The Optimal Timing of Adoption of a Green Technology

Abstract

We study the optimal timing of adoption of a cleaner technology and its effects on the rate of growth of an economy in the context of an AK endogenous growth model. We show that the results depend upon the behavior of the marginal utility of environmental quality with respect to consumption. When it is increasing, we derive the capital level at the optimal timing of adoption. We show that this capital threshold is independent of the initial conditions on the stock of capital, implying that capital-poor countries tend to take longer to adopt. Also, country-specific characteristics, as the existence of high barriers to adoption, may lead to different capital thresholds for different countries. If the marginal utility of environmental quality decreases with consumption, a country should never delay adoption; the optimal policy is either to adopt immediately or, if adoption costs are “too high”, to never adopt. The policy implications of these results are discussed in the context of the international debate surrounding the environmental political agenda.

Appendices

Appendix

Determination of V(T)

Derivation of (18)

 

Following Seierstad and Sydsaeter (1997), p. 196, an additional condition at the jump point has to hold. Given the magnitude of the jump, we may write that

$$ K_{T_{+}}-K_{T_{-}}=g(K_{T_{-}},\beta,T)=(\beta-1)K_{T_{-}} $$

Then, at the jump point, we have

$$ \lambda_{T_{+}}-\lambda_{T_{-}}=-\lambda_{T_{+}}\frac{\partial g(K_{T_{-}},\beta, T)}{\partial K_{T_{-}}} $$

that is,

$$ \lambda_{T_{+}}-\lambda_{T_{-}}=-\lambda_{T_{+}}(\beta-1) $$

which is equivalent to (18).

Expression for K _t .

Utility can be represented by a function of K _t and ${C_{t}/K_{t}}$, that is,

$$ V=\int_{0}^{\infty}\frac{C_{t}^{1-\sigma}(AK_{t})^{-\alpha\mu(1-\sigma)} a^{\mu (1-\sigma)}-1}{1-\sigma}e^{-\rho t}\,dt=\int_{0}^{\infty}\frac{\left(C_{t}/K_{t}\right)^{1-\sigma}K_{t} ^{(1-\alpha\mu)(1-\sigma)}A^{-\alpha\mu(1-\sigma)}a^{\mu(1-\sigma)}-1} {1-\sigma}e^{-\rho t}\,dt $$

From (2), and substituting (20), we may write the level of K _t at any moment before adoption as

$$ K_{t}=K_{0}e^{\int_{0}^{t}\left(A-\frac{\frac{\sigma-1}{\sigma}A+ \frac{\rho}{\sigma(1-\alpha\mu)}}{1+e^{-\Lambda(T-\tau)}\left[\delta ^{\frac{(1-\sigma)}{\sigma}}-1\right]}\right)d\tau} $$

Solving the integral we obtain

$$ K_{t}=K_{0}e^{At}\left[\frac{1+e^{-\Lambda(T-t)}\left[\delta^ {\frac{(1-\sigma)}{\sigma}}-1\right]}{e^{-\Lambda(T-t)}\left[\delta ^{\frac{(1-\sigma)}{\sigma}}-1\right]+e^{\Lambda t}}\right] ^{1/(1-\alpha\mu)} $$

(A1)

Notice that the primitive of ${\frac{\frac{\sigma-1}{\sigma}A+ \frac{\rho}{\sigma(1-\alpha\mu)}}{1+e^{-\Lambda(T-\tau)} \left[\delta^{\frac{(1-\sigma)}{\sigma}}-1\right]}}$ is given by ${\left[-\frac{1}{1-\alpha\mu}\ln\left(e^{-\Lambda T}\left[\delta^ {\frac{(1-\sigma)}{\sigma}}-1\right]+e^{-\Lambda\tau}\right)\right]}$ Thus, with adoption at T, total welfare is given by

$$ \eqalign{ V(T)=& \int_{0}^{T}\frac{(C_{t}/K_{t})^{1-\sigma}K_{t}^{(1-\alpha\mu) (1-\sigma)}A^{-\alpha\mu(1-\sigma)}a_{0}^{\mu(1-\sigma)}-1}{1-\sigma}e^{-\rho t}\,dt+ \cr & +\int_{T}^{\infty}\frac{\left(C_{t}/K_{t}\right)^{1-\sigma}K_{t} ^{(1-\alpha\mu)(1-\sigma)}(A)^{-\alpha\mu(1-\sigma)}a_{1}^{\mu (1-\sigma)}-1}{1-\sigma}e^{-\rho(t-T)}\,dt- \cr & -e^{-\rho T}(\gamma\upsilon+f) }<!endaligned> $$

where K _t and ${C_{t}/K_{t}}$ in the first integral must be substituted by the expressions in (A1) and (20), respectively. The second integral is given by ${\varphi(K_{T_{+}})}$ where ${K_{T_{+}}=\beta K_{T_{-}}}$ and ${K_{T_{-}}}$ is given by (A1). After some algebra we obtain expression (21).

Proposition 1

 

Proof The derivative of V(T) is ${\frac{\partial V}{\partial T}=a_{0}^{\mu(1-\sigma)}K_{0}^{(1-\alpha\mu)(1-\sigma)}L\Gamma +e^{-\rho T}\rho(\gamma\upsilon+f)}$, where Γ is defined as follows

$$ \Gamma=\sigma\left[1+e^{-\Lambda T}\left[\delta^{\frac{(1-\sigma)} {\sigma}}-1\right]\right]^{\sigma-1}e^{-\Lambda T}(-\Lambda) \left[\delta^{\frac{(1-\sigma)}{\sigma}}-1\right] $$

Recall that ${\Lambda > 0}$ to satisfy the transversality condition on capital (9), and that ${\left[1+e^{-\Lambda T}\left[\delta ^{\frac{(1-\sigma)}{\sigma}}-1\right]\right] > 0.}$

For ${\delta < 1}$ and

• (i) ${\sigma < 1: L > 0}$ and ${\Gamma > 0\Longrightarrow \frac{\partial V(T)} {\partial T} > 0}.$

• (ii) ${\sigma > 1:L < 0}$ and ${\Gamma < 0\Longrightarrow \frac{\partial V(T)}{\partial T} > 0}$.

Therefore, the country never adopts. □

Proposition 2

 

Proof The derivative of V(T) is ${\frac{\partial V}{\partial T}=e^{-\rho T}F(T)}$ where

$$ F(T)=a_{0}^{\mu(1-\sigma)}K_{0}^{(1-\alpha\mu)(1-\sigma)}L\Gamma e^{\rho T}+\rho(\gamma \upsilon+f) $$

and ${\gamma}$ is defined above. Also, recall that costs decrease at the rate ${\rho}$ in present value terms.

Let ${\delta > 1}$. In this case, ${\frac{\partial V}{\partial T}}$ cannot be unambiguously signed, as the first term in F(T) is negative and the second is positive. Thus, we look at the behavior of ${\frac{\partial^{2}V}{\partial T^{2}}=e^{-\rho T} \left(\frac{\partial F}{\partial T}-\rho F(T)\right)}$. Moreover,

$$ \frac{\partial F}{\partial T}=e^{\rho T}\Gamma a_{0}^{\mu(1-\sigma)} K_{0}^{(1-\alpha\mu)(1-\sigma)}L\frac{1-\sigma}{\sigma}\left [((1-\alpha\mu)A-\rho)+\sigma\Lambda\frac{e^{-\Lambda T}\left(\delta^ {\frac{(1-\sigma)}{\sigma}}-1\right)}{1+e^{-\Lambda T}\left(\delta^{\frac{(1-\sigma)} {\sigma}}-1\right)}\right] $$

where ${((1-\alpha\mu)A-\rho) > 0}$, ${\Lambda > 0}$ and ${\left[1+e^{-\Lambda T}\left[\delta^{\frac{(1-\sigma)}{\sigma}}-1\right] \right] > 0.}$

• (i) ${\sigma < 1\Rightarrow L > 0, \Gamma < 0}$ and ${(\delta^{\frac{(1-\sigma)}{\sigma}}-1) > 0}$. Thus, in this case, ${{\partial F}/{\partial T} < 0}$, implying that benefits decrease slower than costs. Also, F(0) can be either positive or negative and ${\lim\limits_{T\rightarrow+\infty}F(T)=-\infty.}$ Thus, ${{\partial V}/{\partial T}}$ may be positive or negative for small values of T but it is always negative for large enough T and goes to 0 as ${T\rightarrow+\infty}$. Therefore, we conclude that:

  • (a) if ${\partial V/\partial T > 0}$ at T = 0 this implies that ${F(0) > 0}$ and F(T) decreases with T. At some $t=T^{\ast}, F(T^{\ast})=0{\Rightarrow\frac{\partial V(T^{\ast})}{\partial T}=0}$ and ${({\partial^{2}V(T^{\ast})}/{\partial T^{2}}) < 0,}$ then ${\partial V/\partial T}$ becomes negative. Thus, ${V(T^{\ast})}$ is a maximum, as in Figure 1, and delaying is optimal;

  • (b) if ${\partial V/\partial T < 0}$ at T = 0, then as F(T) decreases with T, ${\partial V/\partial T < 0}$ for all T, as illustrated in Figure 2, and the country adopts immediately.

• (ii) ${\sigma > 1\Rightarrow L < 0}$, ${\Gamma > 0}$ and ${(\delta ^{\frac{(1-\sigma)}{\sigma}}-1) < 0}$. Thus, the term outside the square brackets in ${{\partial F}/{\partial T}}$ is positive but the terms inside the square brackets have opposite signs. Recall that ${0 < (1+e^{-\Lambda T}(\delta^{\frac{(1-\sigma)}{\sigma}}-1)) < 1}$, and ${(1-\alpha\mu)A-\rho > 0}$. We show that the term inside the square brackets increases with T,

  $$ \frac{\partial\left[((1-\alpha\mu)A-\rho)+\sigma\Lambda\frac{e^{-\Lambda T}(\delta^{\frac{(1-\sigma)}{\sigma}}-1)}{1+e^{-\Lambda T}(\delta^{\frac{(1-\sigma)}{\sigma}}-1)}\right]}{\partial T}=\frac{-\sigma\left(-\Lambda\right)^{2}e^{-\Lambda T}(\delta^{\frac{(1-\sigma)}{\sigma}}-1)}{\left[1+e^{-\Lambda T}(\delta^{\frac{(1-\sigma)}{\sigma}}-1)\right]^{2}} > 0 $$

  At T = 0, we have

  $$ \frac{\partial F}{\partial T}_{\mid T=0}=(1-\delta^{\frac{-(1-\sigma)}{\sigma}})A(1-\alpha\mu) \sigma+(1-\alpha\mu)A-\rho))\delta^{\frac{-(1-\sigma)}{\sigma}} $$

  From (19), in order to have ${\frac{C}{K} < A}$, this last expression has to be positive, implying that ${\frac{\partial F}{\partial T} > 0}$ for all T. Thus, benefits decrease faster than costs.

  F(0) can be either positive or negative and ${\lim\limits_{T\rightarrow+\infty}F(T)=\rho(\gamma\upsilon+f) > 0.}$ Thus, ${{\partial V}/{\partial T}}$ may be positive or negative for small values of T and goes to 0 as ${T\rightarrow+\infty}$. Therefore, we conclude that:

  • (a) if ${\frac{\partial V}{\partial T} > 0}$ at T = 0, then it will always be positive, decreasing to zero as ${T\rightarrow\infty}$, and the country will never adopt;

  • (b) if ${\frac{\partial V}{\partial T} < 0}$ at T = 0, as F(T) increases with T, it will keep increasing. At some ${t=T^{\ast},0 < T^{\ast} < \infty}$, ${F(T^{\ast})=0}$ and ${\frac{\partial ^{2}V(T^{\ast})}{\partial T^{2}} > 0}$, before becoming positive. Therefore, V(T) reaches its minimum at ${T^{\ast}}$. Therefore, the country adopts immediately if ${V(0) > V(\infty)}$ or never adopts. □

Proposition 3

 

Proof. Evaluating (A1) at ${t=T^{\ast}}$ and rearranging terms, we obtain that at ${T^{\ast}}$,

$$ {\small K}_{T_{-}^{\ast}}^{\ast}=K_{0}{e}^{AT}\left[\frac{e^{-\Lambda T^{\ast}}\delta^{\frac{(1-\sigma)}{\sigma}}}{1+e^{-\Lambda T^{\ast}}\left[\delta^{\frac{(1-\sigma)}{\sigma}}-1\right]}\right] ^{\frac{1}{(1-\alpha\mu)}}{\small=K}_{0}\left[\frac{e^{\frac{(1-\sigma)} {\sigma}\left[(1-\alpha\mu)A-\rho\right]T^{\ast}}\delta^{\frac{ (1-\sigma)^{2}}{\sigma}}}{\left[1+e^{-\Lambda T^{\ast}}\left[\delta^ {\frac{(1-\sigma)}{\sigma}}-1\right]\right]^{(1-\sigma)}}\right] ^{\frac{1}{(1-\sigma)(1-\alpha\mu)}} $$

Let ${\delta > 1}$, ${\sigma < 1}$, and ${0 < T^{\ast} < \infty}$. Rearranging (23), we obtain

$$ \frac{e^{\frac{(1-\sigma)}{\sigma}\left[(1-\alpha\mu)A-\rho\right] T^{\ast}}}{\left[1+e^{-\Lambda T^{\ast}}\left[\delta^{\frac{(1-\sigma)} {\sigma}}-1\right]\right]^{1-\sigma}}=\frac{\rho(\gamma\upsilon+f)} {a_{0}^{\mu(1-\sigma)}K_{0}^{(1-\alpha\mu)(1-\sigma)}L\left[\delta^ {\frac{(1-\sigma)}{\sigma}}-1\right]\Lambda\sigma} $$

Substituting in the expression for ${K_{T_{-}^{\ast}}^{\ast}}$, we obtain (24).

Moreover, let us denote ${\Phi}$ by

$$ \Phi=\left[L^{-1}\frac{\rho(\gamma\upsilon+f)\delta^{\frac{(1-\sigma) ^{2}}{\sigma}}}{\left[\delta^{\frac{(1-\sigma)}{\sigma}}-1\right] \sigma \Lambda}\right]^{\frac{1}{(1-\sigma)(1-\alpha\mu)}} > 0 $$

Then,

$$ \frac{\partial K_{T_{-}^{\ast}}^{\ast}}{\partial f}=a_{0}^{\frac{-\mu}{(1-\alpha\mu)}}\Phi\frac{(\gamma\upsilon+f) ^{-1}}{(1-\sigma)(1-\alpha\mu)} > 0 $$

$$ \frac{\partial K_{T_{-}^{\ast}}^{\ast}}{\partial \upsilon}= a_{0}^{\frac{-\mu}{(1-\alpha\mu)}}\Phi\left[\frac{(\gamma\upsilon+f) ^{-1}\gamma}{(1-\sigma)(1-\alpha\mu)}\right] > 0 $$

$$ \frac{\partial K_{T_{-}^{\ast}}^{\ast}}{\partial \gamma}=a_{0}^ {\frac{-\mu}{(1-\alpha\mu)}}\Phi\frac{(\gamma\upsilon+f)^{-1}\upsilon} {(1-\sigma)(1-\alpha\mu)} > 0 $$

$$ \frac{\partial K_{T_{-}^{\ast}}^{\ast}}{\partial a_{0}}=a_{0}^ {\frac{-\mu}{(1-\alpha\mu)}-1}\left(-\frac{\mu}{(1-\alpha\mu)}\right) \Phi < 0 $$