Adapting to Climate Change: Equilibrium Welfare Implications for Large and Small Economies

Abstract

We show that the availability of adaptation can be welfare-reducing in the non-cooperative equilibrium in a setting with multiple countries. Adaptation is a private good while abatement is a public good. This means that substitution out of abatement and into adaptation by any one country imposes a negative externality on all other countries. The potentially deleterious impact of adaptation is asymmetric: small economies are most likely to be hurt by the availability of adaptation because they control a small fraction of global emissions relative to the biggest emitters.

Appendix

1.1 Proof of Proposition 1

The first-order conditions for $x_i$ and $a_i$ are, respectively,

$$\begin{aligned} 2x_i y_i =y_i \delta \sum _{j=1}^n {y_j (1-a_j )} \quad \forall i \end{aligned}$$

(22)

and

$$\begin{aligned} 2\theta a_i y_i =y_i \delta \sum _{j=1}^n {y_j (1-x_j )} \quad \forall i \end{aligned}$$

(23)

From (23) we obtain

$$\begin{aligned} a_i =\frac{\delta }{2\theta }\sum _{j=1}^n {y_j (1-x_j )} \equiv a \quad \forall i \end{aligned}$$

(24)

Thus, adaptation is identical across countries. Substituting $a$ from (24) for $a_j$ in (22), and rearranging, we obtain

$$\begin{aligned} x_i =\frac{\delta (1-a)Y}{2}\equiv x \quad \forall i \end{aligned}$$

(25)

Thus, technologies are identical across countries. Substituting $x$ from (25) for $x_j$ in (24), and rearranging, we obtain

$$\begin{aligned} a^{**}=\frac{\delta }{2\theta }\left( {1-\frac{\delta (1-a^{**})Y}{2}} \right) Y \end{aligned}$$

(26)

Solving for $a^{**}$ yields

$$\begin{aligned} a^{**}=\frac{\delta (2-\delta Y)Y}{4\theta -\delta ^{2}Y^{2}} \end{aligned}$$

(27)

Finally, substituting $a^{**}$ for $a$ in (25) yields

$$\begin{aligned} x^{**}=\frac{\delta (2\theta -\delta Y)Y}{4\theta -\delta ^{2}Y^{2}} \end{aligned}$$

(28)

Solving for the conditions under which $a^{**}\in [0,1]$ and $x^{**}\in [0,1]$ yields the three parts of Proposition 1. It is straightforward to show that second-order conditions are satisfied.   $\square$

1.2 Proof of Proposition 2

The first-order conditions for $x_i$ and $a_i$ are, respectively,

$$\begin{aligned} 2x_i y_i =\delta y_i^2 (1-a_i ) \end{aligned}$$

(29)

and

$$\begin{aligned} 2\theta a_i y_i =\delta (y_i (1-x_i )+E_{-i} )y_i \end{aligned}$$

(30)

Solving (29) and (30) yields best-response functions for $x_i$ and $a_i$. These are, respectively,

$$\begin{aligned} x_i (E_{-i} )=\frac{\delta (2\theta -\delta y_i -\delta E_{-i} )y_i }{4\theta -\delta ^{2}y_i^2 } \end{aligned}$$

(31)

and

$$\begin{aligned} a_i (E_{-i} )=\frac{\delta (2y_i +2E_{-i} -\delta y_i^2 )}{4\theta -\delta ^{2}y_i^2 } \end{aligned}$$

(32)

From (31) we can obtain the best-response function in terms of emissions:

$$\begin{aligned} e_i =[1-x_i (E_{-i} )]y_i=\frac{(4\theta -2\theta \delta y_i +\delta ^{2}y_i E_{-i} )}{4\theta -\delta ^{2}y_i^2 } \end{aligned}$$

(33)

Setting $E_{-i} = E-e_i$ in (33) and rearranging to make $e_i$ the subject, we have

$$\begin{aligned} e_i=y_i-\frac{(2\theta -\delta E)\delta y^{2}_{i}}{4\theta } \end{aligned}$$

(34)

Summing across i allows us to solve for the equilibrium $E$:

$$\begin{aligned} \hat{{E}}=\frac{2\theta (2Y-\delta S)}{4\theta -\delta ^{2}S} \end{aligned}$$

(35)

Setting $E=E^{\wedge }$ in (34), we can then solve for the equilibrium $\hat{{e}}_i$:

$$\begin{aligned} \hat{{e}}_i =\frac{[4\theta -\delta ^{2}S-y_i (2\delta \theta -\delta ^{2}Y)]y_i }{4\theta -\delta ^{2}S} \end{aligned}$$

(36)

Then from (36) we can obtain

$$\begin{aligned} \hat{{x}}_i =1-\frac{\hat{{e}}_i }{y_i }=\frac{\delta (2\theta -\delta Y)y_i }{4\theta -\delta ^{2}S} \end{aligned}$$

(37)

Substituting $\hat{{x}}_i$ for $x_i$ in (29) then allows us to solve for $a_i$:

$$\begin{aligned} \hat{{a}}_i =\frac{\delta (2Y-\delta S)}{4\theta -\delta ^{2}S} \quad \forall i \end{aligned}$$

(38)

Assumption 1 (from Sect. 4) ensures that these solutions are interior, that second-order conditions hold, and that the equilibrium is stable. We can then express $\hat{{e}}_i =(1-\hat{{x}}_i )y_i$ as $\hat{{e}}_i =y_i -\phi y_i^2$, where

$$\begin{aligned} \phi =\frac{2\delta \theta -\delta ^{2}Y}{4\theta -\delta ^{2}S}>0 \end{aligned}$$

(39)

as reported in the text.$\square$

1.3 Proof of Proposition 3

(a) Equilibrium domestic cost for country $i$ is

$$\begin{aligned} C(y_i ,\theta )=x_i^2 y_i +\theta a_i^2 y_i +\delta Ey_i (1-a_i ) \end{aligned}$$

(40)

In an interior equilibrium, $x_i , a_i$ and $E$ are evaluated at their equilibrium values given by (37), (38) and (35) respectively. It will prove useful to express this cost function in terms of $\phi$:

$$\begin{aligned} C(y_i ,\theta )=\phi ^{2}y_i^3 +S\left( {\phi +\frac{\delta }{2}} \right) \left( {\frac{Y}{S}-\phi } \right) y_i \end{aligned}$$

(41)

where $\phi$ is given by (39). The second bracketed term can be expressed as

$$\begin{aligned} \frac{Y}{S}-\phi =\left( {\frac{2\theta }{\delta S}} \right) \hat{{a}} \end{aligned}$$

(42)

so it must be positive if the equilibrium is interior.

Differentiating $C(y_i ,\theta )$ with respect to $\theta$ yields

$$\begin{aligned} \frac{\partial C(y_i ,\theta )}{\partial \theta }=2\phi y_i (y_i^2 -\tilde{s})\frac{\partial \phi }{\partial \theta } \end{aligned}$$

(43)

where

$$\begin{aligned} \tilde{s}=S-\frac{2Y-\delta S}{4\phi } \end{aligned}$$

(44)

and

$$\begin{aligned} \frac{\partial \phi }{\partial \theta }=\frac{2\delta ^{2}(2Y-\delta S)}{(4\theta -\delta ^{2}S)^{2}}>0 \end{aligned}$$

(45)

at an interior equilibrium. Thus, $\partial C/\partial \theta >0$ for $y_i^2 >\tilde{s}$ and $\partial C/\partial \theta <0$ for $y_i^2 <\tilde{s}$.

(b) Making adaptation universally unavailable is equivalent to taking the limit

$$\begin{aligned} {\begin{array}{l} {\lim } \\ {\theta \rightarrow \infty } \\ \end{array} }\left( \phi \right) =\frac{\delta }{2} \end{aligned}$$

(46)

Making this substitution for $\phi$ in (41) yields cost for country $i$ when adaptation is universally unavailable :

$$\begin{aligned} C_0 (y_i )=\left( {\frac{\delta ^{2}}{4}} \right) y_i^3 +\delta \left( {\frac{2Y-\delta S}{2}} \right) y_i \end{aligned}$$

(47)

where the second bracketed term must be positive in an interior equilibrium. Setting $C(y_i ,\theta )=C_0 (y_i )$ and solving for $y_i^2$ yields a critical threshold denoted

$$\begin{aligned} \bar{{s}}\equiv S-\frac{2Y-\delta S}{2\phi +\delta } \end{aligned}$$

(48)

Taking the difference $\tilde{s}-\bar{{s}}$ yields

$$\begin{aligned} \tilde{s}-\bar{{s}}=\left( {\phi -\frac{\delta }{2}} \right) \left( {\frac{2Y-\delta S}{4\phi ^{2}+2\phi \delta }} \right) \end{aligned}$$

(49)

This is strictly positive at an interior equilibrium for any finite $\theta$ since $\partial \phi /\partial \theta >0$. Thus, $\bar{{s}}<\tilde{s}$. Since $\partial C/\partial \theta <0$ for $y_i^2 <\tilde{s}$ (by part (a) above), it follows that $\partial C/\partial \theta <0$ at $y_i^2 =\bar{{s}}$. That is, $C(y_i ,\theta )$ crosses $C_0 (y_i )$ at $y_i^2 =\bar{{s}}$ from above. Thus, $C_0 (y_i )<C(y_i ,\theta )$ for $y_i^2 <\bar{{s}}$ and $C_0 (y_i )>C(y_i ,\theta )$ for $y_i^2 >\bar{{s}}$.$\square$

1.4 Proof of Proposition 4

(a) Total cost for country $i$ is given by (41). Summing across $i$ yields total global cost:

$$\begin{aligned} G(Y,\theta )=\phi ^{2}Q+S\left( {\phi +\frac{\delta }{2}} \right) \left( {\frac{Y}{S}-\phi } \right) Y \end{aligned}$$

(50)

where $Q=\sum _{i=1}^N {y_i^3 }$. Differentiating $G(Y,\theta )$ with respect to $\theta$ yields

$$\begin{aligned} \frac{\partial G(Y,\theta )}{\partial \theta }=2\phi (Q-\tilde{Q})\frac{\partial \phi }{\partial \theta } \end{aligned}$$

(51)

where

$$\begin{aligned} \tilde{Q}=YS-\frac{(2Y-\delta S)Y}{4\phi } \end{aligned}$$

(52)

and $\partial \phi /\partial \theta >0$ is given by (45). Thus, $\partial G/\partial \theta >0$ for $Q>\tilde{Q}$ and $\partial G/\partial \theta <0$ for $Q<\tilde{Q}$.

(b) Taking the limit of $G(Y,\theta )$ as $\theta \rightarrow \infty$ yields total global cost when adaptation is universally unavailable:

$$\begin{aligned} G_0 (Y)=\left( {\frac{\delta ^{2}}{4}} \right) Q+\delta \left( {\frac{2Y-\delta S}{2}} \right) Y \end{aligned}$$

(53)

Setting $G(Y,\theta )=G_0 (Y)$ and solving for $Q$ yields a critical threshold denoted

$$\begin{aligned} \bar{{Q}}=YS-\frac{(2Y-\delta S)Y}{2\phi +\delta } \end{aligned}$$

(54)

Taking the difference $\tilde{Q}-\bar{{Q}}$ yields

$$\begin{aligned} \tilde{Q}-\bar{{Q}}=Y\left( {\phi -\frac{\delta }{2}} \right) \left( {\frac{2Y-\delta S}{4\phi ^{2}+2\phi \delta }} \right) \end{aligned}$$

(55)

This is strictly positive at an interior equilibrium for any finite $\theta$ since $\partial \phi /\partial \theta >0$. Thus, $\bar{{Q}}<\tilde{Q}$. Since $\partial G/\partial \theta <0$ for $Q<\tilde{Q}$ (by part (a) above), it follows that $\partial G/\partial \theta <0$ at $Q=\bar{{Q}}$. That is, $G(Y,\theta )$ crosses $G_0 (Y)$ at $Q=\bar{{Q}}$ from above. Thus, $G_0 (Y)<G(Y,\theta )$ for $Q<\bar{{Q}}$ and $G_0 (Y)>G(Y,\theta )$ for $Q>\bar{{Q}}$.$\square$