Pollution standards, technology investment and fines for non-compliance

Abstract

In this paper, we analyze whether it is socially desirable that fines for exceeding pollution standards depend not only on the degree of non-compliance but also on technology investment efforts by the polluting firms. For that purpose, we consider a partial equilibrium framework where a representative firm chooses the investment effort and the pollution level in response to an environmental policy composed of a pollution standard, an inspection probability and a fine for non-compliance. We find that the fine should strictly decrease with the investment effort when (i) there are administrative costs of sanctioning; (ii) the optimal policy induces non-compliance; and (iii) either the fine is sufficiently convex in the degree of non-compliance or the investment effort decreases marginal abatement costs significantly.

Appendix

Proof of Lemma 1

First note that $e<s$ is not an optimal decision, since the firm can save on abatement costs by infinitesimally increasing pollution without being penalized. This reduces the set of possible firm’s pollution choices to $e\ge s$. Then, the problem to be solved is:

$$\begin{aligned} \mathop {{Min}}_{e}&c\left( e,z\right) +z+pf\left( e-s,z\right) , \\ s.t.&e\ge s. \end{aligned}$$

The optimality conditions of this problem are:

$$\begin{aligned}&c_{e}\left( e,z\right) +pf_{e-s}\left( e-s,z\right) -\lambda =0\end{aligned}$$

(9)

$$\begin{aligned}&\lambda \left( e-s\right) =0;\,\lambda \ge 0;\,e\ge s, \end{aligned}$$

(10)

where $\lambda \ge 0$ is the Kuhn–Tucker multiplier of the inequality restriction $e\ge s$. From (9) and (10), $\lambda \ge 0$ implies $e=s$ and $c_{e}\left( s,z\right) +pf_{e-s}\left( 0,z\right) \ge 0$. If conversely, $\lambda =0$, we then have $e>s\$and $c_{e}\left( e,z\right) +pf_{e-s}\left( e-s,z\right) =0$. Combining both possibilities, we obtain the desired result.

Proof of Lemma 2

At the optimum, we have $c_{e}+pf_{e-s}=0$, which can be rewritten as $p=-\frac{c_{e}}{f_{e-s}}$. Thus, the expression $c_{ez}+pf_{e-s,z}$ in the numerator of (4) can be written as $c_{ez}-\frac{c_{e}}{ f_{e-s}}f_{e-s,z}$. Since $c_{e}<0$, we have $c_{ez}-\frac{c_{e}}{f_{e-s}} f_{e-s,z}>0$ if and only if $\frac{c_{ez}}{c_{e}}-\frac{f_{e-s,z}}{f_{e-s}} <0$. Multiplying both sides of this expression by $z$, we have $\frac{c_{ez} }{c_{e}}z-\frac{f_{e-s,z}}{f_{e-s}}z<0$, and applying the definitions $\varepsilon _{c_{e},z}=c_{ez}\frac{z}{c_{e}}$ and $\varepsilon _{f_{e-s},z}=f_{e-s,z}\frac{z}{f_{e-s}}$, we obtain the desired result.

Proof of Lemma 4

Fully differentiating expressions (3) and (6) with respect to $\left( e,z,s,p\right)$, we obtain:

$$\begin{aligned} \left( c_{ee}+pf_{e-s,e-s}\right) de+\left( c_{ez}+pf_{e-s,z}\right) dz+f_{e-s}dp-pf_{e-s,e-s}&= 0; \\ \left( c_{ez}+pf_{e-s,z}\right) de+\left( c_{zz}+pf_{zz}\right) dz+f_{z}dp-pf_{e-s,z}ds&= 0; \end{aligned}$$

or put differently,

$$\begin{aligned} \left( \begin{array}{ll} c_{ee}+pf_{e-s,e-s} &{} c_{ez}+pf_{e-s,z} \\ c_{ez}+pf_{e-s,z} &{} c_{zz}+pf_{zz} \end{array} \right) \left( \begin{array}{l} de \\ dz \end{array} \right) =-\left( \begin{array}{cc} f_{e-s} &{} -pf_{e-s,e-s} \\ f_{z} &{} -pf_{e-s,z} \end{array} \right) \left( \begin{array}{l} dp \\ ds \end{array} \right) . \end{aligned}$$

Let $A=\left( \begin{array}{ll} c_{ee}+pf_{e-s,e-s} &{} c_{ez}+pf_{e-s,z} \\ c_{ez}+pf_{e-s,z} &{} c_{zz}+pf_{zz} \end{array} \right) .$ Note that $\left| A\right| >0$ (this ensures that sufficient conditions for the firm’s optimal choices hold). Applying Cramer’s rule, we then have:

$$\begin{aligned} \frac{\partial e}{\partial p}&= -\frac{\left| \begin{array}{ll} f_{e-s} &{} c_{ez}+pf_{e-s,z} \\ f_{z} &{} c_{zz}+pf_{zz} \end{array} \right| }{\left| A\right| }=-\frac{1}{\left| A\right| } \left\{ f_{e-s}\left( c_{zz}+pf_{zz}\right) -f_{z}\left( c_{ez}+pf_{e-s,z}\right) \right\} ; \\ \frac{\partial z}{\partial p}&= -\frac{\left| \begin{array}{ll} c_{ee}+pf_{e-s,e-s} &{} f_{e-s} \\ c_{ez}+pf_{e-s,z} &{} f_{z} \end{array} \right| }{\left| A\right| }\\&= -\frac{1}{\left| A\right| } \left\{ f_{z}\left( c_{ee}+pf_{e-s,e-s}\right) -f_{e-s}\left( c_{ez}+pf_{e-s,z}\right) \right\} ; \\ \frac{\partial e}{\partial s}&= -\frac{\left| \begin{array}{ll} -pf_{e-s,e-s} &{} c_{ez}+pf_{e-s,z} \\ -pf_{e-s,z} &{} c_{zz}+pf_{zz} \end{array} \right| }{\left| A\right| }\\&= -\frac{p}{\left| A\right| } \left\{ f_{e-s,z}\left( c_{ez}+pf_{e-s,z}\right) -f_{e-s,e-s}\left( c_{zz}+pf_{zz}\right) \right\} ; \\ \frac{\partial z}{\partial s}&= -\frac{\left| \begin{array}{ll} c_{ee}+pf_{e-s,e-s} &{} -pf_{e-s,e-s} \\ c_{ez}+pf_{e-s,z} &{} -pf_{e-s,z} \end{array} \right| }{\left| A\right| }\\&= -\frac{p}{\left| A\right| } \left\{ f_{e-s,e-s}\left( c_{ez}+pf_{e-s,z}\right) -f_{e-s,z}\left( c_{ee}+pf_{e-s,e-s}\right) \right\} ; \end{aligned}$$

as desired.

Proof of Lemma 5

We consider expressions $\frac{ \partial e}{\partial p}$ and $\frac{\partial e}{\partial s}$ from Lemma 4 above to obtain:

$$\begin{aligned} \left. \frac{dp}{ds}\right| _{e}=-\frac{\frac{\partial e}{\partial s}}{ \frac{\partial e}{\partial p}}=-\frac{p\left\{ f_{e-s,z}\left( c_{ez}+pf_{e-s,z}\right) -f_{e-s,e-s}\left( c_{zz}+pf_{zz}\right) \right\} }{ f_{e-s}\left( c_{zz}+pf_{zz}\right) -f_{z}\left( c_{ez}+pf_{e-s,z}\right) }. \end{aligned}$$

(11)

Evaluating (11) at $e\!=\!s$, we have $\left. \frac{ dp}{ds}\right| _{e=s}\!=\!-\frac{p\left\{ f_{e-s,z}\left( c_{ez}+pf_{e-s,z}\right) -f_{e-s,e-s}\left( c_{zz}+pf_{zz}\right) \right\} }{ f_{e-s}\left( c_{zz}+pf_{zz}\right) }$, since $f_{z}=0$ when $e=s$. By Lemma 2, $c_{ez}+pf_{e-s,z}\ge 0$ if and only if $\varepsilon _{c_{e},z}\le \varepsilon _{f_{e-s},z}$. Therefore, $\left. \frac{dp}{ds} \right| _{e=s}\ge 0$ if $\varepsilon _{c_{e},z}\le \varepsilon _{f_{e-s},z}$, as desired.

Proof of Proposition 1

The first order conditions of problem (7) are the following:

$$\begin{aligned} \left( w.r.e\right)&c_{e}+d^{\prime }\!+\!ptf_{e-s}+\lambda \left( c_{ee}+pf_{e-s,e-s}\right) \!+\!\gamma \left( c_{ez}+pf_{e-s,z}\right) +\mu \!=\!0;\qquad \end{aligned}$$

(12)

$$\begin{aligned} \left( w.r.z\right)&c_{z}+1+ptf_{z}+\lambda \left( c_{ez}+pf_{e-s,z}\right) +\gamma \left( c_{zz}+pf_{zz}\right) =0;\end{aligned}$$

(13)

$$\begin{aligned} \left( w.r.p\right)&m+tf+\lambda f_{e-s}+\gamma f_{z}=0;\end{aligned}$$

(14)

$$\begin{aligned} \left( w.r.s\right)&-ptf_{e-s}-\lambda pf_{e-s,e-s}-\gamma pf_{e-s,z}-\mu =0; \end{aligned}$$

(15)

where $\left( \lambda , \gamma , \mu \right)$ are the respective Kuhn–Tucker multipliers associated with the constraints in problem (7).

Under compliance, we have $e=s, \,\mu \le 0$ and $c_{z}+1=0$, by (6). The optimality conditions reduce to:

$$\begin{aligned} \left( w.r.e\right) \,&c_{e}+d^{\prime }+ptf_{e-s}+\lambda \left( c_{ee}+pf_{e-s,e-s}\right) +\gamma \left( c_{ez}+pf_{e-s,z}\right) +\mu =0;\\ \left( w.r.z\right) \,&\lambda \left( c_{ez}+pf_{e-s,z}\right) +\gamma \left( c_{zz}+pf_{zz}\right) =0; \\ \left( w.r.p\right) \,&m+\lambda f_{e-s}=0; \\ \left( w.r.s\right) \,&\mu =-ptf_{e-s}-\lambda pf_{e-s,e-s}-\gamma pf_{e-s,z}; \end{aligned}$$

from which we obtain $\lambda =-\frac{m}{f_{e-s}}$ and $\gamma =\frac{ m\left( c_{ez}+pf_{e-s,z}\right) }{f_{e-s}\left( c_{zz}+pf_{zz}\right) }$. Thus, $\mu \le 0$ (or equivalently, $e=s$) if and only if

$$\begin{aligned} tf_{e-s}+\frac{m}{f_{e-s}}f_{e-s,e-s}-\frac{m\left( c_{ez}+pf_{e-s,z}\right) }{f_{e-s}\left( c_{zz}+pf_{zz}\right) }f_{e-s,z}\ge 0, \end{aligned}$$

(16)

evaluated at $e=s$.

Conversely, $e>s$ if and only if:

$$\begin{aligned} tf_{e-s}+\frac{m}{f_{e-s}}f_{e-s,e-s}-\frac{m\left( c_{ez}+pf_{e-s,z}\right) }{f_{e-s}\left( c_{zz}+pf_{zz}\right) }f_{e-s,z}<0, \end{aligned}$$

(17)

evaluated at $e=s$.

From expression (11), we have $\left. \frac{dp}{ ds}\right| _{e}=-\frac{p\left\{ f_{e-s,z}\left( c_{ez}+pf_{e-s,z}\right) -f_{e-s,e-s}\left( c_{zz}+pf_{zz}\right) \right\} }{f_{e-s}\left( c_{zz}+pf_{zz}\right) }$ when $e=s$. Therefore, (17) can be rewritten as:

$$\begin{aligned} ptf_{e-s}<m\left. \frac{dp}{ds}\right| _{e=s}, \end{aligned}$$

as desired.

Proof of Proposition 3

Consider the solution of problem (7), $\left( s^{*},p^{*}\right)$. Substituting this policy in the Lagrangian of problem (7), we have:

$$\begin{aligned} L\left( s^{*},p^{*}\right)&= c\left( e^{*},z^{*}\right) +z^{*}+d\left( e^{*}\right) +p^{*}\left[ m+tf\left( e^{*}-s^{*},z^{*}\right) \right] \nonumber \\&+\ \lambda ^{*}\left[ c_{e}\left( e^{*},z^{*}\right) +p^{*}f_{e-s}\left( e^{*}-s^{*},z^{*}\right) \right] \nonumber \\&+\ \gamma ^{*}\left[ c_{z}\left( e^{*},z^{*}\right) +1+p^{*}f_{z}\left( e^{*}-s^{*},z^{*}\right) \right] +\mu ^{*}\left( e^{*}-s^{*}\right) ,\qquad \end{aligned}$$

(18)

where $\left( \lambda ^{*},\gamma ^{*},\mu ^{*}\right)$ are the respective Kuhn–Tucker multipliers associated with the constraints in problem (7). Since $e^{*}\ge s^{*}$, we then have $\mu ^{*}=0$.

From (13) and (15), the corresponding expressions for the Kuhn–Tucker multipliers $\left( \lambda ^{*},\gamma ^{*}\right)$ evaluated at $f_{z}=f_{e-s,z}=0$ reduce to:

$$\begin{aligned} \lambda ^{*}&= -t\frac{f_{e-s}}{f_{e-s,e-s}}<0; \end{aligned}$$

(19)

$$\begin{aligned} \gamma ^{*}&= -\lambda ^{*}\frac{c_{ez}}{c_{zz}}=t\frac{f_{e-s}}{ f_{e-s,e-s}}\frac{c_{ez}}{c_{zz}}>0, \end{aligned}$$

(20)

since $c_{z}+1=0$ under $f_{z}=0$, see (6).

Now, consider the following polynomial approximation of the fine:

$$\begin{aligned} f\left( e-s,z\right)&\approx f\left( e-s,0\right) +f_{z}\left( e-s,0\right) z \nonumber \\&\approx \left[ f\left( 0,0\right) +f_{e-s}\left( 0,0\right) \left( e-s\right) +f_{e-s,e-s}\left( 0,0\right) \frac{\left( e-s\right) ^{2}}{2} \right] \nonumber \\&\quad +\left[ f_{z}\left( 0,0\right) +f_{e-s,z}\left( 0,0\right) \left( e-s\right) +f_{e-s,e-s,z}\left( 0,0\right) \frac{\left( e-s\right) ^{2}}{2} \right] z \nonumber \\&=\left[ f_{e-s}\left( 0,0\right) \left( e-s\right) +f_{e-s,e-s}\left( 0,0\right) \frac{\left( e-s\right) ^{2}}{2}\right] \nonumber \\&\quad +\left[ f_{e-s,z}\left( 0,0\right) \left( e-s\right) +f_{e-s,e-s,z}\left( 0,0\right) \frac{\left( e-s\right) ^{2}}{2}\right] z, \end{aligned}$$

(21)

since we have assumed $f\left( 0,0\right) =f_{z}\left( 0,0\right) =0$.

From (21), the corresponding expressions for $f_{e-s}\left( e-s,z\right)$ and $f_{z}\left( e-s,z\right)$ are the following:

$$\begin{aligned} f_{e-s}\left( e-s,z\right)&\approx f_{e-s}\left( 0,0\right) +f_{e-s,e-s}\left( 0,0\right) \left( e-s\right) \nonumber \\&+\left[ f_{e-s,z}\left( 0,0\right) +f_{e-s,e-s,z}\left( 0,0\right) \left( e-s\right) \right] z; \end{aligned}$$

(22)

$$\begin{aligned} f_{z}\left( e-s,z\right)&\approx f_{e-s,z}\left( 0,0\right) \left( e-s\right) +f_{e-s,e-s,z}\left( 0,0\right) \frac{\left( e-s\right) ^{2}}{2}. \end{aligned}$$

(23)

We substitute (21), (22) and (23) in (18), and we now consider the effect on social costs of an infinitesimal change of the fine associated with the change in investment effort $z$. Through (21), (22) and (23), we note that the fine has two components that measure the effect of $z, \,f_{e-s,z}\left( 0,0\right)$ and $f_{e-s,e-s,z}\left( 0,0\right)$. We then compute the partial derivatives of the Lagrangian expression (18) with respect to these two terms evaluated at $f_{z}=f_{e-s,z}=0$:

$$\begin{aligned} \frac{\partial L\left( s^{*},p^{*}\right) }{\partial f_{e-s,z}\left( 0,0\right) }&= p^{*}\left\{ t\left( e^{*}-s^{*}\right) z^{*}+\lambda ^{*}z^{*}+\gamma ^{*}\left( e^{*}-s^{*}\right) \right\} ; \end{aligned}$$

(24)

$$\begin{aligned} \frac{\partial L\left( s^{*},p^{*}\right) }{\partial f_{e-s,e-s,z}\left( 0,0\right) }&= p^{*}\left( e^{*}-s^{*}\right) \left\{ t\frac{\left( e^{*}-s^{*}\right) }{2}z^{*}+\lambda ^{*}z^{*}+\gamma ^{*}p^{*}\frac{\left( e^{*}-s^{*}\right) }{2}\right\} \!;\qquad \quad \end{aligned}$$

(25)

where $\left( \lambda ^{*},\gamma ^{*}\right)$ are given by (19) and (20), respectively.

Since $f_{e-s,z}\left( 0,0\right) \le 0$ and $f_{e-s,e-s,z}\left( 0,0\right) \le 0$, social costs decrease when the fine depends on the investment effort as long as (24) or (25) are strictly positive. Since $tz^{*}+\gamma ^{*}>0$ (see 20), the requirement for (24) to be positive is weaker than that for (25), since $tz^{*}+\gamma ^{*}>\frac{tz^{*}+\gamma ^{*}}{2}$. Therefore, substituting (19) and (20) in (24), we then have:

$$\begin{aligned} \frac{\partial L\left( s^{*},p^{*}\right) }{\partial f_{e-s,z}\left( 0,0\right) }&= p^{*}t\left\{ \left( e^{*}-s^{*}\right) z^{*}-\frac{f_{e-s}}{f_{e-s,e-s}}z^{*}+\frac{f_{e-s}}{f_{e-s,e-s}}\frac{ c_{ez}}{c_{zz}}\left( e^{*}-s^{*}\right) \right\} \\&= p^{*}t\left( e^{*}-s^{*}\right) \left\{ \left[ 1-\frac{f_{e-s} }{\left( e^{*}-s^{*}\right) f_{e-s,e-s}}\right] z^{*}+\frac{ f_{e-s}}{f_{e-s,e-s}}\frac{c_{ez}}{c_{zz}}\right\} >0 \end{aligned}$$

if and only if $t>0, \,e^{*}>s^{*}$and $\left[ 1-\frac{f_{e-s}}{ \left( e^{*}-s^{*}\right) f_{e-s,e-s}}\right] z^{*}+\frac{f_{e-s} }{f_{e-s,e-s}}\frac{c_{ez}}{c_{zz}}>0$, as desired.