Sequential claim games

Abstract

We consider the estate division or bankruptcy problem and assume that players sequentially put claims on the estate. Each part of the estate is then divided proportionally with respect to the number of claims on it. We focus on myopic play: players first claim the hitherto least claimed parts, and on subgame perfect equilibria. Our main result is that myopic strategies constitute a subgame perfect equilibrium if punishments for deviators are included.

1 Introduction

Consider an estate of size $E > 0$ and players $j=1,\ldots ,n$ who have entitlements (rights) $c_1,\ldots ,c_n$ on this estate, where $0< c_j < E$ for every j. In general, the sum of these entitlements is larger than the size of the estate, and the question then is how to divide the estate. In the literature, this problem is also known as the bankruptcy problem. It was first formally studied in O’Neill (1982) and Aumann and Maschler (1985).

Apart from dividing an estate among heirs or distributing the left-overs of a bankrupt firm among different stakeholders and creditors, the estate division problem has many other applications in economics and operations research. The museum pass problem is the problem how to divide the revenues of the sale of museum passes among the participating museums. In this case, for an entitlement $c_j$ we can take the number of visitors to museum j using the museum pass times the price of a regular ticket asked by j; see Ginsburgh and Zang (2003), Casas-Méndez et al. (2011), and Bergantiños and Moreno-Ternero (2016). In a bipartite rationing problem, the nodes on one side represent resources and the nodes on the other side correspond to the agents, who have entitlements to the resources to which they are linked. Dividing a resource or set of resources among the linked agents is again an estate division problem; see Moulin and Sethuraman (2013) and İlkiliç and Kayi (2014). Similarly, the estate division problem is a stepping stone of a more general multi-issue allocation problem; see Calleja et al. (2005), González-Alcón et al. (2007), Bergantiños et al. (2010), or Acosta-Vega el al. (2022). In a taxation problem, an amount of taxes has to be collected from a number of citizens with different incomes: also this problem can be reformulated as an estate division problem; see Thomson (2003, 2015), Young (1987a, 1987b, 1988), and Stovall (2020). Still another application goes back to the work of Hotelling (1929): the estate is a market (e.g., a street), the players are firms, and the entitlements are advertising budgets which are spent on segments of the market in the competition for customers.

Most of the approaches to the estate division problem in the literature, including the seminal papers of O’Neill (1982) and Aumann and Maschler (1985), are non-strategic or axiomatic, or both: division rules and their properties are studied, as well as cooperative games based on the bankruptcy or estate division problem. For a recent survey of this literature see Thomson (2019). O’Neill (1982), however, also proposes a strategic, noncooperative approach to the bankruptcy problem. In this approach, players simultaneously and independently put claims on parts of the estate, and each part is divided proportionally among those players who put a claim on it. This approach was thoroughly studied in Atlamaz et al. (2011) for proportional sharing of claimed parts, and in Peters et al. (2019) for other sharing rules from the literature on the bankruptcy problem. It was extended to include multiple estates by Pálvölgyi et al. (2014).

Such a noncooperative approach makes sense, in particular, in competitive situations. Think, for instance, of the estate as the aforementioned consumer market and of the players as firms who spend their advertising budget by focusing on segments of this market; or of the estate as representing a continuum of voters and of the players as political parties endowed with a campaigning budget,¹ which takes an axiomatic approach to solving a sequence of bankruptcy problems. Clearly, applications like these call for a noncooperative approach. In this paper, we also take a noncooperative approach, but now consider a dynamic game in which the players make their claims consecutively, rather than simultaneously, in a given order: in the mentioned applications, firms may not enter the market simultaneously, and political parties may not start campaigning at the same time. We analyze this game for a(ny) given order, and this analysis therefore also applies to the case where the game starts with a chance move determining the order, but we do not address the question how in practice this order is established: this depends on the application under consideration.

Our dynamic game is an extensive form game of perfect information, defined as follows. We assume that $E=1$, the estate is identified with the interval [0, 1], and each player j claims a finite number of (disjoint) closed subintervals of [0, 1] with total length equal to j’s entitlement $c_j$. Player 1 starts, player 2 observes player 1’s claims and moves next, then player 3 observes the claims of players 1 and 2 and moves next, etc., until the last player n. This gives rise to a division of the interval [0, 1] into subintervals, and on each subinterval a number of claims – a so-called claims profile. Each subinterval is then divided proportionally among those players who put a claim on that interval, and for each player the payoff is equal to the sum of the shares which that player obtains this way. The chosen order $1,\ldots ,n$ in which the players put claims on the interval is arbitrary, but the analysis for other orders is analogous. It should be noted, however, that the equilibria, in particular the subgame perfect equilibrium identified in Theorem 3.1, and also the associated outcome (shares) will in general depend on the order of play, but it seems difficult to make general statements about this. Therefore, the order of play is important, comparable to for instance the order of claims in real-life bankruptcy problems. Within the context of this paper, one could also start with a chance move determining the order – this would again not essentially change the analysis.² In general, and as already mentioned, determining the order of play is not an issue that we address in the present paper.³

For this so-called sequential claim game, it is natural to consider (as we will) strategy profiles that are subgame perfect equilibria (Selten, 1965): for any j, players j up to n play a Nash equilibrium (Nash 1951), for any claims of the preceding players. In particular, this rules out threats which would not be carried out because there is a player who can improve by deviating from the threat.

We focus on myopic play: this means that a player always first claims the hitherto least claimed parts, next the second least claimed parts, and so on and so forth, until the player’s entitlement is exhausted. Such a strategy is always optimal for the last player n: since every part of the interval is divided proportionally among the players who have put a claim on it, it is optimal for player n to put claims on the hitherto least claimed parts, in order to maximize total shares (this is proved formally in Lemma 2.1). For the other players, myopic play may not be optimal since their actions may influence the actions of later players (in fact, myopic strategies may not even form a Nash equilibrium, as we will see), but the myopic strategy is nevertheless a simple and natural way to play the game. This is one reason for our focus on myopic play. A second reason is that myopic play always results in a claims profile where the difference between the most and least claimed part of the interval is at most one. If this difference would be more than one, then there is always a player who with hindsight could improve by shifting some claim to a lesser claimed part of the interval: in fact, myopic play implies that the final division is also a Nash equilibrium division in the simultaneous (one-shot) version of the claim game (see Sect. 4 for details). Moreover, this difference being at most one means that no part of the interval is over-claimed – which seems natural since the estate is homogenous. Third, a practical reason is that this sequential claim game is simply too complicated to achieve a complete analysis of all Nash or subgame perfect equilibria. This concerns not so much the computation of any particular equilibrium, but rather characterizing all such equilibria or establishing their properties. For instance, it is an open problem whether the difference between the most and least claimed part of the interval is at most one for every subgame perfect equilibrium. Thus, non-myopic strategies are not only complex for the players, but also for the analyst of the game.

The main question we then answer in the paper is: how can we turn a myopic strategy profile into a subgame perfect equilibrium? To answer this question we assume that in the subgame perfect equilibrium profile players play myopically from left to right: if no player preceding a player j has deviated, then player j claims the intervals least claimed by $1,\ldots ,j-1$ from left to right, i.e., starting with the left end point of the left most of these intervals (the end point closest to zero). This is, basically, without loss of generality: myopic strategies are not unique and in order to examine how a myopic strategy profile can be turned into a subgame perfect equilibrium we need to make an assumption about the exact strategies; assuming other play than from left to right would work as well, see Remark 3.3. We then introduce the possibility of punishment: whenever, in some subgame starting with a player j, some preceding player has deviated, then j and all following players punish the last deviating player r by claiming as much as possible the intervals also claimed by r, as long as this is consistent with myopic play. Thus, in order to do so, they no longer adhere to left-right play. One can think of this situation as of the players agreeing that each part of the estate should be over-claimed as little as possible – then players should be kept from doing so by credibly threatening with punishment, as described. The main result in the paper is that the associated strategies indeed form a subgame perfect equilibrium (Theorem 3.1).

Just as simultaneous claim games, as in O’Neill (1982) and Atlamaz et al. (2011), are related to Blotto games (Borel 1921), sequential claim games can be seen as sequential Blotto games. These have been and are studied in the literature, but usually as a series of simultaneous Blotto games or more generally, contests. See, e.g., Klumpp , Konrad (2018) and the references therein.

Sect. 2 is devoted to preliminaries and illustrative examples, and Sect. 3 contains our main result, Theorem 3.1, that myopic strategies with punishment constitute a subgame perfect equilibrium. Section 4 contains a few remarks on equilibrium payoffs in comparison with the static approach in Atlamaz et al. (2011), and Sect. 5 concludes. The proof of Theorem 3.1 can be found in the Appendix.

2 Model and preliminaries

There are $n \ge 2$ players, and an estate of size 1 is to be distributed among these players. The size of the estate is inessential: all our results can easily be adapted for estates of any size. Players care only about their shares of the estate (the more, the better). Each player $j=1,\ldots ,n$ has an entitlement $c_j\in {\mathbb {R}}$ with $0< c_j < 1$. Throughout, unless stated otherwise, our definitions and results are with respect to the vector of entitlements $c=(c_1,\ldots ,c_n)$.

The estate is represented by the interval from 0 to 1, and distributed by means of a sequential claim game: there is a fixed order according to which the players claim parts of the interval, where player j can claim $c_j$ in total. This leads to a partition of the interval, and each element of this partition is divided equally among those players who put a claim on it. This partition plus the claims will be called a ‘claims profile’.

In the following subsections, we formally define actions and claims profiles, strategies, payoffs, best replies, Nash equilibrium, and subgame perfect equilibrium. We also introduce the concept of a myopic strategy, which plays a central role in the paper. The section is concluded by some examples.

2.1 Actions and claims profiles

A claim is a closed subinterval of the interval [0, 1]. An action of player j is a finite number of claims with disjoint interiors and total length equal to $c_j$. Throughout this paper, unless stated otherwise, we assume that claims are made consecutively by the players, in the order $1,\ldots ,n$. We write $[j]=\{1,\ldots ,j\}$ for every $j \in \{1,\ldots ,n\}$.

For any player j, a claims profile for [j] is a triple $(y,\beta ,m)$ where (i) $m \in {\mathbb {N}}$, (ii) $y = (y_0,\ldots ,y_m) \in {\mathbb {R}}^{m+1}$ with $0 = y_0< y_1< \ldots< y_\mathrm{m-1} < y_m = 1$, and (iii) $\beta = (\beta _i)_{i\in [j]}$, where $\beta _i :\{1,\ldots ,m\} \rightarrow \{0,1\}$ such that

$$\begin{aligned} \sum _\mathrm{t=1}^m \beta _i(t)(y_t - y_{t-1}) = c_i \end{aligned}$$

for every $i \in [j]$. We use the notation $\beta _{[j]}(t)$ to denote the total claim on the interval t between $y_{t-1}$ and $y_t$ by the players in [j], i.e., $\beta _{[j]}(t) = \sum _\mathrm{i=1}^j \beta _i(t)$. Thus, a claims profile for [j] tells us which parts of the interval [0, 1] have been claimed by whom of the players $1,\ldots ,j$. A claims profile for [j] can be obtained by considering the coarsest common refinement of the claims of the players in [j].

For a claims profile $(y,\beta ,m)$ for [j] and a player $h < j$, the triple $(y,(\beta _i)_{i\in [h]},$ m) is a claims profile for [h].

2.2 Strategies and payoffs

Consider any player j. A strategy $\sigma _j$ of player j is a collection of actions of player j, exactly one at each claims profile for $[j-1]$.⁴ The set of all strategies of player j is denoted by $\mathcal {S}_j$.

A strategy profile is an n-tuple of strategies $\sigma =(\sigma _1,\ldots ,\sigma _n) \in \mathcal {S}= \mathcal {S}_1 \times \ldots \times \mathcal {S}_n$. A strategy profile $\sigma$ results in a sequence of claims profiles for $[1],\ldots ,[n]$.

A strategy profile $\sigma$, resulting in a claims profile $(y,\beta ,m)$ for [n], leads to the payoff $u_j(\sigma )$, defined by

$$\begin{aligned} u_j(\sigma ) = \sum _{t \in \{1,\ldots m\} :\beta _j(t) = 1} \frac{\beta _j(t)}{\beta _{[n]}(t)}(y_t-y_{t-1}) \end{aligned}$$

for every player $j\in [n]$. In words, every part of the final partition of the interval [0, 1] is divided equally among those players who have claimed it. Instead of $u_j(\sigma )$ we also sometimes write $u_j(y,\beta ,m)$.

2.3 Best replies, Nash equilibrium, and subgame perfect equilibrium

Given a strategy profile $\sigma \in \mathcal {S}$, strategy $\sigma _j\in \mathcal {S}_j$ of player j is a best reply against the strategy profile $\sigma _{-j} = (\sigma _1,\ldots , \sigma _{j-1},$ $\sigma _{j+1},\ldots ,\sigma _n)$ of the other players if for every strategy $\sigma '_j\in \mathcal {S}_j$ of player j, we have

$$\begin{aligned} u_j(\sigma ) \ge u_j(\sigma _{-j},\sigma '_j), \end{aligned}$$

where $u_j(\sigma _{-j},\sigma '_j) = u_j(\sigma _1,\ldots , \sigma _{j-1},\sigma '_j, \sigma _{j+1},\ldots ,\sigma _n)$.

A strategy profile $\sigma \in \mathcal {S}$ in which every player plays a best reply, is a Nash equilibrium.

Let $\sigma \in \mathcal {S}$ be a strategy profile, let $j \in [n]$, and let $(y,\beta ,m)$ be a claims profile for $[j-1]$. We identify $(y,\beta ,m)$ with a subgame for $\{j,\ldots ,n\}$ in the obvious way; and $\sigma$ induces a strategy profile in this subgame, again in the obvious way, which, with a slight abuse of notation, will be denoted by $(\sigma _{j},\ldots ,\sigma _n)$.

A strategy profile $\sigma \in \mathcal {S}$ is a subgame perfect equilibrium if for every player $j \in [n]$ and every claims profile $(y,\beta ,m)$ for $[j-1]$, the strategy profile $(\sigma _{j},\ldots ,\sigma _n)$ is a Nash equilibrium in the subgame $(y,\beta ,m)$.

2.4 Myopic strategies

In view of our definition of the player payoffs, an obvious action for a player j would be to start with claiming the (thus far) least claimed parts of the interval [0, 1], then the second least claimed parts, and so on, until the total entitlement $c_j$ is exhausted. Such an action, of course, does not take the actions of the players $k>j$ into account.

Formally, at some claims profile for $[j-1]$, an action of player j is myopic if it results in a claims profile $({\bar{y}},{\bar{\beta }},{\bar{m}})$ for [j] satisfying the following condition: for all $t,t' \in \{1,\ldots ,{\bar{m}}\}$ with ${\bar{\beta }}_{[j-1]}(t) < {\bar{\beta }}_{[j-1]}(t')$ and ${\bar{\beta }}_j(t')=1$, we have ${\bar{\beta }}_j(t)=1$. A strategy of player j is myopic if all its actions are myopic. That a player’s strategy is myopic means that this player always plays optimally as if he were the last player to move in the game: this is quite intuitive and follows formally from the following lemma.

Lemma 2.1

Let $\sigma$ be a strategy profile. Then $\sigma _n$ is a best reply against $\sigma _{-n}$ if, and only if, the action of player n at the claims profile for $[n-1]$ resulting from $\sigma$, is myopic.

Proof

Let $(y,\beta ,m)$ be the claims profile for [n] resulting from $\sigma$, and suppose that the action of player n is not myopic at the claims profile for $[n-1]$ resulting from $\sigma$. Since player n’s action is not myopic, there are $t,t'\in \{1,\ldots ,m\}$ such that $\beta _{[n-1]}(t) < \beta _{[n-1]}(t')$, $\beta _n(t')=1$, and $\beta _n(t)=0$. Let $\varepsilon = \min \{y_t-y_{t-1},y_{t'}-y_{t'-1}\}$. A change of player n’s action by including the claim $[y_{t-1},y_{t-1}+\varepsilon ]$ and excluding the claim $[y_{t'-1},y_{t'-1}+\varepsilon ]$, results in a payoff loss of $\varepsilon /(\beta _{[n-1]}(t')+1)$ and a payoff gain of $\varepsilon /(\beta _{[n-1]}(t)+1)$. Since $\beta _{[n-1]}(t) < \beta _{[n-1]}(t')$, this implies that player n has a net payoff gain. Hence, a non-myopic action is not a best reply. Moreover, by applying a finite number of changes like this one, each time increasing player n’s payoff, we obtain a myopic action in the end. Since every myopic action of player n results in the same payoff for that player, we conclude that an action of player n is a best reply if and only if it is myopic. $\square$

Besides Lemma 2.1, as mentioned in the Introduction there are three main reasons to focus on myopic strategies, as we will in this paper. First, myopic play is simple and intuitive, not only for the last player. Second, myopic play results in a claims profile where the difference between the least and most claimed intervals is at most one, which is a desirable property: this means that with hindsight no player would want to shift any claim, and the final division is also a Nash equilibrium division in the simultaneous (one-shot) version of the claim game (see Sect. 4). It also means that the interval is distributed in a balanced manner, e.g., in the application where firms put claims on the consumer interval no segment is over-claimed compared to another one. Third, a complete analysis of the Nash and/or subgame perfect equilibria seems too difficult if not impossible, as demonstrated by the examples below, but focusing on myopic strategies makes the analysis more tractable.

2.5 Examples

We start with an example of a Nash equilibrium which is not subgame perfect. In this equilibrium, the least claimed interval is claimed by only one player, whereas the most claimed interval is claimed by three players. This seems counterintuitive and undesirable since obviously there must be player with a claim in the latter interval who would improve by shifting this claim or part of it to the only once claimed interval (cf. the proof of Lemma 2.1).

Example 2.2

Let $n=4$, $c_1=\frac{3}{8}$, $c_2=\frac{2}{8}$, $c_3=\frac{6}{8}$, and $c_4=\frac{5}{8}$. We define a strategy profile. (Cf. Figure 1.)

Player 1’s strategy is to claim $\left[ 0,\frac{3}{8}\right]$. In this case, player 2’s action is to claim $\left[ 0,\frac{1}{8}\right] \cup \left[ \frac{3}{8},\frac{4}{8}\right]$. Otherwise, i.e., if player 1’s action is not $\left[ 0,\frac{3}{8}\right]$, then player 2 claims in total $c_2=\frac{2}{8}$ of the interval(s) also claimed by player 1, in a left to right manner, i.e., starting with the most left (closest to 0) part claimed by player 1, and continuing to the right until player 2’s entitlement of $\frac{2}{8}$ is exhausted.

Player 3’s strategy is as follows. If player 1 has claimed $[0,\frac{3}{8}]$ and player 2 has claimed $\left[ 0,\frac{1}{8}\right] \cup \left[ \frac{3}{8},\frac{4}{8}\right]$, then player 3 claims $\left[ 0,\frac{2}{8}\right] \cup \left[ \frac{4}{8},1\right]$. In all other cases, player 3 claims all intervals also claimed by player 1 and/or by player 2, and uses what is left of $c_3=\frac{6}{8}$ to claim the part of [0, 1] not claimed by player 1 or 2, from left to right.

Finally, we describe player 4’s strategy. If player 1 has claimed $\left[ 0,\frac{3}{8}\right]$, player 2 has claimed $\left[ 0,\frac{1}{8}\right] \cup \left[ \frac{3}{8},\frac{4}{8}\right]$, and player 3 has claimed $\left[ 0,\frac{2}{8}\right] \cup \left[ \frac{4}{8},1\right]$, then player 4 claims $\left[ \frac{2}{8},\frac{7}{8}\right]$, which in fact means that player 4 acts myopically from left to right. If player 1 has claimed $\left[ 0,\frac{3}{8}\right]$, player 2 has claimed $\left[ 0,\frac{1}{8}\right] \cup \left[ \frac{3}{8},\frac{4}{8}\right]$, but player 3’s action was not $\left[ 0,\frac{2}{8}\right] \cup \left[ \frac{4}{8},1\right]$, then player 4 claims $\left[ \frac{4}{8},1\right]$ and puts the remainder $c_4-\frac{4}{8}=\frac{1}{8}$ from left to right, until exhaustion, on those intervals of $\left[ 0,\frac{4}{8}\right]$ claimed by player 3. In all other cases, player 4 claims all intervals claimed by player 1 and/or player 2, and distributes the remainder of $c_4=\frac{5}{8}$ myopically, from left to right.

If the players play according to these strategies, then the game ends in the claims profile in which $\left[ 0,\frac{1}{8}\right]$ is claimed by players 1, 2, and 3; $\left[ \frac{1}{8},\frac{2}{8}\right]$ is claimed by players 1 and 3; $\left[ \frac{2}{8},\frac{3}{8}\right]$ is claimed by players 1 and 4; $\left[ \frac{3}{8},\frac{4}{8}\right]$ is claimed by players 2 and 4; $\left[ \frac{4}{8},\frac{7}{8}\right]$ is claimed by players 3 and 4; and $\left[ \frac{7}{8},1\right]$ is claimed by player 3. See Fig. 1a.

The described strategy profile is a Nash equilibrium. If players 1, 2, and 3 play according to these strategies, then player 4’s action is myopic and therefore a best reply. If players 1 and 2 play according to these strategies, then playing $\left[ 0,\frac{2}{8}\right] \cup \left[ \frac{4}{8},1\right]$ by player 3 is followed by myopic play of player 4 and gives player 3 a payoff of $\frac{1}{8}\cdot \frac{1}{3} + \frac{1}{8}\cdot \frac{1}{2} + \frac{3}{8}\cdot \frac{1}{2} + \frac{1}{8}\cdot 1 = \frac{5}{12}$. Deviating results in at most $\frac{4}{8} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{3} + \frac{1}{8} \cdot \frac{1}{2} = \frac{17}{48} < \frac{5}{12}$ (see Fig. 1(b) for an optimal deviation by player 3). It can be similarly checked that players 1 and 2 cannot gain from deviating.

Clearly, this is not a subgame perfect equilibrium, e.g., if player 3 deviates as in Fig. 1b, then player 4 does not act myopically and therefore by Lemma 2.1 does not play a best reply.

As announced, in this equilibrium the most claimed part, namely the interval $\left[ 0,\frac{1}{8}\right]$, is claimed thrice, while the least claimed part, the interval $\left[ \frac{7}{8},1\right]$, is claimed once. It is easy to see that with myopic play this cannot happen. $\lhd$

Fig. 1

Part a exhibits the claims profile of the Nash equilibrium of Example 2.2, and part b the claims profile if player 3 deviates optimally

The next example shows that myopic strategies do not have to constitute a Nash equilibrium. Since such myopic strategies are not unique and for ease of presentation, it is convenient to consider actions and strategies where players put their claims, as much as possible, from left to right, as we already did to some extent in Example 2.2.

Formally, the left-right myopic action (or strategy) of player 1 is simply to claim $[0,c_1]$. The left-right myopic action of player $j>1$ at some claims profile for $[j-1]$ is the myopic action such that in the resulting claims profile $(y,\beta ,m)$ for [j] we have: for all $t,t'\in \{1,\ldots ,m\}$ such that $t'>t$ and $\beta _{[j-1]}(t') = \beta _{[j-1]}(t)$, if $\beta _j(t')=1$ then $\beta _j(t)=1$. In words, player j plays myopically such that, on every level of the claims profile for $[j-1]$, player j’s claims are put from left to right without ‘holes’.

The strategy of player j in which all actions are left-right myopic, is called player j’s left-right myopic strategy, and is denoted by $\mu _j$. The left-right myopic strategy profile is denoted by $\mu$, i.e., $\mu = (\mu _1,\ldots ,\mu _n)$.

Example 2.3

(Cf. Fig. 2.) Let $n=3$, $c_1=c_3=\frac{4}{8}$, and $c_2=\frac{3}{8}$. Now consider the strategy profile $\mu = (\mu _1,\mu _2,\mu _3)$. By Lemma 2.1, $\mu _3$ is a best reply against $(\mu _1,\mu _2)$. Also, it is easy to see that $\mu _2$ is a best reply against $(\mu _1,\mu _3)$: player 2 obtains a payoff of $\frac{3}{8}$, which is equal to player 2’s entitlement and therefore maximal. By playing $\mu _1$ against $(\mu _2,\mu _3)$, player 1 obtains a payoff of $\frac{5}{16}$, whereas claiming $\left[ \frac{4}{8},1\right]$ yields $\frac{8}{16}$. Hence $\mu$ is not a Nash equilibrium.$\lhd$

Fig. 2

Example 2.3. Part a corresponds to $\mu$ and part b corresponds to the case when player 1 deviates to claim $\left[ \frac{4}{8},1\right]$

Remark 2.4

Intuitively, that (for instance) a myopic strategy profile, and in particular the left-right myopic strategy profile, may fail to be a Nash equilibrium is caused by the fact that players are completely free where to deposit their claims. Suppose we put the left-right restriction, described before Example 2.3 above, on all (!) admissible actions (which implies that we restrict the strategy sets of the players). Hence, if a player j claims two parts of the interval [0, 1] on which the total claims of his predecessors are equal, then player j should start with claiming the left most part. In particular, player 1 can only claim $[0,c_1]$ and thus becomes a strategic dummy. Under this restriction one can prove that the left-right myopic strategy profile $\mu$ is always a Nash equilibrium. For such a proof, see Kong (2021). There, it is also shown that, under the left-right restriction on actions, $\mu$ is also the unique subgame perfect equilibrium for $n<4$, but $\mu$ still does not have to be a subgame perfect equilibrium for $n \ge 4$. The intuition for this result is as follows. In case there are only three players, since player 3, being the last player, will always act myopically, and player 2 faces only one subgame and knows how player 3 will act, it is straightforward to show that player 2’s left-right myopic action provides maximal payoff to that player. This logic breaks down if there are four or more players. $\lhd$

3 Myopic strategies as a subgame perfect equilibrium

We have already seen (Example 2.3) that a myopic strategy profile does not have to be a Nash equilibrium, let alone a subgame perfect equilibrium. Even under a restriction on actions such as the left-right restriction, the left-right myopic strategy profile $\mu$ does not have to be a subgame perfect equilibrium if there are at least four players (see Remark 2.4). In this section, we show that $\mu$ can be modified to a strategy profile that is a subgame perfect equilibrium, by including punishments of deviating players. In this modified strategy profile, all actions are still myopic.

In words, this modification is as follows. Suppose a player i deviates by not acting left-right myopically. This means that player i does not act myopically – claims some interval whereas intervals with lower total claims of i’s predecessors are still available – or acts myopically but with claims not always from left to right, that is, leaving ‘holes’. Then we will assume that all players $j > i$ play myopically, and as much as possible from left to right, but with the exception that, when choosing between intervals with equal total preceding claims, they first claim – from left to right – the intervals claimed by the deviator i.

Formally, let $i,j \in [n]$ with $i < j$ and consider a(ny) claims profile for $[j-1]$. The lrm-i action (‘lrm’ stands for ‘left-right myopic’) of player j is the myopic action of j, resulting in the claims profile $(y,\beta ,m)$ for [j], such that for all $t',t \in \{1,\ldots ,m\}$ with $t<t'$ and $\beta _{[j-1]}(t') = \beta _{[j-1]}(t)$ we have:

• (i) if $\beta _i(t)=0$ and $\beta _i(t')=1$, then $\beta _j(t)=1$ implies $\beta _j(t')=1$,

• (ii) in all (three) other cases, $\beta _j(t')=1$ implies $\beta _j(t)=1$.

Part (i) can be interpreted as player j ‘punishing’ player i: if player i has claimed interval $t'$ and not t although t is to the left of $t'$, then player j claims $t'$ before claiming t. Part (ii) just describes the remaining cases, where player i does not have to be ‘punished’.

Next, for any claims profile for $[j-1]$, we define the red card player $\hat{r}$ as follows. If every player $1,\ldots ,j-1$ has acted left-right myopically, then $\hat{r}= 0$. Otherwise, let $r_1$ be the first player who has not acted left-right myopically. If every player $r_1+1,\ldots ,j-1$ has used the lrm-$r_1$ action, then $\hat{r}= r_1$. Otherwise, let $r_2$ be the first player after $r_1$ who has not used the lrm-$r_1$ action. If every player $r_2+1,\ldots ,j-1$ has used the lrm-$r_2$ action, then $\hat{r}= r_2$. And so on and so forth. In words, $\hat{r}$ is the last player who has not acted left-right myopically, or who has deviated from the punishment action.

By ${\hat{\mu }}_j \in \mathcal {S}_j$ we denote the strategy of player j according to which j plays the lrm-$\hat{r}$ action at each claims profile for $[j-1]$, $\hat{r}$ being the red card player of that profile, with lrm-0 being the left-right myopic action. Let ${\hat{\mu }} = ({\hat{\mu }}_1,\ldots ,{\hat{\mu }}_n)$.

The main result of this section and of the paper says that ${\hat{\mu }}$ is always a subgame perfect equilibrium.

Theorem 3.1

For every vector of entitlements c, ${\hat{\mu }}$ is a subgame perfect equilibrium in $\mathcal {S}$.

The proof of Theorem 3.1 can be found in the Appendix. According to this result, myopic play, resulting, in particular, in a balanced distribution in the sense that the difference in claims between different parts of the estate is at most one, can be achieved as a subgame equilibrium. Subgame perfection means that even after a deviation of some player the later players still play an equilibrium. In order to achieve this, players should be able to punish a deviator while still playing in their own interest.

We conclude this section with a few remarks about Theorem 3.1.

Remark 3.2

The proof of Theorem 3.1 is quite involved. In this proof, we consider a subgame for players $j,\ldots ,n$ and let $\hat{r}\in \{0,\ldots ,j-1\}$ be the red card player, i.e., the last player $i<j$ who has deviated from ${\hat{\mu }}_i$. We first show that the payoff to j from playing the lrm-$\hat{r}$ action, followed by $j+1,\ldots ,n$ also playing this action, is at least as large as when $j+1,\ldots ,n$ play the lrm-j action. Therefore, it is sufficient to prove that for player j the lrm-$\hat{r}$ action is a best reply against all players $j+1,\ldots ,n$ playing the lrm-j action. This constitutes the by far major part of the proof. Throughout the proof, we often rearrange the claims of an action of player j from left to right without ‘holes’: this may change the payoffs of j’s predecessors but does not affect j’s payoff, and therefore is without loss of generality. $\lhd$

Remark 3.3

Theorem 3.1 is about myopic play from left to right, but this is just to obtain an easy description of the strategies. It is, for instance, obvious that the theorem can be adapted for myopic play from right to left. More generally, as long as all actions in the strategies under consideration are myopic, an associated version of Theorem 3.1 can be proved: as we mentioned in Remark 3.2, we may always rearrange claims from left to right. Nevertheless, we chose to present the left-right version of the strategies in order to avoid even more cumbersome notations and technicalities. $\lhd$

Remark 3.4

Obviously, player 1’s strategy ${\hat{\mu }}_1$ is the same as $\mu _1$, and it is also easy to see that player 2’s strategy ${\hat{\mu }}_2$ is the same as $\mu _2$. For player 3, this is no longer the case, and as can be seen in Example 2.3, $(\mu _1,\mu _2,\mu _3)$ does not have to be a subgame perfect equilibrium, in fact not even a Nash equilibrium. One may wonder if it is sufficient that only the last player n plays the myopic strategy with punishment ${\hat{\mu }}_n$ in order to have a subgame perfect equilibrium, but also this is not true. For instance, in Example 2.3, add a player 4 with entitlement $c_4=\frac{1}{8}$. Then the strategy profile $(\mu _1,\mu _2,\mu _3,{\hat{\mu }}_4)$ is not a subgame perfect equilibrium (again, not even a Nash equilibrium): player 1 obtains $\frac{1}{4}$ by not deviating as in panel (a), but $\frac{1}{16}+\frac{3}{8} = \frac{7}{16}$ by deviating as in panel (b). Thus, punishment by only the last player 4 is too light to keep (in this case) player 1 from deviating. $\lhd$

4 Static claim games and payoffs

In Atlamaz et al. (2011) a static version of the claim game is studied. In this static game, a strategy of a player is simply a collection of claims, and players play simultaneously and independently. Apart from the static nature, another difference with our approach is that in Atlamaz et al. (2011) multiple claims are allowed: an interval can be claimed more than once by the same player. Here, we reconsider this static game, but without allowing for multiple claims, thus, exactly following O’Neill (1982). A Nash equilibrium profile in the static game is a claims profile such that no player can gain by deviating to a different strategy. We omit the (relatively straightforward) proof of the following theorem – it is the single-claim version of Theorem 1 in Atlamaz et al. (2011). The theorem says that the Nash equilibrium profiles are exactly those with difference between minimal and maximal claims at most one.

Proposition 4.1

Claims profile $(y,\beta ,m)$ is a Nash equilibrium profile in the static claim game if and only if $|\beta _{[n]}(t)-\beta _{[n]}(t')| \le 1$ for all $t,t' \in \{1,\ldots ,m\}$.

From Proposition 4.1 the possible payoffs from Nash equilibrium profiles can be characterized. Let $r = \sum _\mathrm{i=1}^n c_i \mod 1$, hence $\sum _\mathrm{i=1}^n c_i = K + r$ for some $K \in {\mathbb {N}} \cup \{0\}$. Then, at a Nash equilibrium profile $(y,\beta ,m)$ we have $\beta _{[n]}(t) \in \{K,K+1\}$ for every $t=1,\ldots ,m$. For every player j, let $r_j$ be the part of $c_j$ invested in intervals t with $\beta _{[n]}(t) = K+1$. Then $r_j \le c_j$, $r_j \le r$, and $c_j-r_j \le 1-r$, hence

$$\begin{aligned} \sum _\mathrm{i=1}^n r_i = (K+1)r, \text{ and } \max \{0, c_j-(1-r)\} \le r_j \le \min \{c_j,r\} \text{ for } \text{ every } j\in [n]. \end{aligned}$$

(1)

It is not hard to show that, conversely, for every vector $(r_1,\ldots ,r_n)$ satisfying (1) there is an associated Nash equilibrium claims profile. In particular, the set of Nash equilibrium profile payoff vectors is a convex set.

Returning to the sequential claim game, since ${\hat{\mu }}$ is a subgame perfect equilibrium in $\mathcal {S}$ (Theorem 3.1), it is also a Nash equilibrium in $\mathcal {S}$, and the resulting claims profile is a Nash equilibrium profile in the static claim game, as follows from Proposition 4.1 and the fact that the claims profile resulting from ${\hat{\mu }}$ has maximal difference of one between the least and most claimed intervals. To obtain a meaningful general comparison, however, we at least need to consider all possible orders in which the players could make their claims. The next example shows that in that case there still may be payoffs associated with Nash equilibrium profiles in the static game that cannot be obtained by sequential left-right myopic play in any order, nor even by convex combinations of these.

Example 4.2

Let $n=4$, $c_1=c_2=\frac{3}{7}$, $c_3=\frac{4}{7}$, and $c_4=\frac{5}{7}$, and consider the Nash equilibrium profile of the static game given by $(y,\beta ,m)$ in Fig. 3. The payoff to player 4 in this Nash equilibrium profile is equal to $\frac{5}{14}$. In all 24 possible orders of left-right myopic play in the sequential game, however, player 4 always obtains $\frac{7}{21}$, as is straightforward to verify. $\lhd$

Fig. 3

Example 4.2

It is an open question whether payoffs from Nash equilibrium profiles of the static game can always be obtained by some Nash equilibrium of the sequential game, or as convex combinations of such payoffs, and considering all n! orders in which the players can make claims. On the other hand, there can be Nash equilibrium payoffs of the sequential game that cannot be obtained as payoffs of Nash equilibrium profiles of the static game, as the following example shows.

Example 4.3

Consider the (sequential) Nash equilibrium in Example 2.2. In this example, there is only one payoff vector that can be obtained in a Nash equilibrium of the static game, since the sum of the entitlements is equal to 2, namely the payoff vector $\left( \frac{9}{48},\frac{6}{48},\frac{18}{48},\frac{15}{48}\right)$. The payoffs from the Nash equilibrium in Example 2.2, however, are equal to $\left( \frac{8}{48},\frac{5}{48},\frac{20}{48},\frac{15}{48}\right)$. $\lhd$

5 Concluding remarks

In this paper, we studied noncooperative dynamic games, so-called sequential claim games, in order to solve the estate division problem. The main result (Theorem 3.1) is that the left-right myopic strategy profile with punishment is always a subgame perfect equilibrium.

Many open questions are left. These include characterizing all Nash equilibria and subgame perfect Nash equilibria, but answering these questions may be intractable. This concerns not so much the computational complexity of computing an(y) equilibrium, but rather establishing general properties of equilibria. For instance, our conjecture is that in any subgame perfect equilibrium the maximal difference in the final claims profile is at most one. This is trivial for less than three players, easy to prove for three players, and it can also still be proved for four players (see Kong 2021), but for more than four players, so far we did not manage to (dis)prove this.

Other open questions or, rather, possible extensions, include: (i) allowing for multiple claims by the same player on an interval, as studied in Atlamaz et al. (2011) for the static claim game; (ii) considering sharing functions other than proportional sharing, as in Peters et al. (2019) for the static claim game; (iii) considering multiple estates, as in Pálvölgyi et al. (2014) for the static claim game; and (iv) considering heterogenous preferences on the estate – which part a player gets matters to that player (this is related to (iii)).

Appendix: Proof of Theorem 3.1

We have to show that ${\hat{\mu }}$ is a subgame perfect equilibrium in $\mathcal {S}$.

Let $j \in [n]$ and, throughout the proof, fix an arbitrary claims profile for $[j-1]$, with numbers of claims $\beta _i$, $i \in [j-1]$. Let $\hat{r}$ be the red card player of this claims profile, possibly $\hat{r}= 0$. We have to show that for player j the lrm-$\hat{r}$ action is a best reply against all players $h>j$ playing according to ${\hat{\mu }}_h$.

If $j=1$, then $\hat{r}=0$. Any action of player 1 is myopic and, by definition of ${\hat{\mu }}_2,\ldots {\hat{\mu }}_n$, results in the same payoff for player 1 as playing lrm-0, i.e., claiming $[0,c_1]$. So, in particular, lrm-0 is a best reply.

Now let $j \ge 2$. Without loss of generality we may assume that the claims profile for $[j-1]$ is decreasing, otherwise we can rearrange the claims of the players in $[j-1]$ (this may effect the payoffs of the players in $[j-1]$, but not those of the later players).

We first prove that if j plays the lrm-$\hat{r}$ action, then j’s payoff does not increase if every player $h>j$, instead of the lrm-$\hat{r}$ action, plays the lrm-j action. We denote the strategy for player $h>j$, prescribing always the lrm-j action, by ${{\bar{\mu }}}_h$. Let $(y',\beta ',m')$ be the claims profile for [n] resulting from $({\hat{\mu _i}})_{i\in \{j,\ldots ,n\}}$ (i.e., player j and all later players always playing the lrm-$\hat{r}$ action) and let $({\bar{y}},{\bar{\beta }},{\bar{m}})$ be the claims profile for [n] resulting from $({\hat{\mu _j}},({{\bar{\mu }}}_i)_{i\in \{j+1,\ldots ,n\}})$. Without loss of generality (otherwise rearrange) we may assume that $y'={\bar{y}}$ and $m'={\bar{m}}$, and we write y and m instead.

Claim 1. $u_j(y,\beta ',m) \ge u_j(y,{\bar{\beta }},m)$.

Proof

Clearly, ${{\bar{\beta }}}_{[j]}(t) = \beta '_{[j]}(t)$ for all $t\in \{1,\ldots ,m\}$. Since every $h>j$ plays the lrm-j action in ${{\bar{\beta }}}$ and plays the lrm-${\hat{r}}$ action in $\beta '$, it follows that for all $t\in \{1,\ldots ,m\}$ with $\beta '_j(t) = 1$ (and hence ${\bar{\beta }}_j(t)=1$) and $\beta '_{h}(t)=1$ we have ${{\bar{\beta }}}_{h}(t)=1$. Therefore, ${{\bar{\beta }}}_{[n]}(t)\ge \beta '_{[n]}(t)$ for all $t\in \{1,\ldots ,m\}$ with $\beta '_j(t) = 1$. This implies the inequality in Claim 1.

By Claim 1 it is sufficient to show that for player j the lrm-$\hat{r}$ action is a best reply against all later players h playing according to ${\bar{\mu }}_h$, that is, playing always the lrm-j action. Since the profile for $[j-1]$ is decreasing, we may also assume, by rearranging if necessary, that if player j plays the lrm-$\hat{r}$ action, then for all $t,t'$ with $t'>t$ such that $\beta _{[j-1]}(t)=\beta _{[j-1]}(t')$ and j claims $t'$, then j also claims t. (In words, if player j plays the lrm-$\hat{r}$ action, then j’s claims on each level of the claims profile for $[j-1]$ are ordered from left to right, without ‘holes’. Hence, the lrm-$\hat{r}$ action leads to the claims profile for [j] resulting from j playing left-right myopically.) $\square$

Suppose that player j plays an action different from the lrm-$\hat{r}$ action. For the rest of this proof, we denote the resulting claims profile for [n] by $(y,\beta ,m)$, and let $M=\{t\in \{1,\ldots ,m\} :\beta _j(t)=1\}$. The proof proceeds in two parts. In the first part (a) we show that if, on each level of $(y,\beta ,m)$, j rearranges claims so that they are located from left to right (i.e., gets rid of ‘holes’), then j’s payoff does not change. In the second part (b) we show that if j, if possible, shifts claims to lower levels (which means, in particular, to the right of the interval), then again j’s payoff does not decrease.

(a) Suppose there is $r_0\in \{1,\ldots ,m\}$ such that $\beta _{[j-1]}(r_0) = \beta _{[j-1]}(r_0+1)$, $\beta _j(r_0)=0$, and $\beta _j(r_0+1)=1$. Without loss of generality we may assume that $y_{r_0+1}-y_{r_0} = y_{r_0}-y_{r_0-1}$ (otherwise appropriately refine the end partition). Let ${\hat{\beta }}$ be such that ${\hat{\beta }}_i = \beta _i$ for all $i \in [j-1]$, ${\hat{\beta }}_j(r_0)=1$, ${\hat{\beta }}_j(r_0+1)=0$, ${\hat{\beta }}_j(t) = \beta _j(t)$ for all $t\ne r_0,r_0+1$, and ${\hat{\beta }}_i$, like $\beta _i$, is determined by ${\bar{\mu }}_{j+1},\ldots ,{\bar{\mu }}_n$ for all $i = j+1,\ldots ,n$. (In words, the change from $\beta$ to ${\hat{\beta }}$ consists of player j claiming $r_0$ instead of $r_0+1$, plus the consequence for the claims of the later players.) For this situation we prove the following claim.

Claim 2. ${\hat{\beta }}_{[n]}(r_0) = \beta _{[n]}(r_0+1)$, ${\hat{\beta }}_{[n]}(r_0+1) = \beta _{[n]}(r_0)$, and ${\hat{\beta }}_{[n]}(t) = \beta _{[n]}(t)$ for all $t \ne r_0,r_0+1$ with $t\in M$.

Proof

If $\beta _h(r_0)=0$ for all $h \in [n]$ with $h>j$, then ${\hat{\beta }}_{[n]}(r_0) = \beta _{[j]}(r_0)+1=\beta _{[j]}(r_0+1)=\beta _{[n]}(r_0+1)$, ${\hat{\beta }}_{[n]}(r_0+1) = \beta _{[j]}(r_0+1)-1=\beta _{[j]}(r_0)=\beta _{[n]}(r_0)$, and ${\hat{\beta }}_{[n]}(t) = \beta _{[n]}(t)$ for all $t \ne r_0,r_0+1$ with $t\in M$.

Otherwise, let $h_0 = \min \{h \in [n] :h >j, \beta _h(r_0)=1\}$. Then $h_0 = \min \{h \in [n] :h >j, {\hat{\beta }}_h(r_0+1)=1\}$. Also, ${\hat{\beta }}_{[h_0]}(t) = \beta _{[h_0]}(t)$ for all $t \in M\backslash \{r_0, r_0+1\}$. Consider player $h_0+1$. Since this player plays the lmr-j action, following the claims $\beta _j$ of player j, we have: if $\beta _{h_0+1}(r_0) = \beta _{h_0+1}(r_0+1) = 0$ then ${\hat{\beta }}_{h_0+1}(r_0) = {\hat{\beta }}_{h_0+1}(r_0+1) = 0$; if $\beta _{h_0+1}(r_0) = \beta _{h_0+1}(r_0+1) = 1$ then ${\hat{\beta }}_{h_0+1}(r_0) = {\hat{\beta }}_{h_0+1}(r_0+1) = 1$; if $\beta _{h_0+1}(r_0) = 0$ and $\beta _{h_0+1}(r_0+1) = 1$ then ${\hat{\beta }}_{h_0+1}(r_0) = 1$ and ${\hat{\beta }}_{h_0+1}(r_0+1) = 0$. The fourth case, $\beta _{h_0+1}(r_0) = 1$ and $\beta _{h_0+1}(r_0+1) = 0$, can only occur, due to the lmr-j action of player $h_0+1$, if $\beta _{h_0}(r_0) = \beta _{h_0}(r_0+1) =1$ and hence $\beta _{[h_0]}(r_0) = \beta _{[h_0]}(r_0+1)-1$ and ${\hat{\beta }}_{[h_0]}(r_0) = {\hat{\beta }}_{[h_0]}(r_0+1)+1$; in that case, ${\hat{\beta }}_{h_0+1}(r_0) = 0$, ${\hat{\beta }}_{h_0+1}(r_0) = 1$, and ${\hat{\beta }}_{[h_0+1]}(r_0) = {\hat{\beta }}_{[h_0+1]}(r_0+1) = \beta _{[h_0+1]}(r_0) = \beta _{[h_0+1]}(r_0+1)$.

Hence, in all cases we have ${\hat{\beta }}_{[h_0+1]}(r_0) = \beta _{[h_0+1]}(r_0+1)$ and ${\hat{\beta }}_{[h_0+1]}(r_0+1) = \beta _{[h_0+1]}(r_0)$, and ${\hat{\beta }}_{[h_0+1]}(t) = \beta _{[h_0+1]}(t)$ for all $t \ne r_0, r_0+1$ with $t\in M$. By repeating this argument for the players $h_0+2,\ldots ,n$, the statements in the Claim 2 follow.

As a consequence of Claim 2, player j’s payoff from changing his action by claiming $r_0$ instead of $r_0+1$ does not change. By performing consecutive changes like this, we may therefore assume, just keeping, with some abuse, the notation $\beta$, that the following holds: for all $t,t' \in \{1,\ldots ,m\}$, with $t'>t$, if $\beta _{[j-1]}(t) = \beta _{[j-1]}(t')$ and $\beta _j(t')=1$, then $\beta _j(t)=1$. (In words, on each level of the claims profile for $[j-1]$ the claims of player j are ordered from left to right without ‘holes’.) $\square$

We next introduce notations, describing the situation reached in $(**)$ in detail. Since the claims profile for $[j-1]$ is decreasing, there are $\{t_0,\ldots ,t_k\} \subseteq \{0,\ldots ,m\}$ with $0 = t_0< t_1< \cdots < t_k = m$ such that (i) $\beta _{[j-1]}(s) = \beta _{[j-1]}(t)$ for all $t_\ell < s,t \le t_{\ell +1}$, for all $\ell = 0,\ldots ,k-1$; and (ii) $\beta _{[j-1]}(t) > \beta _{[j-1]}(t')$ for all $t,t'$, and $\ell ,\ell ' \in \{0,\ldots ,k-1\}$ with $t_\ell< t \le t_{\ell +1} \le t_{\ell '} < t' \le t_{\ell '+1}$. (In words, this is a description of the levels of the decreasing claims profile for $[j-1]$.) Then by $(**)$, for every $\ell = 0,\ldots ,k-1$, there is $t'$ with $t_\ell \le t' \le t_{\ell +1}$ such that $\beta _j(t)=1$ for every $t_\ell < t \le t'$ and $\beta _j(t)=0$ for every $t' < t \le t_{\ell +1}$. (In words, at every level of the claims profile for $[j-1]$, if player j has claims between $t_\ell$ and $t_{\ell +1}$, that is, if $t' > t_\ell$, then these claims are arranged from left to right without ‘holes’.) With this profile for [j] fixed, we now proceed with the final part of the proof, in which, possibly, claims of player j are shifted to lower levels.

(b) Suppose there exist $s_0,s_1 \in \{1,\ldots ,m\}$ and $\ell \in \{0,\ldots ,k-1\}$ such that $\beta _{[j-1]}(s_1) < \beta _{[j-1]}(s_0)$ (and hence $s_1 > s_0$), $s_1 \in \{t_{\ell -1}+1,\ldots ,t_{\ell }\}$, and such that $\beta _j(s_0)=1$ and $\beta _j(s_1)=0$. Without loss of generality we assume that $y_t-y_{t-1} = y_{t'}-y_{t'-1}$ for all $t,t'\in \{1,\ldots ,m\}$ (otherwise, since payoffs depend continuously on the lengths of the intervals, we may refine the end partition such that it becomes arbitrarily close to a partition with all intervals equal). Also, we can take $s_1$ such that, first, $\beta _j(t)=1$ for all $t_{\ell -1}+1\le t <s_1$, or $s_1=t_{\ell -1}+1$ if $\beta _j(t)=0$ for all $t_{\ell -1}+1 \le t \le t_{\ell }$, and, second, $\beta _j(t) = 1$ for all $t > t_{\ell }$. (In words, we can take $s_1$ such that player j claims everything to the left of $s_1$ at the same level, and, moreover, every lower level interval.) Moreover, we can take $s_0$ such that $\beta _j(t)=0$ for all $t< s_0$. (In words, we take $s_0$ such that player j claims nothing to the left of $s_0$, i.e., $s_0$ is the left most interval claimed by j at $\beta$.)

Let $\tilde{\beta }$ be such that $\tilde{\beta }_i = \beta _i$ for all $i \in [j-1]$, $\tilde{\beta }_j(s_0)=0$, $\tilde{\beta }_j(s_1)=1$, $\tilde{\beta }_j(t) = \beta _j(t)$ for all $t\ne s_0,s_1$, and $\tilde{\beta }_i$ is determined by ${\bar{\mu }}_{j+1},\ldots ,{\bar{\mu }}_n$ for all $i = j+1,\ldots ,n$. (In words, player j claims $s_1$ instead of $s_0$.)

Claim 3. $\tilde{\beta }_{[n]}(s_1)\le \beta _{[n]}(s_0)$, and $\tilde{\beta }_{[n]}(t) = \beta _{[n]}(t)$ for all $t\in M$ with $t>s_1$.

Proof

Since ${\tilde{\beta }}_{[j-1]}(s_0) >\tilde{\beta }_{[j-1]}(s_1)$, $\tilde{\beta }_j(s_0)=0$ and $\tilde{\beta }_j(s_1)=1$, we have ${\tilde{\beta }}_{[j]}(s_0) \ge \tilde{\beta }_{[j]}(s_1)$. Hence, since every $h>j$ plays the lrm-j action, and therefore, in particular, myopically, we have $\tilde{\beta }_{[n]}(s_1) \le \tilde{\beta }_{[n]}(s_0)+1$. Also, $\tilde{\beta }_{[n]}(s_0) \le \beta _{[n]}(s_0)$. We consider two cases. (i) If $\tilde{\beta }_{[n]}(s_1) \le \tilde{\beta }_{[n]}(s_0)$, then $\tilde{\beta }_{[n]}(s_1) \le \beta _{[n]}(s_0)$. (ii) Otherwise, $\tilde{\beta }_{[n]}(s_1) = \tilde{\beta }_{[n]}(s_0) +1$, but in that case $\tilde{\beta }_{[n]}(s_0)=\beta _{[n]}(s_0)-1$, so that $\tilde{\beta }_{[n]}(s_1) = \tilde{\beta }_{[n]}(s_0) +1 = \beta _{[n]}(s_0)$. This proves the first statement in Claim 3.

Since every $h>j$ plays the lrm-j action, we have $\tilde{\beta }_h(t)=\beta _h(t)$ for all $t>s_1$ with $t\in M$. Together with $\tilde{\beta }_{[j]}(t) = \beta _{[j]}(t)$, we have $\tilde{\beta }_{[n]}(t)={\beta }_{[n]}(t)$ for all $t > s_1$ with $t\in M$. This is the second statement in the claim. Hence, the proof of Claim 3 is now complete.

The following claim is an intermediate result used in the final part below. It says that, when going from $\beta$ to $\tilde{\beta }$, there is at most one interval on which the total claim is increased (by one). $\square$

Claim 4. Let $M'_n=\{t\in \{1,\ldots ,m\} :\tilde{\beta }_{[n]}(t) = \beta _{[n]}(t)+1\}$. Then $|M'_n| \le 1$.

Proof

We use induction on the number of players. For $n=j$, the claim holds since by definition of $\beta$ and $\tilde{\beta }$ we have $M'_j=\{s_1\}$. Assume that $|M'_{h'}|\le 1$ for all $j \le h' \le h$ for some $j< h < n$. We prove that $|M'_{h+1}|\le 1$. We distinguish two cases.

(1) $|M'_h|=0$. In this case, $\tilde{\beta }_{[h]}(t)=\beta _{[h]}(t)$ for all $t\in \{1,\ldots ,m\}$. The lrm-j action of player $h+1$ results in a sequence of claims, starting with the least claimed intervals, giving precedence to the intervals claimed by j, and from left to right without ‘holes’. Exactly one of the following two claiming sequences will result. $\square$

• $M_1\rightarrow \{s_1\}\rightarrow M_2 \rightarrow \{s_0\}\rightarrow M_3$. Here, $M_1,M_2,M_3,\{s_0\},\{s_1\}$ form a partition of $\{1,\ldots ,m\}$. Player $h+1$ first claims intervals in $M_1$, next (if there is some entitlement of the total $c_{h+1}$ left) $s_1$, next intervals from $M_2$, etc, until $c_{h+1}$ is exhausted.

• $M_1\rightarrow \{s_0\}\rightarrow M_2 \rightarrow \{s_1\}\rightarrow M_3$, with a similar interpretation.

We only consider the first claiming sequence, the other one is similar. There are essentially three possibilities, where we use the assumption that all intervals have equal length, denoted by $\tau$. Also, the number of intervals in some set $M_i$ is denoted by $m_i$.

• (1.1) $c_{h+1} \le m_1 \tau$. Then $\tilde{\beta }_{[h+1]}(t) = \beta _{[h+1]}(t)$ for all $t\in \{1,\ldots ,m\}$. Thus, $M'_{h+1}=\emptyset$.

• (1.2) $(m_1+1) \tau \le c_{h+1} \le (m_1+m_2+1) \tau$. If $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, we have $M'_{h+1}=\emptyset$. If $\tilde{\beta }_{h+1}(s_1)=1$ and $\beta _{h+1}(s_1)=0$, then $M'_{h+1}=\{s_1\}$.

• (1.3) $(m_1+m_2+2) \tau \le c_{h+1} \le (m_1+m_2+m_3+2)\tau$. If $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$ and $\tilde{\beta }_{h+1}(s_0) = \beta _{h+1}(s_0)$, then $M'_{h+1}=\emptyset$. If $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, $\tilde{\beta }_{h+1}(s_0) = 0$ and $\beta _{h+1}(s_0)=1$, then $M'_{h+1}$ consists of the last interval claimed by player $h+1$ according to $\tilde{\beta }$. If $\tilde{\beta }_{h+1}(s_1)=1$ and $\beta _{h+1}(s_1)=0$, we have $M'_{h+1}=\{s_1\}$.

(2) $|M'_{h}|=1$. Suppose that $M'_{h}=\{s'_1\}$. Then there is an interval $s'_0$ with $\tilde{\beta }_{[n]}(s'_0) = \beta _{[n]}(s'_0)-1$, and $\tilde{\beta }_{[h]}(t) = \beta _{[h]}(t)$ and $\tilde{\beta }_{j}(t) = \beta _{j}(t)$ for all $t\in \{1,\ldots ,m\}\backslash \{s_0,s'_0,s_1,s'_1\}$. Thus, there are 4! claiming sequences for player $h+1$. We only consider the sequence $M_1\rightarrow s'_1\rightarrow M_2\rightarrow s_1\rightarrow M_3\rightarrow s'_0\rightarrow M_4\rightarrow s_0\rightarrow M_5$ with $s_0,s'_0,s_1,s'_1\notin M_i$, $i=1,\ldots ,5$ (the arguments for the other sequences are again similar). We have the following possibilities.

• (2.1) $c_{h+1} \le m_1 \tau$. Then $M'_{h+1}=M'_{h}$.

• (2.2) $(m_1+1) \le c_{h+1} \le (m_1+m_2+1)\tau$. If $\tilde{\beta }_{h+1}(s'_1) = \beta _{h+1}(s'_1) = 1$, then $M'_{h+1}=M'_{h}$. If $\tilde{\beta }_{h+1}(s'_1) = 0$ and $\beta _{h+1}(s'_1) = 1$, then $M'_{h+1}$ consists of the last interval claimed by player $h+1$ according to $\tilde{\beta }$.

• (2.3) $(m_1+m_2+2)\tau \le c_{h+1}\le (m_1+m_2+m_3+2)\tau$. If $\tilde{\beta }_{h+1}(s'_1) = \beta _{h+1}(s'_1)$, then $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$ by the lrm-j action of player $h+1$ and thus, $M'_{h+1} = \{s'_1\}$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$ and $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, then $M'_{h+1}$ consists of the last interval claimed by $h+1$ according to $\tilde{\beta }$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = 1$ and $\beta _{h+1}(s_1)=0$, then $M'_{h+1} = \{s_1\}$.

• (2.4) $(m_1+m_2+m_3+3)\tau \le c_{h+1} \le (m_1+\cdots +m_4+3)\tau$. If $\tilde{\beta }_{h+1}(s'_1) = \beta _{h+1}(s'_1)$ and $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, then $M'_{h+1} = \{s'_1\}$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$ and $\tilde{\beta }_{h+1}(s'_0) = \beta _{h+1}(s'_0)$, then $M'_{h+1}$ consists of the last interval claimed by $h+1$ according to $\tilde{\beta }$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, $\tilde{\beta }_{h+1}(s'_0) = 1$ and $\beta _{h+1}(s'_0)=0$, then $M'_{h+1}=\emptyset$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = 1$ and $\beta _{h+1}(s_1)=0$, then $M'_{h+1}=\{s_1\}$.

• (2.5) $(m_1+\cdots +m_4+4)\tau \le c_{h+1} \le (m_1+\cdots +m_5+4)\tau$. If $\tilde{\beta }_{h+1}(s'_1) = \beta _{h+1}(s'_1)$ and $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, then $M'_{h+1} = \{s'_1\}$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$ and $\tilde{\beta }_{h+1}(s'_0) = \beta _{h+1}(s'_0)$, we have $\tilde{\beta }_{h+1}(s_0) = \beta _{h+1}(s_0) = 1$. Then $M'_{h+1}$ consists of the last interval claimed by $h+1$ according to $\tilde{\beta }$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, $\tilde{\beta }_{h+1}(s'_0) = 1$, $\beta _{h+1}(s'_0)=0$ and $\tilde{\beta }_{h+1}(s_0) = \beta _{h+1}(s_0)$, then $M'_{h+1} = \emptyset$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = \beta _{h+1}(s_1)$, $\tilde{\beta }_{h+1}(s'_0) = 1$, $\beta _{h+1}(s'_0)=0$, $\tilde{\beta }_{h+1}(s_0)=0$ and $\beta _{h+1}(s_0)=1$, then $M'_{h+1}$ consists of the last interval claimed by $h+1$ according to $\tilde{\beta }$. If $\tilde{\beta }_{h+1}(s'_1) =0$, $\beta _{h+1}(s'_1) =1$, $\tilde{\beta }_{h+1}(s_1) = 1$ and $\beta _{h+1}(s_1)=0$, then $M'_{h+1}=\{s_1\}$.

This concludes the proof of Claim 4.

Based on Claim 3, we consider two cases.

Case(i): $\tilde{\beta }_{[n]}(s_1) = \beta _{[n]}(s_0)$.

We consider two subcases.

Fig. 4

Case (i.1): $\tilde{\beta }_{[n]}(s_0) = \beta _{[n]}(s_0)$

(i.1) $\tilde{\beta }_{[n]}(s_0) = \beta _{[n]}(s_0)$. If there would be an $r\in M$, $s_0<r<s_1$, with $\tilde{\beta }_{[n]}(r) = \tilde{\beta }_{[n]}(s_1) + 1$, then by the lrm-j actions of the players $h>j$ we would have $\tilde{\beta }_{[n]}(s_0) = \beta _{[n]}(s_0)-1$. Therefore, $\tilde{\beta }_{[n]}(t)= \tilde{\beta }_{[n]}(s_1)$ for all $s_0< t < s_1$ with $t\in M$. (Cf. Fig. 4.)

(i.2) Otherwise, $\tilde{\beta }_{[n]}(s_0) + 1 = \beta _{[n]}(s_0)$. If there would be an $r'\in M$, $s_0< r' < s_1$ with $\tilde{\beta }_{[n]}(r')\ge \tilde{\beta }_{[n]}(s_1)+1$, then $\tilde{\beta }_{[n]}(r')-\tilde{\beta }_{[n]}(s_0)\ge \tilde{\beta }_{[n]}(s_1) - \tilde{\beta }_{[n]}(s_0) + 1 = 2$, which contradicts with the myopic strategies of all players $h>j$. (Cf. Fig. 5.) Hence, we again have $\tilde{\beta }_{[n]}(t)= \tilde{\beta }_{[n]}(s_1)$ for all $s_0< t < s_1$ with $t\in M$.

Fig. 5

Case (i.2): $\tilde{\beta }_{[n]}(s_0) = \beta _{[n]}(s_0)-1$

Now suppose that there would exist $s_0< p < s_1$ with $p\in M$ such that $\tilde{\beta }_{[n]}(p)= \beta _{[n]}(p)+1$. Then, since $\tilde{\beta }_{[n]}(t)= \tilde{\beta }_{[n]}(s_1)$ for all $s_0< t < s_1$ with $t\in M$, we would have $\tilde{\beta }_{[n]}(t)= \beta _{[n]}(t)+1$ for all $p< t \le s_1$ (cf. Fig. 6), but this contradicts Claim 4. Hence, $\tilde{\beta }_{[n]}(t)\le \beta _{[n]}(t)$ for all $s_0< t < s_1$ with $t\in M$. Together with Claim 3 we conclude that in Case (i) the payoff to player j does not decrease by going from $\beta$ to $\tilde{\beta }$.

Fig. 6

Illustrating the final argument in Case (i)

Case (ii): $\tilde{\beta }_{[n]}(s_1) \le \beta _{[n]}(s_0)-1$.

In this case, by Claim 4 there is again at most one $t' \in M$, $s_0 < t' \le s_1$, such that $\tilde{\beta }_{[n]}(t') = \beta _{[n]}(t')+1$. If $t'\ne s_1$, then $\tilde{\beta }_{[n]}(s_1)=\beta _{[n]}(s_1) \le \beta _{[n]}(t')=\tilde{\beta }_{[n]}(t')-1$. Now, by Claim 3, when going from $\beta$ to $\tilde{\beta }$, player j has a gain of

$$\begin{aligned} \frac{1}{\tilde{\beta }_{[n]}(s_1)}, \end{aligned}$$

and a loss of at most

$$\begin{aligned} \frac{1}{\beta _{[n]}(s_0)} + \frac{1}{\beta _{[n]}(t')} - \frac{1}{\tilde{\beta }_{[n]}(t')}. \end{aligned}$$

We have

$$\begin{aligned} \frac{1}{\tilde{\beta }_{[n]}(s_1)} - \frac{1}{\beta _{[n]}(s_0)}\ge & {} \frac{1}{\tilde{\beta }_{[n]}(s_1)} - \frac{1}{\tilde{\beta }_{[n]}(s_1)+1}\\\ge & {} \frac{1}{\tilde{\beta }_{[n]}(t')-1} - \frac{1}{\tilde{\beta }_{[n]}(t')} \\= & {} \frac{1}{\beta _{[n]}(t')} - \frac{1}{\tilde{\beta }_{[n]}(t')}, \end{aligned}$$

hence player j’s gain is at least as large as the maximal loss. Again, together with Claim 3, it follows that also in Case (ii) player j’s payoff, when going from $\beta$ to $\tilde{\beta }$, cannot decrease.

By repeating the argument of part (b), by $(**)$ we end up with the profile resulting from player j playing the lrm-$\hat{r}$ action, while in each step player j’s payoff cannot decrease. Together with $(*)$, this completes the proof of the theorem.$\square$