Symmetric Renegotiation-Proof Climate Agreements

Abstract

In this paper, we study the impacts and effects of the number of countries and green technology on climate agreements. In a repeated climate game framework, we focus on the symmetric and self-enforcing weakly renegotiation-proof equilibrium. The set of every country’s feasible abatements is bounded and independent of the number of signatory countries. We derive the maximal number of signatory countries that can reach a fully cooperative (most efficient) climate agreement. We also analyze the maximal equilibrium abatement level when there are too many countries to sustain the fully cooperative agreement. The maximal equilibrium abatement level changes non-monotonically in the number of the countries; it first increases and then decreases in the number of countries involved. Our findings demonstrate a trade-off between the breadth and the depth for international climate agreements.

Appendix

In the proof of Proposition 1 we will use the following lemma, which generalizes Lemma 2 of Farrell and Maskin [7] to n-player repeated games with our symmetry requirements. We denote by $\sigma ^{\prime }\left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$ a simple strategy profile (Abreu [1]) that is defined by $n+1$ paths $\pi ^{0},\pi ^{1},...,$ and $\pi ^{n}$.

Lemma 1

Let $\sigma$ be a SWRP equilibrium with the cooperative path $\pi ^{0}$ for discount factor $\delta \in (0,1)$. Then there exists a simple strategy profile $\sigma ^{\prime }\left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$, which is a SWRP equilibrium, for the same discount factor $\delta$, such that for all $i=1,\ldots ,n,$ we have $U_{i}\left( \sigma ^{\prime }\left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right) \right) =U_{i}\left( \sigma \right)$ and

$$\begin{aligned} U_{i}\left( \pi ^{i}\right)= & {} \inf \left\{ U_{i}\left( \sigma _{h}^{\prime }\right) \mid \sigma _{h}^{\prime } continuation \ equilibria \ of \sigma ^{\prime }\left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right) \right\} \\= & {} \inf \left\{ U_{i}\left( \sigma _{h}\right) \mid \sigma _{h}continuation \ equilibria \ of \sigma \right\} \text {.} \end{aligned}$$

Proof

For $i=1,...,n,$ set $k_{i}=\inf \left\{ U_{i}\left( \sigma _{h}\right) \mid \sigma _{h}\text { continuation equilibria of }\sigma \text { }\right\}$. Then, there exists a sequence $\left\{ \sigma _{w}^{i}\right\} _{w=1}^{\infty }$ of continuation equilibria of $\sigma$ such that $\lim _{w\rightarrow \infty }U_{i}\left( \sigma _{w}^{i}\right) =k_{i}.$

The first period abatement sequence $\left\{ \sigma _{w}^{i}\left( 1\right) \right\} _{w=1}^{\infty }\subseteq \left[ 0,M\frac{b}{c}\right] ^{n}$, which is a compact set. Hence there exists a subsequence which converges to some point of $\left[ 0,M\frac{b}{c}\right] ^{n}$. That is, there exists $\left\{ \sigma _{w,1}^{i}\right\} _{w=1}^{\infty }\subset \left\{ \sigma _{w}^{i}\right\} _{w=1}^{\infty }$ such that $\lim _{w\rightarrow \infty }\sigma _{w,1}^{i}\left( 1\right) =\left( q_{1}^{i}\left( 1\right) ,\ldots ,q_{n}^{i}\left( 1\right) \right) ,$ and since u is continuous, then $\lim _{w\rightarrow \infty }u\left( \sigma _{w,1}^{i}\left( 1\right) \right) =u\left( q_{1}^{i}\left( 1\right) ,\ldots ,q_{n}^{i}\left( 1\right) \right) .$

Proceeding inductively, there exists a subsequence $\left\{ \sigma _{w,t}^{i}\right\} _{w=1}^{\infty }$, such that

$$\begin{aligned} \left\{ \sigma _{w,t}^{i}\right\} _{w=1}^{\infty }\subseteq \left\{ \sigma _{w,t-1}^{i}\right\} _{w=1}^{\infty }\subseteq \cdots \subseteq \left\{ \sigma _{w}^{i}\right\} _{w=1}^{\infty }, \end{aligned}$$

and for all $r=1,\ldots ,t$,

$$\begin{aligned} \lim _{w\rightarrow \infty }\sigma _{w,t}^{i}\left( r\right) =\left( q_{1}^{i}\left( r\right) ,\ldots ,q_{n}^{i}\left( r\right) \right) , \\ \lim _{w\rightarrow \infty }u\left( \sigma _{w,t}^{i}\left( r\right) \right) =u\left( q_{1}^{i}\left( r\right) ,\ldots ,q_{n}^{i}\left( r\right) \right) . \end{aligned}$$

Now we define the simple strategy profile $\sigma ^{\prime }\left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$ as follows: take $\pi ^{0}$ as the outcome path of $\sigma$ and $\pi ^{i}=\left\{ \left( q_{1}^{i}\left( t\right) ,\ldots ,q_{n}^{i}\left( t\right) \right) \right\} _{t=1}^{\infty }$ for $i=1,...,n$. Following similar steps than Farrell and Maskin [1989] in their lemma, we can conclude that $\sigma ^{\prime }\left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$ holds the properties as claimed.

Proof (Proof of Proposition 1)

If abatement cooperation at $v>1$ can be reached for some discount factor $\delta \in (0,1)$, then, from Lemma 1, there exists a SWRP equilibrium $\sigma \left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$ with cooperative path $\pi ^{0}=\left\{ \left( q^{v},\ldots ,q^{v}\right) ,\left( q^{v},\ldots ,q^{v}\right) ,\ldots \right\}$. Denote by $\sigma ^{i}\left( \pi ^{1},...,\pi ^{n}\right)$ the continuation strategy profile induced by $\sigma \left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$ after any unilateral deviation of country i. It holds that $U_{i}\left( \sigma ^{i}\left( \pi ^{1},...,\pi ^{n}\right) \right) =U_{i}(\pi ^{i})$ is the worst continuation payoff among all the continuation payoffs of $\sigma \left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right)$ for country i.

Following similar steps as in Farrell and Maskin [7] for the necessity of their Theorem 1 (page 336), and having in mind that $u_{i}^{BR}\left( q^{v},\ldots ,q^{v}\right) >u_{i}\left( q^{v},\ldots ,q^{v}\right)$ since $v>1$, we obtain that the first-period abatement profile $\left( q_{1}^{i}\left( 1\right) ,\ldots ,q_{n}^{i}\left( 1\right) \right)$ of the path $\pi ^{i}$ holds

$$\begin{aligned} u_{i}^{BR}\left( q_{1}^{i}(1),\ldots ,q_{n}^{i}(1)\right) <U_{i}\left( \sigma \left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right) \right) , \end{aligned}$$

(5)

and

$$\begin{aligned} u_{j}\left( q_{1}^{i}(1),\ldots ,q_{n}^{i}(1)\right) \ge U_{j}\left( \sigma \left( \pi ^{0},\pi ^{1},...,\pi ^{n}\right) \right) \text { for all }j\ne i. \end{aligned}$$

(6)

Symmetry requires that there exists $p\in [0,M]$ and $s\in [0,M]$ such that $q_{i}^{i}(1)=q^{p}$ and $q_{j}^{i}(1)=q^{s}$ for $j\ne i$, and from (5) and (6), the result follows.

Proof (Proof of Proposition 2)

Consider the strategy profile $\sigma (\pi ^{0},\pi ^{1},\ldots ,\pi ^{n})$ defined by (3). This strategy has only $nT+1$ continuations despite the great variety of histories. It is easy to see that conditions (2) imply that there is not one continuation which all the countries strictly prefer to another. To illustrate, since

$$\begin{aligned} u_{1}^{BR}(q^{p},q^{s}\ldots ,q^{s})<u_{1}(q^{v},q^{v},\ldots ,q^{v})\le u_{2}(q^{p},q^{s}\ldots ,q^{s}), \end{aligned}$$

it follows that

$$\begin{aligned} U_{1}\left( \pi ^{1}\right) =(1-\delta ^{T})u_{1}(q^{p},q^{s}\ldots ,q^{s})+\delta ^{T}u_{1}(q^{v},q^{v},\ldots ,q^{v})<U_{1}\left( \pi ^{0}\right) \end{aligned}$$

but $U_{2}\left( \pi ^{0}\right) \le U_{2}\left( \pi ^{1}\right) ,$ and $U_{1}\left( \pi ^{1}\right) <U_{2}\left( \pi ^{1}\right) =U_{1}\left( \pi ^{2}\right)$ but $U_{2}\left( \pi ^{2}\right) =U_{1}\left( \pi ^{1}\right) <U_{2}\left( \pi ^{1}\right)$. Note that this conclusion is independent of T and $\delta$.

We next show that there exists a finite T such that for $\delta$ sufficiently close to 1, the strategy profile $\sigma (\pi ^{0},\pi ^{1},\ldots ,\pi ^{n})$ is SPE. This strategy profile is a simple strategy profile and therefore only one-shot deviations must be checked to ensure this strategy is SPE, where a one-shot deviation from a strategy consists of a single-period unilateral deviation from the strategy, and sticking to it subsequently. Given the symmetry of the strategy, only the following one-shot deviations must be checked:

Case 1: Country 1 has no incentive to take any deviation from $(q^{v},\ldots ,q^{v})$ if

$$\begin{aligned} f(\delta )\equiv (1-\delta )u_{1}^{BR}(q^{v},\ldots ,q^{v})+\delta U_{1}(\pi ^{1})-U_{1}(\pi ^{0})\le 0. \end{aligned}$$

(7)

Note that $\lim _{\delta \rightarrow 1}f(\delta )=0$ and

$$\begin{aligned} \lim _{\delta \rightarrow 1}f^{\prime }(\delta )=-u_{1}^{BR}(q^{v},\ldots ,q^{v})+u_{1}(q^{v},\ldots ,q^{v})-Tu_{1}(q^{p},q^{s},\ldots ,q^{s})+Tu_{1}(q^{v},\ldots ,q^{v}). \end{aligned}$$

Thus taking

$$\begin{aligned} T> & {} \frac{u_{1}^{BR}(q^{v},\ldots ,q^{v})-u_{1}(q^{v},\ldots ,q^{v})}{u_{1}(q^{v},\ldots ,q^{v})-u_{1}(q^{p},q^{s},\ldots ,q^{s})} \\= & {} \frac{\left( \frac{1}{2}+(n-1)v\right) -\left( nv-\frac{v^{2}}{2}\right) }{\left( nv-\frac{v^{2}}{2}\right) -\left( p+(n-1)s-\frac{p^{2}}{2}\right) }, \nonumber \end{aligned}$$

(8)

then $\lim _{\delta \rightarrow 1}f^{\prime }(\delta )>0$ and (7) holds for $\delta$ close enough to 1.

Case 2: If country 1 deviates from $(q^{p},q^{s},\ldots ,q^{s})$ in any period of its punishment phase, then its payoff is at most $(1-\delta )u_{1}^{BR}(q^{p},q^{s},\ldots ,q^{s})+\delta U_{1}(\pi ^{1})$. If country 1 does not deviate in any period of its punishment phase, then its payoff is at least $U_{1}(\pi ^{1})$. Country 1 has no incentive to take any deviation from $(q^{p},q^{s},\ldots ,q^{s})$ if $(1-\delta )u_{1}^{BR}(q^{p},q^{s},\ldots ,q^{s})+\delta U_{1}(\pi ^{1})\le U_{1}(\pi ^{1})$, that is,

$$\begin{aligned} g(\delta )\equiv u_{1}^{BR}(q^{p},q^{s},\ldots ,q^{s})-(1-\delta ^{T})u_{1}(q^{p},q^{s},\ldots ,q^{s})-\delta ^{T}u_{1}(q^{v},\ldots ,q^{v})\le 0. \end{aligned}$$

(9)

Note that $\lim _{\delta \rightarrow 1}g(\delta )=u_{1}^{BR}(q^{p},q^{s},\ldots ,q^{s})-u_{1}(q^{v},\ldots ,q^{v})<0$, from the left-hand side of (2). Therefore, for any T, (9) holds for $\delta$ close enough to 1.

Case 3: If country 2 deviates from $(q^{p},q^{s},\ldots ,q^{s})$ in any period of the punishment phase of country 1, then its payoff is at most $(1-\delta )u_{2}^{BR}(q^{p},q^{s},\ldots ,q^{s})+\delta U_{2}(\pi ^{2})$. If country 2 does not deviate in any period of the punishment phase of country 1, then its payoff is at least $(1-\delta )u_{2}(q^{p},q^{s},\ldots ,q^{s})+\delta u_{2}(q^{v},\ldots ,q^{v})$, due to the right-hand side of (2). Country 2 has no incentive to take any deviation from $(q^{p},q^{s},\ldots ,q^{s})$ if $h(\delta )\le 0$, where

$$\begin{aligned} h(\delta )\equiv & \ (1-\delta )u_{2}^{BR}(q^{p},q^{s},\ldots ,q^{s})+\delta U_{2}(\pi ^{2})\\ &- \ \left[ (1-\delta )u_{2}(q^{p},q^{s},\ldots ,q^{s})+\delta u_{2}(q^{v},\ldots ,q^{v})\right] . \end{aligned}$$

(10)

Note that $\lim _{\delta \rightarrow 1}h(\delta )=0$ and

$$\begin{aligned} \lim _{\delta \rightarrow 1}h^{\prime }(\delta )=&-u_{2}^{BR}(q^{p},q^{s},\ldots ,q^{s})-Tu_{2}(q^{s},q^{p},q^{s},\ldots ,q^{s})\\&+Tu_{2}(q^{v},\ldots ,q^{v})+u_{2}(q^{p},q^{s},\ldots ,q^{s}). \end{aligned}$$

Thus taking

$$\begin{aligned} T> & {} \frac{u_{2}^{BR}(q^{p},q^{s},\ldots ,q^{s})-u_{2}(q^{p},q^{s},\ldots ,q^{s})}{u_{2}(q^{v},\ldots ,q^{v})-u_{2}(q^{s},q^{p},q^{s},\ldots ,q^{s})} \\= & {} \frac{\left( p+(n-2)s+\frac{1}{2}\right) -\left( p+(n-1)s-\frac{s^{2}}{2} \right) }{\left( nv-\frac{v^{2}}{2}\right) -\left( p+(n-1)s-\frac{p^{2}}{2} \right) }, \nonumber \end{aligned}$$

(11)

then $\lim _{\delta \rightarrow 1}h^{\prime }(\delta )>0$ and (10) holds for $\delta$ close enough to 1. Summarizing, the strategy $\sigma (\pi ^{0},\pi ^{1},\ldots ,\pi ^{n})$is SWRP for $\delta$ close enough to 1 when T holds (8) and (11).

Proof (Proof of Proposition 3)

Let’s fix $v\in (1,n]$ and consider the discount factor is close enough to 1. We will see below that the existence of values p and s holding conditions (2) is equivalent to $p\in (\bar{p},M]$, where $\bar{p }=\frac{1}{2\left( n-1\right) ^{2}}\left( nv-\frac{1}{2}v^{2}-\frac{1}{2} \right) ^{2}+\frac{1}{2}$, and then the result follows.

The left-hand side of conditions (2) holds

$$\begin{aligned} \frac{1}{2}+(n-1)s<nv-\frac{1}{2}v^{2}\text { if and only if }s<\hat{s}, \end{aligned}$$

(12)

where $\hat{s}=\frac{1}{n-1}\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right)$. The value $\hat{s}>1$ for $v\in (1,n]$ since $\hat{s}=1$ for $v=1$ and $\hat{ s}$ is strictly increasing on [1, n].

The right-hand side of conditions (2) holds

$$\begin{aligned} nv-\frac{1}{2}v^{2}\le p+(n-1)s-\frac{1}{2}s^{2}\text { if and only if }\bar{s}_{p}\le s\le \hat{s}_{p}\text {,} \end{aligned}$$

(13)

when the roots of the quadratic inequality $\bar{s}_{p}=n-1-\sqrt{ (n-1)^{2}+2p-2nv+v^{2}}$ and $\hat{s}_{p}=n-1+\sqrt{(n-1)^{2}+2p-2nv+v^{2}}$ are real numbers. When $p<p^{*}=nv-\frac{1}{2}v^{2}-\frac{1}{2}(n-1)^{2}$ then the roots are not real numbers and $p+(n-1)s-\frac{1}{2}s^{2}<nv-\frac{1 }{2}v^{2}$ for all $s\in \mathbb {R}$. Therefore conditions (2) never hold when $p<p^{*}$.

Next, we study if cooperation v is possible when $p\ge p^{*}$. For it we have to combine (12) and (13), that is, we need to know when $\bar{s}_{p}<\hat{s}$. Note that $\hat{s}\le n-1$ when $n\ge 3$ and $v\in (1,n]$. Let’s consider $\bar{p}=\frac{1}{2\left( n-1\right) ^{2}} \left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) ^{2}+\frac{1}{2}$. It follows that

$$\begin{aligned} \bar{s}_{\bar{p}}= & \ n-1-\sqrt{(n-1)^{2}+2\left( \frac{1}{2\left( n-1\right) ^{2}}\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) ^{2}+\frac{1}{2}\right) -2nv+v^{2}} \\= & \ n-1-\frac{1}{\left( n-1\right) }\sqrt{(n-1)^{4}+\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) ^{2}-2(n-1)^{2}\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) } \\= & \ n-1-\frac{1}{\left( n-1\right) }\left( (n-1)^{2}-\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) \right) \\= & \ \frac{1}{\left( n-1\right) }\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) =\hat{s}. \end{aligned}$$

Note that $(n-1)^{2}-\left( nv-\frac{1}{2}v^{2}-\frac{1}{2}\right) \ge 0$ when $n\ge 3$ and $v\in (1,n]$. Hence, $\bar{s}_{p}<\hat{s}$ if and only if $p>\bar{p}$.

Summarizing for $n\ge 3$, if the punishment level $p\le \bar{p}$ then there is not s holding conditions (2). If $M>\bar{p}$, then for any $p\in (\bar{p},M]$ there is a range of values, $s\in \left[ \max \left\{ 1,\bar{s}_{p}\right\} ,\hat{s}\right)$, which holds conditions (2).

For the particular case $n=2$, it is easy to see that the punishment level p must be at least $p^{*}=2v-\frac{1}{2}v^{2}-\frac{1}{2}$, in which case $s=1$ holds conditions (2).

Proof (Proof of Proposition 4)

If the upper bound M holds $M\le \frac{1}{8}\left( n+1\right) ^{2}+\frac{1}{2}$, then full cooperation is not reached. The inequality (4) is equivalent to

$$\begin{aligned} v^{2}-2nv+1+2(n-1)\sqrt{2M-1}>0. \end{aligned}$$

Equation $v^{2}-2nv+1+2(n-1)\sqrt{2M-1}=0$ has the roots

$$\begin{aligned} v=n\pm \sqrt{n^{2}-1-2(n-1)\sqrt{2M-1}}, \end{aligned}$$

which are real numbers when $M\le \frac{1}{8}\left( n+1\right) ^{2}+\frac{1}{2}$. It holds that

$$\begin{aligned} v^{2}-2nv+1+2(n-1)\sqrt{2M-1}>0 \end{aligned}$$

for $v\in (1,n]$ if $M>\frac{1}{8}\left( n+1\right) ^{2}+\frac{1}{2}$, and for

$$\begin{aligned} v\in (1,n-\sqrt{n^{2}-1-2(n-1)\sqrt{2M-1}}) \end{aligned}$$

if $1<M\le \frac{1}{8}\left( n+1\right) ^{2}+\frac{1}{2}$, when $n\ge 3$. If $n=2$, the maximal cooperation is $2-\sqrt{3-2M}$ when $1<M\le 3/2$. Remember that the cooperation $v=1$ is reached for any discount factor.

Moreover, it is the case that

$$\begin{aligned} lim_{n\rightarrow \infty }\left( n-\sqrt{n^{2}-1-2(n-1)\sqrt{2M-1}}\right) =\sqrt{2M-1}, \end{aligned}$$

that is, the cooperation upper bound decreases asymptotically to $\sqrt{2M-1}$ when the number of countries becomes large.