Pollution Control Under Uncertainty and Sustainability Concern

Abstract

We analyze the implications of environmental policy on pollution in a stochastic framework with finite horizon and sustainability concern. The social planner seeks to minimize the social (environmental and economic) costs associated with pollution. We allow for the planner to attach different relative weights to the discounted and end-of-planning-horizon costs in order to assess how sustainability concern might affect the optimal level of policy intervention. We show that the optimal environmental policy increases with the degree of sustainability concern, reducing thus the amount of pollution the society is forced to bear. A calibration based on world $\textit{CO}_2$ data supports our conclusions, further highlighting the importance of higher degrees of sustainability concern to achieve greener long run outcomes. It also allows us to show that under a realistic model’s parametrization the optimal environmental policy tends to rise with higher degrees of uncertainty in a precautionary manner.

Appendices

Appendix 1: Optimal Solution and Sufficiency

By denoting with $\mathcal {J}(t,p_t)$ the value function associated to our stochastic problem (4), (5) and (6) and by omitting the time subscript for sake of clarity, the HJB equation reads as:

$$\begin{aligned} -\frac{\partial \mathcal {J}}{\partial t}= & {} \min _{\tau } \left\{ \frac{1}{2}p^2(1+\tau ^2)e^{-\rho t}+[(1-\tau )\alpha -\delta ]p\frac{\partial \mathcal {J}}{\partial p}+\frac{1}{2}\sigma ^2p^2\frac{\partial ^2 \mathcal {J}}{\partial p^2} \right\} \end{aligned}$$

(20)

while the corresponding terminal condition as:

$$\begin{aligned} \mathcal {J}(T,p_T)= & {} \left( \frac{1-\theta }{\theta }\right) \frac{1}{2}p_T^2e^{-\rho T} \end{aligned}$$

(21)

The first order necessary (and sufficient; see below) condition for $\tau$ yields:

$$\begin{aligned} \tau= & {} \frac{\eta e^{-\rho t} }{p}\frac{\partial \mathcal {J}}{\partial p} \end{aligned}$$

(22)

We proceed by guessing the form of the value function and verifying that our guess is correct. Our sophisticated guess is:

$$\begin{aligned} \mathcal {J}(t,p)= & {} \frac{1}{2}p^2V e^{-\rho t} \end{aligned}$$

(23)

where V is a variable to be determined. By computing its derivatives:

$$\begin{aligned} \frac{\partial \mathcal {J}}{\partial t}= & {} \frac{1}{2}p^2 \bigg [\frac{\partial V}{\partial t}-\rho V \bigg ]e^{-\rho t} , \end{aligned}$$

(24)

$$\begin{aligned} \frac{\partial \mathcal {J}}{\partial p}= & {} pV e^{-\rho t} \end{aligned}$$

(25)

$$\begin{aligned} \frac{\partial ^2 \mathcal {J}}{\partial p^2}= & {} V e^{-\rho t}, \end{aligned}$$

(26)

and substituting (25) into (22), we obtain:

$$\begin{aligned} \tau= & {} \eta V \end{aligned}$$

(27)

By plugging (24), (25) and (26) into (20) and simplifying the expression, we obtain the following ordinary differential equation in V:

$$\begin{aligned} \frac{\partial V}{\partial t}= & {} V^2 \eta ^2 + V[\rho - 2(\eta -\delta )-\sigma ^2] -1, \end{aligned}$$

(28)

with the boundary condition $V_T=\frac{1-\theta }{\theta } \ge 0$, from evaluating (21) and (23) at T. The solution of the above differential equation can be used to derive the path of the optimal tax rate [from (27)] and finally the expected path of pollution [from (5)]. Indeed, by solving (28) along with its boundary condition for $V_t$ and substituting into (27) we get the optimal dynamics of the tax rate:

$$\begin{aligned} \tau ^*= & {} \frac{1}{2\eta } \Bigg \{ 2(\eta -\delta ) -\rho +\sigma ^2 + \tanh \Bigg [ \frac{\sqrt{M}(T-t)}{2}\nonumber \\&+ {{\mathrm{arctanh}}}\Bigg ( \frac{2(1-\theta )\eta ^2-2(\eta -\delta )\theta +\rho \theta -\sigma ^2 \theta }{\theta \sqrt{M}} \Bigg ) \Bigg ] \sqrt{M} \Bigg \}, \end{aligned}$$

(29)

where $M=[2(\eta -\delta )+ \sigma ^2-\rho ]^2+4\eta ^2$, $\tanh (z)=(e^z-e^{-z})/(e^z+e^{-z})$ is the hyperbolic tangent function and ${{\mathrm{arctanh}}}(z)=\frac{1}{2}[\log (1+z)-log(1-z)]$, with $-1<z<1$, its inverse. By plugging the above expression in (5), which describes a geometric Brownian motion with time-dependent coefficients, it is possible to determine the time evolution of pollution, whose closed form expression is given in Eq. (11).

In order to verify the correctness of our guess, we use the stochastic maximum principle proposed by Framstad et al. (2004) to show that the policy rule identified in (29) is optimal. By defining $m_t \equiv \partial \mathcal {J}/ \partial p$ and $n_t \equiv \partial ^2 \mathcal {J} / \partial p^2$, it is possible to rewrite (20) as:

$$\begin{aligned} -\frac{\partial \mathcal {J}}{\partial t}= & {} \min _{\tau }\left\{ \frac{1}{2}p^2(1+\tau ^2)e^{-\rho t }+ [(1-\tau )\eta -\delta ]pm+ \frac{1}{2}\sigma ^2p^2n \right\} = \min _{\tau } \ \mathcal {H}, \end{aligned}$$

where:

$$\begin{aligned} \mathcal {H}= & {} \frac{1}{2}p^2(1+\tau ^2)e^{-\rho t }+ [(1-\tau )\eta -\delta ]pm+ \frac{1}{2}\sigma ^2p^2n \end{aligned}$$

(30)

where denotes the the stochastic Hamiltonian. Theorem 2.1 of Framstad et al. (2004) states that, for an admissible set of state and controls, if the minimized Hamiltonian $\hat{\mathcal {H}}$ (that is the Hamiltonian $\mathcal {H}$ evaluated at the value of the optimal control $\tau ^*$) is convex in p for all t in [0, t], then the pair $(\tau ^*, p)$ represents an optimal pair for the problem. Note that $\mathcal {H}$ is strictly convex in $\tau$ since $\partial ^2 \mathcal {H}/ \partial \tau ^2 =p^2e^{-\rho t } > 0$. The control which minimizes $\mathcal {H}$ is given by Eq. (22) and so the minimized Hamiltonian is:

$$\begin{aligned} \hat{\mathcal {H}}= & {} \frac{1}{2}p^2 \left( 1+ \frac{m^2\eta ^2}{p^2(e^{- \rho t})^2} \right) e^{- \rho t}+pm \left[ \left( 1-\frac{m\eta }{pe^{- \rho t}} \right) \eta -\delta \right] +\frac{1}{2}\sigma ^2p^2n \end{aligned}$$

which is strictly convex in p, since $\partial ^2 \hat{\mathcal {H}}/ \partial p^2= e^{-\rho t}+ n\sigma ^2 >0$.

Appendix 2: Proof of Proposition 2

The derivative of the optimal policy $\tau ^*$ with respect to $\theta$ reads as:

$$\begin{aligned} \frac{\partial \tau _t^*}{\partial \theta }= & {} \frac{\sqrt{M}}{2}\left\{ 1- \tanh \left[ \frac{\sqrt{M}}{2}(T-t)\right. \right. \nonumber \\&\left. \left. +{{\mathrm{arctanh}}}\left( \frac{B}{\sqrt{M}}\right) \right] ^{2}\right\} \left\{ \left( \frac{-2\eta ^2-\sigma ^2+2\delta -2\eta +\rho }{\theta \sqrt{M}}-\frac{B}{\theta ^2 \sqrt{M}}\right) \right. \nonumber \\&\left. \times \left[ \left( 1-\frac{B}{\theta ^2 (\sqrt{M})^2}\right) \eta \right] ^{-1}\right\} \end{aligned}$$

(31)

where $M=[2(\eta -\delta )+ \sigma ^2-\rho ]^2+4\eta ^2$ and $B=2(1-\theta )\eta ^{2} -2(\eta -\delta )\theta +\rho \theta -\sigma ^2\theta$. The sign of the above derivative is determined by the product of three terms, $\frac{\sqrt{M}}{2}$ and the two terms in the curly brackets. Provided that the hyperbolic tangent function is well defined, as per condition (9), the first term, $\frac{\sqrt{M}}{2}$, is clearly non-negative. The second term, namely $\left\{ 1- \tanh \left[ \frac{\sqrt{M}}{2}(T-t) +{{\mathrm{arctanh}}}\left( \frac{B}{\sqrt{M}}\right) \right] ^{2}\right\}$, is non-negative too since the hyperbolic tangent takes values in $[-1,1]$. After some algebra, the third term, $\left\{ \left( \frac{-2\eta ^2-\sigma ^2+2\delta -2\eta +\rho }{\theta \sqrt{M}}-\frac{B}{\theta ^2 \sqrt{M}}\right) \left[ (1-\frac{B}{\theta ^2 (\sqrt{M})^2})\eta \right] ^{-1}\right\}$, can be rearranged to obtain:

$$\begin{aligned} -\frac{\sqrt{\sigma ^4-4\delta \sigma ^2+4\eta \sigma ^2- 2\rho \sigma ^2+4\delta ^2-8\delta \eta +4\delta \rho +8\eta ^2-4\eta \rho +\rho ^2}}{2[\eta \theta ^2 (-\eta ^2-\sigma ^2+2\delta -2\eta +\rho +1)+\eta \theta (2\eta ^2+\sigma ^2-2\delta + 2\eta -\rho )-\eta ^3]} \end{aligned}$$

(32)

Since the numerator in (32) is clearly non-negative, the sign of its denominator determines the sign of (31): if this is positive then the whole derivative will be negative, while it will be positive otherwise. It turns out that the conditions for the hyperbolic tangent function to be well defined, as in Eq. (9), ensure that the denominator of the above expression is positive, such that the sign of (31) is overall negative.

Appendix 3: Proof of Proposition 3

From Eq. (10) it is clear that what complicates the determination of the sign of $\frac{\partial \tau ^*}{\partial \sigma ^2}$ is the argument of the inverse hyperbolic tangent. Indeed, apart from this term whose derivative has an uncertain sign, all other terms suggest the existence of a monotonically increasing relationship between the optimal taxation and the degree of uncertainty. Thus, we can undoubtedly assess the sign of $\frac{\partial \tau ^*}{\partial \sigma ^2}$ only whenever also the argument of the inverse hyperbolic tangent rises with $\sigma ^2$. In the following we denote the argument of the inverse hyperbolic tangent with $\Omega$, which from Eq. (10) reads as:

$$\begin{aligned} \Omega = \frac{2(1-\theta )\eta ^2-2(\eta -\delta )\theta +\rho \theta -\sigma ^2 \theta }{\theta \sqrt{[2(\eta -\delta )+ \sigma ^2-\rho ]^2+4\eta ^2}} \end{aligned}$$

After some algebra the derivative of the above term with respect to $\sigma ^2$ yields:

$$\begin{aligned} \frac{\partial \Omega }{\partial \sigma ^2}=\frac{-2\eta ^2 \left( 2\delta \theta -2\eta \theta +\rho \theta -\theta \sigma ^2-2\delta +2\eta -\rho +2\theta +\sigma ^2 \right) }{\theta \left( 4\delta ^2-8\delta \eta +4\delta \rho -4\delta \sigma ^2+8\eta ^2-4\eta \rho +4\eta \sigma ^2+ \rho ^2-2\rho \sigma ^2+\sigma ^2\right) ^{\frac{3}{2}}} \end{aligned}$$

Since the denominator in above expression is clearly non-negative, the sign of its numerator determines the sign of $\frac{\partial \Omega }{\partial \sigma ^2}$: whenever this is negative then the whole derivative will be positive. This happens whenever the condition stated in Proposition 3, $\sigma ^2 \le \rho -2(\eta -\delta )-\frac{2\theta }{1-\theta }$, holds.

Appendix 4: Proof of Proposition 4

Let us first notice that the following equation:

$$\begin{aligned} \dot{A_t} = A_t^{2}\eta ^{2}\gamma _1^2 + A_t [2(\eta \gamma _2-\delta )+\sigma ^2-\rho ] -1 \end{aligned}$$

(33)

admits a closed-form solution given by:

$$\begin{aligned} A_t= & {} \frac{1}{2(\eta \gamma _1)^2} \left\{ 2(\eta \gamma _2-\delta ) -\rho +\sigma ^2\right. \\&\left. +\, \sqrt{[2(\eta \gamma _2-\delta )+ \sigma ^2-\rho ]^2+4(\eta \gamma _1)^2} \tanh \left[ \frac{\sqrt{[2(\eta \gamma _2-\delta )+ \sigma ^2-\rho ]^2+4(\eta \gamma _1)^2}(T-t)}{2} \right. \right. \nonumber \\&\left. \left. +\, {{\mathrm{arctanh}}}\left( \frac{2(1-\theta )(\eta \gamma _1)^2-2(\eta \gamma _2-\delta )\theta +\rho \theta -\sigma ^2 \theta }{\theta \sqrt{[2(\eta \gamma _2-\delta )+ \sigma ^2-\rho ]^2+4(\eta \gamma _1)^2}} \right) \right] \right\} \end{aligned}$$

By repeating the same calculations as in Sect. 3, one can reduce the problem of solving the HJB equation to the following ordinary differential equation:

$$\begin{aligned} \dot{V}_t = V_t^2 (1+\gamma _t)^2 \eta ^2 + V_t[\rho - 2(\eta (1+\gamma _t) -\delta )-\sigma ^2] -1, \end{aligned}$$

where $\gamma _t$ measures the excess of capital growth. Some algebra yields to the following inequalities:

$$\begin{aligned} \dot{V}_t= & {} V_t^2 (1+\gamma _t)^2 \eta ^2 + V_t[\rho - 2(\eta (1+\gamma _t) -\delta )-\sigma ^2] -1 \\\le & {} V_t^2 \overline{\gamma }^2 \eta ^2 + V_t [\rho - 2(\eta \underline{\gamma } -\delta )-\sigma ^2] -1 \end{aligned}$$

and, in an analogous way:

$$\begin{aligned} \dot{V}_t \ge V_t^2 \underline{\gamma }^2 \eta ^2 + V_t [\rho - 2(\eta \overline{\gamma } -\delta )-\sigma ^2] -1 \end{aligned}$$

By using the above Eq. (33) and the classical comparison theorem for differential equations, we obtain the lower and upper bounds as in (18) and (19).