Revising claims and resisting ultimatums in bargaining problems

Johannes Spinnewijn; Frans Spinnewyn

doi:10.1007/s10058-015-0168-7

Review of Economic Design

June 2015, Volume 19, Issue 2, pp 91–116 | Cite as

Revising claims and resisting ultimatums in bargaining problems

Authors
Authors and affiliations

Johannes SpinnewijnEmail author
Frans Spinnewyn

Original Paper

First Online: 14 February 2015

Received: 20 July 2012

Accepted: 03 February 2015

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Abstract

We propose a simple mechanism which implements a unique solution to the bargaining problem with two players in subgame-perfect equilibrium. The mechanism incorporates two important features of negotiations; players can revise claims in an attempt to reach a compromise or pursue their claims in an ultimate take-it-or-leave-it offer. Players restrain their claims to avoid a weak bargaining position or their resistance to uncompromising behavior to acquire leadership. The Nash solution and the Kalai–Smorodinsky solution are implemented in the extreme cases when respectively no and all revisions are allowed.

Keywords

Bargaining solutions Nash program Ultimatums

JEL Classification

C78 D74

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Appendix

The subgame-perfect equilibrium

Let $\sigma =\left( \sigma _{1},\sigma _{2}\right) $ be a strategy profile in $\Gamma $ for $m\in \mathcal {M}.$ The history $h^{\tau -1}\in \mathcal {H} ^{\tau -1}$ at stage $\tau =1,\ldots ,4$ is recursively defined by $h^{\tau }=\left( a^{\tau },h^{\tau -1}\right) $ and $h^{0}\in \emptyset $, where $ a^{1}=p\in D$, $a^{2}=q\in D$, $a_{L}^{3}=c\in C\left( p\right) $ and $ a_{F}^{4}\in ${Y,N}. The strategy of player $i$ at stage $\tau $ in the subgame for the history $h^{\tau -1}$ in $\sigma $ is denoted by $ a_{i}^{\tau ,\sigma }(h^{\tau -1})$. We denote by $\sigma $ the strategy profile in subgame-perfect equilibrium in $\Gamma $.

Assuming that $F$ accepts a compromise in a tie when $c_{F}=(1-q_{L})p_{F}$, by the definition of the risk limit for $h^{3}\in \mathcal {H}^{3}$,

$$\begin{aligned} a_{F}^{4,\sigma }(h^{3})=\left\{ \! \begin{array}{l@{\quad }l} \text {Y} &{} \text {if }q_{L}\ge r_{F}(c_{F},p_{F})\text { and }p_{F}>0, \\ \text {N}&{} \text {otherwise.} \end{array} \right. \end{aligned}$$

The leader $L$ proposes the compromise $c\in C\left( p\right) $, which is accepted by $F$ and gives $L$ the largest payoff, so that for $h^{2}\in \mathcal {H}^{2}$,

$$\begin{aligned} a_{L}^{3,\sigma }\left( h^{2}\right) \in \arg \max _{u\in C\left( p\right) }\left\{ \left. c_{L}\right| a_{F}^{\sigma }(c,h^{2})=\text {Y}\right\} . \end{aligned}$$

Since $r_{F}(.,p_{F})$ is decreasing, $c=a_{L}^{3,\sigma }\left( h^{2}\right) \in PO\left( S\right) $ for $q_{L}=r_{F}(c_{F},p_{F})$. The choices $a^{2,\sigma }(h^{1})$ for $h^{1}\in \mathcal {H}^{1}$ and $ a^{1,\sigma }(h^{0})$ in subgame-perfect equilibrium for $\Gamma ^{R}$ are given in Lemmas 1, 2 and Proposition 3.

Proofs of propositions

Proof of Proposition 1

The mechanism $\Gamma ^{M}$ belongs to the class of mechanisms considered in Proposition 3. In $\Gamma ^{M}$, $C_{1}\left( 1\right) =C_{2}\left( 1\right) =D$, so that $u^{KS}\left( \mathbf {1}\right) \in C\left( 1\right) $ for $ i\in N$ and that case $\left( i\right) $ of Proposition 3 holds. We refer to the proof of the first case of Lemma 1 to show that players make the bids $ q=q^{KS}\left( p\right) $ and the first case in Proposition 3 showing that $ u^{KS}\left( \mathbf {1}\right) $ is implemented in subgame-perfect equilibrium. $\square $

Proof of Proposition 2

The mechanism $\Gamma ^{N}$ belongs to the class of mechanisms considered in Proposition 3. In $\Gamma ^{N}$, $m_{-i}^{i}\left( p_{i}\right) =u_{-i}^{P}\left( p_{i}\right) $ for $i\in N$. The extended Nash product of a claim of a player in Proposition 3 is equal to the Nash product in that case. It is maximized for $\hat{p}_{i}=u_{i}^{N}$ for $i\in N$ and the maximized values are equal for the two players. For $p=\hat{p}=u^{N}$, $ u^{N}=m^{1}\left( p_{1}\right) =m^{2}\left( p_{2}\right) =u^{KS}(p).$ We refer to the proof of Lemma 1 for the bidding strategies with $ m_{1}^{1}=p_{1}$ and to the fourth case in Proposition 3 showing that $u^{N}$ is implemented in subgame-perfect equilibrium. $\square $

Proof of Proposition 3

We distinguish between four solutions in subgame-perfect equilibrium implemented in one of the following four cases.

In the first case, exception $\left( i\right) $ of Proposition 3 holds. The proportional solution is a feasible compromise for $p=\mathbf {1}$ and, by Lemma 1, $u^{KS}\left( \mathbf {1}\right) $ is implemented. Let $s$ be the strong player for $p=\mathbf {1}$. By Lemma 2, $u_{s}^{KS}\left( \mathbf {1} \right) $ for $p_{w}=1$ is a lower bound for $s$’s payoff. By its monotonicity, the proportional solution would remain feasible and would be implemented by Lemma 1 for a lower claim of player $w$, but would reduce $w$ ’s payoff. For $p_{w}=1,$ the payoff of player $s$ is bounded from above by $ u_{s}^{KS}\left( \mathbf {1}\right) $. Hence, $u=u^{KS}\left( \mathbf {1} \right) $ is the unique solution for $p=\mathbf {1}$ in the first case. In the remaining cases, the proportional solution is not feasible for $p= \mathbf {1}$.

In the second case, subcase $\left( a\right) $ or $\left( c\right) $ of $ \left( ii\right) $ of Proposition 3 holds. In subcase $\left( a\right) $, player 1 is strong for his maximal claim and obtains the payoff $\max \left\{ u_{1}^{KS}\left( 1,p_{2}\right) ,m_{1}^{1}\left( 1\right) \right\} $ by Lemma 1. Since he is strong for $p_{1}=1$ and for all claims of player 2, his claim is maximal in equilibrium by Lemma 2. In subcase $\left( c\right) $, player $s$ is strong for $\tilde{p}_{s}=1$ and the claim $\tilde{p}_{w}.$ He obtains the payoff $\max \left\{ u_{s}^{KS}\left( \tilde{p}\right) ,m_{s}^{s}\left( 1\right) \right\} $ by Lemma 1, which is a lower bound of his payoff for $\tilde{p}_{w}$ by Lemma 2. Remark that the conditions $ m^{s}\left( 1\right) =m^{w}\left( \tilde{p}_{w}\right) =u^{KS}(\tilde{p})$ uniquely define $\tilde{p}\ge \hat{p}$ by the properties of the proportional solution. In both subcases, $m^{s}\left( 1\right) $ is implemented iff $m_{s}^{s}\left( 1\right) \ge u_{s}^{KS}\left( p\right) $ for $p_{s}=1.$ The proportional solution remains feasible and would be implemented for claims below $p_{w}$ of the weak player, but would reduce his payoff below $m_{w}^{s}\left( 1\right) $. Hence $u=m^{s}\left( 1\right) $ is the unique solution implemented when the maximal revisions are incompatible or meet for $p_{s}=1$ in the second case.

In the third case, condition $\left( b\right) $ of $\left( ii\right) $ of Proposition 3 holds. The extended Nash products are equal for $p=\left( \check{p}_{1},\hat{p}_{2}\right) ,$ so that player 1 is strong for $p$. Since the proportional solution is not feasible for $p$, $m^{1}\left( \check{ p}_{1}\right) $ is implemented by Lemma 1. Since $p_{2}=\hat{p}_{2}$, player 1 remains strong for $\check{p}_{1}$ and all claims of player 2, so that player 1 never claims less than $\check{p}_{1}$ by Lemma 2. Remark that player 2 becomes strong for $\hat{p}_{2}$ and any claim of player 1 exceeding $\check{p}_{1}$. Player 1’s payoff cannot be improved upon for the claim $\hat{p}_{2}$. Hence, $u=m^{1}\left( \check{p}_{1}\right) $ is the unique solution for $\check{p}_{1}$ and $\hat{p}_{2}$ in the third case.

In the fourth case, the exceptions of Proposition 3 do not hold and there exists $\left( \check{p}_{1},\hat{p}_{2}\right) $ defining equal extended Nash products for which $C_{1}\left( \check{p}_{1}\right) \cap C_{2}(\hat{p} _{2})\ne \emptyset $. If $C_{1}\left( p_{1}\right) \cap C_{2}(p_{2})\subseteq C_{1}\left( \check{p}_{1}\right) \cap C_{2}(\hat{p} _{2})$ and player $s$ is strong for $p\ge \hat{p},$ his payoff is equal to $ \max \left\{ u_{s}^{KS}\left( p\right) ,m_{s}^{s}\left( p_{s}\right) \right\} $ by Lemma 1, which is a lower bound for the payoff of player $s$ for $p_{w}$ by Lemma 2. This lower bound is strictly decreasing in $p_{w}$ if $u_{s}^{KS}\left( p\right) >m_{s}^{s}\left( p_{s}\right) $ by the monotonicity of the proportional solution. This lower bound cannot be reduced and player $w$ cannot gain by increasing his claim as the weak player iff $C_{1}\left( p_{1}\right) \cap C_{2}(p_{2})=\left\{ u^{KS}\left( p\right) \right\} $, implying that in the solution $m^{1}\left( p_{1}\right) =m^{2}\left( p_{2}\right) =u^{KS}\left( p\right) $ for claims defining equal extended Nash products. These conditions uniquely identify $p\ge \hat{p}$ by the properties of the proportional solution. None of the players can gain by changing his claim. For a larger claim, the other player is strong and implements his maximal revision without changing the utility allocation. For a lower claim, his payoff is reduced in the proportional solution, which remains feasible and would be implemented. Hence, $u=m^{1}\left( p_{1}\right) =m^{2}\left( p_{2}\right) =u^{KS}\left( p\right) $ is the unique solution in the fourth case. $\square $

Proof of Proposition 4

Assume that condition $\left( a\right) $ of $\left( ii\right) $ of Proposition 3 holds. In the subgame for the claims $p=\left( 1,\hat{p} _{2}\right) $, both players bid equal resistance probabilities in $\left[ \rho _{1}\left( p\right) ,\rho _{2}\left( p\right) \right] $ and player 1 proposes $c\in C_{1}\left( 1\right) $ such that $c_{2}=\left( 1-q_{1}\right) \hat{p}_{2}$ which is as good as player 2’s ultimatum. If $q_{1}=\rho _{2}\left( p\right) =1-m_{1}^{2}\left( \hat{p}_{2}\right) /p_{1}$, then $ c_{2}=m_{1}^{2}\left( \hat{p}_{2}\right) \hat{p}_{2}$ bounds player 2’s payoff from below. If $q_{1}=\rho _{1}\left( p\right) $, $m^{1}\left( 1\right) $ is imposed, which bounds player 1’s payoff from below. There are no profitable deviations. Player 1 remains leader and proposes the same compromise for a higher resistance probability of player 2. Player 2 would lead if he lowers his or if player 1 increases his resistance probability. Either $q_{2}=q_{1}=\rho _{2}\left( p\right) $, $m^{2}\left( \hat{p} _{2}\right) $ is implemented and player 1 looses by increasing $q_{1}$. Or player 2 is unable to impose a compromise in $C_{2}\left( \hat{p}_{2}\right) $ and player 1 gives his ultimatum, giving a zero payoff to player 2 and, by concavity of $u_{1}^{P}$, $\left( 1-q_{1}\right) u_{1}^{P}\left( \hat{p} _{2}\right) \le u_{1}^{P}\left( \left( 1-q_{1}\right) \hat{p}_{2}\right) =u_{1}^{P}\left( c_{2}\right) $ as payoff to player 1. Player 1 remains leader by lowering $q_{1}$, but can only impose compromises with higher payoff for player 2. Hence for the claims $p$, all proposals $\tilde{u}$ for which $\tilde{u}_{2}\in \left[ m_{1}^{2}\left( \hat{p}_{2}\right) \hat{p} _{2},m_{2}^{1}\left( 1\right) \right] $ can be implemented in subgame-perfect equilibrium. Any claim $p_{1}^{\prime }<1$ of player 1 would reduce the lower bound $m_{1}^{1}\left( 1\right) $ on his payoff for $ q_{1}=q_{2}=\rho _{2}\left( p_{1}^{\prime },\hat{p}_{2}\right) .$ Any other claim than $\hat{p}_{2}$ of player 2 would reduce the lower bound $ m_{1}^{2}\left( \hat{p}_{2}\right) \hat{p}_{2}$ of player 2. Hence, $p$ are the equilibrium claims when condition $\left( a\right) $ of $\left( ii\right) $ of Proposition 3 holds.

Assume that condition $\left( a\right) $ of $\left( ii\right) $ of Proposition 3 does not apply. Then either $u^{KS}\left( \mathbf {1}\right) \in C_{1}\left( 1\right) \cap C_{1}\left( 1\right) $ for $p=\mathbf {1}$ or $ \rho _{2}\left( p\right) =\rho _{1}\left( p\right) $ in $\Gamma $ for $m\in \mathcal {M}$ and $\tilde{u}=u$ can be implemented in $\tilde{\Gamma }$ for $m$. Since $\tilde{u}\in PO\left( S\right) $, these lower bounds cannot be improved upon. $\square $

Proof of the lemma’s

Proof of Lemma 1

For any subgame with claims $p$ which are not strictly compatible, $ u^{KS}\left( p\right) $ is well defined. By definition, $q_{1}^{KS}\left( p\right) =q_{2}^{KS}\left( p\right) =r_{i}\left( u_{i}^{KS}\left( p\right) ,p_{i}\right) $ and $u_{i}^{KS}\left( p\right) \!/\!p_{i}=1-q_{i}^{KS}\left( p\right) $ for $i\in N$. A proposal $c$ of $L$ is proposed and accepted for $ q_{L}$ if and only if $q_{L}\ge r_{F}\left( c_{F},p_{F}\right) $. By the monotonicity of $r_{F}\left( .,p_{F}\right) $, $F$ rejects $c^{\prime }$ if he strictly prefers $c$ to $c^{\prime }$. We derive the bids in subgame-perfect equilibrium for any subgame for $p$. We distinguish between two cases when $s$ is strong for $p$.

In the first case, $u_{s}^{KS}\left( p\right) \ge m_{s}^{s}\left( p_{s}\right) ,$ so that $u^{KS}\left( p\right) \in C_{s}\left( p_{s}\right) \subseteq C\left( p\right) $ and the proportional solution is feasible. We distinguish between two subcases.

In the first subcase, $u_{s}^{KS}\left( p\right) <p_{s}$, so that $ r_{i}\left( u_{i}^{KS}\left( p\right) ,p_{i}\right) >0$ and $u_{i}^{P}\left( p_{-i}\right) <u_{i}^{KS}\left( p\right) <p_{i}$ for $i\in N$. For the bidding $q=q^{KS}\left( p\right) $, $q_{1}=q_{2}.$ The allocation would remain unchanged for a higher bid of player $i\in N$, since player $-i$ would be the leader for $q_{-i}$ and would propose $u^{KS}\left( p\right) $ which would be accepted by player $i$. The utility of a lower bidder $i$ would be reduced. As a leader, either he proposes an acceptable offer which reduces his payoff by the monotonicity of $r_{-i}\left( .,p_{-i}\right) $ or he proposes an unacceptable offer yielding $\left( 1-q_{i}\right) u_{i}^{P}\left( p_{-i}\right) \le u_{i}^{P}\left( p_{-i}\right) <u_{i}^{KS}\left( p\right) $. Since no player has a profitable deviation, $ q=q^{KS}\left( p\right) $ is an equilibrium for $p$. Moreover, player $i\in N $ ensures a payoff which is bounded below by $u_{i}^{KS}\left( p\right) $ for the bid $q_{i}=q_{i}^{KS}\left( p\right) $. Since $u^{KS}\left( p\right) \in PO\left( S\right) $, the lower bound for one player sets an upper bound on the payoff for the other player. Hence $L$ proposes $u^{KS}\left( p\right) $, $F$ accepts and both players bid $q_{i}^{KS}\left( p\right) $ in equilibrium for the claims $p$.

In the second subcase, $u_{s}^{KS}\left( p\right) =p_{s}$, so that $ u_{i}^{KS}\left( p\right) =p_{i}$ and $r_{i}\left( u_{i}^{KS}\left( p\right) ,p_{i}\right) =0$ for $i\in N$. If $u_{w}^{KS}\left( p\right) >m_{w}^{w}\left( p_{s}\right) ,$ $q_{w}=0$ is the only way for $w$ to avoid that $s$ acquires leadership for $q_{s}>0$ and makes a proposal in $ C_{w}\left( p_{w}\right) $ which $s$ would prefer to $u^{KS}\left( p\right) $ and which $w$ would accept as a follower$.$ Hence, $u^{KS}\left( p\right) $ is implemented for $q_{w}=0$ and $q_{s}\in \left[ 0,1\right] $. If $ u_{w}^{KS}\left( p\right) =m_{w}^{w}\left( p_{s}\right) $, that is $C\left( p\right) =\left\{ u^{KS}\left( p\right) \right\} $, $L$ has no other option than to propose $u^{KS}\left( p\right) $ and leadership is valuable for none of the players. Hence, $q_{i}\in \left[ 0,1\right] $ for $i\in N$ implements $u^{KS}\left( p\right) $. It follows that $u^{KS}\left( p\right) $ is implemented in equilibrium. Conclude that in the first case, the proportional solution is the unique solution implemented in equilibrium whenever it is feasible for claims $p$.

In the second case $m_{s}^{s}\left( p_{s}\right) >u_{s}^{KS}\left( p\right) $, so that $u^{KS}\left( p\right) \notin C_{s}\left( p_{s}\right) $ and $ r_{s}\left( m_{s}^{s}\left( p_{s}\right) ,p_{s}\right) <q_{1}^{KS}\left( p\right) =q_{2}^{KS}\left( p\right) $. By (2), it follows that $ \rho _{w}\left( p\right) \ge \rho _{s}\left( p\right) >q_{s}^{KS}\left( p\right) $, so that $r_{s}\left( m_{s}^{s}\left( p_{s}\right) ,p_{s}\right) <\rho _{s}\left( p\right) $ and $u^{KS}\left( p\right) \notin C_{w}\left( p_{w}\right) $. It follows that $u^{KS}\left( p\right) \notin C\left( p\right) $, so that the proportional solution is not feasible. We show that $ m^{s}\left( p_{s}\right) $ is implemented for the equilibrium bids

$$\begin{aligned} q_{w}&\in [r_{s}(m_{s}^{s}\left( p_{s}\right) ,p_{s}),\rho _{s}\left( p\right) ], \\ q_{s}&\in \left\{ \! \begin{array}{c} \left[ \rho _{s}\left( p\right) ,\rho _{w}\left( p\right) \right] \\ \left[ \rho _{s}\left( p\right) ,\rho _{w}\left( p\right) \right) \end{array} \right. \left. \begin{array}{c} \text {if}\quad s=1, \\ \text {if}\quad s=2. \end{array} \right. \end{aligned}$$

If $w=L$, then $q_{w}\le q_{s}<\rho _{w}\left( p\right) $ or $ q_{2}<q_{1}=\rho _{2}\left( p\right) $ and $s$ rejects proposals in $ C_{w}\left( p_{w}\right) .$ As a result, $w$ cannot do better than by proposing $m^{s}\left( p_{s}\right) $ in $C_{s}\left( p_{s}\right) $ which is accepted by $s$ for $q_{w}\ge r_{s}\left( m_{s}^{s}\left( p_{s}\right) ,p_{s}\right) $. If $s=L$, then $q_{s}=\rho _{s}\left( p\right) $ for $ q_{w}\le \rho _{s}\left( p\right) $ implies that $m^{s}\left( p_{s}\right) $ is accepted by $w$ and that any better proposal for $s$ in $C_{s}\left( p_{s}\right) \!{\setminus }\{m^{s}\left( p_{s}\right) \}$, if any, is rejected by $w$. The payoff of player $i\in N$ is bounded below by $m_{i}^{s}\left( p\right) $ for these bids. Since $m^{s}\left( p\right) \in PO\left( S\right) $, the lower bound for one player sets an upper bound on the payoff for the other player. Hence, $m^{s}\left( p_{s}\right) $ is implemented for the bids $q$ in equilibrium. Remark that $m^{s}\left( p_{s}\right) $ would also be implemented for $q_{w}\in [0,1]$ if $C_{s}\left( p_{s}\right) =\{m^{s}\left( p_{s}\right) \}$ and $s$ has no other choice than to propose $ m^{s}\left( p_{s}\right) $ as a leader. We show that some player has a profitable deviation for all other bidding strategies. For $ q_{w}<r_{s}\left( m_{s}^{s}\left( p_{s}\right) ,p_{s}\right) $, $w=L$ and proposes $u\in PO\left( C_{s}\left( p_{s}\right) \right) $ for which $ u_{s}=\left( 1-q_{w}\right) p_{s}>m_{s}^{s}\left( p_{s}\right) $. For $ q_{w}>\rho _{s}\left( p\right) $ and $C_{s}\left( p_{s}\right) \ne \{m^{s}\left( p_{s}\right) \}$, $s=L$ for $q_{w}>q_{s}>\rho _{s}\left( p\right) $ and can impose a preferred compromise in $C_{s}\left( p_{s}\right) {\setminus }\{m^{s}\left( p_{s}\right) \}$. For $q_{s}<\rho _{s}\left( p\right) $, $s=L$ when player $w$ chooses $q_{w}=\rho _{s}\left( p\right) $ and player $s$ must propose in $C_{w}\left( p_{w}\right) $. Finally, for $q_{s}>\rho _{w}\left( p\right) $ if $s=1$ and $q_{s}\ge \rho _{w}\left( p\right) $ if $s=2$, $w=L$ for $q_{w}=\rho _{w}\left( p\right) $ and can propose in $C_{w}\left( p_{w}\right) $. Hence, if any player were to change his bidding strategy, his payoff would be lower than the one in $ m^{s}\left( p_{s}\right) $. We conclude that $m^{s}\left( p_{s}\right) $ is implemented in equilibrium when the proportional solution is not feasible for $p$. $\square $

Proof of Lemma 2

For strictly compatible claims $p$, $m_{1}^{1}\left( p_{1}\right) \le p_{1}<u_{1}^{P}\left( p_{2}\right) \le m_{1}^{2}\left( p_{2}\right) $. Player 1 is leader by bidding $q_{1}=0$. For this bid, player 2’s ultimatum and player 1’s proposal $\left( u_{1}^{P}\left( p_{2}\right) ,p_{2}\right) $ are equivalent. Let player 2 bid $q_{2}\in \left[ 0,1\right] $ if $ m_{2}^{2}\left( p_{2}\right) =p_{2}$ and $q_{2}=0$ if $m_{2}^{2}\left( p_{2}\right) <p_{2}$. In the former, player 2 accepts $\left( u_{1}^{P}\left( p_{2}\right) ,p_{2}\right) $, player 1’s preferred outcome in $C\left( p\right) .$ In the latter, player 2 would be leader for $q_{1}>0$ and $\left( p_{1},u_{2}^{P}\left( p_{1}\right) \right) $ would be implemented, reducing player 1’s payoff. Hence, $\left( u_{1}^{P}\left( p_{2}\right) ,p_{2}\right) $ is implemented in a subgame with strictly compatible claims $p$. Formulating strictly compatible claims cannot occur in subgame-perfect equilibrium, since the strictly compatible claims $ p^{\prime }$, $p_{1}^{\prime }=p_{1}$ and $p_{2}^{\prime }>p_{2}$ increase player 2’s payoff and player 2 has a profitable deviation.

Assume that $s$ is strong in the subgame for claims $p$ which are not strictly compatible. By Lemma 1, $u$ is the strong player’s preferred option in $\hat{C}_{s}\left( p\right) $ for the claims $p$. By the monotonicity of the proportional solution and the comprehensiveness of the revision procedure, $\max \left\{ m_{i}^{i}\left( p_{i}\right) ,u_{i}^{KS}\left( p\right) \right\} $ is strictly increasing in $p_{i}$. If $p_{s}<\hat{p}_{s}$, then by claiming $\hat{p}_{s}$, player $s$ would remain strong and increase his payoff for given $p_{w}$. Since profitable deviations of one player are excluded, $p_{s}\ge \hat{p}_{s}$ in subgame-perfect equilibrium. If $p_{w}<\hat{p}_{w}$ and $u_{s}^{KS}\left( p\right) \ge m_{s}^{s}\left( p_{s}\right) $, then the proportional solution is implemented and player $w$ could increase his payoff for a larger claim for given $p_{s}$. If $p_{w}< \hat{p}_{w}$ and $u_{s}^{KS}\left( p\right) <m_{s}^{s}\left( p_{s}\right) $, then $m^{s}\left( p_{s}\right) $ is implemented. By claiming $\hat{p}_{w}$, either player $w$ becomes strong for $p^{\prime }$, $p_{w}^{\prime }=\hat{p} _{w}$ and $p_{s}^{\prime }=p_{s}$ and would obtain $\max \left\{ m_{w}^{w}\left( \hat{p}_{w}\right) ,u_{w}^{KS}\left( p^{\prime }\right) \right\} \ge u_{w}^{KS}\left( p^{\prime }\right) >u_{w}^{KS}\left( p\right) $. Or player $s$ remains strong for $\hat{p}_{w}$, $m^{s}\left( p_{s}\right) $ is implemented and $m_{w}^{s}\left( p_{s}\right) p_{s}\ge m_{s}^{w}\left( \hat{p}_{w}\right) \hat{p}_{w}>m_{s}^{w}\left( p_{w}\right) p_{w}$. Player $ s $ would remain strong and would gain for a claim larger than $p_{s}\ge \hat{p}_{s}$ for given $p_{w}$. Since profitable deviations of one player are excluded, $p_{w}\ge \hat{p}_{w}$ in subgame-perfect equilibrium.

Consider any subgame with claims $\bar{p}\ge \hat{p}$ implementing $\bar{u}$. If player $s$ is strong for $\bar{p}$, he remains strong for $ p_{s}^{\prime }\in \left[ \hat{p}_{s},\bar{p}_{s}\right] $ and for $ p_{w}^{\prime }=\bar{p}_{w}$. Hence, $\bar{u}_{s}=\max \left\{ m_{s}^{s}\left( \bar{p}_{s}\right) ,u_{s}^{KS}\left( \bar{p}\right) \right\} \ge \max \left\{ m_{s}^{s}\left( p_{s}^{\prime }\right) ,u_{s}^{KS}\left( p^{\prime }\right) \right\} $. Moreover, if $s$ is strong for $\bar{p}$ and for all claims of player $w$, then he will never claim less than $\bar{p} _{s} $ in subgame-perfect equilibrium. $\square $

Proof of the corollary

Proof of Corollary 1

We distinguish between two cases. In the first case, condition $\left( i\right) $ of Proposition 3 holds for $m$, so that $u=u^{KS}\left( \mathbf {1} \right) \in C\left( \mathbf {1}\right) $. Since $C\left( \mathbf {1}\right) \subseteq \breve{C}\left( \mathbf {1}\right) $, $u^{KS}\left( \mathbf {1} \right) \in \breve{C}\left( \mathbf {1}\right) $ for $\breve{m}$. It follows that $\breve{u}=u$ and $\breve{p}=p=\mathbf {1}$. In the second case $ u^{KS}\left( \mathbf {1}\right) \notin \breve{C}\left( \mathbf {1}\right) $, so that $u^{KS}\left( \mathbf {1}\right) \notin C\left( \mathbf {1}\right) $ and condition $\left( i\right) $ of Proposition 3 does not hold for $m$ as well as for $\breve{m}$. It suffices to consider three subcases. Remark that $\max _{p_{1}}\breve{m}_{2}^{1}\left( p_{1}\right) p_{1}\ge m_{2}^{1}\left( \hat{p}_{1}\right) \hat{p}_{1}\ge m_{1}^{2}\left( \hat{p}_{2}\right) \hat{p} _{2}$, so that the labeling of the players is the same for $\breve{m}$ as for $m$.

In the first subcase, subcase $\left( a\right) $ of Proposition 3 holds for $ m$. Since $m^{1}\left( 1\right) =\breve{m}^{1}\left( 1\right) $ and $m^{2}= \breve{m}^{2}$, this subcase also holds for $\breve{m}$, so that $\breve{u} _{1}=u_{1}$ and $\breve{p}_{1}=p_{1}=1$.

In the second subcase, subcase $\left( b\right) $ of Proposition 3 holds for $m$. Since $\breve{m}_{2}^{1}\left( \check{p}_{1}\right) \check{p}_{1}\ge m_{2}^{1}\left( \check{p}_{1}\right) \check{p}_{1}=m_{1}^{2}\left( \hat{p} _{2}\right) \hat{p}_{2},$ player 1 is strong for any claim of player 2 when claiming $\check{p}_{1}$ for $\breve{m}$, so that $\breve{p}_{1}\ge \check{p}_{1}$ by Lemma 2 and $\breve{u}_{1}\ge u_{1}$ when $\breve{m} ^{1}\left( \breve{p}_{1}\right) $ is implemented

In the third subcase, the exceptions of Proposition do not hold or subcase $ \left( c\right) $ of $\left( ii\right) $ holds. There exists $p$ such that $ m^{1}\left( p_{1}\right) =m^{2}\left( p_{2}\right) =u^{KS}\left( p\right) $ for $m$. Since the extended Nash products are equal, player 1 is the strong player. Since $\breve{m}_{2}^{1}\left( p_{1}\right) \ge m_{2}^{1}\left( p_{1}\right) $ and $\breve{m}^{2}=m^{2}$, player 1 remains strong for $p$ and for $\breve{m}$. By Lemma 2 for $\breve{m}$, player 1’s payoff is not smaller than $\max \left\{ \breve{m}_{1}^{1}\left( p_{1}\right) ,u_{1}^{KS}\left( p\right) \right\} $ for $p_{2}$. If $\breve{m}^{1}\left( p_{1}\right) =m^{1}\left( p_{1}\right) ,$ $\breve{u}_{1}=u_{1}$ and there exist a claim for player 1 for which $\breve{p}_{1}=p_{1}$. If $\breve{m} ^{1}\left( p_{1}\right) >m^{1}\left( p_{1}\right) $, then $\breve{m} _{1}^{1}\left( p_{1}\right) >u_{1}^{KS}\left( p\right) $ and $\breve{m} _{1}^{1}\left( p_{1}\right) $ is implemented for $p$ by Lemma 1. Since player 2 remains weak for larger claims than $p_{2}$, $\breve{u}_{1}>u_{1}$. For $\breve{p}$, the equality of the extended Nash products must be restored, either for $\breve{p}_{2}=\hat{p}_{2},$ as in the second subcase or for $\breve{m}^{1}\left( \breve{p}_{1}\right) =\breve{m}^{2}\left( \breve{ p}_{2}\right) =u^{KS}\left( \breve{p}\right) $. In both cases, $\breve{p} _{1}>p_{1}$. $\square $

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Johannes Spinnewijn
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- 2
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Frans Spinnewyn
- 3
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1.Department of Economics, STICERDLondon School of EconomicsLondonUK
2.LSE and CEPRLondonUK
3.Department EconomicsKatholieke Universiteit LeuvenLouvainBelgium
4.KULeuven and ECORESLeuvenBelgium

Revising claims and resisting ultimatums in bargaining problems

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