On Environmental Regulation of Oligopoly Markets: Emission versus Performance Standards

Abstract

By specializing Montero’s (J Environ Econ Manag 44:23–44, 2002) model of environmental regulation under Cournot competition to an oligopoly with linear demand and quadratic abatement costs, we extend his comparison of firms incentives to invest in R&D under emission and performance standards by solving for a closed form solution of the underlying two-stage game. This allows for a full comparison of the two instruments in terms of their resulting propensity for R&D and equilibrium industry output. In addition, we incorporate an equilibrium welfare analysis. Finally, we investigate a three-stage game wherein a welfare-maximizing regulator sets a socially optimal emission cap under each policy instrument. For the latter game, while closed-form solutions for the subgame-perfect equilibrium are not possible, we establish numerically that the resulting welfare is always larger under a performance standard.

Appendix

This section contains the proofs and computations for all the results of the paper.

Proof of Lemma 1

In the last stage, firm i solves the optimization problem (2), with first order condition (or FOC)

$$\begin{aligned} -2bq_{i}+a-bq_{j}-c+x_{i}=0. \end{aligned}$$

(8)

The reaction function of firm i is given by

$$\begin{aligned} q_{i}(q_{j},x_{i})= {\left\{ \begin{array}{ll} \frac{a-c+x_{i}}{2b}-\frac{q_{j}}{2}, &{} \text {if }\frac{a-c+x_{i}}{b}\ge q_{j} \\ 0, &{} \text {otherwise. } \end{array}\right. } \end{aligned}$$

The objective function is strictly concave, thus, a unique maximum is attained (the second-order condition (SOC) is $-2b<0$). Solving simultaneously the two non-zero reaction functions of the firms yields (for $i\ne j$)

$$\begin{aligned} {q}_{i}^{{}}(x_{i},x_{j})=\frac{a-c+2x_{i}-x_{j}}{3b}. \end{aligned}$$

(9)

In the first stage, firm i chooses the level of R&D, $x_{i}$, given $x_{j}$, e and ${q}_{i}^{{}}(x_{i},x_{j})$ in Eq. (9).

The FOC of (3) is then given by

$$\begin{aligned} \frac{2}{3b}[-2b{q}_{i}^{{}}(x_{i},x_{j})+a-b{q} _{j}^{{}}(x_{i},x_{j})-c+x_{i}]+\frac{4}{3}{q}_{i}^{{}}(x_{i},x_{j})-e- \gamma x_{i}=0. \end{aligned}$$

(10)

The first term in (10) vanishes by the FOC in (8). By (A1), $9b\gamma >8$, the SOC holds. Plugging Eq. (9) into Eq. (10) and using the fact that the firms are symmetric and subject to the same standard e, we get that every firm invests

$$\begin{aligned} {x}^{{}}(e)=\frac{4(a-c)-9be}{9b\gamma -4} \end{aligned}$$

(11)

in R&D, yielding an output of

$$\begin{aligned} {q}^{{}}(e)=\frac{3\gamma (a-c)-3e}{9b\gamma -4}. \end{aligned}$$

(12)

Since e is selected exogenously, we want to choose values that are economically meaningful. In particular, we want $0<x(e)<c$ and $0<e<q(e)$; we show that $0<e<\frac{3\gamma (a-c)}{9b\gamma -1}$ is sufficient to guarantee that these two inequalities hold.

First notice that e, such that $0<e<\frac{3\gamma (a-c)}{9b\gamma -1},$ implies that $\frac{(a-c)}{9b\gamma -1}<{x}^{{}}(e)=\frac{4(a-c)-9be}{ 9b\gamma -4}<\frac{4(a-c)}{9b\gamma -4}$ by multiplying the first inequality by $-9b$, adding $4(a-c)$ and dividing by $9b\gamma -4,$ which is positive by (A1). Assumption (A1) implies that the left hand side of the second inequality is strictly positive, hence, $0<x(e)$. The proof of $x(e)<c$ also follows from (A1); to see it rewrite $9b\gamma c>4a$ as $9b\gamma c-4c>4(a-c),$ which implies that $\frac{4(a-c)}{9b\gamma -4}<c$, hence, that ${x}^{{}}(e)<c.$

To show that $e<q(e)$ (which immediately implies that $0<q(e)$) consider the following implications:

$e<\frac{3\gamma (a-c)}{9b\gamma -1},$ hence, $(9b\gamma -1)e<3\gamma (a-c)$ by (A1) and, further, $9b\gamma e-4e+3e<3\gamma (a-c)$, thus, $(9b\gamma -4)e<3\gamma (a-c)-3e,$ consequently leading to $e<\frac{3\gamma (a-c)-3e}{ 9b\gamma -4}=q(e)$. This completes the proof of Lemma 1. $\square$

Proof of Lemma 2

In the last stage, firms solve problem (4), with FOC:

$$\begin{aligned} a-2bq_{i}-bq_{j}-(c-x_{i})(1-h)=0, \end{aligned}$$

(13)

with SOC $-2b<0$.

Solving simultaneously for the positive parts of the reaction curves of the firms, gives us

$$\begin{aligned} {q}_{i}(x_{i},x_{j},h)=\frac{(2x_{i}-x_{j}-c)(1-h)+a}{3b}. \end{aligned}$$

(14)

Using Eq. (14) in problem (5) helps derive the following FOC for the first stage:

$$\begin{aligned}&\frac{2}{3b}(1-h)[a-2b{q}_{i}^{{}}(x_{i},x_{j},h)-b{q} _{j}^{{}}(x_{i},x_{j},h)-(c-x_{i})(1-h)]\nonumber \\&\quad +\frac{4}{3}{q} _{i}^{{}}(x_{i},x_{j},h)(1-h)-\gamma x_{i}=0, \end{aligned}$$

(15)

with SOC $8(1-h)^{2}/(9b)-\gamma <0$, which holds by (A1), $8<9b\gamma$, and $0<h<1$.

The first term in Eq. (15) vanishes by FOC (13). By the symmetry of the model, Eqs. (14) and (15) imply our result, ${x}(h)=\frac{4(1-h)[c(1-h)-a]}{4(1-h)^{2}-9b\gamma }$ and ${q}(h)=\frac{3\gamma [c(1-h)-a]}{4(1-h)^{2}-9b\gamma }.$

Note that the following assumption:(A’) $8(1-h)^{2}<9b\gamma$, $a>c(1-h)$ and $9bc\gamma >4a(1-h)$ for all $0<h<1$, guarantees that $0<{x}^{{}}(h)<c$ and ${q}^{{}}(h)>0$. It is easy to verify that $0<h<1$ and (A1) ($8<9b\gamma$, $a>c$ and $9bc\gamma >4a$ ) imply (A’), since the lowest side of the (positive) inequalities is multiplied by a number between 0 and 1. This completes the proof of Lemma 2. $\square$

Proof of Proposition 3

To avoid confusion between the equilibrium variables, we add the superscript e for the equilibrium under the emission standard, and p for the equilibrium under the performance standard. Thus, $x^{e}(e)$ and $q^{e}(e)$ correspond to x(e) and q(e) in Lemma 1, respectively; and $x^{p}(h)$ and $q^{p}(h)$ correspond to x(h) and q(h) in Lemma 2, respectively.

Now, let us fix $e=hq^{p}(h)$, so that both instruments generate the same level of emissions. Hence, e is a function of h and we can write $x^{e}$ in terms of h, $x^{e}(e)=x^{e}(hq^{p}(h))=x^{e}(h)$. Similarly, we can write $q^{e}$ as a function of h (this feature will be used in part (c)).

For parts (a) and (b), we find the values of h for which $x^{e}(h)=x^{p}(h)$ holds. For these values, we can analyze the functions of interest to see which function takes a larger value on given intervals. It turns out that there are only two solutions: $h_{1}=0$, which is trivial and implies zero level of emissions, and $\overline{h}= \frac{5(9b\gamma )-(9b\gamma +16)a/c}{9b\gamma -16a/c}$, which is the threshold of interest. Since there are only two roots, we analyze the slopes of $x^{e}(h)$ and $x^{p}(h)$ around zero, to prove our result. We have to distinguish between two cases, which depend on the sign of $9b\gamma -16a/c$.

• (a) Suppose that $a/c<9b\gamma /16$ (i.e., $9b\gamma -16a/c>0$). Two possibilities arise; either $\overline{h}>0$ or $\overline{h}<0$. First, let us assume that $\overline{h}>0$. Since, by assumption, its denominator is positive, the numerator must be positive as well, i.e., $5(9b\gamma )-(9b\gamma +16)a/c>0$. Further, it can be shown that ${x^{e}}^{\prime }(0)-{ x^{p}}^{\prime }(0)=-c\frac{5(9b\gamma )-(9b\gamma +16)a/c}{(4-9b\gamma )^{2} }$ is negative by the statement above, thus, $x^{p}(h)$ is above $x^{e}(h)$ for all $0<h<\overline{h}$. A similar argument shows that ${x^{e}}^{\prime }(0)-{x^{p}}^{\prime }(0)>0$ when $\overline{h}<0$, which implies that $x^{e}(h)>x^{p}(h)$ for all $0<h<1$. The reader can easily verify that (i) $a/c<4$ implies $\overline{h}>1$, (ii) $a/c>5\frac{9b\gamma }{9b\gamma +16}$ implies $\overline{h}<0$ and (iii) $4<a/c<5\frac{9b\gamma }{9b\gamma +16}$ implies $0<\overline{h}<1$; which clearly defines regions for where an instrument is superior that the other in terms of R&D. To complete our result, it suffices to show that ${x^{e}}^{\prime }( \overline{h})-{x^{p}}^{\prime }(\overline{h})\ne 0$, which holds by our assumptions, since ${x^{e}}^{\prime }(\overline{h})-{x^{p}}^{\prime }( \overline{h})=\frac{c(5(9b\gamma )-(9b\gamma +16)a/c)(16a-9bc\gamma )^{2}}{ 9b\gamma (-64+9b\gamma )(-4+9b\gamma )(-4a^{2}+9bc^{2}\gamma )}$.

• (b) Now suppose that $a/c>9b\gamma /16$. The proof of this part is analogous to that in part a, and is therefore omitted.

• (c) This proof follows the same strategy as the one implemented in parts a and b. We look for h that satisfies the equality $q^{e}(e)=q^{p}(h)$, with $e=hq^{p}(h)$. Again two solutions are obtained; the trivial root $h_{1}=0,$ and the root of interest, $h_{0}= \frac{5a-9bc\gamma -c}{4a-c}$. Note that (A1) implies that $h_{0}<1$, yet, it remains to establish whether its value is positive or negative. First, let us suppose that $h_{0}>0$ and analyze the slopes of $q^{e}(h)$ and $q^{p}(h)$ at $h_{1}=0$. If $h_{0}>0$, then $5a-9bc\gamma -c>0$, by (A1), $a>c$. Further rearrangements lead to ${q^{e}} ^{\prime }(0)-{q^{p}}^{\prime }(0)=\frac{3\gamma (5a-9bc\gamma -c)}{ (4-9b\gamma )^{2}}$, which is positive by the assumptions $h_{0}>0$ and $\gamma >0$. Hence, $q^{e}(h)$ is always above $q^{p}(h)$ (for all $0<h<h_{0}$ ). If $h_{0}<0$, then $5a-9bc\gamma -c<0$ and ${q^{e}}^{\prime }(0)-{q^{p}} ^{\prime }(0)<0$; thus, $q^{p}(h)$ is above $q^{e}(h)$ for all $0<h<1$. Notice that (i) $a/c<(9b\gamma +1)/5$ implies $h_0<0$ and (ii) $a/c>(9b\gamma +1)/5$ implies $0<h_0<1$. The fact that ${q^{e}}^{\prime }(h_{0})-{q^{p}}^{\prime }(h_{0})=\frac{3(-4a+c)^{2}\gamma (5a-9bc\gamma -c)}{ (-4+9b\gamma )(-1+36b\gamma )(-4a^{2}+9bc^{2}\gamma )}$ differs from zero (given our assumptions in any case), completes our proof.$\square$

Proof of Lemma 4

Given the results in Lemma 1, we solve the problem of the regulator presented in Eq. (6). Here, the FOC is given by the following equation:

$$\begin{aligned} \frac{d{q}^{{}}(e)}{de}[a-2b{q}^{{}}(e)-c+{x}^{{}}(e)]+\frac{d{x}^{{}}(e)}{de }[{q}^{{}}(e)-e-\gamma {x}^{{}}(e)]+c-{x}^{{}}(e)-2se=0. \end{aligned}$$

(16)

The SOC of this problem is $\frac{2(9b+8s-18b\gamma s)}{9b\gamma -4}$, which is negative by (A1) and (A2). Using the FOCs (8) and (10), Eq. (16) becomes

$$\begin{aligned} b{q}^{{}}(e)\frac{d{q}^{{}}(e)}{de}-\frac{1}{3}{q}^{{}}(e)\frac{d{x}^{{}}(e) }{de}+c-{x}^{{}}(e)-2se=0. \end{aligned}$$

(17)

Combining Eqs. (11), (12) and (17), leads to the results outlined in Lemma 4. (A1) and (A2) guarantee that in equilibrium, $0<x^{*}<c$ and $0<e^{*}<q^{*}$. $\square$