Strategic Divide and Choose1 Antonio Nicolò2

and Yan Yu3

April 11, 2007

1 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. 2 Department of Economics, University of Padua, via del Santo 33, 35123 Padova, Italy. Email: [email protected]; home page: http://www.decon.unipd.it/~nicolo. 3 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.

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 "end-of-state" 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 (“end-of-state”) 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 “end-of-state” 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 example1 . 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 1

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.

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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 over Ω. We say that an element of F is a portion and that a F -measurable subset of a portion is a morsel. 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.2 (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 a
continuous 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
b of the interval, that is 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 2

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.

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if interchanging the preferences means interchanging the assigned portions, that is for any (u1 , u
2 ) ∈ U, fi (u1 , u
2 ) = fj (u
1 , 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. The Iterated Divide and Choose Mechanism The mechanism consists in 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 he3 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

• take one of the two portions, X1t or X2t . Agent 1 gets the other portion and the game ends; 3

We refer to agent 1 as a female agent and to agent 2 as a male agent

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• 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 ] 1 if s ∈ ( 13 , 23 ] . vB (s) =  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 we illustrate agents’ equilibrium proposals (concessions). In this example, the outcome of the iterated divide and choose mechanism is the single cut 8

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 ˆ 1 = ([0, xˆ1 ], (ˆ 0.7083. If Anna proposes a single cut allocation X x1 , 1]) with xˆ1 > 0.7083, and hence concedes to Bob less in stage 1, Bob will take the portion [0, xˆ1 ] (if Bob chooses to propose Y 1 , in any subgame perfect Nash ˆ 1 and any possible Y 1 , Bob gets a utility level equilibrium contingent on X lower than the utility level obtained by consuming the portion (0.5, 1]) and Anna gets (ˆ x1 , 1] and is worse off than when proposing X 1 . If Anna proposes ˆ 1 = ([0, xˆ1 ], (ˆ a single cut allocation X x1 , 1]) with xˆ1 < 0.7083, and hence concedes to Bob more in stage 1, Bob will choose to continue, but the final outcome ([0, sˆ], (ˆ s1 , 1]) in the subgame perfect Nash equilibrium will have sˆ < 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 to cut the cake at 0.7083. The ordering of agents has no impact on equilibrium payoff in this procedure. 9

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 2 y = 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 in the formal definition below. Let (Pi1 , ..., PiT ) be a partition of agent i’s portion Pi in T morsels.

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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 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 11

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)4 . 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
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 4

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

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] ui a, Fi = ui Fi , b = 12 ui (a, b). We call Fi agent i’s indifference   [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].   [0,1] [0,1] [0,1] [0,1] Lemma 2 For any preference profile u ∈ U sc , if F1 ≤ F2 F1 ≥ F2   [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. Proof: We provide the intuition of the proof here. The formal proof is [0,1] [0,1] in the 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. [0,1] [0,1] Without 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 [0,1] of the cake. For any c ∈ [0, F1 ), ([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 equivalent to an efficient REEF for any c ∈ (F2[0,1] , 1]. Hence, if ([0, c), [c, 1]) [0,1] [0,1] is equivalent to an efficient REEF allocation, then c ∈ [F1 , F2 ]. [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 13

(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 relevant 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 previous 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 approches 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 significately at t increases. In our leading example, outcome of the 2-truncated version is very close to the REEF allocation.

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 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 16

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.

References [1] Alkan, A., G. Demange and D. Gale (1991), Fair Allocation of Indivisible Goods and Criteria of Justice, Econometrica, 59(4), 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 (3), 251-270. [3] Berliant, M., K. Dunz, and W. Thomson (1992), On the Fair Division of a Heterogeneous Commodity, Journal of Mathematical Economics, 21(3), 201-16. [4] Brams, S. and A. Taylor (1996), Fair Division: From cake-cutting to Dispute Resolution, Cambridge UK, Cambridge University Press. 17

[5] Brams, S. and P. Fishburn (2000), Fair Division of Indivisible Items between Two People with Identical Preferences: Envy-Freeness, ParetoOptimality and Equity, Social Choice and Welfare, 17(2), 247-267. [6] Brams, S., M. Jones and C. Klamler (2004), Perfect Cake-Cutting Procedures with Money", preprint. [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, Review of Economics Studies 71, 975-1000. [9] Crawford, V. (1977), A Game of Fair Division, Review of Economic Studies, 44, 235-47. [10] Crawford, V. and W. Heller (1979), Fair Division with Indivisible Commodities, Journal of Economic Theory, 21(1), 10-27. [11] Demko, S. and T. Hill (1988), Equitable Distribution of Indivisible Objects, Mathematical Social Sciences, 16(2), 145-158. [12] Edelman, P. and P. Fishburn (2001), Fair Division of Indivisible Items among People with Similar Preferences, Mathematical Social Sciences, 41(3), 327-347. [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), 1-17. [15] Maniquet, F. and Y. Sprumont (2000), On Resource Monotonicity in the Fair Division Problem, Economics Letters, 68(3), 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

[18] Thomson W. (1994a) Resource-Monotonic Solutions to the Problem of Fair Division When Preferences Are Single-Peaked, Social Choice and Welfare, 11(3), 205-223. [19] Thomson W. (1994b), Consistent Solutions to the Problem of Fair Division When Preferences Are Single-Peaked, Journal of Economic Theory, 63(2), 219-45. [20] Thomson W. (1996), Concepts of Implementation, Japanese Economic Review, 47(2), 133-143. [21] Thomson W. (2005), Divide-and-Permute, Games and Economic Behavior, 52, 186-200.

4

Appendix

Proposition 1 For any preference profile u ∈ U sc , there exists a unique efficient residual-equivalent envy-free allocation. [0,1] [0,1] Proof. 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 (it is also the unique, in terms of utility, envy-free allocation). Without loss of [0,1] [0,1] generality, assume F1 < F2 , then in any efficient single cut allocation, agent 1 gets the left part and agent 2 gets the right part of the cake. For any [0,1] c ∈ [0, F1 ), ([0, c), [c, 1]) is not equivalent to an efficient residual-equivalent envy-free allocation because agent 1 envies agent 2. Similarly, for any c ∈ [0,1] (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 REEF, [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] Claim 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] Proof of the Claim Because u ∈ U sc and F1 < F2 , by Lemma 2, 1 1 1 1 [y ,x ] [y ,x ] [y1 ,x1 ] F1 c c ≤ F2 c c . Since x1c and yc1 are decreasing and continuous in c, F1 c c 19

[y1 ,x1 ]

and 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 ]. [0,1] [y1 ,x1 ] [0,1] [0,1] When c = F1 , F2 c c achieves its maximum on [F1 , F2 ]. 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 [0,1] [0,1]
functions of c from [F1 , F2 ] 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 [0,1] [y1 ,x1 ] [y 1 ,x1 ] any c ∈ [F1 , c1 ), c < c1 ≤ F1 c c ≤ F2 c c ; hence, ([0, c), [c, 1]) is not [0,1] an efficient REEF 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 [y1 ,x1 ] [0,1] [y1 ,x1 ] is not a fixed point of F2 c c , such that F2 c c = c. Note that c = F2 [0,1] [0,1] [y1 ,x1 ] [y1 ,x1 ] therefore c1 < F2 . Similarly, for any c ∈ (c1 , F2 ], c > F2 c c ≥ F1 c c , [0,1] therefore, ([0, c), [c, 1]) is not an efficient REEF 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, 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, (yc1 , yc2 ] is equivalent to agent 1’s morsel in stage 2, P12 , and [y 2 ,x2 ] [y2 ,x2 ] [x2c , x1c ) is equivalent to agent 2’s morsel in stage 2, P22 . Let F1 c c (F2 c c )

20

[y 2 ,x2 ]

denote agent 1’s (agent 2’s) indifference point over [yc2 , x2c ], i.e., u1 ([yc2 , F1 c c ]) = [y2 ,x2 ] u1 ([F1 c c , x2c ]). [y2 ,x2 ] Similar to the proof of above claim, we can establish that F1 c c 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 [0,1] [0,1] (in terms of utility) efficient REEF allocation when F1 < F2 . 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 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 at this stage. [0,1] [0,1] From now on we suppose, without loss of generality, that F1 ≤ F2 . Lemma 4 Consider any subgame starting at step 1 of some stage t. Let Pti denote agent i’s portion in the subgame perfect equilibrium allocation of the

t t a ,b [ ] subgame Ωt . Then, ui (Pti ) ≥ ui Fi = 12 ui (Ωt ), for both i = 1, 2, and for all t = {1, 2, ..., T } . 21

Proof. We actually prove a stronger claim, that is for all t = {1,

2, ..., T} [at ,bt ] each agent has a strategy that guarantees her to obtain ui (Pti ) ≥ ui Fi (not only in equilibrium). Let X t be the allocation proposed by agent 1 at stage t and Ωt ⊂ [0, 1] the cake still to be assigned at this stage. Either t) t) or u2 (X2t ) > u2 (Ω . Hence, agent 2 by taking his preferred u2 (X1t ) ≥ u2 (Ω 2 2 t) u (Ω portion obtains u2 (Pt2 ) ≥ 2 2 . Agent 1 can also guarantee herself at least her half-cake-equivalent utility of stage t. Suppose that agent 1 announces [at ,bt ] xt = F1 and proposes for herself the portion [at , xt ]. Either agent 2 takes one of the two portions, and then the claim is proved, or he announces an t) allocation Y t and the proof is completed noticing that either u1 (Y1t ) > u1 (Ω 2 t) or u1 (Y2t ) > u1 (Ω . 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. Claim 1: Agent 1 never chooses to end the game at step 3. Suppose that t X1 = [at , xt ]. We already proved that 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 . In this case agent 1’s best response is to take the portion [at , y t ]. But then agent 2’s best response at step 2 cannot be to propose Y t , because taking [xt , bt ] he would obtain a higher payoff; (ii) y t < xt . We suppose that agent 1 takes one of the two portions and we show that this cannot occurs along the equilibrium path. If agent 1 takes the portion [y t , bt ], then to announce y t cannot be a best response for agent 2 since he could obtain a portion [at , xt ] at step 2 which contains the portion [a, y t ]. 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 in the ensuing stages (by Lemma 4). The proof of the case X1t = [xt , bt ] follows the same argument. Claim 2: Agent 2 chooses to end the game only if agent 1 partitions the cake in two portions which are indifferent for agent 2. Let xt denote the single 22

cut point proposed by agent 1. If agent 2 takes a portion which is strictly [at ,bt ] preferred by him to the other one, then it must be that either xt > F2 and t ,bt a [ ] he takes the portion [at , xt ] or xt < F2 and he takes the portion [xt , bt ] . The first case contradicts Lemma 4 since agent 1 receives less than her half[at ,bt ] [at ,bt ] cake-equivalent utility . Hence, it must be that F1 ≤ xt < F2 . If t t agent 2’s best response is to take the portion [x , b ] , then it must be the [at ,bt ] [at ,bt ] case that F1 = xt . Suppose not, and let F1 < xt . By proposing t t [a ,b ] Y1t = [at , y t ] with F1 < y t < xt agent 2 will obtain a higher payoff. In fact either agent 1 takes the portion [at , y t ] or she does not take anything. But then agent 2 either will receive the entire cake (if X1t = [xt , bt ]) or he will receive the morsel [xt , bt ] at stage t and some morsel with positive Lebesgue measure in the ensuing stages. It follows that if agent 2 takes his strictly [at ,bt ] [at ,bt ] preferred portion [xt , bt ] , then it must be that F1 = xt < F2 . We t ,bt a [ ] is not a best response now prove that to propose X1t = [at , xt ] = at , F1 ˜ t = [at , x˜t ] with for agent 1. By continuity there exists a different proposal X 1 [at ,bt ] [at ,bt ] t t t < x˜ < F2 such that either agent 2 picks up [˜ x , b ] (and therefore F1 t t t t agent 1 receives the portion [a , x˜ ] ⊃ [a , x ]), or he proposes a different [at ,bt ] single-cut at y t . If y t = F1 then agent 1 by accepting the proposal [at ,bt ] receives the morsel at , F1 at stage t and some morsels with positive

[at ,bt ] Lebesgue measure in the ensuing stages. If y t = F1 then there exists a [at ,bt ] portion that agent 1 can take whose value is strictly higher than at , F1 . 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. We have already proved that the mechanism ends [at ,bt ] at stage t only if xt = F2 . Suppose that agent 1 proposes X t with t t [a ,b ] [at ,bt ] [at ,bt ] X1t = at , F2 . We prove that if F1 < F2 , then there ex23

ists a proposal Y t which gives agent 2 a higher payoff than the half-cake utility level. Suppose, in fact, that agent 2 proposes Y1t = [at , y t ] with [at ,bt ] [at ,bt ] F1 ≤ y t < F2 . If agent 1 takes a portion, she will take the portion [at , y t ] . If she does not take any portion, agent 2 will receive the morsel [xt , bt ] at stage t and some morsel with positive Lebesgue measure

in the t ,bt a [ ] ensuing stages. Suppose that agent 1 proposes X1t = F2 , bt . Then

[at ,bt ] t t forcing agent 1 to take this portion agent 2 will announce Y1 = a , F1

[at ,bt ] at , F1 at stage 3 (otherwise agent 2 receives all the cake). But obvi t t

[a ,b ] t t ously to propose X1 = F2 , b is not a best response since, as we just

[at ,bt ] t t . proved, agent 1 obtains a higher payoff by announcing X1 = a , F2

Sufficiency. Suppose now that at some T ∈ Z+ there exists a unique envy [at ,bt ] [at ,bt ] free allocation, characterized by the point z = F1 = F2 . 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.

[at ,bt ] [at ,bt ] Proof. Consider any stage t < T. By Lemma 2 F1 < F2 for t ,bt a [ ] all t < T. Suppose X1t = [xt , bt ] . If xt ≤ F1 then agent 2 will pick the [at ,bt ] t portion [xt , bt ] , contradicting Lemma 6. If x > F then agent 2 can 1



t ,bt a [ ] [at ,bt ] t t t propose Y1 = a , F1 forcing agent 1 to take the portion a , F1

[at ,bt ] i t t t at step 3 (note that ∩i∈N P1 (Ω ) = ∅). But by proposing X1 = a , F2 agent 1 can obtain a payoff strictly higher than the half-cake utility level. [at ,bt ] Hence X1t = [at , xt ] with xt > F1 . 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

24

[xt , bt ] . Hence P12 (Ωt ) = [at , y t ] with y t < xt .5 Corollary 1 If there exists a subgame perfect equilibrium outcome of the game, then it is efficient. Since the SPNE outcome is efficient, and the game is a sequential game with perfect information, if there exists a subgame perfect equilibrium outcome of the game, then it is unique in terms of utility. Suppose, in fact, that there are at least two SPNEs which are not unique in terms of utility. Since the game is of perfect information, then any information set is a singleton. Since the game is also sequential, then there exists at least one player who is indifferent between the two SPNE outcomes, otherwise he is not playing a best response in at least one of his information sets. If one agent is indifferent between the two SPNE equilibria and they are not unique in terms of utility, then the other player strictly prefers one SPNE outcome to the other, and efficiency is violated in at least one case. 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] . Consider any subgame starting at stage t ∈ {1, ..., T } and let Ωt = [at , bt ] be [0,1] [0,1] the cake still to be assigned. Since we assumed F1 ≤ F2 , then by Lemma [at ,bt ] [at ,bt ] [at ,bt ] [at ,bt ] ≤ F2 for all t = 1, 2, ..., T. By Lemma 4, S ∈ [F1 , F2 ] for 2, F1 all t = 1, 2, ..., T. We now show that the following strategy profile (T1,T2) is a subgame perfect equilibrium: Let R[at ,bt ] denote the single-cut point of the unique efficient REEF allo[at ,bt ] cation of the cake [at , bt ]. As shown in the proof of Proposition 1, F1 < [at ,bt ] R[at ,bt ] < F2 . T1: In any t = 1, 2, ..., T, [at ,bt ] Step 1: agent 1 proposes X1t = [at , xt ] with xt ≥ F2 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 5

[at ,bt ]

It is straightforward to note that if F1 with xt < yt for all t = T.

25

[at ,bt ]

> F2

then X1t = [xt , bt ] and Y1t = [yt , bt ]

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 ; T2: In any t = 1, 2, ..., T, Step 2: agent 2 [at ,bt ]

1. takes [xt , bt ] if xt ≤ F1 ;  [at ,bt ] [at ,bt ] [at ,bt ] t if xt > F1 , X1t = [xt , bt ] and u2 (F1 ,b ) ≥ 2. proposes Y1t = at , F1 t t u2 (a , x ) 3. proposes Y1t = [at , y t ] such that u1 (Y2t ) = u1 (at , R[yt ,xt ] ), if X1t = [at , xt ] ,   [at ,bt ] xt > F1 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 principle6 , 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. 6

See Theorem 4.1 and Theorem 4.2 in Fudenberg and Tirole "Game Theory" page 109-110.

26

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]. It must be true that for any ˜b < b, R[a,˜b] < R[a,b] and for any a ˜ > a, R[˜a,b] > R[a,b] . Proof. By definition of the REEF allocation there exists a partition of Pi such that ui (Pit ) = ui (Ptj ) for all t = {1, 2, ..., T } and for both i = 1, 2. We can find a sequence of points (p11 , p21 , ..., pT1 ), where a < p11 < p21 < ... < pT1 −1 < pT1 = R[a,b] , such that u1 (P11 ) = u1 ([a, p11 ]), u1 (P12 ) = u1 (p11 , p21 ), ..., u1 P1T = u1 (pT1 −1 , R[a,b] ). We can also find a sequence of points (pT2 , pT2 −1 , ..., p12 ), where R[a,b] = pT2 < pT2 −1 < pT2 −2 ... < p12 <  b, such that  u2 (P21 ) = u2 ([p12 , b]), u2 (P22 ) = u2 ([p22 , p12 ]), ..., u2 (P2T ) = u2 ( R[a,b] , pT2 −1 , ). (Pi1 , ..., PiT )

Define (˜ pQ similarly for the REEF allocation for the ˜11 ) and p˜12 , ..., p˜Q 1 , ..., p 2 interval [a, ˜b]. Suppose that R[a,˜b] > R[a,b] for some ˜b < b. By definition of the REEF allocation, u2 (p12 , b) = u2 (a, R[a,b] ) and u2 (˜ p12 , ˜b) = u2 (a, R[a,˜b] ). Since R[a,˜b] > R[a,b] and ˜b < b, then p˜12 < p12 . Moreover, by definition of the REEF allocation, u1 (a, p11 ) = u1 (R[a,b] , b) and u1 (a, p˜11 ) = u1 (R[a,˜b] , ˜b); therefore p˜11 < p11 . Similarly, we can show that p˜t2 < pt2 and p˜t1 < pt1 for all t ≤ min{T, Q}. If T ≤ Q, p˜2Q−1 < pT2 −1 and p˜1Q−1 < pT1 −1 . By definition of REEF allocation, u2 (R[a,b] , pT2 −1 ) = u2 (pT1 −1 , R[a,b] ) and u2 (R[a,˜b] , p˜2Q−1 ) = u2 (˜ p1Q−1 , R[a,˜b] ). If R[a,˜b] > R[a,b] , these two equalities are not compatible. Similarly, we can find contradiction when T ≥ Q. The proof for the part “for any a ˜ > a, S[a,˜b] > S[a,b] ” follows the same argument.

Now we are ready to prove the existence and uniqueness of the SPNE allocation. We first note that if the strategy profile (T1,T2) is a SPNE of the  t   game, then the equilibrium outcome is the allocation [a , R[at ,bt ] ), R[at ,bt ] , bt . Now we prove that T1 is best response to strategy T2 in all subgames. Consider step 3 of any stage t < T. It is easy to check that strategy T1 is best response in cases (1), (2) . Suppose that Xjt ∩ Yjt = ∅ for both j = 1, 2. We consider two cases: (3a) If agent 1 decides to continue the game, then she does not take any portion and, by one-stage deviation principle, obtains a final payoff equal to u1 (at , y t ) + u1 (y t , R[yt ,xt ] ). If she decides to stop the game, then her best response is clearly to take the portion [y t , bt ] . According to strategy T1 she compares these two payoffs and therefore it is straightforward that she is playing a best response. A similar argument holds for the (3c) case. 27

(3b) In this case by deviating and not taking anything agent 1 receives the current morsel [at , xt ] and some portion of the remaining cake [xt , y t ]. Since by Lemma 4 and by the one-stage deviation principle she does not receive the entire cake [xt , y t ] in the ensuing stages, then by deviating she receives a final portion that it is strictly contained in the portion [at , y t ]. Therefore there is no deviation better than the response prescribed in strategy T1. The same argument holds for case (3d) Consider now step 1 of any stage t. If agent 1 deviates and proposes t t any allocation with single-cut point xt ≤ F1[a ,b ] , then agent 2 takes the portion [xt , bt ] and therefore agent 1 obtains a payoff equal to u1 (at , xt ) < [at ,bt ] t t t t u1 (at , R[at ,bt ] ). If she proposes , then agent  X1 = [x , b ] with x > F1 t

t

2 at step 2 proposes Y1t = at , F1[a ,b ] and agent 1 obtains a payoff equal t

t

to u1 (at , F1[a ,b ] ) < u1 (at , R[at ,bt ] ). Finally, if she proposes X1t = [at , x˜t ] with x˜t < xt , by Lemma 8 she cannot obtain a higher payoff, while if she offers t t x˜t > xt ≥ F2[a ,b ] then agent 2 will take the portion [at , xt ] and therefore t t agent 1 obtains a payoff u1 (xt , bt ) < u1 (at , F1[a ,b ] ) < u1 (at , R[at ,bt ] ).

We prove now that the strategy T2 is a best response to strategy T1 in all subgames. We consider the following three cases: [at ,bt ] (i) X1t = [xt , bt ] and xt ≤ F1 . T2 strategy prescribes that 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 her half-cake equivalent utility. 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. In this case agent 1 according to T1 at step 3 chooses an action which is a best response in all subgames. By Lemma 4 agent 1’s utility is equal or higher than the half-cake equivalent utility. Therefore by proposing any allocation agent 2 lowers his payoff. t t (ii) X1t = [xt , bt ] and xt > F1[a ,b ] . According to strategy T2 agent 2 [at ,bt ] [at ,bt ] t proposes Y1t = at , F1 if u2 (F1 , b ) ≥ u2 (at , xt ), takes the portion  [at ,bt ] t t t t [a , x ] otherwise. First, note that in case agent 2 proposes Y1 = a , F1   t t according to strategy T1 agent 1 takes the portion at , F1[a ,b ] and therefore [at ,bt ]

agent 2 receives the portion (F1 , bt ]. If agent 2 deviates from strategy T2, [at ,bt ] t then he either takes the portion [xt , bt ] ⊂ (F1 , b ] or makes a different proposal. But then agent 1’s utility is strictly higher than her half-cake 28

[at ,bt ]

utility level and therefore agent 2’s utility will be lower than u2 (F1 , bt ]). t (iii) X1t = [at , xt ] . According to strategy T2 then agent 2 proposes  Y1 = t  t t t t t t t t [a , y ] such that u1 (Y2 ) = u1 (a , R[yt ,xt ] ) if u2 (a , x ) ≤ u2 (x , b )+u2 R[yt ,xt ] , x , 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 ] provides to him a higher payoff than the utility of the final payoff induced by his best proposal, and therefore strategy T2 is a best response. Proof of Lemma 3 The K-truncated version is a finite game with complete information, so 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. C 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 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 In all subgame perfect equilibria X1t = [at , xt ] and Y1t = [at , y t ] with xt > y t for all t < K < T ∗ (u). At stage K step 1, agent 1 proposes the single [yK−1 ,xK−1 ] [yK−1 ,xK−1 ] cut allocation X K = ([y K−1 , F2 ), [F2 , 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 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 2 the game ends in equilibrium at stage K. By C 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 continue if and only if       [yK−1 ,xK−1 ] K−2 u1 [y , F2 ) ≥ max u1 [y K−1 , xK−2 ) , u1 [y K−2 , y K−1 ] , that 29

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, smaller is y K−1 , 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 ≡ 1 and ≡0)  yK−1    [y ,xK−1 ] [yK−1 ,xK−1 ] K−1 K−1  u [F [y , x ) = u , F )  2 2 2 2       K−1 K−2  K−1 ,xK−1 ]  [y K−2  , F2 ) = u1 [y ,x ) u1 [y         K−1 K−1  [y ,x ] u2 [F2 , xK−2 ) = u2 [y K−2 , xK−1 ) 2K−1 equations  ···      K−1 ,xK−1 ]  [y  u1 [0, F2 ) = u1 ([y 1 , 1))        [yK−1 ,xK−1 ]  u2 [F2 , 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. Finally, let Ωt (K) denote the cake to be divided at stage t when agents play the K − truncated divide and choose mechanism. First, suppose that x1 (K) < x1 (K + 1). In equilibrium u2 (s(Q), 1) = u2 (0, x1 (Q)) for all Q ≤ T ∗ (u). It follows that u2 (s(K + 1), 1) > u2 (s(K), 1). Second, suppose that x1 (K) = x1 (K + 1) (and therefore u2 (s(K + 1), 1) = u2 (s(K + 1), 1)). By efficiency of the SPNE outcome u1 (0, s(K + 1)) = u1 (0, s(K))). Hence, for agent 1 proposing the single cut allocation X t (K) = X t (K + 1) for all t = 2, ..., K, and for agent 2 proposing Y t (K) = Y t (K + 1) for all t = 2, ..., K − 1, and taking the portion X2K in stage K should be 30

(part of) a SPNE strategy profile. However, taking the portion X2K is not a best response foragent 2 at stage K step 2, since he obtains  a larger payoff [y K−1 ,xK−1 ] [yK−1 ,xK−1 ] K K−1 K−1 proposing Y = [y , F1 ), [F1 ,x ) .

Suppose, hence, that x1 (K) > x1 (K + 1), and therefore u2 (s(K + 1), 1) < u2 (s(K), 1). Consider agent 2 best response. By efficiency of the SPNE outcome and the fact that u1 (0, s(K + 1)) = u1 (y 1 (K + 1), 1) it follows that y 1 (K + 1) < y 1 (K). Hence, Ω2 (K + 1) ≡ [y 1 (K + 1), x1 (K + 1)] . Consider now a K − truncated divide and choose game played over this cake Ω2 (K + 1). Let s(K, Ω) denote the single cut characterizing the single cut allocation SPNE outcome of the K −truncated divide and choose game when agents, endowed with utility profile u ∈ U sc , have to divide cake Ω. With a small abuse in notation let s(K) ≡ s(K, [0, 1]). The following C (whose proof is straightforward) is the last ingredient we need. C 4: Let Ω = [a, b] ⊂ [0, 1] and Ω
= [a
, b
] ⊂ [0, 1] with a ≤ a
and b ≤ b
. Then, s(K, Ω) ≤ s(K, Ω
). The portion that agent 2 receives as the SPNE outcome of the K + 1 truncated divide and choose game over the cake [0,1] cannot be smaller than the portion s(K, Ω2 (K + 1)) ∪ [x1 (K + 1), 1] , contradicting the fact that u2 (s(K + 1), 1) < u2 (s(K), 1).


31

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

[ y c '1 , x c '1 ]

Figure 1: 32

xc1'

y1c

x1c c

0

F1[ 0,1]

F2[ 0,1]

1

1

1

F1[ y c , xc ]

y

c

x1c

y

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

F2[ y c , xc ]

1 c

0

1

F1[ 0,1]

c = F2[ 0,1]

Figure 2: 33

1