A Model of Benchmarking Regulation: Revisiting the Efficiency of Environmental Standards

Abstract

The conventional economic argument favors the use of market-based instruments over ‘command-and-control’ regulation. This viewpoint, however, is often limited in the description and characteristics of the latter; namely, environmental standards are often portrayed as lacking structured abatement incentives. Yet contemporary forms of command-and-control regulation, such as standards stipulated via benchmarking, have the potential to be efficient. We provide a first formal analysis of environmental standards based on performance benchmarks. We show that under specific conditions, standards can provide efficient incentives to improve environmental performance.

Appendices

Mathematical Appendix

1.1 Proof of Proposition 1

Proof

To solve (13) we derive the following set of first-order conditions:

$$\begin{aligned} \frac{\partial \pi _i}{\partial c_i}= \frac{\mu P \left( \frac{3\left( \frac{1}{\mu I_i + \gamma _i}\right) ^{\frac{3}{2}}}{\mu I_{-i}+\gamma _{-i}} + \frac{\sqrt{\frac{1}{\mu I_{-i}+\gamma _{-i}}}}{\left( \mu I_i+\gamma _i\right) ^2}\right) }{4 \left( \sqrt{\frac{1}{\mu I_i+\gamma _i}}+\sqrt{\frac{1}{\mu I_{-i}+\gamma _{-i}}}\right) ^3} - 1 =0 \quad \text {for} \quad i,-i\in \Phi . \end{aligned}$$

(34)

Substitution of the first FOC into the second yields after some simplification

$$\begin{aligned} \mu I_X + \gamma _X = \mu I_Y+\gamma _Y. \end{aligned}$$

As $k_i = \frac{1}{\gamma _i + \mu I_i}$, it follows

$$\begin{aligned} k_X^*=k_Y^*. \end{aligned}$$

(35)

Substituting (35) into (34) yields the optimal level of investment:

$$\begin{aligned} I_i^*=\frac{\mu P-8 \gamma _i}{8 \mu }. \end{aligned}$$

(36)

As $\frac{\partial ^2 \pi _i}{\partial I_i^2}<0$, (36) denotes indeed a maximum.

With $k_i = \frac{1}{\gamma _i + \mu I_i}$, (36) ultimately defines the cost parameter $k_i^* = \frac{8}{\mu P}$. Finally, substituting (15) into (13) yields the corresponding payoff:

$$\begin{aligned} \pi _i^* = \frac{P}{4}+\frac{\gamma _i}{\mu }. \end{aligned}$$

$\square$

1.2 Proof of Proposition 2

Proof

The proof requires solving the game represented by Eq. (17) via backward induction. The solution to the second stage is defined by the following first-order condition:

$$\begin{aligned} \frac{\partial \pi _i}{\partial c_i} = \frac{P q c_X^{\frac{q}{2}} c_Y^{\frac{q}{2}-1} k_X^{\frac{q}{2}} k_Y^{\frac{q}{2}}}{2 \left( c_Y^{\frac{q}{2}} k_X^{\frac{q}{2}}+c_X^{\frac{q}{2}} k_Y^{\frac{q}{2}}\right) ^2} - 1 = 0. \end{aligned}$$

Equating both first-order conditions, we find $c_i = c_{-i}$. Solving the game is then straightforward and yields:

$$\begin{aligned} \tilde{a}_{i,q}&= \sqrt{\frac{P q k_i^{\frac{q}{2}-1} k_{-i}^{\frac{q}{2}}}{2 \left( k_i^{\frac{q}{2}}+k_{-i}^{\frac{q}{2}}\right) ^2}}, \end{aligned}$$

(37)

$$\begin{aligned} \tilde{A}_q&= \frac{\sqrt{P q k_i^{\frac{q}{2}} k_{-i}^{\frac{q}{2}}} \left( \sqrt{k_i}+\sqrt{k_{-i}}\right) }{\sqrt{2 k_i k_{-i}} \left( k_i^{\frac{q}{2}}+k_{-i}^{\frac{q}{2}}\right) }, \end{aligned}$$

(38)

$$\begin{aligned} \tilde{c}_{i,q}&= \frac{P q k_i^{\frac{q}{2}} k_{-i}^{\frac{q}{2}}}{2 \left( k_{-i}^{\frac{q}{2}}+k_i^{\frac{q}{2}}\right) ^2}, \end{aligned}$$

(39)

$$\begin{aligned} \tilde{C}_q&= \frac{P q k_i^{\frac{q}{2}} k_{-i}^{\frac{q}{2}}}{\left( k_{-i}^{\frac{q}{2}}+k_i^{\frac{q}{2}}\right) ^2}, \end{aligned}$$

(40)

where $k_i = k_i(I_i) = \frac{1}{\gamma _i + \mu I_i}$.

With the solution to the second stage defined by (37)–(40), the objective function reduces to (18). Equating the first-order conditions for $X$ and $Y$ yields

$$\begin{aligned} \left( \frac{1}{\mu I_Y+\gamma _Y}\right) ^{\frac{q}{2}} \left( \frac{q+2}{\mu I_X+\gamma _X} \!+\! \frac{q-2}{\mu I_Y+\gamma _Y}\right)&= \left( \frac{1}{\mu I_X + \gamma _X}\right) ^{\frac{q}{2}} \left( \frac{q-2}{\mu I_X+\gamma _X}\!+\!\frac{q+2}{\mu I_Y+\gamma _Y}\right) , \nonumber \\ \text{ which }&\text{ implies } \nonumber \\ \frac{1}{\gamma _X + \mu I_X}&= \frac{1}{\gamma _Y + \mu I_Y}. \end{aligned}$$

(41)

To see this, note that there exists no $\delta \ne 1$ with $\frac{1}{\gamma _X + \mu I_X} = \delta \cdot \frac{1}{\gamma _Y + \mu I_Y}$, for which the first-order condition holds. From (41) it directly follows that $k_X = k_Y$.

With this finding, the equilibrium level of total abatement reduces to $\tilde{A}_q = \sqrt{\frac{Pq}{2k_X}}$ and first-best abatement is given by $A_I^* = \frac{\bar{E}}{b+k_X}$. Equating $\tilde{A}_q$ and $A_I^*$, solving for $q^*$ is straightforward and results in (19).

Note that to ensure consistency of the mechanism’s abatement incentives, $q$ is constrained such that total abatement under benchmarking regulation, $\tilde{A}$, increases with higher levels of $q$; i.e., abatement incentives are consistent iff $\frac{\partial \tilde{A}_q}{\partial q}>0$:

$$\begin{aligned} \frac{\partial \tilde{A}_q}{\partial q} = -\frac{P (\sqrt{k_X}+\sqrt{k_Y}) \sqrt{k_X^{\frac{q}{2}} k_Y^{\frac{q}{2}}} \left( k_X^{\frac{q}{2}} \left( -2 + q \cdot \ln \left( \frac{k_X}{k_Y} \right) \right) - k_Y^{\frac{q}{2}} \left( 2 + q \cdot \ln \left( \frac{k_X}{k_Y} \right) \right) \right) }{4 \sqrt{2 P q k_X k_Y} \left( k_X^{\frac{q}{2}}+k_Y^{\frac{q}{2}}\right) ^2}. \end{aligned}$$

Hence, for $k_Y=\sigma k_X, \frac{\partial \tilde{A}_q}{\partial q}{>}0$ iff

$$\begin{aligned} k_X^{\frac{q}{2}} \left( 2 - q \cdot \ln \left( \frac{k_X}{\sigma k_X} \right) \right)&> (\sigma k_X)^{\frac{q}{2}} \left( -2 - q \cdot \ln \left( \frac{k_X}{\sigma k_X} \right) \right) \nonumber \\ \Longleftrightarrow \left( \frac{1}{\sigma }\right) ^{\frac{q}{2}}&> \frac{q \cdot \ln (\sigma )-2}{q \cdot \ln (\sigma )+2}. \end{aligned}$$

(42)

A sufficient condition for the mechanism to yield consistent incentives to abate pollution is $q \cdot \ln (\sigma )-2 <0$ and thus $q < \frac{2}{\ln (\sigma )}$. $\square$

1.3 Proof of Proposition 4

Proof

To derive Proposition 4, we proceed with an unconstrained optimization of (28)—where $k_Y = \sigma k_X$—and then derive conditions for potential corner solutions. The unconstrained optimization yields

$$\begin{aligned} \alpha ^*=\frac{2 \bar{E}^2}{P \sqrt{\sigma k_X^2} \left( -2 + \frac{b}{\sqrt{\sigma k_X^2}} + \frac{4 \sqrt{\sigma }}{(1+ \sqrt{\sigma })^2}\right) ^2}. \end{aligned}$$

(43)

Hence, for $\bar{E}>0, \alpha ^* > 0$, thus there is no corner solution at the lower bound of the domain of $\alpha$. It is also straightforward from (43) that a corner solution $\alpha ^*=1$ exists iff

$$\begin{aligned} \bar{E} \ge \sqrt{P \cdot \Omega }, \end{aligned}$$

(44)

where

$$\begin{aligned} \Omega =\frac{\sqrt{\sigma k_X^2}}{2} \left( -2 + \frac{b}{\sqrt{\sigma k_X^2}} + \frac{4\sqrt{\sigma }}{\left( 1+\sqrt{\sigma }\right) ^2}\right) ^2. \end{aligned}$$

(45)

$\square$

1.3.1 Proof of Proposition 5

Proof

To prove Proposition 5, we first show that for an interior solution of $\alpha ^*, A(\alpha ^*)>A^*$. For $\alpha ^* \in (0,1), A(\alpha ^*)< A^*$ iff

$$\begin{aligned} \underbrace{\frac{\bar{E} (1+\sigma )}{b+b\sigma +2\sigma k_X}}_{A^*}&> \underbrace{\frac{\bar{E}}{\sqrt{\sigma } k_X \left( -2+\frac{b}{\sqrt{\sigma k_X^2}} + \frac{4 k_X \sqrt{\sigma }}{\left( \sqrt{\sigma k_X}+\sqrt{k_X}\right) ^2}\right) }}_{A(\alpha ^*)}. \end{aligned}$$

(46)

Condition (46) simplifies to $1<-\frac{(1+\sigma )^2}{\sqrt{\sigma } (1+\sqrt{\sigma })^2}$, which never holds. Hence, $\forall \alpha ^* \in (0,1), A(\alpha ^*)>A^*$.

Second, it is straightforward from (44) that $\alpha ^* = 1$ iff $\Omega \le \frac{\bar{E}^2}{P}$, where $\Omega$ is defined in Eq. (32). Finally, from (12) we know that $A(\alpha ^* \mid \alpha ^*=1)<A^*$ iff $\Psi < \frac{\bar{E}^2}{P}$, where $\Psi = \frac{\left( b \sigma + b +2\sigma k_X\right) ^2}{2k_X(1+\sigma )^2 \sqrt{\sigma }}$. This completes the proof. $\square$

A Generalized Model with $n$ Agents

1.1 The Maximization Problem of the Agents

For completeness, we show the results of the model with rent seeking in a generalized framework with $i = 1, \ldots , n$ firms. For $i \in \Phi$ and $-i \in \Phi \backslash \{i\}$, the payoff function of the individual is

$$\begin{aligned} \pi _i = \left\{ \begin{array}{c@{\quad }l} \left( \alpha \frac{m_i c_i^{\frac{1}{2}}}{m_i c_i + \sum _{-i} m_{-i} c_{-i}^{\frac{1}{2}}} + (1-\alpha )\frac{s_i}{s_i+\sum _{-i} s_{-i}}\right) P-c_i-s_i &{} \text {if } max\{c_i, c_{-i}, s_i, s_{-i}\}>0, \\ 0 &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

(47)

First- and second-order conditions allow us to solve for the individual and total equilibrium levels of abatement costs, abatement, and rent seeking.

The first derivative of the payoff function (47) with respect to $c_i$ is given by

$$\begin{aligned} \frac{\partial \pi _i}{\partial c_i} = \frac{\alpha P c_i^{- \frac{1}{2}} m_i \left( \sum _{-i} c_{-i}^{\frac{1}{2}} m_{-i}\right) }{2 \left( c_i^{\frac{1}{2}} m_i + \sum _{-i} c_{-i}^{\frac{1}{2}} m_{-i}\right) ^2} -1 = 0. \end{aligned}$$

As $c_1 = c_2 = \cdots = c_i = \cdots = c_n$, we can calculate $c_i$, which is given by

$$\begin{aligned} \hat{c}_{i,n}&= \frac{\alpha P m_i\left( \sum _{-i} m_{-i}\right) }{2 \left( m_i + \sum _{-i} m_{-i}\right) ^2}. \end{aligned}$$

Total abatement costs $C$ are simply given by $\hat{C}_n = n \cdot \hat{c}_{i,n} = \sum _{i=1}^n \hat{a}_{i,n}^2 k_i$.

As we have established that $a_i = \left( \frac{c_i}{k_i}\right) ^{\frac{1}{2}}$, we can easily compute individual and total abatement levels, i.e., $\hat{a}_{i,n}$ and $\hat{A}_n = \sum _{i=1}^{n} \hat{a}_{i,n}$.

The equilibrium level of rent seeking $\hat{s}_{i,n}$ is derived by using the first-order condition of the payoff function with respect to $s_i$:

$$\begin{aligned} \frac{\partial \pi _i}{\partial s_i}&= (1-\alpha )P \frac{\sum _{-i} s_{-i}}{s_i+\left( \sum _{-i} s_{-i}\right) ^2}-1 = 0. \end{aligned}$$

Given that $s_1 = s_2 = \cdots = s_i = \cdots = s_n$, we have

$$\begin{aligned} \hat{s}_{i,n}&= \frac{(1-\alpha )P(n-1)}{n^2}. \end{aligned}$$

Aggregate rent seeking is given by the sum of the $\hat{s}_{i,n}$: $\hat{S}_n = \sum _{i=1}^n \hat{s}_{i,n} = n \cdot \hat{s}_{i,n}$.

To summarize, if payoffs are defined by (47), the equilibrium is defined by

$$\begin{aligned} \hat{c}_{i,n}&= \frac{\alpha P \left( \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) }{2 \sqrt{k_i} \left( \frac{1}{\sqrt{k_i}}+\sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}, \\ \hat{C}_n&= \sum _{i=1}^n \frac{\alpha P \sum _{-i} \frac{1}{\sqrt{k_{-i}}}}{2 \sqrt{k_i} \left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}, \\ \hat{a}_{i,n}&= \sqrt{\frac{\alpha P \left( \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) }{2k_i^{\frac{3}{2}} \left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}}, \\ \hat{A}_n&= \sqrt{\frac{\alpha P}{2}} \cdot \sum _{i=1}^n \sqrt{\frac{\left( \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) }{k_i^{\frac{3}{2}}\left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}}, \\ \hat{s}_{i,n}&= \frac{(1-\alpha )P(n-1)}{n^2}, \\ \hat{S}_n&= \sum _{i=1}^n \hat{s}_{i,n} = n \cdot \hat{s}_{i,n} = \frac{(1-\alpha )P(n-1)}{n}. \end{aligned}$$

1.2 Welfare Optimization

We now determine the optimal $\alpha _n$ by minimizing the costs to society:

$$\begin{aligned} \min _{\alpha }&\left[ D(\hat{A}_n)+ \hat{C}_n + \hat{S}_n \right] , \\ \text {s.t. } 0&\le \alpha \le 1, \nonumber \end{aligned}$$

(48)

where

$$\begin{aligned} D(\hat{A}_n)&= \frac{\left( \bar{E} - \sqrt{\frac{\alpha P}{2}} \cdot \sum _{i=1}^n \sqrt{\frac{\left( \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) }{k_i^{\frac{3}{2}}\left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}}b\right) ^2}{2b}, \\ \hat{C}_n&= \frac{\alpha P}{2} \sum _{i=1}^n \frac{\sum _{-i} \frac{1}{\sqrt{k_{-i}}}}{\sqrt{k_i} \left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}, \\ \hat{S}_n&= \frac{(1-\alpha )P(n-1)}{n}. \end{aligned}$$

Rearranging the first-order condition allows one to solve for the interior solution of the optimal $\alpha$ in the $n$-player case:

$$\begin{aligned} \alpha _n^* = \frac{2 \bar{E}^2 \left( \sum _{i=1}^n \sqrt{\frac{\sum _{-i} \frac{1}{\sqrt{k_{-i}}}}{k_i^{\frac{3}{2}} \left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}}\right) ^2}{P\left( -4 + \frac{4}{n} + b \left( \sum _{i=1}^n \sqrt{\frac{\sum _{-i} \frac{1}{\sqrt{k_{-i}}}}{k_i^{\frac{3}{2}} \left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2}}\right) ^2 + 2 \sum _{i=1}^n \frac{\sum _{-i} \frac{1}{\sqrt{k_{-i}}}}{\sqrt{k_i} \left( \frac{1}{\sqrt{k_i}} + \sum _{-i} \frac{1}{\sqrt{k_{-i}}}\right) ^2} \right) ^2}. \end{aligned}$$

It is easy to check that this is equivalent to the results of the case with two firms.  Therefore, the underlying incentives remain unchanged.