Renewable Resource Use and Nonseparable Amenity Benefits

Gregory Amacher; Markku Ollikainen; Mikko Puhakka

doi:10.1007/s10640-016-0095-2

Environmental and Resource Economics

April 2018, Volume 69, Issue 4, pp 637–659 | Cite as

Renewable Resource Use and Nonseparable Amenity Benefits

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Authors and affiliations

Gregory Amacher
Markku Ollikainen
Mikko Puhakka

Article

First Online: 30 November 2016

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Abstract

We incorporate amenity benefits into an overlapping generations model with a renewable resource as a factor of production, source of amenity benefits and store of value. Unlike the conventional renewable resource problems studied under the assumption of additive consumption and amenity benefits, we let amenity benefits affect the utility of consumers in a nonseparable fashion. We examine the role that weights given to consumption and amenities have for harvesting and the resource stock. We characterize dynamics and stability of steady state equilibria with a logistic resource growth function. We demonstrate in parametric and numerical models that the weights given to consumption and amenities in the utility function matter substantially for the steady state equilibrium stock and its stability and dynamics. Both conventional saddle point equilibria and indeterminacy with infinite number of equilibria and saddle-node bifurcation is possible depending on the weights given to consumption and amenities. In addition, we show that for each inefficient equilibrium stock, there is a unique subsidy rate that can move the economy from an inefficient equilibrium to an efficient one. The presence of indeterminacy provides a challenge to resource policies, because the system becomes unpredictable. Therefore, expectations and market psychology may play an important role in resource utilization and provision of amenities.

Keywords

Overlapping generations Renewable resources Nonseparable preferences Indeterminacy

JEL Classification

D90 Q20 C62

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Appendix

The derivation of Eq. (12) with the general growth function:

We first reproduce Eq. (8) here

$$\begin{aligned}&f^{\prime }(h_t )u_1 \left[ {f(h_t )-h_t f^{\prime }(h_t )-f^{\prime }(h_t )x_{t+1} ,x_t } \right] \nonumber \\&\quad = f^{\prime }(h_{t+1} )(1+g^{\prime }(x_{t+1} ))\beta u_1 \left[ {f^{\prime }(h_{t+1})(x_{t+1} +g(x_{t+1} )),x_{t+1} } \right] \nonumber \\&\quad \quad +\,\beta u_2 \left[ {f^{\prime }(h_{t+1} )(x_{t+1} +g(x_{t+1} ),x_{t+1}}\right] \end{aligned}$$

(35)

With the utility function $u(c,x)=c^{\gamma }x^{1-\gamma }$ with $0<\gamma <1$, (35) can be written as

$$\begin{aligned} \frac{\gamma f^{\prime }(h_t )(c_1^t )^{\gamma }x_t ^{1-\gamma }}{c_1^t }=\gamma f^{\prime }(h_{t+1} )(1+g^{\prime }(x_{t+1} ))\frac{\beta (c_2^t )^{\gamma }x_{t+1} ^{1-\gamma }}{c_2^t }+\frac{\beta (1-\gamma )(c_2^t )^{\gamma }x_{t+1} ^{1-\gamma }}{x_{t+1}}.\nonumber \\ \end{aligned}$$

(36)

We now proceed step by step substituting first for consumptions in the denominators to get

$$\begin{aligned} \frac{\gamma f^{\prime }(h_t )(c_1^t )^{\gamma }x_t ^{1-\gamma }}{f(h_t )-h_t f^{\prime }(h_t )-f^{\prime }(h_t )x_{t+1}}= & {} \gamma f^{\prime }(h_{t+1} )(1+g^{\prime }(x_{t+1} ))\frac{\beta (c_2^t )^{\gamma }x_{t+1}^{1-\gamma }}{f^{\prime }(h_{t+1} )(x_{t+1} +g(x_{t+1} ))}\nonumber \\&+\frac{\beta (1-\gamma )(c_2^t )^{\gamma }x_{t+1} ^{1-\gamma }}{x_{t+1}} \end{aligned}$$

(37)

Cancelling terms out and rearranging we get

$$\begin{aligned} \frac{\gamma (c_1^t )^{\gamma }x_t ^{1-\gamma }}{\frac{f(h_t )}{f^{\prime }(h_t )}-h_t -x_{t+1} }=\frac{\gamma (1+g^{\prime }(x_{t+1} ))\beta (c_2^t )^{\gamma }x_{t+1} ^{1-\gamma }}{(x_{t+1} +g(x_{t+1} ))}+\frac{\beta (1-\gamma )(c_2^t )^{\gamma }x_{t+1} ^{1-\gamma }}{x_{t+1} }. \end{aligned}$$

(38)

Now we assume the following production function $f(h)=h^{\alpha }$ with $0<\alpha <1$. Then we get

$$\begin{aligned} \frac{\alpha \gamma }{(1-\alpha )h_t -\alpha x_{t+1} }(c_1^t )^{\gamma }x_t ^{1-\gamma }=\beta \left[ {\gamma \frac{1+g^{\prime }(x_{t+1} )}{x_{t+1} +g(x_{t+1} )}+\frac{1-\gamma }{x_{t+1} }} \right] (c_2^t )^{\gamma }x_{t+1} ^{1-\gamma }. \end{aligned}$$

(39)

We recall the transition equation $x_{t+1} =x_t +g(x_t )-h_t$. We solve for the harvest, and substitute it to (39). We do that first on the left-hand side of (39) to get

$$\begin{aligned} \frac{\alpha \gamma }{(1-\alpha )(x_t +g(x_t )-x_{t+1} )-\alpha x_{t+1} }(c_1^t )^{\gamma }x_t ^{1-\gamma }=\beta \left[ {\gamma \frac{1+g^{\prime }(x_{t+1} )}{x_{t+1} +g(x_{t+1} )}+\frac{1-\gamma }{x_{t+1}}}\right] (c_2^t )^{\gamma }x_{t+1} ^{1-\gamma },\nonumber \\ \end{aligned}$$

(40)

Table 3

Land area under FSC certification in millions of hectares

	Africa	Asia	Europe	Latin America	North America
Land area 2016	7.60	8.40	90.70	12.87	68.80
Land area 2012	7.42	5.0	63.97	9.51	60.54
Four year change in land area (%)	2.4	68	41.8	35.3	13.6

Source: Forest Stewardship Council (2012, 2016). Facts and Figures. Available at https://us.fsc.org/en-us/what-we-do/facts-figures

which by cancelling terms yields

$$\begin{aligned} \frac{\alpha \gamma }{(1-\alpha )(x_t +g(x_t ))-x_{t+1} }(c_1^t )^{\gamma }x_t^{1-\gamma }=\beta \left[ {\gamma \frac{1+g^{\prime }(x_{t+1})}{x_{t+1}+g(x_{t+1} )}+\frac{1-\gamma }{x_{t+1}}}\right] (c_2^t)^{\gamma }x_{t+1}^{1-\gamma }.\nonumber \\ \end{aligned}$$

(41)

We now use the budget constraints (11a) and (11b) from the text, and plug them into (41) to get

$$\begin{aligned}&\frac{\alpha \gamma }{(1-\alpha )(x_t +g(x_t ))-x_{t+1} }\left( \left[ {(1-\alpha )(x_t +g(x_t ))-x_{t+1} } \right] (x_t +g(x_t )-x_{t+1} )^{\alpha -1}\right) ^{\gamma }x_t ^{1-\gamma }\nonumber \\&\quad =\alpha ^{\gamma }\beta \left[ {\gamma \frac{1+g^{\prime }(x_{t+1})}{x_{t+1}+g(x_{t+1})}+\frac{1-\gamma }{x_{t+1}}}\right] \left[ {x_{t+1}+g(x_{t+1})-x_{t+2}} \right] ^{(\alpha -1)\gamma }\left[ {x_{t+1} +g(x_{t+1})} \right] ^{\gamma }x_{t+1} ^{1-\gamma }.\nonumber \\ \end{aligned}$$

(42)

We collect terms and rewrite

$$\begin{aligned}&\frac{\alpha \gamma x_t ^{1-\gamma }}{\left[ {(1-\alpha )(x_t +g(x_t ))-x_{t+1}}\right] ^{1-\gamma }}(x_t +g(x_t )-x_{t+1} )^{(\alpha -1)\gamma }\nonumber \\&\quad =\alpha ^{\gamma }\beta \left[ {\gamma \frac{1+g^{\prime }(x_{t+1} )}{x_{t+1}+g(x_{t+1})}+\frac{1-\gamma }{x_{t+1} }} \right] \left[ {x_{t+1} +g(x_{t+1})-x_{t+2}} \right] ^{(\alpha -1)\gamma }\left[ {x_{t+1}+g(x_{t+1})} \right] ^{\gamma }x_{t+1} ^{1-\gamma }.\nonumber \\ \end{aligned}$$

(43)

This is a second-order nonlinear difference equation, which governs the behavior of the stock over time. Rewriting (43) slightly we obtain Eq. (12) in the text (Table 3).

The proof of Corollary 1

We reproduce the condition, $\textit{RHS}(0)>\textit{LHS}(0)$, here with a slight abuse of notation as $\textit{LHS}(0;\gamma )\equiv \alpha ^{1-\gamma }\gamma \left( {\frac{1+a}{a-\alpha (1+a)}} \right) ^{1-\gamma }<\alpha ^{\gamma }\beta (1+a)\equiv \textit{RHS}(0;\gamma )$. We first note that $\textit{RHS}(0;0)=\beta (1+a)>0=\textit{LHS}(0;0), \frac{\lim }{\gamma \rightarrow 1}{} \textit{LHS}(0;\gamma )=1$ and $\frac{\lim }{\gamma \rightarrow 1}{} \textit{RHS}(0;\gamma )=\alpha \beta (1+a)<1$. Thus we have $\frac{\lim }{\gamma \rightarrow 1}{} \textit{RHS}(0;\gamma )<\frac{\lim }{\gamma \rightarrow 1}{} \textit{LHS}(0;\gamma )$. Since both $\textit{RHS}(0;\gamma )$ and $\textit{LHS}(0;\gamma )$ are continuous functions of $\gamma $, there must exist at least one value of $\hat{{\gamma }}\in (0,1)$ such that $\textit{RHS}(0;\hat{{\gamma }})=\textit{LHS}(0;\hat{{\gamma }})$. We define $\gamma ^{*}=\inf (\hat{{\gamma }})$. From above it follows that $\gamma ^{*}>0$, and furthermore that $\textit{LHS}(0;\gamma )>\textit{LHS}(0;\gamma )$ for all $\gamma \in \left[ {0,\gamma ^{*}} \right) $. We differentiate $\textit{RHS}_\gamma (0;\gamma )=\ln \alpha \left[ {\alpha ^{\gamma }\beta (1+a)} \right] <0$. Differentiating the left-hand side we get $\textit{LHS}_\gamma (0;\gamma )=\left\{ {-\ln \alpha +\frac{1}{\gamma }-\ln \left[ {\frac{\alpha (1+a)}{a-\alpha (1+a)}} \right] } \right\} \alpha ^{1-\gamma }\gamma \left( {\frac{1+a}{a-\alpha (1+a)}} \right) ^{1-\gamma }$. $\textit{LHS}_\gamma (0;\gamma )>0$, if $\frac{1}{\gamma }-\ln \left[ {\frac{\alpha (1+a)}{a-\alpha (1+a)}} \right] >0$, and the fact that $0<\alpha <1$. Since $\textit{RHS}(0;0)>\textit{LHS}(0;0)$, there is thus a unique $0<\gamma ^{**}<1$ such that $\textit{RHS}(0;\gamma ^{**})=\textit{LHS}(0;\gamma ^{**})$ and for $\gamma ^{**}<\gamma<~1 \quad \textit{RHS}(0;\gamma )<\textit{LHS}(0;\gamma )$. Q.E.D.

The characterization of the function $\gamma =z(\alpha )$:

We first note the following facts:

$$\begin{aligned}&\frac{\lim }{\alpha \rightarrow a/(1+a)}z(\alpha )=0, \frac{\lim }{\alpha \rightarrow a/(2(1+a))}z(\alpha )=\infty \\&z^{\prime }(\alpha )=-\frac{1}{\left\{ {\ln \left[ {1\bigg /{\left( \frac{a}{\alpha (1+a)}-1\right) }}\right] }\right\} ^{2}}\left( {\frac{1}{\alpha }+\frac{1+a}{a-\alpha (1+a}} \right) <0 \end{aligned}$$

and

$$\begin{aligned} z''(\alpha )=+\frac{2}{\left\{ {\ln \left[ {1\bigg /{\left( \frac{a}{\alpha (1+a)}-1\right) }} \right] } \right\} ^{3}}\left( {\frac{1}{\alpha }+\frac{1+a}{a-\alpha (1+a}} \right) \left( {-\frac{1}{\alpha ^{2}}+\frac{(1+a)^{2}}{(a-\alpha (1+a))^{2}}} \right) . \end{aligned}$$

We rewrite the second derivative as

$$\begin{aligned} z''(\alpha )=+\frac{2}{\left\{ {\ln 1\bigg /{\left( \frac{a}{\alpha (1+a)}-1\right) }} \right\} ^{3}}\left( {\frac{1}{\alpha }+\frac{1+a}{a-\alpha (1+a}} \right) \left( {\frac{a(-a+2\alpha (1+a)}{\alpha ^{2}(a-\alpha (1+a))^{2}}} \right) . \end{aligned}$$

Since $\alpha $ must lie on the open interval $\left( {a/(2(1+a),a/(1+a)} \right) , z''(\alpha )>0$, the function $z(\alpha )$ is decreasing and strictly convex as depicted in Fig. 2.

The partial derivatives for the equation (22):

We compute the partial derivatives from (22)

$$\begin{aligned} F_1 (x_{t+1} ,x_t )= & {} 1+a-bx_{t+1} -A^{\prime }(x_{t+1} )B(x_{t+1} ,x_t )-A(x_{t+1} )B_1 (x_{t+1} ,x_t ) \end{aligned}$$

(44)

$$\begin{aligned} F_2 (x_{t+1} ,x_t )= & {} -A(x_{t+1} )B_2 (x_{t+1} ,x_t ). \end{aligned}$$

(45)

What is then left to compute is the derivative, $A^{\prime }(x_{t+1})$, and the partial derivatives $B_1 (x_{t+1} ,x_t)$ and $B_2 (x_{t+1}, x_t)$ from above.

We start the computations with $A^{\prime }(x_{t+1})$, and after some manipulations end up with

$$\begin{aligned} A^{\prime }(x_{t+1})= & {} \beta ^{\frac{1}{(1-\alpha )\gamma }}\frac{1}{(1-\alpha )\gamma }\left\{ {\left[ {\frac{1+a-(1+\gamma )(1/2)bx_{t+1}}{\left[ {1+a-(1/2)bx_{t+1}}\right] ^{1-\gamma }}} \right] } \right\} ^{\frac{1-\gamma (1-\alpha )}{(1-\alpha )\gamma }}\nonumber \\&\times \left\{ {\frac{-2(1+a)+(1+\gamma )(1/2)bx_{t+1} }{\left[ {1+a-(1/2)bx_{t+1}}\right] ^{2-\gamma }}}\right\} (1/2)\gamma b. \end{aligned}$$

(46)

Given the carrying capacity, the resource can exhibit either positive or zero growth permanently over time, so that we can safely restrict the stock to provide nonnegative growth: $g(x)=ax-(1/2)bx^{2}\ge 0$. It then follows that $A^{\prime }(x_{t+1})$ is unambiguously negative. The following steps prove this assertion: $g(x)=ax-(1/2)bx^{2}\ge 0\Rightarrow x\left[ a-(1/2)bx\right] \ge 0\Rightarrow -a+(1/2)bx\le 0$. We rewrite $-2(1+a)+(1+\gamma )(1/2)bx= -2-a-a+(1/2)bx+\gamma (1/2)bx=-a+(1/2)bx-a+\gamma (1/2)bx-2.$ Since $0<\gamma <1$, it is clear that $-2(1+a)+(1+\gamma )(1/2)bx<0$. This shows that $A^{\prime }(x_{t+1} )<0$, since $a-(1/2)bx_{t+1}>0$ for all $x<\tilde{x}$.

Next we compute $B_1 (x_{t+1} ,x_t)$

$$\begin{aligned} B_1 (x_{t+1} ,x_t )= & {} \left\{ {\frac{\alpha ^{1-\gamma }\gamma x_t ^{1-\gamma }}{\left[ {(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1} }\right] ^{1-\gamma }}}\right\} ^{\frac{1}{(\alpha -1)\gamma }}\nonumber \\&\times \left\{ {\frac{1-\gamma }{(\alpha -1)\gamma }\frac{(1+a)x_t -(1/2)bx_t ^{2}-x_{t+1}}{(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1} }-1} \right\} . \end{aligned}$$

(47)

We check out that $B_1 (x_{t+1} ,x_t )<0$. The term, $(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1} $, is the first period consumption, which must be positive at the optimum because of the Inada conditions in the utility function. Then it clearly follows that also the term, $(1+a)x_t -(1/2)bx_t ^{2}-x_{t+1} $, must be positive. Since $\alpha <1, B_1 (x_{t+1} ,x_t )$ must be negative.

And finally we compute $B_2 (x_{t+1} ,x_t )$

$$\begin{aligned}&B_2 (x_{t+1} ,x_t )=\left\{ {\frac{\alpha ^{1-\gamma }\gamma x_t ^{1-\gamma }}{\left[ {(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1} } \right] ^{1-\gamma }}} \right\} ^{\frac{1}{(\alpha -1)\gamma }}\left[ {1+a-bx_t } \right] \nonumber \\&\quad +\frac{1}{(\alpha -1)\gamma }\left\{ {\frac{\alpha ^{1-\gamma }\gamma x_t^{1-\gamma }}{\left[ {(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1}}\right] ^{1-\gamma }}} \right\} ^{\frac{1+(1-\alpha )\gamma }{(\alpha -1)\gamma }} \left[ {(1+a)x_t -(1/2)bx_t ^{2}-x_{t+1}}\right] \nonumber \\&\quad \times \left\{ {\frac{(1-\gamma )\alpha ^{1-\gamma }\gamma x_t ^{-\gamma }}{\left[ {(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1}} \right] ^{1-\gamma }}-\frac{(1-\gamma )((1-\alpha )((1+a)-bx_t )\alpha ^{1-\gamma }\gamma x_t^{1-\gamma }}{\left[ {(1-\alpha )((1+a)x_t -(1/2)bx_t ^{2})-x_{t+1}}\right] ^{2-\gamma }}}\right\} .\nonumber \\ \end{aligned}$$

(48)

It seems that we cannot sign this without further analysis/assumptions.

The evaluation of the functions A and B, and their derivatives in the steady state:

$$\begin{aligned} A(x^{*})=\beta ^{\frac{1}{(1-\alpha )\gamma }}\left\{ {\left[ {\frac{1+a-(1+\gamma )(1/2)bx^{*}}{\left[ {1+a-(1/2)bx^{*}} \right] ^{1-\gamma }}}\right] }\right\} ^{\frac{1}{(1-\alpha )\gamma }} \end{aligned}$$

(49a)

$$\begin{aligned} B(x^{*},x^{*})=\left\{ {\frac{\alpha ^{1-\gamma }\gamma }{\left[ {a-(1/2)bx^{*}-\alpha (1+a-(1/2)bx^{*})} \right] ^{1-\gamma }}\left( {\frac{1}{ax^{*}-(1/2)b(x^{*})^{2}}}\right) ^{(1-\alpha )\gamma }} \right\} ^{\frac{1}{(\alpha -1)\gamma }}.\nonumber \\ \end{aligned}$$

(49b)

The derivatives in steady state can be expressed as follows

$$\begin{aligned} A^{\prime }(x^{*})= & {} \beta ^{\frac{1}{(1-\alpha )\gamma }}\frac{1}{(1-\alpha )\gamma }\left\{ {\left[ {\frac{1+a-(1+\gamma )(1/2)bx^{*}}{\left[ {1+a-(1/2)bx^{*}} \right] ^{1-\gamma }}} \right] } \right\} ^{\frac{1-\gamma (1-\alpha )}{(1-\alpha )\gamma }}\nonumber \\&\times \left\{ {\frac{-2(1+a)+(1+\gamma )(1/2)bx^{*}}{\left[ {1+a-(1/2)bx^{*}} \right] ^{2-\gamma }}} \right\} (1/2)\gamma b \end{aligned}$$

(50a)

$$\begin{aligned} B_1 (x,x)= & {} \left\{ {\frac{\alpha ^{1-\gamma }\gamma x^{1-\gamma }}{\left[ {(1-\alpha )((1+a)x-(1/2)bx^{2})-x} \right] ^{1-\gamma }}} \right\} ^{\frac{1}{(\alpha -1)\gamma }}\nonumber \\&\quad \times \left\{ {\frac{1-\gamma }{(\alpha -1)\gamma }\frac{(1+a)x-(1/2)bx^{2}-x}{(1-\alpha )((1+a)x-(1/2)bx^{2})-x}-1} \right\} \end{aligned}$$

(50b)

$$\begin{aligned} B_2 (x,x)= & {} \left\{ {\frac{\alpha ^{1-\gamma }\gamma }{\left[ {(1-\alpha )((1+a)-(1/2)bx)-1} \right] ^{1-\gamma }}} \right\} ^{\frac{1}{(\alpha -1)\gamma }}\left[ {1+a-bx} \right] \nonumber \\&+\frac{1}{(\alpha -1)\gamma }\left\{ {\frac{\alpha ^{1-\gamma }\gamma }{\left[ {(1-\alpha )((1+a)-(1/2)bx)-1} \right] ^{1-\gamma }}} \right\} ^{\frac{1+(1-\alpha )\gamma }{(\alpha -1)\gamma }}\left[ {ax-(1/2)bx^{2}} \right] \nonumber \\&\times \left\{ {\frac{(1-\gamma )\alpha ^{1-\gamma }\gamma x}{\left[ {(1-\alpha )((1+a)-(1/2)bx)-1} \right] ^{1-\gamma }}+\frac{(1-\gamma )((1-\alpha )((1+a)-bx)\alpha ^{1-\gamma }\gamma x}{\left[ {(1-\alpha )((1+a)-(1/2)bx)-1} \right] ^{2-\gamma }}} \right\} .\nonumber \\ \end{aligned}$$

(50c)

The proof of Lemma 1

The first period consumption is

$$\begin{aligned} c_1^t =w_t -p_t x_{t+1}. \end{aligned}$$

(51)

Given the pricing relations (the subsidy included) this can be written as

$$\begin{aligned} c_1^t =f(h_t )-h_t f^{\prime }(h_t )-\frac{f^{\prime }(h_{t+1} )}{1-\mu _t }x_{t+1}. \end{aligned}$$

(52)

Manipulating this expression we obtain

$$\begin{aligned} c_1^t =f^{\prime }(h_t )\left[ {\frac{f(h_t )}{f^{\prime }(h_t )}-h_t -\frac{x_{t+1} }{1-\mu _t }} \right] . \end{aligned}$$

(53)

Given the production function, $f(h)=h^{\alpha }$, and expressing (53) in a steady state we get

$$\begin{aligned} c_1 =\alpha h^{\alpha -1}\left[ {\frac{h^{\alpha }}{\alpha h^{\alpha -1}}-h-\frac{x}{1-\mu }} \right] , \end{aligned}$$

(54)

and finally we end up with

$$\begin{aligned} c_1 =h^{\alpha -1}\left[ {(1-\alpha )h-\frac{\alpha }{1-\mu }x} \right] . \end{aligned}$$

(55)

Since the optimal first period consumption is positive we have proved the claim. Furthermore, since$(1-\alpha )g(x)-\alpha x/(1-\mu )\ge 0$, the following holds for the subsidy rate

$$\begin{aligned} \mu \le 1-\frac{\alpha }{1-\alpha }\frac{x}{g(x)}. \end{aligned}$$

(56)

We define $\bar{{\mu }}$ such that (56) holds as an equality.

The Proof of Proposition 2

We first note the following properties for the functions in (34):

$$\begin{aligned}&\textit{RHS}'(\mu )<0, \textit{RHS}''(\mu )<0\frac{\lim }{\mu \rightarrow 0}{} \textit{RHS} (\mu )>0, \frac{\lim }{\mu \rightarrow 1} \textit{RHS}(\mu )=0,\\&\quad \textit{LHS}^{\prime }(\mu )=(\gamma -1)\alpha \gamma \left[ {(1-\alpha )g(x)-\frac{\alpha }{1-\mu }x} \right] ^{\gamma -2}\left( {-\frac{\alpha }{(1-\mu )^{2}}x} \right) >0. \end{aligned}$$

The second derivative is

$$\begin{aligned} \textit{LHS}''(\mu )= & {} (\gamma -1)(\gamma -2)\alpha \gamma \left[ (1-\alpha ) g(x)-\frac{\alpha x}{1-\mu })\right] ^{\gamma -3}\left( \frac{\alpha x}{(1-\mu )^{2}}\right) ^{2}\\&+\,(\gamma -1)\alpha \gamma \left[ (1-\alpha ) g(x)-\frac{\alpha x}{1-\mu })\right] ^{\gamma -3} \left( -\frac{2\alpha x}{(1-\mu )^{3}}\right) \end{aligned}$$

which in turn yields $\textit{LHS}''(\mu )>0$. Furthermore, we get, and $\frac{\lim }{\mu \rightarrow \bar{{\mu }}}{} \textit{LHS}(\mu )=\infty $. It then follows that for any feasible level of the stock there is a unique subsidy rate. Figure 5 describes the behaviour of the equilibrium functions. Q.E.D.

Fig. 5
The uniqueness of the subsidy rate

The subsidy rate and the level of the stock are inversely related meaning that raising the subsidy will decrease the level of the stock. Differentiating the functions $\textit{LHS}(\mu )$ and $\textit{RHS}(\mu )$ with respect to x we get

$$\begin{aligned} \frac{\partial \textit{LHS}(\mu )}{\partial x}= & {} \alpha \gamma (\gamma -1)\left[ {(1-\alpha )g^{\prime }(x)-\frac{\alpha }{1-\mu }} \right] \left[ {(1-\alpha )g(x)-\frac{\alpha }{1-\mu }x} \right] ^{\gamma -2}>0\nonumber \\ \end{aligned}$$

(57)

$$\begin{aligned} \frac{\partial \textit{RHS}(\mu )}{\partial x}= & {} \gamma \alpha ^{\gamma }(1-\mu )^{1-\gamma }\beta \left[ {\gamma \frac{1+g^{\prime }(x)}{x+g(x)}+\frac{1-\gamma }{x}} \right] \left[ {x+g(x)} \right] ^{\gamma -1}\left[ {1+g^{\prime }(x)} \right] \nonumber \\&\quad +\,\,\alpha ^{\gamma }(1-\mu )^{1-\gamma }\beta \left[ {\gamma \frac{g'' (x)}{x+g(x)}-\gamma \frac{\left[ {1+g^{\prime }(x)} \right] ^{2}}{\left[ {x+g(x)} \right] ^{2}}-\frac{1-\gamma }{x^{2}}} \right] \left[ {x+g(x)} \right] ^{\gamma }.\nonumber \\ \end{aligned}$$

(58)

Collecting the joint terms, and cancelling we can rewrite (58) as

$$\begin{aligned} \frac{\partial \textit{RHS}(\mu )}{\partial x}=\gamma \alpha ^{\gamma }(1-\mu )^{1-\gamma }\beta \left[ {x+g(x)} \right] ^{\gamma }\left\{ {\frac{1-\gamma }{x^{2}}\left[ {\gamma \frac{1+g^{\prime }(x)}{1+\frac{g(x)}{x}}-1} \right] +\gamma \frac{g''(x)}{x+g(x)}} \right\} <0.\nonumber \\ \end{aligned}$$

(59)

The last sign follows from the fact that the growth function is a strictly concave function, which means $g^{\prime }(x)<g(x)/x$. The signs of the derivatives of these functions with respect to the stock mean that the $\textit{LHS}(\mu )$ function shifts up, when the level of the stock is increased, and the $\textit{RHS}(\mu )$ function shifts down, when the stock is raised. All in all, this means that there is a negative relationship in equilibrium with the rate of subsidy and the level of the stock. Figure 5 clarifies this result. Q.E.D.

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Gregory Amacher
- 1
Markku Ollikainen
- 2
Mikko Puhakka
- 3
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1.College of Natural ResourcesVirginia Polytechnic Institute and State UniversityBlacksburgUSA
2.Department of Economics and ManagementUniversity of HelsinkiHelsinkiFinland
3.Department of EconomicsUniversity of OuluOuluFinland

Renewable Resource Use and Nonseparable Amenity Benefits

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