Poverty, social preference for employment, and natural resource depletion

Abstract

We show that in poor resource-based communities, the socio-psychological preference for employment, which arises from a strong desire to follow the communal norm of sharing in harvesting efforts, can lead to the optimality of full-employment harvesting until resource extinction. We show that such communities may be able to sustain both their natural resources and full employment by using outside-the-community employment opportunities or by economic diversification. However, to be effective, such policies must ensure that the outside wage rate and the initial capital stock are above certain minimum levels which depend on the existing size of the resource stock, the characteristics of the community’s harvesting technology, and the biological growth characteristics of the resource in question, and which will be higher the longer the remedial policies are delayed.

Appendices

Appendix

A1. Proof of Proposition 1

We first prove the following lemmas:

Lemma 1

If an interior effort level $E = \tilde{E} \in (0,\bar{E})$ is optimal for a stock level S > 0, then

$$\varPsi_{EE} \left( {\tilde{E},S,\frac{{w_{E} (\tilde{E},S)}}{{H_{E} (\tilde{E},S)}}} \right) \le 0 .$$

(23)

Proof

By Pontryagin’s maximum principle, the following two conditions hold at $\tilde{E}$:

$$\varPsi_{E} (\tilde{E},S,\lambda ) = w_{E} (\tilde{E},S) - \lambda H_{E} (\tilde{E},S) = 0,\;\varPsi_{EE} (\tilde{E},S,\lambda ) \le 0.$$

(24)

Combine these to get the inequality in the claim. ■

Lemma 2

Let $\lambda (E,S) = w_{E} (E,S)/H_{E} (E,S)$ . Assume that problem 2 has a solution for each $S \in [0,k]$ . If $\varPsi_{EE} (E,S,\lambda (E,S)) > 0$ for all S > 0 and all $E \in (0,\bar{E})$ , the optimal path is full employment obsession path.

Proof

The contraposition of Lemma 1 is that if $\varPsi_{EE} \left( {\tilde{E},S,\lambda (\tilde{E},S)} \right) > 0$ for all S > 0 and all $\tilde{E} \in (0,\bar{E})$, then optimal effort cannot be interior for any stock level S > 0. Therefore, with the condition $\varPsi_{EE} \left( {\tilde{E},S,\lambda (\tilde{E},S)} \right) > 0$, the optimal policy $E^{*} (S)$ for the problem 2 satisfies $E^{*} (S) \in \{ 0,\bar{E}\}$ for all $S \in (0,k)$. Note that an interior stock level $S \in (0,k)$ is sustained by interior effort level $E \in (0,\bar{E})$such that $G(S) - H(E,S) = 0$. Therefore, there is no interior optimal steady state. Dockner and Nishimura (2005, Lemma 2) prove that an optimal state path for a continuous time one-state variable optimal control problem is monotone. Since the state space $[0,k]$ is bounded, every monotone optimal state path converges to an optimal corner steady state $S_{ss}^{*} = 0$ or $k$. We eliminate $S_{ss}^{*} = k$, since it implies $E^{*} (S_{ss}^{*} ) = 0$ in all periods, which is obviously suboptimal. Therefore, every optimal state path monotonically decreases to the zero stock level. The associated optimal policy is $E^{*} (S) = \bar{E}$ for all $S \in (0,k]$.■

Proof of Proposition 1

Since $\varPsi_{EE} \left( {\tilde{E},S,\lambda (\tilde{E},S)} \right) > 0$ is equivalent to 6, the statement is true by Lemma 2.■

A2. Proof of finite time exhaustion

Proposition A1

Let $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{S}$ be the supremum such that if $0 < S < \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{S}$, $\dot{S} = G(S) - H(\bar{E},S) < 0$ . When initial stock S ₀ is in $(0,\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{S} )$ , the path $S(t;S_{0} )$ described by

$$\dot{S} = G(S) - H(\bar{E},S) = rS\left( {1 - \frac{S}{k}} \right) - \gamma \bar{E}^{\alpha } S^{\beta } , \, S(0) = S_{0} ,$$

(25)

goes to zero within a finite time.

Proof

Notice that $\dot{S} = G(S) - H(\bar{E},S) < rS - \gamma \bar{E}^{\alpha } S^{\beta }$. Let $\sigma (t,S_{0} )$ be the solution of $\dot{\sigma } = r\sigma - \gamma \bar{E}^{\alpha } \sigma^{\beta } , \, \sigma (0) = S_{0}$. This Bernoulli’s differential equation is solved as $\sigma (t;S_{0} ) = \left[ {\left( {S_{0}^{1 - \beta } - \frac{{\gamma \bar{E}^{\alpha } }}{r}} \right)e^{r(1 - \beta )t} + \frac{{\gamma \bar{E}^{\alpha } }}{r}} \right]^{{^{{\frac{1}{1 - \beta }}} }}$. From this, $\sigma (t;S_{0} )$ goes to zero within a finite time if the initial stock satisfies $S_{0}^{1 - \beta } < \gamma \bar{E}^{\alpha } /r$. By construction, $S(t;S_{0} ) < \sigma (t;S_{0} )$ and thus $S(t;S_{0} )$ also go to zero within a finite time. Note that since $\dot{S}(t;S_{0} ) < 0$, $\lim_{t \to \infty } S(t;S_{0} ) = 0$ is obvious, and we can arbitrarily choose a sufficiently small initial value.■

Proposition A2

If $\gamma > \frac{{r[(1 - \beta )k]^{1 - \beta } }}{{(\bar{E})^{\alpha } (2 - \beta )^{2 - \beta } }},$ then $\dot{S} = G(S) - H(\bar{E},S) < 0$ for all $S \ge 0$.

Proof

Let $g(S;\gamma ) = rS\left( {1 - S/k} \right) - \gamma \bar{E}^{\alpha } S^{\beta }$. If there is a unique pair ($\tilde{\gamma }$, $\tilde{S}$) such that $g(\tilde{S};\tilde{\gamma }) = 0, \, g_{S} (\tilde{S};\tilde{\gamma }) = 0,{\text{ and }}g_{SS} (\tilde{S};\tilde{\gamma }) < 0$, then $g(S;\gamma ) < 0$ for all S > 0 and $\gamma > \tilde{\gamma }$. Solve the system of equations $g(\tilde{S};\tilde{\gamma }) = 0{\text{ and }}g_{S} (\tilde{S};\tilde{\gamma }) = 0$ and we have the unique solution

$$\tilde{S} = \frac{1 - \beta }{2 - \beta }k, \, \tilde{\gamma } = \frac{{r[(1 - \beta )k]^{1 - \beta } }}{{(\bar{E})^{\alpha } (2 - \beta )^{2 - \beta } }}$$

(26)

It is easily seen that $g_{SS} (\tilde{S};\tilde{\gamma }) = - r(2 - \beta )/k < 0$.■

A3. Proof of the existence of optimal paths

We apply the Romer’s Theorem (Romer 1986) to prove the existence, because the theorem is applicable to a non-convex problem as in our model.

The reduced form welfare function for our model is written as:

$$\omega (S,\dot{S}) = \left\{ {\begin{array}{*{20}c} {\gamma^{{ - \eta_{E} /\alpha }} [G(S) - \dot{S}]^{{\eta_{C} + \eta_{E} /\alpha }} S^{{ - \eta_{E} \beta /\alpha }} } & {{\text{if }}\dot{S} \in [G(S) - H(\bar{E},S),G(S)],} \\ { - \infty } & {{\text{otherwise}}.} \\ \end{array} } \right.$$

Since we have assumed $\eta_{E} - \alpha (1 - \eta_{C} ) > 0$, $\omega$ is convex in $\dot{S}$. Hence, to apply the theorem, we introduce an “artificial” argument,

$$\mathop S\limits^{..} (t){\text{ such that}}\text{ }\dot{S}(t) - \dot{S}(s) \equiv \int\limits_{s}^{t} {\mathop S\limits^{..} (\tau )d\tau } ,$$

and rewrite $\omega (S,\dot{S})$ as $\omega (S,\dot{S},\mathop S\limits^{..} )$. Notice that for $\mathop S\limits^{..}$ to exist almost everywhere, $\dot{S}$ must be absolutely continuous. This restriction is a cost of applying the Romer’s Theorem. Another restriction is that we have to assume that $\mathop S\limits^{..}$ is essentially bounded, i.e., there exists a real number $M$ such that $\mathop {{\text{ess}} \cdot \sup }\limits_{{t^{3} 0}} |\mathop S\limits^{..} (t)| < M$. The Romer’s Theorem ensures the existence of optimal paths if the following two conditions are met:

• (1) $\omega (S,\dot{S},\mathop S\limits^{..} )$ is upper-semi continuous and $\omega (S,\dot{S}, \cdot )$ is concave for all $(S,\dot{S}) \in {\mathbb{R}}^{2} .$

• (2) There is a real number $m$ such that $\omega (S(t),\dot{S}(t),\mathop S\limits^{..} (t)) \le m - |\mathop S\limits^{..} (t)|$ holds almost everywhere on $[0,\infty )$ and for all $(S(t),\dot{S}(t),\mathop S\limits^{..} (t)).$

Our model obviously satisfies the first condition. For the second condition, notice that there is an upper bound of $\omega (S,\dot{S},\mathop S\limits^{..} )$, say $\bar{\omega }$ ($< \infty$). Thus, choose $m = M + \bar{\omega }$ and then condition 2 is satisfied.■

A4. Proof of $d\tilde{S}/d\omega > 0$ for the problem 12

By Pontryagin’s maximum principle, an interior optimal path satisfies, with costate $\lambda \ge 0$:

$$\begin{aligned} & \varPsi_{E} (E,S,\lambda ) = W_{C} (H_{E} - \omega ) - \lambda H_{E} = 0, \\ & \dot{\lambda } = \rho \lambda - W_{C} H_{S} - \lambda (G^{\prime} - H_{S} ). \\ \end{aligned}$$

(14)

The first equation implies that $\left( {W_{C} - \lambda } \right) = \omega /H_{E} > 0$. At a steady state ($\dot{\lambda } = 0$), the second equation becomes $\lambda (\rho - G^{\prime}) - (W_{C} - \lambda )H_{S} = 0$. Therefore, we have $\rho - G^{\prime} > 0$ and $\lambda > 0$. The latter and the first equation in 14 imply $H_{E} - \omega > 0$. We use these inequalities below.

Let $(\tilde{E},\tilde{S})$ be the effort and the resource stock level at a steady state. As in the main text, $(\tilde{E},\tilde{S})$ satisfies

$$(H_{E} - \omega )(\rho - G^{\prime} + H_{S} ) - H_{E} H_{S} = 0{\text{ and }}G - H = 0.$$

(15)

Totally differentiate the second equation in 15 to have $\left( {G^{\prime} - H_{S} } \right)d\tilde{S} - H_{E} d\tilde{E} = 0$. Thus,

$$\frac{{\partial \tilde{E}}}{{\partial \tilde{S}}} = \frac{{G^{\prime} - H_{S} }}{{H_{E} }}.$$

Using this, totally differentiate the first equation in 15

$$\begin{aligned} & - (\rho - G^{\prime} + H_{S} )d\omega + \{ (H_{ES} + H_{EE} \partial \tilde{E}/\partial \tilde{S})(\rho - G^{\prime} + H_{S} ) \\ & \, + (H_{E} - \omega )( - G^{\prime\prime} + H_{SS} + H_{SE} \partial \tilde{E}/\partial \tilde{S}) \\ & \, - \left( {H_{ES} + H_{EE} \partial \tilde{E}/\partial \tilde{S}} \right)H_{S} \\ & \, - H_{E} \left( {H_{SS} + H_{SE} \partial \tilde{E}/\partial \tilde{S}} \right)\} = 0. \\ \end{aligned}$$

(27)

Observe the following calculations:

$$\begin{gathered} H_{ES} + H_{EE} \frac{{\partial \tilde{E}}}{{\partial \tilde{S}}} = \frac{{\alpha H_{S} }}{{\tilde{E}}} + \frac{{\tilde{E}H_{EE} }}{{H_{E} }}\frac{{G^{\prime} - H_{S} }}{{\tilde{E}}} = \frac{{H_{S} - (1 - \alpha )G^{\prime}}}{{\tilde{E}}},{\text{ and}} \hfill \\ H_{SS} + H_{SE} \frac{{\partial \tilde{E}}}{{\partial \tilde{S}}} = \frac{{(\beta - 1)H_{S} }}{{\tilde{S}}} + \frac{{\beta H_{E} }}{{\tilde{S}}}\frac{{G^{\prime} - H_{S} }}{{H_{E} }} = \frac{{\beta G^{\prime} - H_{S} }}{{\tilde{S}}}. \hfill \\ \end{gathered}$$

(28)

Substitute 28 into 27 and we have:

$$\begin{gathered} (\rho - G^{\prime} + H_{S} )d\omega \hfill \\ = \{ \left( {H_{S} - (1 - \alpha )G^{\prime}} \right)(\rho - G^{\prime})/\tilde{E} - \omega (\beta G^{\prime} - H_{S} )/\tilde{S} - G^{\prime\prime}(H_{E} - \omega )\} d\tilde{S}. \hfill \\ \end{gathered}$$

(29)

As shown above, $\rho - G^{\prime} > 0$ and $H_{E} - \omega > 0$, and thus $\rho - G^{\prime} + H_{S} > 0$. We have $\beta G^{\prime} - H_{S} < 0$ because

$$\beta G^{\prime} - H_{S} = \frac{\beta }{S}\left[ {rS\left( {1 - \frac{2S}{k}} \right) - H} \right] = \frac{\beta }{S}\left[ { - rS\frac{S}{k} + G - H} \right] = \frac{ - rS\beta }{k} < 0.$$

(30)

Finally, by Assumption A5, $H_{S} - (1 - \alpha )G^{\prime} = H_{S} - \beta G^{\prime} > 0$. Therefore, from 29, we have $d\tilde{S}/d\omega > 0$.■