International Cooperation and Institution Formation: A Game Theoretic Perspective

Abstract

Game theory presents a useful analytical tool for addressing the problem of international cooperation and the formation of institutions. We first examine four problems that must be solved to achieve international cooperation: the common knowledge problem, agreement problem, compliance problem, and participation problem. An institution is a mechanism used to enforce participants to cooperate for collective benefits. We consider a multi-stage game model of institution formation and show that a group of participants voluntarily forms an institution for international cooperation in a strict subgame perfect equilibrium if and only if the group satisfies the criticality condition. Some of the implications on the international frameworks that attempt to prevent the proliferation of nuclear weapons in East Asia are finally discussed.

I would like to thank Motoshi Suzuki and members of the Workshop on International Political and Economic Analysis for their valuable suggestions and comments. Financial support from the Japan Society for the Promotion of Science under Grant No. (A)26245020 is gratefully acknowledged.

Appendix

Proof

The first part of the proposition can be easily seen by the definition of individual rationality. Suppose that a group S of cooperators is Pareto optimal. Then, the largest group N of cooperators is not Pareto superior to S. If S = N, then the condition $n -\frac{1} {a} \leq s$ trivially holds. If S ≠ N, then a defector outside S is no better off when N is formed than when S is formed. This finding implies that ω + as ω ≥ an ω, that is, $n -\frac{1} {a} \leq s$. Conversely, suppose that S is not Pareto optimal. Then, there exists some group T ≠ S such that T is Pareto superior to S. Then, there exists some i ∈ T − S. Otherwise, T is never Pareto superior to S according to (6.2). By supposition, it holds that ω + as ω < at ω. Since at ω ≤ an ω, it holds that ω + as ω < an ω, that is, $n -\frac{1} {a}> s$. This is a contradiction. Q.E.D.

Proposition 6.1 can be generalized to the case that countries have different MPCRs. A group S of cooperators is Pareto optimal if and only if for every T with T − S ≠ ∅ there exists some i ∈ T − S such that $s \geq t - \frac{1} {a_{i}}$.

Proof

For an individually rational group S, consider the group-trigger strategy such that all members of S cooperate in the first period and keep cooperating as long as all members of S do so (otherwise they contribute nothing forever). All the non-members of S contribute nothing in all periods. The discounted payoff of every member of S for this strategy is thus $\frac{a_{i}s\omega } {1-\delta }$. If a member defects, then its discounted payoff is at most $\omega +a_{i}(s - 1)\omega + \frac{\delta } {1-\delta }\omega.$ Thus, the group-trigger strategy for S is a subgame perfect equilibrium of the repeated game G ^∞ if $\frac{a_{i}s\omega } {1-\delta } \geq \omega +a_{i}(s - 1)\omega + \frac{\delta } {1-\delta }\omega$. This condition is equivalent to (6.3). Q.E.D.

Proof

 

• (1) Suppose that a set of participants S is individually rational. If all participants i ∈ S accept the implementation of S, then they receive the per-period payoff a _i s ω. Otherwise, the institution is rejected and thus they receive the per-period payoff ω. Since a _i s ω > ω for all i ∈ S, the action profile that all participants accept the implementation of S is a Nash equilibrium of the implementation stage. According to the rule of unanimous voting, the action profiles that at least two participants reject the implementation of S are also Nash equilibria.

• (2) Suppose that a set of participants S is not individually rational. Then, there exists at least one participant that is strictly better off by rejecting the institution S than when S is implemented. Note that the case of a _i s ω = ω is omitted for all i ∈ S since $\frac{1} {a_{i}}$ is not an integer. Thus, the action profile that all participants accept the implementation of S is not a Nash equilibrium of the implementation stage. Q.E.D.

Proof

If an institution S is implemented in a subgame perfect equilibrium, then S must be individually rational from Proposition 6.1. Conversely, let S be an individually rational group. Consider the following strategy profile. All members of S participate in an institution and no other countries participate. In the implementation stage, S is accepted and all other sets of participants are rejected. It can be shown without much difficulty that this strategy profile induces a Nash equilibrium in the participation stage. For any number of participants s (1 ≤ s ≤ n), consider the strategy profile that all participants reject an institution, regardless of the number of participants. This strategy trivially composes the status-quo equilibrium. Q.E.D.

Proof of Proposition 6.5

According to Proposition 6.3, every individually rational group is accepted to form in a strict subgame perfect equilibrium. Given this equilibrium outcome of the implementation stage, it is sufficient to prove that an action profile of the participation stage is a strict Nash equilibrium if and only if a set of participants S is a critical group. First, suppose that S is a critical group. If any member i of S does not participate in S, then S −{ i} is not individually rational and thus it is not implemented. Country i is worse off by deviating from S and thus the action profile with S is a strict Nash equilibrium. Second, suppose that S is individually rational but not critical. Then, there exists some member i of S that is not critical to S. If country i deviates from S, then it is better off because S −{ i} is individually rational and is implemented. This finding means that the action profile with S is not a Nash equilibrium. Finally, suppose that S is not individually rational. Then, S is not implemented according to Proposition 6.3. Choose any j ∉ S. Country j is better off by joining S if S ∪{ j} is implemented. Otherwise, the per-period payoff of country j does not change. This means that the action profile with S is not a strict Nash equilibrium. Q.E.D.