 2 and strategy T2 in the ensuing stages,
then for agent 1 to continue is the unique best response and agent 2 gets
u2 (R[yt ,xt ] , b) > u2 (R[a,b] , b) > v2 contradicting the assumption that v2 is a
SPNE payoff. Suppose now that v1 < u1 (a, R[a,b] ). There exists xt such that
u1 (a, R[yt ,xt ] ) > v1 , and
u2 (R[yt ,xt ] , b) > u2 (a, xt ),
where y t is agent 2’s proposal according to T2 when xt has been proposed
by agent 1. If agent 1 plays xt at stage t step 1 and T1 at step 3 and in
the following stages, the unique best response for agent 2 is to propose y t ;
hence agent 1 gets at least a payoff equal to u1 (y t , b) = u1 (a, R[yt ,xt ] ) > v1
contradicting the assumption that v1 is a SPNE payoff.
Uniqueness follows from Proposition 1 and Claim 2.3.
Proof of Lemma 3 The K-truncated version is a finite game with complete information, so a SPNE exists. Note that the result of Lemma 4 still
holds for the truncated divide and choose mechanism. The following claims
can also be easily proved by using the same arguments used in previous Lemmata 5, 6 and 7, and applying well-know arguments on the divide and choose
rule (namely on agent 1’s proposal at the last stage K of the truncated game).
Let u ∈ U sc be the agents’ utility profile. Without loss of generality, assume
[0,1]
[0,1]
F1 ≤ F2 .
C 3.1 In all subgame perfect equilibria of the game the mechanism ends
only if at stage t < K < T ∗ (u), agent 2 chooses one portion of the allocation
proposed by agent 1 and both portions of the allocation are indifferent for
agent 2.
C 3.2 In any subgame perfect equilibrium path, the mechanism ends at
stage t < K < T ∗ (u) if and only if there exists a unique envy free allocation
of the cake Ωt .
C 3.3 In all subgame perfect equilibria X1t = [at , xt ] and Y1t = [at , y t ]
with xt > y t for all t such that t < K < T ∗ (u). At stage K step 1, agent
1 proposes the single cut allocation X K = ([y K−1 , F2K−1 ), [F2K−1 , xK−1 ]) and
at step 2 agent 2 takes the portion X2K .
It follows that the SPNE outcome of the truncated divide and choose
game is efficient and envy-free, and when K > 1 the utility of both agents
27
is higher than the utility level that the chooser achieves if agents play the
divide and choose mechanism (part (i) of Lemma 3). Moreover by C
3.2 the game ends in equilibrium at stage K. By C 3.3 in equilibrium
agent 1 at the last stage K proposes the allocation X K which makes agent
2 indifferent between the two portions and agent 2 takes the portion X2K . In
equilibrium,
at stage
K −1 step
3 agent 1 chooses
to continueif and only if
K−2 K−1
K−1
K−2
K−2
u1 [y
, F2 ) ≥ max u1 [y
,x
) , u1 [y
, y K−1 ] , that is the
utility which she obtains by continuing is not lower than the utility she gets
by taking her preferred portion of the allocation Y K−1 . It follows that in
any subgame perfect equilibrium, at stage K − 1 step 2 agent 2, conditioned
to the fact that he prefers to not take any portion of the allocation X K−1 ,
proposes a single cut allocation Y K−1 with the smallest single cut y K−1 so
long as agent 1 would weakly prefer continuing to taking [y K−1 , y K−2 ]. In
fact, the smaller is y K−1 , the larger is the overall portion agent 2 receives.
Similarly, in any subgame perfect equilibrium, at stage K − 1, step 1 agent
1 proposes a single cut allocation X K−1 with the largest single cut xK−1 so
long as agent 2 weakly prefers proposing an allocation to taking his preferred
portion of the allocation X K−1 .
It follows that for any K−truncated divide and choose game with 1 ≤
K ≤ T ∗ (u) the single cuts which characterize agents’ proposals along the
equilibrium path can be computed by means of the following system of 2K −1
0
equations (where x0 ≡
≡0) yK−1
K−1 K−1  1 and
K−1
u
[F
,
x
)
=
u

2
2
2

K−2 K−1 [y K−1, F2K−2 )


u1 [y
, F2 ) = u1 [y
,x
)



K−1
K−2
K−2
K−1
u2 [F2 , x
) = u2 [y
,x
)
2K − 1 equations
·
·
·




u1 [0, F2K−1 ) = u1 ([y 1 , 1))



u2 [F2K−1 , 1) = u2 ([0, x1 ))
Consider now a K − truncated divide and choose mechanism and a (K +
1) − truncated mechanism for any K < T ∗ (u). Let xj (K) denote agent 1’s
SPNE proposal at stage j ≤ K when the mechanism is truncated at stage
K and s(K) denote the single cut characterizing the single cut allocation,
SPNE outcome of the K − truncated divide and choose mechanism. With a
little abuse of notation let s(K, [a, b]) the single cut SPNE allocation of the
K − truncated game when the interval [a, b] is the amount to be divided.
First we show that for any interval [a, b] ⊆ [0, 1], s(2, [a, b]) ≤ s(1, [a, b])
and the equality holds only when F1 (a, b) = F2 (a, b). When k = 1, it’s the
28
standard divide and choose, s(1, [a, b]) = F2 (a, b). When k = 2, by Lemma 4
and the efficiency of SPNE, s(2, [a, b]) ≤ F2 (a, b). In the SPNE, x1 (2, [a, b])
is such that u2 (0, s(2, [a, b]) = u2 (x1 (2, [a, b]), 1). If s(2, [a, b]) = F2 (a, b), then
x1 (2, [a, b]) = F2 (a, b). If so, when F1 (a, b) = F2 (a, b), agent 2 can get strictly
more than (F2 (a, b), 1] by proposing y 1 (2, [a, b]) = F1 (a, b). Contradictory to
the fact that s(2, [a, b]) is the SPNE outcome.
Suppose that s(K + 1) ≥ s(K), then the equilibrium proposals x1 (K +
1), x1 (K) are such that u2 (x1 (K +1), 1) = u2 (0, s(K +1)) and u2 (x1 (K), 1) =
u2 (0, s(K)). Therefore, x1 (K + 1) ≤ x1 (K). Similarly, y 1 (K + 1) ≤ y 1 (K).
Similarly, y t (K + 1) ≤ y t (K) and xt (K + 1) ≤ xt (K) for all t < K. Hence,
F2 (y k−1 (K + 1), xk−1 (K + 1)) ≤ F2 (y k−1 (K), xk−1 (K)) = s(K). Let [a, b] =
[y k−1 (K + 1), xk−1 (K + 1)], we have shown that s(2, [a, b]) < s(1, [a, b]) (since
K + 1 < T ∗ (u), F1 (a, b) = F2 (a, b)). Therefore, s(K + 1) < s(1, [a, b]) =
F2 (y k−1 (K + 1), xk−1 (K + 1)) ≤ s(K). Contradictory to the assumption that
s(K + 1) ≥ s(K). Hence, s(K + 1) < s(K), i.e., agent 2’s utility is strictly
increasing in K so long as K ≤ T ∗ (u).
29
x1c
y1c
c
0
F1[ 0,1]
F2[ 0,1]
1
1
F1[ y c , xc ]
y
1
1
1
F2[ y c , xc ]
c
x1c
1
c
y1c '
xc1'
c’
0
1
c’
y1c '
[ y c '1 , x c '1 ]
F1
F2
[x1c ,yc1 ]
Figure 1: x1c , yc1 , F1
[ y c '1 , x c '1 ]
[x1c ,yc1 ]
,and F2
30
xc1'
are decreasing in c
y1c
x1c
c
0
F1[ 0,1]
F2[ 0,1]
1
1
1
F1[ y c , xc ]
y
1
F2[ y c , xc ]
c
x1c
1
c
y
0
1
F1[ 0 ,1]
_
_
F11
F21
c = F1[ 0,1]
y1F [ 0 ,1]
F
x1F [ 0 ,1]
1
F2[ 0,1]
1
F
−1
1
1
− 2
x1F [ 0 ,1]
2
2
0
1
F1[ 0,1]
c = F2[ 0,1]
1
[0,1]
Figure 2: F 1 = F1
31
[0,1]
and F 12 = F2
1
No changes have been made to the content of the
paper from the accepted version.
32