Global Emission Ceiling Versus International Cap and Trade: What is the Most Efficient System to Solve the Climate Change Issue?

Abstract

We model the climate change issue as a pollution control game with the purpose of comparing two possible departures from the business as usual (BAU) where countries noncooperatively choose their emission levels. In the first scenario, players have to agree on a global emission cap (GEC) that is enforced by a uniform taxation scheme. They still behave strategically when choosing emission levels but are now subject to the coupled constraint imposed by the cap. The second scenario consists of the implementation of an international cap and trade (ICT) system. In this case, players decide on their emission quotas, and emission trading is allowed. A three heterogenous player quadratic game serves as a basis for the analysis. When the cap is binding, among all the coupled constraints Nash equilibria, we select a particular normalized equilibrium by solving a variational inequality. Comparing the normalized equilibrium with the Nash equilibria of the BAU and the ICT, we first show that if the cap is appropriately chosen, then the GEC system improves all players’ payoffs, relative to the BAU. The GEC system may thus be unanimously approved whereas the ICT is not, because moving from the BAU to the ICT is costly for one player. Second, for some values of the cap, all players get a higher payoff under the GEC than under the ICT. Therefore, the GEC outperforms the ICT both in terms of feasibility and efficiency.

Appendix A

1.1 A.1 Proof of Theorem 1

$\bar {e} ^{R}(\alpha ,\bar {r}) = \big (e_{1}^{R}(\alpha ,\bar {r}),\ldots , e_{n}^{R}(\alpha , \bar {r})\big ) \in \mathcal {E(\alpha )}$ is a $\bar {r}$-normalized equilibrium for the pseudogame ${\Gamma }(\alpha )$ if and only if:

$$ \sum\limits_{i=1}^{n} \, r_{i}W_{i}\big(\bar{e}^{R}(\alpha,\bar{r})\big) = \mbox{Max}_{\bar{e} \in \mathcal{E(\alpha)} }\; \sum\limits_{i=1}^{n} \; r_{i}W_{i}\big(e_{i}, \bar{e}_{-i}^{R}(\alpha,\bar{r})\big) $$

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that is, if and only if: $\bar {e}^{R}(\alpha ,\bar {r})$ is a fixed point of the set-valued function $L_{\alpha }$ defined on $\mathcal {E(\alpha )}$ by:

$$\begin{array}{rll} \bar{u} &=& (u_{1}, \ldots, u_{n}) \in \mathcal{E(\alpha)} \longrightarrow L_{\alpha}(\bar{u}) \\&=& \mbox{ArgMax}_{\bar{e} \in \mathcal{E(\alpha)} } \; \sum\limits_{i=1}^{n} \; r_{i}W_{i}(e_{i}, \bar{u}_{-i}). \end{array} $$

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For $\alpha > 0$, $\mathcal {E(\alpha )}$ and $L_{\alpha }$ satisfy the assumptions of Kakutani’s theorem which guarantees the existence of such a fixed point.

In fact:

• 1. $\mathcal {E(\alpha )}$ is a nonempty, convex and compact subset of $\mathbf {R}^{n}$;

• 2. $L_{\alpha }$ has closed graph;

• 3. $L_{\alpha }(\bar {u})$ is a nonempty and convex subset of $\mathcal {E(\alpha )}$ for all $\bar {u} \in \mathcal {E(\alpha )}.$

The proofs is as follows:

• 1. The first point is obvious.

• 2. Let us show that, for all $\alpha >0$, $L_{\alpha }$ has a closed graph over $\mathcal {E(\alpha )}$.

  Let $(u_{1,k}, \ldots ,u_{n,k})_{k \in \bf N}$ be a sequence such that $\bar {u}_{k}=(u_{1,k}, \ldots ,u_{n,k}) \in \mathcal {E(\alpha )}$ for all ${k \in \bf N}$ and such that $\bar {u}_{k}$ converges to $\bar {u}$ in $\mathcal {E(\alpha )}$ as $k \to \infty$. Then, for all $i \in N, u_{i,k}$ converges to $u_{i}$ as $k \to \infty$ and, since $\mathcal {E(\alpha )}$ is a closed set, $\bar {u}=(u_{1},\ldots , u_{n}) \in \mathcal {E(\alpha )}$.

  Moreover, let $(e^{R}_{1,k}, \ldots ,e^{R}_{n,k})_{k \in \bf N}$ be a sequence such that:

  $$\begin{array}{rll} \bar{e}^{R}_{k}&=& \big(e^{R}_{1,k}, \ldots,e^{R}_{n,k}\big)\,\, \mbox{converges to}\, \,\, \bar{e}^{R} \\&=&\big (e^{R}_{1}, \ldots, e^{R}_{n}\big)\,\,\, \mbox{as}\, k \to \infty \,\,\,\,\mbox{and} \end{array} $$

  (25)
  $$\bar{e}^{R}_{k}\in L_{\alpha}( \bar{u}_{k}), \, \mbox{for all} \; k \in \bf N. $$

  (26)

  Then, for all $i \in N$, $e^{R}_{i,k}$ converges to $e^{R}_{i}$ as $k \to \infty$.

  One has to prove that $\big (e_{1}^{R},\ldots \,, e_{n}^{R}\big ) \in L_{\alpha }(u_{1},\ldots \;, u_{n})$, that is:

  $$ \bar{e}^{R}=\big(e_{1}^{R},\ldots \,, e_{n}^{R}\big) \in \mathcal{E(\alpha)} \,\,\,\mbox{and}\\ $$

  (27)
  $$\begin{array}{rll}\bar{e}{\kern-2.5pt}={\kern-2.5pt}(e_{1}, \ldots \,, e_{n}){\kern-2.2pt}\in{}\mathcal{E{}({}\alpha{})} {} \;{} &\Longrightarrow &\;{} \sum\limits_{i=1}^{n}r_{i}W_{i}\big(e^{R}_{i}, \bar{u}_{-i}\big)\\&\geq& \sum\limits_{i=1}^{n}r_{i}W_{i}(e_{i}, \bar{u}_{-i})\\ \end{array} $$

  (28)

  Due to Eq. (25) and Eq. (26), we obtain condition Eq. (27) using a limit process.

  Regarding implication Eq. (28), let $\bar {e} = (e_{1}, \ldots \,, e_{n}) \in \mathcal {E(\alpha )}$. From Eq. (26), one has:

  $$ \sum\limits_{i=1}^{n}r_{i}W_{i}\big(e_{i,k}^{R}, \bar{u}_{-i,k}\big)\geq \sum\limits_{i=1}^{n}r_{i}W_{i}(e_{i}, \bar{u}_{-i,k}) \\ $$

  The function $W_{i}$ is continuous on $\mathbf {R}^{n}_+$, for all $i \in N$, therefore

  $$\begin{array}{lll}&& \sum\limits_{i=1}^{n} r_{i}\lim_{k \to \infty} W_{i}(e_{i,k}^{R}, \bar{u}_{-i,k})\\ &&\quad\; \geq\sum\limits_{i=1}^{n}r_{i}\lim\limits_{k \to \infty}W_{i}(e_{i}, \bar{u}_{-i,k}) \end{array} $$

  This implies

  $$ \sum\limits_{i=1}^{n}r_{i}W_{i}\big(e^{R}_{i}, \bar{u}_{-i}\big)\geq \sum\limits_{i=1}^{n}r_{i}W_{i}(e_{i}, \bar{u}_{-i}) $$

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  This is true for all $\bar {e} = (e_{1}, \ldots \,, e_{n}) \in \mathcal {E(\alpha )}$. Therefore, $L_{\alpha }$ has closed graph over $\mathcal {E(\alpha )}$.

• 3. For all $\bar {u} \in \mathcal {E(\alpha )}, L_{\alpha }(\bar {u})$ is a nonempty set since $\mathcal {E(\alpha )}$ is compact and, for all $i \in N$, the function $W_{i}$ is continuous $\mathbf {R}^{n}_+$.

  For all $\bar {u} \in \mathcal {E(\alpha )}, L_{\alpha }(\bar {u})$ is a convex set since the function defined by $F_{\bar {u}}(\bar {e})=\sum _{i=1}^{n} \; r_{i}W_{i}(e_{i}, \bar {u}_{-i})$ is assumed to be quasiconcave.

1.2 A.2 Proof of Theorems 2 and 3

Let $\alpha > 0$ and $\alpha _{k} \in \mathbf {R}_+$ such that $\alpha _{k} \rightarrow \alpha$ as $k \to \infty$. We have to show that $\bar {e}^{R}(\alpha _{k},\bar {r}) \rightarrow \bar {e}^{R}(\alpha ,\bar {r})$ as $k \to \infty$.

• There exists a compact set $K_{\alpha } \subset \mathbf {R}^{n}_+$ such that $\mathcal {E}(\alpha _{k}) \subset K_{\alpha }$ for all $k \in \bf N$ sufficiently large. So there exists a subsequence $(\bar {e}^{R}(\alpha _{k_{j}},\bar {r}))_{j \in \bf N}$ and a vector $\bar {e}^{R} \in K_{\alpha }$ such that: $\bar {e}^{R}(\alpha _{k_{j}},\bar {r}) \rightarrow \bar {e}^{R}$. Since $\bar {e}^{R}(\alpha _{k_{j}},\bar {r}) \in \mathcal {E}(\alpha _{k_{j}})$, for all $k_{j}$, and $\alpha _{k_{j}} \rightarrow \alpha$, we have: $\bar {e}^{R} \in \mathcal {E(\alpha )}.$

• Now, let $\bar {e} \in \mathcal {E(\alpha )}$. Then, for all $j \in \bf N$, there exists $\bar {e_{j}} = (e_{1,j}, \cdots \,,e_{n,j}) \in \mathcal {E}(\alpha _{k_{j}})$ such that $\bar {e_{j}} \rightarrow \bar {e}$ as $j \to \infty$. But $\bar {e}^{R}(\alpha _{k_{j}},\bar {r})$ satisfies:

  $$\begin{array}{rll} \sum\limits_{i=1}^{n}r_{i}W_{i}\big(\bar{e}^{R}(\alpha_{k_{j}},\bar{r})\big)&\geq& \sum\limits_{i=1}^{n}r_{i}W_{i}\big(f_{i}, \bar{e}_{-i}^{R}(\alpha_{k_{j}},\bar{r})\big)\;\\ &&{}\mbox{for all} \; \bar{f}= (f_{1},...,f_{n}) \in \mathcal{E}(\alpha_{k_{j}}) \end{array} $$

  Then:

  $$\sum\limits_{i=1}^{n}r_{i}W_{i}\big(\bar{e}^{R}(\alpha_{k_{j}},\bar{r})\big)\geq \sum\limits_{i=1}^{n}r_{i}W_{i}\big(e_{i,j}, \bar{e}_{-i}^{R}(\alpha_{k_{j}},\bar{r})\big).$$

  Passing to the limit for $j \to \infty$, we obtain:

  $$\sum\limits_{i=1}^{n}r_{i}W_{i}\big(\bar{e}^{R}\big)\geq \sum\limits_{i=1}^{n}r_{i}W_{i}\big(e_{i}, \bar{e}_{-i}^{R}\big). $$

  But this is true for all $\bar {e} \in \mathcal {E(\alpha )}$ then $\bar {e}^{R}$ coincides with $\bar {e}^{R}(\alpha ,\bar {r})$, the unique $\bar {r}$-normalized equilibrium and we have:

  $$\bar{e}^{R}(\alpha_{k_{j}},\bar{r}) \rightarrow \bar{e}^{R}(\alpha,\bar{r})\, \,\mbox{as} \,\, j \to \infty. $$

• This is true for all convergent subsequences of $\bar {e}^{R}(\alpha _{k},\bar {r})$ so one can prove that this is true for the sequence $(\bar {e}^{R}(\alpha _{k},\bar {r}))_{k \in N}$ and at the end we obtain:

  $$\bar{e}^{R}(\alpha_{k},\bar{r}) \rightarrow \bar{e}^{R}(\alpha,\bar{r}) \,\mbox{as} \, k \to \infty, \,\mbox{for all}\, \alpha_{k} \to \alpha $$

  which completes the proof.

1.3 A.3 Proof of Proposition 1

The proof follows directly from the comparison between (a) Eqs. (13) and (15) and (b) Eqs. (10) and (15). Indeed, $\omega _{i}^{N}<e_{i}^{*}(\omega ^{N})\Leftrightarrow \gamma _{-i}-2\gamma _{i}<0$ and $\omega _{i}^{N}<e_{i}^{N}\Leftrightarrow \gamma _{-i}-2\gamma _{i}<0$. The same inequality is involved in the two comparisons. Under assumption (9), this inequality is satisfied for player 1 whereas the converse holds for players 2 and 3.

1.4 A.4 Proof of Proposition 2

One obtains the Nash equilibrium payoffs in the two scenarios, $W_{i}^{N}$ and $V_{i}^{N}$, by substituting respectively the solutions Eqs. (10) and (15) in Eqs. (8) and (14). Then, one can verify that $W_{i}^{N}<V_{i}^{N}\Leftrightarrow -\alpha ^{2}(\gamma _{-i}-2\gamma _{i})^{2}<4\alpha (\gamma _{-i}-2\gamma _{i})(\kappa _{i}+\kappa _{-i}-2\alpha )$. Under assumption (9), this inequality is satisfied for players 2 and 3 but not for player 1. In addition, $W^{N}=\sum W_{i}^{N}<V^{N}=\sum V_{i}^{N}$ always holds.

1.5 A.5 Proof of Proposition 3

Consider $\bar {e}$ such that $e_{i}+e_{-i}= \alpha \Leftrightarrow e_{-i}^{R}-e_{-i}=-(e_{i}^{R}-e_{i})$ and use this relation to remove $e_{2}$ and $e_{2}^{R}$ from the variational inequality Eq. (19), then:²⁰

$$\begin{array}{r} \sum_{i\neq2}\big(\kappa_{i} - \kappa_{2} + (2 + \gamma_{2} - \gamma_{i})\alpha - 4e_{i}^{R} - 2e_{-i\setminus2}^{R}\big)\big(e_{i}^{R} - e_{i}\big)\geq0\\ \mathrm{\,for\,all }\,\bar{e}\,\mathrm{such\,that }\, e_{i}+e_{-i}= \alpha\, \mathrm{and}\, e_{i},e_{-i}\geq\,0 \end{array} $$

If the equation $\kappa _{i}-\kappa _{2}+(2+\gamma _{2}-\gamma _{i})\alpha -4e_{i}^{R}-2e_{-i \setminus 2}^{R}=0$ holds for $i=1,3$, then using the feature that $e_{2}^{R}=\alpha -e_{-2}^{R}$, one gets the interior solution

$$\begin{array}{r} e_{i}^{R}(\alpha)=e_{i}(\alpha)=\frac{2\kappa_{i}-\kappa_{-i}+\alpha\big(2(1-\gamma_{i})+\gamma_{-i}\big)}{6}\\ \mathrm{for\,all}\, i=1,2,3 \end{array} $$

Otherwise, it is possible to obtain corner solutions with $e_{i}^{R}(\alpha )=0$ or $e_{i}^{R}(\alpha )=\alpha$. Note that, by definition, $e_{2}(\alpha )\in [0,\alpha ]$ when $e_{1}(\alpha )$ and $e_{3}(\alpha )$ also lie in the interval $[0,\alpha ]$. From assumption (12), one can easily check that $e_{1}(\alpha )\geq 0$ and $e_{3}(\alpha )\leq \alpha$ for all $\alpha \in (0,\bar {\alpha }]$. Define two boundaries $\hat { \alpha }$ and $\tilde { \alpha }$ as:

$$\begin{array}{rll} \hat{ \alpha}&=&\frac{\kappa_{3}-\kappa_{1}}{2(-2+\gamma_{3}-\gamma_{1})}\;\big (\Leftrightarrow e_{1}(\alpha)=\alpha\big) \mathrm{and}\\ \tilde{\alpha}&=&\frac{2(\kappa_{1}-\kappa_{3})}{2+\gamma_{1}-\gamma_{3}} \;\big(\Leftrightarrow e_{3}(\alpha)=0\big), \end{array} $$

these boundaries satisfy: $0<\hat { \alpha }<\tilde { \alpha }<\bar { \alpha }$. Putting all these elements together, one obtains the strategies as defined by Eq. (20).

1.6 A.6 Proof of Proposition 4

To give the proof of proposition 4, expressions of individual payoffs in both scenarios are needed.

Using the property that strategies, and consequently individual payoffs, are continuous at the critical bound $\bar {\alpha }$, $e_{i}^{R}(\bar {\alpha })=e_{i}^{N}$, one obtains

$$\begin{array}{rll} W_{i}^{R}(\alpha)& = & \frac{1}{36}\big(\big(4\kappa_{i} + \kappa_{-i} - \alpha\big(2(1 - \gamma_{i}) + \gamma_{-i}\big)\big)\\ && \times \big(2\kappa_{i} - \kappa_{-i} + \alpha\big(2(1 - \gamma_{i}) + \gamma_{-i}\big)\big)\big) - \frac{\gamma_{i}}{2}\alpha^{2}, \end{array} $$

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with $W_{i}^{R}(\bar {\alpha })=W_{i}^{N}$ and expressions are symmetric for players ${-i}$.

The proof then uses the features of the value function $W_{i}^{R}(\alpha )$ in the neighborhood of the critical bound $\bar {\alpha }$. For the sake of simplicity, attention is paid to the interval $[\tilde {\alpha },\bar {\alpha }]$. Properties of $W_{i}^{R}(\alpha )$: $(W_{i}^{R})'(\alpha )\geq 0\Leftrightarrow \alpha \leq \alpha _{i}$ with $\alpha _{i}= \frac {(2(1-\gamma _{i})+\gamma _{-i})(\kappa _{i}+\kappa _{-i})}{(2(1-\gamma _{i})+\gamma _{-i})^{2}+18\gamma _{i}}$. In addition, $\alpha _{i}<\bar {\alpha }\Leftrightarrow -\gamma _{-i}+2\gamma _{i} +4>0$, which holds under assumptions (9) and (12) for all i. The function $W_{i}^{R}(\alpha )$ is decreasing in the neighborhood of $\bar {\alpha }$ and, since it is continuous at $\bar {\alpha }$, $\exists I_{i}\subseteq [\tilde {\alpha },\bar {\alpha }]$ such that $\forall \alpha \in I_{i}$, $W_{i}^{R}(\alpha )\geq W_{i}^{N}$. This feature holds for $i=1,2,3$. Finally notice that $\cap I_{i} \neq \phi$.

1.7 A.7 Proof of Proposition 5

Here, we also need the expression of player i’s payoff at the Nash equilibrium of the ICT system:

$$\begin{array}{rll} V_{i}^{N}&=&\frac{1}{36}\big((4\kappa_{i}+\kappa_{-i}-2\bar{\alpha})(2\kappa_{i}-\kappa_{-i}+2\bar{\alpha})\big)\\ && +\frac{1}{6}(\kappa_{i}+\kappa_{-i}-2\bar{\alpha})(\gamma_{-i}-2\gamma_{i})\bar{\alpha} -\frac{\gamma_{i}}{2}\bar{\alpha}^{2}, \end{array} $$

with $\bar {\alpha }$ the BAU aggregate emission level as defined by Eq. (18).

Let us first compute the difference between the maximum possible payoff under the GEC alone, $W_{i}^{R}(\alpha _{i})$, and the payoff at the Nash equilibrium of BAU, $W_{i}^{N}=W_{i}^{R}(\bar {\alpha })$

$$W_{i}^{R}\left(\alpha_{i}^{W}\right)-W_{i}^{R}(\bar{\alpha})=\frac{9\gamma_{i}^{2}\big(\gamma_{-i}-2\gamma_{i}-4\big)^{2}}{36\big((2+\gamma_{-i}-2\gamma_{i})^{2}+18\gamma_{i}\big)}\bar{\alpha}^{2}>0 $$

Next, compute the difference between the Nash equilibrium payoffs of the BAU and the ICT system:

$$V_{i}^{N}-W_{i}^{N}=\frac{(\gamma_{-i}-2\gamma_{i})(5\gamma_{-i}+2\gamma_{i})}{36}\bar{\alpha}^{2}>0 $$

To complete the proof, it is sufficient to impose $W_{i}^{R}(\alpha _{i})>V_{i}^{N}$ which is equivalent to Eq. (22). This implies that there exist values of $\alpha$ such that players 2 and 3 are better off under the GEC system alone.