Environmental Quality, Public Debt and Economic Development

Abstract

This article analyzes the consequences on capital accumulation and environmental quality of environmental policies financed by public debt. A public sector of pollution abatement is financed by a tax or by public debt. We show that if the initial capital stock is high enough, the economy monotonically converges to a long-run steady state. On the contrary, when the initial capital stock is low, the economy is relegated to an environmental poverty trap. We also explore the implications of public policies on the trap and on the long-run stable steady state. In particular, we find that government should decrease debt and increase pollution abatement to promote capital accumulation and environmental quality at the stable long-run steady state. Finally, a welfare analysis shows that there exists a level of public debt that allows a long run steady state to be optimal.

Appendices

Appendix A: An Example of Longevity Function

Assume $\beta (e)=\frac{\alpha +x}{a+x}$ with $x=e-\underline{e}$ and $0<\alpha <a$. In our paper, we take the limit case where $\underline{e}$ tends to $-\infty$. This simplification allows to have that $\beta (e)$ is defined for all $e\in \mathbb{R }$. This function needs to satisfy: $\underline{\beta }<\beta (e)<1$, $0<\beta ^{\prime }(e)<\widetilde{\beta }$ and $|\beta ^{\prime \prime }(e)/\beta ^{\prime }(e)|<\overline{\beta }$. We hence have:

$$\begin{aligned} \beta ^{\prime }(e)&= \frac{a-\alpha }{(a+x)^{2}}>0 \\ \beta ^{\prime \prime }(e)&= \frac{-2(a-\alpha )}{(a+x)^{3}}<0 \end{aligned}$$

which implies that:

$$\begin{aligned} \left|\frac{\beta ^{\prime \prime }(e)}{\beta ^{\prime }(e)}\right|=-\frac{\beta ^{\prime \prime }(e)}{\beta ^{\prime }(e)}=\frac{2}{a+x} \end{aligned}$$

We assume $\left( i\right) a=\frac{2}{\overline{\beta }}; \left( ii\right) \alpha =\frac{2}{\overline{\beta }}\left( 1-\frac{2\widetilde{ \beta }}{\overline{\beta }}\right)$ and $\left( iii\right) \underline{\beta }$ sufficiently low, such that $\underline{\beta }<1-\frac{2\widetilde{ \beta }}{\overline{\beta }}.$ Of course, this requires $\overline{\beta }>2\widetilde{\beta }$. We deduce that:

$$\begin{aligned}&\underline{\beta }<\frac{\alpha }{a}<\beta (e)<1 \\&0<\beta ^{\prime }(e)<\frac{a-\alpha }{a^{2}}=\widetilde{\beta } \\&\left|\frac{\beta ^{\prime \prime }(e)}{\beta ^{\prime }(e)}\right|<\frac{2}{a}=\overline{\beta } \end{aligned}$$

Appendix B: Proof of Proposition 2

We differentiate the dynamic system (13)–(14) in the neighborhood of a steady state:

$$\begin{aligned} de_{t+1}&= \frac{1-m}{n}de_t -\frac{\alpha sk^{s-1}}{n} dk_t \\ dk_{t+1}&= (k+b)\frac{\beta ^{\prime }(e)}{\beta (e)(1+\beta (e))}de_t +\frac{\beta (e)}{n(1+\beta (e))} (1-s)sk^{s-2}(k+b) dk_t \end{aligned}$$

The trace $T$ and the determinant $D$ of the associated characteristic polynomial $P(\lambda )\equiv \lambda ^2 -T\lambda +D=0$ are given by:

$$\begin{aligned} T&= \frac{1-m}{n}+\frac{\beta (e)}{n(1+\beta (e))}(1-s)sk^{s-2}(k+b) \end{aligned}$$

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$$\begin{aligned} D&= \frac{1-m}{n}\frac{\beta (e)}{n(1+\beta (e))} (1-s)sk^{s-2}(k+b) +\frac{\alpha sk^{s-1}}{n}(k+b)\frac{\beta ^{\prime }(e)}{\beta (e)(1+\beta (e))} \end{aligned}$$

(32)

Since $T>0$ and $D>0$, we have $P(-1)=1+T+D>0$. We now determine $P(1)=1-T+D$. After some computations, we get:

$$\begin{aligned} P(1)=\frac{n+m-1}{n}\frac{\beta (e)}{1+\beta (e)}(k+b)\left( \Omega ^{\prime }(k)- \Phi ^{\prime }(k)\right) \end{aligned}$$

(33)

with $\Omega ^{\prime }(k)=\frac{\alpha sk^{s-1}}{n-1+m}\frac{\beta ^{\prime }(e)}{\beta (e)^2}$ and $\Phi ^{\prime }(k)=\frac{g-nb-h(k)}{n(k+b)^2}$. Therefore, at the steady state $k_1$, we have $P(1)<0$, because $\Omega ^{\prime }(k)< \Phi ^{\prime }(k)$. This shows that this is a saddle. At the steady state $k_2$, we have $\Omega ^{\prime }(k)>\Phi ^{\prime }(k)$, implying that $P(1)>0$. This steady state is stable or unstable, depending on the value of $D$ with respect to $1$.

Choose ${\widetilde{\beta }}$ sufficiently small. Since $\beta ^{\prime }(e)<\widetilde{\beta }, D<1$ requires:

$$\begin{aligned} \frac{1-m}{n}\frac{\beta (e)}{n(1+\beta (e))} (1-s)sk^{s-2}(k+b)<1 \end{aligned}$$

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We can show that this is equivalent to $\Phi ^{\prime }(k)<0$, which shows that the steady state $k_2$ is stable. $\square$

Appendix C: Proof of Lemma 1

Using the definition of the matrix $A$, we can compute:

$$\begin{aligned} det A&= \xi \zeta -\chi \kappa \\&= \beta (e) (n-1+m)\left[ (1-s)sk^{s-2}(k+b)-n\frac{1+\beta (e)}{\beta (e)} \right. \\&\left. -n(k+b)\frac{\beta ^{\prime }(e)}{\beta (e)^2}\frac{\alpha sk^{s-1}}{n-1+m} \right] \end{aligned}$$

One may easily see that this expression is equivalent to:

$$\begin{aligned} det A = \beta (e) (n-1+m)n(k+b)(\Phi ^{\prime }(k)-\Omega ^{\prime }(k)) \end{aligned}$$

Recall that $\Phi ^{\prime }(k_1)>\Omega ^{\prime }(k_1)$, while $\Phi ^{\prime }(k_2)<\Omega ^{\prime }(k_2)$. We deduce that $detA>0$ when the matrix $A$ is evaluated at the steady state $\left( k_{1},e_{1}\right)$, while $detA<0$ when the matrix $A$ is evaluated at the steady state $\left( k_{2},e_{2}\right)$. $\square$

Appendix D: Second Order Conditions for the Optimal Solution

The social planner solves $Max_{k,d} H(d, k)$, with:

$$\begin{aligned} H(d, k)\equiv \ln \left[ k^s -nk -g-\beta \left( \frac{\gamma g-\alpha k^s}{n-1+m} \right) \frac{d}{n} \right] + \beta \left( \frac{\gamma g-\alpha k^s}{n-1+m} \right) \ln d \end{aligned}$$

The first order conditions are give by¹⁷:

$$\begin{aligned} H_d&= -\frac{\beta (e)\frac{1}{n}}{k^s-nk-g-\beta (e)\frac{d}{n}}+\beta (e)\frac{1}{d}=0 \end{aligned}$$

(35)

$$\begin{aligned} H_k&= \frac{sk^{s-1}-n+\frac{\alpha sk^{s-1}}{n-1+m}\beta ^{\prime }(e)\frac{d}{n}}{k^s-nk-g-\beta (e)\frac{d}{n}}-\beta ^{\prime }(e)\frac{\alpha sk^{s-1} }{n-1+m} \ln d =0 \end{aligned}$$

(36)

The Hessian matrix is given by:

$$\begin{aligned} \left[ \begin{array}{cc} H_{dd}&H_{dk} \\ H_{kd}&H_{kk} \end{array} \right] \end{aligned}$$

with:

$$\begin{aligned} H_{dd}&= -\beta (e)(1+\beta (e))\frac{1}{d^2}<0 \\ H_{dk}&= \beta (e) \frac{n}{d^2} \left( sk^{s-1}-n+\frac{\alpha sk^{s-1}}{n-1+m}\beta ^{\prime }(e)\frac{d}{n} \right) = H_{kd} \\ H_{kk}&= -\left[ \frac{n}{d}\left( sk^{s-1}-n+\frac{\alpha sk^{s-1}}{n-1+m} \beta ^{\prime }(e)\frac{d}{n} \right) \right] ^2 \\&-\frac{n}{dk} \left[ (1-s)n +\frac{\beta ^{\prime \prime }(e)}{\beta ^{\prime }(e)} \frac{\alpha sk^{s-1}}{n-1+m}(n-sk^{s-1})\right] \end{aligned}$$

We deduce that

$$\begin{aligned} H_{dd}H_{kk}-H_{dk}^2&= \beta (e)(1+\beta (e))\frac{n}{d^3 k}\left[ (1-s)n +\frac{\alpha sk^{s}}{n-1+m}\frac{\beta ^{\prime \prime }(e)}{\beta ^{\prime }(e)}(n-sk^{s-1}) \right] \\&+\beta (e)\frac{n^2}{d^4}\left[ sk^{s-1}-n+\frac{\alpha sk^{s-1}}{n-1+m} +\beta ^{\prime }(e)\frac{d}{n} \right] ^2 \end{aligned}$$

The second order conditions require $H_{dd}<0$, $H_{kk}<0$ and $H_{dd}H_{kk}-H_{dk}^2>0$. They are satisfied for $|\beta ^{\prime \prime }(e)/\beta ^{\prime }(e)|<\overline{\beta }$, with $\overline{\beta }>0$ a constant sufficiently low. $\square$