Trade, Transboundary Pollution, and Foreign Lobbying


In this paper, we explore the use of trade policy in addressing transboundary stock pollution problems such as acid rain and water pollution. We show that a tariff determined by the current level of accumulated pollution can induce the time path of emissions optimal for the downstream (polluted) country. But if the upstream (polluting) country can lobby the downstream government to impose lower tariffs, distortions brought by corruption and foreign lobbying lead to a rise in the upstream country’s social welfare, and to a decrease in social welfare in the downstream country. Thus, the usefulness of trade policy as a tool for encouraging cooperation and internalizing transboundary externalities depends critically on the degree of governments’ susceptibility to foreign political influence.

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  1. 1.

    For example, several upwind countries, including the UK, refused to ratify the 1985 Helsinki protocol on the Reduction of Sulfur Emissions on their Transboundary Fluxes as it was estimated that the costs of abatement for these countries would exceed the domestic environmental benefits (Nordstrom and Vaughan 1999).

  2. 2.

    It is well documented that as a result of their unfortunate downwind location, much of the U.S. production of SO2 and NOx is deposited in Canada and much of the UK production is deposited in Scandinavian countries (e.g., Newbery 1990). It has also been shown that Chinese emissions of SO2 cause acid rain in Japan (Nagase and Silva 2007).

  3. 3.

    See Bayramoglu (2006). Among other cases that have historically been important are heavy metals and chloride pollution suffered by downstream countries on the Danube and Rhine Rivers (Bernauer and Moser 1996; McGlade 2000), and salinity problems in the lower basin of the Colorado River where it crosses the Mexican-American border (Maler and de Zeeuw 1998).

  4. 4.

    See, for example, Baumol and Oates (1988), Copeland (1996) and Markusen (1975). It has also been suggested that trade policy may serve as a mechanism for promoting cooperation between countries linked by the externality (Nordstrom and Vaughan 1999). A more recent thread in the literature investigates the impact of changes in trade openness upon the level of pollution taxes set by countries who are impacted by a transboundary pollutant (Baksi and Chaudhuri 2009; Burguet and Sempre 2003; Nkuiya 2013). The flavor of results in this strand of the literature is that increased openness of trade, as reflected by lower tariffs, can raise welfare in both the emitting country and the impacted country, if the original tariff is sufficiently large. This result seems to hang on the presence of imperfectly competitive firms in the markets that operate within the countries in question, as well as less than complete transmission of the pollutant across boundaries (i.e., the pollutant cannot be purely transboundary, as for example with greenhouse gases). Our analysis contains neither of these assumptions, and so provides an interesting contrast.

  5. 5.

    These include the connection between foreign lobbying and 2008 U.S. presidential campaign fundraising (Evans and Patel 2008; McElhatton and Seper 2008), allegations of foreign contributions during fundraising for the 1996 U.S. presidential campaign (Miller 1996; Woodward and Duffy 1997); the controversy surrounding the Chinese government’s recent hiring of top lobbying firms to represent Chinese interests in the U.S. (Guevara 2005) or the foreign lobby industry in Washington DC (Silverstein 2007).

  6. 6.

    For example, the LobbyWatch project of the Center for Public Integrity ( reports that BP, one of the biggest foreign spenders on Washington DC lobbying, lobbied almost as much on environmental issues and Superfund as it did on matters related to oil and gas.

  7. 7.

    This modeling approach is in line with contributions to environmental economics literature that study transboundary pollution problems using differential games (Dockner and van Long 1993; List and Mason 2001; Fernandez 2002).

  8. 8.

    Here we restrict our attention to the comparison of two second-best scenarios and do not consider the socially optimal solution to the problem. The comparison of our results to the optimum would involve a contrast between the effects of the environmental tax (first-best) and a tariff (not two different tariffs), which would lead to obvious results.

  9. 9.

    From now on, unless otherwise stated, we will suppress the time argument t.

  10. 10.

    These structural assumptions give rise to a linear-quadratic model, which facilitates the analysis. While a more general framework can be employed in the simple version of the model analyzed in this section, it is generally difficult to make headway in differential games without imposing stark structural assumptions, such as the linear-quadratic framework. To keep the discussion of this section on a parallel footing to that of the later sections, we opt to restrict attention to the linear-quadratic framework here.

  11. 11.

    Details of the derivations for the linear-quadratic model are provided in Appendix 1.

  12. 12.

    The optimal tariff must also satisfy the inequality constraint \({\uptau }\ge {\uptau }^0\), which will automatically be satisfied so long as the shadow value on pollution is negative. In addition, \({\uptau }\le {\uptau }^p\), where \({\uptau }^p = \frac{(b+c)a - ba_u}{b(b+c)}\) is the tariff level that forces downstream demand to zero.

  13. 13.

    The optimal tariff must also satisfy the two inequality constraints: \(\frac{Y_0}{b(1-A)} \ge {\uptau }\ge \frac{AY_0}{b(1-A^2)} \equiv {\uptau }_0\), where \(Y_0\) is the level of exports from U to D when \({\uptau }= {\uptau }_0\). The second of these will automatically be satisfied so long as the shadow value on pollution is negative, which is a mild restriction; for the first constraint to hold the shadow value can not be too large in magnitude: \({\uptheta }\ge -\frac{Y_0}{{\uplambda }}\). (In the event this restriction does not hold, Downstream would set the tariff at the prohibitive level, and there would be no trade.)

  14. 14.

    That players might believe such lobbying could influence behavior seems highly likely: in the U.S., billions of dollars are devoted to lobbying activities annually. Moreover, a recent U.S. Supreme Court decision (Citizens United v. Federal Election Commission, 558 U.S. 08-205 (2010), 558 U.S. ——–, 130 S.Ct. 876, January 21, 2010) explicitly protects organizations’ right to pursue a political agenda; the majority opinion specifically avoids determining whether there is a “compelling governmental interest in limiting foreign influence over the Nation’s political process.” Apparently it is fair game for foreign interest to engage in political lobbying, at least in the U.S.

  15. 15.

    A referee notes that firms may anticipate the U government’s lobbying activities, and that this could impact their choices. Such a perspective is reasonable with strategic firms, as when the industry is imperfectly competitive; it seems less natural in a market with atomistic sellers (as in our case). In addition, having the market take actions on a static, myopic basis—as we do—is in keeping with much of the extant literature.

  16. 16.

    We use the optimal control approach here in part to ease exposition, and to retain a parallel structure to the preceding section. There is a significant thread of the literature that uses such an approach; see for example Dockner et al. (2000), Karp (1992), Mason (1997), Mason and Polasky (1997, 2002). An alternative approach would be to use dynamic programming, as in Dockner and van Long (1993), List and Mason (2001). The two approaches are basically dual (Dockner et al. 2000), so the choice of technique is something of a matter of personal preference.

  17. 17.

    This, of course, is the whole point of the lobbying efforts in the first place. If for some reason \(\frac{\partial {\uptau }}{\partial P} < 0\) the decision-maker in U would not waste resources on lobbying, i.e., \(I = 0\), so that \(P = 0\).

  18. 18.

    The Supreme Court decision we discussed in footnote 14 implicitly views lobbying actions as beneficial to the typical citizen. That decision alleges value in the flow of information resulting from such lobbying efforts, i.e., the informational benefits are a function of I.

  19. 19.

    The derivation of this vector is neither illuminating nor straightforward, and so we relegate details to Appendix 2. There, we derive six equations that characterize these six parameters. The system need not have a unique solution. For example, if D anticipates U will set \(I = 0\) (i.e., \({\upgamma }_k = 0, k = 1, 2, 3\)) there may be a Markov equilibrium in which \({\upalpha }_3 = 0\); in that case, the scenario collapses to the case analyzed in Sect. 2. But since the decision-maker in D enjoys benefits from positive P, it is unclear why U would expect D to ignore P in setting the tariff. An additional complication arises if \({\upgamma }_3 < 0\), as we find in the example discussed below; in that case, the non-negativity constraint on I will bind for \(P \ge |{\upgamma }_1/{\upgamma }_3| + |{\upgamma }_2/{\upgamma }_3| Z.\) But in such a scenario \(I = 0\), implying (because \({\updelta }> 0\)) that P would fall below this critical level. For the constraint to continue to bind, then, Z must rise just enough to offset the decline in P; this implies a knife-edge solution. A similar consideration applies in D’s choice of tariff: there are combinations of Z and P that would force the constraint \({\uptau }\ge {\uptau }^0\) to bind; again, continued play at this point would require a knife-edge solution. While these special outcomes may be mathematically possible they seems unlikely to obtain in practice, and so we focus on interior solutions in the pursuant discussion.

  20. 20.

    The resultant solution is (\({\upalpha }_1, {\upalpha }_2, {\upalpha }_3, {\upgamma }_1, {\upgamma }_2, {\upgamma }_3) = (48.14, 0.042, -0.0004, 4.10, 0.014, -0.392)\). We derived the solution using the fmincon subroutine in MATLAB. This subroutine identifies a constrained optima; both equality and inequality constraints are allowed. We supplied constraints corresponding to the six equations (26)–(28) and (30)–(32), along with non-negativity constraints on IP and Z; the function to be “maximized” is a constant. Since any parameter combination maximizes the (constant) objective function, a feasible solution must satisfy the six equality constraints. As such, the vector (\({\upalpha }_1, {\upalpha }_2, {\upalpha }_3, {\upgamma }_1, {\upgamma }_2, {\upgamma }_3\)) selected by this subroutine must satisfy the system of six equations; i.e., it represents an equilibrium.

  21. 21.

    For two of the parameters, \({\upbeta }\) and f, an equilibrium existed through the largest value assessed; we indicated this feature by appending an asterisk to the tabulated maximum values.

  22. 22.

    This construct is “naive” in the sense that the reaction of the Upstream player is not taken into consideration, underscoring the importance of the strategic interaction here. Because \(I^{\textit{MP}} = {\updelta }P^{\textit{MP}}\) in steady state, it follows that \({\upeta }^{\textit{MP}}_P\) can be expressed in terms of \(P^{\textit{MP}}\); accordingly, \({\upeta }^{\textit{MP}}_Z\) can be expressed in terms of \(Z^{\textit{MP}}\) and \(P^{\textit{MP}}\).

  23. 23.

    In the example discussed above, these steady state values are \(Z^{\textit{MP}} = 1569.4, P^{\textit{MP}} = 55.65\). The long run tariff and political investment levels associated with this steady state are \({\uptau }^{\textit{MP}} = 114.8, P^{\textit{MP}} = 4.17\).

  24. 24.

    Details are available in the Appendix 2. If \({\upsigma }_2 > 0\) then \({\upnu }_2 = 0\). In general, illustration of our model would require a four-variable phase diagram. Because both the optimal tariff and investment paths are linearly dependent on Z and P, one can reduce the four-dimensional representation to a two-dimensional phase diagram.

  25. 25.

    All other parameters are the asme as in the base case: (\(k, {\updelta }, {\upbeta }, f\)) = (0.1, 0.1, 1750, 10).

  26. 26.

    It is important to note here that we are not using this empirical approach to discern causality, as in a conventional empirical investigation. Rather, our aim is pattern recognition: we want to see how the various parameters correlate to improved welfare for D. We note that the units of measurement for \({\upbeta }\) and f were adjusted, so as to make the resultant values similar in magnitude to the other parameters: we divided \({\upbeta }\) by 1000 and we dividied f by 10.

  27. 27.

    Starting with Eq. (28), one can solve for \({\upgamma }_3\); then inserting into Eq. (27) yields a solution for \({\upgamma }_2\); then inserting these values into Eq. (26) one may obtain \({\upgamma }_1\). Because the coefficients in Eq. (28) are real there is a real solution for \({\upgamma }_3\), which is used throughout (Press et al. 2007). One way to think of this process is that we are deriving a best-reply to the linear Markov strategy for D defined by \(({\upalpha }_1, {\upalpha }_2, {\upalpha }_3)\). As we noted in footnote 19, this approach implicitly assumes that \(I \ge 0\); for \(P \ge |{\upgamma }_1/{\upgamma }_3| + |{\upgamma }_2/{\upgamma }_3| Z\) we impose \(I=0\).

  28. 28.

    This assumes that the roots are real. Noting that \(Q({\upsigma })\) is a convex function, a necessary and sufficient condition for this to obtain is \(Q(\hat{{\upsigma }}) < 0\), where \(Q^{\prime }(\hat{{\upsigma }}) = 0\), i.e., Q is negative at its minimum point. It is straightforward to see that \(\hat{{\upsigma }} = -({\updelta }- {\upgamma }_3 + k + {\uplambda }{\upalpha }_2)\), so the condition \(Q(\hat{{\upsigma }}) < 0\) can be reduced to the requirement that \((k + {\uplambda }{\upalpha }_2 + {\upgamma }_3 - {\updelta })^2 > 4 {\uplambda }{\upalpha }_3 {\upgamma }_2\); this condition will hold under the assumption that \({\upgamma }_2 > 0\) and \({\upalpha }_3 < 0\).


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Corresponding author

Correspondence to Charles F. Mason.

Additional information

The authors are grateful to Richard Arnott, Hassan Benchekroun, Tom Crocker, Ariel Dinar, Linda Fernandez, George Frisvold, Robert Godby, Jason Shogren and two anonymous referees for their comments on earlier drafts. We also thank seminar participants at 6th Meeting on Game Theory and Practice in Zaragoza, Spain, Heartland Environmental and Resource Economics Workshop in Ames, IA, Western Economic Association International 82nd Annual Conference in Seattle,WA, 10th Occasional Workshop on Environmental and Natural Resource Economics in Santa Barbara, CA, for their helpful comments. All remaining errors are our own.


Appendix 1: Details of Linear Quadratic Model

In this appendix we provide calculations leading to various equations in the text. We start with the description of payoffs. With linear demand the amount purchased when the price in D is p and the tariff is \({\uptau }\) equals \(Q_i^d = a_i - bP\), where \(P = p\) in D and \(P = p - {\uptau }\) in U. At each point in time, market clearing requires that combined quantity demanded, \(Q_u + Q_d\), equals quantity supplied, \(Q = c(p-{\uptau })\). The equilibrium quantity is then \(Q = Q_u + Q_d = a_u + a_d - 2bp +b{\uptau }\). It is easy to see that the resultant equilibrium price is a linear function of the tariff:

$$\begin{aligned} p = \hat{p} + (1-A) {\uptau }, \end{aligned}$$

where \(\hat{p} = \frac{a_d + a_u}{2b+c}\) and \(1-A = \frac{b+c}{2b+c}\).

Consumer surplus in country \(i, \textit{CS}_i\), can be written as

$$\begin{aligned} \textit{CS}_i = \frac{(Q_i)^2}{2b} = \frac{(a_i - bP)^2}{2b}. \end{aligned}$$

Because the supply relation in U is \(Q = cP\), marginal costs in U are Q / c and so total costs are \(\frac{Q^2}{2c} = \frac{c}{2}(p-{\uptau })^2\). Thus, profits earned by sellers in U are \(\frac{c}{2}(p-{\uptau })^2\). We note that this characterization implicitly embodies tariff payments, via the impact of \({\uptau }\) upon the net price received by sellers in U. Accordingly,\(W_u\)—the sum of consumer and profits in U, less tariff payments—equals

$$\begin{aligned} W_u = \frac{\left( a_u - b(p-{\uptau })\right) ^2}{2b} + \left( \frac{c}{2} \right) (p-{\uptau })^2. \end{aligned}$$

Straightforward algebraic manipulation of this expression yields Eq. (3) in the main body of the text. Tariff payments collected by D are \({\uptau }Q_d = {\uptau }(a_d - bp)\), so that the sum of consumer surplus and tariff payments are

$$\begin{aligned} \frac{\left( a_d - bp\right) ^2}{2b} + {\uptau }(a_d - bp). \end{aligned}$$

The optimal static tariff \({\uptau }^0\) maximizes \(W_d\); making use of Eq. (21) we have

$$\begin{aligned} {\uptau }^0 = \frac{AY_0}{b(1-A^2)}, \end{aligned}$$

where \(Y_0 = a_d - b \hat{ p}\), the exports to D in the absence of a tariff. The tariff \({\uptau }^0\) induces a price \(p^0\), and an associated downstream demand \(Q_d^0\); making use of Eq. (21), we have

$$\begin{aligned} p = p^0 + (1-A) ({\uptau }- {\uptau }^0). \end{aligned}$$

Let \(V_d^0\) denote the value of gross payoffs to D based on the tariff \({\uptau }^0\), which we can write in terms of the quantity demanded \(Q_d^0\) (which results from the tariff \({\uptau }^0\)):

$$\begin{aligned} V_d^0 = \frac{(Q_d^0)^2}{2b} + {\uptau }Q_d^0. \end{aligned}$$

Then we can express gross payoffs at any tariff \({\uptau }\) as:

$$\begin{aligned} V_d&= \frac{(Q_d)^2}{2b} + {\uptau }Q_d \\&= V_d^0 + \frac{Q_d^2 - (Q_d^0)^2}{2b} + ({\uptau }- {\uptau }^0) Q_d +{\uptau }^0(Q_d - Q_d^0) \\&= V_d^0 + \frac{(Q_d - Q_d^0)^2}{2b} + \frac{Q_d^0(Q_d - Q_d^0)}{b} + {\uptau }^0(Q_d - Q_d^0) + ({\uptau }- {\uptau }^0) Q_d \end{aligned}$$

It is straightforward to show that the static optimal tariff satisfies \(Q_d^0 + b{\uptau }^0 =\frac{Q_d^0}{1-A}\); making use of this fact, and noting also that \(Q_d - Q_d^0 = -b(1-A)({\uptau }- {\uptau }^0)\), the expression for \(V_d\) can be simplified to

$$\begin{aligned} V_d&= V_d^0 + \frac{b(1-A)^2}{b}\left( {\uptau }- {\uptau }^0\right) ^2 - b(1-A)({\uptau }- {\uptau }^0)^2 \\&= V_d^0 - \frac{b}{2}(1-A^2)({\uptau }- {\uptau }^0)^2. \end{aligned}$$

Subtracting damages \(sZ^2/2\) then yields Downstream flow payoffs \(W_d\), as given by Eq. (2) in the main body of the text.

Appendix 2: Deriving the Coefficients in the Markov Strategies

We obtain the Markov-perfect equilibrium strategies by solving each country’s optimization problem. To solve U’s dynamic optimization problem, we time-differentiate the first-order condition (8), assuming \(I \ge 0\), and D plays the linear strategy given in Eq. (14); we then use Eq. (8) to substitute for \({{\upxi }}_P\) and Eq. (10) to substitute for \({\dot{{\upxi }}}_P\). This approach yields

$$\begin{aligned} \dot{I}=\left( r+{\updelta }-{\uplambda }{{\upalpha }}_3\right) I-{{\upalpha }}_3(Y_0-b(1-A){\uptau }). \end{aligned}$$

Then, using Eqs. (42) and (43), we express \(\dot{I}\) in terms of Z and P:

$$\begin{aligned} \dot{I}&= \left( r+{\updelta }-{\uplambda }{\upalpha }_3\right) ({\upgamma }_1 + {\upgamma }_2Z + {\upgamma }_3P) - {\upalpha }_3[Y_0-b(1-A)({\upalpha }_1 + {\upalpha }_2 Z + {\upalpha }_3P)] \nonumber \\&=\left( r+{\updelta }-{\uplambda }{\upalpha }_3\right) {\upgamma }_1 - {\upalpha }_3[Y_0-b(1-A){\upalpha }_1] + ({\upgamma }_2 + b(1-A){\upalpha }_2 {\upalpha }_3)Z \nonumber \\&\quad + ({\upgamma }_3 + b(1-A){\upalpha }_3^2) P. \end{aligned}$$

The expression for \(\dot{I}\) also satisfies a second equation, which can be obtained by time-differentiating Eq. (15) and then using Eqs. (17)–(18) to substitute for \(\dot{Z}\) and \(\dot{P}\):

$$\begin{aligned} \dot{I} = {\upgamma }_2 (E^0 - {\uplambda }{\upalpha }_1) + {\upgamma }_1 {\upgamma }_3 + {\upgamma }_2 \bigl ( {\upgamma }_3 - ({\uplambda }{\upalpha }_2 + k) \bigr )Z + \bigl ( {\uplambda }{\upalpha }_3 {\upgamma }_2 + {\upgamma }_3 ({\upgamma }_3 - {\updelta }) \bigr ) P. \end{aligned}$$

Comparing Eqs. (24) and (25), one see there are three restrictions corresponding to equating the intercepts, the slopes on Z and the slopes on P:

$$\begin{aligned} {\upgamma }_1&= \frac{(E^0 - {\uplambda }{\upalpha }_1){\upgamma }_2}{r+{\updelta }- {\upgamma }_3 -{\uplambda }{\upalpha }_3} + \frac{ {\upalpha }_3[Y_0-b(1-A){\upalpha }_1]}{r+{\updelta }- {\upgamma }_3 -{\uplambda }{\upalpha }_3} ; \end{aligned}$$
$$\begin{aligned} {\upgamma }_2&= -\frac{(b(1-A){\upalpha }_2 {\upalpha }_3}{1 + {\uplambda }{\upalpha }_2 + k -{\upgamma }_3}; \end{aligned}$$
$$\begin{aligned} {\upgamma }_3^3&- (2 + {\uplambda }{\upalpha }_2 + k + {\updelta }) {\upgamma }_3^2 + \bigl ( ({\updelta }+ 1) (1+ {\uplambda }{\upalpha }_2 + k) - b(1-A) {\upalpha }_3^2 \bigr ) {\upgamma }_3 \nonumber \\&\qquad + b(1-A){\upalpha }_3^2 (1 + 2{\uplambda }{\upalpha }_2 + k) = 0. \end{aligned}$$

For a given vector of parameters \(({\upalpha }_1, {\upalpha }_2, {\upalpha }_3)\) describing D’s strategy, one can solve for the vector of parameters describing U’s strategy.Footnote 27

To solve D’s dynamic optimization problem, we time-differentiate the first-order condition (11), assuming U plays the linear strategy given in Eq. (15); we then use Eq. (12) to replace \(\dot{{\upeta }}_Z\) and Eq. (11) to replace \({\upeta }_Z\):

$$\begin{aligned} \dot{{\uptau }}&= -\frac{{\uplambda }}{b\left( 1-A^2\right) }\dot{{\upeta }}_Z \nonumber \\&= -\frac{{\uplambda }}{b\left( 1-A^2\right) } \bigl \{(r+k) {\upeta }_Z + sZ - [{\upbeta }+ {\upeta }_P] {\upgamma }_2 \bigr \} \nonumber \\&= -(r+k){\uptau }_0 + \frac{{\uplambda }{\upbeta }{\upgamma }_2}{b\left( 1-A^2\right) } + (r+k) {\uptau }- \frac{{\uplambda }s}{b\left( 1-A^2\right) }Z + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } {\upeta }_P. \end{aligned}$$

The next step is to eliminate \({\upeta }_P\) from this expression. To that end, we first time-differentiate Eq. (29) and use Eq. (13) to substitute for \(\dot{{\upeta }}_P\):

$$\begin{aligned} \ddot{{\uptau }}&= (r+k) \dot{{\uptau }}-\frac{{\uplambda }s}{b\left( 1-A^2\right) } \dot{Z} + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } \dot{{\upeta }}_P \\&= (r+k) \dot{{\uptau }}-\frac{{\uplambda }s}{b\left( 1-A^2\right) } \dot{Z} + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } \bigl ( (r+{\updelta }){\upeta }_P - f + P - [{\upbeta }+{\upeta }_P] {\upgamma }_3 \bigr ). \end{aligned}$$

Then, we use Eq. (29) to substitute for \({\upeta }_P\):

$$\begin{aligned} \ddot{{\uptau }}&= (r+k) \dot{{\uptau }}-\frac{{\uplambda }s}{b\left( 1-A^2\right) } \dot{Z} - \frac{{\uplambda }(f + {\upbeta }{\upgamma }_3) {\upgamma }_2}{b\left( 1-A^2\right) } + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } P \\&\quad + (r + {\updelta }- {\upgamma }_3) \bigl [ \dot{{\uptau }} + (r+k) ({\uptau }_0 - {\uptau }) - \frac{{\uplambda }{\upbeta }{\upgamma }_2}{b\left( 1-A^2\right) } + \frac{{\uplambda }s}{b\left( 1-A^2\right) } Z \bigr ] . \end{aligned}$$

Next, we use the linear Markov strategies, in Eqs. (14)–(15) to rewrite I and \({\uptau }\) in terms of Z and P, and \(\dot{I}\) and \(\dot{{\uptau }}\) in terms of \(\dot{Z}\) and \(\dot{P}\); then recalling the equations of motion for Z and P, we obtain an expression for \(\ddot{{\uptau }}\) in terms of Z and P:

$$\begin{aligned} \ddot{{\uptau }}&= \left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] (E^0 - {\uplambda }{\upalpha }_1) + {\upalpha }_3 {\upgamma }_1 (2r + {\updelta }+ k - {\upgamma }_3) \\&\quad - \frac{{\uplambda }{\upgamma }_2}{b (1-A^2 )} \bigl (f + {\upbeta }[1+{\upgamma }_3] \bigr )+ (r+k)(r+{\updelta }- {\upgamma }_3)({\uptau }_0 - {\upalpha }_1) \\&\quad + \left\{ \frac{{\uplambda }s(1 + k + {\uplambda }{\upalpha }_2)}{b\left( 1-A^2\right) } - {\upalpha }_2 \bigr [2r + {\updelta }+ k - {\upgamma }_3 + (r+k)(r+{\updelta }- {\upgamma }_3) \bigr ] \right. \\&\quad +\left. {\upalpha }_3 {\upgamma }_2 (2r + {\updelta }+ k - {\upgamma }_3) \right\} Z \\&\quad + \left\{ \frac{ {\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } - {\upalpha }_3 \left( {\uplambda }\left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] \right. \right. \\&\quad \left. \left. + ({\upgamma }_3 - {\updelta })(2r + {\updelta }+ k - {\upgamma }_3) - (r+k)(r+{\updelta }- {\upgamma }_3) \right) \right\} P. \end{aligned}$$

To complete the solution, we obtain a second expression for \(\ddot{{\uptau }}\) in terms of Z and I. Starting from Eq. (14) and time-differentiating, substituting for \(\dot{Z}\) and \(\dot{P}\), and then time-differentiating again and substituting for \(\dot{Z}\) and \(\dot{P}\) again yields:

$$\begin{aligned} \ddot{{\uptau }}&= {\upalpha }_3 \bigl [ {\upgamma }_2(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1) - {\upgamma }_1( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ] -{\upalpha }_2( {\uplambda }{\upalpha }_2 + k)(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1) \\&\quad - \bigl ( ({\uplambda }{\upalpha }_2 + k)({\upalpha }_3 {\upgamma }_2 - {\upalpha }_2({\uplambda }{\upalpha }_2 + k)) + {\upalpha }_2 {\upalpha }_3 ({\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ) Z \\&\quad - {\upalpha }_3 \bigl ({\uplambda }[ {\upalpha }_3 {\upgamma }_2 - {\upalpha }_2({\uplambda }{\upalpha }_2 + k)] + ({\upalpha }_3 - {\updelta })( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ) P. \end{aligned}$$

The two expressions for \(\ddot{{\uptau }}\) must both apply for all values of Z and P; accordingly, one obtains three restrictions corresponding to equating the intercepts, the slopes on Z and the slopes on P):

$$\begin{aligned}&\left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] (E^0 - {\uplambda }{\upalpha }_1) + (r+k)(r+{\updelta }- {\upgamma }_3)({\uptau }_0 - {\upalpha }_1) \\&\qquad + {\upalpha }_3 {\upgamma }_1 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }{\upgamma }_2}{b (1-A^2 )} \bigl (f + {\upbeta }[1+{\upgamma }_3] \bigr ) \\&\quad ={\upalpha }_3 \bigl [ {\upgamma }_2(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1) - {\upgamma }_1( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ] -{\upalpha }_2( {\uplambda }{\upalpha }_2 + k)(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1); \\ \nonumber \\&\quad \frac{{\uplambda }s(1 + k + {\uplambda }{\upalpha }_2)}{b\left( 1-A^2\right) } - {\upalpha }_2 \bigr [2r + {\updelta }+ k - {\upgamma }_3 + (r+k)(r+{\updelta }- {\upgamma }_3) \bigl ] \,+\, {\upalpha }_3 {\upgamma }_2 (2r + {\updelta }+ k - {\upgamma }_3)\\&\quad = -(({\uplambda }{\upalpha }_2+k)({\upalpha }_3{\upgamma }_2-{\upalpha }_2({\uplambda }{\upalpha }_2+k))+{\upalpha }_2{\upalpha }_3({\uplambda }{\upalpha }_2+{\updelta }-{\upgamma }_3))\\&\quad {\upalpha }_3 \left( {\uplambda }\left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] \right. \nonumber \\&\quad +\left. ({\upgamma }_3 - {\updelta })(2r + {\updelta }+ k - {\upgamma }_3) + (r+k)(r+{\updelta }- {\upgamma }_3) \right) \nonumber \\&\quad - \frac{ {\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } = {\upalpha }_3 \bigl ({\uplambda }\left[ {\upalpha }_3 {\upgamma }_2 - {\upalpha }_2({\uplambda }{\upalpha }_2 + k)\right] + ({\upalpha }_3 - {\updelta })( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ). \end{aligned}$$

These three conditions can be manipulated to isolate expressions for the intercept (\({\upalpha }_1\)), slope on Z (\({\upalpha }_2\)) and the slope on P (\({\upalpha }_3\)) in D’s Markov strategy:

$$\begin{aligned} {\upalpha }_1&= \frac{ {\upalpha }_2 \bigl [2(r+k) + {\updelta }- {\upgamma }_3 + {\uplambda }{\upalpha }_2 \bigr ] E^0 + (r+k)(r+ {\updelta }- {\upgamma }_3) {\uptau }_0}{ {\uplambda }\bigl [2(r+k) + {\updelta }- {\upgamma }_3 + {\uplambda }{\upalpha }_2 \bigr ] {\upalpha }_2 - \frac{{\uplambda }s}{b\left( 1-A^2\right) } - {\upalpha }_3{\upgamma }_2 + (r+k)(r+ {\updelta }- {\upgamma }_3) } \nonumber \\&\quad - \frac{{\uplambda }\bigl (f {+} {\upbeta }(1{+}{\upgamma }_3) \bigr ) {\upgamma }_2}{b(1\!-A^2) \bigl ( {\uplambda }\bigl [2(r\!+k) {+} {\updelta }{-} {\upgamma }_3 {+} {\uplambda }{\upalpha }_2 \bigr ] {\upalpha }_2 \!- \frac{{\uplambda }s}{b\left( 1-A^2\right) } - {\upalpha }_3{\upgamma }_2 + (r+k)(r+ {\updelta }- {\upgamma }_3) \bigr )} \nonumber \\&\quad + \frac{{\upalpha }_3 {\upgamma }_1 [2(r+{\updelta }- {\upgamma }_3) + k(1+{\upgamma }_2) ] - {\upalpha }_2 {\upgamma }_1 k( {\uplambda }{\upalpha }_2 + k)}{ {\uplambda }\bigl [2(r+k) + {\updelta }- {\upgamma }_3 + {\uplambda }{\upalpha }_2 \bigr ] {\upalpha }_2 - \frac{{\uplambda }s}{b\left( 1-A^2\right) } - {\upalpha }_3{\upgamma }_2 + (r+k)(r+ {\updelta }- {\upgamma }_3) } ; \end{aligned}$$
$$\begin{aligned} {\upalpha }_2^2&- \bigl ({\upalpha }_3 + {\upgamma }_3 - 2(r + k + {\updelta }) \bigr ) {\upalpha }_2 + \frac{{\uplambda }s - [b (1-A^2){\upalpha }_3^2 + 1] {\upgamma }_2}{{\uplambda }b {\upalpha }_3(1-A^2)} \nonumber \\&\qquad \qquad + \frac{({\upgamma }_3 - {\updelta })(r + {\upalpha }_3 - {\upgamma }_3) + r(r+k)}{{\uplambda }^2} = 0; \end{aligned}$$
$$\begin{aligned} {\upalpha }_3&= \frac{ {\uplambda }s(1 + k + {\uplambda }{\upalpha }_2) }{b(1-A^2)\bigl (( {\upalpha }_2 + {\upgamma }_2) ({\upgamma }_3 - ({\uplambda }{\upalpha }_2 - {\updelta }) - 2(r+k) {\upgamma }_2\bigr )} \nonumber \\&\quad - \left( \frac{(r+k)(r+{\updelta }- b(1-A^2){\upgamma }_3) + ({\uplambda }{\upalpha }_2 + k)^2 + b(1-A^2)(2r + {\updelta }+ k )}{ b(1-A^2)( {\upalpha }_2 + {\upgamma }_2) ({\upgamma }_3 - ({\uplambda }{\upalpha }_2 - {\updelta }) - 2(r+k) {\upgamma }_2} \right) {\upalpha }_2. \end{aligned}$$

As for was the case when solving for the parameters in U’s strategy, once the rival’s strategy is specified it is straightforward to solve for the parameters in D’s strategy.

The equilibrium Markov strategies for U and D are described by the six coefficients \({\upalpha }_1, {\upalpha }_2, {\upalpha }_3, {\upgamma }_1, {\upgamma }_2\) and \({\upgamma }_3\), which are determined by the six equations (26)–(28) and (30)–(32).

As the system governing the state variables, (17)–(18), is a pair of linear first-order differential equations, its solution is given by the sum of a particular solution, (\(Z^{\textit{MP}}, P^{\textit{MP}}\))—which correspond to the steady values—and the general solution to the system of homogenous differential equations

$$\begin{aligned} \dot{Z}&= -({\uplambda }{\upalpha }_2 + k) Z - {\uplambda }{\upalpha }_3P; \end{aligned}$$
$$\begin{aligned} \dot{P}&= {\upgamma }_2Z + ({\upgamma }_3 - {\updelta })P. \end{aligned}$$

The solution to the system (33)–(34) is of the form (ZP) = (\({\upnu }e^{{\upsigma }t}, {\upmu }e^{{\upsigma }t}\)). Inserting these functions into Eqs. (33) and (34) leads to the characteristic equation

$$\begin{aligned} Q({\upsigma }) \equiv ({\upsigma }+{\uplambda }{{\upalpha }}_2+k)\left( {\upsigma }+{\updelta }-{{\upgamma }}_3\right) + {\uplambda }{{\upalpha }}_3{{\upgamma }}_2=0, \end{aligned}$$

which the parameter \({\upsigma }\) must satisfy. Denote the roots of this equation by \({\upsigma }_1\) and \({\upsigma }_2\), with \({\upsigma }_1\) the smaller root. It is straightforward to see that \({\upsigma }_1<\) min {\(-({\uplambda }{\upalpha }_2 + k), {\upgamma }_3 - {\updelta }\)} and \({\upsigma }_2>\) max{\(-({\uplambda }{\upalpha }_2 + k), {\upgamma }_3 - {\updelta }\)}; accordingly, \({\upsigma }_1\) is negative while \({\upsigma }_2\) could be either negative or positive. If \({\upsigma }_2 < 0\), the solution is a stable node, while if \({\upsigma }_2 > 0\) the solution is saddlepoint stable.Footnote 28 The state variables are subject to transversality conditions that ensure they converge to a steady state; it follows that the solution is (\(Z_g, P_g\)) = (\({\upnu }_1 e^{{\upsigma }_1t}, {\upmu }_1 e^{{\upsigma }_1t}\)) if \({\upsigma }_2> 0 > {\upsigma }_1\), and (\(Z_{g}, P_g\)) = (\({\upnu }_1 e^{{\upsigma }_1t} + {\upnu }_2 e^{{\upsigma }_2t}, {\upmu }_1 e^{{\upsigma }_1t} + {\upmu }_2 e^{{\upsigma }_2t}\)) if \(0> {\upsigma }_{2} > {\upsigma }_{1}\).

The particular solution is found by setting \(\dot{Z}=0=\dot{P}\) in Eqs. (17) and (18), and then solving the resultant system of two equations. By construction, then, the particular solution corresponds to the pair of steady state values for the two state variables:

$$\begin{aligned} Z^{\textit{MP}}&= [({\updelta }- {\upgamma }_3)(E^0 - {\uplambda }{\upalpha }_1) - {\uplambda }{\upalpha }_3 {\upgamma }_1] /\left[ ({\uplambda }{\upalpha }_2 + k)({\updelta }- {\upgamma }_3) + {\uplambda }{\upalpha }_3 {\upgamma }_2 \right] ; \end{aligned}$$
$$\begin{aligned} P^{\textit{MP}}&= [{\upgamma }_2 (E^0 - {\uplambda }{\upalpha }_1) + {\upgamma }_1 ({\uplambda }{\upalpha }_2 + k )/\left[ ({\uplambda }{\upalpha }_2 + k)({\updelta }- {\upgamma }_3) + {\uplambda }{\upalpha }_3 {\upgamma }_2 \right] . \end{aligned}$$

Altogether, the complete solution is

$$\begin{aligned} Z(t)&= {\upnu }_1 e^{{\upsigma }_1t} + {\upnu }_2 e^{{\upsigma }_2t} + Z^{\textit{MP}}, \end{aligned}$$
$$\begin{aligned} P(t)&= {\upmu }_1 e^{{\upsigma }_1t} + {\upmu }_2 e^{{\upsigma }_2t} + P^{\textit{MP}}, \end{aligned}$$

where \({\upnu }_2 = 0 = {\upmu }_2\) if \({\upsigma }_2 > 0\). Upon time-differentiating the right-hand side of Eqs. (38)–(39), and combining with Eqs. (33)–(34), one may derive

$$\begin{aligned} h_1\equiv \frac{{\upnu }_1}{{\upmu }_1} = \frac{{\upgamma }_2}{{\upsigma }_1 + {\updelta }- {\upgamma }_3}; \end{aligned}$$

Because \({\upsigma }_1<\) min{\(-({\uplambda }{\upalpha }_2 + k\)), \({\upgamma }_3 - {\updelta }\)}, it follow that the sign of \(h_1 < 0\) is opposite that of the sign of \({\upgamma }_2\). If in addition \({\upsigma }_2 < 0\), we have

$$\begin{aligned} h_2 \equiv \frac{{\upnu }_2}{{\upmu }_2} = \frac{{\upgamma }_2}{{\upsigma }_2 + {\upgamma }_3 + {\updelta }}. \end{aligned}$$

As \({\upsigma }_2>\) max{\(-({\uplambda }{\upalpha }_2 + k\)), \(-({\upgamma }_3 + {\updelta }\))}, the sign of \(h_2 > 0\) is the same as the sign of \({\upgamma }_2\).

Using these proportionality coefficients, the formulae for the state variables may be rewritten as

$$\begin{aligned} Z(t)&= {\upnu }_1 e^{{\upsigma }_1t} + {\upnu }_2 e^{{\upsigma }_2t} + Z^{\textit{MP}}, \end{aligned}$$
$$\begin{aligned} P(t)&= h_1 {\upnu }_1 e^{{\upsigma }_1t} + h_2 {\upnu }_2 e^{{\upsigma }_2t} + P^{\textit{MP}}. \end{aligned}$$

These expressions must hold at all points in time, in particular at t = 0. Noting that \(Z(0) = Z_0\) and \(P(0) = 0,\) it follows that:

$$\begin{aligned} {\upnu }_1&= [h_2\left( Z_0-Z^{\textit{MP}}\right) +P^{\textit{MP}}]/(h_2 - h_1); \end{aligned}$$
$$\begin{aligned} {\upnu }_2&= -[h_1\left( Z_0-Z^{\textit{MP}}\right) +P^{\textit{MP}}]/(h_2 - h_1). \end{aligned}$$

Combining Eqs. (40)–(45) then completes the solution for (Z, P).

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Mason, C.F., Umanskaya, V.I. & Barbier, E.B. Trade, Transboundary Pollution, and Foreign Lobbying. Environ Resource Econ 70, 223–248 (2018).

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  • Transboundary pollution
  • Differential game
  • Lobbying

JEL Classification

  • D72
  • F18
  • F59
  • Q56