Green Trade Unions: Structure, Wages and Environmental Technology

Abstract

This paper investigates the effect of trade union structure on firms’ technological choices when unions care about environmental protection. We compare a decentralized with a centralized union structure in a Cournot duopoly. Our results suggest that a decentralized structure provides higher incentives for the investment in cleaner technologies, although emissions may be lower under a centralized structure. The effect of the environmental damage parameter on wages and output may be non-monotonic.

Appendix

1.1 Proofs

1.1.1 Proofs to Lemmata and Propositions in Sect. 4, 5 and 6

Proof of Lemma 1

It follows immediately from the analysis of the first derivatives with respect to $k_{i}$ and $k_{j}: \frac{\partial w_{i}^{D}}{\partial k_{i}}= \frac{8e}{15}>0, \frac{\partial w_{i}^{D}}{\partial k_{j}}=\frac{2e}{15} >0. \square$

Proof of Lemma 2

It is straightforward to calculate that $\frac{\partial y_{i}^{D}}{\partial a }=\frac{90\gamma (28e(e-2a)+405\gamma )}{(28e^{2}-405\gamma )^{2}}$. Setting $\frac{\partial y_{i}^{D}}{\partial a}=0$ and solving with respect to $a$ we calculate the $a_{cv}^{D}=\frac{28e^{2}+405\gamma }{56e}.$ The second order derivative is $\frac{\partial \overline{y}_{i}^{D}}{\partial a}=\frac{ -90\gamma (56ea))}{(28e^{2}-405\gamma )^{2}}<0$. Hence $y_{i}^{D}$ reaches its maximum at $a_{cv}^{D}.$

Proof of Proposition 1

It is straightforward to check that $\frac{\partial q_{i}^{D}}{\partial e}= \frac{90(56ae-28e^{2}-405\gamma )\gamma }{(28e^{2}-405\gamma )^{2}}$ and $\frac{\partial w_{i}^{D}}{\partial e}=-\frac{270(56ae-28e^{2}-405\gamma )\gamma }{(28e^{2}-405\gamma )^{2}}.$ Hence, the signs of $\frac{\partial q_{i}^{D}}{\partial e}$ and $\frac{\partial w_{i}^{D}}{\partial e}$ depend on the sign of $(56ae-28e^{2}-405\gamma )$, which is continuous in $e$. Recall that $a>e>0$. At $e=0,$ this term is negative$.$ Hence, initially $\frac{\partial q_{i}^{D}}{\partial e}<0$ and $\frac{\partial w_{i}^{D}}{ \partial e}>0$. The derivative of $(56ae-28e^{2}-405\gamma )$ with respect to $e$ is $56(a-e)$, which is positive. Thus, $(56ae-28e^{2}-405\gamma )$ may turn positive (and therefore $\frac{\partial q_{i}^{D}}{\partial e}$ and $\frac{\partial w_{i}^{D}}{\partial e}$ positive and negative respectively) as $e$ increases. In fact, at $a=e, \frac{\partial q_{i}^{D}}{\partial e}= \frac{90\gamma }{(28e^{2}-405\gamma )^{2}}>0$ and $\frac{\partial w_{i}^{D}}{\partial e}=-\frac{270\gamma }{(28e^{2}-405\gamma )^{2}}<0.$ Setting $56ae-28e^{2}-405\gamma =0$, we find $e_{cv}^{D}=\frac{1}{14}(14a-\sqrt{7} \sqrt{28a^{2}-405\gamma }$) (we can discard the other root as it implies $e>a$). $e_{cv}^{D}$ is only a real root if $a^{2}>405/28=14.46$. Thus, if $a^{2}\le 14.46, \frac{\partial q_{i}^{D}}{\partial e}<0$ and $\frac{ \partial w_{i}^{D}}{\partial e}>0$. If $a^{2}>14.46, \frac{\partial q_{i}^{D}}{\partial e}$ and $\frac{\partial w_{i}^{D}}{\partial e}$ move from negative to positive and from positive to negative respectively as $e$ increases, with the turning point at $e_{cv}^{D}$. $\square$

Proof of Lemma 3

It follows immediately from the analysis of the first derivatives with respect to $k_{i}$ and $k_{j}: \frac{\partial w_{i}^{c}}{{\partial k_i}}= \frac{1e}{2}>0, \frac{\partial w_{i}^{D}}{\partial k_{j}}=0. \square$

Proof of Lemma 4

It is easy to check that $\frac{\partial y_{i}^{C}}{\partial a}=\frac{ 3\gamma (e^{2}-2ae+18\gamma )}{(e^{2}-18\gamma )^{2}}$. Setting $\frac{\partial y_{i}^{C}}{\partial a}=0$, we find $a_{cv}^{D}=\frac{e^{2}+18\gamma }{2e}$. The second derivative is $\frac{\partial ^{2}y_{i}^{C}}{\partial a^{2}}=-\frac{6\gamma e}{(e^{2}-18\gamma )^{2}}<0$. Hence, $y_{i}^{C}$ reaches a maximum at $a_{cv}^{D}$. $\square$

Proof of Proposition 2

Calculating the derivatives of the equilibrium levels of output and wages with respect to $e$ yields $\frac{\partial q_{i}^{C}}{\partial e}=\frac{ 3(2ae-e^{2}-18\gamma )\gamma }{(e^{2}-18\gamma )^{2}}$ and $\frac{\partial w_{i}^{C}}{\partial e}=-\frac{9\gamma (2ae-e^{2}-18\gamma )}{(e^{2}-18\gamma )^{2}}.$ Hence, the signs of $\frac{\partial q_{i}^{C}}{\partial e}$ and $\frac{\partial w_{i}^{C}}{\partial e}$ depend on the sign of $(2ae-e^{2}-18\gamma )$, which is continuous in $e$. Recall that $a>e>0$. At $e=0,$ this term is negative$.$ Hence, initially $\frac{\partial q_{i}^{C}}{ \partial e}<0$ and $\frac{\partial w_{i}^{C}}{\partial e}>0.$ The derivative of $(2ae-e^{2}-18\gamma )$ with respect to $e$ is $2(a-e)>0$, which implies that $(2ae-e^{2}-18\gamma )$ may potentially turn positive (and therefore $\frac{\partial q_{i}^{C}}{\partial e}>0$ and $\frac{\partial w_{i}^{C}}{ \partial e}<0$) for a sufficiently large $e$. In fact, at $e=a, \frac{\partial q_{i}^{C}}{\partial e}=\frac{ 3\gamma }{(e^{2}-18\gamma )^{2}}>0$ and $\frac{\partial w_{i}^{C}}{\partial e }=-\frac{9\gamma }{(e^{2}-18\gamma )^{2}}<0.$ Setting $(2ae-e^{2}-18\gamma )=0$, we find $e_{cv}^{C}=a-\sqrt{a^{2}-18\gamma }$ (we can discard the other root as it implies $e>a$). Note that $e_{cv}^{D}$ is only a real root if $a^{2}>18\gamma$. Thus, if $a^{2}\le 18\gamma , \frac{\partial q_{i}^{D}}{\partial e}<0$ and $\frac{\partial w_{i}^{C}}{\partial e}>0$. If $a^{2}>18\gamma , \frac{\partial q_{i}^{D}}{\partial e}$ and $\frac{\partial w_{i}^{D}}{\partial e}$ move from negative to positive and from positive to negative respectively as $e$ increases, with the turning point taking place at $e_{cv}^{C}$. $\square$

Proof of Proposition 3

It is easy to check that $q_{i}^{C}-q_{i}^{D}=\frac{3(a-e)(2e^{2}-135\gamma )\gamma }{(28e^{2}-405\gamma )(e^{2}-18\gamma )} <0$ and $w_{i}^{C}-w_{i}^{D}=\frac{3(a-e)(135\gamma -2e^{2})\gamma }{ (28e^{2}-405\gamma )(e^{2}-18\gamma )}>0$ given that $a>e>0, \gamma >0$ and $e^{2}<9\gamma$. $\square$

Proof of Proposition 4

It is immediate to check that $k_{i}^{C}-k_{i}^{D}=(99(a-e)e\gamma )/((28e^{2}-405\gamma )(e^{2}-18\gamma )) >0$ since $a>e>0$ and $e^{2}<9\gamma . \square$

Proof of Proposition 5

Note that $y_{i}^{C}-y_{i}^{D} =\frac{3(a-e)\gamma R}{(28e^{2}-405\gamma )^{2}(e^{2}-18\gamma )^{2}}$ where $R=56ae^{5}-7{,}560ae^{3}\gamma +1{,}962e^{4}\gamma +108{,}135ae\gamma ^{2}+29{,}160e^{2}\gamma ^{2}-984{,}150\gamma ^{3}$. Given that $a>e>0$, the sign of $y_{i}^{C}-y_{i}^{D}$ depends on the sign of $R$. Note that at $a=0, R=1{,}962e^{4}\gamma +29{,}160e^{2}\gamma ^{2}-984{,}150\gamma ^{3}<0$ for $e^{2}<9\gamma$. Moreover, $\frac{\partial R}{ \partial a}=56e^{5}-7{,}560e^{3}\gamma +108{,}135e\gamma ^{2}>0$ for $e^{2}<9\gamma$. Thus, $R$ may change sign as $a$ increases. Setting $R=0$ and solving for $a$, we get $a=35\gamma A/eB$ where $A=(501.6\gamma ^{2}-e^{2}(e^{2}+14.86\gamma ))$ and $B=(e^{2}(e^{2}-135\gamma )+1{,}930.9\gamma ^{2})$ where both $A>0$ and $B>0$ for $e^{2}<9\gamma$. All in all, $R<0$ if $a<35\gamma A/eB$ and positive if $a>35\gamma A/eB.$ Therefore, $y_{i}^{C}<y_{i}^{D}$ for $a<35\gamma A/eB$ and $y_{i}^{C}>y_{i}^{D}$ for $a>35\gamma A/eB$. At $35\gamma A/eB, y_{i}^{C}=y_{i}^{D}. \square$

Proof of Proposition 6

Note that $\pi ^{C}-\pi ^{D}=-\frac{27(a-e)^{2}\gamma ^{2}(244e^{4}-6{,}573e^{2}\gamma +42{,}525\gamma ^{2})}{(28e^{2}-405\gamma )^{2}(e^{2}-18\gamma )^{2}}$. Given that $a>e, \gamma >0$ and $e^{2}<9\gamma$, the sign of $\pi ^{C}-\pi ^{D}$ depends on the sign of $(244e^{4}-6{,}573e^{2}\gamma \,+\,42{,}525\gamma ^{2})$. Moreover, at $e=0$, this term is positive. This term will be zero if $e^{2}=16.14\gamma$ or $e^{2}=10.79\gamma$. Given that $e^{2}<9\gamma$, we know $(244e^{4}-6{,}573e^{2} \gamma +42{,}525\gamma ^{2})>0$. As a consequence, we can state that, $\pi ^{C}-\pi ^{D}<0.$

Furthermore, $U_{i}^{C}-\sum U_{i}^{D}=\frac{54(a-e)^{2}\Psi }{(405\gamma -28e^{2})^{2}(18\gamma -e^{2})^{2}}$ where $\Psi =(334e^{4}-6{,}480\gamma e^{2}+18{,}225\gamma ^{2}).$ Hence, the sign of $U_{i}^{C}-\sum U_{i}^{D}$ is determined by the sign of $\Psi$. At $e=0, \Psi =18{,}225\gamma ^{2}>0$. Evaluating $\Psi$ for $e^{2}=9\gamma$, we have $-13{,}031\gamma ^{2}<0.$ The derivative of $\Psi$ with respect to $e$ is $e(13{,}36e^{2}-12{,}960\gamma )$ which is negative for any $e>0$ such that $e^{2}<9\gamma$. Hence we know that $\Psi$ moves from positive to negative and will cross only once in the interval $e\in (0,\sqrt{9\gamma }$). Setting $\Psi =0$ if $e^{2}=3.41\gamma$. Thus, if $e^{2}<3.41\gamma , U_{i}^{C}-\sum U_{i}^{D}>0$ and if $e^{2}>3.41\gamma , U_{i}^{C}-\sum U_{i}^{D}<0$. At $e^{2}=3.41\gamma , U_{i}^{C}-\sum U_{i}^{D}=0$. $\square$

Proof of Proposition 7

It is straightforward to check that $SW^{D}-SW^{C}=2(a-e)^{2}\gamma \varsigma$ where $\varsigma =-\frac{99\gamma (140e^{4}+909e^{2}\gamma -18{,}225\gamma ^{3})}{(405\gamma -28e^{2})^{2}(18\gamma -e^{2})^{2}}>0$ for any $e$ such that $e^{2}<9\gamma$. Hence, $SW^{D}- SW^{C}>0$. $\square$

Proof of Lemma 5

It is straightforward to see that $k_{i}^{D}-k_{i}^{O}=\frac{293(a-e)e\gamma }{(405\gamma -28e^{2})(4\gamma -e^{2})}>0$ since $a>e>0$ and $(4\gamma -e^{2})>0$ (otherwise $q_{i}^{O}<0$). Hence, we also know that $k_{i}^{C}-k_{i}^{O}>0$ since from proposition 4, we know that $k_{i}^{C}-k_{i}^{D}>0$. Hence, $k_{i}^{C}>k_{i}^{D}>k_{i}^{O}$. Likewise, it is straightforward to see that $q_{i}^{D}-q_{i}^{O}=\frac{2(a-e)\gamma (17e^{2}+225\gamma )}{(405\gamma -28e^{2})(4\gamma -e^{2})}>0$. Hence, we also know that $q_{i}^{C}-q_{i}^{O}<0$ since from proposition 5, we know that $q_{i}^{C}-q_{i}^{D}<0$. Hence, $q_{i}^{C}<q_{i}^{D}<q_{i}^{O}$.

As for emissions: $y_{i}^{C}-y_{i}^{O}=\frac{(a-e)\gamma }{(18\gamma -e^{2})^{2}(4\gamma -e^{2})^{2}}\chi$ where $\chi =-59ae^{2}+96ae^{3}\gamma +62e^{4}\gamma -696ae\gamma ^{2}-720e^{2}\gamma ^{2}+3{,}456\gamma ^{3}$. Hence, the sign of $y_{i}^{C}-y_{i}^{O}$ depends on the sign of $\chi$. The derivative of $\chi$ with respect to $a$ is $e(-5e^{4}+96e^{2}\gamma -696\gamma ^{2})<0$ if $4\gamma -e^{2}>0$. Hence, $\chi$ is decreasing in $a$. Recall that $a>e$. At the limit ($a=e$), $\chi =62e^{4}\gamma -720e^{2}\gamma ^{2}+3{,}456\gamma ^{3}>0$. Hence, $\chi$ is positive in the begining and may turn negative at a given value of $a$. Setting $\chi =0$ and solving for $a$, we find: $a=\frac{2(31e^{4}\gamma -360e^{2}\gamma ^{2}+1{,}728\gamma ^{3})}{e(5e^{4}-96e^{2}\gamma +696\gamma ^{2}).}$. If $a< \frac{2(31e^{4}\gamma -360e^{2}\gamma ^{2}+1{,}728\gamma ^{3})}{ e(5e^{4}-96e^{2}\gamma +696\gamma ^{2}).}, y_{i}^{C}>y_{i}^{O}$ and if $a> \frac{2(31e^{4}\gamma -360e^{2}\gamma ^{2}+1{,}728\gamma ^{3})}{ e(5e^{4}-96e^{2}\gamma +696\gamma ^{2}).}, y_{i}^{C}<y_{i}^{O}.$ If $a= \frac{2(31e^{4}\gamma -360e^{2}\gamma ^{2}+1{,}728\gamma ^{3})}{ e(5e^{4}-96e^{2}\gamma +696\gamma ^{2}).}, y_{i}^{C}=y_{i}^{O}$.

As for emissions: $y_{i}^{D}-y_{i}^{O}=\frac{(a-e)\gamma }{(405\gamma -28e^{2})^{2}(4\gamma -e^{2})^{2}}\Gamma$ where $\Gamma =-2{,}044ae^{5}+32{,}760ae^{3}\gamma +21{,}361e^{4}\gamma -184{,}185ae\gamma ^{2}-236{,}520e^{2}\gamma ^{2}+947{,}700\gamma ^{3}$. Hence, the sign of $y_{i}^{D}-y_{i}^{O}$ depends on the sign of $\Gamma$. The derivative of $\Gamma$ with respect to $a$ is $-2{,}044e^{5}+32{,}760e^{3}\gamma -184{,}185e\gamma ^{2}<0$ if $4\gamma -e^{2}>0$. Hence, $\Gamma$ is decreasing in $a$. Recall that $a>e$. At the limit ($a=e$), $\Gamma =(585\gamma -73e^{2})(405\gamma -28e^{2})(4\gamma -e^{2})>0$. Hence, $\chi$ is positive in the beginning and may turn negative at a given value of $a$. Setting $\Gamma =0$ and solving for $a$, we find: $a=\frac{\gamma (21{,}361e^{4}-236{,}520e^{2}\gamma +947{,}700\gamma ^{2})}{2{,}044e^{5}-32{,}760e^{3} \gamma +184{,}185e\gamma ^{2}}$. Thus, if $a<\frac{\gamma (21{,}361e^{4}-236{,}520e^{2}\gamma +947{,}700\gamma ^{2})}{2{,}044e^{5}-32{,}760e^{3} \gamma +184{,}185e\gamma ^{2}}, y_{i}^{D}>y_{i}^{O}$ and if $a>\frac{\gamma (21{,}361e^{4}-236{,}520e^{2}\gamma +947{,}700\gamma ^{2})}{2{,}044e^{5}-32{,}760e^{3} \gamma +184{,}185e\gamma ^{2}}, y_{i}^{D}<y_{i}^{O}$. If $a=\frac{\gamma (21{,}361e^{4}-236{,}520e^{2}\gamma +947{,}700\gamma ^{2})}{2{,}044e^{5}-32{,}760e^{3} \gamma +184{,}185e\gamma ^{2}}, y_{i}^{D}=y_{i}^{O}$. Hence, in both regimes, for sufficiently small $a$, the equilibrium emissions are larger than the socially optimal level of emissions.

1.1.2 Proofs of Results in Sect. 7.1

Proof of remark (i)

Note that $q_{i}^{C}-q_{i}^{UN}=\frac{9(a-e)e^{2}\gamma }{(e^{2}-72\gamma )(e^{2}-18\gamma )}$ and $w_{i}^{UN}-w_{i}^{C}=\frac{27(a-e)e^{2}\gamma }{ (e^{2}-72\gamma )(e^{2}-18\gamma )}$. Given that $a>e>0, \gamma >0$ and $e^{2}<9\gamma$, it is immediate to see that $q_{i}^{C}-q_{i}^{UN}>0$ and $w_{i}^{UN}-w_{i}^{C}>0$. Moreover from Proposition 1, we know that $q_{i}^{C}-q_{i}^{D}<0$ and $w_{i}^{C}-w_{i}^{D}>0$. It follows that $q_{i}^{D}>q_{i}^{C}>q_{i}^{UN}$ and $w_{i}^{D}<w_{i}^{C}<w_{i}^{UN} \square$

Proof of remark (ii)

Note that $k_{i}^{UN}-k_{i}^{C}=(54(a-e)e\gamma )/((72\gamma -e^{2})(18\gamma -e^{2}))$. Given that $a>e, \gamma >0$ and $e^{2}<9\gamma$, it is easy to see that $k_{i}^{UN}-k_{i}^{C}>0$. Moreover, we know that $k_{i}^{C}>k_{i}^{D}$, it follows that $k_{i}^{UN}>k_{i}^{C}>k_{i}^{D}. \square$

Proof of remark (iii)

Note that $y_{i}^{UN\ }-y_{i}^{C} =-3(a-e)\gamma e\Upsilon /((72\gamma -e^{2})^{2}(18\gamma -e^{2})^{2}),$ where $\Upsilon =(ae^{4}-90e^{3}\gamma -1{,}296(a-2e)\gamma ^{2})$. Since $a>e>0$ and $\gamma >0$, the sign of $y_{i}^{UN\ }-y_{i}^{C}$ depends on the sign of $\Upsilon$ (if $\Upsilon$ is negative (positive), $y_{i}^{UN\ }-y_{i}^{C}>(<)0$). Note that at $a=e, \Upsilon =e^{5}-90e^{3}\gamma +1{,}296e\gamma ^{2}>0$ for $e^{2}<9\gamma$. Moreover, $\frac{\partial \Upsilon }{\partial a}=e^{4}-1{,}296\gamma ^{2}<0$ for $e^{2}<9\gamma$. Thus, $\Upsilon$ may change sign as $a$ increases. Setting $\Upsilon =0$ and solving for $a$, we get $a=\frac{ 18e(5e^{2}-144\gamma )\gamma }{e^{4}-1{,}296\gamma ^{2}}>0$ given that $e^{2}<9\gamma$. Therefore, $y_{i}^{UN\ }-y_{i}^{C}<0$ for $a<\frac{ 18e(5e^{2}-144\gamma )\gamma }{e^{4}-1{,}296\gamma ^{2}}$ and $y_{i}^{UN}-y_{i}^{C}>0$ for $a>\frac{18e(5e^{2}-144\gamma )\gamma }{ e^{4}-1{,}296\gamma ^{2}}$.

On the other hand, $y_{i}^{UN\ }-y_{i}^{D} = -6(a-e)\gamma \digamma /((28e^{2}-405\gamma )^{2}(e^{2}-72\gamma )^{2}),$ where $\digamma =(1{,}148ae^{5}+15{,}120ae^{3}\gamma -106{,}821e^{4}\gamma -1{,}849{,}230ae\gamma ^{2}+2{,}391{,}120e^{2}\gamma ^{2}+7{,}873{,}200\gamma ^{3})$. Since $a>e>0$ and $\gamma >0$, the sign of $y_{i}^{UN\ }-y_{i}^{D}$ depends on the sign of $\digamma$ (if $\Upsilon$ is negative (positive), $y_{i}^{UN\ }-y_{i}^{C}>(<)0$).). At $a=e, F=1{,}148e^{6}-91{,}701e^{4}\gamma +541{,}890e^{2}\gamma ^{2}+7{,}873{,}200\gamma ^{3}>0$ for $e^{2}<9\gamma$. Moreover, $\frac{\partial F}{\partial a} =1{,}148e^{5}+15{,}120e^{3}\gamma -1{,}849{,}230e\gamma ^{2}<0$ for $e^{2}<9\gamma$. Thus, $F$ may change sign as $a$ increases. Setting $\digamma =0$ and solving for $a$, we get $a=\frac{9(11{,}869e^{4}\gamma -265{,}680e^{2}\gamma ^{2}-874{,}800\gamma ^{3})}{2e(574e^{4}+7{,}560e^{2}\gamma -924{,}615\gamma ^{2})}>0$ given that $e^{2}<9\gamma$. Therefore, $y_{i}^{UN\ }-y_{i}^{C}<0$ for $a< \frac{9(11{,}869e^{4}\gamma -265{,}680e^{2}\gamma ^{2}-874{,}800\gamma ^{3})}{ 2e(574e^{4}+7{,}560e^{2}\gamma -924{,}615\gamma ^{2})}$ and $y_{i}^{UN}-y_{i}^{C}>0$ for $a>\frac{9(11{,}869e^{4}\gamma -265{,}680e^{2}\gamma ^{2}-874{,}800\gamma ^{3})}{ 2e(574e^{4}+7{,}560e^{2}\gamma -924{,}615\gamma ^{2})}$.

The rest of the result follows. $\square$

Proof of remark (iv)

It is immediate to see that $\pi ^{UN}-\pi ^{C}=\frac{27(a-e)^{2}\gamma ^{2}(e^{2}+36\gamma )}{(e^{2}-72\gamma )^{2}(e^{2}-18\gamma )^{2}} >0\$ and $\pi ^{UN}-\pi ^{D}=-\frac{81(a-e)^{2}\gamma ^{2}(180e^{2}+22{,}231e^{2}\gamma -226{,}800\gamma ^{2})}{(e^{2}-72\gamma )^{2}(28e^{2}-405\gamma )^{2}}<0$ given that $e^{2}<9\gamma$. Thus, $\pi ^{D}>\pi ^{UN}>\pi ^{C}$. Likewise, $U_{i}^{UN}-U_{i}^{C}=\frac{ 162(a-e)^{2}e^{2}(5e^{2}-144\gamma )\gamma }{(e^{2}-72\gamma )^{2}(e^{2}-18\gamma )^{2}}<0$. Hence, irrespective of the relative ranking between $U^{C}$ and $U^{D}$, we know that $U_{i}^{UN}<\max [U^{C}$,$U^{D}] \square$

Proof of remark (v)

It is easy to see that $SW^{C}-SW^{UN}=162(a-e)^{2}\gamma ^{2}\omega$ where $\omega =\left[ \frac{180e^{2}\gamma -7e^{4}}{(72\gamma -e^{2})^{2}(18\gamma -e^{2})^{2}}\right] >0$ for any $e$ such that $e^{2}<9\gamma$. Hence, $SW^{C}-SW^{UN}>0$. Given that we know that $SW^{D}>SW^{C}$, it follows that $SW^{D}>SW^{C}> SW^{UN}. \square$

1.2 Illustrations of Result in Proposition 5

In the main text, we have shown that emissions are higher under the decentralized structure than in the centralized structure for low market sizes but the opposite applies to large market sizes (see Proposition 5). This is a general result which we illustrate here with some numerical examples. In the tables below, we present the equilibrium results for given $e$ and $\gamma$ under a relatively small and a relatively large market size. As the reader can see from the three tables, when the market is relatively small, emissions are lower under a centralized structure than under a decentralized structure, although the opposite applies when the market is relatively large.³³ See Tables 1, 2 and 3.

Table 1 Equilibrium results $(e=4,\; \gamma =2$)

Table 2 Equilibrium results $(e=1.5, \gamma =0.5)$

Table 3 Equilibrium results $(e=1, \gamma =0.3)$