Using real options theory to a country’s environmental policy: considering the economic size and growth

Abstract

The aim of this paper is to consider how the economic size and growth of a country affect its environmental policy under uncertainty in a real options framework. In contrast to the prior literature, this work explicitly takes into account the link between the development of an economy and the pollution state of the environment. Policy implementation is found to be determined by the levels of the economic size and the disutility of the pollution. We illustrate how to apply our method to the implementation of an environmental policy in an actual situation and show with numerical calculations that the optimal threshold is sensitive only to the subjective time preference, while the expected implementation time is affected by other parameters.

Appendices

Appendix 1: Proof of Proposition 1

First, we calculate W _A  = W _A (θ, Y, M), the value function for the “adopt” region ${\mathcal{A}} \subset {\bf R}R^2$ $((\theta, Y) \in {\mathcal{A}} \Leftrightarrow \beta_t = \beta_A).$ When M _t follows the dynamics (2), we have for t > 0

$$ M_t = \hbox{e}^{-\delta t} M_0 + \beta_A \int\limits_0^t \hbox{e}^{\delta(s-t)} Y_s^{\xi} \hbox{d} s. $$

Then, substituting it into (4) and applying Fubini’s theorem, we can directly calculate the value function as

$$ \begin{aligned} W_A &= {\mathbb E}\left[\left.\int\limits_0^{\infty} (-\theta_t M_t) \hbox{e}^{-\phi t} \hbox{d} t\right|Y_0=Y,\;\theta_0=\theta,\;M_0=M\right] \\ &= - {\mathbb E}_0 \left[\int\limits_0^{\infty} \theta_t \left(\hbox{e}^{-\delta t}M_0 + \int\limits_0^t \hbox{e}^{\delta (s - t)}\beta_A Y_s^{\xi} \hbox{d} s \right) \hbox{e}^{- \phi t} \hbox{d} t\right]\\& = -\frac{\theta M}{\phi + \delta - \mu_{\theta}}- \beta_A \int\limits_0^{\infty} \left(\int\limits_0^t \hbox{e}^{\delta s} {\mathbb E}_0 \left[\theta_t Y_s^{\xi}\right] \hbox{d} s\right) \hbox{e}^{-(\phi + \delta) t} \hbox{d} t.\\ \end{aligned}$$

(12)

Using the tower property of conditional expectations (t ≥ s), we have

$$ {\mathbb E}_0[\theta_t Y_s^{\xi}]={\mathbb E}_0[\theta_s\hbox{e}^{\mu_{\theta}(t-s)}Y_s^{\xi}] =\theta Y^{\xi}\hbox{e}^{\mu_{\zeta}s}\hbox{e}^{\mu_{\theta}(t-s)} =\theta Y^{\xi}\hbox{e}^{\left(\xi\mu_Y+\rho\xi\sigma_{\theta}\sigma_Y+\frac{1}{2} \xi(\xi-1)\sigma_Y^2\right)s}\hbox{e}^{\mu_{\theta} t}. $$

Thus, we can derive the third term on the right-hand side of Eq. 12 as

$$ \begin{aligned} & \int\limits_0^{\infty} \left(\int\limits_0^t\hbox{e}^{\delta s}{\mathbb E}_0\left[\theta_t Y_s^{\xi}\right] \hbox{d} s\right)\hbox{e}^{- (\phi + \delta) t} \hbox{d} t \\ & = \theta Y^{\xi} \int\limits_0^{\infty}\left(\int\limits_0^t \hbox{e}^{\left( \xi \mu_Y + \rho \xi \sigma_{\theta} \sigma_Y + \frac{1}{2} \xi (\xi - 1)\sigma_Y^2 + \delta \right) s} \hbox{d} s\right) \hbox{e}^{- (\phi + \delta - \mu_{\theta}) t} \hbox{d} t \\ & = \frac{\theta Y^{\xi}}{\xi \mu_Y + \rho \xi \sigma_{\theta} \sigma_Y + \frac{1}{2} \xi (\xi - 1) \sigma_Y^2 + \delta} \\ & \times \int\limits_0^{\infty} \left(\hbox{e}^{\left(\xi \mu_Y + \rho \xi \sigma_{\theta} \sigma_Y + \frac{1}{2} \xi (\xi - 1) \sigma_Y^2 + \delta\right) t} - 1\right) \hbox{e}^{- (\phi + \delta - \mu_{\theta}) t} \hbox{d} t\\ & = \frac{\theta Y^{\xi}}{(\phi + \delta -\mu_{\theta}) ( \phi - \mu_{\theta} - \xi \mu_Y - \rho \xi \sigma_{\theta} \sigma_Y -\frac{1}{2} \xi (\xi - 1) \sigma_Y^2)}.\\ \end{aligned} $$

As a result, we obtain the value function (8).

Next, we consider the case of the “nonadopt” region. Let W _N  = W _N (θ, Y, M) denote the value function for the nonadopt region $((\theta, Y,) \in {\bf R}R^2 \setminus {\mathcal{A}} \Leftrightarrow \beta_t = \beta_N).$ W _N (θ,Y, M) must satisfy the Bellman equation

$$ \begin{aligned} \phi W_N & = - \theta M + (\beta_N Y^{\xi} - \delta M) \frac{\partial W_N}{\partial M} + \mu_{\theta} \theta \frac{\partial W_N}{\partial \theta} +\mu_Y Y \frac{\partial W_N}{\partial Y}\\ &+\frac{1}{2} \sigma_{\theta}^2 \theta^2 \frac{\partial^2 W_N} {\partial \theta^2} + \rho \sigma_{\theta} \sigma_Y \theta Y \frac{\partial^2 W_N}{\partial \theta \partial Y} + \frac{1}{2} \sigma_Y^2 Y^2 \frac{\partial^2 W_N}{\partial Y^2}.\\ \end{aligned} $$

(13)

Now we conjecture the solution of the above ODE as

$$ W_N = a (\theta Y^{\xi})^{\gamma } + b \theta M + c \theta Y^{\xi}, $$

where a, b, c and γ are some constants to be derived. Substituting (13) into (14), we have

$$ \begin{aligned} & \phi [ a (\theta Y^{\xi})^{\gamma} + b \theta M + c \theta Y^{\xi}]\\ = & - \theta M + (\beta_N Y^{\xi} - \delta M) c \theta\\ & + \mu_{\theta} \theta ( a \gamma \theta^{\gamma - 1} Y^{\gamma \xi} + c M + d Y^{\xi}) + \mu_Y Y (a \gamma \xi \theta^{\gamma} Y^{\gamma \xi - 1} + b + d \xi \theta Y^{\xi - 1})\\ & + \frac{1}{2} a \sigma_{\theta}^2 \theta^2 \gamma (\gamma - 1) \theta^{\gamma - 2} Y^{\gamma \xi} + \rho \sigma_{\theta} \sigma_Y \theta Y (a \gamma^2 \xi \theta^{\gamma - 1} Y^{\gamma \xi - 1} + d \xi Y^{\xi - 1})\\ & + \frac{1}{2} \sigma_Y^2 Y^2 \left[ a \gamma \xi (\gamma \xi - 1) \theta^{\gamma} Y^{\gamma \xi - 2} + \frac{1}{2} \xi (\xi - 1) d \theta Y^{\xi - 2} \right].\\ \end{aligned} $$

(14)

By setting the coefficient of each term to be zero, we obtain the constants a to γ.

First, we deal with the term (Y θ^ξ)^γ in Eq. (14). Setting the coefficient of (Y θ^ξ)^γ to be zero, we obtain

$$\frac{1}{2} (\sigma_{\theta}^2 + 2 \rho \xi \sigma_Y \sigma_{\theta} + \xi^2 \sigma_Y^2) \gamma^2 - \left[ \frac{1}{2} (\sigma_{\theta}^2 + \xi \sigma_Y^2) - (\mu_{\theta} + \xi \mu_Y) \right] \gamma - \phi=0 $$

By the standard real options arguments, γ must be greater than 1. Hence, we have

$$ \gamma = \frac{\frac{1}{2} (\sigma_{\theta}^2 + \xi \sigma_Y^2) - (\mu_{\theta} + \xi \mu_Y) + \sqrt{\left[\frac{1}{2} (\sigma_{\theta}^2 + \xi \sigma_Y^2) - (\mu_{\theta} +\xi \mu_Y) \right]^2 + 2 \phi (\sigma_{\theta}^2 + 2 \rho \xi \sigma_Y \sigma_{\theta} + \xi^2 \sigma_Y^2)}}{\sigma_{\theta}^2 + 2 \rho \xi \sigma_Y \sigma_{\theta} + \xi^2 \sigma_Y^2}. $$

Second, calculating the term of θM yields

$$ b = -\frac{1}{\phi + \delta - \mu_{\theta}}. $$

Third, for the term θY ^ξ, we get

$$ c = \frac{\beta_N}{(\phi + \delta - \mu_{\theta}) (\phi - \mu_{\zeta})}. $$

Consequently, we obtain

$$ W_N = a (\theta Y^{\xi})^{\gamma} - \frac{\theta M}{\phi + \delta - \mu_{\theta}} - \frac{\beta_N \theta Y^{\xi}}{(\phi + \delta - \mu_{\theta}) (\phi - \mu_{\zeta})}. $$

(15)

We still have two unknown parameters, the constant a and the threshold. They can be derived by the two boundary conditions, i.e., the value-matching and the smooth-pasting conditions.

By the definition of ${\mathcal{A}},$ we know that when (θ _t , Y _t ) first reaches $\partial {\mathcal{A}},$ the policy maker will exercise his or her option to adopt the policy. The value-matching condition says that W _N (θ _B , Y _B , M) = W _A (θ _B , Y _B , M) − K for $(\theta_B, Y_B) \in \partial {\mathcal{A}}.$ Therefore, we derive

$$ a (\zeta_B)^{\gamma} - \frac{\beta_N \zeta_B}{(\phi + \delta - \mu_{\theta})(\phi - \mu_{\zeta})} = - \frac{\beta_A \zeta_B}{(\phi + \delta - \mu_{\theta})(\phi - \mu_{\zeta})}-K, $$

where ζ _B : = θ _B Y ^ξ _B . Solving with respect to a, we obtain

$$ a = \frac{1}{(\zeta_B)^{\gamma}} \left( \frac{(\beta_N - \beta_A) \zeta_B}{(\phi + \delta - \mu_{\theta}) (\phi - \mu_{\zeta})} - K \right). $$

The smooth-pasting condition implies that W _N is maximized with respect to the boundary condition ζ _B . In this case, we get

$$ \frac{(1 - \gamma) (\beta_N - \beta_A)}{(\phi + \delta - \mu_{\theta}) (\phi - \mu_{\zeta})} (\zeta_B)^{-\gamma} + \gamma K (\zeta_B)^{- \gamma - 1} = 0. $$

We finally obtain

$$ \zeta_B = \frac{\gamma (\phi + \delta - \mu_{\theta}) (\phi - \mu_{\zeta}) K}{(\gamma - 1) (\beta_N - \beta_A)} $$

(16)

and

$$ a = \left(\frac{\gamma - 1}{K}\right)^{\gamma - 1} \left( \frac{(\beta_N - \beta_A)}{\gamma (\phi + \delta - \mu_{\theta}) (\phi - \mu_{\zeta})} \right)^{\gamma}. $$

(17)

Substituting (16) and (17) into (15), we obtain (7). $\hfill\square$

Appendix 2: Proof of Corollary 1

Note that γ satisfies the characteristic equation as

$$ \frac{1}{2} (\underbrace{\sigma_{\theta}^2 + 2 \rho \xi \sigma_Y \sigma_{\theta} + \xi^2 \sigma_Y^2}_{= \sigma_{\zeta}^2}) \gamma^2 - \left[\underbrace{\frac{1}{2} (\sigma_{\theta}^2 + \xi \sigma_Y^2) - (\mu_{\theta} + \xi \mu_Y)}_{= \mu_{\zeta} - \frac{1}{2} \sigma_{\zeta}^2} \right] \gamma - \phi=0. $$

(18)

Then, by rearranging (18), we have the relation

$$ \frac{\gamma}{\gamma - 1} (\phi - \mu_{\zeta}) = \frac{1}{2} \sigma_{\zeta}^2 \gamma + \phi, $$

and

$$ \zeta_B = \left(\frac{1}{2} \sigma_{\zeta}^2 \gamma + \phi \right) (\phi + \delta - \mu_{\theta}) \frac{K}{\beta_N - \beta_A}. $$

(19)

The derivative with respect to ϕ From (19), we calculate

$$ \begin{aligned} \frac{\partial \zeta_B}{\partial \phi} = & \frac{K} {\beta_{N}-\beta_{A}} \frac{\partial}{\partial \phi} \left\{ \left(\frac{1}{2}\sigma^{2}_{\zeta}\gamma+\phi \right) (\phi+\delta-\mu_{\theta}) \right\} \\ = & \frac{K}{\beta_{N}-\beta_{A}} \left\{ \left(\frac{1}{2}\sigma^{2}_{\zeta} \frac{\partial \gamma} {\partial \phi} \right) (\phi + \delta - \mu_{\theta})+\left( \frac{1}{2}\sigma^{2}_{\zeta}\gamma +\phi\right) \right\}.\\ \end{aligned} $$

Since γ is a positive root of (18), we verify that $\frac{\partial \gamma}{\partial \phi} > 0$ , and that ζ _B is increasing in ϕ.

The derivative with respect to μ _Y : We have

$$ \frac{\partial \zeta_B}{\partial \mu_Y} =\frac{1} {2}\sigma^{2}_{\zeta} (\phi+\delta-\mu_{\theta}) \frac{K} {\beta_{N}-\beta_{A}} \frac{\partial \gamma}{\partial \mu_Y}. $$

Now the second inequality of (9) readily follows from noting that $\frac{\partial \gamma}{\partial \mu_Y} < 0.$

The derivative with respect to μ_θ: Eq. (19) yields

$$ \frac{\partial \zeta_B}{\partial \mu_{\theta}} = \frac{1}{2}\sigma^{2}_{\zeta} (\phi+\delta-\mu_{\theta}) \frac{K}{\beta_{N}-\beta_{A}} \frac{\partial \gamma}{\partial \mu_{\theta}} - \left(\frac{1}{2} \sigma_{\zeta}^2 \gamma + \phi \right) \frac{K}{\beta_N - \beta_A}. $$

Since ${\frac{{\partial \gamma }}{{\partial \mu _{\theta } }}} < 0,$ we have the third inequality of (9).

Inequalities in (10): When the correlation coefficient ρ is positive, we observe that

$$ \frac{\partial \gamma}{\partial \sigma_Y} > 0 \quad \hbox{and} \quad \frac{\partial \gamma}{\partial \sigma_{\theta}} > 0. $$

By a similar procedure to the above, we get (10).$\square$