Abatement Technology and the Environment–Growth Nexus with Education

Abstract

The present paper adopts a Yaari (Rev Econ Stud 32:137–150, 1965) and Blanchard (J Polit Econ 93(2):223–247, 1985) overlapping generations model with Lucas’s (J Monet Econ 22(1):3–42, 1988) human capital accumulation in order to examine how environmental taxation influences growth. It challenges conventional result that tighter environmental tax has either no long term effect at all, or has a positive long term effect. In the case of Cobb–Douglas production functions and logarithmic utilities, this paper demonstrates that the technology used in the abatement sector determines the existence and direction of the growth effect. In the case of output taxation, if the abatement sector is relatively more (respectively less) intensive in human capital than the final output sector, tighter environmental tax increases (resp. reduces) growth in the long term. This result always holds true for finite life expectancy but for infinite life expectancy it only holds true when the labour supply is endogenous. The transitional impact of tighter environmental policy is also investigated. We find that the magnitude of the short term effects of environmental taxation varies according to the technology used in the abatement and final output sectors, and that there is a reverse trade-off between the short and the long term effects.

Appendices

Appendix A

In this appendix, we solve the model.

The program of the households is

$$\begin{aligned} \begin{array}{rl} \displaystyle {\max _{c(s,t),l(s,t),a(s,t),h(s,t),u(s,t)}}&{}\displaystyle \int \limits _s^\infty \left[ \log c(s,t)+\xi _{l}\log l(s,t)- \zeta \log S\right] e^{-(\varrho +\beta )(t-s)}dt\\ s.t.&{} \dot{a}(s,t)=\left[ r_{n}+\beta \right] a(s,t) +u(s,t)h(s,t) w(t)-c(s,t)\\ &{}\dot{h}(s,t)=B \left[ 1-u(s,t)-l(s,t)\right] h(s,t)\\ &{} a(s,s)=0\qquad h(s,s)= H(s)>0 \end{array} \end{aligned}$$

The Hamiltonian of the program may be written as

$$\begin{aligned} \mathcal {H}&= \log c(s,t)+\xi _{l}\log l(s,t)- \zeta \log S+\pi _{1}(t)\\&\times \left[ \left( r_{n}+\beta \right) a(s,t) +u(s,t)h(s,t) w(t)-c(s,t)\right] \\&+\,\,\pi _{2}(t)B \left[ 1-u(s,t)-l(s,t)\right] h(s,t) \end{aligned}$$

where $\pi _{1}(t)$ and $\pi _{2}(t)$ are the co-state variables.

The first-order conditions are

$$\begin{aligned} \frac{\displaystyle \partial \mathcal {H}}{\displaystyle \partial c(s,t)}&= 0 \qquad \Rightarrow \qquad \dfrac{1}{c(s,t)}=\pi _{1}(t)\end{aligned}$$

(30)

$$\begin{aligned} \frac{\displaystyle \partial \mathcal {H}}{\displaystyle \partial u(s,t)}&= 0 \qquad \Rightarrow \qquad \pi _{1}(t)w(t)=\pi _{2}(t)B\end{aligned}$$

(31)

$$\begin{aligned} \frac{\displaystyle \partial \mathcal {H}}{\displaystyle \partial l(s,t)}&= 0 \qquad \Rightarrow \qquad \dfrac{\xi _{l}}{l(s,t)}=\pi _{2}(t)B h(s,t)\end{aligned}$$

(32)

$$\begin{aligned} \frac{\displaystyle \partial \mathcal {H}}{\displaystyle \partial a(s,t)}&= -\dot{\pi }_{1}(t)+(\varrho +\beta )\pi _{1}(t) \qquad \Rightarrow \qquad \pi _{1}(t)(r(t)+\beta )=-\dot{\pi }_{1}(t)+(\varrho +\beta )\pi _{1}(t)\nonumber \\ \end{aligned}$$

(33)

$$\begin{aligned} \frac{\displaystyle \partial \mathcal {H}}{\displaystyle \partial h(s,t)}&= -\dot{\pi }_{2}(t)+(\varrho +\beta )\pi _{2}(t) \nonumber \\&\Rightarrow \pi _{1}(t)w(t)u(s,t)+\pi _{2}(t)B (1\!-\!u(s,t)\!-\!l(s,t))=-\dot{\pi }_{2}(t)+(\varrho +\beta )\pi _{2}(t)\nonumber \\ \end{aligned}$$

(34)

From (30) and (33), we obtain

$$\begin{aligned} \dot{c}(s,t)=(r(t)-\varrho )c(s,t) \end{aligned}$$

(4)

Equations (30), (31) and (32) give

$$\begin{aligned} l(s,t)= \dfrac{\xi _{l}c(s,t)}{w(t)h(s,t)} \end{aligned}$$

(35)

From Eq. (33): $\varrho +\beta =r(t)+\beta +\frac{\dot{\pi }_{1}(t)}{\pi _{1}(t)}$. From Eqs. (31), (33) and (34): $\varrho +\beta =B (1-l(s,t))+\frac{\dot{\pi }_{2}(t)}{\pi _{2}(t)}$. Equalizing both expression we obtain

$$\begin{aligned} \frac{B (1-l(s,t))\pi _{2}(t)+\dot{\pi }_{2}(t)}{\pi _{2}(t)}=\frac{(r(t)+\beta )\pi _{1}(t)+\dot{\pi }_{1}(t)}{\pi _{1}(t)} \end{aligned}$$

(36)

First, this equation means that $l(s,t)$ is independent from $s$: $l(s,t)=l(t)$ defined by Eq. (35). Second, this equation is the “fundamental principle of valuation” (Miller and Modigliani 1961, p. 412) according to which the rate of return on different assets (dividends plus capital gains) must be equal.²⁰ $B (1-l(t))\pi _{2}(t)$ is the dividend on human capital and $(r(t)+\beta )\pi _{1}(t)$ is the dividend on physical capital (or financial assets) valued in terms of utility. The dividend on human capital is explained as follows (see Eq. 34): Investing in one unit of human capital rises the stock of human capital by $B(1-u-l)$ valued in terms of utility at $\pi _{2}$ the shadow price of human capital, and increases earnings from production by $uw$ valued in terms of utility at $\pi _{1}$ the shadow price of physical capital (or financial assets). Therefore the dividend on human capital is $B(1-u-l)\pi _{2}+uw\pi _{1}$. Because time must be equally valuable in its two uses (physical capital accumulation and human capital accumulation), we have $\pi _{1} w=\pi _{2}B$ (see Eq. 31). Therefore, the dividend on human capital is equal to $B(1-l)\pi _{2}$. Differentiating (31) with respect to time, we obtain $\frac{\dot{\pi }_{1}(t)}{\pi _{1}(t)}+\frac{\dot{w}(t)}{w(t)}= \frac{\dot{\pi }_{2}(t)}{\pi _{2}(t)}$. Replacing by the expressions of $\frac{\dot{\pi }_{1}(t)}{\pi _{1}(t)}$ and $\frac{\dot{\pi }_{2}(t)}{\pi _{2}(t)}$, it gives

$$\begin{aligned} \frac{\dot{w}(t)}{w(t)}+B( 1-l(t))=r(t)+\beta \end{aligned}$$

which is an another way to write the fundamental principle of valuation. Because the return to physical capital is independent of $s$, the return to human capital is also independent of $s$ and, therefore, agents allocate the same effort to schooling: $u(s,t)=u(t)$.

Differentiating the expression of aggregate consumption, $C(t)=\int _{-\infty }^{t}c(s,t)\beta e^{-\beta (t-s)}ds$, with respect to time, using the expression of $dK/dt,\,d\Omega /dt$ and Eq. (4) we obtain

$$\begin{aligned} \dot{C}/C=r-\varrho - \beta \left( \varrho +\beta \right) K/C\end{aligned}$$

(37)

The market clearing condition is written as

$$\begin{aligned} \dot{K}(t)=(1-\tau )Y(t)-C(t) \end{aligned}$$

Recalling that $\dot{h}(s,t)=B(1-u(t)-l(t))h(s,t)$ and $h(s,s)= H(s)$, differentiating with respect to time $H(t)=\int _{-\infty }^t h(s,t) \beta e^{-\beta (t-s)} ds$ gives the aggregate accumulation of human capital:

$$\begin{aligned} \dot{H}(t)=B(1-u(t)-l(t)) H(t) \end{aligned}$$

(17)

Finally, from Eq. (6), from the expression of the wage rate (Eq. 13) and the fact that $\int _{-\infty }^t u(s,t) h(s,t) \beta e^{-\beta (t-s)} ds=u(t)H(t)$ because $u(s,t)=u(t)$, we obtain

$$\begin{aligned} \dfrac{\dot{u}(t)}{u(t)}=\dfrac{\dot{K}(t)}{K(t)}-\dfrac{\dot{H}(t)}{H(t)}-\alpha ^{-1}\left[ r(t)+\beta -B (1-l(t))\right] \end{aligned}$$

Defining $x\equiv C/K$ and $z\equiv H/K$, and using previous results, the dynamical system is summarized by

$$\begin{aligned} \dot{x}/x=\left[ \alpha (1-\tau ) -\Phi (\tau )\right] \Psi (\tau )^{\alpha -1}(z\ u)^{1-\alpha }-\varrho -\beta (\beta +\varrho )x^{-1}+x\end{aligned}$$

(19)

$$\begin{aligned} \dot{z}/z=B\left[ 1-u-l\right] -\Phi (\tau )\Psi (\tau )^{\alpha -1}(z\ u)^{1-\alpha }+x\end{aligned}$$

(20)

$$\begin{aligned} \dot{u}/u=\alpha ^{-1}\left[ B(1-l)-\beta -\alpha (1-\tau )\Psi (\tau )^{\alpha -1}(z\ u)^{1-\alpha } \right] -\dot{z}/z\end{aligned}$$

(21)

Appendix B

1.1 B.1 The Case $\beta >0$

From (21), $\dot{u}=0$, we obtain

$$\begin{aligned} (z^{\star }\ u^{\star })^{1-\alpha }=\dfrac{\Psi (\tau )^{1-\alpha }}{\alpha (1-\tau )}\left[ B(1-l^{\star })-\beta \right] \end{aligned}$$

(38)

Because $z^{\star }\ u^{\star }>0$, it implies

$$\begin{aligned} B(1-l^{\star })-\beta >0 \qquad \Rightarrow \qquad B>\beta \end{aligned}$$

(39)

The returns to education in the long term $B(1-l^{\star })$ net of the probability to die $\beta$ must be positive in order to incite agents to invest in human capital accumulation, and therefore to have a positive rate of growth along the BGP.

Furthermore, from (19) and (20), $\dot{x}-\dot{z}=0$ at the steady-state gives (with 38):

$$\begin{aligned} x^{\star }=\dfrac{\beta (\beta +\varrho )}{B u^{\star }- \beta -\varrho } \end{aligned}$$

(40)

Because $x^{\star }>0$, it is required that

$$\begin{aligned} u^{\star }>\dfrac{\beta +\varrho }{B}\qquad \text{ with }~~B>\beta +\varrho \end{aligned}$$

(41)

$B>\beta +\varrho$ means that the productivity in education $B$ must be higher than the subjective discount rate $\varrho +\beta$ (the rate of time preference plus the instantaneous probability of death) in order to incite agents to give up one unit of consumption today and invest it in education. Conditions (39) and (41) are conventional in the Lucas (1988) human capital accumulation model.²¹ Therefore, $x^{\star }\ u^{\star }$ is an increasing function of $u^{\star }$.

Using (23) and (38), we obtain $l^{\star }=\frac{\xi _{l }\alpha }{(1-\alpha )\Psi (\tau )}\times \frac{ x^{\star }\ u^{\star }}{B(1-l^{\star })-\beta }$ which is a quadratic equation of $l^{\star }\in [0,1[$ with only one solution verifying $B(1-u^{\star }-l^{\star })>0$:

$$\begin{aligned} l^{\star }=\dfrac{1}{2}\left( \dfrac{B-\beta -\sqrt{(B-\beta )^{2}-4\frac{\xi _{l }\alpha }{(1-\alpha )\Psi (\tau )}B\ x^{\star }\ u^{\star }}}{B}\right) \end{aligned}$$

(42)

with $\partial {l^{\star }}/\partial {u^{\star }}>0$ (because $x^{\star }u^{\star }$ is increasing in $u^{\star }$ as demonstrated above) and $\partial {l^{\star }}/\partial {\Psi (\tau )}<0$.

$\dot{z}=0$ leads to, using (20), (38) and (22):

$$\begin{aligned} x^{\star }=B u^{\star }- \left( 1+\dfrac{1-\alpha }{\alpha }\Psi (\tau )\right) \beta + \dfrac{1-\alpha }{\alpha }\Psi (\tau )B\left[ 1-l^{\star }\right] \end{aligned}$$

(43)

Equating (40) and (43) enables to express $u^{\star }$ as the solution of

$$\begin{aligned} \Gamma (u,\tau )\equiv B u - \beta + \dfrac{1-\alpha }{\alpha }\Psi (\tau )\left( B\left( 1-l^{\star }\right) -\beta \right) -\dfrac{\beta (\beta +\varrho )}{B u- \beta -\varrho } =0 \end{aligned}$$

with $l^{\star }$ defined by Eq. (42), and $\Gamma _{u}(u,\tau )>0$ and $\Gamma _{\Psi (\tau )}(u,\tau )>0$ because $\Gamma _{u}(u,\tau )=B+\frac{B\beta (\beta +\varrho )}{(Bu-\beta -\varrho )^{2}}+\xi _{l} \frac{\frac{B\beta (\beta +\varrho )^{2}}{(Bu-\beta -\varrho )^{2}}}{\sqrt{(B-\beta )^{2}-\frac{4 \alpha \xi _{l}}{(1-\alpha )\Psi (\tau )} B\ x^{\star }u^{\star }}}>0$. From (41), $u^{\star }\in ](\beta +\varrho )/B,1[$ with $B>\beta +\varrho$. We have $\lim _{u\rightarrow (\beta +\varrho )/B}=-\infty$ and $\lim _{u\rightarrow 1}>0$ because $B-\beta -\varrho >\beta (\beta +\varrho )$ (sufficient condition). Therefore, $u^{\star }\in ](\beta +\varrho )/B,1[$ is unique.

From the theorem of the implicit function, $u^{\star }$ is a decreasing function of $\Psi (\tau )$ and because from (22), we have $\Psi _{\tau }(\tau )\gtreqqless 0$ when $\alpha \gtreqqless \varepsilon$, therefore it comes

$$\begin{aligned} u^{\star }={\mathcal {U}}(\tau )\qquad \text{ with } \qquad \mathcal {U}_{\tau }(\tau )\le 0 ~~ \text{ when }~~ \alpha \ge \varepsilon \end{aligned}$$

From (42), we obtain that

$$\begin{aligned} l^{\star }={\mathcal {L}}(\tau )\qquad \text{ with }~~{\mathcal {L}}_{\tau }(\tau )\le 0 ~~ \text{ when }~~ \alpha \ge \varepsilon \end{aligned}$$

and

$$\begin{aligned} g^{\star }=B(1-\mathcal {U}(\tau )-{\mathcal {L}}(\tau )) \qquad \text{ with }~~g^{\star }_{\tau }\ge 0 ~~ \text{ when }~~ \alpha \ge \varepsilon \end{aligned}$$

1.2 B.2 The Case $\beta =0$

When life expectancy is infinite, $\beta =0$ and the five Eqs. (19–21, 22, 23) becomes:

$$\begin{aligned} \dot{x}/x&= \left[ \alpha (1-\tau ) -\Phi (\tau )\right] \Psi (\tau )^{\alpha -1}(z\ u)^{1-\alpha }-\varrho +x\end{aligned}$$

(44)

$$\begin{aligned} \dot{z}/z&= B(1-u-l) - \Phi (\tau )\Psi (\tau )^{\alpha -1}(z\ u)^{1-\alpha }+x\end{aligned}$$

(45)

$$\begin{aligned} \dot{u}/u&= \alpha ^{-1}\left[ B(1-l) -(1-\tau )\alpha \Psi (\tau )^{\alpha -1}(z\ u)^{1-\alpha } \right] -\dot{z}/z \end{aligned}$$

(46)

To obtain the expression of the BGP rate of growth, just remember that $g^{\star }=r^{\star }-\varrho$ where $r^{\star }$ is the interest rate along the BGP defined as $r^{\star }= (1-\tau )\alpha \Psi (\tau )^{\alpha -1}(z^{\star }\ u^{\star })^{1-\alpha }$. Therefore, along the BGP $\dot{u}=0$, implies that

$$\begin{aligned} r^{\star }=B(1-l^{\star })\qquad \qquad \Rightarrow \qquad \qquad l^{\star }=1-\dfrac{r^{\star }}{B} \end{aligned}$$

Furthermore, from (44),

$$\begin{aligned} x^{\star }=\varrho -\left[ 1-\dfrac{\Phi (\tau )}{\alpha (1-\tau )}\right] \ r^{\star }=\varrho +\Psi (\tau )r^{\star }\end{aligned}$$

and $\dot{x}-\dot{z}=0$ leads to

$$\begin{aligned} u^{\star }=\dfrac{\varrho }{B} \end{aligned}$$

Therefore Eq. (23) gives the implicit expression of $r^{\star }$:

$$\begin{aligned} 1-\dfrac{r^{\star }}{B}=\xi _{l}\dfrac{\left[ \varrho +\Psi (\tau )r^{\star }\right] \ \varrho /B}{\left( \frac{1-\alpha }{\alpha }\right) \Psi (\tau )\ r^{\star }} \end{aligned}$$

that is

$$\begin{aligned} B-r^{\star }=\dfrac{\xi _{l}}{\left( \frac{1-\alpha }{\alpha }\right) \ \Psi (\tau )}\left[ \dfrac{\varrho }{r^{\star }}+\Psi (\tau )\right] \ \varrho \end{aligned}$$

whose the unique solution such that $r^{\star }-\varrho >0$ is:

$$\begin{aligned} r^{\star }=\dfrac{(1-\alpha )B -\xi _{l}\alpha \varrho +\sqrt{((1-\alpha )B-\xi _{l} \alpha \varrho )^{2}-4\xi _{l} \alpha (1-\alpha )\varrho ^{2}/\Psi (\tau )}}{2(1-\alpha )}>0 \end{aligned}$$

It is straightforward that $\partial r^{\star }/\partial \Psi (\tau )>0,\,\forall \xi _{l}> 0$. Because $\partial \Psi (\tau )/\partial \tau \ge 0$ if and only if $\alpha \ge \varepsilon$ then $\forall \xi _{l}>0,\,\partial r^{\star }/\partial \tau \ge 0$ if and only if $\alpha \ge \varepsilon$. When $\xi _{l}=0,\,r^{\star }=B$ independent from $\tau$.

Appendix C: Environmental Tax is Applied to Physical Capital Income

From (28), $\dot{u}=0$, we obtain

$$\begin{aligned} (z^{\star }\ u^{\star })^{1-\alpha }=\dfrac{\Upsilon (\tau _{k})^{1-\alpha }}{\alpha (1-\tau _{k})}\left[ B(1-l^{\star })-\beta \right] \end{aligned}$$

(47)

with $\Upsilon (\tau _{k})\equiv 1+\frac{\alpha -\varepsilon }{1-\alpha }\tau _{k}$. Because $z^{\star }\ u^{\star }>0$, condition (39) $B>\beta$ always holds.

From (26) and (27), $\dot{x}-\dot{z}=0$ at the steady-state gives (with 47):

$$\begin{aligned} x^{\star }=\dfrac{\beta (\beta +\varrho )}{B u^{\star }- \beta -\varrho } \end{aligned}$$

(48)

and because condition (41) always holds, $x^{\star }\ u^{\star }$ is an increasing function of $u^{\star }$

Using (23) and (47), we obtain $l^{\star }=\frac{\xi _{l}\alpha (1-\tau _{k})}{(1-\alpha )\Upsilon (\tau _{k})}\ \times \frac{ x^{\star }\ u^{\star }}{B(1-l^{\star })-\beta }$ which is a quadratic equation of $l^{\star }\in ]0,1[$ with only one solution verifying $B(1-u^{\star }-l^{\star })>0$:²²

$$\begin{aligned} l^{\star }= \dfrac{1}{2}\left( \dfrac{B-\beta -\sqrt{(B-\beta )^{2}-4\frac{ \alpha \xi _{l} (1-\tau _{k})}{(1-\alpha )\Upsilon (\tau _{k})} B\ x^{\star }\ u^{\star }}}{B}\right) \end{aligned}$$

(49)

with $\partial {l^{\star }}/\partial {u^{\star }}>0$ and $\partial {l^{\star }}/\partial {\tau _{k}}\le 0$.

$\dot{z}=0$ leads to, using (27) and (47):

$$\begin{aligned} x^{\star }=Bu^{\star }-\beta + \left( \dfrac{1+(\alpha -\varepsilon )\tau _{k}}{\alpha (1-\tau _{k})}-1\right) \left[ B\left( 1-l^{\star }\right) -\beta \right] \end{aligned}$$

(50)

Equating (48) and (50) enables to express $u^{\star }$ as the solution of

$$\begin{aligned} Bu^{\star }-\beta + \left( \dfrac{1+(\alpha -\varepsilon )\tau _{k}}{\alpha (1-\tau _{k})}-1\right) \left[ B\left( 1-l^{\star }\right) -\beta \right] -\dfrac{\beta (\beta +\varrho )}{B u- \beta -\varrho }=0 \nonumber \end{aligned}$$

with $l^{\star }$ defined by Eq. (49). The LHS is increasing in $u^{\star }$ and $\tau _{k}\,\forall ~(\alpha ,\varepsilon )\in ]0,1[$, because $\partial \left( \frac{1+(\alpha -\varepsilon )\tau _{k}}{\alpha (1-\tau _{k})}-1\right) /\partial \tau _{k}=\frac{1+\alpha -\varepsilon }{\alpha (1-\tau _{k})^{2}}>0$. From (41), $u^{\star }\in ](\beta +\varrho )/B,1[$ with $B>\beta +\varrho$. We have $\lim _{u\rightarrow (\beta +\varrho )/B}=-\infty$ and $\lim _{u\rightarrow 1}>0$ because $B-\beta -\varrho >\beta (\beta +\varrho )$ (sufficient condition). Therefore, $u^{\star }\in ](\beta +\varrho )/B,1[$ is unique.

From the theorem of the implicit function we obtain:

$$\begin{aligned} u^{\star }={\mathcal {U}}^{K}(\tau _{k})\quad \text{ with } \quad {\mathcal {U}}^{K}_{\tau _{k}}(\tau _{k})<0 \end{aligned}$$

From (49), we have

$$\begin{aligned} l^{\star }={\mathcal {L}}^{K}(\tau _{k})\qquad \text{ with }~~{\mathcal {L}}^{K}_{\tau _{k}}(\tau _{k})< 0 \end{aligned}$$

and therefore

$$\begin{aligned} g^{\star }=B(1-\mathcal {U}^{K}(\tau _{k})-{\mathcal {L}}^{K}(\tau _{k})) \qquad \text{ with }~~g^{\star }_{\tau _{k}}> 0,~\forall (\alpha ,\varepsilon )\in ]0,1[ \end{aligned}$$