Optimal strategic regulations in international emissions trading under imperfect competition

Abstract

This paper analyzes governments’ strategic regulations in an imperfectly competitive market of international emissions trading (IET). Whether and how governments intervene in IET is explored. If regulations are decided, it is optimal for price-influencing countries to subsidize but for price-taking countries to tax permit trading. Conducting simulations of the Annex-1 emissions trading, we discover that no-intervention of all countries cannot be supported by any equilibrium. In contrast, all or some countries would regulate at equilibrium. In the latter case, price-influencing countries would not regulate but price-taking countries would. This justifies the necessity of considering no-intervention as a policy choice, and shows that a country’s decisions about strategically regulating IET may be affected by other countries’ intervention resolutions.

Appendix

Proof of Eq. (11)

Denote H the Hessian matrix of L equations in Eqs. (9) and (10) with

$$ H = \left[ {\begin{array}{*{20}c} {C^{\prime\prime}_{1} (e_{1}^{*} ) + 2A} & A & \cdots & A \\ A & {C^{\prime\prime}_{2} (e_{2}^{*} ) + 2A} & \cdots & A \\ \vdots & \vdots & \cdots & \vdots \\ A & A & \cdots & {C^{\prime\prime}_{L} (e_{L}^{*} ) + 2A} \\ \end{array} } \right] = A \, \left[ {\begin{array}{*{20}c} {K_{1} } & 1 & \cdots & 1 \\ 1 & {K_{2} } & \cdots & 1 \\ \vdots & \vdots & \cdots & \vdots \\ 1 & 1 & \cdots & {K_{L} } \\ \end{array} } \right], $$

where $A = \frac{\partial \, p}{{\partial \, e_{m} }} = \frac{ - 1}{{\sum\nolimits_{n = L + 1}^{N} {(\partial \, e_{n}^{*} /\partial \, p)} }} > 0$ and $K_{m} = \frac{{C^{\prime\prime}_{m} (e_{m}^{*} ) + 2A}}{A}$ with K _m  > 2 given $C^{\prime\prime}_{m} (e_{m}^{*} ) > 0$, m = 1, …, L.

Letting $\hat{H} = \, \left[ {\begin{array}{*{20}c} {K_{1} } & 1 & \cdots & 1 \\ 1 & {K_{2} } & \cdots & 1 \\ \vdots & \vdots & \cdots & \vdots \\ 1 & 1 & \cdots & {K_{L} } \\ \end{array} } \right]$, we can show $\left| {\hat{H}} \right| = [\prod\nolimits_{m = 1}^{L} {(K_{m} - 1} ][1 + \sum\nolimits_{m = 1}^{L} { \, \frac{1}{{K_{m} - 1}}} ] > 1$ by K _m  > 2. Applying Cramer’s rule, we acquire

$$ \frac{{\partial e_{r}^{*} }}{{\partial t_{r} }} = \frac{ - 1}{{A\left| {\hat{H}} \right|}}C_{r\,r} < 0, $$

where $C_{r\,r} = ( - 1)^{r + r} [\prod\nolimits_{m \ne r}^{L} {(K_{m} - 1)} ][1 + \sum\nolimits_{m \ne r}^{L} { \, \frac{1}{{K_{m} - 1}}} ] > 0$ is the (r, r)th cofactor of $\hat{H}$.

On the other hand, the market-clearing condition for the IET system at equilibrium is

$$ \sum\limits_{m = 1}^{L} {e_{m}^{*} (t_{1} ,..,t_{{L^{\prime}}} ;t_{L + 1} ,..,t_{{N^{\prime}}} )} + \sum\limits_{u = L + 1}^{{N^{\prime}}} {e_{u}^{*} } (p^{*} (t_{1} ,..,t_{{L^{\prime}}} ;t_{L + 1} ,..,t_{{N^{\prime}}} ), \, t_{u} ) + \sum\limits_{{v = N^{\prime} + 1}}^{N} {e_{v}^{*} } (p^{*} (t_{1} ,..,t_{{L^{\prime}}} ;t_{L + 1} ,..,t_{{N^{\prime}}} )) = \sum\limits_{i = 1}^{N} { \, \bar{w}_{i} }. $$

Differentiating this equation with respect to t _r yields

$$ \sum\nolimits_{m = 1}^{L} { \, \frac{{\partial e_{m}^{*} }}{{\partial t_{r} }}} + \sum\nolimits_{n = L + 1}^{N} {\frac{{\partial e_{n}^{*} }}{\partial p}\frac{{\partial p^{ *} }}{{\partial t_{r} }}} = 0. $$

Rearranging the above equation gives $\frac{{\partial p^{*} }}{{\partial t_{r} }} = - (\sum\nolimits_{n = L + 1}^{N} {\frac{{\partial e_{n}^{*} }}{\partial p}} )^{ - 1} (\sum\nolimits_{m = 1}^{L} { \, \frac{{\partial e_{m}^{*} }}{{\partial t_{r} }}} )$. Substituting Eq. (8), i.e., $\frac{\partial p}{{\partial e_{r} }} = \frac{ - 1}{{\sum\nolimits_{n = L + 1}^{N} {(\partial e_{n}^{*} /\partial p)} }} > 0$, into this equation produces

$$ \frac{{\partial p^{*} }}{{\partial t_{r} }} = \left( {\frac{\partial p}{{\partial e_{r} }}} \right)\left( {\sum\limits_{m = 1}^{L} {\frac{{\partial e_{m}^{*} }}{{\partial t_{r} }}} } \right) < 0\;{\text{since}}\;\left( {\sum\limits_{m = 1}^{L} {\frac{{\partial e_{m}^{*} }}{{\partial t_{r} }}} } \right) = \frac{ - 1}{{A\left| {\hat{H}} \right|}}\left[ {\prod\limits_{m \ne r}^{L} {(K_{m} - 1)} } \right] < 0. $$

Proof of Eq. (12)

Differentiating the market-clearing condition with respect to t _u and rearranging it yields

$$ \frac{{\partial p^{*} }}{{\partial t_{u} }} = A \, \left( {\frac{{\partial e_{u}^{*} }}{\partial p} + \sum\limits_{m = 1}^{L} {\frac{{\partial e_{m}^{*} }}{{\partial t_{u} }}} } \right) = A \, \left( {\frac{{\partial e_{u}^{*} }}{\partial p}} \right)\left( {1 + \sum\limits_{m = 1}^{L} {\frac{1}{{K_{m} - 1}}} } \right)^{ - 1} . $$

Since $- 1 < A{ (}\frac{{\partial e_{u}^{*} }}{\partial p} )< 0$, we have $- 1 < \frac{{\partial p^{*} }}{{\partial t_{u} }} < 0$. Differentiating Eq. (4) with respect to t _u gives

$$ \frac{{\partial e_{u}^{*} }}{{\partial t_{u} }} = \left( {\frac{{\partial e_{u}^{*} }}{\partial p}} \right)\left( {1 + \frac{{\partial p^{*} }}{{\partial t_{u} }}} \right) < 0. $$

Proof of Eq. (19)

Following the proof of Eq. (11), we have $\frac{{\partial p^{*} }}{{\partial t_{r} }} = (\frac{\partial p}{{\partial e_{r} }})(\sum\nolimits_{m = 1}^{L} {\frac{{\partial e_{m}^{*} }}{{\partial t_{r} }}} )$. Thus,

$$ \frac{{\partial p^{*} }}{{\partial t_{r} }} - \left( {\frac{\partial p}{{\partial e_{r} }}} \right)\left( {\frac{{\partial e_{r}^{*} }}{{\partial t_{r} }}} \right) = \left( {\frac{\partial p}{{\partial e_{r} }}} \right)\left( {\sum\limits_{m \ne r}^{L} {\frac{{\partial e_{m}^{*} }}{{\partial t_{r} }}} } \right)\;\left\{ {\begin{array}{*{20}c} { = 0} & {{\text{if}}\quad L = 1} \\ { > 0} & {{\text{if}}\quad L \ge 2} \\ \end{array} } \right. $$

due to $(\frac{\partial p}{{\partial e_{r} }}) > 0$ and $\frac{{\partial e_{m}^{*} }}{{\partial t_{r} }} = \frac{ - 1}{{A\left| {\hat{H}} \right|}}C_{r\,m} > 0$ for m = 1, …, L, r = 1, …, L′, and m ≠ r, where $C_{r\,m} = ( - 1)^{2(r + m) - 3} [\prod\nolimits_{t \ne m,r}^{L} {(K_{t} - 1)} ] < 0$ is the (r, m)th cofactor of $\hat{H}$. Combining these results with Eq. (11), Eq. (19) is obvious.