The Paradox of Compliance: Does Legalization Increase Compliance in Trade Disputes?

Keisuke Iida

doi:10.1007/978-4-431-56466-9_9

Games of Conflict and Cooperation in Asia pp 207-223 | Cite as

The Paradox of Compliance: Does Legalization Increase Compliance in Trade Disputes?

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Keisuke Iida

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First Online: 08 February 2017

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Part of the The Political Economy of the Asia Pacific book series (PEAP)

Abstract

The theory of legalization postulates that compliance increases with the degree of legalization, facilitated by binding adjudication. However, a simple comparison of dispute settlement under the former General Agreement on Tariffs and Trade (GATT) and the current World Trade Organization (WTO), which has absorbed GATT into its fold, shows that there is no apparent increase in compliance despite a considerable increase in legalization. This chapter argues that such a paradoxical result could come about because of the endogeneity of behavior in bringing disputes to the dispute settlement system and, hence, the endogeneity of adjudication. In part due to legalization, a greater number of cases have been filed under the WTO. Furthermore, data analysis shows that more difficult cases have been filed under the WTO. Hence, the proportion of disputes that have failed to be resolved after adjudication has increased, leading to the abovementioned paradox.

Keywords

Dispute settlement Legalization Compliance World Trade Organization (WTO) Asia

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Appendix 9.1: Proof of the Proposition

Lemma 9.1:

At their last node, the compliant type will comply and the non-compliant type will not comply in any perfect Bayesian equilibrium (PBE).

Proof:

Since it is the last node in the game, the defendant does not have to worry about his reputation. Therefore, he will decide whether to comply by comparing the payoff of compliance with that of non-compliance. Since h_C > 0 and consequently h_C – k> − k, the compliant type will comply. Owing to the assumption that h_N < 0, h_N – k< − k, the non-compliant type will not comply.

Lemma 9.2:

In any PBE, given the posterior belief p¹, the following is the optimum acceptance rule for the complainant: accept any offer b if b ≥ p¹–k, otherwise reject.

Proof:

From Lemma 9.1, the defendant’s expected payoff of rejecting an offer is p¹(1–k)+(1–p¹)( −k) = p¹–k. Therefore, it is the complainant’s best response to accept an offer b, which yields a payoff greater than or equal to p¹–k.

Proof of the Proposition

The first half of the proposition follows directly from Lemmas 9.1 and 9.2. In the following, we show that the second half of the proposition holds. In order to show that the equilibrium exists under given values of the parameters, we need only show that the defendant has no incentive to deviate from his equilibrium offer and that the beliefs of the equilibrium path are based on the Bayes rule.

(a)

Pooling equilibrium I (settlement equilibrium)

First, we show that the pooling equilibrium (settlement equilibrium) exists if and only if p⁰ ≤ 2 k − h_C. In a pooling equilibrium, the compliant type’s offer, denoted by b_C, and that of the non-compliant type, denoted by b_N, are the same. After this offer is issued, the posterior belief of the complainant will be p¹ = p⁰ by the Bayes rule.

If the compliant type (or non-compliant type) of defendant follows this equilibrium offer strategy, p⁰ − k will be accepted by the complainant, and the defendant’s payoff will be k − p⁰. If an offer greater than p⁰ − k is made, the posterior belief will be p¹ = p⁰ by assumption, and will be accepted according to Lemma 9.2. However, the higher the value of the offer, the lower the payoff of the defendant will be. Therefore, such a deviation is not profitable for either type of defendant.

On the other hand, if an offer less than p⁰ − k is made, the posterior belief will be p¹ = p⁰ by assumption, and it will be rejected by the complainant according to Lemma 9.2. Then, the compliant type will receive h_C − k, while the non-compliant type will receive −k. Now, given the condition p⁰ ≤ 2k − h_C, h_C − k ≤ k − p⁰ and –k < k − p⁰, neither type can deviate profitably from this equilibrium offer.

Next, suppose that p⁰ > 2k − h_C. We show that pooling equilibrium I does not exist under this condition.

As shown above, the compliant type will receive k − p⁰ if he follows the strategy of pooling equilibrium, while if he were to offer less than p⁰ − k, he would receive h_C − k. In the latter case, because p⁰ > 2k − h_C, h_C − k > k − p⁰, he would be able to deviate profitably from this equilibrium. Therefore, this pooling equilibrium does not exist if p⁰ > 2k − h_C.

(b)

Pooling equilibrium II (escalation equilibrium)

First, we show that if p⁰ ≧ 2 k, this pooling equilibrium exists. Since the equilibrium offer of the compliant type b_C and that of the non-compliant type b_N are the same, the posterior belief of the complainant will be p¹ = p⁰ by the Bayes rule.

If the compliant type (or non-compliant type) of defendant follows the equilibrium strategy, the offer b_C = b_N (< p⁰ − k) will be rejected, according to Lemma 9.2. From Lemma 9.1, the compliant type will receive h_C – k, while the non-compliant type will receive-k. If the compliant type (or non-compliant type) offers less than p⁰ − k and is different from the equilibrium offer b_C, the posterior belief will p¹ = p⁰ by assumption, and it will be rejected. Thus, his payoff will be the same as in the equilibrium. On the other hand, if he were to offer more than p⁰ − k, again the posterior belief will be p¹ = p⁰ by assumption, and it would be accepted by Lemma 9.2. The defendant would receive less than k − p⁰. However, since p⁰ ≥ 2k, h_C − k > k − p⁰ and −k ≥ k − p⁰, neither type of defendant could deviate profitably from the equilibrium offer.

Next, we show that this pooling equilibrium does not exist if p⁰ < 2k. Under this condition, the non-compliant type will receive −k if he follows the equilibrium strategy, while he would receive k − p⁰ if he offers p⁰ − k. Because p⁰ < 2k, −k < k − p⁰, the non-compliant type can deviate profitably from the equilibrium offer and, hence, pooling equilibrium II does not exist.

(c)

Separating equilibrium

First, we show that if h_C ≥ k, separating equilibrium exists. Under this equilibrium, the compliant type’s offer b_C and the non-compliant type’s offer b_N are distinct, and by the Bayes rule, the posterior belief will be p¹ =1 if b_C is offered while the posterior belief will be p¹ = 0 if b_N is offered.

If the compliant type follows the equilibrium strategy, his offer b_C < p¹ − k = 1 − k will be rejected from Lemma 9.2. Thus, from Lemma 9.1, we know that he will receive h_C − k. If the non-compliant type follows the equilibrium strategy, b_N = 0 > p¹ − k = −k will be accepted. In this case, he will receive-b_N = 0.

If the compliant type were to make an offer that is rejected, his payoff will remain the same. The highest payoff he could hope to get is to offer zero by mimicking the non-compliant type. However, by the condition that h_C ≥ k, he cannot profit from making that offer and getting it accepted. Next, all we need to show is that the non-compliant type cannot profitably make an offer that is rejected. From Lemma 9.1, we know that the best he could get is -k. Since k > 0, it would not make him better off.

Next, we show that if h_C < k, separating equilibrium does not exist. From the abovementioned discussion, we know that the compliant type will receive h_C − k if he follows the equilibrium strategy while he would get zero payoff if he were to make an offer equal to zero. From the condition h_C < k, the compliant type could deviate profitably from separating equilibrium. Therefore, separating equilibrium cannot exist if h_C < k.

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Keisuke Iida
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1.Faculty of Law and Graduate Schools for Law and PoliticsThe University of TokyoTokyoJapan

The Paradox of Compliance: Does Legalization Increase Compliance in Trade Disputes?

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