No. 2004–119 A SILENT BATTLE OVER A CAKE By J. González-Díaz, P.E.M. Borm, H.W. Norde November 2004 ISSN 0924-7815 A Silent Battle Over a Cake Julio Gonz´ alez-D´ ıaz∗ , Peter Borm† , Henk Norde† November 22, 2004 Abstract: The division of a cake by n players is modeled as a game of timing. We show that such games admit a unique Nash equilibrium. Key Words: Game of timing, Nash equilibrium. JEL Classiﬁcation: C72, D61, D63. 1 Introduction We consider the situation that n players have to divide a cake of size S. At time 0 player i has the initial right to receive the amount αi, where it is assumed that i∈N αi < S. If player i claims his part at time t > 0 then he receives the discounted part δtαi of the cake, unless he is the last claimant in which case he receives the discounted remaining part of the cake δt(S − j =i αj). Cake sharing games are examples of games of timing, ﬁrst analyzed by S.Karlin (1959) in the zero sum context. Other examples of games of timing can be found in Baston and Garnaev (2000) and Laraki et al. (2003). Hamers (1993) showed that 2-person cake sharing games always admit a unique Nash equilibrium. In this paper we consider cake sharing games that are slightly diﬀerent from the games introduced in Hamers (1993). We will ﬁrst provide an al- ternative, but more direct, existence and uniqueness result for 2-player cake sharing games and we will generalize this result to cake sharing games with more players. This paper is organized as follows. In Section 2 we will introduce cake sharing games. Sections 3 and 4 deal with 2-player cake sharing games and more player cake sharing games respectively. ∗Faculty of Mathematics, Department of Statistics and OR, University of Santiago de Compostela, 15771 Santiago de Compostela, Spain. Supported by the Spanish Ministry for Science and Technology and FEDER through project BEC2002-04102-C02-02, and from the Xunta de Galicia under project PGIDT03PXIC20701PN. †CentER and Department of Econometrics and OR, Tilburg University, P.O. Box 90153, 5000 LE Tilburg, The Netherlands. 12 The Model In this section we formally introduce the cake sharing games. Let N = {1,...,n} be a set of players with n ≥ 2, let S > 0, let α = (α1,...,αn) ∈ RN + be such that α1 + + αn < S and let δ ∈ (0,1). Throughout this paper we assume that 0 < α1 < α2 < < αn. The number S is called the size of the cake, the vector α the initial right vector and δ the discount factor. The cake sharing game with pure strategies, associated with S, α and δ, is the triple Γ pure S,α,δ =< N,{Ai}i∈N,{πi}i∈N >, where • Ai = [0,∞) is the set of pure strategies of player i ∈ N, • πi is the payoﬀ function of player i ∈ N, deﬁned by: πi(t1,...,tn) =    (S − j =i αj)δti if ti > max j =i tj αiδti otherwise. So, if there is a unique last claimant this player will receive the discounted value of the cake that remains after that the other players have taken their initial rights. If there is not a unique last claimant all players receive the discounted value of their initial rights. Note that the payoﬀ functions deﬁned above diﬀer slightly from the payoﬀ functions introduced in Hamers (1993), where, in case there is not a unique last claimant, the discounted value of the remaining cake is shared equally between the last claimants. One easily veriﬁes that Γ pure S,α,δ has no Nash equilibria: if there is a unique last claimant this player can improve his payoﬀ by claiming a little bit earlier (remaining the last claimant of course), if there is no unique last claimant then one of the last claimants can improve his payoﬀ by claiming a little bit later (becoming the unique last claimant in this way). So, for an appropriate analysis of cake sharing games we need to consider mixed strategies. Formally, a mixed strategy is a function G : [0,∞) → [0,1] satisfying: • G(0) = 0, • G is a nondecreasing function, • G is left-continuous, • limx→∞ G(x) = 1. For a mixed strategy G we can always ﬁnd a probability measure P on [0,∞)1 such that G(x) = P [0,x) (1) 1See V.K.Rohatgi (1976) for more details. 2for all x ∈ [0,∞). On the other hand, every probability measure P on [0,∞) deﬁnes by formula (1) a mixed strategy G. So the set of mixed strategies coincides with the set of probability measures on [0,∞)2. Let G denote the set of all mixed strategies. We introduce now some other notations related to mixed strategy G: • for x ∈ [0,∞) we denote limy↓x G(y), the probability of choosing an element in the closed interval [0,x], by G(x+). • if for an x > 0 it holds that for every a,b ∈ [0,∞) with a < x < b we have G(b) > G(a+), i.e., the probability of choosing an element in (a,b) is positive, then x is an element of the support of G. If for every b > 0 it holds that G(b) > 0, i.e., the probability of choosing an element in [0,b) is positive, then 0 is an element of the support of G. The support of the distribution function G will be denoted by S(G). One easily veriﬁes that S(G) is a closed set. • the set of jumps (discontinuities) of G is J(G) := {x ∈ [0,∞) | G(x+) > G(x)}, i.e., the set of pure strategies which are chosen with positive probability. If player i chooses pure strategy t and all other players choose mixed strate- gies {Gj}j =i then the expected payoﬀ for player i is πi(G1,...,Gi−1,t,Gi+1,...,Gn) = j =i Gj(t) δt (S − j =i αj) + (1 − j =i Gj(t)) δt αi = δt (αi + (S − j∈N αj) j =i Gj(t)). If player i also chooses a mixed strategy Gi, whereas all other players stick to mixed strategies {Gj}j =i then the expected payoﬀ for player i can be computed by use of the Lebesgue-Stieltjes integral: πi(G1,...,Gn) = πi(G1,...,Gi−1,t,Gi+1,...,Gn)dGi(t). (2) Note that, with a slight abuse of notation, the functions πi do not only de- note payoﬀs to players when pure strategies are played, but also when mixed strategies are used. The cake sharing game, associated with S, α and δ, is the triple ΓS,α,δ =< N,{Xi}i∈N,{πi}i∈N >, where 2An alternative way of deﬁning mixed strategies G is as a nondecreasing, right- continuous function from [0,∞) to [0,1] with limx→∞ G(x) = 1. For such a function we can always ﬁnd a probability measure P on [0,∞) such that G(x) = P [0,x] for all x ∈ [0,∞), i.e., G is the (cumulative) distribution function corresponding to P. Although this equivalent approach seems more natural, this would lead to technical problems when computing Lebesgue-Stieltjes integrals later on. 3• Xi = G is the set of mixed strategies of player i ∈ N, • πi, deﬁned by (2), is the (expected) payoﬀ function of player i ∈ N. Given a strategy proﬁle G = (G1,G2,...,Gn) ∈ Gn, let πG i (t) denote the payoﬀ πi(G1,...,Gi−1,t,Gi+1,...,Gn). So πG i (t) is the expected payoﬀ of player i when he chooses the pure strategy t and all the other players act in accordance with G. 3 Two Players In this section we provide an alternative proof of the result of Hamers (1993) for 2-player cake sharing games. Our incentives for doing this job are three- fold. First of all we want to recall that our model is slightly diﬀerent from the model of Hamers (1993), so a new proof is required. Secondly, our proof is more direct than Hamers’ proof. Finally, our proof will form the basis for the results in Section 4 for cake sharing games with three or more players. First we derive a number of properties for Nash equilibria of n-player cake sharing games. The following lemma shows that in a Nash equilibrium players do not put positive probability on a pure strategy t > 0. Lemma 1. Let ΓS,α,δ be an n-player cake sharing game and let the proﬁle G = (Gi)i∈N ∈ GN be a Nash equilibrium of ΓS,α,δ. Then J(Gi)∩(0,∞) = ∅ for every i ∈ N. Proof. Let i ∈ N. We will show that J(Gi) ∩ (0,∞) = ∅. Without loss of generality we may assume that i = 1. Suppose that u ∈ J(G1) ∩ (0,∞). If Gi(u+) = 0 for some i = 1 then πG 1 (t) = δtα1 whenever t ∈ [0,u]. Since the function πG 1 ( ) is strictly de- creasing on [0,u], player 1 would be better oﬀ moving the probability in u to 0. So Gi(u+) > 0 for all i ∈ N. Now consider the functions πG i (t) = δt(αi + (S − j∈N αj) j =i Gj(t)) for i ∈ N\{1}. Since G1 is discontinuous at u, i.e., G1(u+) > G1(u), there exist u1 < u, u2 > u and ε > 0 such that for every i = 1 πG i (u2) − πG i (t) ≥ ε for every t ∈ [u1,u]. If player i ∈ N\{1} puts positive probability on [u1,u], i.e., if Gi(u+) > Gi(u1), then he can increase his payoﬀ by at least ε(Gi(u+) − Gi(u1)) by moving all this probability to u2. So for all i ∈ N\{1} we have Gi(u+) = Gi(u1) and hence Gi(t) = Gi(u) for every t ∈ [u1,u]. Therefore the function πG 1 (t) = δt(α1 + (S − j∈N αj) j =1 Gj(t)) 4is strictly decreasing on [u1,u]. So player 1 can improve his payoﬀ by moving some probability from u to u1. Lemma 1 implies that in a Nash equilibrium G players use mixed strate- gies which are continuous on (0,∞). Therefore we may write Gi(t+) = Gi(t) for every i ∈ N and t > 0. Moreover, the functions πG i ( ) are continuous on (0,∞). Lemma 2. Let ΓS,α,δ be an n-player cake sharing game and let the proﬁle G = (Gi)i∈N ∈ GN be a Nash equilibrium of ΓS,α,δ. Let i ∈ N and t ∈ S(Gi). There exists j ∈ N\{i} such that t ∈ S(Gj). Proof. Suppose that t / ∈ ∪j =iS(Gj). We will distinguish between two cases: Case (i): t > 0. There exist t1,t2 > 0 with t1 < t < t2 such that Gj(t2) = Gj(t1) for all j = i.3 Hence Gj(u) = Gj(t2) for every u ∈ [t1,t2] and every j = i. Therefore the function πG i (u) = δu(αi + (S − j∈N αj) j =i Gj(u)) is strictly decreasing on [t1,t2]. Since t ∈ S(Gi) we have Gi(t2) > Gi(t+ 1 ), i.e., player i puts positive probability on (t1,t2). Now player i can strictly improve his payoﬀ by moving all this probability to t1. Case (ii): t = 0. Let b > 0 be the smallest element in ∪j =iS(Gj) (recall that all the S(Gj) are closed). Clearly Gj(b) = 0 for all j = i. Again, if Gi(b) > Gi(0+), i.e., if player i puts positive probability on (0,b), then similar arguments as in Case (i) can be used to show that player i can strictly improve his payoﬀ by moving this probability to 0. So Gi(b) = Gi(0+) and hence, since 0 ∈ S(Gi), we have Gi(0+) > 0. Moreover Gi(t) = Gi(b) for every t ∈ (0,b] (this is relevant only for the case n = 2), so πG j (t) = δt(αj + (S − k∈N αk) k =j Gk(t)) is strictly decreasing on (0,b] for every j ∈ N\{i}. Let a ∈ (0,b) and let j ∈ N\{i} be a player such that b ∈ S(Gj). Clearly ε := πG j (a) − πG j (b) > 0. Since the function πG j ( ) is continuous on (0,∞) we have, for δ > 0 suﬃciently small, πG j (a) − πG j (t) > 1 2 ε for every t ∈ [b,b + δ]. Since b ∈ S(Gj) we have Gj(b+δ) > 0 = Gj(b). So, player j can improve his payoﬀ by moving the probability he assigns to [b,b+δ) to a. Contradiction. 3For every j ∈ N\{i} there exist t j 1,t j 2 > 0 with t j 1 < t < t j 2 such that Gj(t j 2) = Gj(t j 1). We therefore take t1 = maxj∈N\{i} t j 1 and t2 = minj∈N\{i} t j 2. 5The following lemma shows that if some pure strategy t does not belong to the support of any of the equilibrium strategies, then no pure strategy t′ > t belongs to the support of any of the equilibrium strategies either. Lemma 3. Let G = (Gi)i∈N be a Nash equilibrium of the n-player cake sharing game ΓS,α,δ. Suppose t ∈ [0,∞) is such that t / ∈ S(Gj) for every j ∈ N. Then (t,∞) ∩ S(Gj) = ∅ for every j ∈ N. Proof. Deﬁne K = ∪j∈NS(Gj). Clearly K is closed and t / ∈ K. We have to show that K∩(t,∞) = ∅. Suppose that K∩(t,∞) = ∅. Deﬁne t∗ = min{u ∈ K|u > t}. Let j∗ ∈ N be such that t∗ ∈ S(Gj∗). Because [t,t∗)∩S(Gj) = ∅ for every j ∈ N we have Gj(t) = Gj(t∗) for every j ∈ N. So the functions Gj are constant on [t,t∗]. Since πG j∗(u) = δu(αj∗ + (S − j∈N αj) j =j∗ Gj(u)) for every u ∈ [0,∞), the function πG j∗( ) is strictly decreasing on [t,t∗]. By continuity of πG j∗( ) at t∗ we infer that πG j∗(t) > πG j∗(u) for every u ∈ [t∗,t∗+ε], with ε > 0 suﬃciently small. Hence Gj∗ is constant on [t∗,t∗ + ε] as well, contradicting the fact that t∗ ∈ S(Gj∗). Now we provide speciﬁc results for 2-player cake sharing games. The fol- lowing lemma shows that in a Nash equilibrium players use mixed strategies of which the supports coincide. Lemma 4. Let ΓS,α,δ be a 2-player cake sharing game and let (G1,G2) ∈ G × G be a Nash equilibrium of ΓS,α,δ. Then S(G1) = S(G2). Proof. This result is just a consequence of Lemma 2. In the following lemma we will show that the supports of the strategies in a Nash equilibrium are compact intervals. Lemma 5. Let ΓS,α,δ be a 2-player cake sharing game and let G = (G1,G2) ∈ G × G be a Nash equilibrium of ΓS,α,δ. Let k := logδ α2 S−α1. Then S(G1) = S(G2) = [0,k]. Proof. First we will show that S(G1) = S(G2) ⊆ [0,k]. For every t ∈ (k,∞) we have πG 2 (t) = δt(α2 + (S − α1 − α2)G1(t)) ≤ δt(α2 + (S − α1 − α2)) = δt(S − α1) < δk(S − α1) = α2 = πG 2 (0). 6If G2(k) = G2(k+) < 1, i.e., if player 2 puts positive probability on (k,∞), he can improve his payoﬀ strictly by moving all this probability to 0. Therefore G2(k) = 1 and hence S(G1) = S(G2) ⊆ [0,k]. Let k∗ be the largest element in the closed set S(G1). Clearly k∗ ≤ k. If k∗ = 0 then (G1,G2) would be an equilibrium in pure strategies, a contradiction. So k∗ > 0. According to Lemma 3 we have S(G1) = S(G2) = [0,k∗]. The only thing which remains to be shown is that k∗ = k. Suppose that k∗ < k. For τ ∈ (0,k − k∗) we have πG 1 (k∗ + τ) = δk∗+τ(α1 + (S − α1 − α2)G2(k∗ + τ)) = δk∗+τ(α1 + (S − α1 − α2)) = δk∗+τ(S − α2) > δk(S − α2) = α2(S − α2) S − α1 ≥ α1 = πG 1 (0), where at the weak inequality we used that α2(S − α2) ≥ α1(S − α1). So, if G1(0+) > 0, i.e., if player 1 plays pure strategy 0 with positive probability, then he can improve his payoﬀ by moving some probability from 0 to pure strategy k∗ + τ. Therefore G1(0+) = 0. For some t ∈ (k∗,k) we have πG 2 (t) = δt(α2 + (S − α1 − α2)G1(t)) = δt(α2 + (S − α1 − α2)) = δt(S − α1) > δk(S − α1) = α2 = πG 2 (0). Since 0 ∈ S(G1) and πG 2 ( ) is continuous at 0 (due to G1(0+) = 0) player 2 can improve his payoﬀ strictly by moving some probability from the neigh- borhood of 0 to t. Contradiction, so k∗ = k. Now we are able to prove the main theorem of this section. Theorem 1. Let ΓS,α,δ be a 2-player cake sharing game and k := logδ α2 S−α1. Deﬁne G∗ = (G∗ 1,G∗ 2) ∈ G × G by G∗ 1(t) =    α2 − α2δt δt(S − α1 − α2) if 0 ≤ t ≤ k 1 if t > k G∗ 2(t) =        0 if t = 0 α2(S − α2) − α1(S − α1)δt δt(S − α1)(S − α1 − α2) if 0 < t ≤ k 1 if t > k 7Then G∗ is the unique Nash equilibrium of ΓS,α,δ with equilibrium payoﬀs π1(G∗ 1,G∗ 2) = α2(S − α2) S − α1 π2(G∗ 1,G∗ 2) = α2. Proof. One easily veriﬁes that πG∗ 1 (t) =        α1 if t = 0 α2(S − α2) S − α1 if 0 < t ≤ k δt(S − α2) if t > k and πG∗ 2 (t) = α2 if 0 ≤ t ≤ k δt(S − α1) if t > k. Therefore π1(G∗ 1,G∗ 2) = α2(S − α2) S − α1 and π2(G∗ 1,G∗ 2) = α2. Since π1(t,G∗ 2) ≤ α2(S − α2) S − α1 and π2(G∗ 1,t) ≤ α2 for every t ∈ [0,∞) we infer that G∗ is a Nash equilibrium of ΓS,α,δ. In order to show that there are no other Nash equilibria let (G1,G2) be a Nash equilibrium of ΓS,α,δ. According to Lemma 1 the strategies G1 and G2 are continuous on (0,∞). In the same way as in the proof of Lemma 5 we can show that G1(0+) = 0. So, the function πG 1 ( ) is continuous on (0,∞) and the function πG 2 ( ) is continuous on [0,∞). By Lemma 5 we have S(G1) = S(G2) = [0,k]. So, there exist constants c and d such that c = πG 1 (t) = δt(α1 + (S − α1 − α2)G2(t)) for every t ∈ (0,k] d = πG 2 (t) = δt(α2 + (S − α1 − α2)G1(t)) for every t ∈ [0,k]. Since G1(0) = 0 we have d = πG 2 (0) = α2, from which we derive that G1(t) = α2 − α2δt δt(S − α1 − α2) = G∗ 1(t) for every t ∈ [0,k]. Clearly, G1(t) = 1 = G∗ 1(t) for every t > k, so G1 = G∗ 1. Moreover, since G2(k) = 1 we have c = πG 1 (k) = δk(S − α2) = α2(S−α2) S−α1 , from which we derive that G2(t) = α2(S − α2) − α1(S − α1)δt δt(S − α1)(S − α1 − α2) = G∗ 2(t) for every t ∈ (0,k]. Clearly G2(0) = 0 = G∗ 2(0) and G2(t) = 1 = G∗ 2(t) for every t > k, so G2 = G∗ 2. This ﬁnishes the proof. 84 More players In this section we consider cake sharing games with more than two players. Again we will show that such games admit a unique Nash equilibrium. First we show that mixed strategies in a Nash equilibrium have a bounded support. Lemma 6. Let ΓS,α,δ be an n-player cake sharing game with n ≥ 3 and let G = (Gi)i∈N ∈ GN be a Nash equilibrium of ΓS,α,δ. For every i ∈ N let ki := logδ αi 1 − j =i αj . Then k1 > k2 > > kn and S(Gi) ⊂ [0,ki] for every i ∈ N. Moreover S(G1) ⊂ [0,k2]. Proof. Let i,j ∈ N be such that i > j. Deﬁne γ := l =i,j αl. Then αi(S − γ − αi) − αj(S − γ − αj) = αi(S − γ − αi) − αiαj + αiαj − αj(S − γ − αj) = (αi − αj)(S − l∈N αl) > 0. So δki δkj = αi S − γ − αj αj S − γ − αi = αi(S − γ − αi) αj(S − γ − αj) > 1 (3) and hence ki < kj. Now, for every i ∈ N and t ∈ (ki,∞) πG i (t) = δt(αi + (S − j∈N αj) j =i Gj(t)) ≤ δt(αi + (S − j∈N αj)) = δt(S − j =i αj) < δki(S − j =i αj) = αi = πG i (0). Repeating the reasoning in Lemma 5, if Gi(ki) = Gi(k+ i ) < 1, then player i can strictly improve his payoﬀ by moving all the probability in (ki,∞) to 0. Therefore Gi(ki) = 1 for all i ∈ N. For any j ∈ N\{1} we have k2 ≥ kj and hence Gj(k2) ≥ Gj(kj) = 1. So Gj(k2) = 1 and hence S(Gj) ⊂ [0,k2]. Now, because of Lemma 2 we conclude that S(G1) ⊂ [0,k2] as well. In the following lemma we show that pure strategy 0 belongs to the support of every equilibrium strategy. Moreover, players 2,...,n play this strategy with positive probability. 9Lemma 7. Let ΓS,α,δ be an n-player cake sharing game and let G = (Gi)i∈N ∈ GN be a Nash equilibrium of ΓS,α,δ. Then 0 ∈ S(Gj) for every j ∈ N. More- over Gj(0+) > 0 for every j ∈ N\{1}. Proof. Suppose 0 / ∈ S(Gi) for some i ∈ N. Let s > 0 be the smallest element in the closed set S(Gi). Then [0,s) ∩ S(Gi) = ∅ and hence Gi(t) = 0 for every t ∈ [0,s]. Consequently the function πG j (t) = δt(αj + (S − k∈N αk) k =j Gk(t)) = αjδt is strictly decreasing on [0,s] for every j ∈ N\{i}. Hence (0,s)∩S(Gj) = ∅ for every j ∈ N\{i}. Choose s∗ ∈ (0,s). Then s∗ / ∈ S(Gj) for every j ∈ N. According to Lemma 3 we have (s∗,∞) ∩ S(Gi) = ∅, a contradiction with s ∈ S(Gi). So, 0 ∈ S(Gj) for every j ∈ N. Now suppose i ∈ N\{1} is such that Gi(0+) = 0. This implies that the function πG 1 (t) = δt(α1 + (S − j∈N αj) j =1 Gj(t)) is continuous at 0. Let k2 := logδ α2 S − j =2 αj . According to Lemma 6 we have Gj(k2) = 1 for every j ∈ N. Hence πG 1 (k2) = δk2(S − j =1 αj) = α2 S − j =2 αj (S − j =1 αj) > α1 = πG 1 (0). Due to continuity of πG 1 ( ) at 0 we have πG 1 (k2) > πG 1 (t) for every t ∈ [0,ε] with ε > 0 suﬃciently small. So, [0,ε) ∩ S(G1) = ∅, contradicting the fact that 0 ∈ S(G1). The following lemma provides the equilibrium payoﬀs in a Nash equilib- rium. Lemma 8. Let G = (Gi)i∈N be a Nash equilibrium of the n-player cake sharing game ΓS,α,δ and let η = (ηi)i∈N be the corresponding vector of equi- librium payoﬀs. Then η1 = α2(S − j =1 αj) S − j =2 αj , and ηj = αj for every j ∈ N\{1}. Proof. According to Lemma 7 we have Gj(0+) > 0 for every j ∈ N\{1}. Hence ηj = πG j (0) = αj for every j ∈ N\{1}. Again, let k2 := logδ α2 S− P j =2 αj. According to Lemma 6 we have Gj(k2) = 1 for all j ∈ N. Hence η1 ≥ πG 1 (k2) = α2(S − j =1 αj) S − j =2 αj . 10If η1 > πG 1 (k2) then, by the continuity of πG 1 ( ) at k2, we infer that for γ > 0 suﬃciently small we have η1 > πG 1 (t) for every t ∈ [k2 − γ,k2]. Hence S(G1) ⊆ [0,k2 − γ]. Now player 2 can obtain more than α2 by putting all his probability at k2 − γ + ǫ for ǫ > 0 small enough. In the following lemma we show that in a Nash equilibrium players 3,...,n claim their initial right immediately, i.e., they play pure strategy 0. Lemma 9. Let G = (Gi)i∈N be a Nash equilibrium of the n-player cake sharing game ΓS,α,δ with n ≥ 3. Then for every i ∈ N\{1,2} we have Gi(t) = 0 if t = 0 1 if t > 0 i.e., Gi corresponds to pure strategy 0. Proof. Let i ∈ N\{1,2} and suppose that it is not true that Gi(t) = 0 if t = 0 1 if t > 0 Deﬁne t∗ = inf{t | Gi(t) = 1}. Note that t∗ > 0 and t∗ ∈ S(Gi). Moreover, by continuity of Gi in t∗ we have Gi(t∗) = 1. Now we have πG 2 (t∗) = δt∗ (α2 + (S − j∈N αj) j =2 Gj(t∗)) ≤ α2, (4) since otherwise player 2 could deviate to pure strategy t∗ obtaining strictly more than his equilibrium payoﬀ α2. Moreover, due to t∗ ∈ S(Gi), πG i (t∗) = δt∗ (αi + (S − j∈N αj) j =i Gj(t∗)) = αi, (5) From (4) and (5) we obtain δt∗ (S − j∈N αj) j =2 Gj(t∗) ≤ α2(1 − δt∗ ), δt∗ (S − j∈N αj) j =i Gj(t∗) = αi(1 − δt∗ ). According to Lemma 7 we have Gj(t∗) > 0 for every j ∈ N. So, dividing these two expressions we get Gi(t∗) G2(t∗) ≤ α2 αi < 1 which leads to the conclusion that G2(t∗) > Gi(t∗) = 1, a contradiction. 11As a consequence of the last result, the only possible Nash equilibrium in a cake sharing game is one in which players 3,...,n play pure strategy 0 and players 1 and 2 play the game with total cake size S − n i=3 αi. Theorem 2. Let ΓS,α,δ be an n-player cake sharing game with n ≥ 3 and let k2 := logδ α2 S− P j =2 αj. Deﬁne G∗ = (G∗ i)i∈N ∈ GN by G∗ 1(t) =    α2 − α2δt δt(S − j∈N αj) if 0 ≤ t ≤ k2 1 if t > k2 G∗ 2(t) =                  0 if t = 0 α2(S − j =1 αj) − α1δt(S − j =2 αj) δt(S − j∈N αj)(S − j =2 αj) if 0 < t ≤ k2 1 if t > k2 G∗ i(t) = 0 if t = 0 1 if t > 0 for every i ∈ {3,...,n}. Then G∗ is the unique Nash equilibrium of ΓS,α,δ. Proof. Suppose G = (Gi)i∈N ∈ GN is a Nash equilibrium of ΓS,α,δ. As a consequence of Lemma 9 we have Gi = G∗ i for every i ∈ {3,...,n}, so players 3,...,n claim their initial rights immediately. Now (G1,G2) is a Nash equilibrium of the 2-player cake sharing game with cake size S − n i=3 αi and initial right vector (α1,α2). According to Theorem 1 we have G1 = G∗ 1 and G2 = G∗ 2. So G = G∗. Now we show that G∗ is indeed a Nash equilibrium. Since, according to Theorem 1, (G∗ 1,G∗ 2) is a Nash equilibrium of the 2-player cake sharing game with cake size S − n i=3 αi and initial right vector (α1,α2), players 1 and 2 can not gain by deviating unilaterally. Now we show that players 3,...,n are not interested in deviating either. For this it is suﬃcient to show that πG∗ i (t) ≤ αi for every i ∈ {3,...,n} and every t ∈ [0,∞). So, let i ∈ {3,...,n}. Then for every t ∈ [k2,∞) we have πG∗ i (t) = δt(αi + (S − j∈N αj) j =i G∗ j(t)) = δt(αi + (S − j∈N αj)) = δt(S − j =i αj) ≤ δk2(S − j =i αj) ≤ δki(S − j =i αj) = αi, where ki := logδ αi S− P j =i αj and at the last inequality we used the fact that ki < k2 (see Lemma 6). Hence it is suﬃcient to show that πG∗ i (t) ≤ αi for every t ∈ [0,k2]. Note that for every t ∈ [0,k2] (also for t = 0!) we have 12πG∗ i (t) = δt(αi + (S − j∈N αj) j =i G∗ j(t)) = δt(αi + (S − j∈N αj)G∗ 1(t)G∗ 2(t)) = δtαi + (α2 − α2δt) α2(S − j =1 αj) − α1δt(S − j =2 αj) δt(S − j∈N αj)(S − j =2 αj) = δt(αi + α1α2 S − j∈N αj ) + δ−t α2 2(S − j =1 αj) (S − j∈N αj)(S − j =2 αj) −( α1α2 S − j∈N αj + α2 2(S − j =1 αj) (S − j∈N αj)(S − j =2 αj) ) = aδt + bδ−t + c, where a = αi + α1α2 1 − j∈N αj b = α2 2(1 − j =1 αj) (1 − j∈N αj)(1 − j =2 αj) c = − α1α2 1 − j∈N αj − α2 2(1 − j =1 αj) (1 − j∈N αj)(1 − j =2 αj) . Making the change of variables x = δt note that it is suﬃcient to show that for the function f : (0,∞) → I R, deﬁned by f(x) = ax + b x + c we have f(x) ≤ αi for every x ∈ [δk2,1]. Since f′′(x) = 2b x3 > 0 for every x ∈ [δk2,1], the function f is convex on [δk2,1]. Consequently, f(x) ≤ max{f(δk2),f(1)}. Since f(1) = a + b + c = αi and f(δk2) = πG∗ i (k2) ≤ αi, this ﬁnishes the proof. Remark 1. Throughout this paper we assumed that α1 < α2 < < αn. Scrutinizing the proofs of Lemmas 1-5 and Theorem 1 we may conclude that for 2-player cake sharing games the same result (existence and uniqueness of a Nash equilibrium) also holds in case α1 = α2. For cake sharing games with at least three players the existence result is still valid in the more general case α1 ≤ α2 ≤ ≤ αn (and a Nash equilibrium is still provided by the proﬁle described in Theorem 2). With few additional eﬀorts we can show that this Nash equilibrium is unique if and only if α2 < α3. 13References Baston, V. J., A. Y. Garnaev. 2000. On a game in manufacturing. Mathe- matical Methods of Operations Research 52 237–249. Hamers, H. 1993. A Silent Duel over a Cake. Methods and Models of Oper- ations Research 37 119–127. Laraki, R., E. Solan, N. Vieille. 2003. Continuous-time Games of Timing. Discussion Paper 1363, Kellogg School of Management, Northwestern Uni- versity. S.Karlin. 1959. Mathematical Methods and Theory in games. Programming and Economics. Addison-Wesley. V.K.Rohatgi. 1976. An introduction to probability theory and mathematical statistics. Wiley series in probability and mathematical statistics. 14

A Silent Battle over a Cake

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