A dynamic analysis of the income–pollution relationship in a two-country setting

Abstract

This research explores the presence of the income–pollution relationship in a dynamic two-country setting. We argue that the initial conditions of an economy, as described by its initial stock of physical capital, its total factor productivity, and the degree of environmental awareness, are critical determinants of the income–pollution paths. More importantly, these initial conditions have a more pronounced effect in a world where economies interact with one another and transboundary and global pollutants are affecting more than one country. We show that, in the presence of pollution externalities between two countries, the income–pollution paths these countries follow depend on the stage of development of each country, on their relative and absolute productivities, on the degree of environmental awareness and on the transboundary nature of the pollutants. In a two-country setting with pollution externalities, even more complex patterns may arise compared to single country models, thereby confirming the nonlinear effect of growth on environmental outcomes. It is shown that even small asymmetries span income–pollution relationships that are significantly different than the prescribed inverted U-shaped patterns.

Appendices

Appendix 1

See Table 4.

Table 4 Combinations of emissions intervals

Appendix 2

Maximizing conditions for optimization problem (10). Optimization problem (10) can be expressed as

$$\begin{aligned} \begin{array}{ll} \underset{x_{i,t},s_{i,t},\lambda _{i,t},\mu _{i,t}}{\max } &{} L_{i}\left( x_{i,t},s_{i,t},\lambda _{i,t},\mu _{i,t}\right) =\left( A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t}\right) - \\ &{} -\frac{\beta }{2}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t}\right) ^{2}+\left( s_{i,t}-\frac{\beta }{2}s_{i,t}^{2}\right) -\frac{\gamma _{i}}{2} \left( x_{i,t}+\rho _{ij}x_{j,t}\right) ^{2} \\ &{} +\lambda _{i,t}\left( A_{i}B^{2}k_{i,t}^{1/2}-x_{i,t}\right) +\mu _{i,t}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t}\right) \end{array} \end{aligned}$$

(28)

For $i=1,2$, the necessary Kuhn–Tucker conditions are the first-order conditions

$$\begin{aligned}&\begin{array}{ll} x_{i,t}: &{} A_{i}k_{i,t}^{1/2}-\gamma _{i}\left( x_{i,t}+\rho _{ij}x_{j,t}\right) -A_{i}\beta k_{i,t}^{1/2}(A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t})\\ &{}\quad =\lambda _{i,t}-\mu _{i,t}A_{i}k_{i,t}^{1/2}, \end{array} \end{aligned}$$

(29)

$$\begin{aligned}&\begin{array}{ll} s_{i,t}:&\beta (A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t})-\beta s_{i,t}=\mu _{i,t}, \end{array} \end{aligned}$$

(30)

along with complementarity slackness conditions

$$\begin{aligned}&\lambda _{i,t}\left( A_{i}B^{2}k_{i,t}^{1/2}-x_{i,t}\right) =0, \end{aligned}$$

(31)

$$\begin{aligned}&\mu _{i,t}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t}\right) =0, \end{aligned}$$

(32)

positivity of the Lagrange multipliers

$$\begin{aligned}&\lambda _{t,i}\ge 0, \end{aligned}$$

(33)

$$\begin{aligned}&\mu _{t,i}\ge 0, \end{aligned}$$

(34)

and feasibility conditions

$$\begin{aligned} h_{x}\left( x_{i,t},s_{i,t}\right) =A_{i}B^{2}k_{i,t}^{1/2}-x_{i,t}\ge 0, \end{aligned}$$

(35)

and

$$\begin{aligned} h_{s}\left( x_{i,t},s_{i,t}\right) =A_{i}k_{i,t}^{1/2}x_{i,t}-s_{i,t}-k_{i,t}\ge 0, \end{aligned}$$

(36)

where $h_{i}\left( x_{i,t},s_{i,t}\right)$, $i=x,s$, denotes the respective constraint function. Note that the objective function is defined over the compact set $\left( x_{i,t},s_{i,t}\right) \subset {\mathbb {R}} _{+}^{2}$. Moreover, $U_{i}\left( \cdot \right)$ is everywhere continuous. Therefore, according to the Weierstrass Theorem, this problem has a global minimum and a global maximum. Furthermore, the constraint qualification is satisfied since the rank of the matrix below is equal to the number of the constraints:

$$\begin{aligned} rank\left[ \begin{array}{cc} \dfrac{\partial h_{x}}{\partial x_{i,t}} &{} \dfrac{\partial h_{x}}{\partial s_{i,t}} \\ \dfrac{\partial h_{s}}{\partial x_{i,t}} &{} \dfrac{\partial h_{s}}{\partial s_{i,t}} \end{array} \right] =rank\left[ \begin{array}{cc} -1 &{} 0 \\ A_{i}k_{i,t}^{1/2} &{} -1 \end{array} \right] =2. \end{aligned}$$

According to Proposition 6.5 in Sundaram (1996), this ensures that conditions (29)–(36) define the solution $\left( x_{i,t}^{*},s_{i,t}^{*}\right)$ in the case of a single country, or the best responses $\left( x_{i,t}^{*}\left( x_{j,t},s_{j,t}\right) ,s_{i,t}^{*}\left( x_{j,t},s_{j,t}\right) \right)$ in the case of two countries. $\square$

Proof of Proposition 1

For the proof of this proposition, we have to examine the sign of

$$\begin{aligned} \partial x^{*}\left( t\right) /\partial t=\partial x^{*}\left[ k\left( t\right) \right] /\partial k\left( t\right) \times \partial k\left( t\right) /\partial t. \end{aligned}$$

We start with the sign of the $\partial x^{*}\left[ k\left( t\right) \right] /\partial k\left( t\right)$. Differentiating equation (18) with respect to the capital yields

$$\begin{aligned} \frac{\partial x^{*}\left[ k\left( t\right) \right] }{\partial k\left( t\right) }=\frac{A\left( A^{2}\beta ^{2}k\left( t\right) ^{2}-2A^{2}\beta k\left( t\right) +6\gamma \beta k\left( t\right) +4\gamma \right) }{2\sqrt{ k\left( t\right) }\left( \beta k\left( t\right) A^{2}+2\gamma \right) ^{2}} \end{aligned}$$

Note that this is a continuous function for all $k\left( t\right) \ge 0$. Setting the derivative equal to zero under the assumptions that

$$\begin{aligned}&\frac{A\left( A^{2}\beta ^{2}\left( \frac{A^{2}-3\gamma }{A^{2}\beta } \right) ^{2}-2A^{2}\beta \left( \frac{A^{2}-3\gamma }{A^{2}\beta }\right) +6\gamma \beta \left( \frac{A^{2}-3\gamma }{A^{2}\beta }\right) +4\gamma \right) }{2\left( \frac{A^{2}-3\gamma }{A^{2}\beta }\right) ^{1/2}\left( \beta \left( \frac{A^{2}-3\gamma }{A^{2}\beta }\right) A^{2}+2\gamma \right) ^{2}} \\&\quad -\frac{A^{4}-10A^{2}\gamma +9\gamma ^{2}}{2\left( A^{2}-\gamma \right) ^{2}\sqrt{\frac{A^{2}-3\gamma }{ \beta }}}\\&\beta \ne 0\wedge \frac{1}{A^{2}\beta }\left( -3\gamma +\sqrt{9\gamma ^{2}-10A^{2}\gamma +A^{4}}+A^{2}\right) \in {\mathbb {C}} \setminus \left\{ 0,-\frac{2}{A^{2}\beta }\gamma \right\} \wedge \\&\quad \wedge A \ne 0\wedge -\frac{1}{A^{2}\beta }\left( 3\gamma +\sqrt{9\gamma ^{2}-10A^{2}\gamma +A^{4}}-A^{2}\right) \in {\mathbb {C}} \setminus \left\{ 0,-\frac{2}{A^{2}\beta }\gamma \right\} , \end{aligned}$$

yields two solutions:

$$\begin{aligned}&\frac{A\left( A^{2}\beta ^{2}k^{2}-2A^{2}\beta k+6\gamma \beta +4\gamma \right) }{2k^{1/2}\left( \beta kA^{2}+2\gamma \right) ^{2}} \\&{\underline{k}}\left( t\right) =\frac{1}{A^{2}\beta }\left( A^{2}-3\gamma - \sqrt{9\gamma ^{2}-10A^{2}\gamma +A^{4}}\right) , \\&{\overline{k}}\left( t\right) =\frac{1}{A^{2}\beta }\left( A^{2}-3\gamma + \sqrt{9\gamma ^{2}-10A^{2}\gamma +A^{4}}\right) . \end{aligned}$$

It is straightforward to show that

$$\begin{aligned} \left. \frac{\partial x^{*}\left[ k\left( t\right) \right] }{\partial k\left( t\right) }\right| _{k\left( t\right) =0}\rightarrow \infty \end{aligned}$$

and that, under (13),

$$\begin{aligned} \left. \frac{\partial x^{*}\left[ k\left( t\right) \right] }{\partial k\left( t\right) }\right| _{\left( 1/2\right) \left( {\underline{k}}\left( t\right) +{\overline{k}}\left( t\right) \right) }=-\frac{\beta ^{1/2}\left( A^{4}-10A^{2}\gamma +9\gamma ^{2}\right) }{2\left( A^{2}-\gamma \right) ^{2}\left( A^{2}-3\right) ^{1/2}} <0 \end{aligned}$$

Therefore, ${\underline{k}}\left( t\right)$ is a local maximum and $\overline{ k}\left( t\right)$ is a local minimum of the $x^{*}\left[ k\left( t\right) \right] .$ We compare the extrema to the steady-state value of capital defined by equation (14). Note that $k^{SS}-{\overline{k}} \left( t\right) =-\left( A^{4}-10A^{2}\gamma +9\gamma ^{2}\right) ^{1/2}/\beta A^{2}<0$. This shows that the local minima can never be reached. On the other hand,$~k^{SS}-{\underline{k}}\left( t\right) =\left( A^{4}-10A^{2}\gamma +9\gamma ^{2}\right) ^{1/2}/\beta A^{2}>0$, so that a maximum is always reached. We now turn on the sign of $\partial k\left( t\right) /\partial t$ that by assumption is nonnegative. There are two cases. When the $k^{SS}~$is reached before $\partial x^{*}\left[ k\left( t\right) \right] /\partial k\left( t\right)$ turns negative, the dynamics of emissions over time are positively monotonic. If, on the other hand, $k^{SS}$ is reached once $\partial x^{*}\left[ k\left( t\right) \right] /\partial k\left( t\right) <0$, the dynamics of $x^{*}\left( t\right)$ will depend only on the initial condition. $\square$

Lemma 5

For any positive values of domestic and foreign emissions, country i’ s optimal savings equals half the disposable income, i.e.,

$$\begin{aligned} s_{i,t}^{*}=\frac{1}{2}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) \text {.} \end{aligned}$$

Therefore, the following optimization problem is equivalent to problem (28):

$$\begin{aligned} \begin{array}{ll} \underset{x_{i,t},~\lambda _{i,t}}{\max } &{} L_{i}\left( x_{t,i},\lambda _{i,t}\right) =\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) -\frac{\beta }{4}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) ^{2} \\ &{} -\frac{\gamma _{i}}{2}\left( x_{i,t}+\rho _{ij}x_{j,t}\right) ^{2}+\lambda _{i,t}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) . \end{array} \end{aligned}$$

(37)

Proof of Lemma 5

Consider some $x_{i,t}^{o}\in [0,{\overline{x}}_{i,t}]$, $x_{j,t}^{o}\in [0,{\overline{x}}_{j,t}]$, and also $s_{i,t}^{*}= \frac{1}{2}\left( A_{i}k_{t,i}^{1/2}x_{t,i}-k_{t,i}\right)$. Evaluating ( 7) at $\left( x_{i,t}^{o},x_{j,t}^{o},s_{i,t}^{*}\right)$ yields

$$\begin{aligned} {\widetilde{U}}_{i}\left( x_{i,t}^{o},x_{i,t}^{o},s_{i,t}^{*}\right)= & {} \left( A_{i}k_{i,t}^{1/2}x_{i,t}^{o}-k_{i,t}\right) -\frac{\beta }{4}\left( A_{i}k_{i,t}^{1/2}x_{i,t}^{o}-k_{i,t}\right) ^{2}\\&-\gamma _{i}\left( x_{i,t}^{o}+\rho _{ij}x_{j,t}^{o}\right) ^{2}\text {.} \end{aligned}$$

Consider now any $s_{i,t}^{**}\ne s_{i,t}^{*}$. Evaluating (7) at $\left( x_{i,t}^{o},x_{j,t}^{o},s_{i,t}^{**}\right)$ yields

$$\begin{aligned} {\widehat{U}}_{i}\left( x_{i,t}^{o},x_{j,t}^{o},s_{i,t}^{**}\right)= & {} \left( A_{i}k_{i,t}^{1/2}x_{i,t}^{o}-k_{i,t}-s_{i,t}^{**}\right) - \frac{\beta }{2}\left( A_{i}k_{i,t}^{1/2}x_{i,t}^{o}-k_{i,t}-s_{i,t}^{**}\right) ^{2} \\&+\left( s_{i,t}^{**}-\frac{\beta }{2}\left( s_{i,t}^{**}\right) ^{2}\right) -\gamma _{i}\left( x_{i,t}^{o}+\rho _{ij}x_{j,t}^{o}\right) ^{2} \end{aligned}$$

Taking the difference

$$\begin{aligned} {\widetilde{U}}_{i}\left( x_{i,t}^{o},x_{i,t}^{o},s_{i,t}^{*}\right) - {\widehat{U}}_{i}\left( x_{i,t}^{o},x_{j,t}^{o},s_{i,t}^{**}\right) = \frac{\beta }{4}\left( k_{i,t}+2s_{i,t}^{**}-A_{i}k_{i,t}^{1/2}x_{i,t}\right) >0 \end{aligned}$$

implying that $s_{i,t}^{*}=\left( A_{i}k_{t,i}^{1/2}x_{t,i}-k_{t,i}\right)$ is indeed the maximizer of (28). Therefore, by substituting $s_{i,t}^{*}$the optimization problem (28) transforms to

$$\begin{aligned} \begin{array}{ll} \underset{x_{i,t},~\lambda _{i,t}~}{\max } &{} L_{i}\left( x_{i,t},\lambda _{i,t}\right) =\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) -\frac{\beta }{4}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) ^{2} \\ &{} -\frac{\gamma _{i}}{2}\left( x_{i,t}+\rho _{ij}x_{j,t}\right) ^{2}+\lambda _{i,t}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) . \end{array} \end{aligned}$$

$\square$

Maximization of problem (37). Maximization of problem (37) requires the following conditions:

$$\begin{aligned}&A_{i}k_{i,t}^{1/2}-\frac{\beta }{2}A_{i}k_{i,t}^{1/2}\left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) -\gamma _{i}\left( x_{i,t}+\rho _{ij}x_{j,t}\right) =\lambda _{i,t}, \end{aligned}$$

(38)

$$\begin{aligned}&A_{i}B^{2}k_{i,t}^{1/2}-x_{i,t}\ge 0, \end{aligned}$$

(39)

$$\begin{aligned}&\lambda _{i,t}\left( A_{i}B^{2}k_{i,t}^{1/2}-x_{i,t}\right) =0, \end{aligned}$$

(40)

and

$$\begin{aligned} \lambda _{i,t}\ge 0. \end{aligned}$$

(41)

Conditions (38)–(41) for $i=1,2$ characterize the solution of the pollution game. $\square$

Proof of Lemma 3

First, note that $\forall i\in \left\{ 1,2\right\}$, the welfare functions are defined as $U_{i}:X_{i}\rightarrow {\mathbb {R}}$, where $X_{i}\subset {\mathbb {R}}$ and they are continuously differentiable. Function $P:X_{i}\rightarrow {\mathbb {R}}$ is also continuously differentiable and $\forall i\in \left\{ 1,2\right\}$ we have

$$\begin{aligned} \frac{\partial P}{\partial x_{i}}=A_{i}k_{i,t}^{1/2}\left( 1-\frac{\beta }{2} \left( A_{i}k_{i,t}^{1/2}x_{i,t}-k_{i,t}\right) \right) -\gamma \left( x_{i,t}+x_{j,t}\right) =\frac{\partial {\overline{U}}_{i}}{\partial x_{i}}. \end{aligned}$$

Therefore, according to Lemma 4.4 in Monderer and Shapley (1996), (22) is a potential function of the pollution game. Moreover, the Hessian of P(x) is given by

$$\begin{aligned} H= \begin{bmatrix} \frac{\partial ^{2}P}{\partial x_{1}^{2}} &{} \frac{\partial ^{2}P}{\partial x_{1}\partial x_{2}} \\ \frac{\partial ^{2}P}{\partial x_{1}\partial x_{2}} &{} \frac{\partial ^{2}P}{ \partial x_{2}^{2}} \end{bmatrix} = \begin{bmatrix} -\frac{\beta }{2}A_{1}k_{1,t}^{1/2}-\gamma &{} -\gamma \\ -\gamma &{} -\frac{\beta }{2}A_{2}k_{2,t}^{1/2}-\gamma \end{bmatrix} \end{aligned}$$

Consider ${\mathbf {z}}\in {\mathbb {R}} ^{2}\diagup \{{\mathbf {0}}_{2\times 1}\}$ and construct the quadratic form

$$\begin{aligned}&{\mathbf {z}}^{T}H{\mathbf {z}}=\\&\quad = \begin{bmatrix} z_{1} \\ z_{2} \end{bmatrix} ^{T} \begin{bmatrix} -\frac{\beta }{2}A_{1}^{2}k_{1,t}-\gamma &{} -\gamma \\ -\gamma &{} -\frac{\beta }{2}A_{2}^{2}k_{1,t}-\gamma \end{bmatrix} \begin{bmatrix} z_{1} \\ z_{2} \end{bmatrix}\\&\quad =-\sum _{i=1}^{2}\left[ \left( \frac{\beta }{2}A_{i}^{2}k_{i,t}+\gamma \right) z_{i}^{2}\right] -2\gamma z_{1}z_{2}<0 \end{aligned}$$

Thus the Hessian matrix of the potential function is negative definite and, therefore, the potential function $P({\mathbf {x}})$  is strictly concave. $\square$