A Tradable Permit System in an Intertemporal Economy

Abstract

It is known that in an intertemporal competitive economy, a tradable permit system may not achieve efficiency without setting appropriate permit interest rates (i.e., rewards for holding permits). To find the rates, however, we need to know in advance the path of efficient permit prices, which is difficult to obtain. This study intends to solve this problem in two ways. First, we analyze a special case in which the permit interest rates are given by a simple rule. For example, if the marginal abatement cost of pollution emission is constant, then the appropriate rate is to equal the monetary interest rate. As is the case for global warming, if the damage is caused in the future far beyond the planning period of the environmental program, the appropriate rate coincides with the marginal self-recovery of environmental stock under certain conditions. As a second approach, we propose a tradable permit system with a permit bank, as a mechanism by which the permit interest rates are generated endogenously without governmental intervention other than the issuance of permits. However, we also show that this approach raises the problem of indeterminacy of the equilibrium.

Appendix: Proofs

1.1 Proposition 2.1

The proof proceeds in three steps. First, we characterize the choices of individual firms at equilibrium. Second, we relate the aggregated values of individual firms’ choices to the representative firm. Finally, we verify that the aggregated values are the solution to the profit maximization problem of the representative firm.

Lemma 6.1

At equilibrium, the following hold:

$$\begin{aligned}&\text{ Arbitrage} \text{ condition}\text{:} a(t)=r(t)-\frac{\dot{p}(t)}{p(t)},\end{aligned}$$

(6.1)

$$\begin{aligned}&\text{ Market} \text{ clearing} \text{ condition}\text{:} \sum _{i=1}^{m}y^{ie}(t)=0, \end{aligned}$$

(6.2)

and, for each firm $i$, profit maximization condition:

$$\begin{aligned} F_{k}^{i}(k^{ie}(t),x^{ie}(t))&= r(t),\end{aligned}$$

(6.3)

$$\begin{aligned} F_{x}^{i}(k^{ie}(t),x^{ie}(t))&= p(t),\end{aligned}$$

(6.4)

$$\begin{aligned} \int \limits _{0}^{\infty }p(t)y^{ie}(t)\exp \left( -\int \limits _{0}^{t}r(s)ds\right) dt&= \int \limits _{0}^{\infty }p(t)x^{ie}(t)\exp \left( -\int \limits _{0}^{t}r(s)ds\right) dt-p(0)B_{0}^{i}, \nonumber \\ \end{aligned}$$

(6.5)

where (6.5) implies that the no-Ponzi game condition in (2.12) holds with equality.

Proof

For firm $i$, the current value Hamiltonian is given by:

$$\begin{aligned} H^{i}(k,x,y,b,\zeta ^{i})=\left[ F^{i}(k,x)-r(t)k-p(t)y\right] +\zeta ^{i}\left[ a(t)b-x+y\right] . \end{aligned}$$

The first order conditions are:

$$\begin{aligned} \text{(a)} F_{k}^{i}(k^{ie}(t),x^{ie}(t))&= r(t); \text{(b)} F_{x} ^{i}(k^{ie}(t),x^{ie}(t))=\zeta ^{i}(t), \text{(c)} \zeta ^{i} (t)=p(t),\end{aligned}$$

(6.6)

$$\begin{aligned} \dot{\zeta }^{i}(t)&= \zeta ^{i}(t)\left[ r(t)-a(t)\right] . \end{aligned}$$

(6.7)

The transversality condition is:

$$\begin{aligned} \lim _{t\rightarrow \infty }\zeta ^{i}(t)b^{i}(t)\exp \left( -\int \limits _{0} ^{t}r(s)ds\right) =0. \end{aligned}$$

(6.8)

(6.6) implies (6.3) and (6.4). (6.6 c) and (6.7) imply the arbitrage condition (6.1). (6.1) is rewritten as:

$$\begin{aligned} \exp \left( -\int \limits _{0}^{t}a(s)ds\right) =\frac{p(t)}{p(0)}\exp \left( -\int \limits _{0}^{t}r(s)ds\right) . \end{aligned}$$

(6.9)

Substitute (6.6 c) and (6.9) into (6.8) to obtain:

$$\begin{aligned} \lim _{t\rightarrow \infty }p(0)b^{i}(t)\exp \left( -\int \limits _{0}^{t}a(s)ds\right) =0, \end{aligned}$$

(6.10)

which shows that the no-Ponzi game condition in (2.12) holds at equilibrium. The initial value problem (2.11) is solved as:

$$\begin{aligned} b^{i}(T)=\left[ \int \limits _{0}^{T}(-x^{ie}(t)+y^{ie}(t))\exp \left( -\int \limits _{0} ^{t}a(s)ds\right) dt+B_{0}^{i}\right] \exp \left( \int \limits _{0}^{T}a(t)dt\right) .\qquad \end{aligned}$$

(6.11)

Then, substitute (6.11) into (6.10) to obtain:

$$\begin{aligned} \lim _{t\rightarrow \infty }p(0)\left[ \int \limits _{0}^{T}(-x^{ie}(t)+y^{ie}(t))\exp \left( -\int \limits _{0}^{t}a(s)ds\right) dt+B_{0}^{i}\right] =0. \end{aligned}$$

(6.12)

Then, substitute (6.9) into (6.12) and arrange to obtain (6.5). Finally, the market clearing condition (6.2) must hold at equilibrium. $\square$

Lemma 6.2

For each point in time $t$, (i) the equilibrium firms’ choice $\{k^{ie} (t),x^{ie}(t)\}_{i=1}^{m}$ is the solution for the problem (2.1). (ii) The aggregate production function satisfies:

$$\begin{aligned} F_{K}\left( K^{e}(t),X^{e}(t\right) )=r(t) \text{ and} F_{X}\left( K^{e}(t),X^{e}(t\right) )=p(t), \end{aligned}$$

(6.13)

where $K^{e}(t)=\sum _{i=1}^{m}k^{ie}(t)$ and $X^{e}(t)=\sum _{i=1}^{m} x^{ie}(t)$.

Proof

(i) The Lagrangian associated with the problem on the right-hand side of (2.1) is defined by:

$$\begin{aligned} L(K,X)=\sum _{i=1}^{m}F^{i}(k^{i},x^{i})+\xi \left( K-\sum _{i=1}^{m} k^{i}\right) +\eta \left( X-\sum _{i=1}^{m}x^{i}\right) . \end{aligned}$$

(6.14)

By the concavity of $F^{i}$ and the Inada condition, the necessary and sufficient condition for the solution is given by $K=\sum _{i=1}^{m}k^{i}$, $X=\sum _{i=1}^{m}x^{i}$,

$$\begin{aligned} F_{k}^{i}(k^{i},x^{i})&= F_{k}^{j}(k^{j},x^{j})=\xi \text{ and} \nonumber \\ F_{x}^{i} (k^{j},x^{j})&= F_{x}^{j}(k^{j},x^{j})=\eta \text{ for} \text{ all} i,j\in \{1,2,\ldots ,m\}. \end{aligned}$$

(6.15)

This in fact holds by (6.3) and (6.4). (ii) By the envelope theorem:

$$\begin{aligned} F_{K}\left( K^{e}(t),X^{e}(t\right) )=\xi \text{ and} F_{X}\left( K^{e}(t),X^{e}(t\right) )=\eta . \end{aligned}$$

(6.16)

Then, (6.13) follows from (6.3), (6.4), (6.15) and (6.16).

Proof of Proposition 2.1

(a) The representative firm’s problem (2.13) is formally the same as the individual firm’s problem (2.9). As the problem is a concave program, the first order condition and the transversality condition are sufficient for optimality. As shown in (6.13) in Lemma 6.2, $\left( K^{e}(t),X^{e}(t\right) )$ satisfies the first order condition. The aggregate balance of permits $B^{e}(t)=\sum _{i=1}^{m}b^{ie}(t)$ satisfies the no-Ponzi game condition in (2.13) with equality because each firm satisfies the same condition. This implies that the transversality condition is satisfied as follows:

$$\begin{aligned} \lim _{t\rightarrow \infty }p(0)\exp \left( -\int \limits _{0}^{t}a(s)ds\right) B^{e}(t)=\lim _{t\rightarrow \infty }p(t)\exp \left( -\int \limits _{0}^{t}r(s)ds\right) B^{e}(t)=0,\qquad \quad \end{aligned}$$

(6.17)

where the first equality follows from (6.9). The market clearing condition determines the aggregate permit trade as $Y^{e}(t)=\sum _{i=1} ^{m}y^{ie}(t)=0$. (b) From the proof so far, it is obvious that the results hold independent of the permit distribution, as far as all $m$ firms exist at equilibrium. This condition holds because of the concavity of the production function and the Inada condition, even if a firm is not endowed with any permits and has to buy permits at each point in time. That is, such a firm can gain a nonnegative profit as seen from:

$$\begin{aligned} F^{i}(k,x)=F^{i}(k,x)-F^{i}(0,0)\ge F_{K}^{i}(k,x)k+F_{X}^{i}(k,x)x=rk+px. \end{aligned}$$

(6.18)

1.2 Proposition 2.2

When the equilibrium path is optimal, i.e., $(c^{e}(t),X^{e}(t),K^{e} (t))=(c^{*}(t),X^{*}(t),K^{*}(t))$ and the equilibrium prices $(r^{e}(t),p^{e}(t))$ are given as in the first order conditions for the firm (2.19), the first order conditions for the households (2.21) are satisfied with $\lambda (t)=\lambda ^{*}(t)$. Then, what we have to prove here is that the household satisfies the transversality conditions (2.21 c) and the no-Ponzi game condition in (2.17). Because $m^{e}(t)=K^{*}(t)$ at equilibrium, this follows from the transversality condition for the social planner’s problem (2.8) and from (2.8):

$$\begin{aligned} \lim _{t\rightarrow \infty }e^{-\rho t}\lambda ^{*}(t)K^{*}(t)=\lim _{t\rightarrow \infty }e^{-\rho t}\lambda ^{*}(t)m(t)=\lambda ^{*}(0)\lim _{t\rightarrow \infty }m(t)\exp \left[ -\int \limits _{0}^{t}r(s)ds\right] =0, \nonumber \\ \end{aligned}$$

(6.19)

where the second equality is obtained from (2.21 b).

1.3 Proposition 2.3

The proof is quite similar to the proof of Proposition 2.2. Therefore, we address the differences. Suppose that an equilibrium path is optimal, i.e., $(c^{e}(t),X^{e}(t),K^{e}(t),S^{e}(t))=(c^{*}(t),X^{*}(t),K^{*}(t),S^{*}(t))$. By (2.19 b) and (2.34), we have:

$$\begin{aligned} p^{e}(t)=F_{X}(K^{*}(t),X^{*}(t))=\frac{\xi ^{*}(t)}{\lambda ^{*}(t)}A_{X}(S^{*}(t),X^{*}(t)). \end{aligned}$$

(6.20)

Substitute this and (2.21 b) into (2.19 c) to obtain:

$$\begin{aligned} a(t)=\left( \rho -\frac{\dot{\lambda }^{*}(t)}{\lambda ^{*}(t)}\right) -\left( \frac{\dot{\xi }^{*}(t)}{\xi ^{*}(t)}+\frac{d\ln \left[ A_{X}(S^{*}(t),X^{*}(t))\right] }{dt}-\frac{\dot{\lambda }^{*} (t)}{\lambda ^{*}(t)}\right) . \end{aligned}$$

(6.21)

From (2.36), we have (2.39). Notice that $\xi ^{*} (t)=-V_{S}(K^{*}(t),S^{*}(t))$ by Léonard (1987).

1.4 Proposition 2.4

Note first that (2.50) is the finite time version of (2.15). Regarding (2.51), by the same argument as the proof of Proposition 2.3, we obtain (6.21) for the global warming model. Substitution of (2.43) and (2.44) into (6.21) yields (2.51). Finally, for this model, (2.19 b and c) and (2.21 a and b) hold. They imply (2.27) in Proposition 2.2.

1.5 Proposition 3.1

Let $r^{*}$ and $p^{*}$ be the equilibrium prices. Suppose that the household expects the path of pollution $\tilde{X}(t)=X^{*}(t)$ for the flow pollution model and $\tilde{S}(t)=S^{*}(t)$ for the stock pollution model. For the flow pollution model, there are the costates $\lambda ^{*}(t)$ and $\mu ^{*}(t)$ with which the optimal controls $c^{*}(t)$ and $X^{*}(t)$ satisfy (2.5–2.8). Then, it is easy to see that the optimal path satisfies the sufficient conditions for the firm (3.4) and the household (3.5). Notice that the firm’s transversality condition is satisfied:

$$\begin{aligned} Q(T)\exp \left( -\int \limits _{0}^{T}r^{*}(t)dt\right) =Q(0)-\int \limits _{0}^{T}p^{*}(t)X^{*}(t)\exp \left( -\int \limits _{0}^{t}r^{*}(s)ds\right) dt=0,\nonumber \\ \end{aligned}$$

(6.22)

where the first equality follows from (3.1) and the second equality follows from (3.6). The same argument is applied to the stock pollution model.

1.6 Proposition 4.1

Recall that the steady state $\left( K_{\infty },z_{ss}\right)$ of the system of differential equations (4.7 a) and (4.7 c) depends on $\omega$. If $\omega =K_{ss}/z_{ss}^{\left( 1-\alpha +\alpha /\sigma \right) }$, then $K_{\infty }=K_{ss}=K_{0}$ and the equilibrium path keeps the initial stock level, which is socially optimal. If $\omega >K_{ss}/z_{ss}^{\left( 1-\alpha +\alpha /\sigma \right) }$, then $K_{\infty }>K_{ss}$. This case is illustrated in Fig. 1 where point A is $K_{ss}$. Because the equilibrium path increases along the stable manifold, $z(t)<z_{ss}$ and $K(t)>K_{ss}$ for all $t\ge 0$. This implies that (4.7 d) is not satisfied:

$$\begin{aligned} \int \limits _{0}^{\infty }\left[ K(t)/z(t)\right] e^{-\rho t}dt>\int \limits _{0}^{\infty }\left( K_{ss}/z_{ss}\right) e^{-\rho t}dt=B_{ss}. \end{aligned}$$

(6.23)

Therefore, this case is ruled out. Consider the case of $\omega <K_{ss} /z_{ss}^{\left( 1-\alpha +\alpha /\sigma \right) }$. This time, point B on Fig. 1 is $K_{ss}$. The symmetric argument concludes that this case is also ruled out. Therefore, we have a unique $\omega =K_{ss}/z_{ss}^{\left( 1-\alpha +\alpha /\sigma \right) }$ and a unique equilibrium path satisfying (4.7) that coincides with the optimal path of the social planner’s problem.

1.7 Proposition 4.2

We prove Proposition 4.2 in a heuristic way. Take arbitrarily a path of interest rates $r(t)$ ($t>0$) such that $\lim _{t\rightarrow \infty }r(t)=\rho$. Note that we do not determine $r(0)$, but it does not invalidate the argument below. Using $r(t)$, $F(K(t),X(t))$ is written as:

$$\begin{aligned} F(K(t),X(t))=\frac{r(t)}{\alpha }K(t). \end{aligned}$$

(6.24)

From (4.10 a, c), there is $c_{0}>0$ and it holds that:

$$\begin{aligned} c(t)=\exp \left[ \int \limits _{0}^{t}\frac{r(s)-\rho }{\sigma }ds\right] c_{0}=-\dot{K}(t)+\frac{r(t)}{\alpha }K(t), K(0)=K_{0}. \end{aligned}$$

(6.25)

This equation is solved as:

$$\begin{aligned} K(t)=\exp \left[ \int \limits _{0}^{t}\frac{r(s)}{\alpha }ds\right] \left\{ K_{0}-c_{0}\int \limits _{0}^{t}\exp \left[ -\int \limits _{0}^{s}\left( \frac{r(\nu )}{\alpha }-\frac{r(\nu )-\rho }{\sigma }\right) d\nu \right] ds\right\} . \nonumber \\ \end{aligned}$$

(6.26)

The no-Ponzi game condition (4.10 b) is satisfied only if:

$$\begin{aligned} c_{0}=\frac{K_{0}}{\int _{0}^{\infty }\exp \left[ -\int _{0}^{t}\left( \frac{r(s)}{\alpha }-\frac{r(s)-\rho }{\sigma }\right) ds\right] dt}. \end{aligned}$$

(6.27)

Note that the denominator on the right-hand side of (6.27) is finite because

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{r(t)}{\alpha }-\frac{r(t)-\rho }{\sigma } =\frac{\rho }{\alpha }>0 \end{aligned}$$

(6.28)

(see Lemma 6.4 below). With (6.27), the no-Ponzi game condition (4.10 b) holds as follows:

$$\begin{aligned}&\liminf _{t\rightarrow \infty }K(t)\exp \left( -\int \limits _{0}^{t}F_{K} (K(s),X(s))ds\right) \nonumber \\&=\liminf _{t\rightarrow \infty }\exp \left[ -\int \limits _{0}^{t}r(s)ds\right] \exp \left[ \int \limits _{0}^{t}\frac{r(s)}{\alpha }ds\right] \nonumber \\&\qquad \qquad \times \left\{ K_{0}-c_{0}\int \limits _{0}^{t}\exp \left[ -\int \limits _{0} ^{s}\left( \frac{r(\nu )}{\alpha }-\frac{r(\nu )-\rho }{\sigma }\right) d\nu \right] ds\right\} \nonumber \\&=\lim _{t\rightarrow \infty }\exp \left[ \int \limits _{0}^{t}\frac{\left( 1-\alpha \right) r(s)}{\alpha }ds\right] \left\{ K_{0}-c_{0}\int \limits _{0}^{t} \exp \left[ -\int \limits _{0}^{s}\left( \frac{r(\nu )}{\alpha }-\frac{r(\nu )-\rho }{\sigma }\right) d\nu \right] ds\right\} \nonumber \\&=\lim _{t\rightarrow \infty }\frac{K_{0}-c_{0}\int _{0}^{t}\exp \left[ -\int _{0}^{s}\left( \frac{r(\nu )}{\alpha }-\frac{r(\nu )-\rho }{\sigma }\right) d\nu \right] ds}{-\exp \left[ \int _{0}^{t}\frac{\left( 1-\alpha \right) r(s)}{\alpha }ds\right] }\nonumber \\&=\lim _{t\rightarrow \infty }\frac{c_{0}\exp \left[ -\int _{0}^{t}\left( \frac{r(s)}{\alpha }-\frac{r(s)-\rho }{\sigma }\right) ds\right] }{\frac{\left( 1-\alpha \right) r(t)}{\alpha }\exp \left[ -\int _{0}^{t} \frac{\left( 1-\alpha \right) r(s)}{\alpha }ds\right] }\nonumber \\&=\lim _{t\rightarrow \infty }\frac{c_{0}\alpha }{\left( 1-\alpha \right) r(t)}\exp \left[ -\int \limits _{0}^{t}\left( \frac{r(s)}{\alpha }-\frac{r(s)-\rho }{\sigma }\right) -\frac{\left( 1-\alpha \right) r(s)}{\alpha }ds\right] \nonumber \\&=\frac{c_{0}\alpha }{\left( 1-\alpha \right) \rho }\lim _{t\rightarrow \infty }\exp \left[ -\int \limits _{0}^{t}\frac{\rho -(1-\sigma )r(s)}{\sigma }ds\right] \nonumber \\&=0, \end{aligned}$$

(6.29)

because $\lim _{t\rightarrow \infty }\rho -(1-\sigma )r(t)=\sigma \rho >0$. Letting $Q_{0}=F_{X}(K_{0},X(0))B_{0}$, the differential equation in (4.10 d) is rewritten as:

$$\begin{aligned} \dot{Q}(t)=r(t)Q(t)-(1-\alpha )F(K(t),X(t)),\quad Q(0)=Q_{0} , \end{aligned}$$

(6.30)

and this is solved as:

$$\begin{aligned} Q(t)\exp \left( -\int \limits _{0}^{t}r(s)ds\right) -Q_{0}=-(1-\alpha )\int \limits _{0} ^{t}F(K(s),X(s))\exp \left( -\int \limits _{0}^{s}r(\nu )d\nu \right) ds. \nonumber \\ \end{aligned}$$

(6.31)

Using the following result:

$$\begin{aligned}&\int \limits _{0}^{t}F(K(s),X(s))\exp \left( -\int \limits _{0}^{s}r(\nu )d\nu \right) ds\nonumber \\&=\int \limits _{0}^{t}\left( \dot{K}(s)+c(s)\right) \exp \left( -\int \limits _{0}^{s} r(\nu )d\nu \right) ds\nonumber \\&=\left[ K(s)\exp \left( -\int \limits _{0}^{s}r(\nu )d\nu \right) \right] _{s=0}^{t}\nonumber \\&\qquad \qquad +\alpha \int \limits _{0}^{t}\alpha ^{-1}r(s)K(s)\exp \left( -\int \limits _{0} ^{s}r(\nu )d\nu \right) ds+\int \limits _{0}^{t}c(s)\exp \left( -\int \limits _{0}^{s}r(\nu )d\nu \right) ds\nonumber \\&=K(t)\exp \left( -\int \limits _{0}^{t}r(\nu )d\nu \right) -K_{0}\nonumber \\&\qquad \qquad +\alpha \int \limits _{0}^{t}F(K(s),X(s))\exp \left( -\int \limits _{0}^{s} r(\nu )d\nu \right) ds+\int \limits _{0}^{t}c(s)\exp \left( -\int \limits _{0}^{s}r(\nu )d\nu \right) ds, \nonumber \\ \end{aligned}$$

(6.32)

we have:

$$\begin{aligned} Q(t)&= \exp \left[ \int \limits _{0}^{t}r(s)ds\right] \nonumber \\&\quad \times \left[ Q_{0}+K_{0}-K(t)\exp \left( -\int \limits _{0}^{t}r(\nu )d\nu \right) -\int \limits _{0}^{t}c(s)\exp \left( -\int \limits _{0}^{s}r(\nu )d\nu \right) ds\right] . \nonumber \\ \end{aligned}$$

(6.33)

The no-Ponzi game condition (4.10 e) holds if and only if:

$$\begin{aligned} Q_{0}=\int \limits _{0}^{\infty }c(t)\exp \left( -\int \limits _{0}^{t}r(s)ds\right) dt-K_{0}. \end{aligned}$$

(6.34)

Using (6.27), we have:

$$\begin{aligned} Q_{0}&= \frac{K_{0}\int _{0}^{\infty }\exp \left[ \int _{0}^{t}\frac{r(s)-\rho }{\sigma }ds\right] \exp \left( -\int _{0}^{t}r(s)ds\right) dt}{\int _{0}^{\infty }\exp \left[ -\int _{0}^{t}\left( \frac{r(s)}{\alpha } -\frac{r(s)-\rho }{\sigma }\right) ds\right] dt}-K_{0}\nonumber \\&= K_{0}\left\{ \frac{\int _{0}^{\infty }\exp \left[ -\int _{0}^{t}\frac{\rho -(1-\sigma )r(s)}{\sigma }ds\right] }{\int _{0}^{\infty }\exp \left[ -\int _{0}^{t}\left( \frac{\rho -(1-\sigma )r(s)}{\sigma }+\frac{1-\alpha }{\alpha }r(s)\right) ds\right] dt}-1\right\} . \end{aligned}$$

(6.35)

As:

$$\begin{aligned} Q_{0}=F_{X}(K_{0},X(0))B_{0}=(1-\alpha )(K_{0}/X(0))^{\alpha }B_{0}, \end{aligned}$$

(6.36)

when we choose:

$$\begin{aligned} X(0)&= K_{0}^{-\frac{1-\alpha }{\alpha }}\left[ (1-\alpha )B_{0}\right] ^{\frac{1}{\alpha }}\nonumber \\&\quad \times \left\{ \frac{\int _{0}^{\infty }\exp \left[ -\int _{0}^{t}\frac{\rho -(1-\sigma )r(s)}{\sigma }ds\right] }{\int _{0}^{\infty }\exp \left[ -\int _{0}^{t}\left( \frac{\rho -(1-\sigma )r(s)}{\sigma }+\frac{1-\alpha }{\alpha }r(s)\right) ds\right] dt}-1\right\} ^{-\frac{1}{\alpha }}, \end{aligned}$$

(6.37)

the no-Ponzi game condition holds and (4.10 e) is satisfied. Finally, $X(t)$ ($t>0$) is given by:

$$\begin{aligned} X(t)=\left( \frac{r(t)}{\alpha }\right) ^{\frac{1}{1-\alpha }}K(t). \end{aligned}$$

(6.38)

Summing up the above results, we have:

Lemma 6.3

For any positive measurable function $r:(0,\infty )\rightarrow \mathbb R _{++}$ satisfying $\lim _{t\rightarrow \infty }r(t)=\rho$, there is an equilibrium path on which $r(t)$ is the equilibrium interest rate. The equilibrium path is characterized by: (a) (6.25) for the consumption path with the initial value (6.27), (b) (6.26) for the capital stock path, (c) (6.38) for the pollution path with the initial value (6.37), and (d) (6.33) for the path of the permit values with the initial value given by (6.36).

Proof of Proposition 4.2

The statement in Proposition 4.2 is an immediate result from the above lemma.

Lemma 6.4

If measurable function $\xi (t)$ satisfies $\lim _{t\rightarrow \infty }\xi (t)=\xi >0$, then

$$\begin{aligned} \int \limits _{0}^{\infty }\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt<\infty . \end{aligned}$$

Proof

As $\lim _{t\rightarrow \infty }\xi (t)=\xi >0$, there are positive numbers $\varepsilon$ and $T$ such that $|\xi (t)-\xi |<\varepsilon$ for all $t>T$. Then,

$$\begin{aligned}&\int \limits _{0}^{\infty }\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt\nonumber \\&=\int \limits _{0}^{T}\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt+\int \limits _{T}^{\infty }\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt.\nonumber \\&=\int \limits _{0}^{T}\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt+\exp \left[ -\int \limits _{0}^{T}\xi (s)ds\right] \int \limits _{T}^{\infty }\exp \left[ -\int \limits _{T}^{t} \xi (s)ds\right] dt\nonumber \\&<\int \limits _{0}^{T}\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt+\exp \left[ -\int \limits _{0}^{T}\xi (s)ds\right] \int \limits _{T}^{\infty }\exp \left[ -\int \limits _{T}^{t} |\xi -\varepsilon |ds\right] dt\nonumber \\&=\int \limits _{0}^{T}\exp \left[ -\int \limits _{0}^{t}\xi (s)ds\right] dt+\exp \left[ -\int \limits _{0}^{T}\xi (s)ds\right] \frac{\exp \left( |\xi -\varepsilon |T\right) -1}{|\xi -\varepsilon |}\nonumber \\&<\infty . \end{aligned}$$

(6.39)

$\square$