Environmental Policy for Non-Point Source Pollutions in a Bertrand Duopoly

This study investigates the effectiveness of ambient charges under non-point source (NPS) pollutions in an imperfect competition framework. To this end, following Ganguli and Raju [1], it constructs a one-stage game and a two-stage game in which Bertrand duopolistic firms choose their best prices and abatement technology, respectively. It is demonstrated in both games that an increase in the ambient charge can lead to a decrease in pollution. This finding indicates that the ambient charge can efficiently control pollution in a Bertrand duopoly.


Introduction
The main purpose of this study is to demonstrate that the ambient charges can control the total amount of NPS pollutions under ála Bertrand imperfect competition. Based on the analysis of Raju and Ganguli (2013) in a Cournot duopoly setting, Matsumoto et al. (2017) show a "goodnatured" e¤ect of ambient charges in an N -…rm Cournot framework that an increase in the ambient charge leads to decreases of the total level of industry pollutions. On the other hand, Ganguli and Raju (2012) consider the same subject in a Bertrand duopolistic market and numerically exhibits a "perverse" e¤ect that an increase of ambient charge may lead to an increase in pollution in two distinct settings, one in the one-stage game and the other in the two-stage game. This study, under the same framework, shows the good-natured e¤ect of ambient charges in Bertrand duopoly, analytically in one-stage game and numerically in two-stage game.
The rest of the paper is organized as follows. In section 2, the optimal price strategies of Bertrand duopolistic …rms are derived in one-stage game in which all actions take place simultaneously. In Section 3, the optimal choices of abatement technology at the …rst stage and prices at the second stage are considered in two-stage game in which the actions take place in sequence. Concluding remarks and further extension of this study are given in Section 4

One-stage Game
In this section we consider the e¤ect of the ambient charge in one stage game in which the regulator has announced the ambient charge and a cut-o¤ ambient standard while two …rms have …xed their pollution abatement technologies. Under this circumstance the …rms choose their optimal prices to maximize their pro…ts. Each …rm produces a di¤erentiated product. Market demand function for …rm i for i; j = 1; 2 and i 6 = j is where q i denotes good i produced by …rm i; p i is the price of q i , p j is the price for the good j and is a parameter with 0 < < 1 measuring the substitutability between two goods. 1 We exlude two extreme cases, one with = 1 where the two goods are homogenous and the other with = 0 where they are independent. The total amount of pollution E generated by the two …rms is given by where i and j represent pollution abatement technologies of …rms i and j. i = 1 means the worst technology with 100% pollution while i = 0 means the best technology with no (0%) pollution. Accordingly, it is assumed that 0 i 1: The pro…t function of …rm i is where E > 0 denotes ambient standard speci…ed by the regulator, cq i is the production cost where c > 0 is the common marginal cost of production and t is an ambient charge or tax with 0 t 1.
According to the spirit of the ambient charge, although the two …rms'contributions to pollutions might be di¤erent, each …rm will pay the identical …ne t E E if E E > 0 and receive the identical subsidy t E E if E E < 0. Substituting (1) into (3) and di¤erentiating the resultant pro…t function with respect to p i give the …rst order condition for an interior solution maximizing pro…t of …rm i, Maximizing j with respect to p j presents a similar …rst-oder condition for …rm j. Hence solving the following simultaneous system, which is obtained from …rst order conditions for …rms i and j, that are, after arranging the terms, Concerning the positivity of the Bertrand price, we have the following results.
Theorem 1 If i j or if i < j and a + c 1=3; then p B i > 0: Proof. If i j holds, then Then the …rst equation of (5) implies p B i > 0: Now suppose that i < j . If the right hand side of the …rst equation of (5) is equal to zero, then solving it for j gives the form of Assumption t 1 implies (2 + )(a + c) t (2 + )(a + c) : Since =(2 + ) < 1=3 for < 1; the term on the right hand side is greater than unity if a + c 1=3 under which, for all j 1; The inequality implies that p B i > 0: Di¤erentiating the Bertrand price of …rm i with respect to t reveals that the sign of the derivative is the same as the sign of the terms in the square brackets in (5), Hence the e¤ect caused by a change in the ambient charge on the Bertrand prices can summarized as follows.
Theorem 2 A …rm with a larger or equal abatement technology positively responds to a change in the ambient charge whereas the response of a …rm with a smaller abetment technology is ambiguous, Proof. If i j ; then for …rm i; the bracketed terms in (5) are and for …rm j; from the bracketed terms in the second equation of (5) to be equal to zero, we can de…ne the ratio of the abetment technologies This ratio is less than unity implying dp B j dt Q 0 according to The same procedure can be applied for the case of i < j : Substituting the Bertrand prices into the demand functions in (1) presents the Bertrand outputs of …rm i and j; To check whether q B i is positive, we subtract the second equation of (6) from the …rst equation to obtain where, in the same way, from (5) Concerning the positivity of the Bertrand output, we have the following, Proof. If i j holds, then (8) leads to p B i p B j with which (7) implies On the other hand, the …rst equation of (6) with the forms of p B i and p B j given in (5) is reduced to The direction of the last inequality is due to a c + 1 and t i 1: Let the numerator of (9) be f ( ): Then due to Descartes'rule of sign, f ( ) = 0 has only one positive root, 0 > 0: Substituting = 1 gives The last inequality means that f ( ) < 0 = f ( 0 ) for < 1; with which then (9) leads to q B i > 0; implying that q B j > 0 as well. If j > i ; then interchanging the two …rms generates the same result. Now consider the e¤ect of a change in the ambient charge on each output level.
Concerning the ambient charge e¤ect on output, we have the following results.
Theorem 4 A …rm with a larger or equal abetment technology negatively responds to a change in the ambient charge whereas the response of a …rm with a smaller abetment technology is ambiguous, Proof. If i = j ; then (3 2 ) < 2 for < 1 immediately implies the results. If i > j ; then (3 2 ) > 0 for > 0 indicates that for …rm i; and for …rm j, there is a threshold value of the parameter ratio such as The same procedure can be applied for the case of i < j : It should be noticed …rst that …rm i with a larger i has an ine¢ cient technology because it generates larger emission. Theorems 2 and 4 imply that, the …rm with ine¢ cient technology exhibits natural response to the change in the ambient charge, that is, it increases price and decreases output. On the other hand the …rm with e¢ cient technology responds ambiguously. It should be noticed second that these …rm-speci…c responses are non-observable for the regulator which can see only the total amount in the case of NPS pollution.
The total level of pollution at the Bertrand equilibrium is obtained by substituting q B i and q B j into (2) E B = i q B i + j q B j : Concerning the e¤ect of a change in the ambient charge on the total pollution, we have the following result.
Theorem 5 An increase in the ambient charge decreases the total level of pollution, Notice that (3 2 ) 2 and the equality holds when = 1: Therefore we arrive at the result where the strict inequality is due to the assumption, < 1: Although Theorem 4 implies a possibility of the perverse e¤ect on emission of the individual …rm with the e¢ cient abetment technology, Theorem 5 implies that the total e¤ect is always negative, implying that the negative e¤ect of the ine¢ cient …rm dominates the positive e¤ect of the e¢ cient …rm.

Two-stage Game
In this section the …rms and the regulator take actions in two stages. At the …rst stage, each …rm determines its optimal abatement technology whereas the regulator announces the ambient charge and the cut-o¤ level of total pollution. Then at the second stage, the …rms choose their prices to maximize their pro…ts, given the ambient charge, the cut-oof level and their abatement technologies.
The decision-making at the second stage have been already considered in the one-stage game. Given the Bertrand prices and outputs in (5) and (6), we consider the actions of choosing abatement technology at the …rst stage. The Bertrand pro…t function of …rm i under Bertrand prices and Bertrand output is de…ned as for i; j = 1; 2 and i 6 = j. Notice that there is a small di¤erence between the de…nitions of (3) and (11). There is a term (1 i ) 2 in (11) and no such a term in (3). This term re ‡ects the cost associated with selecting the abatement technology. 2 At the second stage, the abatement technology has been already selected somehow and thus it does not a¤ect the determination of the optimal price whether this cost is included or not. However this cost function is e¤ective for choosing technology at the …rst stage where the …rm i determines its abatement technology, i ; so as to maximize its Bertrand pro…t. Di¤erentiating (11) with respect to i gives the …rst-order condition for the optimal level of i ; Arranging the terms in @ B i =@ i = 0 with taking account of the forms of the Bertrand prices in (5) yield the modi…ed FOC at the …rst stage, In the same way, arranging the terms in @ B j =@ j = 0 yields the modi…ed FOC for …rm j 4 9 2 + 16 t 2 i + 2 (2t) 2 (4 2 ) 2 j = 2(4 2 ) 2 + td: Solving (12) and (13) simultaneously for i and j presents the optimal abatement technology for both …rms, Although the form of the solution seems to be highly complicated, the following result is obtained.
Proposition 1 If 2=3 a > c; then the optimal abatement technology ( ; t) is positive for 0 < < 1 and 0 t 1: Proof. It is numerically con…rmed that the denominator of (14) is negative for 0 < < 1 and 0 t 1: 3 The numerator is rewritten as where f ( ) = (2 ) 2 (2 + ) and 3 < f ( ) < 8 for 0 < < 1: Assumption a > c implies a (1 )c > 0: The sign of the terms in the braces of (15) is negative for t = 0 (i.e., f ( ) < 0) and is also negative for t = 1 if 2a < 3. As the terms satis…es the inequality and the lower bound of f ( ) is 3; then the right hand side is negative for 0 < < 1 if 2a 3. Thus the sign of the left hand side is also negative. Hence 2 [a (1 )c] t f ( ) < 0 for 0 t 1: Therefore (14) with the negative denominator and the negative numerator implies ( ; t) > 0.
Notice that the following set is not empty, f(a; c) j 0 < a 3=2 and 0 < c a 1g ; implying that the conditions imposed on the values of a and c given in Theorem 3 and Proposition 1 are compatible. Substituting into the Bertrand prices in (5) gives the optimal Bertrand price that is clearly positive for 0 < < 1 and 0 t 1; which is summarized as follows.
Proposition 2 If 3=2 a > c; then the optimal Bertrand price p ( ; t) is positive for 0 < < 1 and 0 t 1: The optimal Bertrand output is obtained by substituting p ( ; t) into the demand function, (1), Concerning the optimal output, we can have the following, where the numerator is positive and the denominator is negative. This completes the proof.
Proposition 3 implies that q ( ; t) is inevitably negative for a small neighborhood of point (0; 1). Numerical con…rmation of Proposition 3 is given in Figure 1 in which the Bertrand output is negative in the shaded region and positive otherwise. 4 The shaded region should be eliminated from further considerations. Finally, the optimal level of total pollution is that is apparently negative if q ( ; t) < 0; that is, in the shaded region in Figure 1. Its derivative with respect to t is dE dt = 2 q t t @ @t + t q @q @t where @ =@t and @q =@t can be of either sign and thus the sign of the terms in the square bracket seems ambiguous. Numerically, as as shown in Figure 2 Figure 2(B). The dotted downward-sloping curve in Figure 2(A) is described by the q = 0 curve. The shaded region is included in the region with q < 0 so that the derivative of the optimal technology is negative in the feasible region with q > 0. Hence the sign of the tax-derivative of the total emission level is de…nitely negative in the white region in Figure 2(B) as both derivatives are negative. On the other hand, it is sensitive to the relative magnitude between the elasticities of the optimal ambient technology and the optimal output with respect to t in the shaded region in Figure 2 However, Figure 3 numerically shows that dE =dt < 0 for 0 < < 1 and 0 t 1: In other word, the elasticity of in absolute value is larger that that of q in the shaded region. Therefore we have our main result that an increase in ambient charge always decreases the total level of optimal pollution. 5 We summarize this result, Proposition 4 It is numerically con…rmed that a change in the ambient charge has the goodnatured e¤ ect, dE dt < 0 for 0 < < 1 and 0 t 1: (20)

Concluding Remark
In this paper we reconsider the "perverse" e¤ect caused by a change in ambient charges shown by Ganguli and Raju (2012). To this end, following their basic framework, we …rst re-examine the e¤ect in one-stage game in which the Bertrand …rms determine their prices so as to maximize their pro…ts, given the abatement technology. Our …rst result analytically demonstrates that an increase of ambient charges decreases the total level of NPS pollutions. We then turn attention to the e¤ect in two-stage game in which the optimal abatement technology is selected at the …rst stage and the optimal prices are determined at the second stage. Our second result numerically show the good-natured e¤ect on the total level of pollution. With these results we conclude that the ambient charge might be an e¢ cient method to control NPS pollutions in a duopoly market.