Dynamic Game of International Pollution Control with General Oligopolistic Equilibrium: Neary Meets Dockner and Long

Abstract

This study develops a two-country differential game model of transboundary pollution control with a continuum of polluting and oligopolistic industries. Governments choose the paths of their pollution permits, the market-clearing prices of which are determined endogenously, as in a general oligopolistic equilibrium model. We consider both autarky and bilateral free trade of polluting goods and derive solutions under international cooperation, open-loop Nash equilibrium, and linear Markov-perfect Nash equilibrium in each regime. Under autarky, we obtain similar results to those in the literature; that is, international cooperation achieves the least pollution, and the linear Markov-perfect Nash equilibrium results in higher pollution than the open-loop Nash equilibrium. Under free trade, each country’s welfare depends on the emissions of both countries, and if trade is frictionless, the same solution can be achieved by international cooperation, the open-loop Nash equilibrium, and the linear Markov-perfect Nash equilibrium. Moreover, when all industries have the same number of firms and trade is frictionless, free trade is better for the environment than autarky. However, if trade costs exist, international trade may result in higher pollution than autarky.

Appendices

Appendix A Proofs

1.1 A.1 Proof of Proposition 1

In the matrix form, the linear dynamic system consisting of (5) and (19) is

$$\begin{aligned} \begin{bmatrix} \dot{TE} \\ \dot{S} \end{bmatrix} = \begin{bmatrix} \rho +\delta &{} \frac{4(A_2)^2\beta }{B_2} \\ 1 &{} -\delta \end{bmatrix} \begin{bmatrix} TE \\ S \end{bmatrix} + \begin{bmatrix} -\frac{2(\rho +\delta )a(A_1 B_2+A_2 B_1)}{B_2} \\ 0 \end{bmatrix} . \end{aligned}$$

(A1)

In the steady state, $\dot{TE}=\dot{S}=0$. By solving (A1) for the steady-state solutions for the pollution stock and sum of emissions, we obtain

$$\begin{aligned} S^{ac}=\frac{2(\rho +\delta )a(A_1 B_2+A_2 B_1)}{(\rho +\delta )\delta B_2+4\beta (A_2)^2}, \quad TE^{ac}=E^{ac}+E^{*ac}=\delta S^{ac}. \end{aligned}$$

The eigenvalues of the coefficient matrix in (A1) are $\frac{\rho +\sqrt{(\rho +2\delta )^2+16(A_2)^2 \beta /B_2}}{2}$ and $\frac{\rho -\sqrt{(\rho +2\delta )^2+16(A_2)^2 \beta /B_2}}{2}$, which are positive and negative, respectively. As the dynamic system has one predetermined variable, S, and one jump variable, TE, the steady state is a saddle point.

For future reference, we derive the explicit solutions of (A1) with the initial condition $S(0)=S_0$ and the terminal condition $\lim _{t\rightarrow \infty } S(t)=S^{ac}$. The solutions are

$$\begin{aligned}&S(t)=\left( S_0-S^{ac}\right) e^{\eta ^{-} t}+S^{ac}, \end{aligned}$$

(A2a)

$$\begin{aligned}&TE(t)=\left( S_0-S^{ac}\right) \frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16(A_2)^2\beta /B_2}}{2}e^{\eta ^{-}t}+TE^{ac}, \end{aligned}$$

(A2b)

where $\eta ^-$ is the negative eigenvalue of the coefficient matrix in (A1). By substituting (A2a) into (A2b), the total emissions along the optimal path can be represented in the feedback form:

$$\begin{aligned} TE(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16(A_2)^2\beta /B_2}}{2}\left[ S(t)-S^{ac}\right] +TE^{ac}. \end{aligned}$$

(A3)

Along the saddle path toward the steady state, the total emissions are decreasing in the stock of pollution. $\square$

1.2 A.2 Proof of Proposition 2

From (5), (24) and its Foreign counterpart, the matrix form of the dynamic system of the open-loop Nash equilibrium is given by

$$\begin{aligned} \begin{bmatrix} \dot{E} \\ \dot{E}^* \\ \dot{S} \end{bmatrix} = \begin{bmatrix} \rho +\delta &{} 0 &{} \frac{(A_2)^2 \beta }{B_2} \\ 0 &{} \rho +\delta &{} \frac{(A_2)^2 \beta }{B_2} \\ 1 &{} 1 &{} -\delta \end{bmatrix} \begin{bmatrix} E \\ E^* \\ S \end{bmatrix} + \begin{bmatrix} -\frac{(\rho +\delta )a(A_1 B_2+A_2 B_1)}{B_2} \\ -\frac{(\rho +\delta )a(A_1 B_2+A_2 B_1)}{B_2} \\ 0 \end{bmatrix} . \end{aligned}$$

(A4)

In the steady state, $\dot{E}=\dot{E}^*=\dot{S}=0$. By solving (A4) for the steady-state solutions for the pollution stock and emissions in each country, we obtain

$$\begin{aligned} S^{ao}=\frac{2(\rho +\delta )a(A_1 B_2+A_2 B_1)}{(\rho +\delta )\delta B_2+2\beta (A_2)^2}, \quad E^{ao}=E^{*ao}=\frac{\delta }{2} S^{ac}. \end{aligned}$$

The eigenvalues of the coefficient matrix in (A4) are $\rho +\delta >0$, $\frac{\rho +\sqrt{(\rho +2\delta )^2+8(A_2)^2 \beta /B_2}}{2}>0$, and $\frac{\rho -\sqrt{(\rho +2\delta )^2+8(A_2)^2 \beta /B_2}}{2}<0$. As the dynamic system has one predetermined variable, S, and two jump variables, E and $E^*$, the steady state is a saddle point.

Deriving the explicit solutions of (A4) and rearranging the terms to obtain the total emissions along the open-loop Nash equilibrium path in the feedback form,

$$\begin{aligned} E(t)+E^*(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+8(A_2)^2\beta /B_2}}{2}\left[ S(t)-S^{ao}\right] +E^{ao}+E^{*ao} \end{aligned}$$

(A5)

is obtained. $\square$

1.3 A.3 Proof of Proposition 3

From the first-order condition for maximizing the right-hand side of (26), it follows that

$$\begin{aligned} E= {\left\{ \begin{array}{ll} \displaystyle \frac{a(A_1 B_2+A_2 B_1)+(A_2)^2V'(S)}{B_2} &{} \text {if} \quad a(A_1 B_2+A_2 B_1)+(A_2)^2V'(S) \ge 0, \\ 0 &{} \text {if} \quad a(A_1 B_2+A_2 B_1)+(A_2)^2V'(S)< 0. \end{array}\right. } \end{aligned}$$

(A6)

For the moment, let us focus on the case where $a(A_1 B_2+A_2 B_1)+(A_2)^2V'(S) \ge 0$, that is,

$$\begin{aligned} E=\frac{a(A_1 B_2+A_2 B_1)+(A_2)^2V'(S)}{B_2} \equiv E(S), \quad E'(S)=\frac{(A_2)^2V''(S)}{B_2}. \end{aligned}$$

(A7)

By substituting (A7) into (26), differentiating by S, and rearranging the terms, we obtain

$$\begin{aligned}&-\left[ \rho +\delta -E^{*\prime }(S)\right] \left[ a(A_1 B_2+A_2 B_1)-B_2 E(S)\right] \nonumber \\&\quad = B_2 E'(S)\left[ E(S)+E^*(S)-\delta S\right] -(A_2)^2\beta S. \end{aligned}$$

(A8)

Letting $E(S)=E^*(S)=E^a(S)$, (A8) can be rewritten as (27) in the text.

By substituting $E^a(S)=\psi _0 +\psi _1 S$ into (27), we obtain

$$\begin{aligned} \psi _1= \frac{(\rho +\delta )a(A_1 B_2+A_2 B_1)-(\rho +\delta )B_2 (\psi _0 +\psi _1 S)-(A_2)^2\beta S}{a(A_1 B_2+A_2 B_1)-3B_2 (\psi _0 +\psi _1 S)+B_2 \delta S} \end{aligned}$$

(A9)

The above equation holds for any $S\ge 0$ that satisfies $E^a(S)\ge 0$ if and only if the following two equations are satisfied:

$$\begin{aligned}&(\rho +\delta -3\psi _1)B_2 \psi _0 -(\rho +\delta -\psi _1)a(A_1 B_2+A_2 B_1)=0, \end{aligned}$$

(A10)

$$\begin{aligned}&(\rho +2\delta -3\psi _1)B_2\psi _1+ (A_2)^2\beta =0. \end{aligned}$$

(A11)

By solving (A11) for $\psi _1$, we obtain

$$\begin{aligned} \psi _1= \frac{\rho +2\delta \pm \sqrt{(\rho +2\delta )^2+12(A_2)^2\beta /B_2}}{6}. \end{aligned}$$

However, the positive root for $\psi _1$ violates the stability of the steady state because

$$\begin{aligned} \frac{d\dot{S}}{dS}=2\psi _1-\delta =\frac{\rho -\delta +\sqrt{(\rho +2\delta )^2+12(A_2)^2\beta /B_2}}{3}>0 \end{aligned}$$

(A12)

in this case. Thus, we must have

$$\begin{aligned} \psi _1= \frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+12(A_2)^2\beta /B_2}}{6}<0. \end{aligned}$$

(A13)

By solving (A10) for $\psi _0$ and using (A13), we obtain

$$\begin{aligned} \psi _0&=\frac{(\rho +\delta -\psi _1)a(A_1 B_2+A_2 B_1)}{(\rho +\delta -3\psi _1)B_2} \nonumber \\&=\frac{a(A_1 B_2+A_2 B_1)\left[ 5\rho +4\delta +\sqrt{(\rho +2\delta )^2+12(A_2)^2\beta /B_2}\right] }{3B_2\left[ \rho +\sqrt{(\rho +2\delta )^2+12(A_2)^2\beta /B_2}\right] }>0. \end{aligned}$$

(A14)

Recall that the strategy $E^a(S)=\psi _0 +\psi _1 S$ is valid only if $E^a(S)\ge 0$, or equivalently, $S\le -\psi _0/\psi _1\equiv \hat{S}$. If $S>\hat{S}$, the optimal emissions are zero for both countries. Substituting $E=E^*=0$ into (26) yields $\rho V(S)=K-(\beta /2)S^2-\delta V'(S)S$. Solving this differential equation of S, we can obtain the value function V(S) for $S>\hat{S}$.

To summarize, we obtain the linear Markov-perfect Nash equilibrium under autarky as follows:

$$\begin{aligned} E^a(S)= {\left\{ \begin{array}{ll} \psi _0 +\psi _1 S &{} \text {for} \quad S\le -\psi _0/\psi _1, \\ 0 &{} \text {for} \quad S>-\psi _0/\psi _1, \end{array}\right. } \end{aligned}$$

(A15)

where $\psi _0$ and $\psi _1$ are given by (A14) and (A13), respectively. From (A15) and the fact that $\dot{S}=0$ means a positive relationship between emissions and the pollution stock in the steady state, we conclude that there exists a unique steady state that satisfies $\dot{S}=2E^a(S)-\delta S=2(\psi _0+\psi _1 S)-\delta S=0$. From this condition, we obtain the steady-state pollution stock (28). $\square$

1.4 A.4 Proof of Proposition 4

It is straightforward from (20) and (25) that $S^{ac}<S^{ao}$. Note that we can generalize this result in the case in which the two countries do not necessarily share the same preferences, production structures, or technologies. In this case, the steady-state pollution stock in each regime can be calculated as

$$\begin{aligned} S^{ac}=\frac{(\rho +\delta )\left[ a(A_1 B_2+A_2 B_1)B_2^*+a^*(A_1^* B_2^*+A_2^* B_1^*)B_2\right] }{(\rho +\delta )\delta B_2 B_2^*+(\beta +\beta ^*)\left[ (A_2)^2 B_2^*+(A_2^*)^2 B_2\right] }, \\ S^{ao}= \frac{(\rho +\delta )\left[ a(A_1B_2+A_2B_1)B_2^*+a^*(A_1^*B_2^*+A_2^*B_1^*)B_2\right] }{(\rho +\delta )\delta B_2B_2^*+(A_2)^2\beta B_2^*+(A_2^*)^2\beta ^* B_2}. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} S^{ao}-S^{ac}=\frac{G_1}{G_2}, \end{aligned}$$

(A16)

where

$$\begin{aligned} G_1&\equiv (\rho +\delta ) \left[ (A_2^*)^2 B_2 \beta +(A_2)^2 B_2^* \beta ^*\right] \left[ a(A_1 B_2+A_2 B_1)B_2^*+a^*(A_1^* B_2^*+A_2^* B_1^*)B_2\right] , \\ G_2&\equiv \left[ (\rho +\delta )\delta B_2B_2^*+(A_2)^2\beta B_2^*+(A_2^*)^2\beta ^* B_2\right] \\&\quad \times \left\{ (\rho +\delta )\delta B_2 B_2^*+(\beta +\beta ^*)\left[ (A_2)^2 B_2^*+(A_2^*)^2 B_2\right] \right\} , \end{aligned}$$

which are both positive.

From (25), (28), and (A14), the difference in the steady-state levels of the pollution stock between the open-loop and linear Markov-perfect Nash equilibria is calculated as

$$\begin{aligned} S^{am}-S^{ao}&=\frac{2\psi _0}{\delta -2\psi _1}-\frac{2(\rho +\delta )a(A_1 B_2+A_2 B_1)}{(\rho +\delta )\delta B_2+2\beta (A_2)^2} \nonumber \\&=\frac{4a\left( A_1 B_2+A_2 B_1\right) \left[ (\rho +2\delta -3\psi _1)B_2\psi _1(\rho +\delta )+\beta (A_2)^2(\rho +\delta -\psi _1)\right] }{(\delta -2\psi _1)\left[ (\rho +\delta )\delta B_2+2\beta (A_2)^2\right] (\rho +\delta -3\psi _1)B_2}. \end{aligned}$$

(A17)

In light of (A13), it follows that

$$\begin{aligned}&(\rho +2\delta -3\psi _1)B_2\psi _1(\rho +\delta )+\beta (A_2)^2(\rho +\delta -\psi _1) \\&=-(A_2)^2\beta (\rho +\delta )+\beta (A_2)^2(\rho +\delta -\psi _1) =-(A_2)^2\beta \psi _1>0. \end{aligned}$$

Thus, the sign of (A17) is positive; the linear Markov-perfect Nash equilibrium results in a larger steady-state pollution stock than the open-loop Nash equilibrium. $\square$

1.5 A.5 Proof of Proposition 5

From (5) and (40), the matrix form of the dynamics of TE and S is given by

$$\begin{aligned} \begin{bmatrix} \dot{TE} \\ \dot{S} \end{bmatrix} = \begin{bmatrix} \rho +\delta &{} \frac{4\beta }{\gamma _2+2\gamma _3+\gamma _5} \\ 1 &{} -\delta \end{bmatrix} \begin{bmatrix} TE \\ S \end{bmatrix} + \begin{bmatrix} -\frac{2(\rho +\delta )(\gamma _1+\gamma _4)}{\gamma _2+2\gamma _3+\gamma _5} \\ 0 \end{bmatrix} . \end{aligned}$$

(A18)

The steady-state solutions for TE and S are derived as

$$\begin{aligned} S^{fc}=\frac{2a(\rho +\delta )(\gamma _1+\gamma _4)}{(\rho +\delta )\delta (\gamma _2+2\gamma _3+\gamma _5)+4\beta }, \quad TE^{fc}=E^{fc}+E^{*fc}=\delta S^{fc}. \end{aligned}$$

Further, the coefficient matrix in (A18) has one positive and one negative eigenvalue, indicating that the steady state is a saddle point.

Deriving the explicit solutions of (A18) and rearranging the terms to obtain the total emissions in the feedback form, we have

$$\begin{aligned} TE(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16\beta /(\gamma _2+2\gamma _3+\gamma _5)}}{2}\left[ S(t)-S^{fc}\right] +TE^{fc}. \end{aligned}$$

(A19)

If $\phi =1$, the coefficient of S(t) in the above equation can be rewritten as

$$\begin{aligned} TE(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16\beta \left( \tilde{A}_2-\tilde{A}_3\right) ^2\big /\tilde{B}_2}}{2}\left[ S(t)-S^f \right] +\delta S^f, \end{aligned}$$

(A20)

where $S^f$ is given by (50). $\square$

1.6 A.6 Proof of Proposition 6

The dynamic system consisting of (5), (43), and (44) can be rewritten in a matrix form as follows:

$$\begin{aligned} \begin{bmatrix} \dot{E} \\ \dot{E}^* \\ \dot{S} \end{bmatrix} = \begin{bmatrix} \rho +\delta &{} 0 &{} \frac{\beta }{\gamma _2+\gamma _3} \\ 0 &{} \rho +\delta &{} \frac{\beta }{\gamma _2+\gamma _3} \\ 1 &{} 1 &{} -\delta \end{bmatrix} \begin{bmatrix} E \\ E^* \\ S \end{bmatrix} + \begin{bmatrix} -\frac{(\rho +\delta )\gamma _1}{\gamma _2+\gamma _3} \\ -\frac{(\rho +\delta )\gamma _1}{\gamma _2+\gamma _3} \\ 0 \end{bmatrix} . \end{aligned}$$

(A21)

The steady-state solutions are

$$\begin{aligned} S^{fo}=\frac{2(\rho +\delta )\gamma _1}{(\rho +\delta )\delta (\gamma _2+\gamma _3)+2\beta }, \quad E^{fo}=E^{*fo}=\frac{\delta }{2}S^{fo}. \end{aligned}$$

We can also verify that the coefficient matrix in (A21) has two positive and one negative eigenvalues, indicating that the steady state is a saddle point.

Deriving the explicit solutions of (A21) and rearranging the terms to obtain the total emissions along the open-loop Nash equilibrium in the feedback form, we have

$$\begin{aligned} E(t)+E^*(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+8\beta /(\gamma _2+\gamma _3)}}{2}\left[ S(t)-S^{fo}\right] +E^{fo}+E^{*fo}. \end{aligned}$$

(A22)

If $\phi =1$, (A22) can be rewritten as

$$\begin{aligned} E(t)+E^*(t) =\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16\beta \left( \tilde{A}_2-\tilde{A}_3\right) ^2\big /\tilde{B}_2}}{2}\left[ S(t)-S^f \right] +\delta S^f, \end{aligned}$$

(A23)

which is the same as the cooperative solution. $\Box$

1.7 A.7 Proof of Proposition 7

From the first-order condition for maximizing the right-hand side of (26), it follows that

$$\begin{aligned} E= {\left\{ \begin{array}{ll} \displaystyle \frac{\gamma _1 -\gamma _3 E^*(S) +V'(S)}{\gamma _2} &{} \text {if} \quad \gamma _1 -\gamma _3 E^*(S) +V'(S) \ge 0, \\ 0 &{} \text {if} \quad \gamma _1 -\gamma _3 E^*(S) +V'(S)< 0. \end{array}\right. } \end{aligned}$$

(A24)

In the case where $\gamma _1 -\gamma _3 E^*(S) +V'(S) \ge 0$, we have

$$\begin{aligned} E=\frac{\gamma _1 -\gamma _3 E^*(S) +V'(S)}{\gamma _2} \equiv E(S), \quad E'(S)=\frac{-\gamma _3 E^{*\prime }(S)+V''(S)}{\gamma _2}. \end{aligned}$$

(A25)

By substituting (A25) into (46), differentiating by S, and rearranging the terms, we obtain

$$\begin{aligned}&\left[ E^{*\prime }(S)-(\rho +\delta )\right] \left[ \gamma _1-\gamma _2 E(S)-\gamma _3 E^*(S)\right] \nonumber \\&=E^{*\prime }(S)\left[ \gamma _4-\gamma _5 E^*(S)\right] -\beta S +\gamma _2 E'(S)\left[ E(S)+E^*(S)-\delta S\right] +\gamma _3 E^{*\prime }(S)\left[ E^*(S)-\delta S \right] . \end{aligned}$$

(A26)

Analogously for the foreign government, we obtain:

$$\begin{aligned}&\left[ E'(S)-(\rho +\delta )\right] \left[ \gamma _1-\gamma _2 E^*(S)-\gamma _3 E(S)\right] \nonumber \\&\quad =E'(S)\left[ \gamma _4-\gamma _5 E(S)\right] -\beta S +\gamma _2 E^{*\prime }(S)\left[ E(S)+E^*(S)-\delta S\right] +\gamma _3 E'(S)\left[ E(S)-\delta S \right] . \end{aligned}$$

(A27)

Letting $E(S)=E^*(S)=E^f(S)$ in (A26) and (A27) and rearranging, we obtain (47).

Substituting $E^f(S)=\tilde{\psi }_0+\tilde{\psi }_1 S$ into (47) yields

$$\begin{aligned} \tilde{\psi }_1= \frac{(\rho +\delta )\left[ \gamma _1-(\gamma _2+\gamma _3)\left( \tilde{\psi }_0+\tilde{\psi }_1 S\right) \right] -\beta S}{\gamma _1-\gamma _4-(3\gamma _2+2\gamma _3-\gamma _5)\left( \tilde{\psi }_0+\tilde{\psi }_1 S\right) +(\gamma _2+\gamma _3)\delta S}. \end{aligned}$$

(A28)

Similar to Proposition 3, we can obtain the values of $\tilde{\psi }_0$ and $\tilde{\psi }_1$ as follows:

$$\begin{aligned} \tilde{\psi }_1&= \frac{(\gamma _2+\gamma _3) \left[ \rho +2\delta -\sqrt{(\rho +2\delta )^2 +\frac{4\beta (3\gamma _2+2\gamma _3-\gamma _5)}{(\gamma _2+ \gamma _3)^2}}\right] }{2(3\gamma _2+2 \gamma _3-\gamma _5)}<0, \end{aligned}$$

(A29)

$$\begin{aligned} \tilde{\psi }_0&= \frac{\gamma _1(\rho +\delta -\tilde{\psi }_1)+\gamma _4\tilde{\psi }_1}{(\gamma _2+\gamma _3) (\rho +\delta )-(3 \gamma _2+2 \gamma _3- \gamma _5)\tilde{\psi }_1} \nonumber \\&= \frac{\Psi }{(3\gamma _2+2 \gamma _3-\gamma _5)\sqrt{(\rho +2\delta )^2 +\frac{4\beta (3\gamma _2+2\gamma _3-\gamma _5)}{(\gamma _2+ \gamma _3)^2}}} >0, \end{aligned}$$

(A30)

where

$$\begin{aligned} \Psi&\equiv \frac{\gamma _1\left[ 2(2\gamma _2+\gamma _3-\gamma _5)\delta + (5\gamma _2+3\gamma _3-2\gamma _5)\rho \right] }{\gamma _2+ \gamma _3} +\gamma _4(\rho +2\delta ) \\&\quad +(\gamma _1-\gamma _4)\sqrt{(\rho +2\delta )^2 +\frac{4\beta (3\gamma _2+2\gamma _3-\gamma _5)}{(\gamma _2+ \gamma _3)^2}}. \end{aligned}$$

Given the nonnegativity constraint on $E^f(S)$, the linear Markov-perfect Nash equilibrium is represented as follows:

$$\begin{aligned} E^f(S)= {\left\{ \begin{array}{ll} \tilde{\psi }_0+\tilde{\psi }_1 S &{} \text {for} \quad S\le -\tilde{\psi }_0/\tilde{\psi }_1, \\ 0 &{} \text {for} \quad S>-\tilde{\psi }_0/\tilde{\psi }_1, \end{array}\right. } \end{aligned}$$

(A31)

where $\tilde{\psi }_0$ and $\tilde{\psi }_1$ are given by (A30) and (A29), respectively. As the steady-state condition is $\dot{S}=2E^f(S)-\delta S=2\left( \tilde{\psi }_0+\tilde{\psi }_1 S\right) -\delta S=0$, the statement of the proposition can be obtained.

If $\phi =1$, (A29) and (A30) can be simplified as

$$\begin{aligned} \tilde{\psi }_1= & {} \frac{\rho +2\delta -\sqrt{(\rho +2\delta )+16\beta \left( \tilde{A}_2-\tilde{A}_3\right) ^2\big /\tilde{B}_2}}{4}, \\ \tilde{\psi }_0= & {} \frac{2(\rho +\delta )a\left( 2\tilde{A}_1\tilde{B}_2+\tilde{A}_2\tilde{B}_1-\tilde{A}_3\tilde{B}_1\right) }{\tilde{B}_2\left[ \rho +\sqrt{(\rho +2\delta )^2+16\beta \left( \tilde{A}_2-\tilde{A}_3\right) ^2\big /\tilde{B}_2}\right] }. \end{aligned}$$

Then, we can verify that $2E^f(S)=2\tilde{\psi }_0+2\tilde{\psi }_1 S$ coincides with the feedback solutions $E(t)+E^*(t)$ in both the cooperative solution and open-loop Nash equilibrium. $\square$

1.8 A.8 Proof of Proposition 10

If $n(z)=n$ for all $z\in [0,1]$ and $\phi =1$, the parameters of the model can be simplified to

$$\begin{aligned}&A_1=\frac{n}{n+1}\mu _1^{\alpha }, \quad A_2=\frac{n}{n+1}\mu _2^{\alpha }, \quad B_1=\frac{n}{(n+1)^2}\mu _1^{\alpha }, \quad B_2=\left( \frac{n}{n+1}\right) ^2\mu _2^{\alpha }, \\&\tilde{A}_1=\frac{n}{2n+1}\mu _1^{\alpha }, \quad \tilde{A}_2=\frac{n(n+1)}{2n+1}\mu _2^{\alpha }, \quad \tilde{A}_3=\frac{n^2}{2n+1}\mu _2^{\alpha }, \\&\tilde{B}_1=\frac{n}{(2n+1)^2}\mu _1^{\alpha }, \quad \tilde{B}_2=\left( \frac{n}{2n+1}\right) ^2\mu _2^{\alpha }. \end{aligned}$$

Substituting these parameters into (13) and (36), the instantaneous utility under autarky is given by

$$\begin{aligned} u=\frac{n(n+2)a^2\sigma ^2}{2(n+1)^2\mu _2^{\alpha }}+a\frac{\mu _1^{\alpha }}{\mu _2^{\alpha }}E-\frac{1}{2\mu _2^{\alpha }}E^2-\frac{\beta }{2}S^2 \end{aligned}$$

(A32)

and the instantaneous utility under free trade is

$$\begin{aligned} u=\frac{2n(n+1)a^2\sigma ^2}{(2n+1)^2\mu _2^{\alpha }}+\frac{a}{2}\frac{\mu _1^{\alpha }}{\mu _2^{\alpha }}(E+E^*)-\frac{1}{8\mu _2^{\alpha }}(E+E^*)^2-\frac{\beta }{2}S^2, \end{aligned}$$

(A33)

where $\sigma ^2\equiv \mu _2^{\alpha }-(\mu _1^{\alpha })^2 \ge 0$ is the variance of technology distribution.

Given the symmetry of countries, $E+E^*=2E$. Substituting this into (A33), we can verify that the coefficients of E and $E^2$ in (A33) are the same as those in (A32). Comparing the constant terms, it is easily verified that the constant term in (A33) is greater than the constant term in (A32).

Substituting the above parameters under the assumption of $n(z)=n$ for all $z\in [0,1]$ and $\phi =1$ into (A3), we obtain the optimal path of total emissions under autarky and international cooperation as follows:

$$\begin{aligned} TE(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16\beta \mu _2^{\alpha }}}{2}\left[ S(t)-S^{ac}\right] +\delta S^{ac}. \end{aligned}$$

(A34)

Under free trade, (A20) and (A23) can be rewritten as

$$\begin{aligned} TE(t)=\frac{\rho +2\delta -\sqrt{(\rho +2\delta )^2+16\beta \mu _2^{\alpha }}}{2}\left[ S(t)-S^f\right] +\delta S^f, \end{aligned}$$

(A35)

which is also identical to $2E^f(S)=2\tilde{\psi }_0+2\tilde{\psi }_1 S$ in the Markov-perfect Nash equilibrium. From Proposition 9, we have $S^{ac}=S^f$. Then, (A34) and (A35) coincide. This means that the discounted present value of (A32) and that of (A33) are identical except for the constant terms, which have higher value under free trade than autarky. Thus, we conclude that free trade achieves higher welfare than autarky when countries cooperate in their pollution control policies.

To consider the comparison between noncooperative equilibria under autarky and the free trade outcome, we focus on the steady-state welfare. Let us denote the steady-state welfare, as a function of the pollution stock, under autarky and free trade by $u_A(S)$ and $u_F(S)$, respectively. Substituting $E=\delta S/2$ into (A32) and $E+E^*=\delta S$ into (A33), respectively, we obtain

$$\begin{aligned}&u_A(S)=\displaystyle \frac{n(n+2)a^2\sigma ^2}{2(n+1)^2\mu _2^{\alpha }} +\frac{a\delta \mu _1^{\alpha }}{2\mu _2^{\alpha }}S -\left( \frac{\delta ^2}{8\mu _2^{\alpha }}+\frac{\beta }{2}\right) S^2, \end{aligned}$$

(A36)

$$\begin{aligned} \displaystyle&u_F(S)=\displaystyle \frac{2n(n+1)a^2\sigma ^2}{(2n+1)^2\mu _2^{\alpha }} +\frac{a\delta \mu _1^{\alpha }}{2\mu _2^{\alpha }}S -\left( \frac{\delta ^2}{8\mu _2^{\alpha }}+\frac{\beta }{2}\right) S^2. \end{aligned}$$

(A37)

From (A36) and (A37), we can verify $u_A(S)<u_F(S)$ for all $S>0$ and that both $u_A(S)$ and $u_F(S)$ attain their maximum at $S=2a\delta \mu _1^{\alpha }/\left( \delta ^2+4\beta \mu _2^{\alpha }\right) \equiv \bar{S}$. It is easily verified that $\bar{S}<S^f=S^{ac}<S^{ao}<S^{am}$, as shown in Fig. 2. Since $u^f=u_F(S^f)$, $u^{ac}=u_A(S^{ac})$, $u^{ao}=u_A(S^{ao})$, and $u^{am}=u_A(S^{am})$, with (52) and (53) we have $u^f>u^{ac}>u^{ao}>u^{am}$. $\square$

Appendix B Cournot–Nash Equilibrium in Product Markets Under Free Trade

From (29), each Home firm’s first-order conditions for profit maximization are

$$\begin{aligned}&\frac{\partial \pi _i(z)}{\partial x_i(z)} =\frac{a-\left[ \sum _{i=1}^{n(z)}x_i(z)+\sum _{j=1}^{n^*(z)}m_j(z)+x_i(z)\right] }{\lambda }-\tau \alpha (z)=0, \\&\frac{\partial \pi _i(z)}{\partial m_i^*(z)} =\frac{a^*-\left[ \sum _{j=1}^{n^*(z)}x_j^*(z)+\sum _{i=1}^{n(z)}m_i^*(z)+m_i^*(z)\right] }{\lambda ^*}-\tau \alpha (z)\phi \le 0, \end{aligned}$$

where the inequality in the second condition arises from the possibility that the trade cost is too high to serve the overseas market. With symmetry, the above profit maximization conditions are simplified to

$$\begin{aligned}&\frac{a-\left[ \left( n(z)+1\right) x(z)+n^*(z)m(z)\right] }{\lambda }-\tau \alpha (z)=0, \end{aligned}$$

(B1)

$$\begin{aligned}&\frac{a^*-\left[ n^*(z)x^*(z)+\left( n(z)+1\right) m^*(z)\right] }{\lambda ^*}-\tau \alpha (z)\phi \le 0. \end{aligned}$$

(B2)

Analogously for Foreign firms, the first-order conditions for profit maximization, with symmetry, are given by

$$\begin{aligned}&\displaystyle \frac{a^*-\left[ \left( n^*(z)+1\right) x^*(z)+n(z)m^*(z)\right] }{\lambda ^*}-\tau ^*\alpha ^*(z)=0, \end{aligned}$$

(B3)

$$\begin{aligned}&\displaystyle \frac{a-\left[ n(z)x(z)+\left( n^*(z)+1\right) m(z)\right] }{\lambda }-\tau ^*\alpha ^*(z)\phi \le 0. \end{aligned}$$

(B4)

The Cournot–Nash equilibrium in the Home market is characterized by (B1) and (B4), which derive each Home firm’s output x(z) and each Foreign firm’s output (export) m(z) as in (30). Analogously, the Cournot–Nash equilibrium in the Foreign market is characterized by (B2) and (B3), which derive each Foreign firm’s output $x^*(z)$ and each Home firm’s output (export) $m^*(z)$ as in (31).