Optimize emission reduction commitments for international environmental agreements

Abstract

In order to restrict global warming to no more than 2 ^∘C, more efforts are needed. Thus, how to attract as more as possible countries to international environment agreements (IEAs) and realize the maximum reduction targets are meaningful. The motivation of this paper is exploring a set of method of designing IEA proposals. The paper built a chance-constrained two-stage cartel formation game model, which can explore whether a country signs an agreement in the first stage and discusses how the countries joining the coalition can make the best emission commitments in the second stage. Based on the model, the real emission data of 45 countries was collected for numerical experiments, which almost completely depict the current global emissions of different countries. A numerical experiment has also been carried out in the paper. Then some interesting results emerge as follows: risk averse, high cost, high emission reduction duty, and external stability impede large coalition formation; transfer scheme and high perceived benefits stimulate countries to join IEAs and make a good commitment; the most influential countries for coalition structure and commitment are those low-cost and low-emission entities. The results also demonstrate that the design of IEA proposals should not only pay attention to those economically developed and high-emission “big” countries, but also attach importance to those low-emission “small” countries.

Appendix

• 1. The difference of payoff between a country being a signatory or not.

Using equilibrium abatement payoffs in Eqs. 5 and 7 gives the following payoffs:

• (1) If country i is a member of coalition C, then the payoff when i leaves the coalition is

  $$ \begin{array}{lll} \widetilde{\pi}^{*}_{i}(C\setminus\{i\})&=&\gamma_{i}\left[\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}-{\lambda_{i}^{2}}\right) *\left( \sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}-\gamma_{i}\right)\right.\\ &&\left.+\sum\limits_{k = 1}^{n}(1-x_{k}){\lambda_{k}^{2}}\gamma_{k} +{\lambda_{i}^{2}}\gamma_{i}\right] -\frac{1}{2}(\lambda_{i}\gamma_{i})^{2}\\ &=&\gamma_{i}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar} -{\lambda_{i}^{2}}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}-\gamma_{i}\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\right.\\ &&\left.+{\lambda_{i}^{2}}\gamma_{i}+\sum\limits_{k = 1}^{n}(1-x_{k}){\lambda_{k}^{2}}\gamma_{k}+{\lambda_{i}^{2}}\gamma_{i}\right)-\frac{1}{2}(\lambda_{i}\gamma_{i})^{2}\\ &=&\gamma_{i}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}+\sum\limits_{k = 1}^{n}(1-x_{k}){\lambda_{k}^{2}}\gamma_{k}\right)\\ &&-{\lambda_{i}^{2}}\gamma_{i}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}-{\gamma_{i}^{2}}\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}+\frac{3}{2}(\lambda_{i}\gamma_{i})^{2}, \forall i\in C. \end{array} $$

  (31)

  Putting Eq. (5) into Eq. (31), we have

  $$ \begin{array}{lll} \widetilde{\pi}^{*}_{i}(C\setminus\{i\})&\,=\,&\widetilde{\pi}^{*}_{i}(C) +\frac{1}{2}{\lambda_{i}^{2}}\left( \sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}\right)^{2} \,-\,{\lambda_{i}^{2}}\gamma_{i}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar} \,-\,{\gamma_{i}^{2}}\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\,+\,\frac{3}{2}(\lambda_{i}\gamma_{i})^{2}\\ &\,=\,&\widetilde{\pi}^{*}_{i}(C)+\frac{1}{2}{\lambda_{i}^{2}}\left( \sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}\,-\,\gamma_{i}\right)^{2}\,-\,{\gamma_{i}^{2}}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}-{\lambda_{i}^{2}}\right), \end{array} $$

  (32)

  which equals to

  $$ \widetilde{\pi}^{*}_{i}(C)-\widetilde{\pi}^{*}_{i}(C\setminus\{i\})={\gamma_{i}^{2}}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}-{\lambda_{i}^{2}}\right) -\frac{1}{2}{\lambda_{i}^{2}}\left( \sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}-\gamma_{i}\right)^{2}, \forall i\in C. $$

  (33)

• (2) If country j is not a member of coalition C, then the payoff when j joins the coalition is

  $$ \begin{array}{lll} \widetilde{\pi}^{*}_{j}(C\cup \{j\})&=&\gamma_{j}\left[\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}+{\lambda_{j}^{2}}\right) *\left( \sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}+\gamma_{j}\right)\right.\\ &&\left.+\sum\limits_{k = 1}^{n}(1-x_{k}){\lambda_{k}^{2}}\gamma_{k} -{\lambda_{j}^{2}}\gamma_{j}\right] -\frac{1}{2}{\lambda_{j}^{2}}\left( \sum\limits_{\hbar= 1}^{n} x_{\hbar}\gamma_{\hbar}+\gamma_{j}\right)^{2}\\ &=&\gamma_{j}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar} +{\lambda_{j}^{2}}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}+\gamma_{j}\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\right.\\ &&\left.+{\lambda_{j}^{2}}\gamma_{j}+\sum\limits_{k = 1}^{n}(1-x_{k}){\lambda_{k}^{2}}\gamma_{k}-{\lambda_{j}^{2}}\gamma_{j}\right)-\frac{1}{2}{\lambda_{j}^{2}}\left( \sum\limits_{\hbar= 1}^{n} x_{\hbar}\gamma_{\hbar}+\gamma_{j}\right)^{2}\\ &=&\gamma_{j}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}+\sum\limits_{k = 1}^{n}(1-x_{k}){\lambda_{k}^{2}}\gamma_{k}\right)\\ &&+{\lambda_{j}^{2}}\gamma_{j}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}+{\gamma_{j}^{2}}\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}-\frac{1}{2}{\lambda_{j}^{2}}\left( \sum\limits_{\hbar= 1}^{n} x_{\hbar}\gamma_{\hbar}+\gamma_{j}\right)^{2}, \forall j\not\in C. \end{array} $$

  (34)

  Inserting Eq. (7) into Eq. (34), we obtain

  $$ \begin{array}{lll} \widetilde{\pi}^{*}_{j}(C\cup\{j\})&=&\widetilde{\pi}^{*}_{j}(C) +\frac{1}{2}(\lambda_{j}\gamma_{j})^{2} +{\lambda_{j}^{2}}\gamma_{j}\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}+{\gamma_{j}^{2}}\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\\ &&-\frac{1}{2}{\lambda_{j}^{2}}\left( \sum\limits_{\hbar= 1}^{n} x_{\hbar}\gamma_{\hbar}+\gamma_{j}\right)^{2}\\ &=&\widetilde{\pi}^{*}_{j}(C)+{\gamma_{j}^{2}}(\sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2})-\frac{1}{2}{\lambda_{j}^{2}}(\sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar})^{2}, \end{array} $$

  (35)

  which equals to

  $$ \widetilde{\pi}^{*}_{j}(C)-\widetilde{\pi}^{*}_{j}(C\cup\{j\})= \frac{1}{2}{\lambda_{j}^{2}}\left( \sum\limits_{\hbar= 1}^{n}x_{\hbar}\gamma_{\hbar}\right)^{2}-{\gamma_{j}^{2}}\left( \sum\limits_{\ell= 1}^{n}x_{\ell}\lambda_{\ell}^{2}\right), \forall j\not\in C. $$

  (36)