A bargaining procedure leading to the serial rule in games with veto players

Abstract

This paper studies an allocation procedure for coalitional games with veto players. The procedure is similar to the one presented by Arin and Feltkamp (J Math Econ 43:855–870, 2007), which is based on Dagan et al. (Games Econ Behav 18:55–72, 1997). A distinguished player makes a proposal that the remaining players must accept or reject, and conflict is solved bilaterally between the rejector and the proposer. We allow the proposer to make sequential proposals over several periods. If responders are myopic maximizers (i.e. consider each period in isolation), the only equilibrium outcome is the serial rule of Arin and Feltkamp (Eur J Oper Res 216:208–213, 2012) regardless of the order of moves. If all players are fully rational, the serial rule still arises as the unique subgame perfect equilibrium outcome if the order of moves is such that stronger players respond to the proposal after weaker ones.

Appendix

1.1 Proof of Theorem 2

We have already shown that $\phi (N,v)$ can be achieved with balanced proposals. It remains to show that the proposer cannot improve upon $\phi$. Let $z=\sum _{t=1}^{T}x^{t}$ be an outcome resulting from balanced proposals. Our objective is to show that $z_{1}\ge \phi _{1}$ implies $z_{i}\ge \phi _{i}$ for all $i$. This result, together with the efficiency of the serial rule, leads to $z=\phi$ being the unique MBRE outcome. We start by establishing the result not for the original game $(N,v)$, but for the sequence of auxiliary games $(N,w^{t})$ (Lemma 9). We then check that the sum of the serial rules of the games $w^{t}$ cannot exceed the serial rule of the original game $(N,v)$ (Lemma 10).

We start with the following property of balanced proposals:

Lemma 6

If $x^{t}$ is a balanced proposal, any player $i$ with $x_{i}^{t}>0$ will be a veto player at $t+1$.

This is because if $x^{t}$ is balanced we have $f_{1i}(x^{t},v^{t})=x_{i}^{t}$, so that all coalitions that have a positive $v^{t}$ but do not involve $i$ have $v^{t}(S)<x^{t}(S)$. Thus, after the payoffs $x^{t}$ are distributed any coalition with positive value must involve $i$.

The following lemma establishes a relationship between balanced proposals in $G^{T}(N,v)$ and the serial rule. Suppose $x^{t}$ is a balanced proposal in period $t$. Consider the game $w^{t}$, where $w^{t}(S)=\min \{v^{t}(S),x^{t}(N)\}$. The serial rule of $w^{t}$ and the balanced proposal $x^{t}$ do not coincide in general. However, the set of players who receive a positive payoff in $x^{t}$ coincides with the set of players who receive a positive payoff according to the serial rule of $w^{t}$.¹⁰

Lemma 7

Let $(N,v)$ be a veto balanced TU game. Consider the associated game $G^{T}(N,v)$. Let $z=\sum _{t=1}^{T}x^{t}$ be the outcome resulting from some strategy profile with balanced proposals. Consider period $t$, its outcome $x^{t}$ and the game $(N,w^{t})$ where $w^{t}(S)=\min \{v^{t}(S),x^{t}(N)\}$. Then it holds that $x_{k}^{t}>0$ if and only if $\phi _{k}(N,w^{t})>0.$

Proof

• (a) If $x_{k}^{t}>0$ we need to prove that $d_{k}(N,w^{t})<x^{t}(N)$, so that the serial rule of $w^{t}$ assigns a positive payoff to $k$.

  Let $S\in \arg \max _{T\subseteq N\backslash \{k\}}v^{t}(T)$. Since $x^{t}$ is balanced we have $f_{1k}(x^{t})=x_{k}^{t}>0$ and that implies $x^{t}(S)>v^{t}(S)$ (otherwise $S$ could have been used to complain against $k$). Hence, $x^{t}(N)\ge x^{t}(S)>v^{t}(S)=d_{k}(v^{t})=d_{k}(w^{t})$, where the last equality follows from Lemma 6.¹¹

• (b) If $x_{k}^{t}=0$ we need to prove that $d_{k}(N,w^{t})=x^{t}(N)$. Since $x^{t}$ is balanced, $f_{1k}(x^{t},v^{t})\le 0$. Let $P$ be a coalition associated to $f_{1k}(x^{t},v^{t})$. Because $f_{1k}(x^{t},v^{t})\le 0$, $x^{t}(P)\le v^{t}(P)$. Coalition $P$ must contain all players receiving a positive payoff at $x^{t}$ (otherwise $x^{t}$ is not balanced since $P$ can be used against any player outside $P$). Therefore $x^{t}(N)=x^{t}(P)\le v^{t}(P)$. Because of the way $w^{t}$ is defined it cannot exceed $x^{t}(N)$ , so $x^{t}(N)=w^{t}(P)=d_{k}(w^{t})$ and $k$ receives 0 according to the serial rule of $w^{t}$.$\square$

The following lemma concerns a property of the serial rule. By definition, the serial rule is such that $d_{k}$ is divided among players $\{j\in N,j<k\}$. Above $d_{k}$, player $k$ and any player $j<k$ get the same payoff.

Lemma 8

For any player $k$ we have $\sum _{i\in \{1,2,\ldots ,k-1\}}\phi _{i}=d_{k}+(k-1)\phi _{k}.$ Hence, $\sum _{i\in \{1,2,\ldots ,k-1\}}\phi _{i}\ge d_{k}+\phi _{k}$. The latter inequality is strict except if $k=2$ or $\phi _{k}=0$.

The next lemma tell us that, given a strategy profile with balanced proposals, the proposer cannot get more than the serial rule of the games $w^{t}$.

Lemma 9

Let $(N,v)$ be a veto balanced TU game. Consider the associated game with $T$ periods $G^{T}(N,v)$. Let $z=\sum _{t=1}^{T}x^{t}$ be an outcome resulting from balanced proposals. Consider period $t$, its outcome $x^{t}$ and the game $(N,w^{t})$ where $w^{t}(S)=\min \{v^{t}(S),x^{t}(N)\}$. Then $x_{1}^{t}\ge \phi _{1}(N,w^{t})$ implies $x_{l}^{t}\ge \phi _{l}(N,w^{t})$ for all $l\in N.$

Proof

Let $T$ be the set of veto players in $(N,w^{t})$, and let $M=\{l_{1},\ldots ,l_{m}\}$ be the ordered (according to the $d$-values of $(N,w^{t})$) set of nonveto players that have received a positive payoff at $x^{t}$. That is, $d_{l_{1}}\le \ldots \le d_{l_{m}}$.¹²

Suppose $x_{1}^{t}\ge \phi _{1}(N,w^{t})$. Since $x^{t}$ is balanced, $x_{1}^{t}=x_{i}^{t}$ for all $i\in T$, thus if $x_{1}^{t}\ge \phi _{1}(N,w^{t})$ it follows that $x_{i}^{t}\ge \phi _{i}(N,w^{t})$ for all $i\in T$.

We now want to prove that $x_{i}^{t}\ge \phi _{i}(N,w^{t})$ for all $i\in M$ . We will do it by induction.

Consider the responder $l_{1}$. Since $x^{t}$ is balanced we have $f_{1l_{1}}(x^{t})=x_{l_{1}}^{t}$. If the coalition associated to $f_{1l_{1}}$ has a value of 0, it follows that $x_{1}^{t}=x_{l_{1}}^{t}$ so $x_{l_{1}}^{t}\ge \phi _{l_{1}}(N,w^{t})$. If the coalition 1 is using has a positive value, all veto players must be in it, so its payoff must be at least $|T|\phi _{1}(N,w^{t})$, and its value (by definition of $d_{l_{1}}$) cannot exceed $d_{l_{1}}$. Hence, $f_{1l_{1}}(x^{t})\ge |T|\phi _{1}(N,w^{t})-d_{l_{1}}$. Because of Lemma 8, $|T|\phi _{1}(N,w^{t})-d_{l_{1}}\ge \phi _{l_{1}}(N,w^{t})$.

Now suppose the result $x_{i}^{t}\ge \phi _{i}(N,w^{t})$ is true for all $i\in \{l_{1},\ldots ,l_{k-1}\}$. We will prove that $x_{l_{k}}^{t}\ge \phi _{l_{k}}(N,w^{t})$. Let $S$ be a coalition such that $f_{1l_{k}}(x^{t})=x^{t}(S)-v^{t}(S)$. As before, the result follows immediately if $v^{t}(S)=0$. If $v^{t}(S)>0$ it must be the case that $T\subseteq S$, but $S$ need not involve all players in $\{l_{1},\ldots ,l_{k-1} \}.$ Denote $\{l_{1},\ldots ,l_{k-1}\}$ by $Q$. We consider two cases, depending on whether $Q\subseteq S$.

If $Q\subseteq S$, we have $x_{l_{k}}^{t}=f_{1l_{k}}(x^{t})=x^{t}(S)-v^{t}(S)\ge \sum _{i\in T\cup Q}\phi _{i}(N,w^{t})-d_{l_{k}}$, where the last inequality uses the induction hypothesis. The set $T\cup Q$ contains all players with $d_{i}<d_{l_{k}}$. Hence, by Lemma 8, $\sum _{i\in T\cup Q}\phi _{i}(N,w^{t})-d_{l_{k}}\ge \phi _{l_{k}}(N,w^{t})$.

If $Q\nsubseteq S$, there is a player $l_{p}<l_{k}$ such that $l_{p}\notin S$ . Because $x^{t}$ is a balanced proposal, $x_{l_{p}}^{t}= f_{1l_{p}}(x^{t})$. Because the veto player can use $S$ to complain against $l_{p}$, $f_{1l_{p}}(x^{t})\le f_{1l_{k}}(x^{t})=x_{l_{k}}$, hence $x_{l_{p}}\le x_{l_{k}}$. By the induction hypothesis, $x_{l_{p}}\ge \phi _{l_{p}}(N,w^{t})$. Since $d_{l_{p}}\le d_{l_{k}}$ we also know that $\phi _{l_{p}}(N,w^{t})\ge \phi _{l_{k}}(N,w^{t})$, so that

$$\begin{aligned} x_{l_{k}}=f_{1l_{k}}(x^{t})\ge f_{1l_{p}}(x^{t})=x_{l_{p}}\ge \phi _{l_{p}}(N,w^{t})\ge \phi _{l_{k}}(N,w^{t}). \end{aligned}$$

So far we have discussed the set of veto players and the set of nonveto players that are getting a positive payoff in $x^{t}$. For players in $N\backslash (T\cup M)$, we have shown in Lemma 7 that $x_{j}^{t}=0$ implies $\phi _{j}(N,w^{t})=0$, hence $x_{j}^{t}\ge \phi _{j}(N,w^{t})$ for all players. $\square$

Corollary 3

If $z=\sum _{t=1}^{T}x^{t}$ is an outcome resulting from balanced proposals, $x_{1}^{t}\ge \phi _{1}(N,w^{t})$ implies $x_{l}^{t}=\phi _{l}(N,w^{t})$ for all $l\in N.$

This corollary follows directly from Lemma 9 and the efficiency of the serial rule. Lemma 9 states that $x_{1}^{t}\ge \phi _{1}(N,w^{t})$ implies $x_{l}^{t}\ge \phi _{l}(N,w^{t})$ for all $l\in N$. By definition of $w^{t}$, $\sum _{l\in N}x_{l}^{t}=w^{t}(N)$. By the efficiency of the serial rule, $\sum _{l\in N}\phi _{l}(N,w^{t})=w^{t}(N)$. Hence, the only way in which player 1 can obtain the serial rule of $(N,w^{t})$ with balanced proposals is that all players in the game obtain their serial rule payoff.

Finally, the sum of the serial rules of the games $w^{t}$ cannot exceed the serial rule of the original game. This is due to the following property of the serial rule:

Lemma 10

Consider the veto balanced TU game $(N,v)$ and a finite set of positive numbers $\left( a_{1},\ldots ,a_{k}\right)$ such that $\underset{l=1}{ \overset{k}{\sum }}a_{l}=v(N)$. Consider the following TU games: $\left( N,w^{1}\right)$, $(N,w^{2})$, ..., $(N,w^{k})$, where

$$\begin{aligned}&w^{1}(S) :=\left\{ \begin{array}{ll} a_{1} &{}\quad \text {if }S=N \\ \min \{a_{1},v(S)\} &{}\quad \text {otherwise} \end{array} \right. \\&w^{2}(S) :=\left\{ \begin{array}{ll} a_{2} &{}\quad \text {if }S=N \\ \min \left\{ a_{2},\max \left[ 0,v(S)-\sum _{i\in S}\phi _{i}\left( N,w^{1}\right) \right] \right\} &{}\quad \text {otherwise} \end{array} \right. \\&w^{l}(S) :=\left\{ \begin{array}{ll} a_{l} &{}\quad \text {if }S=N \\ \min \left\{ a_{l},\max \left[ 0,v(S)-\sum _{m=1}^{l-1}\sum _{i\in S}\phi _{i}\left( N,w^{m}\right) \right] \right\} &{}\quad \text {otherwise} \end{array} \right. \end{aligned}$$

Then $\phi (N,v)=\sum _{m=1}^{k}\phi (N,w^{m}).$

Sketch of proof

In the lemma, we take $v(N)$ and divide it in $k$ positive parts, where $k$ is a finite number. We then compute the serial rule for each of the $k$ games, and see that each player gets the same in total as in the serial rule of the original game. The $k$ games are formed as follows: $w^{l}(N)$ is always $a_{l}$ for each $l:1,\ldots ,k$; the other coalitions have $v(S)$ minus what has been distributed so far according to the serial rule of the previous games, unless this would be negative (in which case the value is 0) or above $w^{l}(N)$ (in which case the value is $a_{l}$). Denote by $d_{i}^{l}$ the $d$-value for player $i$ associated to the auxiliary game $(N,w^{l})$.

The idea of the proof is that player $i$ cannot get anything until $d_{i}$ is distributed and becomes a veto player from then on. For example, if $a_{1}\le d_{i}$, any coalition $S$ associated to $d_{i}$ has $w^{1}(S)=\min (v(S),a_{1})=\min (d_{i},a_{1})=a_{1}$. Hence, $d_{i}^{1}=a_{1}=w^{1}(N)$ and player $i$ gets nothing according to the serial rule of $w^{1}$. The entire payoff $a_{1}$ will go to players with $d_{j}^{1}<a_{1}$. Such players must have $d_{j}<a_{1}\le d_{i}$ and must belong to any $T$ such that $v(T)>a_{1}$ (otherwise $d_{j}^{1}\ge a_{1}$). Hence, any coalition with positive value in $w^{2}$ will have $w^{2}(T)=\min (v(T)-a_{1},a_{2})$ and any player who got a positive payoff according to $\phi (N,w^{1})$ will be a veto player from $w^{2}$ onwards.

The same holds for $w^{2}$ if $a_{1}+a_{2}\le d_{i}$. Any coalition $S$ with $v(S)=d_{i}$ has $w^{2}(S)=\min (v(S)-a_{1}$, $a_{2})=a_{2}$, hence $d_{i}^{2}=a_{2}=w^{2}(N)$ and player $i$ gets nothing according to $\phi (N,w^{2})$. The entire $a_{2}$ goes to players with lower $d_{j}^{2}$ -values, and these players must have $d_{j}<a_{1}+a_{2}\le d_{i}$ and must belong to any coalition $T$ such that $v(T)-a_{1}>a_{2}$. Hence, any coalition with positive value in $w^{3}$ will have $w^{3}(T)=\min (v(T)-a_{1}-a_{2},a_{3})$ and any player who got a positive payoff in $\phi (N,w^{2})$ will be a veto player from $w^{3}$ onwards.

Let $l$ be the first auxiliary game for which $a_{1}+ \cdots +a_{l}>d_{i}$. At this point, the value of any coalition $S$ associated to $d_{i}$ is $w^{l}(S)=\min (v(S)-\sum _{t=1}^{l-1}a_{t},a_{l})=v(S)-\sum _{t=1}^{l-1} a_{t}<a_{l}=w^{l}(N)$ . The value $d_{i}^{l}=w^{l}(S)$ will still be divided between the other players, but above that value player $i$ will receive a share. Note that $\sum _{t=1}^{l}d_{i}^{t}=d_{i}$. Hence, player $i$ finds that $d_{i}$ is divided (not necessarily equally) between players with a lower index. Any player with $\phi _{j}(N,w^{l})>0$ must have $d_{j}^{l}<a_{l}$, and thus must belong to any coalition $T$ such that $v(T)-\sum _{t=1}^{l-1}a_{t}>a_{l}$ . Hence, any $T$ with $w^{l+1}(T)>0$ will have $w^{l+1}(T)=\min (v(T)-\sum _{t=1}^{l}a_{t},a_{l+1})$ and any $j$ with $\phi _{j}(N,w^{l})>0$ -including player $i$- will be a veto player from $w^{l+1}$ onwards and will therefore receive a share of the payoff.

We see that players receive a positive share if and only if $d_{i}$ has already been distributed. Above $d_{i}$, the serial rules of the auxiliary games ensure that the payoff is evenly divided between the players who are getting a positive share. This process replicates the serial rule of the original game. For the same reason, if $\sum _{l=1}^{k}a_{l}<v(N),$ the proposer will get less than $\phi _{1}(N,v)$. $\square$

Note that Lemma 10 refers to a sequence of TU games such that each game is obtained after distributing the serial rule payoffs for the previous game; the games $w^{t}$ in Lemma 9 are obtained by subtracting balanced proposals from $w^{t-1}$. It turns out that the TU games involved are identical in both cases: the sequence $w^{t}$ depends only on the total amounts distributed $x^{1}(N),\ldots ,x^{T}(N)$ (denoted by $a_{1}, \ldots , a_{k}$ in Lemma 10). This is because the set of players that get a positive payoff at period $t$ is the same in both cases (Lemma 7) and all these players become veto players at period $t+1$ (Lemma 6). Hence, any coalition with positive value at $t$ has $w^{t}(S)=\min (w^{t-1}(S)-x^{t-1}(N),x^{t}(N))$ in both cases.

Putting the above lemmas together we can prove Theorem 2. First, any payoff player 1 can achieve in a MBRE can be achieved by balanced proposals (Lemma 4). Second, given that proposals are balanced, the payoff player 1 can get cannot exceed the sum of the serial rules of the games $w^{t}$ (Lemma 9). Since the sum of the serial rules of the games $w^{t}$ cannot exceed the serial rule of the original game (Lemma 10), player 1 can never get more than $\phi _{1}(N,v)$ in a MBRE. Also, player 1 can only get $\phi _{1}(N,v)$ if all other players get their serial rule payoff (Corollary 3). Finally, $\phi (N,v)$ is achievable by the sequential proposals described in Lemma 5.

Note that the assumption $T\ge n$ only plays a role in Lemma 5. For time horizons shorter than $n-1$, all auxiliary results still hold but player 1 may not be able to achieve a payoff as high as $\phi _{1}(N,v).$

1.2 Proof of Theorem 3

We start by pointing out some immediate consequences of the results in Sect. 3.3.

Corollary 4

Let $(N,v)$ be a veto balanced TU game and $G^{T}(N,v)$ its associated extensive form game with $T\ge n$. Let $z={\sum }_{{t=1}}^{{T}}x^{t}$ be an outcome resulting from some SPE of the game $G^{T}(N,v).$ If $z$ differs from $\phi (N,v)$ then $z_{1}$ cannot be achieved by making balanced proposals.

Suppose to the contrary that there is an SPE outcome $z$ that differs from $\phi (N,v)$ and can be achieved by balanced proposals. If $z$ differs from $\phi (N,v)$, $z_{1}\ge \phi _{1}(N,v)$ (otherwise the proposer would prefer to play the strategy described in Lemma 5, which is available since $T\ge n$). If $z$ is achievable by balanced proposals, it is (trivially) achievable under myopic responder behavior. However, under myopic behavior of the responders, the proposer can only achieve at least $\phi _{1}(N,v)$ if all players are getting their serial rule payoffs, that is, if $z=\phi (N,v)$ (Theorem 2), a contradiction.

Corollary 5

Let $(N,v)$ be a veto balanced TU game and $G^{T}(N,v)$ its associated extensive form game with $T\ge n$. Let $z={\sum }_{{t=1}}^{T}x^{t}$ be an outcome resulting from some SPE of the game $G^{T}(N,v).$ If $z$ differs from $\phi (N,v)$ then there exists at least one period $t$ and one player $p$ for which $f_{1p}(x^{t},(N,v^{t}))>x_{p}^{t}\ge 0.$

This is because if $x_{l}^{t}\ge f_{1l}(x^{t},v^{t})$ for all $l$ and $t$, $z$ would be achievable under myopic behavior of the responders by proposing $x^{t}$ in each period $t$ (Lemma 3), but we know from Theorem 2 that the proposer can only get at least $\phi _{1}(N,v)$ if all players are getting their serial rule payoffs.

Since myopic behavior always leads to $f_{1i}(x^{t},v^{t})\le x_{i}^{t}$ for all $i$ and all $t$ (Lemma 3), the presence of a player $p$ for which $f_{1p}(x^{t},(N,v^{t}))>x_{p}^{t}$ indicates non-myopic responder behavior. This responder may be player $p$ (nonmyopically accepting a proposal), or a responder moving after $p$ (nonmyopically rejecting a proposal and transferring payoff to player 1). In Example 3, player 3 rejects a proposal nonmyopically and as a result creates the inequality $f_{12}(x^{t},v^{t})>x_{2}^{t}$ at the end of period $t=4$.

We are now ready to prove Theorem 3. We will show in Lemma 14 that, if responders move following the order of nonincreasing $d$-values of $v^{t}$, any SPE outcome $z$ is such that the proposer can obtain $z_{1}$ by making balanced proposals. Since any SPE outcome $z$ different from $\phi (N,v)$ would be unachievable with balanced proposals according to Corollary 4, this will complete the proof.

In order to prove Lemma 14 we need several auxiliary results.

We denote by $x^{t,i}$ the proposal that emerges in period $t$ immediately after $i$ gets the move. The following lemma establishes a property of $x^{t,i}$ that must be inherited by the final outcome in period $t$, $x^{t}$.

Lemma 11

Suppose after player $i$ responds to the proposal in period $t$ it holds that $f_{1i}(x^{t,i},v^{t})>0$. Then $f_{1i}(x^{t},v^{t})>0$ regardless of the responses of the players moving after $i$.

Proof

Suppose by contradiction that $f_{1i}(x^{t},v^{t})\le 0$. This means that at the end of period $t$ there is a coalition $S^{*}$ such that $i\notin S^{*}$ and $v^{t}(S^{*})\ge x^{t}(S^{*})$. Because $f_{1i}(x^{t,i},v^{t})>0$ immediately after $i$ responds to the proposal, all coalitions excluding $i$ had a positive satisfaction at that point, and in particular $v^{t}(S^{*})<x^{t,i}(S^{*})$. There must be a player $h$ moving after $i$ such that $h\notin S^{*}$ and $h$ has received a payoff transfer from player 1 by rejecting the proposal. At the moment of rejection by $h$ we have $f_{1h}(x^{t,h},v^{t})=x_{h}^{t}>0$. However, since $S^{*}$ can be used by 1 to complain against $h$, at the end of period $t$ we have $f_{1h}(x^{t},v^{t})\le 0$. There must be another player $l$ moving after $h$ that has received a payoff transfer from player 1, and this player cannot be in $S^{*}.$ Then this player is in the same situation as player $h$: he has $f_{1l}(x^{t,l},v^{t})>0$ at the moment of rejection, but at the end of period $t$ he has $f_{1l}(x^{t},v^{t})\le 0$. Thus there must be another player moving after him that has caused this change and would himself be in the same situation as player $h\ldots$ but the number of players is finite. $\square$

Note that Lemma 11 holds for any strategy profile, not necessarily an equilibrium.

The next auxiliary result will allow us to compare the equilibria of games with different characteristic functions. If one of the characteristic functions is “worse” than the other (in the sense of having lower values), player 1 might still have a greater SPE payoff, but only with nonmyopic responder behavior.

Lemma 12

Let $(N,v)$ and $(N,w)$ be two veto balanced games in which player 1 is a veto player. Let $w(S)\ge v(S)$ for any $S.$ Let $G^{T}(N,v)$ and $G^{T}(N,w)$ be the associated extensive form games with $T$ proposals. If the payoff provided to the proposer by an SPE outcome of the game $G^{T}(N,w)$ is strictly lower than the payoff provided to the proposer by an SPE outcome of the game $G^{T}(N,v)$, then the SPE outcome of $G^{T}(N,v)$ is such that $x_{l}^{t}<f_{1l}(x^{t},v^{t})$ for some $l$ and $t$, which implies that at least one responder is behaving nonmyopically.

Proof

By contradiction, suppose the final payoffs in the SPE outcome of $G^{T}(N,v)$ are such that $x_{l}^{t}\ge f_{1l}(x^{t},v^{t})$ for all $l$ and $t$.

Consider the strategy combination (not necessarily an equilibrium) of the game $G^{T}(N,w)$ in which the proposer makes the sequence of proposals $x^{t}$ and responders accept. Because by assumption $w(S)\ge v(S)$ for all $S$, $w^{t}(S)\ge v^{t}(S)$ for all $t$ given this strategy combination and thus $f_{1l}(x^{t},w^{t})\le f_{1l}(x^{t},v^{t})\le x_{l}^{t}$ for all $l$ and $t$.

Given that $f_{1l}(x^{t},w^{t})\le x_{l}^{t}$ for all $l$ and $t$ in this strategy profile, we can use Lemma 4 to construct a sequence of balanced proposals $y^{t}$ for the game $G^{T}(N,w)$ where the payoff of the proposer does not change, that is, $\sum _{t=1}^{T}y_{1}^{t}= \sum _{t=1}^{T}x_{1}^{t}$. Because the proposer always has the option of making the sequence of balanced proposals $y^{t}$ in the game $G^{T}(N,w)$, the proposer’s payoff in any SPE of game $G^{T}(N,w)$ must be at least $\sum _{t=1}^{T}y_{1}^{t}=\sum _{t=1}^{T}x_{1}^{t}$.

We have shown that if an SPE outcome of $G^{T}(N,v)$ is such that $x_{l}^{t}\ge f_{1l}(x^{t},v^{t})$ for all $l$ and $t,$ then the payoff obtained by the proposer in this outcome can also be obtained with balanced proposals (and therefore in any SPE) in $G^{T}(N,w)$. Hence, if the proposer is obtaining strictly less in an SPE of $G^{T}(N,w)$, it must be the case that $x_{l}^{t}<f_{1l}(x^{t},v^{t})$ for some $l$ and $t$. $\square$

The next auxiliary result provides a bound for the payoff difference between player 1 and player $i\ne 1.$

Lemma 13

Let $(N,v)$ be a veto balanced TU game. Consider the associated game with $T$ periods $G^{T}(N,v)$. Fix a period $l\in \{1,\ldots ,T\}$ and a subgame that starts in period $l$ (not necessarily on the equilibrium path), and label the responders according to the nondecreasing order of d-values in the game $v^{l}$. Let $y^{l}=\sum _{t=l}^{T}x^{t}$ be the vector of payoffs accumulated between $l$ and $T$. Then $y_{i}^{l}\ge y_{1}^{l}-d_{i}^{l}$ for all $i\in \{2,\ldots ,n\}$ in any SPE of $G^{T}(N,v)$. Moreover, the inequality is strict if $d_{i}^{l}>d_{2}^{l}$.

Proof

Note that, because period $T$ is the last period of the game, myopic and rational behavior coincide, so $x_{i}^{T}\ge f_{1i}(v^{T},x^{T})$ for all $i$.

Consider player 2. Since players play myopically in period $T$, it must be the case that¹³

$$\begin{aligned} x_{2}^{T}\ge & {} f_{12}(x^{T})\ge x_{1}^{T}-\left( d_{2}^{l}-\sum _{t=l}^{T-1}x_{1}^{t}\right) \\= & {} \sum _{t=l}^{T}x_{1}^{t}-d_{2}^{l}=y_{1}^{l}-d_{2}^{l}. \end{aligned}$$

Since $y_{2}^{l}\ge x_{2}^{T}$, it follows that $y_{2}^{l}\ge y_{1}^{l}-d_{2}^{l}.$

Now consider player $i\ne 2$. There are two possible cases, depending on whether $y_{1}^{l}\le d_{2}^{l}$.

If $y_{1}^{l}\le d_{2}^{l}$, the result follows immediately since

$$\begin{aligned} y_{1}^{l}-d_{i}^{l}\le y_{1}^{l}-d_{2}^{l}\le 0\le y_{i}^{l}. \end{aligned}$$

It is also clear that the inequality is strict if $d_{2}^{l}<d_{i}^{l}$.

From now on we assume $y_{1}^{l}>d_{2}^{l}$. Note that since we have already shown that $y_{2}^{l}\ge y_{1}^{l}-d_{2}^{l}$, it follows that that $y_{2}^{l}>0$ in this case. There are again two possible cases, depending on whether the coalition associated to $f_{1i}(x^{T},v^{T})$ contains 2.

If the coalition contains 2, we have

$$\begin{aligned} y_{i}^{l}\ge & {} x_{i}^{T}\ge f_{1i}\left( x^{T},v^{T}\right) \ge x_{1}^{T}+x_{2}^{T}-\left( d_{i}^{l}-\sum _{t=l}^{T-1}x_{1}^{t}-\sum _{t=l}^{T-1}x_{2}^{t}\right) \\= & {} y_{1}^{l}+y_{2}^{l}-d_{i}^{l}>y_{1}^{l}-d_{i}^{l}, \end{aligned}$$

where the last inequality follows from the fact that $y_{2}^{l}>0$.

If the coalition does not contain 2, we have $f_{1i}(x^{T})\ge f_{12}(x^{T})$. Then

$$\begin{aligned} y_{i}^{l}\ge x_{i}^{T}\ge f_{1i}\left( x^{T},v^{T}\right) \ge f_{12}\left( x^{T},v^{T}\right) \ge y_{1}^{l}-d_{2}^{l}\ge y_{1}^{l}-d_{i}^{l}. \end{aligned}$$

Note that the inequality is strict for $d_{i}^{l}>d_{2}^{l}$. $\square$

The main building block of the proof is the next lemma, which shows that, given the particular order of responders we impose, the proposer cannot do better than with balanced proposals.

Lemma 14

Let $(N,v)$ be a veto balanced TU game. Consider the associated game $G^{T}(N,v)$ in which the responders move following the order of nonincreasing $d$-values of $v^{t}.$ Let $z=\overset{T}{\underset{t=1}{\sum } }x^{t}$ be an outcome resulting from some SPE of the game $G^{T}(N,v).$ Then the proposer can obtain $z_{1}$ by making balanced proposals.

Proof

Suppose on the contrary that $z_{1}$ cannot be obtained with balanced proposals. By Lemma 4 we know that there is a player $k$ and a stage $t$ such that $f_{1k}(x^{t},v^{t})>x_{k}^{t}\ge 0$; otherwise the proposer can obtain $z_{1}$ with balanced proposals.

Let $t$ be the last period¹⁴ in which for some responder it holds that $f_{1k}(x^{t},v^{t})>x_{k}^{t}\ge 0.$ Let $k$ be the last responder at $t$ for whom $f_{1k}(x^{_{t}},v^{t})>x_{k}^{t}\ge 0$. We consider two cases:

Case (a) There is a player $p$ with $d_{p}^{t}\le d_{k}^{t}$ such that $f_{1p}(x^{t},v^{t})\le 0$. Note that $f_{1k}(x^{t},v^{t})>x_{k}^{t} \ge 0$ means that any coalition without player $k$ has a positive satisfaction and, in particular any coalition $S_{k}\in \arg \underset{ S\subseteq N\backslash \left\{ k\right\} }{\max }v^{t}(S).$ On the other hand since $f_{1p}(x^{t},v^{t})\le 0$ then there exists a coalition without player $p$ for which the satisfaction is not positive. Let $S_{p}^{*}$ be one such coalition (it must contain $k$). Then we have the following two inequalities:

$$\begin{aligned} x^{t}\left( S_{k}\right) >d_{k}^{t}\text { and }v^{t}\left( S_{p}^{*}\right) \ge x^{t}\left( S_{p}^{*}\right) . \end{aligned}$$

Combining the two inequalities we obtain

$$\begin{aligned} x^{t}\left( S_{k}\right) -x^{t}\left( S_{p}^{*}\right) >d_{k}^{t}-v^{t}\left( S_{p}^{*}\right) \ge d_{k}^{t}-d_{p}^{t}\ge 0. \end{aligned}$$

The inequality above implies that there are players not in $S_{p}^{*}$ receiving a positive payoff in period $t$.

Consider a new allocation, $y^{t}$, which is identical to $x^{t}$ except that $y_{i}^{t}=0$ for all $i$ not in $S_{p}^{*}$ (thus $y_{i}^{t}=x_{i}^{t}$ for all $i$ in $S_{p}^{*}$). We now show that $f_{1i}(y^{t},v^{t})\le y_{i}^{t}$ for all $i\in N$, so that player 1 can get the same payoff with balanced proposals by Lemma 4.

For any player $i$ it holds that $f_{1i}(y^{t},v^{t})\le f_{1i}(x^{t},v^{t})$.

Because $S_{p}^{*}$ can be used against any player outside $S_{p}^{*}$, for any player outside $S_{p}^{*}$ it holds that $f_{1i}(y^{t},v^{t})\le f_{1p}(y^{t},v^{t})\le f_{1p}(x^{t},v^{t})\le 0$. Thus, $f_{1i}(y^{t},v^{t})\le y_{i}^{t}$ for all $i\notin S_{p}^{*}$.

We can also rule out the existence of a player $l\in \,S_{p}^{*}$ for which $f_{1l}(y^{t},v^{t})>y_{l}^{t}$. If there was such a player, this inequality would have already existed for $x^{t}$, since $f_{1l}(y^{t},v^{t})\le f_{1l}(x^{t},v^{t})$ and $y_{l}^{t}=x_{l}^{t}$. Since player $k$ is the last player satisfying the inequality for $x^{t}$, it must be the case that $d_{l}^{t}\ge d_{k}^{t}\ge d_{p}^{t}$, thus we can repeat the reasoning above with $S_{l}$ and $S_{p}^{*}$ and, given that nothing has changed for $S_{p}^{*}$, we would conclude that $y^{t}(S_{l})-y^{t}(S_{p}^{*})>0$, a contradiction since all players outside $S_{p}^{*}$ have zero payoffs. Thus, $f_{1i}(y^{t},v^{t})\le y_{i}^{t}$ for all $i$.

Note that for this part of the proof no assumption is needed about the order in which the responders move.¹⁵

Case (b) The second case is $f_{1l}(x^{t},v^{t})>0$ for all players moving after $k$ at $t$. By assumption, on the equilibrium path from $t+1$ onwards all proposals have an associated balanced proposal. We distinguish two subcases:

(b1) The last player to act nonmyopically at $t$ has accepted a proposal. This player must be player $k$ or a player moving after $k$. Call this player $p$ ($p$ moving after $k$ is possible if a myopic rejection by a player moving after $p$ has restored $f_{1p}\le x_{p}$).

We will show that it is not in $p$’s interest to accept the proposal. To do this, we need to analyze two subgames: the subgame on the equilibrium path in which $p$ accepts the proposal, and the subgame off the equilibrium path in which $p$ rejects the proposal. We will talk about the $A$-path (the equilibrium path) and the $R$-path. Denote by $x^{A,t}$ and $x^{R,t}$ the final payoffs in period $t$ depending on whether player $p$ accepts or rejects the proposal. If player $p$ rejects the proposal, we take any subgame perfect equilibrium of that subgame. Denote by $v^{A,t+1}$ and $v^{R,t+1}$ the corresponding TU games at $t+1$.

Because $f_{1l}(x^{t},v^{t})>0$ for $l\in \{2,\ldots ,p\}$, on the $A$-path all players in $\{2,\ldots ,p\}$ are veto players at $t+1$.

The game $v^{R,t+1}$ is better than the game $v^{A,t+1}$ (in the sense of Lemma 12). If $v^{A,t+1}(S)>0$, coalition $S$ must contain all players in $\{1,2,\ldots ,p\}$. For this kind of coalition $v^{R,t+1}(S)=v^{A,t+1}(S)$, since any payoff transfers after rejection occur between members of $\{1,\ldots ,p\}$ (here the order of moves is essential). Thus, $v^{A,t+1}(S)\le v^{R,t+1}(S)$ for all $S$.

We now show that $p$ is also veto at $t+1$ on the $R$-path.

Suppose $p$ is not veto at $t+1$ on the $R$-path. Then there is a coalition $S_{p}$ such that $p\notin S_{p}$ and $v^{R,t+1}(S_{p})>0$. This can only happen if $v^{R,t}(S_{p})>x^{R,t}(S_{p})$, or equivalently $f_{1p}(x^{R,t},v^{t})<0$, contradicting Lemma 11.

Thus, player $p$ is a veto player at $t+1$ regardless of whether he accepts or rejects the proposal. Given the order of moves, veto players can secure at least the same payoff as the proposer. There is no reason for veto players to act nonmyopically because the game at $t+2$ will be the same regardless of how the payoff is distributed at $t+1$ between veto players; no payoff can go to anyone else given the order of responders. For the same reason the proposer will never make a proposal that gives another veto player more than he gets himself, so that all veto players must get the same payoff given the order of moves. Let $y_{1}^{R}$ be player 1’s payoff if $p$ rejects the proposal (this is the payoff accumulated between periods $t+1$ and $n$) and $y_{1}^{A}$ be player 1’s payoff if $p$ accepts the proposal. Because of Lemma 12, the only way in which $y_{1}^{A}$ can exceed $y_{1}^{R}$ is if there is a nonmyopic move at $v^{A,t+1}$ that leads to $f_{1i}(x^{A,t+1},v^{A,t+1})>x_{i}^{A,t+1}$ for some $i$. By assumption this is not the case. Thus, it was not in $p$’s interest to accept: rejecting would yield a higher payoff at $t$, and at least the same payoff in the rest of the game.

b2) The last player to act nonmyopically at $t$ has rejected a proposal. Let $p$ be the last player to act nonmyopically at $t$. This player cannot be player $k$ because after any rejection (myopic or otherwise) it holds that $f_{1k}(\cdot )\le x_{k}^{t}$, and given that the remaining responders act myopically this inequality would never be reversed. Someone moving after $k$ must have rejected a proposal nonmyopically (transferring payoff to the proposer) and created the inequality $f_{1k}(x^{t},v^{t})>x_{k}^{t}$, hence player $p$ must be moving after $k$. We will show that it is not in $p$’s interest to reject the proposal. To do this, we need to analyze two subgames: the subgame on the equilibrium path in which $p$ rejects the proposal, and the subgame off the equilibrium path in which $p$ accepts the proposal. We will talk about the $R$-path (the equilibrium path) and the $A$-path.

Because $f_{1l}(x^{t},v^{t})>0$ for $l\in \{2,\ldots ,k\}$, on the equilibrium path all players in $\{2,\ldots ,k\}$ are veto players at $t+1$.

The game $v^{A,t+1}$ is better than the game $v^{R,t+1}$ (in the sense of Lemma 12). If $v^{R,t+1}(S)>0$, coalition $S$ must contain all players in $\{1,2,\ldots ,p\}$. For this kind of coalition $v^{R,t+1}(S)=v^{A,t+1}(S)$, since any payoff transfers on the $A$-path must occur between members of $\{1,2,\ldots ,p\}$ (again, here the order of moves is essential). Thus, $v^{R,t+1}(S)\le v^{A,t+1}(S)$ for all $S$.

Suppose player $p$ is veto also on the $A$-path. Then the reasoning of case b1 applies, and there is no reason for $p$ to act nonmyopically in period $t$ .

Now suppose player $p$ is not veto on the $A$-path at $t+1$. We define $d_{p}^{A,t+1}:=\underset{S\subseteq N\backslash \left\{ p\right\} }{\max } v^{A,t+1}(S)$. Since $p$ is not a veto player at $t+1$ on the $A$-path, $d_{p}^{A,t+1}>0$. There is a coalition $S_{p}$ such that $v^{t}(S_{p})-x^{A,t}(S_{p})=d_{p}^{A,t+1}>0$. Since by assumption $f_{1p}(x^{t},v^{t})>0$ on the $R$-path, we also have $x^{R,t}(S_{p})-v^{t}(S_{p})>0$. From the two inequalities we get $x^{R,t}(S_{p})-x^{A,t}(S_{p})>d_{p}^{A,t+1}.$

Let $\alpha$ be the payoff player $p$ transfers to player 1 when rejecting the proposal (part of this payoff may then go to other players between 2 and $p-1$ if they myopically reject a proposal). We want to show that $\alpha \ge x^{R,t}(S_{p})-x^{A,t}(S_{p})$, which implies $\alpha >d_{p}^{A,t+1}$. This is not completely obvious because part of the difference could be due to a player outside $S_{p}$ myopically rejecting on the $A$-path.

Claim. $\alpha \ge x^{R,t}(S_{p})-x^{A,t}(S_{p})>d_{p}^{A,t+1}$.

Note that $x^{R,t}$ and $x^{A,t}$ are identical for all responders moving before $p$. Because of the order of moves, any payoff transfers due to a change of $p$’s action from R to A occur within the set $\{1,2,\ldots ,p\}$. Denote by $T$ the set $\{2,\ldots ,p-1\}\backslash S_{p}$. We can write $x^{R,t}(S_{p})+x^{R,t}(T)=x^{A,t}(S_{p})+x^{A,t}(T)+\alpha$. All we need to show is that $x_{h}^{R,t}\ge x_{h}^{A,t}$ for all $h\in T$ (this is obvious if $T$ is empty). This implies $x^{R,t}(T)\ge x^{A,t}(T)$ and hence $x^{R,t}(S_{p})-x^{A,t}(S_{p})\le \alpha$.

Suppose there is a player $h$ in $T$ that has $x_{h}^{R,t}<x_{h}^{A,t}$. Since this player is not in $S_{p}$, $S_{p}$ can be used by player 1 against him.

Player $h$ must have rejected in the $A$-path (if he had accepted he would have $x_{h}^{R,t}\ge x_{h}^{A,t}$).¹⁶ After rejecting he is left with $f_{1h}(\cdot )=x_{h}^{A,t}>0$. On the other hand, $v^{t}(S_{p})-x^{A,t}(S_{p})>0$. Thus, $f_{1h}(x^{A,t},v^{t})<0$, contradicting Lemma 11.

Now we are in a position to compare payoffs on the $A$ and $R$-paths and see that $p$ prefers to accept the proposal.

Since player $p$ is veto on the $R$-path, he gets $y_{1}^{R}$. On the $A$ -path, he gets $y_{p}^{A}$, whereas the proposer gets $y_{1}^{A}$. Because Lemma 13 applies to all subgames regardless of whether they are on the equilibrium path, $y_{p}^{A}\ge y_{1}^{A}-d_{p}^{A,t+1}$.

In order to have an equilibrium, $p$ must prefer to reject the proposal, thus we need $x_{p}^{t}+y_{1}^{R}\ge x_{p}^{t}+\alpha +y_{p}^{A}>x_{p}^{t}+d_{p}^{A,t+1}+y_{1}^{A}-d_{p}^{A,t+1}$. Therefore we need $y_{1}^{R}>y_{1}^{A}$. Since the game $v^{A,t+1}$ is better than the game $v^{R,t+1}$ (in the sense of Lemma 12), and by assumption $f_{1i}(x^{l},v^{l})>x_{i}^{l}$ never happens on the $R$-path from $t+1$ onwards, the inequality $y_{1}^{R}>y_{1}^{A}$ cannot hold. $\square$

We have shown that any SPE outcome is such that the proposer can always achieve $z_{1}$ with balanced proposals. From the previous section we know that the best outcome the proposer can achieve with balanced proposals is $\phi (N,v)$, hence this completes the proof.