A Dynamic Game of Emissions Pollution with Uncertainty and Learning

Abstract

We introduce learning in a dynamic game of international pollution, with ecological uncertainty. We characterize and compare the feedback non-cooperative emissions strategies of players when the players do not know the distribution of ecological uncertainty but they gain information (learn) about it. We then compare our learning model with the benchmark model of full information, where players know the distribution of ecological uncertainty. We find that uncertainty due to anticipative learning induces a decrease in total emissions, but not necessarily in individual emissions. Further, the effect of structural uncertainty on total and individual emissions depends on the beliefs distribution and bias. Moreover, we obtain that if a player’s beliefs change toward more optimistic views or if she feels that the situation is less risky, then she increases her emissions while others react to this change and decrease their emissions.

Appendix

1.1 Proof of Proposition 1

Denote by ${v_{i}^{I}}\left( S;\theta ^{*}\right)$ the value function of player $i$, where the superscript $I$ refers to an informed player. The Hamilton–Jacobi–Bellman (HJB) equation is given by

$$\begin{aligned} {v_{i}^{I}}\left( S;\theta ^{*}\right)&= \max _{e_{i}}\left\{ e_{i}\left( \alpha _{i}-e_{i}-\gamma \sum _{j\ne i}^{N}e_{j}\right) -\beta _{i}S\right. \nonumber \\&\left. +\,\delta \int _{\mathcal {H}} \left( {v_{i}^{I}}\left( \eta \left( \sum _{j=1}^{N}e_{j}+dS\right) \right) \right) \phi (\eta |\theta ^{*}){\mathbf {d}}\eta \right\} . \end{aligned}$$

(28)

Considering the linear-state structure of our model, we conjecture that the value function is linear and specified as follows:

$$\begin{aligned} {v_{i}^{I}}\left( S;\theta ^{*}\right) ={\kappa _{i,1}^{I}}S+\kappa _{i,2}^{I}. \end{aligned}$$

Plugging the conjectured value function in (50), we obtain

$$\begin{aligned} {\kappa _{i,1}^{I}}S+\kappa _{i,2}^{I}&= \max _{e_{i}}\left\{ e_{i}\left( \alpha _{i}-e_{i}-\gamma \sum _{j\ne i}^{N}e_{j}\right) -\beta _{i}S\right. \nonumber \\&\left. +\,\delta \int _{\mathcal {H}}\left[ {\kappa _{i,1}^{I}}\eta \left( \sum _{j=1}^{N}e_{j}+dS\right) +\kappa _{i,2}^{I} \right] \phi (\eta |\theta ^{*}){\mathbf {d}}\eta \right\} . \end{aligned}$$

(29)

The first-order equilibrium condition is given by⁸

$$\begin{aligned} \alpha _{i}-2e_{i}-\gamma \sum _{j\ne i}e_{j}+\delta \int _{\mathcal {H}} {\kappa _{i,1}^{I}}\eta \phi (\eta |\theta ^{*}){\mathbf {d}}\eta =0. \end{aligned}$$

(30)

To find the coefficients of the value function, we substitute for ${e_{i}^{I}}$ in (29) and equate the coefficients in order of $S$. This leads to a system of $2N$ equations (2 equations for each player) and $2N$ unknowns (${\kappa _{i,1}^{I}}$ and $\kappa _{i,2}^{I}\,\forall i=1,2$), that is,

$$\begin{aligned} {\kappa _{i,1}^{I}}&= -\beta _{i}+\delta (1-d)\int _{\mathcal {H}} \eta {\kappa _{i,1}^{I}}\phi (\eta |\theta ^{*}){\mathbf {d}}\eta , \end{aligned}$$

(31)

$$\begin{aligned} \kappa _{i,2}^{I}&= \alpha _{i}e_{i}-e_{i}^{2}-\gamma e_{i}\sum _{j\ne i}e_{j}+\delta \int _{\mathcal {H}}\left( \eta {\kappa _{i,1}^{I}}\left( \sum _{j=1}^{N}e_{j}^{I}\right) +\kappa _{i,2}^{I}\right) \left( \phi (\eta | \theta ^{*})\right) {\mathbf {d}}\eta . \end{aligned}$$

(32)

Since $\mu (\theta ^{*})=\int _{\mathcal {H}} \eta \phi (\eta |\theta ^{*}){\mathbf {d}}\eta$, from (31), we get

$$\begin{aligned} {\kappa _{i,1}^{I}}=\frac{-\beta _{i}}{1-\delta d\mu (\theta ^{*})}, \end{aligned}$$

(33)

and from (32)

$$\begin{aligned} {\kappa _{i,2}^{I}}=\frac{1}{1-\delta }\left( \alpha _{i}{e_{i}^{I}} -\left( {e_{i}^{I}}\right) ^{2}-\gamma e_{i}\sum _{j\ne i}e_{j}+ \delta \mu (\theta ^{*}){\kappa _{i,1}^{I}}\left( \sum _{j=1}^{N}e_{j}^{I}\right) \right) . \end{aligned}$$

To solve for $\left\{ {e_{i}^{I}}\right\} _{i=1}^{N}$ rewrite (30) as a function of total emissions, $\sum _{j=1}^{N}e_{j}$, i.e.,

$$\begin{aligned} \alpha _{i}-\left( 2-\gamma \right) e_{i}-\gamma \sum _{j=1}^{N}e_{j}+\delta \int _ {\mathcal {H}}{\kappa _{i,1}^{I}}\eta \phi (\eta |\theta ^{*}){\mathbf {d}} \eta =0, \end{aligned}$$

(34)

and consequently, the best reaction function of player $i$ is

$$\begin{aligned} {e_{i}^{I}}=\frac{1}{2-\gamma }\left( \alpha _{i}-\gamma \sum _{j=1}^{N}e_{j}-\delta \frac{\beta _{i}\mu (\theta ^{*})}{1-\delta d\mu (\theta ^{*})}\right) . \end{aligned}$$

(35)

We first solve for $\sum _{j=1}^{N}e_{j}$. Summing 35 over $i$ yields

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{I}}=\frac{1}{2-\gamma }\left( \sum _{i=1}^{N}\alpha _{i}- \gamma N\sum _{j=1}^{N}e_{j}-\delta \frac{\mu (\theta ^{*}) \sum _{i=1}^{N}\beta _{i}}{1-\delta d\mu (\theta ^{*})}\right) . \end{aligned}$$

(36)

So that total emission is

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{I}}=\frac{1}{2+\left( N-1\right) \gamma } \left( \sum _{i=1}^{N}\alpha _{i}-\delta \frac{\mu (\theta ^{*})}{ 1-\delta d\mu (\theta ^{*})}\sum _{i=1}^{N}\beta _{i}\right) . \end{aligned}$$

(37)

Plugging (37) into 35 yields

$$\begin{aligned} {e_{i}^{I}}=\frac{1}{2-\gamma }\left( \alpha _{i}-\frac{\gamma }{ 2+\left( N-1\right) \gamma }\left( \sum _{i=1}^{N}\alpha _{i}-\delta \frac{ \mu (\theta ^{*})}{1-\delta d\mu (\theta ^{*})} \sum _{i=1}^{N}\beta _{i}\right) -\delta \frac{\beta _{i}\mu \left( \theta ^{*} \right) }{1-\delta d\mu (\theta ^{*})}\right) . \end{aligned}$$

(38)

By rearranging we have

$$\begin{aligned} {e_{i}^{I}}=\frac{1}{2-\gamma }\left( \alpha _{i}-\frac{\gamma \sum _{i=1}^{N} \alpha _{i}}{2+\left( N-1\right) \gamma }-\frac{\delta \mu \left( \theta ^{*} \right) }{1-\delta d\mu (\theta ^{*})}\left( \beta _{i}-\frac{ \gamma \sum _{j=i}^{N}\beta _{j}}{2+\left( N-1\right) \gamma }\right) \right) . \end{aligned}$$

(39)

1.2 Proof of Proposition 2

By virtue of our linear-state model we conjecture that the value function of player $i$ has the linear form of $v_{i}^{L}={\kappa _{i,1}^{L}}(\xi _{i})S+ {\kappa _{i,2}^{L}}(\xi _{1},\ldots ,\xi _{N})$. Replacing this conjectured $v_{i}^{L}$ in the HJB equation, we obtain

$$\begin{aligned} {v_{i}^{L}}\left( S,\xi _{1},\ldots ,\xi _{N}\right)&= \max _{e_{i}}\left\{ e_{i}\left( \alpha _{i}-e_{i}-\gamma \sum _{j\ne i}^{N}e_{j}\right) -\beta _{i}S\right. \nonumber \\&+\,\,\delta \int _{\mathcal {H}}{\kappa _{i,1}^{L}}(\hat{\xi } _{i})\eta \left( \sum _{j=1}^{N}e_{j}+dS\right) \left( \int _{\Theta }\phi (\eta | \theta )\xi _{i}(\theta ){\mathbf {d}}\theta \right) {\mathbf {d}}\eta \nonumber \\&+\,\,\left. \delta \int _{\mathcal {H}}\left( \kappa _{i,2}^{L}\left( \hat{ \xi }_{1}\left( \theta |\eta \right) ,\ldots ,\hat{\xi }_{N}\left( \theta |\eta \right) \right) \right) \left( \int _{\Theta }\phi (\eta |\theta )\xi _{i}(\theta ){\mathbf {d}} \theta \right) {\mathbf {d}}\eta \right\} . \nonumber \\ \end{aligned}$$

(40)

From the first-order conditions we have

$$\begin{aligned} \alpha _{i}-2{e_{i}^{L}}-\gamma \sum _{j\ne i}^{N}e_{j}^{L}+ \delta \int _{\mathcal {H}}{\kappa _{i,1}^{L}}({{\hat{\xi }}_{i}} \left( \theta |\eta \right) )\eta \left( \int _{\Theta }\phi (\eta |\theta ) \xi _{i}(\theta ){\mathbf {d}}\theta \right) {\mathbf {d}}\eta . \end{aligned}$$

(41)

Plugging (41) into (40) and equating the coefficients in order of $S$, we have the following system of equations:

$$\begin{aligned} {\kappa _{i,1}^{L}}(\xi _{i})&= -\beta _{i}+\delta d\int _{\mathcal {H}}{\kappa _{i,1}^{L}}(\hat{\xi _{i}}) \eta \left( \int _{\Theta }\phi (\eta |\theta ) \xi _{i}(\theta ){\mathbf {d}}\theta \right) {\mathbf {d}}\eta \end{aligned}$$

(42)

$$\begin{aligned} {\kappa _{i,2}^{L}}(\xi _{1},\ldots ,\xi _{N})&= \alpha _{i}e_{i}-(e_{i})^{2}-\gamma e_{i}\sum _{j\ne i}^{N}e_{j}\nonumber \\&+\,\delta \int _{\mathcal {H}}\left( {\kappa _{i,1}^{L}}(\hat{ \xi }_{i})\eta \sum _{i}e_{i}+\kappa _{i,2}^{L}\left( \hat{\xi }_{1} \left( \theta |\eta \right) ,\ldots ,\hat{\xi }_{N}\left( \theta |\eta \right) \right) \right) \nonumber \\&\times \,\left( \int _{\Theta }\phi (\eta |\theta )\xi _{i}(\theta ) {\mathbf {d}}\theta \right) {\mathbf {d}}\eta . \end{aligned}$$

(43)

We claim that

$$\begin{aligned} {\kappa _{i,1}^{L}}(\xi _{i})=-\beta _{i}\int _{\Theta } \frac{\xi _{i}(\theta )}{1-\delta d\mu (\theta )}{\mathbf {d}}\theta , \end{aligned}$$

(44)

and consequently

$$\begin{aligned} \begin{aligned}\kappa _{i,2}^{L}(\xi _{1},\ldots ,\xi _{N})=&\alpha _{i}e_{i}-(e_{i})^{2}-\gamma e_{i}\sum _{j\ne i}^{N}e_{j}\\&-\beta _{i}\delta \sum _{i}e_{i}\left( \int _{\Theta }\frac{\mu (\theta ) \xi _{i}(\theta )}{1-\delta d\mu (\theta )}{\mathbf {d}}\theta \right) \\&+\delta \int _{\mathcal {H}}\kappa _{i,2}^{L}\left( {\hat{ \xi }_{1}}\left( \theta |\eta \right) ,\ldots ,\hat{\xi }_{N} \left( \theta |\eta \right) \right) \left( \int _{\Theta }\phi (\eta |\theta ) \xi _{i}(\theta )d\theta \right) {\mathbf {d}}\eta . \end{aligned} \end{aligned}$$

(45)

To show this, let us update (44) for one period to obtain

$$\begin{aligned} {\kappa _{i,1}^{L}}\left( \hat{\xi }_{i}\left( \theta |\eta \right) \right) =- \beta _{i}\int _{\Theta }\frac{\hat{\xi }_{i}\left( \theta |\eta \right) }{1-\delta d\mu (\theta )}{\mathbf {d}}\theta . \end{aligned}$$

(46)

Using (8), we get

$$\begin{aligned} {\kappa _{i,1}^{L}}(\hat{\xi _{i}}\left( \theta |\eta \right) )=-\beta _{i}\int _{ \Theta }\frac{1}{1-\delta d\mu (\theta )}\frac{\phi (\eta |\theta ) \xi _{i}(\theta )}{\int _{\Theta }\phi (\eta |x)\xi _{i}(x) {\mathbf {d}}x}{\mathbf {d}}\theta . \end{aligned}$$

Plugging this back into (42) leads to

$$\begin{aligned} {\kappa _{i,1}^{L}}(\xi _{i})&= -\beta _{i}\left( 1+\delta d\int _{\mathcal {H}}\int _{\Theta }\frac{1}{1-\delta d\mu (\theta )}\frac{\phi (\eta |\theta )\xi _{i}(\theta )}{\int _{\Theta } \phi (\eta |x)\xi _{i}(x)dx} d\theta \eta \right. \\&\left. \times \left( \int _{\Theta }\phi (\eta |\theta )\xi _{i}(\theta ){\mathbf {d}} \theta \right) {\mathbf {d}}\eta \right) , \\&= -\beta _{i}\left( 1+\delta d\int _{\Theta }\frac{\mu (\theta )}{ 1-\delta d\mu (\theta )}\xi _{i}(\theta ){\mathbf {d}}\theta \right) , \\&= -\beta _{i}\left( \int _{\Theta }\left[ \frac{1-\delta d\mu (\theta )}{1-\delta d\mu (\theta )}+\frac{\delta d\mu (\theta )}{1-\delta d\mu (\theta )}\right] \xi _{i}(\theta ){\mathbf {d}}\theta \right) . \end{aligned}$$

This corresponds to our claim. To solve for $\left\{ {e_{i}^{I}}\right\} _{i=1}^{N}$rewrite (41) as a function of total emissions, $\sum _{j=1}^{N}e_{j}$, i.e.,

$$\begin{aligned} \alpha _{i}-\left( 2-\gamma \right) {e_{i}^{L}}- \gamma \sum _{j=1}^{N}e_{j}^{L}-\delta \beta _{i}\int _{\Theta } \frac{\mu (\theta )\xi _{i}(\theta )}{1-\delta d\mu (\theta )} {\mathbf {d}}\theta . \end{aligned}$$

So, for $i=1,\ldots ,N$,

$$\begin{aligned} {e_{i}^{L}}=\frac{1}{2-\gamma }\left( \alpha _{i}- \gamma \sum _{j=1}^{N}e_{j}^{L}-\delta \beta _{i}\int _{\Theta } \frac{\mu (\theta )\xi _{i}(\theta )}{1-\delta d\mu (\theta )} {\mathbf {d}}\theta \right) . \end{aligned}$$

(47)

We first solve for $\left\{ {e_{i}^{L}}\right\} _{i=1}^{N}$, then for ${e_{i}^{L}}$. Summing (47) over $i$ yields

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{L}}=\frac{1}{2-\gamma }\left( \sum _{i=1}^{N} \alpha _{i}-\gamma N\sum _{j=1}^{N}e_{j}^{L}-\delta \sum _{i=1}^{N} \beta _{i}\int _{\Theta }\frac{\mu (\theta )\xi _{i}(\theta )}{1-\delta d\mu (\theta )}{\mathbf {d}}\theta \right) . \end{aligned}$$

So, total emissions is given by

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{L}}=\frac{1}{2+\left( N-1\right) \gamma } \left( \sum _{i=1}^{N}\alpha _{i}-\delta \sum _{i=1}^{N}\beta _{i} \int _{\Theta }\frac{\mu (\theta )\xi _{i}(\theta )}{1-\delta d\mu (\theta )} {\mathbf {d}}\theta \right) . \end{aligned}$$

(48)

Plugging (48) into (47) yields

$$\begin{aligned} {e_{i}^{L}}&= \frac{1}{2-\gamma }\left( \left( \alpha _{i}-\frac{\gamma \sum _{j=1}^{N}\alpha _{j}}{2+\left( N-1\right) \gamma }\right) \right. \nonumber \\&\left. -\,\delta \left( \beta _{i}\int _{\Theta }\frac{\mu (\theta )\xi _{i}(\theta )}{1-\delta d\mu (\theta )}{\mathbf {d}}\theta -\frac{\gamma \sum _{j=1}^{N}\beta _{j} \int _{\Theta }\frac{\mu (\theta )\xi _{j}(\theta )}{1-\delta d\mu (\theta )}{\mathbf {d}}\theta }{2+\left( N-1\right) \gamma } \right) \right) . \end{aligned}$$

(49)

1.3 Proof of Proposition 3

We follow the same procedure as we did for a learner player. We conjecture that ${v_{i}^{ NL}},$ which satisfies (12), has the linear form ${v_{i}^{ NL}}=\kappa _{i,1}^{ NL}(\xi _{i})S+ {\kappa _{i,2}^{ NL}}(\xi _{1},\ldots ,\xi _{N})$. Rewriting (12) with our conjectured value function we have

$$\begin{aligned} {v_{i}^{ NL}}\left( S,\xi _{1},\ldots ,\xi _{N}\right)&= \max _{e_{i}}\left\{ e_{i}\left( \alpha _{i}-e_{i,t}-\gamma \sum _{j\ne i}^{N}e_{j}\right) -\beta _{i}S\right. \nonumber \\&+\left. \delta \left( \int _{\mathcal {H}}\kappa _{i,1}^{ NL}(\xi _{i}) \eta \left( \sum _{j=1}^{N}e_{j}+dS\right) +\kappa _{i,2}^{ NL}(\xi _{1},\ldots , \xi _{N})\right) \right. \nonumber \\&\left. \times \left( \int _{\Theta }\phi (\eta |\theta )\xi _{i}(\theta ) {\mathbf {d}}\theta \right) {\mathbf {d}}\eta \right\} . \end{aligned}$$

(50)

The first-order condition for $i$ is

$$\begin{aligned} \alpha _{i}-2{e_{i}^{ NL}}-\gamma \sum _{j\ne i}^{N}e_{j}^{ NL}+ \delta \int _{\mathcal {H}}{\kappa _{i,1}^{ NL}}(\xi _{i})\eta \left( \int _{\Theta }\phi (\eta |\theta )\xi (\theta ){\mathbf {d}} \theta \right) {\mathbf {d}}\eta , \end{aligned}$$

(51)

and since $\int _{\mathcal {H}}\eta \phi (\eta |\theta ){\mathbf {d}} \eta =\mu (\theta )$, we obtain

$$\begin{aligned} {e_{i}^{ NL}}\left( S,\xi _{1},\ldots ,\xi _{N}\right) =\frac{1}{2} \left( \alpha _{i}-\gamma \sum _{j\ne i}^{N}e_{j}^{ NL}+ \delta \int _{\Theta }\left( \kappa _{i,1}^{ NL}(\xi _{i})\right) \mu (\theta )\xi (\theta ){\mathbf {d}}\theta \right) . \end{aligned}$$

(52)

Plugging (52) into (50) gives the following system of equations

$$\begin{aligned} {\kappa _{i,2}^{ NL}}(\xi _{1},\ldots ,\xi _{N})&= \alpha _{i}e_{i}-e_{i}^{2}-\gamma e_{i}\sum _{j\ne i}^{N}e_{j}+\delta \int _{\mathcal {H}} \left( \kappa _{i,1}^{ NL}(\xi _{i})\eta \left( \sum _{j=1}^{N}e_{j}\right) \right. \nonumber \\&\left. +\,{\kappa _{i,2}^{ NL}}(\xi _{1},\ldots ,\xi _{N})\right) \left( \int _{\Theta } \phi (\eta |\theta )\xi _{i}(\theta ){\mathbf {d}}\theta \right) {\mathbf {d}}\eta , \end{aligned}$$

(53)

and

$$\begin{aligned} {\kappa _{i,1}^{ NL}}(\xi _{i})=-\beta _{i}+\delta d\int _{\Theta }{\kappa _{i,1}^{ NL}}(\xi _{i})\left( \int _{\mathcal {H}} \eta \phi (\eta |\theta ){\mathbf {d}}\eta \right) \xi _{i}(\theta ) {\mathbf {d}}\theta , \end{aligned}$$

(54)

or

$$\begin{aligned} \kappa _{i,1}^{ NL}(\xi _{i})=-\beta _{i}+\delta d\int _{\Theta }\kappa _{i,1}^{ NL}(\xi _{i})\mu (\theta )\xi _{i}(\theta ) {\mathbf {d}}\theta . \end{aligned}$$

(55)

We now show that $\kappa _{i,1}^{ NL}(\xi _{i})=\frac{-\beta _{i}}{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }.$ Inserting this guessed form into (55), we get

$$\begin{aligned} \kappa _{i,1}^{ NL}(\xi _{i})&= -\beta _{i}-\beta _{i}\delta d\int _{\mathcal {H}}\frac{1}{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta } \eta \left( \int _{\Theta }\phi (\eta |\theta )\xi _{i}(\theta ){\mathbf {d}} \theta \right) {\mathbf {d}}\eta ,\end{aligned}$$

(56)

$$\begin{aligned}&= -\beta _{i}-\beta _{i}\delta d\frac{\int _{\Theta }\left( \int _{\mathcal {H}}\eta \phi (\eta |\theta ) {\mathbf {d}}\eta \right) \xi _{i}(\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }, \end{aligned}$$

(57)

$$\begin{aligned}&= -\beta _{i}-\beta _{i}\delta \frac{\int _{\Theta }\mu (\theta ) \xi _{i}(\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }, \end{aligned}$$

(58)

$$\begin{aligned}&= -\beta _{i}\left( \frac{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }{ 1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta } +\frac{\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}} \theta }{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta )\mathbf {d }\theta }\right) , \end{aligned}$$

(59)

$$\begin{aligned}&= \frac{-\beta _{i}}{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }. \end{aligned}$$

(60)

To complete the proof, rewrite (51) as a function of total emissions, i.e.,

$$\begin{aligned} \alpha _{i}-\left( 2-\gamma \right) {e_{i}^{ NL}}- \gamma \sum _{j=1}^{N}{e_{j}^{ NL}}-\delta \frac{\beta _{i}\int _{\Theta } \mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }=0. \end{aligned}$$

(61)

We first solve for $\sum _{j=1}^{N}{e_{j}^{ NL}}$, then for ${e_{i}^{ NL}}$. Summing (61) over $i$ yields

$$\begin{aligned} \sum _{i=1}^{N}\alpha _{i}-\left( 2-\gamma \right) \sum _{i=1}^{N}{e_{i}^{ NL}} -\gamma N\sum _{j=1}^{N}e_{j}^{ NL}-\delta \sum _{i=1}^{N}\frac{ \beta _{i}\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }=0. \end{aligned}$$

(62)

So total emissions is given by

$$\begin{aligned} \sum _{j=1}^{N}e_{j}^{ NL}=\frac{1}{2+\left( N-1\right) \gamma } \left( \sum _{i=1}^{N}\alpha _{i}-\delta \sum _{i=1}^{N}\frac{\beta _{i} \int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta }\right) . \end{aligned}$$

(63)

Plugging this back in (61) yields

$$\begin{aligned} {e_{i}^{ NL}}&= \frac{1}{2-\gamma }\left( \alpha _{i}- \frac{\gamma \sum _{j=1}^{N}\alpha _{j}}{2+\left( N-1\right) \gamma }\right. \nonumber \\&\left. -\delta \left( \beta _{i}\frac{\int _{\Theta }\mu (\theta )\xi _{i}(\theta ) {\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta ){\mathbf {d}} \theta }-\frac{\gamma \sum _{j=1}^{N}\beta _{j}\frac{\int _{\Theta } \mu (\theta )\xi _{j}(\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta ){\mathbf {d}}\theta }}{2+\left( N-1\right) \gamma }\right) \right) \end{aligned}$$

(64)

1.4 Proof of Proposition 4

We have

$$\begin{aligned} \sum _{i=1}^{N}\left( {e_{i}^{L}}-{e_{i}^{ NL}}\right)&= \frac{\delta }{2+\left( N-1\right) \gamma }\left( \sum _{i=1}^{N} \beta _{i}\left[ R\left( \int _{\Theta }\xi _{i}(\theta )\mu (\theta ) {\mathbf {d}}\theta \right) \right. \right. \\&\left. \left. -\int _{\Theta }R(\mu (\theta ))\xi _{i}( \theta ){\mathbf {d}}\theta \right] \right) \end{aligned}$$

Since $R$ is an increasing convex function, by Jensen’s inequality we have

$$\begin{aligned} \int _{\Theta }R(\mu (\theta ))\xi _{i}(\theta ){\mathbf {d}} \theta >R\left( \int _{\Theta }\xi _{i}(\theta )\mu (\theta ){\mathbf {d}} \theta \right) , \end{aligned}$$

that is, $\sum _{i=1}^{N}\left( {e_{i}^{L}}-e_{i}^{A}\right) <0,$ or

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{L}}\left( S,\xi _{1},\ldots ,\xi _{N}\right) < \sum _{i=1}^{N}e_{i}^{A}\left( S,\xi _{1},\ldots ,\xi _{N}\right) . \end{aligned}$$

1.5 Proof of Proposition 5

If $\beta _{i}=\beta$ and $\xi _{i}(\theta )=\xi (\theta ),\,\forall i=1,\ldots ,N$, then

$$\begin{aligned} {e_{i}^{L}}(S;\xi _{i},\xi _{-i})&= \frac{1}{2-\gamma }\left( \alpha _{i}- \frac{\gamma \sum _{j=1}^{N}\alpha _{j}}{2+\left( N-1\right) \gamma }\right) \nonumber \\&-\frac{\delta }{2-\gamma }\beta \left( 1-\frac{\gamma N}{2+\left( N-1\right) \gamma }\right) \int _{\Theta }\frac{\mu (\theta )\xi (\theta ) {\mathbf {d}}\theta }{1-\delta d\mu (\theta )}. \end{aligned}$$

(65)

$$\begin{aligned} {e_{i}^{ NL}}&= \frac{1}{2-\gamma }\left( \alpha _{i}- \frac{\gamma \sum _{j=1}^{N}\alpha _{j}}{2+\left( N-1\right) \gamma }\right) \nonumber \\&-\frac{\delta }{2-\gamma }\beta \left( 1-\frac{\gamma N}{2+\left( N-1\right) \gamma }\right) \frac{\int _{\Theta }\mu (\theta ) \xi (\theta ){\mathbf {d}}\theta }{1-\delta d\int _{\Theta }\mu (\theta ) {\mathbf {d}}\theta }. \end{aligned}$$

(66)

Let $R\left( x\right) =x\left( 1-\delta dx\right) ^{-1}$, then 

$$\begin{aligned} {e_{i}^{L}}\left( S,\xi _{1},\ldots ,\xi _{N}\right)&= \frac{1}{2-\gamma } \left( \alpha _{i}-\frac{\gamma \sum _{j=1}^{N}\alpha _{j}}{2+\left( N-1\right) \gamma }\right) \\&-\frac{\delta \beta }{2-\gamma } \frac{2-\gamma }{2+\left( N-1\right) \gamma }\int _{\Theta }R(\mu (\theta )) \xi (\theta ){\mathbf {d}}\theta , \end{aligned}$$

and

$$\begin{aligned} {e_{i}^{ NL}}\left( S,\xi _{1},\ldots ,\xi _{N}\right)&= \frac{1}{2-\gamma } \left( \alpha _{i}-\frac{\gamma \sum _{j=1}^{N}\alpha _{j}}{2+\left( N-1\right) \gamma }\right) \\&-\frac{\delta \beta }{2-\gamma }\frac{2-\gamma }{ 2+\left( N-1\right) \gamma }\left( R\left( \int _{\Theta }\xi (\theta )\mu \left( \theta \right) {\mathbf {d}}\theta \right) \right) . \end{aligned}$$

Since $0<\delta ,d<1$, and $0<\mu (\theta )<1$, then $R$ is an increasing convex function for acceptable values of our model parameters ($R^{\prime }>0,\,R''>0,\,\forall x\in [0,1]$), then by Jensen’s inequality we have

$$\begin{aligned} \int _{\Theta }R(\mu (\theta ))\xi (\theta ){\mathbf {d}} \theta >R\left( \int _{\Theta }\xi (\theta )\mu (\theta ){\mathbf {d}} \theta \right) , \end{aligned}$$

Since $\frac{1}{2+\left( N-1\right) \gamma }$ is positive, then

$$\begin{aligned} {e_{i}^{L}}\left( S,\xi _{1},\ldots ,\xi _{N}\right) <{e_{i}^{ NL}} \left( S,\xi _{1},\ldots ,\xi _{N}\right) . \end{aligned}$$

1.6 Proof of Proposition 8

Let $G\left( x\right) =\frac{1}{1-\delta dx}$, which is an increasing convex function for $0<x<1$, then we can rewrite ${\kappa _{i,1}^{L}}(\xi _{i})$ and $\kappa _{i,1}^{ NL}(\xi _{i})$ as follows

$$\begin{aligned} {\kappa _{i,1}^{L}}(\xi _{i})&= -\beta _{i}\int _{\Theta }\frac{\xi _{i} (\theta )}{1-\delta d\mu (\theta )}{\mathbf {d}}\theta =-\beta _{i} \int _{\Theta }G(\mu (\theta ))\xi _{i}(\theta ){\mathbf {d}}\theta ,\\ {\kappa _{i,1}^{ NL}}(\xi _{i})&= \frac{-\beta _{i}}{1-\delta d\int _{\Theta }\mu (\theta )\xi _{i}(\theta ){\mathbf {d}}\theta } =-\beta _{i}G\left( \int _{\Theta }\xi _{i}(\theta )\mu (\theta ) \mathbf {d}\theta \right) , \end{aligned}$$

then by Jensen’s inequality we have

$$\begin{aligned} \int _{\Theta }G(\mu (\theta ))\xi _{i}(\theta ){\mathbf {d}} \theta >G\left( \int _{\Theta }\xi _{i}(\theta )\mu (\theta ){\mathbf {d}} \theta \right) , \end{aligned}$$

that yields

$$\begin{aligned} \kappa _{i,1}^{ NL}(\xi _{i})>{\kappa _{i,1}^{L}}(\xi _{i}). \end{aligned}$$

1.7 Proof of Proposition 10

Let $R\left( x\right) =x\left( 1-\delta dx\right) ^{-1}$ and $u\left( x\right) =R\left( \mu \left( x\right) \right)$. Since, $\mu \in \left[ 0,1 \right] ,\,R$ is increasing, and

$$\begin{aligned} \frac{\partial u}{\partial x}=\mu ^{\prime }R^{\prime }\Rightarrow {\left\{ \begin{array}{ll} \begin{array}{cc} \frac{\partial u}{\partial x}>0 &{} \quad \text{ if } \mu ^{\prime }>0 \\ \frac{\partial u}{\partial x}<0 &{} \quad \text{ if } \mu ^{\prime }<0 \end{array} \end{array}\right. }. \end{aligned}$$

(67)

Recall that the individual’s emissions for an non-learner player are given by equation 10, and the total emissions by

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{L}}=\frac{1}{2+\left( N-1\right) \gamma } \left( \sum _{i=1}^{N}\alpha _{i}-\delta \sum _{i=1}^{N}\beta _{i} \int _{\Theta }\xi _{i}(\theta )R(\mu (\theta )){\mathbf {d}}\theta \right) . \end{aligned}$$

(68)

Using definition 1 and conditions (67), confirm the results for an non-learner player.

For the adaptive-learning case, an individual’s emissions are given by (14), and the total emissions are as follows:

$$\begin{aligned} \sum _{i=1}^{N}{e_{i}^{ NL}}=\frac{1}{2+\left( N-1\right) \gamma } \left( \sum _{i=1}^{N}\alpha _{i}-\delta \sum _{i=1}^{N}\beta _{i} R\left( \int _{\Theta }\xi _{i}(\theta )\mu (\theta ){\mathbf {d}} \theta \right) \right) . \end{aligned}$$

(69)

Besides, if ${\xi _{i}^{1}}(\theta )\succ _{1}{\xi _{i}^{2}}\left( \theta \right)$, then $\int _{\Theta }{\xi _{i}^{1}}(\theta )\mu (\theta )d \theta >\int _{\Theta }{\xi _{i}^{2}}(\theta )\mu (\theta )d\theta$ if $\mu ^{\prime }>0$, which gives the results presented in the Proposition for the adaptive learner.

1.8 Proof of Proposition 12

We assumed $\mu ^{\prime }>0$. If $\mu ''\ge 0$, then since $R$ is also increasing and convex, $R(\mu (\theta ))$ would be an increasing convex function. Given that ${\xi _{i}^{1}}(\theta ) \succ _{2}{\xi _{i}^{2}}(\theta )$, we will have

$$\begin{aligned} -\int _{\Theta }R(\mu (\theta ))\xi _{i}(\theta ^{1})\mathbf { d}\theta \ge -\int _{\Theta }R(\mu (\theta ))\xi _{i}( \theta ^{2}){\mathbf {d}}\theta . \end{aligned}$$

(70)

Now, assume that $\mu$ is a concave function, $\mu ^{\prime \prime }\le 0$, and player $i$’s beliefs change from ${\xi _{i}^{1}}(\theta )$ to ${\xi _{i}^{2}}(\theta )$, with ${\xi _{i}^{1}}(\theta ) \succ _{2}{\xi _{i}^{2}}(\theta )$, that is, $\theta ^{1}$ is less volatile than $\theta ^{2}$ meaning that $\theta ^{1}\succ _{2}\theta ^{2}$, then $\mu \left( \theta ^{1}\right) \succ _{2}\mu \left( \theta ^{2}\right)$. Consequently, the inequality (70) is satisfied.