Adaptation for Mitigation

Abstract

This paper develops a two-region (North and South) dynamic model in which regional stocks of effective labor are negatively influenced by the global stock of pollution. By characterizing the equilibrium strategy of each region we show that the regions’ best responses can be strategic complements through a dynamic complementarity effect. The model is then used to analyze the impact of adaptation assistance from North to South. It is shown that North’s unilateral assistance to South (thus enhancing South’s adaptation capacity) can facilitate pollution mitigation in both regions, especially when the assistance is targeted at labor protection. Pollution might increase in the short run, but in the long run the level of pollution will decline. The adaptation assistance we propose is incentive compatible and Pareto improving.

Appendix: Proofs of Propositions

1.1 Proof of Proposition 1

1.1.1 Problem of Period $t=2$

Since $V_{i,3}(Z_{3})=0$, it follows that $E_{i,2}=\bar{E}_{i}$ for both regions. Hence, for each $i \in \{n,s\}$,

$$\begin{aligned} C_{i,2} = Y_{i,2} = \Delta _{i,2}^{Y}A_{i,2} \left( \bar{E}_{i}^{\rho } + L_{i,2}^{\rho }\right) ^{1/\rho } \end{aligned}$$

(25)

and we find that the value function is

$$\begin{aligned} V_{i,2}(Z_{2})&= \ln (C_{i,2}) = \ln \left( \Delta _{i,2}^{Y}A_{i,2} \left( \bar{E}_{i}^{\rho } + L_{i,2}^{\rho } \right) ^{1/\rho }\right) \nonumber \\&= -\,\delta _{i,2}^{Y}M_{2}^{\xi } + \ln (A_{i,2}) + \frac{1}{\rho } \ln \left( \bar{E}_{i}^{\rho } + L_{i,2}^{\rho } \right) , \end{aligned}$$

(26)

where $Z_{2}=(L_{n,2},L_{s,2},M_{2},R_{2})$. We note that

$$\begin{aligned} \frac{d V_{i,2}(Z_{2})}{d L_{i,2}} = \frac{L_{i,2}^{\rho -1}}{\bar{E}_{i}^{\rho } + L_{i,2}^{\rho }}, \qquad \frac{d V_{i,2}(Z_{2})}{d L_{j, 2}} = 0, \end{aligned}$$

(27)

and

$$\begin{aligned} \frac{d V_{i,2}(Z_{2})}{d M_{2}} = -\, \delta _{i,2}^{Y}\xi M_{2}^{\xi -1}. \end{aligned}$$

(28)

1.1.2 Problem of Period $t=1$

Combining Eqs. (28) and (12), we may write the first-order condition as

$$\begin{aligned} \frac{E_{i,1}^{\rho -1}}{E_{i,1}^{\rho }+L_{i,1}^{\rho }} = \beta \delta _{i,2}^{Y}\xi \left( \phi _{M}M_{1} + E_{n,1} + E_{s,1}\right) ^{\xi -1} \quad \forall i \in \{n,s\}, \end{aligned}$$

(29)

which determines $(E_{n,1}, E_{s,1})$ as a function of $Z_{1}=(L_{n,1},L_{s,1},M_{1},R_{1})$. Notice that the left-hand side is decreasing in $E_{i,1}$ and is independent of $E_{j,1}$ whereas the right-hand side is increasing in both $E_{i,1}$ and $E_{j,1}$. It follows that the equilibrium combination of $(E_{n,1},E_{s,1})$ satisfies the stability conditions of Dixit (1986).

Totally differentiating Eq. (29) yields

$$\begin{aligned} \frac{d E_{i,1}}{dM_{1}}= & -\frac{\Psi _{i}}{1+\Psi _{n} + \Psi _{s} }\phi _{M} \le 0, \end{aligned}$$

(30)

$$\begin{aligned} \frac{d E_{i,1}}{dL_{i,1}}= & -\frac{1+\Psi _{j}}{1+\Psi _{n}+\Psi _{s}}\Phi _{i} < 0, \end{aligned}$$

(31)

$$\begin{aligned} \frac{d E_{i,1}}{dL_{j,1}}= & \frac{\Psi _{i}}{1+\Psi _{n}+\Psi _{s}}\Phi _{j} \ge 0, \end{aligned}$$

(32)

where we define

$$\begin{aligned} \Psi _{i}&:= \frac{E_{i,1}\left( E_{i,1}^{\rho }+L_{i,1}^{\rho }\right) }{E_{i,1}^{\rho } + (1-\rho )L_{i,1}^{\rho }}\frac{\xi -1}{\phi _{M}M_{1} + E_{n,1} + E_{s,1}} \ge 0, \end{aligned}$$

(33)

$$\begin{aligned} \Phi _{i}&:= \frac{\rho E_{i,1}L_{i,1}^{\rho -1}}{E_{i,1}^{\rho } + (1-\rho )L_{i,1}^{\rho }} >0. \end{aligned}$$

(34)

Notice that $\lim _{\xi \rightarrow 1}\Psi _{i}=0$ and the substitution yields

$$\begin{aligned} \lim _{\xi \rightarrow 1}\frac{d E_{i,1}}{dM_{1}} = 0, \quad \lim _{\xi \rightarrow 1}\frac{d E_{i,1}}{dL_{i,1}} = -\lim _{\xi \rightarrow 1}\Phi _{i}(Z_{1})<0, \quad \lim _{\xi \rightarrow 1}\frac{d E_{i,1}}{dL_{j,1}} = 0. \end{aligned}$$

(35)

The value function is

$$\begin{aligned} V_{i,1}(Z_{1})&= \ln (C_{i,1}) + \beta V_{i,2}(Z_{2}) \nonumber \\&= \ln \left( \Delta _{i,1}^{Y}A_{i,1}\left( E_{i,1}^{\rho } + L_{i,1}^{\rho }\right) ^{1/\rho }\right) + \beta V_{i,2}(Z_{2}) \nonumber \\&= -\,\delta _{i,1}^{Y}M_{1}^{\xi } + \ln (A_{i,1}) +\frac{1}{\rho }\ln \left( E_{i,1}^{\rho } +L_{i,1}^{\rho }\right) + \beta V_{i,2}(Z_{2}), \end{aligned}$$

(36)

where $E_{i,1}$ is implicitly defined by Eq. (29) and $V_{i,2}$ is given by Eq. (26). Using Eqs. (27), (28), and (29), we obtain

$$\begin{aligned} \frac{d V_{i,1}(Z_{1})}{d M_{1}}&= -\, \delta _{i,1}^{Y}\xi M_{1}^{\xi -1} + \frac{E_{i,1}^{\rho -1}}{E_{i,1}^{\rho } + L_{i,1}^{\rho }} \frac{dE_{i,1}}{dM_{1}} \nonumber \\&\quad +\, \beta \frac{\partial V_{i,2}(Z_{2})}{\partial M_{2}} \frac{d M_{2}}{dM_{1}} + \beta \frac{\partial V_{i,2}(Z_{2})}{\partial L_{i,2}} \frac{d L_{i,2}}{dM_{1}} + \beta \frac{\partial V_{i,2}(Z_{2})}{\partial L_{j,2}} \frac{d L_{j,2}}{dM_{1}} \nonumber \\&= -\, \delta _{i,1}^{Y}\xi M_{1}^{\xi -1} -\beta \delta _{i,2}^{Y}\xi M_{2}^{\xi -1} \left( \phi _{M} + \frac{dE_{j,1}}{dM_{1}}\right) -\beta \frac{L_{i,2}^{\rho }}{\bar{E}_{i}^{\rho } +L_{i,2}^{\rho }}\delta _{i,1}^{L}\xi M_{1}^{\xi -1}, \end{aligned}$$

(37)

where $dE_{j,1}/dM_{1}$ is given by Eqs. (30), (31), and (32). Moreover,

$$\begin{aligned} \frac{d^{2} V_{i,1}(Z_{1})}{d M_{1}^{2}}&= -\,\delta _{i,1}^{Y}\xi (\xi -1)M_{1}^{\xi -2} \nonumber \\&\quad -\, \beta \delta _{i,2}^{Y}\xi (\xi -1)M_{2}^{\xi -2} \left( \phi _{M} + \frac{dE_{j,1}}{dM_{1}}\right) \left( \phi _{M} + \frac{dE_{i,1}}{dM_{1}} +\frac{dE_{j,1}}{dM_{1}} \right) \nonumber \\&\quad -\, \beta \delta _{i,2}^{Y}\xi M_{2}^{\xi -1} \frac{d^{2}E_{j,1}}{dM_{1}^{2}} \nonumber \\&\quad -\, \beta \frac{L_{i,2}^{\rho }}{\bar{E}_{i}^{\rho } + L_{i,2}^{\rho }}\delta _{i,1}^{L}\xi M_{1}^{\xi -2} \left\{ \xi -1 -\frac{\bar{E}_{i}^{\rho }}{\bar{E}_{i}^{\rho } + L_{i,2}^{\rho }}\rho \delta _{i,1}^{L}M_{1}^{\xi } \right\} , \end{aligned}$$

(38)

where

$$\begin{aligned} \frac{d^{2}E_{j,1}}{dM_{1}^{2}} = -\frac{\phi _{M}}{1+\Psi _{n}+\Psi _{s}} \left( \frac{1+\Psi _{j}}{1+\Psi _{n}+\Psi _{s}}\frac{\partial \Psi _{i}}{\partial M_{1}} -\frac{\Psi _{i}}{1+\Psi _{n}+\Psi _{s}}\frac{\partial \Psi _{j}}{\partial M_{1}} \right) \end{aligned}$$

(39)

and

$$\begin{aligned} \frac{\partial \Psi _{i}}{\partial M_{1}}\frac{M_{1}}{\Psi _{i}}&= -\,\phi _{M} - \frac{\partial E_{i,1}}{\partial M_{1}} -\frac{\partial E_{j,1}}{\partial M_{1}} \nonumber \\&\quad + \left( 1 + \rho \frac{E_{i,1}^{\rho }}{E_{i,1}^{\rho } +L_{i,1}^{\rho }} - \rho \frac{E_{i,1}^{\rho }}{E_{i,1}^{\rho } +(1-\rho )L_{i,1}^{\rho }}\right) \frac{\partial E_{i,1}}{\partial M_{1}} \frac{M_{1}}{E_{i,1}}. \end{aligned}$$

(40)

Since $\lim _{\xi \rightarrow 1}\Psi _{i}=0$, it follows that

$$\begin{aligned} \lim _{\xi \rightarrow 1}\frac{\partial \Psi _{i}}{\partial M_{1}} = 0, \quad \lim _{\xi \rightarrow 1}\frac{d^{2}E_{j,1}}{dM_{1}^{2}} = 0, \end{aligned}$$

(41)

and therefore

$$\begin{aligned} \lim _{\xi \rightarrow 1}\frac{d^{2} V_{i,1}(Z_{1})}{d M_{1}^{2}} = \lim _{\xi \rightarrow 1}\beta \rho \bar{E}_{i}^{\rho }L_{i,2}^{\rho } \left( \frac{\delta _{i,1}^{L}}{\bar{E}_{i}^{\rho } + L_{i,2}^{\rho }}\right) ^{2} > 0. \end{aligned}$$

(42)

1.1.3 Problem of Period $t=0$

Combining Eqs. (37) and (12), we may write the first-order condition as

$$\begin{aligned} {\textit{MB}}_{i}(E_{i,0}) = {\textit{MD}}_{i}(E_{i,0},E_{j,0}), \end{aligned}$$

(43)

where ${\textit{MB}}_{i}(E_{i,0})$ is the marginal benefit from emission defined by

$$\begin{aligned} {\textit{MB}}_{i}(E_{i,0}) := \frac{Y_{i,0}}{C_{i,0}} \frac{E_{i,0}^{\rho -1}}{E_{i,0}^{\rho }+L_{i,0}^{\rho }} ={\left\{ \begin{array}{ll} \frac{Y_{n,0}}{Y_{n,0}-R}\frac{E_{n,0}^{\rho -1}}{E_{n,0}^{\rho } +L_{n,0}^{\rho }}, & i = n\\ \frac{E_{s,0}^{\rho -1}}{E_{s,0}^{\rho }+L_{s,0}^{\rho }}, & i = s \end{array}\right. } \end{aligned}$$

(44)

and ${\textit{MD}}_{i}(E_{i,0},E_{j,0})$ is the marginal damage from emission defined by

$$\begin{aligned} {\textit{MD}}_{i}(E_{i,0},E_{j,0}) := -\,\beta \frac{d V_{i,1}(Z_{1})}{d M_{1}} \bigg |_{M_{1}=\phi _{M}M_{0}+E_{i,0}+E_{j,0}}. \end{aligned}$$

(45)

Notice that for any $E_{j,0}\ge 0$, we have

$$\begin{aligned} \lim _{E_{i,0}\rightarrow 0}{\textit{MB}}_{i}(E_{i,0}) = \infty > \lim _{E_{i,0}\rightarrow 0}{\textit{MD}}_{i}(E_{i,0},E_{j,0}) \end{aligned}$$

(46)

and

$$\begin{aligned} \lim _{E_{i,0}\rightarrow \infty }{\textit{MD}}_{i}(E_{i,0},E_{j,0}) > 0 = \lim _{E_{i,0}\rightarrow \infty }{\textit{MB}}_{i}(E_{i,0}). \end{aligned}$$

(47)

Hence, for each $E_{j,0}\ge 0$, there exists $E_{i,0} > 0$ at which the graph of ${\textit{MB}}_{i}$ crosses the graph of ${\textit{MD}}_{i}$ from above. Let $E_{i}^{*}(E_{j,0})>0$ denote the smallest of such points, which is the reaction function of region i. We characterize the equilibrium level of $E_{s,0}$ by $E_{s,0}=E^{*}_{s} (E_{n}^{*}(E_{s,0}))$, which is the smallest solution of

$$\begin{aligned} {\textit{MB}}_{s}(E_{s,0}) = {\textit{MD}}_{s}(E_{s,0},E_{n}^{*}(E_{s,0})). \end{aligned}$$

(48)

The equilibrium level of $E_{n,0}$ is then determined by $E_{n,0}=E_{n}^{*}(E_{s,0})$. We note that, for both regions, the second-order condition is satisfied because the marginal benefit curve crosses the marginal damage curve from above.

To see that this equilibrium is stable, let us define

$$\begin{aligned} {\textit{MB}}_{i,i} := \frac{\partial {\textit{MB}}_{i}}{\partial E_{i,0}}, \quad {\textit{MD}}_{i,j} := \frac{\partial {\textit{MD}}_{i}}{\partial E_{j,0}} \quad \forall j \in \{n,s\} \end{aligned}$$

(49)

for each $i \in \{n,s\}$. The second-order condition implies

$$\begin{aligned} {\textit{MB}}_{i,i} < {\textit{MD}}_{i,i} \quad \forall i \in \{n,s\} \end{aligned}$$

(50)

at equilibrium. Also, by the definition of $E^{*}_{n}$, we have

$$\begin{aligned} {\textit{MB}}_{n}(E_{n}^{*}(E_{s,0})) = {\textit{MD}}_{n}(E_{n}^{*}(E_{s,0}),E_{s,0}) \quad \forall E_{s,0}\ge 0 \end{aligned}$$

(51)

and hence

$$\begin{aligned} {\textit{MB}}_{n,n}\frac{dE^{*}_{n}}{dE_{s,0}} = {\textit{MD}}_{n,n} \frac{dE^{*}_{n}}{dE_{s,0}} + {\textit{MD}}_{n,s}. \end{aligned}$$

(52)

Moreover, since the graph of ${\textit{MB}}_{i}$ crosses the graph of ${\textit{MD}}_{i}$ from above, we must have

$$\begin{aligned} {\textit{MB}}_{s,s} < {\textit{MD}}_{s,s} + {\textit{MD}}_{s,n}\frac{dE^{*}_{n}}{dE_{s,0}} \end{aligned}$$

(53)

at equilibrium. Combining Eqs. (50), (52), and (53) yields

$$\begin{aligned} ({\textit{MD}}_{s,s}-{\textit{MB}}_{s,s})({\textit{MD}}_{n,n}-{\textit{MB}}_{n,n}) - {\textit{MD}}_{s,n}{\textit{MD}}_{n,s} > 0, \end{aligned}$$

(54)

meaning that the stability conditions of Dixit (1986) are satisfied.

1.2 Proof of Proposition 2

To characterize $E_{i,0}$, observe first that the marginal benefit ${\textit{MB}}_{i}(E_{i,0})$ of emission is decreasing in $E_{i,0}$. On the other hand, Eq. (42) implies that at least when $\xi$ is in the neighborhood of $\xi =1$, the marginal damage ${\textit{MD}}_{i}(E_{i,0},E_{j,0})$ of emission is decreasing in $E_{j,0}$, meaning that the marginal damage curve of a region shifts upwards as the other region reduce its emission. Therefore, we conclude that there exists $\bar{\xi }>1$ such that the short-run regional emissions are strategic complements as long as $\xi < \bar{\xi }$.

1.3 Proof of Proposition 3

See text.

1.4 Proof of Proposition 4

Totally differentiating Eq. (43) with respect to R for both $i \in \{n,s\}$ and evaluating every term at $R=0$ yields

$$\begin{aligned} \left( \begin{array}{c} dE_{n,0}/dR \\ dE_{s,0}/dR \end{array}\right) = \frac{D}{\det (D)} \left( \begin{array}{c} \partial {\textit{MB}}_{n}/\partial R \\ -\partial {\textit{MD}}_{s}/\partial R \end{array}\right) , \end{aligned}$$

(55)

where

$$\begin{aligned} \frac{\partial {\textit{MB}}_{n}}{\partial R}&= \frac{E_{n,0}^{\rho -1}}{E_{n,0}^{\rho }+L_{n,0}^{\rho }}\frac{1}{Y_{n,0}} >0, \end{aligned}$$

(56)

$$\begin{aligned} \frac{\partial {\textit{MD}}_{s}}{\partial R}&= -\,\beta \phi _{R} \varepsilon ^{Y} + \beta ^{2}\frac{L_{s,2}^{\rho }}{\bar{E}^{\rho }_{s} +L_{s,2}^{\rho }}\left[ \frac{\bar{E}_{s}^{\rho }}{\bar{E}^{\rho }_{s} +L_{s,2}^{\rho }}(M_{0}+\phi _{R}M_{1})\rho \delta _{s}^{L}-\phi _{R}\right] \varepsilon ^{L}, \end{aligned}$$

(57)

$$\begin{aligned} D&:= \left( \begin{array}{cc} {\textit{MD}}_{s,s}-{\textit{MB}}_{s,s} & -\,{\textit{MD}}_{n,s} \\ -\,{\textit{MD}}_{s,n} & {\textit{MD}}_{n,n}-{\textit{MB}}_{n,n} \end{array}\right) . \end{aligned}$$

(58)

We first note that every element of matrix D is strictly positive: the diagonal elements because of the second-order condition and the off-diagonal elements because of Eqs. (45) and (42). The determinant of D is also strictly positive due to the stability condition (54).

Since every entry of matrix $D/\det (D)$ is strictly positive, Eq. (55) indicates that $dE_{i,0}/dR$ is a positive linear combination of $\partial {\textit{MB}}_{n}/\partial R$ and $-\partial {\textit{MD}}_{s}/\partial R$. We observe from Eq. (56) that $\partial {\textit{MB}}_{n}/\partial R$ is strictly positive. This term represents the income effect we discussed in the main text. The other partial derivative $\partial {\textit{MD}}_{s}/\partial R$ in Eq. (57) may be positive or negative, depending on the relative importance of substitution and complementarity effects.

Notice that in Eq. (57), the terms in square brackets are strictly positive if the damage parameter $\delta _{s}^{L}$ is sufficiently large. A sufficient (but by no means necessary) condition for such a case to be true is

$$\begin{aligned} \rho (\phi _{R}^{-1}+\phi _{M})M_{0}\delta _{s}^{L} > 1 + L_{s,0} \bar{E}_{s}^{-\rho }e^{g_{s,0}+g_{s,1}-(1+\phi _{M})M_{0}\delta _{s}^{L}}, \end{aligned}$$

(59)

which is satisfied by a sufficiently large value of $\delta _{s}^{L}$. Accordingly, under Assumption 1, $\partial {\textit{MD}}_{s}/\partial R$ is a strictly increasing linear function of $\varepsilon ^{L}$. On the other hand, the income effect, $\partial {\textit{MB}}_{n}/\partial R$, is independent of $\varepsilon ^{L}$, as is clear from Eq. (56). Hence, by Eq. (55), we know that $dE_{i,0}/dR$ is a strictly decreasing linear function of $\varepsilon ^{L}$. In particular, if we define

$$\begin{aligned} \varepsilon _{E_{n,0}}^{L} := \frac{\frac{{\textit{MD}}_{s,s} -{\textit{MB}}_{s,s}}{-{\textit{MD}}_{n,s}}\frac{\partial {\textit{MB}}_{n}}{\partial R} +\beta \phi _{R}\varepsilon ^{Y}}{\beta ^{2}\frac{L_{s,2}^{\rho }}{\bar{E}^{\rho }_{s}+L_{s,2}^{\rho }}\left[ \frac{\bar{E}_{s}^{\rho }}{\bar{E}^{\rho }_{s}+L_{s,2}^{\rho }}(M_{0}+\phi _{R}M_{1}) \rho \delta _{s}^{L}-\phi _{R} \right] } > 0 \end{aligned}$$

(60)

and

$$\begin{aligned} \varepsilon _{E_{s,0}}^{L} := \frac{\frac{-{\textit{MD}}_{s,n}}{{\textit{MD}}_{n,n}-{\textit{MB}}_{n,n}} \frac{\partial {\textit{MB}}_{n}}{\partial R} + \beta \phi _{R}\varepsilon ^{Y}}{\beta ^{2} \frac{L_{s,2}^{\rho }}{\bar{E}^{\rho }_{s}+L_{s,2}^{\rho }} \left[ \frac{\bar{E}_{s}^{\rho }}{\bar{E}^{\rho }_{s} +L_{s,2}^{\rho }}(M_{0}+\phi _{R}M_{1})\rho \delta _{s}^{L}-\phi _{R} \right] }> 0, \end{aligned}$$

(61)

it follows that for each region i, $dE_{i,0}/dR<0$ if and only if $\varepsilon ^{L} > \varepsilon ^{L}_{E_{i,0}}$.

Finally, Eq. (54) implies

$$\begin{aligned} \frac{{\textit{MD}}_{s,s}-{\textit{MB}}_{s,s}}{-{\textit{MD}}_{n,s}} > \frac{-{\textit{MD}}_{s,n}}{{\textit{MD}}_{n,n}-{\textit{MB}}_{n,n}}, \end{aligned}$$

(62)

which, together with Eqs. (60) and (61), yields $\varepsilon _{E_{n,0}}^{L} >\varepsilon _{E_{s,0}}^{L}$. For the sake of completeness, we also note that the two thresholds coincide if there is no income effect (i.e., if $\partial {\textit{MB}}_{n}/\partial R = 0$).

1.5 Proof of Proposition 5

First notice

$$\begin{aligned} \frac{dM_{1}}{dR} = \frac{dE_{n,0}}{dR} + \frac{dE_{s,0}}{dR}, \end{aligned}$$

(63)

where, by Proposition 4, the right-hand side is strictly decreasing in $\varepsilon ^{L}$. Also, Proposition 4 shows that there exist thresholds $\varepsilon ^{L}_{E_{n,0}}$ and $\varepsilon ^{L}_{E_{s,0}}$ with $0<\varepsilon _{E_{s,0}}^{L} <\varepsilon _{E_{n,0}}^{L}$ such that

$$\begin{aligned} \frac{dE_{n,0}}{dR} + \frac{dE_{s,0}}{dR} > 0 \quad \forall \varepsilon ^{L} < \varepsilon ^{L}_{E_{s,0}} \end{aligned}$$

(64)

and

$$\begin{aligned} \frac{dE_{n,0}}{dR} + \frac{dE_{s,0}}{dR} < 0 \quad \forall \varepsilon ^{L} > \varepsilon ^{L}_{E_{n,0}}. \end{aligned}$$

(65)

Then there must exist $\varepsilon ^{L}_{M_{1}}$ in the open interval $(\varepsilon ^{L}_{E_{s,0}}, \varepsilon ^{L}_{E_{n,0}})$ such that

$$\begin{aligned} \frac{dM_{1}}{dR} = \frac{dE_{n,0}}{dR} + \frac{dE_{s,0}}{dR} < 0 \iff \varepsilon ^{L} > \varepsilon ^{L}_{M_{1}}. \end{aligned}$$

(66)

Next, observe

$$\begin{aligned} \frac{dM_{2}}{dR} = \phi _{M}\frac{dM_{1}}{dR} + \frac{dE_{n,1}}{dR} + \frac{dE_{s,1}}{dR}, \end{aligned}$$

(67)

where the second term on the right-hand side is zero. By Eq. (17), the third term is strictly negative and strictly increasing in $\varepsilon ^{L}$. Therefore, combined with Eq. (66), this implies that there exists $\varepsilon ^{L}_{M_{2}}<\varepsilon ^{L}_{M_{1}}$ such that

$$\begin{aligned} \frac{dM_{2}}{dR} < 0 \iff \varepsilon ^{L} > \varepsilon ^{L}_{M_{2}}, \end{aligned}$$

(68)

which completes the proof.

1.6 Proof of Proposition 6

By the envelope theorem, a marginal change in the state variables has no first-order effect on $W_{i}(R)$ or $V_{i,t}(Z_{t})$ through the region i’s own control, $E_{i,t}$. Hence,

$$\begin{aligned} \frac{d W_{i}}{d R}&= \frac{1}{C_{i,0}}\frac{\partial C_{i,0}}{\partial R} + \beta \frac{d V_{i,1}(Z_{1})}{d M_{1}}\frac{dE_{j,0}}{dR} \nonumber \\&\quad +\, \beta \frac{d V_{i,1}(Z_{1})}{d L_{i,1}}\frac{dL_{i,1}}{dR} + \beta \frac{d V_{i,1}(Z_{1})}{d L_{j,1}}\frac{dL_{j,1}}{dR} + \beta \frac{d V_{i,1}(Z_{1})}{d R_{1}}\frac{dR_{1}}{dR} \nonumber \\&= {\left\{ \begin{array}{ll} - \frac{1}{Y_{n,0}-R} + \beta \frac{d V_{n,1}(Z_{1})}{d M_{1}} \frac{dE_{s,0}}{dR} + \beta \frac{d V_{n,1}(Z_{1})}{d L_{s,1}} \frac{dL_{s,1}}{dR} + \beta \frac{d V_{n,1}(Z_{1})}{d R_{1}}\phi _{R}, & i = n, \\ -\frac{d\delta _{s}^{Y}(R)}{dR}M_{0} + \beta \frac{d V_{s,1} (Z_{1})}{d M_{1}}\frac{dE_{n,0}}{dR} + \beta \frac{d V_{s,1} (Z_{1})}{d L_{s,1}}\frac{dL_{s,1}}{dR} + \beta \frac{d V_{s,1} (Z_{1})}{d R_{1}}\phi _{R}, & i = s, \end{array}\right. } \end{aligned}$$

(69)

where

$$\begin{aligned} \frac{d V_{n,1}(Z_{1})}{d L_{s,1}} = \beta \frac{d V_{n,2}(Z_{2})}{d M_{2}} \frac{dE_{s,1}}{dL_{s,1}}, \quad \frac{d V_{s,1}(Z_{1})}{d L_{s,1}} =\beta \frac{d V_{s,2}(Z_{2})}{d L_{s,2}}\frac{dL_{s,2}}{dL_{s,1}}, \end{aligned}$$

(70)

and

$$\begin{aligned} \frac{d V_{i,1}(Z_{1})}{d R_{1}}&= \frac{1}{Y_{i,1}} \left( \frac{\partial Y_{i,1}}{\partial R_{1}} +\frac{\partial Y_{i,1}}{\partial E_{i,1}}\frac{d E_{i,1}}{d R_{1}}\right) +\beta \frac{d V_{i,2}(Z_{2})}{d M_{2}}\frac{dE_{j,1}}{dR_{1}}\nonumber \\&\quad +\, \beta \frac{d V_{i,2}(Z_{2})}{d L_{i,2}}\frac{dL_{i,2}}{dR_{1}} +\beta \frac{d V_{i,2}(Z_{2})}{d L_{j,2}}\frac{dL_{j,2}}{dR_{1}} +\beta \frac{d V_{i,2}(Z_{2})}{d R_{2}}\frac{dR_{2}}{dR_{1}} \nonumber \\&= \frac{1}{Y_{i,1}}\frac{\partial Y_{i,1}}{\partial R_{1}} +\beta \frac{d V_{i,2}(Z_{2})}{d L_{i,2}}\frac{dL_{i,2}}{dR_{1}} +\beta \frac{d V_{i,2}(Z_{2})}{d R_{2}}\phi _{R} \nonumber \\&= {\left\{ \begin{array}{ll} \beta \frac{d V_{n,2}(Z_{2})}{d R_{2}}\phi _{R}, & i = n,\\ -\frac{d\delta _{s}^{Y}(R_{1})}{dR_{1}}M_{1} +\beta \frac{d V_{s,2}(Z_{2})}{d L_{s,2}}\frac{dL_{s,2}}{dR_{1}} +\beta \frac{d V_{i,2}(Z_{2})}{d R_{2}}\phi _{R}, & i = s. \end{array}\right. } \end{aligned}$$

(71)

Therefore,

$$\begin{aligned} \frac{d W_{n}}{d R}&= - \frac{1}{Y_{n,0}-R} + \beta \frac{d V_{n,1}(Z_{1})}{d M_{1}}\frac{dE_{s,0}}{dR} +\beta \frac{d V_{n,1}(Z_{1})}{d L_{s,1}}\frac{dL_{s,1}}{dR} +\beta \frac{d V_{n,1}(Z_{1})}{d R_{1}}\phi _{R} \nonumber \\&= - \frac{1}{Y_{n,0}-R} + \beta \frac{d V_{n,1}(Z_{1})}{d M_{1}} \frac{dE_{s,0}}{dR} + \beta ^{2}\frac{d V_{n,2}(Z_{2})}{d M_{2}} \frac{dE_{s,1}}{dL_{s,1}}\frac{dL_{s,1}}{dR} + \beta ^{2} \frac{dV_{n,2}(Z_{2})}{d R_{2}}\phi _{R}^{2} \nonumber \\&= -\frac{1}{Y_{n,0}-R} + \beta \frac{d V_{n,1}(Z_{1})}{d M_{1}} \frac{dE_{s,0}}{dR} + \beta ^{2} \frac{d V_{n,2}(Z_{2})}{d M_{2}}\frac{dE_{s,1}}{dR} \end{aligned}$$

(72)

and

$$\begin{aligned} \frac{d W_{s}}{d R}&= -\frac{d\delta _{s}^{Y}(R)}{dR}M_{0} +\beta \frac{d V_{s,1}(Z_{1})}{d M_{1}}\frac{dE_{n,0}}{dR} +\beta \frac{d V_{s,1}(Z_{1})}{d L_{s,1}}\frac{dL_{s,1}}{dR} +\beta \frac{d V_{s,1}(Z_{1})}{d R_{1}}\phi _{R} \nonumber \\&= -\frac{d\delta _{s}^{Y}(R)}{dR}M_{0} +\beta \frac{d V_{s,1}(Z_{1})}{d M_{1}}\frac{dE_{n,0}}{dR} +\beta ^{2}\frac{d V_{s,2}(Z_{2})}{d L_{s,2}}\frac{dL_{s,2}}{dL_{s,1}} \frac{dL_{s,1}}{dR} \nonumber \\&\quad +\, \beta \left( -\frac{d\delta _{s}^{Y}(R_{1})}{dR_{1}}M_{1} -\beta \frac{d V_{s,2}(Z_{2})}{d L_{s,2}}L_{s,2} \frac{d\delta _{s}^{L}(R_{1})}{dR_{1}}M_{1} + \beta \frac{d V_{i,2}(Z_{2})}{d R_{2}}\phi _{R}\right) \phi _{R}\nonumber \\&= \beta \frac{d V_{s,1}(Z_{1})}{d M_{1}}\frac{dE_{n,0}}{dR} -\frac{d\delta _{s}^{Y}(R)}{dR}M_{0} -\frac{d\delta _{s}^{Y}(R_{1})}{dR_{1}} \beta \phi _{R}M_{1} \nonumber \\&\quad +\, \beta ^{2}\frac{d V_{s,2}(Z_{2})}{d L_{s,2}} \left( \frac{dL_{s,2}}{dL_{s,1}}\frac{dL_{s,1}}{dR} + \phi _{R} \frac{dL_{s,2}}{dR_{1}}\right) \nonumber \\&= \beta \frac{d V_{s,1}(Z_{1})}{d M_{1}}\frac{dE_{n,0}}{dR} -\left( \frac{d\delta _{s}^{Y}(R)}{dR}M_{0} +\frac{d\delta _{s}^{Y} (R_{1})}{dR_{1}}\beta \phi _{R}M_{1}\right) \nonumber \\&\quad -\, \beta ^{2}\frac{d V_{s,2}(Z_{2})}{d L_{s,2}}L_{s,2} \left( \frac{d\delta _{s}^{L}(R)}{dR}M_{0} + \frac{d\delta _{s}^{L} (R_{1})}{dR_{1}}\phi _{R}M_{1}\right) , \end{aligned}$$

(73)

where we use $\phi _{R}^{2}\approx 0$. Evaluating these equations at $R=0$ yields Eqs. (21) and (22).

1.7 Proof of Proposition 7

Since $A_{n}$ does not affect the equilibrium levels of emission (when evaluated at $R=0$), it follows from Eqs. (55) and (56) that for each $i \in \{n,s\}$, $dE_{i,0}/dR$ is strictly decreasing in $A_{n}$. Then, for each given level of $\varepsilon ^{L}$, Eqs. (21) and (22) show that $dW_{i}/dR$ is strictly increasing in $A_{n}$. Since $dW_{i}/dR$ is strictly increasing in $\varepsilon ^{L}$, this implies that the threshold $\varepsilon _{W_{i}}^{L}$ is strictly decreasing in $A_{n}$.