The Impact of Unilateral Commitment on Transboundary Pollution

Abstract

To reach a common target of environmental quality, countries can choose to commit to a stream of pollution abatement right from the beginning of the game or decide upon abatement at each moment of time. Though most of the previous literature studies homogeneous strategies where no country or all countries commit to a (same) predefined policy, reality goes along a different way: some countries make more efforts than others to reduce pollutant emission. The main novelty of this paper resides in the introduction of this kind of heterogeneous strategic behavior currently observed among large pollution nations. We find that the pollution level can be lower under heterogeneous than under homogeneous strategies. A stringent environmental quality target will induce the committed player to produce an abatement effort that more than compensates the free-riding attitude of the non-committed player.

Appendix A: Homogenous strategies:

1.1 A.1 Open-loop strategies

For simplicity, we suppose the two countries are symmetric, commit to an abatement strategy based only upon time t and ignore the actual state of pollution. Player i’s objective function consists in maximizing her utility with respect to u _i (t),

$$ \max_{u_{i}(t)}{\int}_{0}^{\overline{T}} e^{-r_{i} t}\left( -x(t)-{\alpha_{i}\over 2}{u^{2}_{i}} \right)dt+ S(x(\overline{T})),\;\; $$

(12)

subject to the state constraint (1).

It is easy to see that player i’s Hamiltonian function¹⁹ is

$$\begin{array}{@{}rcl@{}} H_{i}(x,u_{i},\lambda_{i}, t)&=&\left( -x(t)-{\alpha_{i}\over 2}{u^{2}_{i}} \right)\\ &&+\lambda_{i}\left[E(t)-(u_{i}+u_{j}^{*}(t))\sqrt{x(t)}-\delta x(t) \right], \end{array} $$

where $u_{j}^{*}$ represents the optimal strategy of variable u _j of player j.

Pontryagin’s maximum principle leads to

$$ u_{i}(t)=-{\lambda_{i}(t)\sqrt{x(t)} \over \alpha_{i}}, $$

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and the costate variable, the shadow value here, is

$$ \dot{\lambda}_{i}(t)=r_{i}\lambda_{i}(t)-{\partial H_{i}\over \partial x}=(r_{i}+\delta)\lambda_{i}+{u_{i}+u_{j}^{*} \over 2\sqrt{x}}\lambda_{i}+1, $$

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with transversality condition

$$ \lim_{t\to \infty} e^{-r_{i} t} \lambda_{i}(t) x(t)=0, \;\;\text{or} \;\; \lambda(T)= S_{x}(x^{*}(T)). $$

(15)

Remark

From Eq. 14, we can see that the natural absorption rate δ has the same effect as the time preference r _i . In the following calculation, we will set δ = 0, i.e., set r _i = r _i + δ, while referring to it when interpreting our results.

Furthermore, supposing symmetry between countries, i.e., r _i = r _j = r, α _i = α _j = α, we have

$$u_{i}(t)=u_{j}(t)=u(t),\; \lambda_{i}(t)=\lambda_{j}(t)=\lambda(t). $$

This symmetry is preserved for the optimal values, such that the above analysis yields the following system

$$ \left\{ \begin{array}{l} u^{*}(t)=-{\lambda(t) \sqrt{x(t)}\over \alpha},\\[.12in] \dot{x}= E(t)+{2\lambda x(t)\over \alpha}-\delta x(t),\\[.12in] \dot{\lambda}= 1+r\lambda -{\lambda^{2}\over \alpha}, \end{array} \right. $$

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with the initial condition and the transversality condition (15) given. The shadow price λ (which comes from a Riccati equation) can be solved explicitly, while pollution state x is a linear equation. Hence, we obtain the explicit solution for the open-loop problem. For the long-run steady state, it yields,

$$ \left\{ \begin{array}{ll} \displaystyle \lambda(t)&= \displaystyle \overline{\lambda}={\alpha r\over 2}-A (<0),\;\;\overline{T}=\infty, \\[.2in]\;\; \displaystyle x_{o}(t) & = \displaystyle e^{{2 \overline{\lambda} \over \alpha}t} \left[x(0)+{{\int}_{0}^{t}} e^{-{2 \overline{\lambda} \over \alpha} s}E(s)ds \right],\;\;\overline{T}=\infty \end{array} \right. $$

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with $A= \sqrt {\alpha + \left ({ r\alpha \over 2}\right )^{2}}$.

While, if $\overline {T}=T<\infty$, the short-run dynamics are given as follows: for 0 < t < T,

$$ \left\{ \begin{array}{l} \displaystyle \lambda(t) = \left\{ \begin{array}{l} \displaystyle {r\alpha\over 2}+A- {2A \over C_{o} e^{2At\over \alpha}+1} , \;\text{if} \;S_{x}(x^{*}(T)) \not= {r\alpha\over 2}-A, \\[.2in] \displaystyle {r\alpha\over 2}-A (=\overline{\lambda} ),\;\text{if} \; S_{x}(x^{*}(T))={r\alpha\over 2}-A, \end{array} \right.\\[.35in] \displaystyle x_{o}(t) \!= \displaystyle e^{{{\int}_{0}^{t}} {2\lambda(s)\over \alpha}ds}\left[ x(0)\,+\, {{\int}_{0}^{t}} e^{-{\int}_{0}^{\tau{2 \lambda(s) \over \alpha} }ds}E(\tau)d\tau\right], \;\;\overline{T}=T\!<\infty, \end{array} \right. $$

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with $\displaystyle C_{o}= {A+\left [S_{x}(x^{*}(T))-{r\alpha \over 2}\right ] \over A- \left [S_{x}(x^{*}(T))-{r\alpha \over 2}\right ]}\;e^{-{2A\over \alpha }T}$.

The above explicit solutions provide the short-run dynamics of the system under the open-loop strategy. Compared with the infinite time horizon, where the shadow price is always a constant over time, if there is a finite time target for the pollution level, S(x(T)), the shadow value, λ(t), changes over time to match this final target. The explicit solution yields

$$\dot{ \lambda}(t) = \left\{ \begin{array}{l} >0, \;\text{if} \;0>S_{x}(x^{*}(T))>{r\alpha\over 2}-A, \\[.2in] \displaystyle 0,\;\text{if} \; S_{x}(x^{*}(T))={r\alpha\over 2}-A, \\[.2in] \displaystyle <0, \;\; \text{if} \;S_{x}(x^{*}(T))<{r\alpha\over 2}-A(<0). \end{array} \right. $$

Recall the shadow value λ measures the unit effect of pollution, hence, it is not surprising to notice that the short-run dynamics depend essentially on the slope of the terminal condition S _x , rather than on the terminal condition itself. Except in a perfect setting where the dynamics are constant and equal to exactly the long-run steady state, we have either over-shooting by imposing a too fast change at the end (∣S _x (x ^∗(T))∣ is too large, $S_{x}(x^{*}(T))<{r\alpha \over 2}-A$) or under-shooting by imposing a too slow change at the end (∣S _x (x ^∗(T))∣ is too small, ${r\alpha \over 2}-A<S_{x}(x^{*}(T))<0$ ). In any of these two cases, the shadow price will not converge to its long-run steady state. Therefore, the short-run target may mislead the shadow value function during the transition, unless it is already at its steady-state level.

We can conclude the above analysis as follows.

Proposition 1

The optimal open-loop strategies yield:

• (I) In the short-run, the dynamics of the shadow value and the level of pollution depend on the terminal condition and the slope of the terminal condition, and are given by Eq. 18 .

• (II) In the long-run, the shadow value is a constant and the level of pollution is changing over time, as noted in Eq. 17 .

  Moreover, there are three possibilities based on the emission of pollution.

  • (II.1) If pollution emission $E=\overline {E}$ is a constant, there is a unique steady state, where states of pollution and abatement are constants and given by

    $$x^{*}_{o}=-{\alpha \overline{E}\over 2\lambda^{*}},\;\; u^{*}_{o}=-{\lambda^{*}\sqrt{x^{*}}\over \alpha}. $$

  • (II.2) If pollution emissions E = E(t)are increasing (or decreasing) over time with constant growth rate g _E ,the pollution state will increase (or decrease) at the same rate g _E while abatement increases(or decreases) at rate ${g_{E}\over 2}$.But the shadow value of the pollution is the same constant as above.

  • (II.3) If pollution emission E = E(t)is neither of the above cases, then there is no balanced growth path nor a steady state.

1.2 A.2 Markovian Nash Equilibrium

Now suppose that both countries can change the abatement strategies based on the pollution state. Then each player is facing the same problem as above by choosing the abatement strategy u _i (t) =Ψ _i (x(t),t) for a given guessing optimal strategy of player j, ${\Phi }_{j}(x, t), \forall (x,t)\in \mathbb {R}_{+}^{2}$. In the sequel, we will employ similar conjecturing methods as in the heterogeneous strategies case to look at one particular symmetric Markovian Nash Equilibrium.

Player i’s Hamiltonian function is

$$\begin{array}{@{}rcl@{}} &&{} H_{i}(x,u_{i}, \phi_{i}, {\Phi}_{j}(x,t), t)=\left( -x(t)-{\alpha_{i}\over 2}{u^{2}_{i}} \right)\\ &&+\phi_{i}\left[E(t)-\left( \!\!\right.u_{i}+{\Phi}_{j}(x,t)\sqrt{x(t)} \right], \end{array} $$

with ϕ _i being player i’s costate variable and Φ _j (x,t) being player i’s guess of player j’s optimal strategy.

Pontryagin’s maximum principle yields

$$ u_{i}(t)=-{\phi_{i}(t)\sqrt{x(t)} \over \alpha_{i}}, $$

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and

$$\begin{array}{@{}rcl@{}} \dot{\phi}_{i}(t)&=&r_{i}\phi_{i}(t)-{\partial H_{i}\over \partial x}=r_{i}\phi_{i}+1+ {u_{i}+{\Phi}_{j}(x,t) \over 2\sqrt{x}}\phi_{i}\\ &&+ \phi_{i} \sqrt{x}{\partial {\Phi}_{j}(x,t)\over \partial x}, \end{array} $$

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and its terminal condition. By conjecturing the optimal strategy of player j, ${\Phi }_{j}(x, t)=-{\phi _{j}(t)\sqrt {x} \over \alpha _{j}}$, together with the symmetry assumption and the search for a symmetric solution, it follows that the optimal strategy and costate variable check

$$ \left\{ \begin{array}{l} u_{i}^{*}(t)=u_{j}^{*}(t)=u^{*}(t)= -{\phi(t)\sqrt{x}\over \alpha} ,\\[.12in] \dot{x}= E(t)+ {2\phi x \over \alpha},\\[.12in] \dot{\phi}= 1+r\phi-{3\phi^{2}\over 2\alpha}, \end{array} \right. $$

(21)

with the initial condition and the transversality condition given.

Using similar methods to the ones used to obtain (18), we could get a similar explicit solution for the short-run and long-run dynamics, especially for the effect of the short-run target on the dynamics.

The long-run explicit solution to the Markovian strategy is given by

$$\begin{array}{@{}rcl@{}} \phi(t)&=&\overline{\phi}={1\over 3}\left( \alpha r- \sqrt{(\alpha r)^{2}+6\alpha} \right)(<0), \;\;\; x_{m}(t)\\ &=&e^{{2 \overline{\phi} \over \alpha}t} \left[x(0)+{{\int}_{0}^{t}} e{-{2 \overline{\phi} \over \alpha} s}E(s)ds \right]. \end{array} $$

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And the short-run dynamics based on the terminal condition are

$$ \left\{ \begin{array}{l} \displaystyle \phi(t) = \left\{ \begin{array}{l} \displaystyle {r\alpha\over 3}+B- {2B \over C_{m} e^{3Bt\over \alpha}+1} , \;\text{if} \;S_{x}(x^{*}(T)) \not= {r\alpha\over 3}-B, \\[.2in] \displaystyle {r\alpha\over 3}-B (=\overline{\lambda} ),\;\text{if} \; S_{x}(x^{*}(T))={r\alpha\over 3}-B, \end{array} \right.\\[.35in] \displaystyle x_{o}(t) = \displaystyle e^{{{\int}_{0}^{t}} {2\phi(s)\over \alpha}ds}\left[ x(0)\,+\, {{\int}_{0}^{t}} e^{-{\int}_{0}^{\tau{2 \phi(s) \over \alpha} }ds}E(\tau)d\tau\right], \;\;\overline{T}=T<\infty, \end{array} \right. $$

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with $\displaystyle C_{m}= {B+\left [S_{x}(x^{*}(T))-{r\alpha \over 3}\right ] \over B- \left [S_{x}(x^{*}(T))-{r\alpha \over 3}\right ]}\;e^{-{3B\over \alpha }T}$ and $B=\sqrt { \left (r\alpha \over 3\right )^{2} +{2\alpha \over 3}}$.

That finishes the proof.