Economic growth and property rights on natural resources

Abstract

We consider two models of economic growth with exhaustible natural resources and agents heterogeneous in their time preferences. In the first model, we assume private ownership of natural resources and show that every competitive equilibrium converges to a balanced-growth equilibrium with the long-run rate of growth being determined by the discount factor of the most patient agents. In the second model, natural resources are public property and the resource extraction rate is determined by majority voting. For this model, we define an intertemporal voting equilibrium and prove that it also converges to a balanced-growth equilibrium. In this scenario, the long-run rate of growth is determined by the median discount factor. Our results suggest that if the most patient agents do not constitute a majority of the population, private ownership of natural resources results in a higher rate of growth than public ownership. At the same time, private ownership leads to higher inequality than public ownership, and if inequality impedes growth, then the public property regime is likely to result in a higher long-run rate of growth. However, an appropriate redistributive policy can eliminate the negative impact of inequality on growth.

Appendices

Appendix 1: Private property regime

1.1 Competitive equilibrium

Suppose we are given an initial state of the economy, $\mathcal {I}_{0} = \{ ({\hat{k}}_{0}^{j})_{j=1}^{L}, ({\hat{R}}_{-1}^{j})_{j=1}^{L} \}$, where $({\hat{k}}_{0}^{j})_{j=1}^{L}$ and $({\hat{R}}_{-1}^{j})_{j=1}^{L}$ are initial distributions of physical capital and natural resources among agents.

We assume that $\mathcal {I}_{0}$ is a non-degenerate initial state,²² i.e.,

$$\begin{aligned} {\hat{k}}_{0}^{j} \ge 0, \quad {\hat{R}}_{-1}^{j} \ge 0, \quad j = 1, \ldots , L; \qquad \frac{1}{L} \sum _{j=1}^L {\hat{k}}_{0}^{j}> 0; \qquad \sum _{j=1}^L {\hat{R}}_{-1}^{j} > 0. \end{aligned}$$

Definition

A competitive equilibrium starting from the initial state $\mathcal {I}_{0}$ is a sequence

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

such that

• 1. For each $j = 1, \ldots , L$, the sequence $\{ c^{j*}_{t}, s^{j*}_{t} \}_{t = 0}^{\infty }$ is a solution to the following utility maximization problem:

  $$\begin{aligned} \begin{array}{cc} \displaystyle \max \sum _{t=0}^\infty \beta _j^t \ln c_{t}^{j}, \\ s. t. \quad c_{t}^{j} + s_{t}^{j} \le \left( 1+r_t \right) s^{j}_{t-1} + w_t, \quad t = 0, 1, \ldots , \\ s_{t}^{j} \ge 0, \quad t = 0, 1, \ldots \end{array} \end{aligned}$$

  (8)

  at $r_t = r_t^*$, $w_t = w_t^*$, and $s^{j}_{-1} = \frac{q_{0}^*}{1+r_{0}^*} {\hat{R}}^{j}_{-1} + {\hat{k}}^{j}_{0}$;

• 2. Capital is paid its marginal product:

  $$\begin{aligned} 1 + r^*_{t} = \alpha _1 A_t (k_t^*)^{\alpha _1 - 1} (e_t^*)^{\alpha _3}, \quad t = 0, 1, \ldots , \end{aligned}$$

  where $k_{0}^{*} = \frac{1}{L} \sum _{j=1}^L {\hat{k}}_{0}^{j}$;

• 3. Labor is paid its marginal product:

  $$\begin{aligned} w^*_t = \alpha _2 A_t (k_t^*)^{\alpha _1} (e_t^*)^{\alpha _3}, \quad t = 0, 1, \ldots ; \end{aligned}$$

• 4. The price of natural resources is equal to the marginal product:

  $$\begin{aligned} q^*_{t} = \alpha _3 A_t (k_t^*)^{\alpha _1} (e_t^*)^{\alpha _3 - 1}, \quad t = 0, 1, \ldots ; \end{aligned}$$

• 5. The Hotelling rule holds:

  $$\begin{aligned} q^*_{t+1} = (1+r_{t+1}^*) q^*_{t}, \quad t = 0, 1, \ldots ; \end{aligned}$$

• 6. The natural balance of exhaustible resources is fulfilled:²³

  $$\begin{aligned} R_{t}^* = R_{t-1}^* - L e^*_{t}, \quad t = 0, 1, \ldots , \end{aligned}$$

  where $R_{-1}^*= \sum _{j=1}^L {\hat{R}}_{-1}^{j}$;

• 7. Aggregate savings are equal to investment into physical capital and natural resources:

  $$\begin{aligned} \sum _{j=1}^L s^{j*}_{t} = \frac{q_{t+1}^*}{1+r_{t+1}^*} R^*_{t} + L k^*_{t+1}, \quad t = 0, 1, \ldots . \end{aligned}$$

Theorem 1

For any initial state $\mathcal {I}_{0}$, there exists a competitive equilibrium starting from $\mathcal {I}_{0}$.

Proof

See Borissov and Pakhnin (2016).²⁴ $\square$

Let us prove two important results about this equilibrium. The following proposition states that if at the initial instant the stocks of physical capital and natural resources are owned by the most patient agents, then the competitive equilibrium starting from this state is unique.

Proposition 1

Suppose that the initial state $\mathcal {I}_{0}$ is such that

$$\begin{aligned} {\hat{k}}_{0}^{j} = 0, \qquad {\hat{R}}_{-1}^{j} = 0, \qquad j \notin J. \end{aligned}$$

Then there exists a unique competitive equilibrium starting from $\mathcal {I}_{0}$,

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots }, \end{aligned}$$

which is given for $t = 0, 1, \ldots$ by

where $R_{-1}^*= \sum _{j=1}^L {\hat{R}}_{-1}^{j}$, $k_{0}^{*} = \frac{1}{L} \sum _{j=1}^L {\hat{k}}_{0}^{j}$, and $s^{j*}_{-1} = \frac{q_{0}^{*}}{1+r_{0}^*} {\hat{R}}^{j}_{-1} + {\hat{k}}^{j}_{0}$.

The following proposition verifies that in every competitive equilibrium from some time onward only the most patient agents can make positive savings. From this time, relatively less patient agents make no savings, and the extraction rate is constant over time and equals $1 - \beta _{1}$.

Proposition 2

Suppose that

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

is a competitive equilibrium starting from an arbitrary initial state $\mathcal {I}_{0}$. Then there exists a point in time T such that for all $t > T$,

$$\begin{aligned}&\displaystyle s^{j*}_{t} = 0, \quad j \notin J, \\&\displaystyle R_{t}^*= \beta _{1} R_{t-1}^{*}, \qquad e_{t+1}^* = \beta _{1} e_{t}^{*}. \end{aligned}$$

Proof of Propositions 1 and 2

Consider a competitive equilibrium

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from a non-degenerate state $\mathcal {I}_{0} = \{ ({\hat{k}}_{0}^{j})_{j=1}^{L}, ({\hat{R}}_{-1}^{j})_{j=1}^{L} \}$. Since for each $j = 1, \ldots , L$, the sequence $\{ c^{j*}_{t}, s^{j*}_{t} \}_{t = 0}^{\infty }$ is a solution to problem (8), it satisfies the first-order conditions,

$$\begin{aligned} c^{j*}_{t+1} \ge \beta _j (1 + r^{*}_{t+1}) c^{j*}_{t} \ ( = \ \text {if} \ s^{j*}_{t} > 0), \quad t = 0, 1, \ldots , \end{aligned}$$

(9)

and the transversality condition,

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{\beta _{j}^{t} s^{j*}_{t}}{c^{j*}_{t}} = 0. \end{aligned}$$

(10)

Lemma 1

Let $\beta > 0$ be such that for some T

$$\begin{aligned} k^{*}_{t+1}> \beta (1 + r^{*}_{t}) k^{*}_{t} = \beta \alpha _1A_t (k_t^{*})^{\alpha _1} (e_t^{*})^{\alpha _3}, \quad t > T. \end{aligned}$$

If $\beta _j < \beta$, then $s^{j*}_{t} = 0$ for all sufficiently large t.

Proof

First let us show that if $\beta _j < \beta$, then $s^{j*}_{t} = 0$ for some $t \ge T$. Assume the converse. By (9), for all $t \ge T$,

$$\begin{aligned} c^{j*}_{t+1} = \beta _j (1 + r^{*}_{t+1}) c^{j*}_{t}, \end{aligned}$$

and hence

$$\begin{aligned} \frac{c^{j*}_{t}}{k^{*}_{t+1}} \le \frac{\beta _j (1 + r^{*}_{t}) c^{j*}_{t-1}}{\beta (1 + r^{*}_{t}) k^{*}_{t}} \le \frac{\beta _{j}}{\beta } \frac{c^{j*}_{t-1}}{k^{*}_{t}}. \end{aligned}$$

By assumption, $\beta _{j} / \beta < 1$, and thus $c^{j*}_{t} / k^{*}_{t+1} \xrightarrow [t \rightarrow \infty ]{}0$. Furthermore, it is clear that $k_{t+1}^{*} \le A_t (k_t^{*})^{\alpha _1} (e_t^{*})^{\alpha _3}$, and therefore

$$\begin{aligned} \displaystyle \frac{c^{j*}_{t}}{w^{*}_{t}} = \frac{c^{j*}_{t}}{\alpha _2 A_t (k_t^{*})^{\alpha _1} (e_t^{*})^{\alpha _3}} \le \frac{c^{j*}_{t}}{\alpha _2 k_{t+1}^{*}} \xrightarrow [t \rightarrow \infty ]{}0. \end{aligned}$$

Thus, for all sufficiently large t, $c^{j*}_{t} < w^{*}_{t}$, which is not optimal for agent j.

Now we know that $s^{j*}_{t} = 0$ at least for some t. Let us show that $s^{j*}_{t} = 0$ for all $t \ge T$. Indeed, assume the converse. Then there are only two possibilities. The first is that there exists $T^{\prime } > T$ such that $s^{j*}_{t} > 0$ for all $t \ge T^{\prime }$. However, applying the same argument as above, we obtain that $s^{j*}_{t} = 0$ for some $t \ge T^{\prime }$. The second possibility is that there are $t_1$ and $t_2$ such that $T \le t_{1} < t_{2} - 1$, and

$$\begin{aligned} s^{j*}_{t_1} = 0, \qquad s^{j*}_{t_2} = 0, \qquad s^{j*}_{t} > 0, \quad t_1< t < t_2. \end{aligned}$$

It follows from the budget constraints of agent j that

$$\begin{aligned} c^{j*}_{t_1 + 1} < w^{*}_{t_1 + 1}, \qquad c^{j*}_{t_2} > w^{*}_{t_2}. \end{aligned}$$

(11)

However, for $t \ge T$,

$$\begin{aligned} \frac{\alpha _1}{\alpha _2} \frac{w^{*}_{t+1}}{1 + r^{*}_{t+1}} = k_{t+1}^{*} > \beta (1 + r^{*}_{t}) k^{*}_{t} = \frac{\alpha _1}{\alpha _2} \beta w^{*}_{t}. \end{aligned}$$

Thus, $w^{*}_{t+1} > \beta (1 + r^{*}_{t+1}) w^{*}_{t}$. Using (9), we get

$$\begin{aligned} c^{j*}_{t_1 + 2} = \beta _j (1 + r^{*}_{t_1 + 2}) c^{j*}_{t_1 + 1}< \beta _j (1 + r^{*}_{t_1 + 2}) w^{*}_{t_1 + 1}< \beta (1 + r^{*}_{t_1 + 2}) w^{*}_{t_1 + 1} < w^{*}_{t_1 + 2}. \end{aligned}$$

Repeating this argument, we obtain

$$\begin{aligned} c^{j*}_{t + 1}< w^{*}_{t + 1}, \quad t_1< t < t_2, \end{aligned}$$

which implies $c^{j*}_{t _2} < w^{*}_{t _2}$, a contradiction of (11). $\square$

Lemma 2

$$\begin{aligned} k^{*}_{t+1} \le \beta _1 (1 + r^{*}_{t}) k^{*}_{t}, \quad t = 0, 1, \ldots . \end{aligned}$$

Proof

Assume the converse. Then there are T and $\zeta > 1$ such that

$$\begin{aligned} k^{*}_{T+1} \ge \zeta \beta _1 (1 + r^{*}_{T}) k^{*}_{T}. \end{aligned}$$

(12)

Let us show that (12) implies

$$\begin{aligned} k^{*}_{t+1} \ge \zeta \beta _1 (1 + r^{*}_{t}) k^{*}_{t}, \quad t \ge T. \end{aligned}$$

(13)

Denote

$$\begin{aligned} J (T) = \left\{ j \in \{ 1, 2, \ldots , L \} \ \mid \ s^{j*}_{T} > 0 \right\} , \end{aligned}$$

and recall that

$$\begin{aligned} \frac{(1 + r^{*}_{t}) k^{*}_{t}}{\alpha _1} = \frac{w^{*}_{t }}{\alpha _2} = \frac{q_t^* e_t^*}{\alpha _3}. \end{aligned}$$

(14)

We have

$$\begin{aligned}&\sum _{j \in J(T)} \left( c_{T+1}^{j*} + s_{T+1}^{j*} \right) - \frac{q^{*}_{T+2}}{1 + r^{*}_{T+2}} R^{*}_{T+1} \\&\quad = \sum _{j \in J(T)} \left( (1 + r^{*}_{T+1}) s_{T}^{j*} + w_{T+1}^{*} \right) - \frac{q^{*}_{T+2}}{1 + r^{*}_{T+2}} R^{*}_{T+1} \\&\quad = q^{*}_{T+1} R^{*}_{T} + (1 + r^{*}_{T+1}) L k^*_{T+1} + | J(T) | w_{T+1}^{*} - q^{*}_{T+1} R^{*}_{T+1} \\&\quad = q^{*}_{T+1} L e_{T+1}^* + (1 + r^{*}_{T+1}) L k^*_{T+1} + | J(T) | w_{T+1}^{*} \\&\quad = (1 + r^{*}_{T+1}) k^*_{T+1} \left( L + L \frac{\alpha _3}{\alpha _1} + | J(T) | \frac{\alpha _2}{\alpha _1} \right) \\&\quad \ge \zeta \beta _1 (1 + r^{*}_{T+1}) (1 + r^{*}_{T}) k^*_{T} \left( L + L \frac{\alpha _3}{\alpha _1} + | J(T) | \frac{\alpha _2}{\alpha _1} \right) \\&\quad = \zeta \beta _1 (1 + r^{*}_{T+1}) \left( L (1 + r^{*}_{T}) k^*_{T} + L q_{T}^* e_{T}^* + | J(T) | w_{T}^{*} \right) \\&\quad = \zeta \beta _1 (1 + r^{*}_{T+1}) \left( (1 + r^{*}_{T}) \left( \sum _{j = 1}^{L} s_{T-1}^{j*} - \frac{q^{*}_{T}}{1 + r^{*}_{T}} R_{T-1} \right) + L q_{T}^* e_{T}^* + \sum _{j \in J(T)} w_{T}^{*} \right) \\&\quad \ge \zeta \beta _1 (1 + r^{*}_{T+1}) \left( (1 + r^{*}_{T}) \sum _{j \in J(T)} s_{T-1}^{j*} + \sum _{j \in J(T)} w_{T}^{*} - q^{*}_{T} R^{*}_{T-1} + L q_{T}^* e_{T}^* \right) \\&\quad = \zeta \beta _1 (1 + r^{*}_{T+1}) \left( \sum _{j \in J(T)} ( c_{T}^{j*} + s_{T}^{j*} ) - \frac{q^{*}_{T+1}}{1 + r^{*}_{T+1}} R^{*}_{T} \right) . \end{aligned}$$

Thus,

$$\begin{aligned}&\sum _{j \in J(T)} \left( c_{T+1}^{j*} + s_{T+1}^{j*} \right) - \frac{q^{*}_{T+2}}{1 + r^{*}_{T+2}} R^{*}_{T+1}\nonumber \\&\quad \ge \zeta \beta _1 (1 + r^{*}_{T+1}) \left( \sum _{j \in J(T)} ( c_{T}^{j*} + s_{T}^{j*} ) - \frac{q^{*}_{T+1}}{1 + r^{*}_{T+1}} R^{*}_{T} \right) . \end{aligned}$$

(15)

For $j \in J (T)$, (9) holds with equality, so

$$\begin{aligned} \sum _{j \in J(T)} c_{T+1}^{j*}= & {} \sum _{j \in J(T)} \beta _j (1 + r^{*}_{T+1}) c^{j*}_{T}\nonumber \\\le & {} \beta _1 (1 + r^{*}_{T+1}) \sum _{j \in J(T)} c^{j*}_{T} < \zeta \beta _1 (1 + r_{T+1}^{*}) \sum _{j \in J(T)} c_{T}^{j*}. \end{aligned}$$

(16)

Clearly, (15) is consistent with (16) only if

$$\begin{aligned} \sum _{j \in J(T)} s_{T+1}^{j*} - \frac{q^{*}_{T+2}}{1 + r^{*}_{T+2}} R^{*}_{T+1} \ge \zeta \beta _1 (1 + r^{*}_{T+1}) \left( \sum _{j \in J(T)} s_{T}^{j*} - \frac{q^{*}_{T+1}}{1 + r^{*}_{T+1}} R^{*}_{T} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} L k_{T+2}^{*}= & {} \sum _{j = 1}^{L} s_{T+1}^{j*} - \frac{q^{*}_{T+2}}{1 + r^{*}_{T+2}} R^{*}_{T+1} \ge \sum _{j \in J(T)} s_{T+1}^{j*} - \frac{q^{*}_{T+2}}{1 + r^{*}_{T+2}} R^{*}_{T+1} \\\ge & {} \zeta \beta _1 (1 + r^{*}_{T+1}) \left( \sum _{j \in J(T)} s_{T}^{j*} - \frac{q^{*}_{T+1}}{1 + r^{*}_{T+1}} R^{*}_{T} \right) \\= & {} \zeta \beta _1 (1 + r^{*}_{T+1}) \left( \sum _{j = 1}^{L} s_{T}^{j*} - \frac{q^{*}_{T+1}}{1 + r^{*}_{T+1}} R^{*}_{T} \right) = \zeta \beta _1 (1 + r^{*}_{T+1}) L k_{T+1}^{*}. \end{aligned}$$

Repeating the argument, we infer that (13) holds for all $t \ge T$.

However, from Lemma 1 it follows that $s^{j*}_{t} = 0$ for all j and for all sufficiently large t. This contradicts the evident positivity of $k_{t}^{*}$ for all $t = 0, 1, \ldots$. $\square$

Lemma 3

$$\begin{aligned} w^{*}_{t+1} \le \beta _1 (1 + r^{*}_{t+1}) w^{*}_{t}, \qquad e^{*}_{t+1} \le \beta _1 e^{*}_{t}, \qquad t = 0, 1, \ldots . \end{aligned}$$

Proof

Both inequalities follow from (14) and Lemma 2. Indeed, for all t

$$\begin{aligned} \frac{w^{*}_{t+1}}{1 + r^{*}_{t+1}} = \frac{\alpha _2 (1 + r^{*}_{t+1}) k_{t+1}^{*}}{\alpha _1 (1 + r^{*}_{t+1})} \le \frac{\beta _1 \alpha _2 (1 + r^{*}_{t}) k^{*}_{t}}{\alpha _1} = \beta _1 w^{*}_{t}. \end{aligned}$$

Moreover, for all t

$$\begin{aligned} \frac{e^{*}_{t+1}}{e^{*}_{t}} = \frac{q_{t}^{*}}{q^{*}_{t+1}} \frac{(1 + r^{*}_{t+1}) k^{*}_{t+1}}{(1 + r^{*}_{t}) k^{*}_{t}} = \frac{(1 + r^{*}_{t+1}) q_{t}^{*}}{q^{*}_{t+1}} \frac{k^{*}_{t+1}}{(1 + r^{*}_{t}) k^{*}_{t}} \le \beta _1, \end{aligned}$$

since $q^{*}_{t+1} = (1 + r^{*}_{t+1}) q_{t}^{*}$ by the Hotelling rule. $\square$

Lemma 4

$$\begin{aligned} s^{j*}_{t+1} \ge \beta _1 (1 + r^{*}_{t+1}) s^{j*}_{t}, \quad j \in J, \quad t = -1, 0, \ldots . \end{aligned}$$

(17)

Proof

Consider $j \in J$. Then by (9),

$$\begin{aligned} \beta _{1}^{t} c_{0}^{j*} \le \frac{c^{j*}_{t}}{(1+r_{1}^{*}) \cdots (1+r_{t}^{*})}, \quad t = 1, 2, \ldots , \end{aligned}$$

and hence

$$\begin{aligned} c_{0}^{j*} (1 + \beta _{1} + \beta _{1}^{2} + \cdots ) \le c_{0}^{j*} + \frac{c^{j*}_{1}}{(1+r_{1}^{*})} + \frac{c^{j*}_{2}}{(1+r_{1}^{*})(1+r_{2}^{*})} + \cdots . \end{aligned}$$

(18)

Adding together all budget constraints of agent j, we obtain

$$\begin{aligned}&c^{j*}_{0} + \frac{c^{j*}_{1}}{(1+r_{1}^{*})} + \frac{c^{j*}_{2}}{(1+r_{1}^{*})(1+r_{2}^{*})} + \cdots \nonumber \\&\quad \le (1+r_{0}^{*}) s^{j}_{-1} + w^{*}_{0} + \frac{w^{*}_{1}}{(1+r_{1}^{*})} + \frac{w^{*}_{2}}{(1 + r_{1}^{*})(1 + r_{2}^{*})} + \cdots . \end{aligned}$$

(19)

Moreover, by Lemma 3 for $t = 0, 1, \ldots$,

$$\begin{aligned} \frac{w^{*}_{t+1}}{(1 + r_{1}^{*}) \cdots (1+r_{t + 1}^{*})} \le \frac{\beta _{1} w^{*}_{t}}{(1+r_{1}^{*}) \cdots (1+r_{t}^{*})} \le \cdots \le \beta _{1}^{t+1} w^{*}_{0}, \end{aligned}$$

which implies

$$\begin{aligned}&(1 + r_{0}^{*}) s^{j*}_{-1} + w^{*}_{0} + \frac{w^{*}_{1}}{(1+r_{1}^{*})} + \frac{w^{*}_{2}}{(1+r_{1}^{*})(1+r_{2}^{*})} + \cdots \nonumber \\&\quad \le (1+r_{0}^{*}) s^{j*}_{-1} + w^{*}_{0} (1 + \beta _{1} + \beta _{1}^{2} + \cdots ). \end{aligned}$$

(20)

Combining (18)–(20), we finally get

$$\begin{aligned} c_{0}^{j*} (1 + \beta _{1} + \beta _{1}^{2} + \cdots ) \le (1 + r_{0}^{*}) s^{j*}_{-1} + w^{*}_{0} (1 + \beta _{1} + \beta _{1}^{2} + \cdots ), \end{aligned}$$

and therefore

$$\begin{aligned} c_{0}^{j*} \le (1 + r_{0}^{*}) (1 - \beta _{1}) s_{-1}^{j*} + w^{*}_{0}. \end{aligned}$$

Thus,

$$\begin{aligned} s_{0}^{j*}= & {} (1 + r_{0}^{*}) s^{j*}_{-1} + w^{*}_{0} - c_{0}^{j*} \ge (1 + r_{0}^{*}) s^{j*}_{-1} + w^{*}_{0} - (1 + r_{0}^{*}) (1 - \beta _{1}) s_{-1}^{j*} - w^{*}_{0}\\= & {} \beta _{1} (1 + r_{0}^{*}) s^{j*}_{-1}. \end{aligned}$$

This proves (17) for $t = -1$. To prove it for $t = 0, 1, \ldots$, it is sufficient to repeat the argument. $\square$

Lemma 5

For any $\delta > 0$, there exists a point in time T such that for all $t > T$,

$$\begin{aligned} k^{*}_{t+1} > \beta _1 (1 - \delta ) (1 + r^{*}_{t}) k^{*}_{t}. \end{aligned}$$

Proof

From (9) and Lemma 3, it is clear that for $j \in J$

$$\begin{aligned} \frac{c^{j*}_{t+1}}{w^{*}_{t+1}} \ge \frac{\beta _1 (1 + r^{*}_{t+1}) c^{j*}_{t}}{\beta _1 (1 + r^{*}_{t+1}) w^{*}_{t}} = \frac{c^{j*}_{t}}{w^{*}_{t}}, \quad t = 0, 1, \ldots . \end{aligned}$$

This means that the sequence $\left\{ \frac{c_{t}^{j*}}{w^{*}_{t}} \right\} _{t=0}^{\infty }$ is non-decreasing. It is also bounded from above, as consumption cannot exceed total output:

$$\begin{aligned} c^{j*}_{t} \le L \frac{w^{*}_{t}}{\alpha _2}, \quad t = 0, 1, \ldots . \end{aligned}$$

Therefore, the sequence $\left\{ \frac{c_{t}^{j*}}{w^{*}_{t}} \right\} _{t=0}^{\infty }$ converges, so the sequence $\left\{ \frac{c_{t}^{j*}}{w^{*}_{t}} \frac{w^{*}_{t+1}}{c_{t+1}^{j*}} \right\} _{t=0}^{\infty }$ converges to 1.

It follows from Lemma 4 that if $s_{-1}^{j} > 0$, then $s_{t}^{j*} > 0$ for all $t \ge 0$ and $j \in J$. Thus,

$$\begin{aligned} \frac{c_{t}^{j*}}{w^{*}_{t}} \frac{w^{*}_{t+1}}{c_{t+1}^{j*}} = \frac{w^{*}_{t+1}}{\beta _1 (1 + r^{*}_{t+1}) w^{*}_{t}}, \quad t = 0, 1, \ldots , \end{aligned}$$

and the sequence $\left\{ \frac{w^{*}_{t+1}}{\beta _1 (1 + r^{*}_{t+1}) w^{*}_{t}} \right\} _{t=0}^{\infty }$ converges to 1 as well. Hence, for any $\delta > 0$ there exists T such that for $t > T$,

$$\begin{aligned} \frac{w^{*}_{t+1}}{\beta _1 (1 + r^{*}_{t+1}) w^{*}_{t}} > (1 - \delta ), \end{aligned}$$

which implies

$$\begin{aligned} k^{*}_{t+1} = \frac{\alpha _1}{\alpha _2} \frac{ w^{*}_{t+1}}{1 + r^{*}_{t+1}} > \frac{\alpha _1}{\alpha _2} \beta _1 (1 - \delta ) w^{*}_{t} = \beta _1 (1 - \delta ) (1 + r^{*}_{t}) k^{*}_{t}. \end{aligned}$$

This proves the lemma. $\square$

Consider $\delta$ that satisfies $\beta _1 (1 - \delta ) > \max _{j \notin J} \beta _{j}$. Applying Lemma 1 with $\beta = \beta _1 (1 - \delta )$, we obtain that for any competitive equilibrium starting from a non-degenerate initial state there exists a point in time T such that for all $t > T$, $s^{j*}_{t} = 0$ for $j \notin J$.

Lemma 6

For all $t > T$,

Moreover,

$$\begin{aligned} e_{t+1}^{*} = \beta _{1} e_{t}^{*}, \qquad R_{t}^{*}= \beta _{1} R_{t-1}^{*}. \end{aligned}$$

Proof

First let us show that

$$\begin{aligned} \lim _{t \rightarrow \infty } R^{*}_{t} = 0. \end{aligned}$$

(21)

To prove this, note that for all j,

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{s^{j*}_{t}}{(1 + r_{1}^{*}) \cdots (1 + r_{t}^{*})} = 0. \end{aligned}$$

Indeed, it follows from (19) that this limit exists. Since savings are nonnegative, this limit is also nonnegative. Suppose that $\lim _{t \rightarrow \infty } \frac{s^{j*}_{t}}{(1 + r_{1}^{*}) \cdots (1 + r_{t}^{*})} > 0$. Then $s^{j*}_{t} > 0$ for all t, and thus by (9),

$$\begin{aligned} c^{j*}_{t} = \beta _j (1 + r^{*}_{t}) c^{j*}_{t-1} = \cdots = \beta _{j}^{t} (1 + r^{*}_{t}) \cdots (1 + r^{*}_{1}) c^{j*}_{0}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{\beta _{j}^{t} s^{j*}_{t}}{c^{j*}_{t}} = \frac{s^{j*}_{t}}{(1 + r^{*}_{t}) \cdots (1 + r^{*}_{1}) c^{j*}_{0}} = \frac{1}{c^{j*}_{0}} \frac{s^{j*}_{t}}{(1 + r^{*}_{1}) \cdots (1 + r^{*}_{t})}. \end{aligned}$$

It follows from (10) that $\lim _{t \rightarrow \infty } \frac{s^{j*}_{t}}{(1 + r^{*}_{1}) \cdots (1 + r^{*}_{t})} = 0$, which is a contradiction. Therefore,

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{\sum _{j=1}^L s^{j*}_{t}}{(1+r_{1}^{*}) \cdots (1+r_{t}^{*})} = 0. \end{aligned}$$

Since $\sum _{j=1}^L s^{j*}_{t} = q_{t}^* R^*_{t} + L k^*_{t+1}$, and both terms are nonnegative, we get

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{q_{t}^* R^*_{t}}{(1+r_{1}^{*}) \cdots (1+r_{t}^{*})} = \lim _{t \rightarrow \infty } q_{0}^* R^*_{t} = 0. \end{aligned}$$

As $q_{0}^* > 0$, (21) indeed holds.

Now suppose that $t > T$. By Lemma 4,

$$\begin{aligned} \beta _1 (1 + r^{*}_{t}) \left( L k^{*}_{t} + \frac{q_{t}^*}{1+r_{t}^*} R^*_{t-1} \right)= & {} \beta _1 (1 + r^{*}_{t}) \sum _{j \in J} s^{j*}_{t-1} \le \sum _{j \in J} s^{j*}_{t}\nonumber \\= & {} L k^{*}_{t + 1} + \frac{q_{t+1}^*}{1+r_{t+1}^*} R^{*}_{t}. \end{aligned}$$

(22)

At the same time, by Lemma 2,

$$\begin{aligned} k^{*}_{t+1} \le \beta _1 (1 + r^{*}_{t}) k^{*}_{t}, \quad t = 0, 1, \ldots . \end{aligned}$$

Therefore for $t > T$,

$$\begin{aligned} \frac{q_{t+1}^*}{1+r_{t+1}^*} R^*_{t} \ge \beta _1 (1 + r^{*}_{t}) \frac{q_{t}^*}{1+r_{t}^*} R^*_{t-1}, \end{aligned}$$

or, equivalently,

$$\begin{aligned} R^*_{t} \ge \beta _1 R^*_{t-1}. \end{aligned}$$

(23)

It follows from the natural balance of exhaustible resources and (21) that

$$\begin{aligned} R_{T}^{*}= & {} R_{T+1}^{*} + L e^{*}_{T+1} = R_{T+2}^{*} + L e^{*}_{T+2} + L e^{*}_{T+1} = \ldots \\= & {} L e^{*}_{T+1} \left( 1 + \frac{e^{*}_{T+2}}{e^{*}_{T+1}} + \frac{e^{*}_{T+3}}{e^{*}_{T+2}} \frac{e^{*}_{T+2}}{e^{*}_{T+1}} + \cdots \right) . \end{aligned}$$

Hence, using Lemma 3, we conclude that

$$\begin{aligned} R_{T}^{*} \le L e^{*}_{T+1} \left( 1 + \beta _1 + \beta ^{2}_1 + \cdots \right) = L e^{*}_{T+1} \frac{1}{1 - \beta _1}. \end{aligned}$$

It follows that $\left( 1 - \beta _1 \right) \left( R_{T+1}^{*} + L e^{*}_{T+1} \right) \le L e^{*}_{T+1}$, or

$$\begin{aligned} R_{T+1}^{*} \le \beta _1 \left( R_{T+1}^{*} + L e^{*}_{T+1} \right) = \beta _1 R_{T}^{*}. \end{aligned}$$

Thus, using (23) we get

$$\begin{aligned} R_{T+1}^{*} = \beta _1 R_{T}^{*}. \end{aligned}$$

Repeating the argument, we obtain that

$$\begin{aligned} R_{t}^{*} = \beta _1 R_{t-1}^{*}, \quad t > T. \end{aligned}$$

Therefore, for all $t > T$, $L e^*_{t} = R_{t-1}^* - R_{t}^* = (1 - \beta _{1}) R^{*}_{t-1}$, and

$$\begin{aligned} \frac{e^{*}_{t+1}}{e^{*}_{t}} = \frac{ R^{*}_{t}}{R^{*}_{t-1}} = \beta _1, \quad t > T. \end{aligned}$$

We have proved that eventually the extraction rate becomes constant over time and equal to $1- \beta _{1}$.

Since for $t > T$,

$$\begin{aligned} \frac{q_{t+1}^*}{1+r_{t+1}^*} R^{*}_{t} = q_{t}^{*} R^{*}_{t} = \beta _1 q_{t}^{*} R_{t-1}^{*}, \end{aligned}$$

it follows from (22) that

$$\begin{aligned} \beta _1 (1 + r^{*}_{t}) k^{*}_{t} \le k^{*}_{t + 1}, \quad t > T. \end{aligned}$$

Using Lemma 2, we obtain that

$$\begin{aligned} k^{*}_{t + 1} = \beta _1 (1 + r^{*}_{t}) k^{*}_{t}, \quad t > T, \end{aligned}$$

and hence for $t > T$, $s^{j*}_{t} = \beta _1 (1 + r^{*}_{t}) s^{j*}_{t-1}$ for $j \in J$, while $s^{j*}_{t} = 0$ for $j \notin J$. $\square$

Proposition 2 is a corollary of Lemma 6. Proposition 1 also easily follows from Lemma 6. If the initial state $\mathcal {I}_{0}$ is such that

$$\begin{aligned} {\hat{k}}_{0}^{j} = 0, \qquad {\hat{R}}_{-1}^{j} = 0, \qquad j \notin J, \end{aligned}$$

then $s_{-1}^{j} = 0$ for $j \notin J$, and we can take $T = -1$. The sequences $\{ r_t^* \}$, $\{ w_t^* \}$, and $\{ q_t^* \}$ are derived from the known sequences $\{ k^*_t \}$ and $\{ e_t^* \}$, described in Lemma 6. $\square$

Thus, in every competitive equilibrium from some time onward only the most patient agents make positive savings, and from this time, resources are extracted at the constant rate $\varepsilon ^{*} = 1- \beta _{1}$.

1.2 Balanced-growth equilibrium

Definition

A competitive equilibrium

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from a non-degenerate initial state $\mathcal {I}_{0}$ is called a balanced-growth equilibrium if there exist an equilibrium rate of balanced growth $\gamma ^{*}$ and an equilibrium extraction rate $\varepsilon ^{*}$ such that for $t = 0, 1, \ldots$,

$$\begin{aligned}&\displaystyle c^{j*}_{t+1} = (1+\gamma ^*) c^{j*}_{t}, \qquad s^{j*}_{t} = (1+\gamma ^*) s^{j*}_{t-1}, \qquad j = 1, \ldots , L, \end{aligned}$$

(24)

$$\begin{aligned}&\displaystyle k^*_{t+1} = (1+\gamma ^*) k^{*}_{t}, \qquad w^*_{t+1} = (1+\gamma ^*) w^{*}_{t}, \end{aligned}$$

(25)

$$\begin{aligned}&\displaystyle 1+r^{*}_{t} = 1 + r^{*}, \qquad q_{t+1}^{*} = \left( 1 + r^* \right) q^{*}_{t}, \end{aligned}$$

(26)

$$\begin{aligned}&\displaystyle e_{t+1}^{*} = \left( 1 - \varepsilon ^* \right) e^{*}_{t}, \qquad R_{t}^{*}= \left( 1 - \varepsilon ^* \right) R^{*}_{t-1}. \end{aligned}$$

(27)

The following proposition proves the existence of a balanced-growth equilibrium and provides its characterization. In particular, it maintains that in every balanced-growth equilibrium less patient agents make no savings.

Proposition 3

A balanced-growth equilibrium

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from a non-degenerate initial state $\mathcal {I}_{0} = \{ ({\hat{k}}_{0}^{j})_{j=1}^{L}, ({\hat{R}}_{-1}^{j})_{j=1}^{L} \}$ exists if and only if

$$\begin{aligned}&\displaystyle {\hat{k}}_{0}^{j} = 0, \qquad {\hat{R}}_{-1}^{j} = 0, \qquad j \notin J, \end{aligned}$$

(28)

$$\begin{aligned}&\displaystyle \alpha _1 A_{0} \left( \frac{1}{L} \sum _{j = 1}^{L} {\hat{k}}_{0}^{j} \right) ^{\alpha _1 - 1} \left( \frac{1 - \beta _{1}}{L} \sum _{j = 1}^{L} {\hat{R}}_{-1}^{j} \right) ^{\alpha _3} = (1 + \lambda )^{\frac{1}{1 - \alpha _1}} \beta _{1}^{\frac{\alpha _1 + \alpha _3 - 1}{1 - \alpha _1}}, \end{aligned}$$

(29)

and (24)–(27) hold.

Proof

Necessity. Suppose that there exists a balanced-growth equilibrium

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from a non-degenerate state $\mathcal {I}_{0}$. It is a competitive equilibrium which satisfies (24)–(27) for some $r^{*}$, $\varepsilon ^*$ and $\gamma ^{*}$.

Repeating a well-known argument by Becker (1980, 2006), we infer that every balanced-growth equilibrium is characterized by the following properties:

$$\begin{aligned}&\displaystyle s_{t-1}^{j*} = 0, \quad j \notin J, \quad t = 0, 1, \ldots , \end{aligned}$$

(30)

$$\begin{aligned}&\displaystyle 1 + \gamma ^{*} = \beta _{1} (1 + r^{*}). \end{aligned}$$

(31)

Moreover, comparing the definitions of competitive and balanced-growth equilibria, we obtain that for every balanced-growth equilibrium the following relationships hold:

$$\begin{aligned} \left( 1 + \gamma ^* \right) ^{1 - \alpha _1}= & {} \left( 1+ \lambda \right) \left( 1 - \varepsilon ^* \right) ^{\alpha _3}, \end{aligned}$$

(32)

$$\begin{aligned} 1 + r^{*}= & {} \frac{1 + \gamma ^*}{1 - \varepsilon ^*}. \end{aligned}$$

(33)

Indeed, (32) follows from the fact that

$$\begin{aligned} 1 = \frac{1 + r_{t+1}^{*}}{1 + r_{t}^{*}} = \frac{A_{t+1}}{A_{t}} \left( \frac{k_{t+1}^{*}}{k_{t}^{*}} \right) ^{\alpha _1 - 1} \left( \frac{e_{t+1}^{*}}{e_{t}^{*}} \right) ^{\alpha _3} = (1 + \lambda ) \left( 1 + \gamma ^{*} \right) ^{\alpha _1 - 1} \left( 1 - \varepsilon ^* \right) ^{\alpha _3}. \end{aligned}$$

We also have

$$\begin{aligned} 1 + r^{*} = \frac{q^{*}_{t+1}}{q^{*}_{t}} = \frac{A_{t+1}}{A_{t}} \left( \frac{k_{t+1}^{*}}{k_{t}^{*}} \right) ^{\alpha _1} \left( \frac{e_{t+1}^{*}}{e_{t}^{*}} \right) ^{\alpha _3 - 1} {=} (1 {+} \lambda ) \left( 1 + \gamma ^{*} \right) ^{\alpha _1} \left( 1 - \varepsilon ^* \right) ^{\alpha _3 - 1}, \end{aligned}$$

which is equivalent to (33).

Using (31)–(33), it is easily checked that

$$\begin{aligned} 1 + r^{*} = (1 + \lambda )^{\frac{1}{1 - \alpha _1}} \beta _{1}^{\frac{\alpha _1 + \alpha _3 - 1}{1 - \alpha _1}}. \end{aligned}$$

(34)

It follows from (30) that $s_{-1}^{j*} = 0$ for $j \notin J$. Since $\mathcal {I}_{0}$ is non-degenerate, ${\hat{k}}_{0}^{j} \ge 0$ and ${\hat{R}}_{-1}^{j} \ge 0$ for all j, and thus (28) holds.²⁵ Furthermore, a constant over time interest rate is consistent with the definition of a competitive equilibrium if and only if

$$\begin{aligned} 1 + r^{*} = 1 + r_{0}^{*} = \alpha _1 A_{0} (k_{0}^*)^{\alpha _1 - 1} (e_{0}^*)^{\alpha _3} = \alpha _1 A_{0}\left( \frac{1}{L} \sum _{j = 1}^{L} {\hat{k}}_{0}^{j} \right) ^{\alpha _1 - 1} \left( \frac{1 - \beta _{1}}{L} \sum _{j = 1}^{L} {\hat{R}}_{-1}^{j} \right) ^{\alpha _3}. \end{aligned}$$

Taking into account (34), we obtain (29).

Sufficiency. Suppose that the initial state $\mathcal {I}_{0}$ is such that (28)–(29) hold. Consider the sequence

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from $\mathcal {I}_{0}$ and determined by (24)–(27).

It is easily checked that this sequence is a competitive equilibrium which is described in Proposition 1, with the constant interest rate

$$\begin{aligned} 1 + r^{*}_{t} = 1 + r^{*}_{0} = \alpha _1 A_{0} \left( \frac{1}{L} \sum _{j = 1}^{L} {\hat{k}}_{0}^{j} \right) ^{\alpha _1 - 1} \left( \frac{1 - \beta _{1}}{L} \sum _{j = 1}^{L} {\hat{R}}_{-1}^{j} \right) ^{\alpha _3}. \end{aligned}$$

Therefore, $\mathcal {E}^{*}$ is a competitive equilibrium which satisfies (24)–(27), i.e., a balanced-growth equilibrium. $\square$

It follows that the interest rate $r^{*}$, the equilibrium extraction rate $\varepsilon ^*$ and the equilibrium rate of balanced growth $\gamma ^{*}$ are uniquely determined by the parameters of the model and are the same for every balanced-growth equilibrium.

Proposition 4

For every balanced-growth equilibrium,

$$\begin{aligned}&\displaystyle 1 + \gamma ^* = \left( 1 + \lambda \right) ^\frac{1}{1-\alpha _1} \beta _{1}^{\frac{\alpha _3}{1-\alpha _1}}, \end{aligned}$$

(35)

$$\begin{aligned}&\displaystyle 1 + r^{*} = \frac{1 + \gamma ^*}{\beta _1}, \end{aligned}$$

(36)

$$\begin{aligned}&\displaystyle \varepsilon ^* = 1 - \beta _{1}. \end{aligned}$$

(37)

Proof

It is sufficient to repeat the argument used in the proof of Proposition 3. Combining (31)–(33), we obtain (35)–(37). $\square$

The following proposition maintains that every competitive equilibrium converges in some sense to a balanced-growth equilibrium.

Proposition 5

Every competitive equilibrium starting from an arbitrary non-degenerate initial state satisfies the following asymptotic properties:

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } 1 + r_{t}^{*} = 1+r^{*} = \frac{1 + \gamma ^{*}}{\beta _{1}}, \end{aligned}$$

(38)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{k^{*}_{t+1}}{k^{*}_{t}} = \lim _{t \rightarrow \infty } \frac{w^{*}_{t+1}}{w^{*}_{t}} = 1 + \gamma ^{*}, \end{aligned}$$

(39)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{s^{j*}_{t+1}}{s^{j*}_{t}} = 1 + \gamma ^{*}, \quad j \in J, \end{aligned}$$

(40)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{c^{j*}_{t+1}}{c^{j*}_{t}} = 1 + \gamma ^{*}, \quad j = 1, \ldots , L, \end{aligned}$$

(41)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{q^{*}_{t+1}}{q^{*}_{t}} = 1+r^{*}, \end{aligned}$$

(42)

where $1 + \gamma ^{*} = (1 + \lambda )^{\frac{1}{1 - \alpha _1}} \beta _{1}^{\frac{\alpha _3}{1 - \alpha _1}}$.

Proof

It follows from Proposition 2 that for $t > T$,

$$\begin{aligned} \frac{1 + r_{t+1}^{*}}{1 + r_{t}^{*}} = \frac{A_{t+1}}{A_{t}} \left( \frac{k_{t+1}^{*}}{k_{t}^{*}} \right) ^{\alpha _1 - 1} \left( \frac{e_{t+1}^{*}}{e_{t}^{*}} \right) ^{\alpha _3} = (1 + \lambda ) \left( \beta _1 (1 + r^{*}_{t}) \right) ^{\alpha _1 - 1} \left( \beta _{1} \right) ^{\alpha _3}, \end{aligned}$$

and thus

$$\begin{aligned} 1 + r^{*}_{t+1} = \left( 1 + \lambda \right) \left( \beta _{1} \right) ^{\alpha _1 + \alpha _3 - 1} \left( 1 + r^{*}_{t} \right) ^{\alpha _1}. \end{aligned}$$

Iterating, we get

$$\begin{aligned} 1 + r^{*}_{t+1+n} = \left( 1 + \lambda \right) ^{1 + \alpha _1 + \cdots + \alpha _1^{n}} \left( \beta _{1} \right) ^{(\alpha _1 + \alpha _3 - 1)(1 + \alpha _1 + \cdots + \alpha _1^{n})} \left( 1 + r^{*}_{t} \right) ^{\alpha _1^{n+1}}, \end{aligned}$$

and

$$\begin{aligned} \lim _{n \rightarrow \infty } 1 + r^{*}_{t+n+1} = (1 + \lambda )^{\frac{1}{1 - \alpha _1}} \beta _{1}^{\frac{\alpha _1 + \alpha _3 - 1}{1 - \alpha _1}}. \end{aligned}$$

From Lemma 6, we know that $\frac{k_{t+1}^{*}}{k_{t}^{*}} = \beta _{1} (1 + r^{*}_{t})$. Moreover, $\frac{w_{t}^{*}}{k_{t}^{*}} = \frac{\alpha _2}{\alpha _1} (1 + r^{*}_{t})$. Now (39) is straightforward. It also follows from Lemma 6 that $\frac{s_{t+1}^{j*}}{s_{t}^{j*}} = \beta _{1} (1 + r^{*}_{t})$ for $j \in J$, which proves (40).

Clearly, for $j \in J$,

$$\begin{aligned} \frac{c_{t}^{j*}}{k_{t}^{*}} = (1 - \beta _{1})(1 + r_{t}^{*}) \frac{s_{t-1}^{j*}}{k_{t}^{*}} + \frac{w_{t}^{*}}{k_{t}^{*}}, \end{aligned}$$

and thus the sequence $c_{t}^{j*}/k_{t}^{*}$ converges to a positive constant as $t \rightarrow \infty$. For $j \notin J$, we have $c_{t}^{j*} = w_{t}^{*}$. Thus, consumption of all agents asymptotically grows at a constant rate. This proves (41).

It remains to note that (42) follows from the Hotelling rule. $\square$

Appendix 2: Public property regime

1.1 Competitive equilibrium under given extraction rates

Suppose that the economy at time $\tau$ is in a state $\mathcal {I}_{\tau -1} = \{ ( \hat{s}^{j}_{\tau -1} )_{j=1}^{L}, {\hat{R}}_{\tau -1} \}$, where $(\hat{s}_{\tau -1}^{j})_{j=1}^{L}$ are agents’ savings and ${\hat{R}}_{\tau -1}$ is the stock of natural resources. We assume that $\mathcal {I}_{\tau -1}$ is a non-degenerate state, i.e.,

$$\begin{aligned} \hat{s}_{\tau -1}^{j} \ge 0, \quad j = 1, \ldots , L; \qquad \frac{1}{L} \sum _{j=1}^L \hat{s}^{j}_{\tau -1}> 0; \qquad {\hat{R}}_{\tau -1} > 0. \end{aligned}$$

Suppose further we are given a sequence of extraction rates $\mathbb {E}_{\tau } = \{ \varepsilon _t \}_{t = \tau }^{\infty }$. We call $\mathbb {E}_{\tau }$ non-degenerate if it is bounded away from 0 and 1, i.e., if there exists $\delta > 0$ such that for all $t \ge \tau$, $\delta \le \varepsilon _{t} \le 1 - \delta$.

Given a sequence of extraction rates $\mathbb {E}_{\tau }$ and the resource stock $R_{\tau -1} = {\hat{R}}_{\tau -1}$, the volume of extraction $e_t$ and the dynamics of the exhaustible resource stock $R_{t}$ are recursively determined for $t \ge \tau$:

$$\begin{aligned} e_{t} = e_{t} (\mathbb {E}_{\tau }) = \frac{\varepsilon _t R_{t-1}}{L}, \quad R_{t} = R_{t} (\mathbb {E}_{\tau }) = (1 - \varepsilon _t) R_{t-1}, \quad t = \tau , \tau +1, \ldots . \end{aligned}$$

(43)

We use this notation to emphasize that the sequence of extraction rates determines the volume of extraction and the dynamics of the resource stock.

Definition

Let $\mathbb {E}_{\tau }$ be a non-degenerate sequence of extraction rates. A sequence

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

is a competitive $\mathbb {E}_{\tau }$-equilibrium starting from $\mathcal {I}_{\tau -1}$ if

• 1. For each $j = 1, \ldots , L$, the sequence $\{ c^{j**}_{t}, s^{j**}_{t} \}_{t = \tau }^{\infty }$ is a solution to the following utility maximization problem:

  $$\begin{aligned} \begin{array}{cc} \displaystyle \max \sum _{t = \tau }^\infty \beta _j^t \ln c_{t}^{j}, \\ s. t. \quad c_{t}^{j} + s_{t}^{j} \le \left( 1 + r_t \right) s^{j}_{t-1} + w_t + v_t, \quad t = \tau , \tau +1, \ldots , \\ s_{t}^{j} \ge 0, \quad t = \tau , \tau +1, \ldots \end{array} \end{aligned}$$

  (44)

  at $r_t=r_t^{**}$, $w_t = w_t^{**}$, $v_t = v_t^{**}$, $s^{j}_{\tau -1} = \hat{s}^{j}_{\tau -1}$;

• 2. Aggregate savings are equal to the capital stock:

  $$\begin{aligned} \sum _{j=1}^L s^{j**}_{t-1} = L k^{**}_t, \quad t = \tau , \tau +1, \ldots ; \end{aligned}$$

• 3. Capital is paid its marginal product:

  $$\begin{aligned} 1 + r^{**}_t = \alpha _1A_t (k_t^{**})^{\alpha _1 - 1} (e_t)^{\alpha _3}, \quad t = \tau , \tau +1, \ldots ; \end{aligned}$$

• 4. Labor is paid its marginal product:

  $$\begin{aligned} w^{**}_t = \alpha _2 A_t (k_t^{**})^{\alpha _1} (e_t)^{\alpha _3}, \quad t = \tau , \tau +1, \ldots ; \end{aligned}$$

• 5. The price of natural resources is equal to the marginal product:

  $$\begin{aligned} q^{**}_t = \alpha _3 A_t (k_t^{**})^{\alpha _1} (e_t)^{\alpha _3 - 1}, \quad t = \tau , \tau +1, \ldots ; \end{aligned}$$

• 6. Resource income is given by:

  $$\begin{aligned} v_t^{**} = q_t^{**} e_t, \quad t = \tau , \tau +1, \ldots . \end{aligned}$$

Here, we do not suppose that the Hotelling rule holds. The Hotelling rule is an equilibrium condition for the asset market. This is the reason why the Hotelling rule holds in the private property regime, where the stock of natural resources is an asset in which agents can invest. In the public property regime, the resource stock is not an asset, so there is no particular reason for the Hotelling rule to hold. Under some circumstances, the rate of change in the resource price is not equal to the interest rate.

It is clear that if

$$\begin{aligned} \mathcal {E}^{**}_{0} = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = 0, 1, \ldots } \end{aligned}$$

is a competitive $\mathbb {E}_{0}$-equilibrium starting from $\{ ( \hat{s}^{j}_{-1} )_{j=1}^{L}, {\hat{R}}_{-1} \}$, then for each $\tau = 1, 2, \ldots$, the sequence

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

is a competitive $\mathbb {E}_{\tau }$-equilibrium starting from $\{ ( s^{j**}_{\tau -1} )_{j=1}^{L}, R_{\tau -1} (\mathbb {E}_{0}) \}$. In other words, competitive equilibria are time consistent.

There always exists a competitive $\mathbb {E}_{\tau }$-equilibrium.

Theorem 2

For any non-degenerate state $\mathcal {I}_{\tau -1}$, there exists a competitive $\mathbb {E}_{\tau }$-equilibrium starting from $\mathcal {I}_{\tau -1}$.

Proof

See Borissov and Pakhnin (2016). $\square$

The issue with uniqueness is more subtle. We can only conjecture that the competitive equilibrium is unique, but we have no proof of this fact. At the same time, the following proposition maintains that the competitive equilibrium starting from the state where the whole stock of physical capital is owned by the most patient agents is unique.

Proposition 6

Suppose that $\mathcal {I}_{\tau -1}$ is such that $\hat{s}^{j}_{\tau -1} = 0$ for $j \notin J$. Then there exists a unique competitive $\mathbb {E}_{\tau }$-equilibrium starting from $\mathcal {I}_{\tau -1}$,

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots }, \end{aligned}$$

which is given for $t = \tau , \tau +1, \ldots$ by

where $s_{\tau -1}^{j**} = \hat{s}^{j}_{\tau -1}$, and $e_{t} = e_{t} (\mathbb {E}_{\tau })$.

This case is important because in every competitive $\mathbb {E}_{\tau }$-equilibrium less patient agents inevitably lose their capital with time. The following proposition verifies that the whole capital stock eventually belongs to the most patient agents.

Proposition 7

Suppose that

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

is a competitive $\mathbb {E}_{\tau }$-equilibrium starting from an arbitrary state $\mathcal {I}_{\tau -1}$. Then there exists a point in time T such that for all $t \ge T$,

$$\begin{aligned} s^{j**}_{t} = 0, \quad j \notin J. \end{aligned}$$

Proof of Propositions 6 and 7

is very similar to that of Propositions 1 and 2. Without loss of generality, let us consider a competitive $\mathbb {E}_{0}$-equilibrium

$$\begin{aligned} \mathcal {E}^{**}_{0} = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from $\mathcal {I}_{-1} = \{ ( \hat{s}^{j}_{-1} )_{j=1}^{L}, {\hat{R}}_{-1} \}$, and give a sketch of the proof.

Lemma 7

Let $\beta > 0$ be such that for some T

$$\begin{aligned} k^{**}_{t+1}> \beta (1 + r^{**}_{t}) k^{**}_{t} = \beta \alpha _1A_t (k_t^{**})^{\alpha _1} (e_t)^{\alpha _3}, \quad t > T. \end{aligned}$$

If $\beta _j < \beta$, then $s^{j**}_{t} = 0$ for all sufficiently large t.

Proof

This lemma can be proved in the same way as Lemma 1. $\square$

Lemma 8

$$\begin{aligned} k^{**}_{t+1} \le \beta _1 (1 + r^{**}_{t}) k^{**}_{t}, \quad t = 0, 1, \ldots . \end{aligned}$$

Proof

It is sufficient to repeat the argument used in the proof of Lemma 2. $\square$

Lemma 9

$$\begin{aligned} w^{**}_{t+1} \le \beta _1 (1 + r^{**}_{t+1}) w^{**}_{t}, \qquad v^{**}_{t+1} \le \beta _1 (1 + r^{**}_{t+1}) v^{**}_{t}, \qquad t = 0, 1, \ldots . \end{aligned}$$

Proof

This statement follows from Lemma 8. $\square$

Lemma 10

$$\begin{aligned} s^{j**}_{t+1} \ge \beta _1 (1 + r^{**}_{t+1}) s^{j**}_{t}, \quad j \in J, \quad t = -1, 0, \ldots . \end{aligned}$$

Proof

This lemma can be proved in the same way as Lemma 4. $\square$

Lemma 11

For any $\delta > 0$, there exists a point in time T such that for all $t > T$,

$$\begin{aligned} k^{**}_{t+1} > \beta _1 (1 - \delta ) (1 + r^{**}_{t}) k^{**}_{t}. \end{aligned}$$

Proof

This lemma can be proved in the same way as Lemma 5. $\square$

Proposition 7 follows from Lemmas 7 and 11. Proposition 6 follows directly from Lemma 12 which explicitly constructs a competitive $\mathbb {E}_{0}$-equilibrium starting from the state $\mathcal {I}_{-1}$ such that $\hat{s}^{j}_{-1} = 0$ for $j \notin J$.

Lemma 12

Suppose that

$$\begin{aligned} k^{**}_{0} = \frac{1}{L} \sum _{j \in J} \hat{s}^{j}_{-1}, \ \text {i.e.}, \ \hat{s}^{j}_{-1} = 0, \ j \notin J. \end{aligned}$$

Then for all $t = 0, 1, \ldots$,

Proof

By Lemma 10,

$$\begin{aligned} \beta _1 (1 + r^{**}_{0}) k^{**}_{0} = \beta _1 (1 + r^{**}_{0}) \frac{1}{L} \sum _{j \in J} \hat{s}^{j}_{-1} \le \frac{1}{L} \sum _{j \in J} s^{j**}_{0} \le k^{**}_{1}. \end{aligned}$$

At the same time, by Lemma 8,

$$\begin{aligned} k^{**}_{1} \le \beta _1 (1 + r^{**}_{0}) k^{**}_{0}. \end{aligned}$$

Therefore, $k^{**}_{1} = \beta _1 (1 + r^{**}_{0}) k^{**}_{0}$, and hence $s^{j**}_{0} = \beta _1 (1 + r^{**}_{0}) \hat{s}^{j}_{-1}$ for $j \in J$, while $s^{j**}_{0} = 0$ for $j \notin J$. We have proved the lemma for $t = 0$. To prove it for $t = 1, 2, \ldots$, it is sufficient to repeat the argument. $\square$

This completes the proof of Propositions 6 and 7. $\square$

1.2 Balanced-growth equilibrium under given extraction rate

Suppose that the sequence of extraction rates is constant,

$$\begin{aligned} \mathbb {E}^{\varepsilon }_{\tau } = \mathbb {E}^{\varepsilon } = \{ \varepsilon , \varepsilon , \varepsilon , \ldots \}. \end{aligned}$$

Then, clearly,

$$\begin{aligned} R_{t} = (1 - \varepsilon )^{t +1 - \tau } {\hat{R}}_{\tau -1}, \qquad e_{t} = (1 - \varepsilon )^{t - \tau } \frac{\varepsilon {\hat{R}}_{\tau -1}}{L}, \qquad t = \tau , \tau +1, \ldots . \end{aligned}$$

Definition

A competitive $\mathbb {E}^{\varepsilon }_{\tau }$-equilibrium

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

starting from $\mathcal {I}_{\tau -1}$ is called a balanced-growth $\mathbb {E}^{\varepsilon }$-equilibrium if there exist an equilibrium rate of balanced growth $\gamma ^{**}$ and the rate of change in the resource price, $\pi ^{**}$, such that for $t = \tau , \tau + 1, \ldots$,

$$\begin{aligned} c^{j**}_{t+1}= & {} (1 + \gamma ^{**}) c^{j**}_{t}, \qquad s^{j**}_{t} = (1 + \gamma ^{**}) s^{j**}_{t-1}, \qquad j = 1, \ldots , L, \end{aligned}$$

(45)

$$\begin{aligned} k^{**}_{t+1}= & {} (1 + \gamma ^{**}) k^{**}_{t}, \qquad w^{**}_{t+1} = (1 + \gamma ^{**}) w^{**}_{t}, \qquad v_{t+1}^{**} = (1 + \gamma ^{**}) v^{**}_{t}, \end{aligned}$$

(46)

$$\begin{aligned} q^{**}_{t+1}= & {} \left( 1 + \pi ^{**} \right) q^{**}_{t}, \qquad 1+r^{**}_{t} = 1+r^{**} . \end{aligned}$$

(47)

The following proposition provides necessary and sufficient conditions for the existence of a balanced-growth $\mathbb {E}^{\varepsilon }$-equilibrium. In particular, this proposition maintains that in a balanced-growth equilibrium only the most patient agents make positive savings and own the whole capital stock.

Proposition 8

Suppose that a constant sequence of extraction rates $\mathbb {E}^{\varepsilon }$ is given. A balanced-growth $\mathbb {E}^{\varepsilon }$-equilibrium

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

starting from a non-degenerate state $\mathcal {I}_{\tau -1} = \{ ( \hat{s}^{j}_{\tau -1} )_{j=1}^{L}, {\hat{R}}_{\tau -1} \}$ exists if and only if

$$\begin{aligned}&\displaystyle \hat{s}^{j}_{\tau -1} = 0, \quad j \notin J, \\&\displaystyle \alpha _1 A_{\tau } \left( \frac{1}{L} \sum _{j = 1}^{L} \hat{s}_{\tau -1}^{j} \right) ^{\alpha _1 - 1} \left( \frac{\varepsilon {\hat{R}}_{\tau -1}}{L} \right) ^{\alpha _3} = \left( 1 + \lambda \right) ^{\frac{1}{1-\alpha _1}} \left( 1 - \varepsilon \right) ^{\frac{\alpha _3}{1-\alpha _1}} \frac{1}{\beta _{1}}, \end{aligned}$$

and (45)–(47) hold.

Proof

It can be proved exactly in the same way as Proposition 3. $\square$

The following proposition maintains that the interest rate, the equilibrium rate of balanced growth and the rate of change in the resource price are determined by the parameters of the model and by the constant over time extraction rate $\varepsilon$.

Proposition 9

Suppose that a constant sequence of extraction rates $\mathbb {E}^{\varepsilon }$ is given. In a balanced-growth $\mathbb {E}^{\varepsilon }$-equilibrium, the interest rate, the equilibrium rate of balanced growth and the rate of change in the resource price are determined as follows:

$$\begin{aligned} 1+\gamma ^{**} = \left( 1 + \lambda \right) ^{\frac{1}{1-\alpha _1}} \left( 1 - \varepsilon \right) ^{\frac{\alpha _3}{1-\alpha _1}}, \qquad 1 + \pi ^{**} = \frac{1+\gamma ^{**}}{1 - \varepsilon }, \qquad 1 + r^{**} = \frac{1+\gamma ^{**}}{\beta _{1}}. \end{aligned}$$

Proof

The proof is similar to that of Proposition 4. $\square$

The following proposition maintains that every competitive $\mathbb {E}^{\varepsilon }_{\tau }$-equilibrium under given constant sequence of extraction rates converges in some sense to a balanced-growth $\mathbb {E}^{\varepsilon }$-equilibrium.

Proposition 10

Every competitive $\mathbb {E}_{\tau }^{\varepsilon }$-equilibrium starting from an arbitrary state $\mathcal {I}_{\tau -1}$ satisfies the following asymptotic properties:

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } 1 + r_{t}^{**} = 1+r^{**} = \frac{1 + \gamma ^{**}}{\beta _{1}}, \\&\displaystyle \lim _{t \rightarrow \infty } \frac{k^{**}_{t+1}}{k^{**}_{t}} = \lim _{t \rightarrow \infty } \frac{w^{**}_{t+1}}{w^{**}_{t}} = \lim _{t \rightarrow \infty } \frac{v^{**}_{t+1}}{v^{**}_{t}} = 1 + \gamma ^{**}, \\&\displaystyle \lim _{t \rightarrow \infty } \frac{s^{j**}_{t+1}}{s^{j**}_{t}} = 1 + \gamma ^{**}, \quad j \in J, \\&\displaystyle \lim _{t \rightarrow \infty } \frac{c^{j**}_{t+1}}{c^{j**}_{t}} = 1 + \gamma ^{**}, \quad j = 1, \ldots , L, \\&\displaystyle \lim _{t \rightarrow \infty } \frac{q^{**}_{t+1}}{q^{**}_{t}} = 1+\pi ^{**}, \end{aligned}$$

where $1 + \gamma ^{**} = \left( 1 + \lambda \right) ^{\frac{1}{1-\alpha _1}} \left( 1 - \varepsilon \right) ^{\frac{\alpha _3}{1-\alpha _1}}$, and $1+\pi ^{**} = \frac{1+\gamma ^{**}}{1 - \varepsilon }$.

Proof

The proof is similar to that of Proposition 5. $\square$

1.3 Time $\tau$ extraction rate

Before making extraction rates endogenous, let us explore the dependence of a competitive $\mathbb {E}_{\tau }$-equilibrium on the time $\tau$ extraction rate.

Suppose we are given a non-degenerate sequence of extraction rates $\mathbb {E}^0_{\tau } = \{ \varepsilon ^0_t\}_{t = \tau }^{\infty }$. Assume that $\varepsilon ^{0}_{\tau }$ is replaced by some other extraction rate $\varepsilon _{\tau }$, while all future extraction rates remain intact. Clearly, the volumes of extraction and the dynamics of the resource stock before and after this replacement are linked in the following way:

$$\begin{aligned} {\tilde{R}}_{t}= & {} \frac{1 - \varepsilon _{\tau }}{1 - \varepsilon ^{0}_{\tau }} R_{t}, \qquad t = \tau , \tau +1, \ldots , \\ {\tilde{e}}_{\tau }= & {} \frac{\varepsilon _{\tau }}{ \varepsilon ^{0}_{\tau }} e_{\tau }, \quad {\tilde{e}}_{t} = \frac{1 - \varepsilon _{\tau }}{1 - \varepsilon ^{0}_{\tau }} e_{t}, \qquad t = \tau +1, \tau +2, \ldots . \end{aligned}$$

A competitive $\mathbb {E}^0_{\tau }$-equilibrium should also change. The change in the competitive $\mathbb {E}^0_{\tau }$-equilibrium and the dependence of a new equilibrium on $\varepsilon _{\tau }$ is characterized in the following lemma.

Lemma 13

Suppose that for a non-degenerate sequence of extraction rates, $\mathbb {E}^0_{\tau } = \{ \varepsilon ^0_t\}_{t = \tau }^{\infty }$, the sequence

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

is a competitive $\mathbb {E}^0_{\tau }$-equilibrium starting from $\mathcal {I}_{\tau -1} = \{ ( \hat{s}^{j}_{\tau -1} )_{j=1}^{L}, {\hat{R}}_{\tau -1} \}$.

Let

$$\begin{aligned} \mathbb {E}_{\tau } = \{ \varepsilon _{\tau }, \varepsilon ^0_{\tau +1}, \varepsilon ^0_{\tau +2}, \ldots \}, \end{aligned}$$

and

$$\begin{aligned} \nu _{\tau } = \frac{1 - \varepsilon _{\tau }}{1 - \varepsilon ^0_{\tau }}. \end{aligned}$$

Consider the sequence

$$\begin{aligned} {\tilde{\mathcal {E}}}_{\tau } (\varepsilon _{\tau })\! = \!\left\{ ({\tilde{c}}^{j}_t (\varepsilon _{\tau }))_{j=1}^L, ({\tilde{s}}^{j}_t (\varepsilon _{\tau }))_{j=1}^L, {\tilde{k}}_t (\varepsilon _{\tau }), {\tilde{r}}_t (\varepsilon _{\tau }), {\tilde{w}}_t (\varepsilon _{\tau }), {\tilde{q}}_t (\varepsilon _{\tau }), {\tilde{v}}_t (\varepsilon _{\tau }) \right\} _{t = \tau , \tau +1, \ldots }\!, \end{aligned}$$

given by

$$\begin{aligned}&\displaystyle {\tilde{k}}_{\tau +1} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} k_{\tau +1}^{**}, \end{aligned}$$

(48)

$$\begin{aligned}&\displaystyle {\tilde{k}}_{t+1} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }} \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})} k_{t+1}^{**}, \quad t = \tau +1, \tau +2, \ldots , \end{aligned}$$

(49)

$$\begin{aligned}&\displaystyle {\tilde{w}}_{\tau } (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} w_{\tau }^{**}, \end{aligned}$$

(50)

$$\begin{aligned}&\displaystyle {\tilde{w}}_{t} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }} \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})} w_t^{**}, \quad t = \tau +1, \tau +2, \ldots , \end{aligned}$$

(51)

$$\begin{aligned}&\displaystyle {\tilde{v}}_{\tau } (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} v_{\tau }^{**}, \end{aligned}$$

(52)

$$\begin{aligned}&\displaystyle {\tilde{v}}_{t} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }} \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})} v_t^{**}, \quad t = \tau +1, \tau +2, \ldots , \end{aligned}$$

(53)

$$\begin{aligned}&\displaystyle {\tilde{c}}^{j}_{\tau } (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} c_{\tau }^{j**}, \quad j = 1, \ldots , L, \end{aligned}$$

(54)

$$\begin{aligned}&\displaystyle {\tilde{c}}^{j}_{t} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }} \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})} c_t^{j**}, \quad t = \tau +1, \tau +2, \ldots , \nonumber \\&\displaystyle j =1, \ldots , L, \end{aligned}$$

(55)

$$\begin{aligned}&\displaystyle {\tilde{s}}^{j}_{\tau } (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} s_{\tau }^{j**}, \quad j = 1, \ldots , L, \end{aligned}$$

(56)

$$\begin{aligned}&\displaystyle {\tilde{s}}^{j}_{t} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }} \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})} s_t^{j**}, \quad t = \tau +1, \tau +2, \ldots , \nonumber \\&\displaystyle j = 1, \ldots , L, \end{aligned}$$

(57)

$$\begin{aligned}&\displaystyle 1 + {\tilde{r}}_{\tau } (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} (1 + r_{\tau }^{**}), \end{aligned}$$

(58)

$$\begin{aligned}&\displaystyle 1 + {\tilde{r}}_{t} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1) \alpha _1^{t - \tau - 1}} \nu _{\tau }^{\alpha _3 \alpha _1^{t - \tau - 1}} (1 + r_t^{**}), \quad t = \tau +1, \tau +2, \ldots , \end{aligned}$$

(59)

$$\begin{aligned}&\displaystyle {\tilde{q}}_{\tau } (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 - 1} q_{\tau }^{**}, \end{aligned}$$

(60)

$$\begin{aligned}&\displaystyle {\tilde{q}}_{t} (\varepsilon _{\tau }) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }} \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1}) - 1} q_t^{**}, \quad t = \tau +1, \tau +2, \ldots . \end{aligned}$$

(61)

The sequence ${\tilde{\mathcal {E}}}_{\tau } (\varepsilon _{\tau })$ is a competitive $\mathbb {E}_{\tau }$-equilibrium starting from the same state $\mathcal {I}_{\tau -1} = \{ ( \hat{s}^{j}_{\tau -1} )_{j=1}^{L}, {\hat{R}}_{\tau -1} \}$.

This lemma plays a very important role in our further considerations. If both the competitive $\mathbb {E}^0_{\tau }$-equilibrium and the competitive $\mathbb {E}_{\tau }$-equilibrium are unique, then (48)–(61) provides formulas of transition from the equilibrium before the change in the time $\tau$ extraction rate to the equilibrium after the change. If we cannot guarantee the uniqueness of a competitive $\mathbb {E}_{\tau }$-equilibrium, then the interpretation of this lemma is slightly different. It maintains that after the change in the time $\tau$ extraction rate, there exists a competitive equilibrium which is given by (48)–(61).

Proof

For the simplicity of exposition, let us slightly abuse the notation and write simply ${\tilde{k}}_{t}$, ${\tilde{w}}_{t}$, etc., instead of ${\tilde{k}}_{t} (\varepsilon _{\tau })$, ${\tilde{w}}_{t} (\varepsilon _{\tau })$, etc.

Obviously, ${\tilde{k}}_{\tau } = k^{**}_{\tau }$, as the initial state is the same. Directly from (48)–(49) and (56)–(57), we get

$$\begin{aligned} {\tilde{k}}_{t+1} = \sum _{j=1}^L {\tilde{s}}^{j}_{t}, \quad t = \tau , \tau +1, \ldots . \end{aligned}$$

Let us show that capital is paid its marginal product:

$$\begin{aligned}&1 + {\tilde{r}}_{\tau } = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} (1 + r_{\tau }^{**}) \\&\quad = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} \alpha _1 A_{\tau } (e_{\tau }^{**})^{\alpha _3} (k_{\tau }^{**})^{\alpha _1 - 1} = \alpha _1 A_{\tau } ({\tilde{e}}_{\tau })^{\alpha _3}({\tilde{k}}_{\tau })^{\alpha _1 - 1},\\&\quad 1 + {\tilde{r}}_{\tau +1} = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1)} \nu _{\tau }^{\alpha _3} ( 1 + r_{\tau +1}^{**}) \\&\quad = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1)} \nu _{\tau }^{\alpha _3} \alpha _1 A_{\tau +1} (e_{\tau +1}^{**})^{\alpha _3} (k_{\tau +1}^{**})^{\alpha _1 - 1} \\&\quad = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1)} \nu _{\tau }^{\alpha _3} \alpha _1 A_{\tau +1} ({\tilde{e}}_{\tau +1})^{\alpha _3}({\tilde{k}}_{\tau +1})^{\alpha _1 - 1} \nu _{\tau }^{- \alpha _3} \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (1 - \alpha _1)} \\&\quad = \alpha _1 A_{\tau +1} ({\tilde{e}}_{\tau +1})^{\alpha _3}({\tilde{k}}_{\tau +1})^{\alpha _1 - 1},\\&\quad 1 + {\tilde{r}}_{t} = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1) \alpha _1^{t - \tau - 1}} \nu _{\tau }^{\alpha _3 \alpha _1^{t - \tau - 1}} (1 + r_t^{**}) \\&\quad = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1) \alpha _1^{t - \tau - 1}} \nu _{\tau }^{\alpha _3 \alpha _1^{t - \tau - 1}} \alpha _1A_t (e_t^{**})^{\alpha _3} (k_t^{**})^{\alpha _1 - 1} \\&\quad = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1) \alpha _1^{t - \tau - 1}} \nu _{\tau }^{\alpha _3 \alpha _1^{t - \tau - 1}} \alpha _1 A_t ({\tilde{e}}_{t})^{\alpha _3}({\tilde{k}}_{t})^{\alpha _1 - 1} \nu _{\tau }^{- \alpha _3} \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{(\alpha _3 \alpha _1^{t - \tau - 1})(1 - \alpha _1)} \times \\&\qquad \times \nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 2}) (1 - \alpha _1)}= \alpha _1 A_t ({\tilde{e}}_{t})^{\alpha _3}({\tilde{k}}_{t})^{\alpha _1 - 1}, \quad t = \tau +2, \tau +3, \ldots . \end{aligned}$$

Similar calculations show that

$$\begin{aligned} {\tilde{w}}_{t} = \alpha _2 A_t ({\tilde{k}}_{t})^{\alpha _1} ({\tilde{e}}_{t})^{\alpha _3}, \qquad {\tilde{q}}_{t} = \alpha _3 A_t ({\tilde{k}}_{t})^{\alpha _1} ({\tilde{e}}_{t})^{\alpha _3 - 1}, \qquad t = \tau , \tau +1, \ldots , \end{aligned}$$

and it is easy to check that

$$\begin{aligned} {\tilde{v}}_{t} = {\tilde{q}}_{t} {\tilde{e}}_{t}, \quad t = \tau , \tau +1, \ldots . \end{aligned}$$

It remains to show that the sequence $\{ ({\tilde{c}}^{j}_{t})_{j=1}^L, ({\tilde{s}}^{j}_{t})_{j=1}^L \}_{t = \tau }^{\infty }$ is a solution to the problem

$$\begin{aligned}&\displaystyle \max \sum _{t=\tau }^{\infty } \beta _j^{t} \ln c^j_t \\&\displaystyle \quad \text {s. t.} \quad c^j_t + s^j_t = (1 + r_t) s^j_{t-1} + w_t + v_t, \quad t = \tau , \tau +1, \ldots , \\&\displaystyle \quad s_t^j \ge 0, \quad t = \tau , \tau +1, \ldots , \end{aligned}$$

at $r_{t} = {\tilde{r}}_t$, $w_{t} = {\tilde{w}}_t$, and $v_{t} = {\tilde{v}}_t$, or, equivalently, that the following conditions hold ($j = 1, \ldots , L$):

$$\begin{aligned}&\displaystyle {\tilde{c}}^j_t + {\tilde{s}}^j_t = (1+{\tilde{r}}_t) {\tilde{s}}^j_{t-1} + {\tilde{w}}_t + {\tilde{v}}_t, \quad t = \tau , \tau +1, \ldots , \end{aligned}$$

(62)

$$\begin{aligned}&\displaystyle {\tilde{c}}_{t+1}^{j} \ge \beta _{j} (1+{\tilde{r}}_{t+1}) {\tilde{c}}_{t}^{j} \ (= \text {if} \ {\tilde{s}}_{t}^{j} > 0), \quad t = \tau , \tau +1, \ldots , \end{aligned}$$

(63)

$$\begin{aligned}&\displaystyle \frac{\beta _j^t {\tilde{s}}_{t}^j}{{\tilde{c}}_{t}^{j}} \xrightarrow [t \rightarrow \infty ]{}0. \end{aligned}$$

(64)

To do this, note that the sequence $\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L \}_{t=\tau }^{\infty }$ is a solution to maximization problem (44) and hence satisfies the following conditions:

Consider (62) for $t = \tau$. We have

$$\begin{aligned} {\tilde{c}}^j_{\tau } + {\tilde{s}}^j_{\tau }= & {} \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} ( c_{\tau }^{j**} + s_{\tau }^{j**} ) = \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} \left( (1+r^{**}_{\tau }) \hat{s}^{j}_{\tau -1} + w^{**}_{\tau } + v^{**}_{\tau } \right) \\= & {} \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} (1+r^{**}_{\tau }) \hat{s}^{j}_{\tau -1} +\left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} w^{**}_{\tau } + \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} v^{**}_{\tau } = (1+{\tilde{r}}_{\tau }) \hat{s}^{j}_{\tau -1} {+} {\tilde{w}}_{\tau } {+} {\tilde{v}}_{\tau }. \end{aligned}$$

Consider (62) for $t = \tau + 1$:

$$\begin{aligned} {\tilde{c}}^j_{\tau +1} + {\tilde{s}}^j_{\tau +1} \\= & {} \left( \frac{\varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1} \nu _{\tau }^{\alpha _3} ( c_{\tau +1}^{j**} + s_{\tau +1}^{j**} )\\= & {} \left( \frac{\varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1} \nu _{\tau }^{\alpha _3} \left( (1+r^{**}_{\tau +1}) s_{\tau }^{j**} + w^{**}_{\tau +1} + v^{**}_{\tau +1} \right) \\= & {} \left( \frac{\varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (\alpha _1 - 1)} \nu _{\tau }^{\alpha _3} (1+r^{**}_{\tau +1}) \left( \frac{\varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3} s_{\tau }^{j**}\\&+\left( \frac{\varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1} \nu _{\tau }^{\alpha _3} (w^{**}_{\tau +1} + v^{**}_{\tau +1}) = (1+{\tilde{r}}_{\tau +1}) {\tilde{s}}^j_{\tau } + {\tilde{w}}_{\tau +1} + {\tilde{v}}_{\tau +1}. \end{aligned}$$

The validity of conditions (62) for $t \ge \tau + 2$ can be proved similarly. The same arguments prove (63) for $t \ge \tau$. Notice also that

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{\beta _j^t {\tilde{s}}_{t}^j}{{\tilde{c}}_{t}^{j}}= & {} \lim _{t \rightarrow \infty } \beta _j^t \frac{\left( \frac{ \varepsilon _{\tau }}{\varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }}}{\left( \frac{\varepsilon _{\tau }}{\varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1^{t - \tau }}} \frac{\nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})}}{\nu _{\tau }^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha _1^{t - \tau - 1})}} \frac{s_{t}^{j**}}{c_t^{j**}}\\= & {} \lim _{t \rightarrow \infty } \frac{\beta _j^t s_{t}^{j**}}{c_{t}^{j**}} = 0. \end{aligned}$$

This completes the proof of the lemma. $\square$

1.4 Time $\tau$ voting equilibrium

We have characterized a competitive equilibrium and a balanced-growth equilibrium under given sequence of extraction rates. Now we make extraction rates endogenous and introduce voting into our model.

Suppose that we start at time $\tau$. The economy is in a non-degenerate state $\mathcal {I}_{\tau -1} = \{ ( \hat{s}^{j}_{\tau -1} )_{j=1}^{L}, {\hat{R}}_{\tau -1} \}$. Suppose further that agents have some expectations about future extraction rates, represented by a non-degenerate sequence $\{ \varepsilon _{t}^{e} \}_{t = \tau +1}^{\infty }$, and they vote on the time $\tau$ extraction rate.

For any $\varepsilon _{\tau } \in (0,1)$, consider the non-degenerate sequence of extraction rates

$$\begin{aligned} \mathbb {E}_{\tau } (\varepsilon _{\tau }) = \{ \varepsilon _{\tau }, \varepsilon _{\tau +1}^{e}, \varepsilon _{\tau +2}^{e}, \ldots \}. \end{aligned}$$

Let us assume that for any $\varepsilon _{\tau } \in (0,1)$ there is a unique competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium starting from $\mathcal {I}_{\tau -1}$,

$$\begin{aligned} \mathcal {E}_{\tau }^{**} (\varepsilon _{\tau }) = \left\{ (c^{j**}_{t} (\varepsilon _{\tau }))_{j=1}^L, (s^{j**}_{t} (\varepsilon _{\tau }))_{j=1}^L, k^{**}_t (\varepsilon _{\tau }), r_t^{**} (\varepsilon _{\tau }), w_t^{**} (\varepsilon _{\tau }), q_t^{**} (\varepsilon _{\tau }), v_t^{**} (\varepsilon _{\tau }) \right\} _{t = \tau , \ldots }. \end{aligned}$$

It is clear that $\mathcal {E}_{\tau }^{**} (\varepsilon _{\tau })$ depends on the expectations and on the parameters of the model as well. However, here we underline its dependence only on $\varepsilon _{\tau }$, as it is the value on which agents vote.

Under the uniqueness assumption, agents’ preferences over the time $\tau$ extraction rate are represented by the following indirect utility functions:

$$\begin{aligned} \mathcal {U}_{\tau }^{j} (\varepsilon _{\tau }) = \ln c^{j**}_{\tau } (\varepsilon _{\tau }) + \beta _j \ln c^{j**}_{\tau +1} (\varepsilon _{\tau }) + \cdots , \quad j = 1, \ldots , L. \end{aligned}$$

Definition

Suppose that for any $\varepsilon _{\tau } \in (0,1)$ there is a unique competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium starting from $\mathcal {I}_{\tau -1}$, $\mathcal {E}_{\tau }^{**} (\varepsilon _{\tau })$. We call a couple $\{\varepsilon ^{**}_{\tau }, \mathcal {E}_{\tau }^{**}\}$ a time $\tau$ voting equilibrium if $\varepsilon ^{**}_{\tau }$ is a Condorcet winner in voting on the time $\tau$ extraction rate, and $\mathcal {E}_{\tau }^{**} = \mathcal {E}_{\tau }^{**} (\varepsilon ^{**}_{\tau })$.

Since the functions $\mathcal {U}^j (\varepsilon _{\tau })$, $j = 1, \ldots , L$, are strictly concave, the agents’ preferences are single-peaked. Hence, the median voter theorem applies, and at each point in time there exists a Condorcet winner. Note that the time $\tau$ voting equilibrium consists of the time $\tau$ voting equilibrium extraction rate $\varepsilon ^{**}_{\tau }$ and the corresponding competitive equilibrium.

In order to determine a Condorcet winner, let us consider the preferred time $\tau$ extraction rate for agent j. This is the value $\varepsilon ^j_{\tau }$ such that

$$\begin{aligned} \mathcal {U}_{\tau }^{j} (\varepsilon ^j_{\tau }) > \mathcal {U}_{\tau }^{j} (\varepsilon _{\tau }) \quad \forall \ \varepsilon _{\tau } \ne \varepsilon ^j_{\tau }. \end{aligned}$$

From Lemma 13, we know how the consumption stream of every agent depends on the time $\tau$ extraction rate, which allows us to obtain agents’ preferred values of time $\tau$ extraction rate.

Proposition 11

Suppose that for any $\varepsilon _{\tau } \in (0,1)$ there is a unique competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium starting from $\mathcal {I}_{\tau -1}$. The preferred time $\tau$ extraction rate for each agent j is given by

$$\begin{aligned} \varepsilon ^j_{\tau } = 1 - \beta _j. \end{aligned}$$

(65)

Proof

Let us take some $\varepsilon ^{0}_{\tau } \in (0,1)$, and consider the non-degenerate sequence

$$\begin{aligned} \mathbb {E}_{\tau } (\varepsilon ^{0}_{\tau }) = \{ \varepsilon ^{0}_{\tau }, \varepsilon _{\tau +1}^{e}, \varepsilon _{\tau +2}^{e}, \ldots \}. \end{aligned}$$

By assumption, there is a unique competitive $\mathbb {E}_{\tau } (\varepsilon ^{0}_{\tau })$-equilibrium

$$\begin{aligned} \mathcal {E}_{\tau }^{**} (\varepsilon ^{0}_{\tau }) = \left\{ (c^{j**}_{t} (\varepsilon ^{0}_{\tau }))_{j=1}^L, (s^{j**}_{t} (\varepsilon ^{0}_{\tau }))_{j=1}^L, k^{**}_t (\varepsilon ^{0}_{\tau }), r_t^{**} (\varepsilon ^{0}_{\tau }), w_t^{**} (\varepsilon ^{0}_{\tau }), \right. \\ \left. q_t^{**} (\varepsilon ^{0}_{\tau }), v_t^{**} (\varepsilon ^{0}_{\tau }) \right\} _{t = \tau , \ldots } \end{aligned}$$

starting from $\mathcal {I}_{\tau -1}$. We use this equilibrium as a benchmark.

Further, for any $\varepsilon _{\tau } \in (0,1)$, there is a unique competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium

$$\begin{aligned} \mathcal {E}_{\tau }^{**} (\varepsilon _{\tau }) = \Bigg \{ (c^{j**}_{t} (\varepsilon _{\tau }))_{j=1}^L, (s^{j**}_{t} (\varepsilon _{\tau }))_{j=1}^L, k^{**}_t (\varepsilon _{\tau }), r_t^{**} (\varepsilon _{\tau }), w_t^{**} (\varepsilon _{\tau }), \\ q_t^{**} (\varepsilon _{\tau }), v_t^{**} (\varepsilon _{\tau }) \Bigg \}_{t = \tau , \ldots } \end{aligned}$$

starting from $\mathcal {I}_{\tau -1}$.

From Lemma 13, we know that if the time $\tau$ extraction rate changes from $\varepsilon ^{0}_{\tau }$ to $\varepsilon _{\tau }$, the benchmark equilibrium $\mathcal {E}_{\tau }^{**} (\varepsilon ^{0}_{\tau })$ transforms to the “new” equilibrium $\mathcal {E}_{\tau }^{**} (\varepsilon _{\tau })$ according to the formulas (48)–(61). In particular, the consumption stream of agent j is given by (54)–(55). Therefore, the indirect utility function of agent j in this equilibrium is given by:

$$\begin{aligned} \mathcal {U}_{\tau }^{j} (\varepsilon _{\tau })= & {} \ln c^{j**}_{\tau } (\varepsilon _{\tau }) + \beta _j \ln c^{j**}_{\tau +1} (\varepsilon _{\tau }) + \cdots \\= & {} \alpha _3 \ln \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) + \ln c_{\tau }^{j**} (\varepsilon ^{0}_{\tau })\\&+\,\beta _j \alpha _3 \alpha _1 \ln \left( \frac{ \varepsilon _{\tau }}{ \varepsilon ^0_{\tau }} \right) + \beta _j \alpha _3 \ln \left( \frac{1 - \varepsilon _{\tau }}{1 - \varepsilon ^0_{\tau }} \right) + \beta _j \ln c_{\tau +1}^{j**} (\varepsilon ^{0}_{\tau }) + \cdots \\= & {} {\varGamma }^j + \alpha _3 \ln \varepsilon _{\tau } (1 + \beta _j \alpha _1 + \beta ^2_j \alpha ^2_1 + \cdots )\\&+ \beta _j \alpha _3 \ln (1 - \varepsilon _{\tau }) \left( 1 + \beta _j (1 + \alpha _1) + \beta ^2_j (1 + \alpha _1 + \alpha ^2_1) + \cdots \right) \\= & {} {\varGamma }^j + \frac{\alpha _3}{1 - \beta _j \alpha _1} \ln \varepsilon _{\tau } + \frac{\beta _j \alpha _3}{1 - \beta _j} (1 + \beta _j \alpha _1 + \beta ^2_j \alpha ^2_1 + \cdots ) \ln (1 - \varepsilon _{\tau }) \\= & {} {\varGamma }^j + \frac{\alpha _3}{1 - \beta _j \alpha _1} \ln \varepsilon _{\tau } + \frac{\alpha _3}{1 - \beta _j \alpha _1} \frac{\beta _j }{1 - \beta _j } \ln (1 - \varepsilon _{\tau }), \end{aligned}$$

where

$$\begin{aligned} {\varGamma }^j= & {} \ln \left[ \left( \frac{1}{\varepsilon ^0_{\tau }} \right) ^{\alpha _3} c_{\tau }^{j**} (\varepsilon ^{0}_{\tau }) \right] + \beta _j \ln \left[ \left( \frac{1}{\varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha _1} \left( \frac{1}{1 - \varepsilon ^0_{\tau }} \right) ^{\alpha _3} c_{\tau +1}^{j**} (\varepsilon ^{0}_{\tau }) \right] + \cdots \\&+\,\beta ^t_j \ln \left[ \left( \frac{1}{\varepsilon ^0_{\tau }} \right) ^{\alpha _3 \alpha ^{t}_1} \left( \frac{1}{1 - \varepsilon ^0_{\tau }} \right) ^{\alpha _3 (1 + \alpha _1 + \cdots + \alpha ^{t-1}_1)} c_{\tau +t}^{j**} (\varepsilon ^{0}_{\tau }) \right] + \cdots . \end{aligned}$$

is the term that depends on the parameters of the model and on the characteristics of the benchmark equilibrium $\mathcal {E}_{\tau }^{**} (\varepsilon ^{0}_{\tau })$ (the extraction rate $\varepsilon ^{0}_{\tau }$ and the consumption stream), but does not depend on $\varepsilon _{\tau }$, on which agents vote. Since the benchmark equilibrium exists, $- \infty< {\varGamma }^j < + \infty$, and hence the indirect utility function of each agent is well defined.

When voting on $\varepsilon _{\tau }$, agent j maximizes her indirect utility $\mathcal {U}_{\tau }^j (\varepsilon _{\tau })$, i.e., solves

$$\begin{aligned} \frac{\partial \mathcal {U}_{\tau }^j (\varepsilon _{\tau })}{\partial \varepsilon _{\tau }} = 0. \end{aligned}$$

This equation can be rewritten as

$$\begin{aligned} \frac{1}{\varepsilon _{\tau }} - \frac{\beta _j }{1 - \beta _j } \frac{1}{1 - \varepsilon _{\tau }} = 0. \end{aligned}$$

The solution to this equation is $\varepsilon ^j_{\tau } = 1 - \beta _j$. $\square$

Proposition 11 maintains that the preferred time $\tau$ extraction rate for every agent is constant over time and depends only on this agent’s discount factor. In particular, the preferred time $\tau$ extraction rate for agent j is time- and expectations-independent.

Now it is straightforward to see that the Condorcet winner in voting on the time $\tau$ extraction rate is

$$\begin{aligned} \varepsilon ^{**}_{\tau } = 1 - \beta _{med}, \end{aligned}$$

where $\beta _{med}$ is the median discount factor. Thus, the following theorem takes place.

Theorem 3

Suppose that for any $\varepsilon _{\tau } \in (0,1)$ there is a unique competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium starting from $\mathcal {I}_{\tau -1}$. Then there exists a unique time $\tau$ voting equilibrium $\{\varepsilon ^{**}_{\tau }, \mathcal {E}_{\tau }^{**}\}$. The equilibrium extraction rate is constant over time and given by

$$\begin{aligned} \varepsilon ^{**}_{\tau } = \varepsilon ^{**} = 1 - \beta _{med}. \end{aligned}$$

(66)

1.5 Intertemporal voting equilibrium

Suppose we are given an initial state $\mathcal {I}_{-1} = \{ ( \hat{s}^{j}_{-1} )_{j=1}^{L}, {\hat{R}}_{-1} \}$ and a non-degenerate sequence of extraction rates $\mathbb {E}^{**} = \mathbb {E}_{0}^{**} = \{ \varepsilon ^{**}_{t} \}_{t=0}^{\infty }$. Therefore, the volumes of extraction and the dynamics of the resource stock are also known:

$$\begin{aligned} e^{**}_t = e_{t} (\mathbb {E}^{**}), \qquad R^{**}_{t} = R_{t} (\mathbb {E}^{**}), \qquad t = 0, 1, \ldots . \end{aligned}$$

Let

$$\begin{aligned} \mathcal {E}^{**}_{0} = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = 0, 1, \ldots } \end{aligned}$$

be a competitive $\mathbb {E}^{**}$-equilibrium starting from $\mathcal {I}_{-1}$. Let for $\tau = 1, 2, \ldots$

$$\begin{aligned} \mathcal {E}^{**}_{\tau } = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

be the corresponding tail of $\mathcal {E}_{0}^{**}$, which is a competitive $\mathbb {E}^{**}_{\tau }$-equilibrium starting from $\mathcal {I}^{**}_{\tau -1} = \{ ( s^{j**}_{\tau -1} )_{j=1}^{L}, R^{**}_{\tau -1} \}$.

Definition

We call a couple $\{ \mathbb {E}^{**}, \mathcal {E}_{0}^{**} \}$ an intertemporal voting equilibrium starting from $\mathcal {I}_{-1}$ if for each time $\tau = 0, 1, \ldots$, a couple $\{ \varepsilon ^{**}_{\tau }, \mathcal {E}_{\tau }^{**} \}$ is a time $\tau$ voting equilibrium starting from $\mathcal {I}^{**}_{\tau -1}$ under perfect foresight about future extraction rates ($\varepsilon _{t}^{e} = \varepsilon _{t}^{**}, \ t = \tau +1, \tau +2, \ldots$).

The following theorem provides the characterization of the sequence of extraction rates in every intertemporal voting equilibrium.

Theorem 4

In every intertemporal voting equilibrium $\{ \mathbb {E}^{**}, \mathcal {E}_{0}^{**} \}$, the sequence of extraction rates $\mathbb {E}^{**}$ is constant over time and given by

$$\begin{aligned} \mathbb {E}^{**} = \mathbb {E}^{\varepsilon ^{**}} = \{ \varepsilon ^{**}, \varepsilon ^{**}, \ldots \}, \end{aligned}$$

(67)

where $\varepsilon ^{**}$ is defined by (66).

Proof

The sequence of extraction rates in every intertemporal voting equilibrium is a sequence of time $\tau$ equilibrium extraction rates. It follows from Theorem 3 that every equilibrium extraction rate is constant and given by (66). $\square$

The answer to the question about the existence and uniqueness of an intertemporal voting equilibrium is provided by the following theorem. It states that if the initial state is such that the whole capital stock belongs to the most patient agents, then an intertemporal voting equilibrium exists and is unique.

Theorem 5

Suppose that the initial state $\mathcal {I}_{-1}$ is such that $\hat{s}^{j}_{-1} = 0$ for $j \notin J$. Then there exists a unique intertemporal voting equilibrium $\{ \mathbb {E}^{**}, \mathcal {E}_{0}^{**} \}$ starting from $\mathcal {I}_{-1}$. The equilibrium sequence of extraction rates $\mathbb {E}^{**}$ is constant over time and given by (67), and $\mathcal {E}_{0}^{**}$ is a unique competitive $\mathbb {E}^{**}$-equilibrium starting from $\mathcal {I}_{-1}$, as described in Proposition 6.

Proof

It follows from Proposition 6 and Theorem 4. $\square$

1.6 Balanced-growth voting equilibrium

Definition

An intertemporal voting equilibrium $\{ \mathbb {E}^{**}, \mathcal {E}_{0}^{**} \}$ starting from $\mathcal {I}_{-1}$ is called a balanced-growth voting equilibrium if $\mathcal {E}_{0}^{**}$ is a balanced-growth $\mathbb {E}^{\varepsilon ^{**}}$-equilibrium starting from $\mathcal {I}_{-1}$, where $\varepsilon ^{**}$ is given by (66).

The following theorem maintains that if at the initial instant the whole capital stock belongs to the most patient agents, then the intertemporal voting equilibrium converges to a balanced-growth voting equilibrium.

Theorem 6

Suppose that the initial state $\mathcal {I}_{-1}$ is such that $\hat{s}^{j}_{-1} = 0$ for $j \notin J$. A unique intertemporal voting equilibrium starting from $\mathcal {I}_{-1}$ satisfies the following asymptotic properties:

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } 1 + r_{t}^{**} = 1+r^{**} = \frac{1 + \gamma ^{**}}{\beta _{1}}, \end{aligned}$$

(68)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{k^{**}_{t+1}}{k^{**}_{t}} = \lim _{t \rightarrow \infty } \frac{w^{**}_{t+1}}{w^{**}_{t}} = \lim _{t \rightarrow \infty } \frac{v^{**}_{t+1}}{v^{**}_{t}} = 1 + \gamma ^{**}, \end{aligned}$$

(69)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{s^{j**}_{t+1}}{s^{j**}_{t}} = 1 + \gamma ^{**}, \quad j \in J, \end{aligned}$$

(70)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{c^{j**}_{t+1}}{c^{j**}_{t}} = 1 + \gamma ^{**}, \quad j = 1, \ldots , L, \end{aligned}$$

(71)

$$\begin{aligned}&\displaystyle \lim _{t \rightarrow \infty } \frac{q^{**}_{t+1}}{q^{**}_{t}} = 1+\pi ^{**}, \end{aligned}$$

(72)

where

$$\begin{aligned} 1 + \gamma ^{**} = \left( 1 + \lambda \right) ^{\frac{1}{1-\alpha _1}} \left( \beta _{med} \right) ^{\frac{\alpha _3}{1-\alpha _1}}, \end{aligned}$$

(73)

and

$$\begin{aligned} 1+\pi ^{**} = \frac{1+\gamma ^{**}}{\beta _{med}}. \end{aligned}$$

(74)

Proof

It follows from Proposition 10 and Theorem 5. $\square$

1.7 Generalized intertemporal voting equilibria

Our definition of an intertemporal voting equilibrium is given under the assumption of uniqueness of a competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium for any $\varepsilon _{\tau } \in (0,1)$. This assumption is crucial in the statement of Theorem 3 about the constant equilibrium extraction rate. Moreover, we obtained the existence and uniqueness of an intertemporal voting equilibrium (Theorem 5) only for the case in which the underlying competitive equilibria are unique. Thus, to guarantee the mere existence of an intertemporal voting equilibrium, we have to prove the uniqueness of a competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium starting from an arbitrary state $\mathcal {I}_{\tau -1}$ for any $\varepsilon _{\tau } \in (0,1)$, which is not an easy task.

Let us discuss the general case in which the competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium starting from an arbitrary state $\mathcal {I}_{\tau -1}$ is not necessarily unique. The difficulty here is that we cannot unambiguously define agents’ indirect utility functions and obtain from them agents’ preferred values of extraction rates. However, if we apply the technique proposed by Borissov et al. (2014b), we can get around this difficulty. Namely, let us impose an additional assumption on the beliefs of agents. Assume that agents simply act as if a competitive $\mathbb {E}_{\tau } (\varepsilon _{\tau })$-equilibrium is unique, and do not take into account the possible multiplicity of equilibria.

Formally, let

$$\begin{aligned} \mathbb {E}^{**} = \mathbb {E}_{0}^{**} = \{ \varepsilon ^{**}_{t} \}_{t=0}^{\infty }, \end{aligned}$$

and

$$\begin{aligned} e^{**}_t = e_{t} (\mathbb {E}^{**}), \qquad R^{**}_{t} = R_{t} (\mathbb {E}^{**}), \quad t = 0, 1, \ldots . \end{aligned}$$

Consider a competitive $\mathbb {E}_{0}^{**}$-equilibrium

$$\begin{aligned} \mathcal {E}_{0}^{**} = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = 0, 1, \ldots } \end{aligned}$$

starting from $\mathcal {I}_{-1} = \left\{ (\hat{s}^{j}_{-1})_{j=1}^{L}, {\hat{R}}_{-1} \right\}$. Let also for $\tau = 1, 2, \ldots$,

$$\begin{aligned} \mathcal {E}_{\tau }^{**} = \left\{ (c^{j**}_{t})_{j=1}^L, (s^{j**}_{t})_{j=1}^L, k^{**}_t, r_t^{**}, w_t^{**}, q_t^{**}, v_t^{**} \right\} _{t = \tau , \tau +1, \ldots } \end{aligned}$$

be the corresponding tail of $\mathcal {E}_{0}^{**}$.

Suppose that the economy has settled on $\mathcal {E}_{0}^{**}$. At each time $\tau$, when the economy is in the state $\mathcal {I}^{**}_{\tau -1} = \{ ( s^{j**}_{\tau -1} )_{j=1}^{L}, R^{**}_{\tau -1} \}$, agents are asked to vote on the time $\tau$ extraction rate. To do this, agents’ indirect utility functions should be unambiguously specified. Originally this was done under the assumption of uniqueness of the competitive $\mathbb {E}^{**}_{\tau }$-equilibrium starting from $\mathcal {I}^{**}_{\tau -1}$. Now let us instead assume that when voting on the time $\tau$ extraction rate, all agents believe that if $\varepsilon ^{**}_{\tau }$ is replaced by the other extraction rate $\varepsilon _{\tau }$, then the economy will settle on the path ${\tilde{\mathcal {E}}}_{\tau } (\varepsilon _{\tau })$, which is linked with the “initial” equilibrium $\mathcal {E}_{\tau }^{**}$ in the way described in Lemma 13.

Recall that under the uniqueness assumption, the interpretation of Lemma 13 is simple. After changing the time $\tau$ extraction rate from $\varepsilon ^{**}_{\tau }$ to $\varepsilon _{\tau }$, a unique competitive $\mathbb {E}^{**}$-equilibrium also changes and becomes a unique competitive $\mathbb {E}_{\tau }$-equilibrium, described in Lemma 13. Here, the interpretation is slightly different. After changing the time $\tau$ extraction rate, the competitive $\mathbb {E}^{**}$-equilibrium can change unpredictably, and the economy can settle on one of multiple $\mathbb {E}_{\tau }$-equilibria. Under our assumption about agents’ beliefs, agents ignore the possible multiplicity of equilibria and believe that after the change in the time $\tau$ extraction rate, the economy settles on the path ${\tilde{\mathcal {E}}}_{\tau } (\varepsilon _{\tau })$, which is described in Lemma 13.

Under this additional assumption, agents’ indirect utility functions, which represent their preferences over the time $\tau$ extraction rate, can be defined unambiguously as follows:

$$\begin{aligned} \mathcal {U}_{\tau }^{j} (\varepsilon _{\tau }) = \ln {\tilde{c}}^{j}_{\tau } (\varepsilon _{\tau }) + \beta _j \ln {\tilde{c}}^{j}_{\tau +1} (\varepsilon _{\tau }) + \cdots , \quad j=1, \ldots , L, \end{aligned}$$

where the sequence $\{ {\tilde{c}}^{j}_{\tau } (\varepsilon _{\tau }), {\tilde{c}}^{j}_{\tau +1} (\varepsilon _{\tau }), \ldots \}$ is constructed according to (54)–(55).

Definition

If for each $t = 0, 1, \ldots$ there is a Condorcet winner in voting on $\varepsilon _{t}$ described above, and it coincides with $\varepsilon ^{**}_{t}$, then we call a couple $\{ \mathbb {E}^{**}, \mathcal {E}_{0}^{**} \}$ a generalized intertemporal voting equilibrium starting from $\mathcal {I}_{-1}$.

Clearly, any intertemporal voting equilibrium is a generalized intertemporal voting equilibrium. Moreover, any generalized intertemporal voting equilibrium starting from an initial state where the whole capital stock belongs to the most patient agents is an intertemporal voting equilibrium. Under the additional assumption about agents’ beliefs, there always exists a generalized intertemporal voting equilibrium starting from an arbitrary initial state.

Theorem 7

For any non-degenerate initial state, there exists a generalized intertemporal voting equilibrium $\{ \mathbb {E}^{**}, \mathcal {E}_{0}^{**} \}$ starting from this state. The equilibrium sequence of extraction rates is constant over time and given by (67).

Proof

It is sufficient to repeat the argument used in the proof of Theorem 4, and refer to Theorem 2. $\square$

Furthermore, every generalized intertemporal voting equilibrium converges to a balanced-growth voting equilibrium.

Theorem 8

Every generalized intertemporal voting equilibrium starting from an arbitrary initial state satisfies asymptotic properties (68)–(72), where $\gamma ^{**}$ and $\pi ^{**}$ are given by (73) and (74), respectively.

Proof

It follows from Proposition 10 and Theorem 7. $\square$

Appendix 3: Private property regime with capital taxation

Consider a competitive equilibrium in the private property regime, and assume in addition that capital income paid by competitive firms to the capital holders is taxed at some constant rate $\theta$ and the revenue is lump-sum redistributed among all agents.

The definition of competitive equilibrium with a capital income tax,

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots }, \end{aligned}$$

repeats the definition of competitive equilibrium in “Appendix 1”, except for the following changes:

$1^{\prime }$.:

For each $j = 1, \ldots , L$, the sequence $\{ c^{j*}_{t}, s^{j*}_{t} \}_{t = 0}^{\infty }$ is a solution to the following utility maximization problem:

$$\begin{aligned} \begin{array}{cc} \displaystyle \max \sum _{t=0}^\infty \beta _j^t \ln c_{t}^{j}, \\ s. t. \quad c_{t}^{j} + s_{t}^{j} \le \left( 1+r_t \right) \left( 1 - \theta \right) s^{j}_{t-1} + w_t + \theta \left( 1+r_t \right) k_{t} \quad t = 0, 1, \ldots , \\ s_{t}^{j} \ge 0, \quad t = 0, 1, \ldots \end{array} \end{aligned}$$

at $r_t = r_t^*$, $w_t = w_t^*$, $k_t = k_t^*$, and $s^{j}_{-1} = \frac{q_{0}^{*}}{(1+r_{0}^{*})(1 - \theta )} {\hat{R}}^{j}_{-1} + {\hat{k}}^{j}_{0}$;

$5^{\prime }$.:

The Hotelling rule takes the form

$$\begin{aligned} q^*_{t+1} = (1+r_{t+1}^*) (1 - \theta ) q^*_{t}, \quad t = 0, 1, \ldots ; \end{aligned}$$

$7^{\prime }$.:

Aggregate savings are equal to investment into physical capital and natural resources:

$$\begin{aligned} \sum _{j=1}^L s^{j*}_{t} = \frac{q_{t+1}^*}{(1+r_{t+1}^{*})(1 - \theta )} R^*_{t} + L k^*_{t+1}, \quad t = 0, 1, \ldots . \end{aligned}$$

Taking into account the new form of the Hotelling rule, we can define a balanced-growth equilibrium with capital taxation along the lines of the definition of a balanced-growth equilibrium in “Appendix 1”. Slightly modifying the arguments from the proof of Propositions 3 and 4, we can provide a characterization of a balanced-growth equilibrium with capital taxation.

Proposition 12

For every balanced-growth equilibrium with capital taxation,

$$\begin{aligned}&\displaystyle 1 + \gamma ^* = \left( 1 + \lambda \right) ^\frac{1}{1-\alpha _1} \beta _{1}^{\frac{\alpha _3}{1-\alpha _1}}, \end{aligned}$$

(75)

$$\begin{aligned}&\displaystyle (1 + r^{*}) (1 - \theta ) = \frac{1 + \gamma ^*}{\beta _{1}}, \end{aligned}$$

(76)

$$\begin{aligned}&\displaystyle \varepsilon ^* = 1 - \beta _{1}. \end{aligned}$$

(77)

Proof

A balanced-growth equilibrium with capital taxation

$$\begin{aligned} \mathcal {E}^{*} = \left\{ (c^{j*}_{t})_{j=1}^L, (s^{j*}_{t})_{j=1}^L, k^*_t, r_t^*, w_t^*, q_t^*, e_t^*, R_{t}^* \right\} _{t = 0, 1, \ldots } \end{aligned}$$

is a competitive equilibrium with capital taxation in which real variables grow at a constant rate $\gamma ^{*}$, while the interest rate $r^{*}$ and extraction rate $\varepsilon ^{*}$ are constant over time. In a competitive equilibria with capital taxation the post-tax interest rate received by agents is equal to the pre-tax gross interest rate $(1 + r^{*})$ multiplied by $(1 - \theta )$. Repeating an argument by Becker (1980, 2006), we obtain that a balanced-growth equilibrium with capital taxation is characterized as follows:

$$\begin{aligned}&\displaystyle s_{t-1}^{j*} = 0, \quad j \notin J, \quad t = 0, 1, \ldots , \nonumber \\&\displaystyle 1 + \gamma ^{*} = \beta _{1} (1 + r^{*}) (1 - \theta ). \end{aligned}$$

(78)

Moreover, since in a balanced-growth equilibrium with capital taxation the extraction rate is constant,

$$\begin{aligned} 1 = \frac{1 + r_{t+1}^{*}}{1 + r_{t}^{*}} = \frac{A_{t+1}}{A_{t}} \left( \frac{k_{t+1}^{*}}{k_{t}^{*}} \right) ^{\alpha _1 - 1} \left( \frac{e_{t+1}^{*}}{e_{t}^{*}} \right) ^{\alpha _3} = (1 + \lambda ) \left( 1 + \gamma ^{*} \right) ^{\alpha _1 - 1} \left( 1 - \varepsilon ^* \right) ^{\alpha _3}, \end{aligned}$$

we get

$$\begin{aligned} \left( 1 + \gamma ^* \right) ^{1 - \alpha _1} = \left( 1+ \lambda \right) \left( 1 - \varepsilon ^* \right) ^{\alpha _3}. \end{aligned}$$

(79)

We also have

$$\begin{aligned} (1 + r^{*})(1 - \theta )= & {} \frac{q^{*}_{t+1}}{q^{*}_{t}} = \frac{A_{t+1}}{A_{t}} \left( \frac{k_{t+1}^{*}}{k_{t}^{*}} \right) ^{\alpha _1} \left( \frac{e_{t+1}^{*}}{e_{t}^{*}}\right) ^{\alpha _3 - 1}\\= & {} (1 + \lambda ) \left( 1 + \gamma ^{*} \right) ^{\alpha _1} \left( 1 - \varepsilon ^* \right) ^{\alpha _3 - 1}, \end{aligned}$$

and it follows that

$$\begin{aligned} (1 + r^{*})(1 - \theta ) = \frac{1 + \gamma ^*}{1 - \varepsilon ^*}. \end{aligned}$$

(80)

Comparing (78) and (80), we immediately obtain (77). Now (76) follows from (77) and (80), while (75) follows from (77) and (79). $\square$

It follows that the long-run growth rate, the post-tax interest rate and the extraction rate in the model with a capital income tax are the same as in the model without capital tax.