Do Social Status Seeking Behaviors Worsen the Tragedy of the Commons?

Abstract

The present paper considers the exploitation of a common-property, nonrenewable resource, by individuals concerned with their social status. Assuming that the social status is reflected by the individuals’ relative consumptions, we formalize this motivation by means of a utility function, depending on the individual’s actual consumption and on the consumption level he aspires, the latter being related to the consumptions in his reference group. We compare the benchmark cooperative solution with a noncooperative Markov-perfect Nash equilibrium. We confirm, under more general conditions than in the existing literature, that the individuals’ concern for social status exacerbates the tragedy of the commons. We finally discuss the policy implications and provide a taxation scheme capable of implementing the cooperative solution as a noncooperative Markov-perfect Nash equilibrium.

Appendix

1.1 A.1 Proof of Lemma 1

Consider any vector of individual consumptions $C ( t ) = ( c_{i} ( t ) ) _{i=1}^{n}$. Recall that $V ( C ( t ) ) =\sum_{i=1}^{n}v ( c_{i} ( t ) ,\phi _{i} ( c_{-i} ( t ) ) )$. Let $\overline {C} ( t ) = ( \overline{c}_{i} ( t ) ) _{i=1}^{n}= ( \sum_{i=1}^{n}c_{i} ( t ) /n ) _{i=1}^{n}$ represent the equal redistribution of $C ( t ) = ( c_{i} ( t ) ) _{i=1}^{n}$. We need to show that if $C ( t ) \neq \overline{C} ( t )$, then $V ( \overline{C} ( t ) ) >V ( C ( t ) )$.

By definition,

$$ V \bigl( \overline{C} ( t ) \bigr) =\sum_{i=1}^{n}v \bigl( \overline{c}_{i} ( t ) ,\phi_{i} \bigl( \overline{c}_{-i} ( t ) \bigr) \bigr). $$

(8)

Since the consumption vector $\overline{C} ( t ) = ( \sum_{i=1}^{n}c_{i} ( t ) /n ) _{i=1}^{n}$ is symmetric, we have (by Assumption 1(b))

$$ \overline{c}_{i} ( t ) =\phi_{i} \bigl( \overline{c}_{-i} ( t ) \bigr) =\sum_{i=1}^{n}c_{i} ( t ) /n\quad \text{for all }i. $$

Substituting into (8), we get

$$\begin{aligned} V \bigl( \overline{C} ( t ) \bigr) =&\sum_{i=1}^{n}v \Biggl( \sum_{i=1}^{n}\frac {c_{i} ( t ) }{n},\sum _{i=1}^{n}\frac{c_{i} ( t ) }{n} \Biggr), \\ =&nv \Biggl( \sum_{i=1}^{n} \frac{c_{i} ( t ) }{n},\sum_{i=1}^{n} \frac{c_{i} ( t ) }{n} \Biggr). \end{aligned}$$

Now, remark that

$$ \Biggl( \sum_{i=1}^{n}\frac{c_{i} ( t ) }{n},\sum_{i=1}^{n}\frac{c_{i} ( t ) }{n} \Biggr) = \sum_{i=1}^{n}\frac{1}{n} \biggl( c_{i} ( t ) ,\sum_{j\neq i}\frac{c_{j} ( t ) }{n-1} \biggr). $$

After substitution, this yields

$$ V \bigl( \overline{C} ( t ) \bigr) =nv \Biggl( \sum _{i=1}^{n}\frac{1}{n} \biggl( c_{i} ( t ) ,\sum_{j\neq i}\frac{c_{j} ( t ) }{n-1} \biggr) \Biggr). $$

Clearly, if $C ( t ) \neq\overline{C} ( t )$, by concavity of v,

$$ v \Biggl( \sum_{i=1}^{n}\frac{1}{n} \biggl( c_{i} ( t ) ,\sum_{j\neq i} \frac{c_{j} ( t ) }{n-1} \biggr) \Biggr) >\sum_{i=1}^{n} \frac{1}{n}v \biggl( c_{i} ( t ) ,\sum _{j\neq i}\frac{c_{j} ( t ) }{n-1} \biggr) \text{} $$

and

$$ V \bigl( \overline{C} ( t ) \bigr) >\sum_{i=1}^{n}v \biggl( c_{i} ( t ) ,\sum_{j\neq i} \frac{c_{j} ( t ) }{n-1} \biggr). $$

Now, recall that, for all i, v is decreasing in y _i , and ϕ _i (c _−i (t))≥∑ _j≠i c _j (t)/(n−1) (by Assumption 1(a)). It follows that

$$ v \biggl( c_{i} ( t ) ,\sum_{j\neq i} \frac{c_{j} ( t ) }{n-1} \biggr) \geq v \bigl( c_{i} ( t ) , \phi_{i} \bigl( c_{-i} ( t ) \bigr) \bigr) \text{} $$

for all i and, summing over i, that

$$ \sum_{i=1}^{n}v \biggl( c_{i} ( t ) ,\sum_{j\neq i}\frac{c_{j} ( t ) }{n-1} \biggr) \geq\sum _{i=1}^{n}v \bigl( c_{i} ( t ) , \phi_{i} \bigl( c_{-i} ( t ) \bigr) \bigr). $$

Finally, this allows us to show that

$$ V \bigl( \overline{C} ( t ) \bigr) >\sum_{i=1}^{n}v \bigl( c_{i} ( t ) ,\phi_{i} \bigl( c_{-i} ( t ) \bigr) \bigr) =V \bigl( C ( t ) \bigr), $$

which shows Lemma 1.

1.2 A.2 Proof of Proposition 1

The social problem is to choose c(⋅) to maximize

$$\begin{aligned} &\int_{0}^{\infty}nu \bigl( c ( t ) \bigr) e^{-\delta t}\,dt, \\ &\text{where:} \\ &\dot{x} ( t ) =-nc ( t ), \qquad x ( 0 ) =x_{0}, \\ &c ( t ) \geq0, \qquad x ( t ) \geq0. \end{aligned}$$

Define the current-value Hamiltonian

$$ H ( x,c,\lambda) =n \bigl( u ( c ) -\lambda c \bigr), $$

where λ is a costate multiplier associated with the state x.

A feasible control path c(⋅) with the corresponding state trajectory x(⋅) is optimal if there exists λ(⋅) such that

$$\begin{aligned} &n \bigl( u^{\prime} \bigl( c ( t ) \bigr) -\lambda( t ) \bigr) \leq 0\text{ and }n \bigl( u^{\prime} \bigl( c ( t ) \bigr) -\lambda( t ) \bigr) c ( t ) =0, \end{aligned}$$

(9)

$$\begin{aligned} &\dot{\lambda} ( t ) =\delta\lambda( t ), \end{aligned}$$

(10)

$$\begin{aligned} &\lim_{t\rightarrow\infty}e^{-\delta t}\lambda( t ) x ( t ) =0. \end{aligned}$$

(11)

Now, f(x) being such that

$$ \varTheta\bigl( f ( x ) \bigr) \equiv\int_{0}^{f ( x ) } \sigma( s )\,ds=\delta x/n\quad \text{for all }x, $$

let c(t), x(t), and λ(t) for all t satisfy

$$\begin{aligned} &c ( t ) =f \bigl( x ( t ) \bigr), \\ &\dot{x} ( t ) =-nc ( t ),\qquad x ( 0 ) =x_{0}, \\ &\dot{\lambda} ( t ) =\delta\lambda( t ),\qquad \lambda( 0 ) =u^{\prime} \bigl( f ( x_{0} ) \bigr). \end{aligned}$$

We show below that the proposed control path c(⋅) is feasible and satisfies conditions (9), (10), and (11). Proposition 1 follows.

(1) Feasibility. We first show, in Lemmas 1 and 2, that the proposed control path c(⋅) is feasible.

Lemma 1

For all x>0, f(x)>0, and f(0)=0.

Proof

If x=0, f(0)=0 follows from Θ(0)=0. Likewise, lim _x→∞ f(x)=∞ follows from lim _c→∞ Θ(c)=∞. If 0<x<∞, as Θ(0)=0<δx/n<∞=lim _c→∞ Θ(c) and Θ(c) is continuous, there exists 0<f(x)<∞ such that Θ(f(x))=δx/n (by the intermediate value theorem). As Θ(c) is increasing (since Θ′(c)=σ(c)>0 for all c), this solution is unique. □

Lemma 2

The individual consumption path c(⋅) generates a trajectory of the resource stock x(⋅) such that x(t)≥0 for all t and lim _t→∞ x(t)=0.

Proof

By definition, the individual consumption path c(⋅) generates a trajectory x(⋅) such that $\dot{x} ( t ) =-nf ( x ( t ) )$ for all t, with initial condition x(0)=x ₀. Lemma 2 follows directly from the properties of f(⋅), which imply that $\dot{x}( t ) =-nf ( x ( t ) ) <0$ when x(t)>0 and $\dot{x} ( t ) =0$ when x(t)=0. □

(2) Necessary conditions. We now check that the necessary conditions (9), (10), and (11) are satisfied.

Let T represent the time of depletion of the resource stock (including the possibility that T=∞).

Lemma 3

Along the path x(⋅) induced by f(⋅), the marginal utility u′(f(x(t))) grows at the rate δ for all t<T and is equal to u′(0) for all t≥T.

Proof

First, remark that Θ(f(x))=δx/n for all x implies that σ(f(x))f′(x)=δ/n for all x (by differentiation).

Define p(t)=u′(f(x(t))) for all t.

For all t<T, differentiation yields

$$ \dot{p} ( t ) =u^{\prime\prime} \bigl( f \bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) \dot{x} ( t ). $$

Substituting $\dot{x} ( t ) =-nf ( x ( t ) )$ and dividing by p(t)=u′(f(x(t))), we get

$$ \frac{\dot{p} ( t ) }{p ( t ) }=n\sigma\bigl( f \bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) =\delta. $$

For all t≥T, x(t)=0 (by definition) and f(0)=0 imply that p(t)=u′(0). □

The maximum condition (9) follows directly from Lemma 3.

The adjoint equation (10) is satisfied by construction of λ(t).

The transversality condition (11) is satisfied since e ^−δt λ(t)=u′(f(x ₀)) by construction and lim _t→∞ x(t)=0 by Lemma 2.

1.3 A.3 Proof of Property 1

For all n, let c ^∘ satisfy Θ(c ^∘)=δx ₀/n, and let C ^∘=nc ^∘. By differentiation we obtain

$$\begin{aligned} \frac{dc^{\circ}}{dn} =&-\frac{\delta x_{0}}{n^{2}\sigma( c{{}^\circ}) }, \\ \frac{dC{{}^\circ}}{dn} =&c{{}^\circ}-\frac{\delta x_{0}}{n\sigma( c{{}^\circ}) }. \end{aligned}$$

By the mean value theorem, given that Θ(0)=0, there exists c, with 0<c<c ^∘, such that

$$ \varTheta\bigl( c{{}^\circ}\bigr) =\sigma( c ) c{{}^\circ}. $$

Hence, we have

$$ c{{}^\circ}=\frac{\delta x_{0}}{n\sigma( c ) }. $$

By substitution it follows that

$$ \frac{dC{{}^\circ}}{dn}=\frac{\delta x_{0}}{n}\frac{\sigma( c{{}^\circ}) -\sigma( c ) }{\sigma( c ) \sigma( c {{}^\circ}) }\text{,} $$

which proves the property.

1.4 A.4 Proof of Proposition 2

Consider any player i. Assume that all individuals j≠i play the Markovian strategy $s_{j}^{\ast}=g ( x )$, where g(x) satisfies

$$ \varPsi\bigl( g ( x ) \bigr) \equiv\int_{0}^{g ( x ) } \bigl( \sigma( s ) - ( n-1+\alpha) /n \bigr) \,ds=\delta x/n\quad \text{for all }x. $$

Then, noting that y _i (t)=ϕ _i (g(x(t)),…,g(x(t)))=g(x(t)) for all t by Assumption 1(b), the problem of the remaining player i is to find a consumption path, c _i (⋅), maximizing

$$\begin{aligned} &\int_{0}^{\infty}v \bigl( c_{i} ( t ) ,g \bigl( x ( t ) \bigr) \bigr) e^{-\delta t}dt, \\ &\text{where:} \\ &\dot{x} ( t ) =- \bigl( c_{i} ( t ) + ( n-1 ) g \bigl( x ( t ) \bigr) \bigr), \qquad x ( 0 ) =x_{0}, \\ &c_{i} ( t ) \geq0, \qquad x ( t ) \geq0. \end{aligned}$$

Define the current-value Hamiltonian

$$ H_{i} ( x,c_{i},\lambda_{i} ) =v \bigl( c_{i},g ( x ) \bigr) -\lambda_{i} \bigl( c_{i}+ ( n-1 ) g ( x ) \bigr), $$

where λ _i is a costate multiplier associated with the state x.

A feasible control path c _i (⋅) with the corresponding state trajectory x(⋅) is optimal if there exists λ _i (⋅) such that

$$\begin{aligned} &v_{1} \bigl( c_{i} ( t ) ,g \bigl( x ( t ) \bigr) \bigr) - \lambda_{i} ( t ) \leq0\text{ and } \bigl( v_{1} \bigl( c_{i} ( t ) ,g \bigl( x ( t ) \bigr) \bigr) -\lambda_{i} ( t ) \bigr) c_{i} ( t ) =0, \end{aligned}$$

(12)

$$\begin{aligned} &\dot{\lambda}_{i} ( t ) = \bigl( \delta+ ( n-1 ) g^{\prime } \bigl( x ( t ) \bigr) \bigr) \lambda_{i} ( t ) -v_{2} \bigl( c_{i} ( t ) ,g \bigl( x ( t ) \bigr) \bigr) g^{\prime} \bigl( x ( t ) \bigr), \end{aligned}$$

(13)

$$\begin{aligned} &\lim_{t\rightarrow\infty}e^{-\delta t}\lambda_{i} ( t ) x ( t ) =0. \end{aligned}$$

(14)

Let c _i (t) and x(t) for all t satisfy

$$\begin{aligned} &c_{i} ( t ) =g \bigl( x ( t ) \bigr), \\ &\dot{x} ( t ) =-ng \bigl( x ( t ) \bigr), \qquad x ( 0 ) =x_{0}. \end{aligned}$$

We show below that the proposed consumption path c _i (⋅)=g(x(⋅)) is feasible, and we find λ _i (⋅) to satisfy conditions (12), (13), and (14). Proposition 2 follows.

(1) Feasibility. It is immediate to adapt Lemmas 1 and 2 to show that the proposed control path c _i (⋅) is feasible.

(2) Necessary conditions. Let T represent the time of depletion of the resource stock (including the possibility that T=∞). We first construct λ _i (⋅) to simultaneously satisfy the maximum condition (12) and the adjoint condition (13). We then prove that the transversality condition (14) holds.

For all t<T, as x(t)>0, c _i (t)=g(x(t))>0. Thus, the maximum condition (12) requires that

$$ \lambda_{i} ( t ) =v_{1} \bigl( c_{i} ( t ) ,g \bigl( x ( t ) \bigr) \bigr) =\frac{1}{1-\alpha}u^{\prime} \bigl( g \bigl( x ( t ) \bigr) \bigr). $$

To show that the adjoint condition (13) holds, differentiate this expression to get

$$ \dot{\lambda}_{i} ( t ) =\frac{1}{1-\alpha}u^{\prime\prime } \bigl( g \bigl( x ( t ) \bigr) \bigr) g^{\prime} \bigl( x ( t ) \bigr) \dot{x} ( t ). $$

Substitute $\dot{x} ( t ) =-ng ( x ( t ) )$ and use λ _i (t)=u′(g(x(t)))/(1−α) to obtain

$$ \dot{\lambda}_{i} ( t ) =n\sigma\bigl( g \bigl( x ( t ) \bigr) \bigr) g^{\prime} \bigl( x ( t ) \bigr) \lambda_{i} ( t ). $$

Now, by differentiation, Ψ(g(x))=δx/n for all x implies that

$$ n\sigma\bigl( g ( x ) \bigr) g^{\prime} ( x ) =\delta+ ( n-1+\alpha) g^{\prime} ( x ) \quad \text{for all }x. $$

Thus, after substitution, we can write

$$ \dot{\lambda}_{i} ( t ) = \bigl( \delta+ ( n-1+\alpha) g^{\prime} \bigl( x ( t ) \bigr) \bigr) \lambda_{i} ( t ). $$

Finally, using λ _i (t)=v ₁(c _i (t),g(x(t))) and v ₂(c _i (t),g(x(t)))=−αv ₁(c _i (t),g(x(t))) (as c _i (t)=g(x(t))), we derive

$$ \dot{\lambda}_{i} ( t ) = \bigl( \delta+ ( n-1 ) g^{\prime } \bigl( x ( t ) \bigr) \bigr) \lambda_{i} ( t ) -v_{2} \bigl( c_{i} ( t ) ,g \bigl( x ( t ) \bigr) \bigr) g^{\prime} \bigl( x ( t ) \bigr), $$

which is the adjoint equation (13).

For t≥T, as x(t)=0, c _i (t)=g(0)=0. Thus, the maximum condition (12) requires that

$$ v_{1} ( 0,0 ) =\frac{1}{1-\alpha}u^{\prime} ( 0 ) \leq \lambda_{i} ( t ). $$

Assume that λ _i (t) satisfies the adjoint equation (13) with λ _i (T)=v ₁(0,0). Then, we can show that

$$ \lambda_{i} ( t ) = \biggl( \lambda_{i} ( T ) - \frac{v_{2} ( 0,0 ) g^{\prime} ( 0 ) }{\delta+ ( n-1 ) g^{\prime} ( 0 ) } \biggr) e^{ ( \delta+ ( n-1 ) g^{\prime} ( 0 ) ) ( t-T ) }+\frac{ v_{2} ( 0,0 ) g^{\prime} ( 0 ) }{\delta+ ( n-1 ) g^{\prime} ( 0 ) }. $$

Substituting v ₂(0,0)=−αv ₁(0,0)=−αλ _i (T) and rearranging, we can write

$$ \lambda_{i} ( t ) =\frac{ ( \delta+ ( n-1+\alpha ) g^{\prime} ( 0 ) ) e^{ ( \delta+ ( n-1 ) g^{\prime} ( 0 ) ) ( t-T ) }-\alpha g^{\prime } ( 0 ) }{\delta+ ( n-1 ) g^{\prime} ( 0 ) }\lambda_{i} ( T ). $$

As e ^{(δ+(n−1)g′(0))(t−T)}≥1, this implies that λ _i (t)≥λ _i (T)=v ₁(0,0), which proves that the maximum condition (12) is true.

To verify the transversality condition (14), we need to separate the cases where T is finite or infinite. If T is finite, the condition is trivially verified since x(t)=0 for all t≥T. If T is infinite, define A(t)=e ^−δt λ _i (t)x(t). By differentiation we obtain

$$ \frac{\dot{A} ( t ) }{A ( t ) }=-\delta+\frac{\dot {\lambda }_{i} ( t ) }{\lambda_{i} ( t ) }+\frac{\dot{x} ( t ) }{x ( t ) }. $$

Using

$$\begin{aligned} \frac{\dot{\lambda}_{i} ( t ) }{\lambda_{i} ( t ) } =&\delta+ ( n-1+\alpha) g^{\prime} \bigl( x ( t ) \bigr), \\ \dot{x} ( t ) =&-ng \bigl( x ( t ) \bigr), \end{aligned}$$

we get

$$ \frac{\dot{A} ( t ) }{A ( t ) }= ( n-1+\alpha) g^{\prime} \bigl( x ( t ) \bigr) -n \frac{g ( x ( t ) ) }{x ( t ) }. $$

Now, by definition, for all x, g(x) satisfies Ψ(g(x))=δx/n, which is equivalent to

$$ \int_{0}^{g ( x ) } \bigl( n \bigl( \sigma( s ) -1 \bigr) +1-\alpha\bigr) \,ds=\delta x. $$

A first implication is that (by differentiation with respect to x)

$$ \bigl( n \bigl( \sigma\bigl( g ( x ) \bigr) -1 \bigr) +1-\alpha\bigr) g^{\prime} ( x ) =\delta. $$

A second implication is that there exists c, with 0<c<g(x), such that (by the mean value theorem)

$$ \bigl( n \bigl( \sigma( c ) -1 \bigr) +1-\alpha\bigr) g ( x ) =\delta x. $$

This piece of information implies that, for all t, there exists c, with 0<c<g(x(t)), such that

$$ \frac{\dot{A} ( t ) }{A ( t ) }= ( n-1+\alpha) \frac{\delta}{n ( \sigma( g ( x ( t ) ) ) -1 ) +1-\alpha}-n\frac{\delta}{n ( \sigma( c ) -1 ) +1-\alpha}. $$

Now, as t tends to infinity, the resource stock x(t) converges to 0, implying that both g(x(t)) and c (as 0<c<g(x(t))) converge to 0. From this we can write

$$ \lim_{t\rightarrow\infty}\frac{\dot{A} ( t ) }{A ( t ) }=- \frac{1-\alpha}{n ( \sigma( 0 ) -1 ) +1-\alpha}\delta<0, $$

which implies that the transversality condition (14) holds.

1.5 A.5 Proof of Property 2

For all n, let c ^∗ satisfy Θ(c ^∗)−(n−1+α)c ^∗/n=δx ₀/n, and let C ^∗=nc ^∗. By differentiation we show

$$\begin{aligned} \frac{dc^{\ast}}{dn} =&-\frac{\varTheta( c^{\ast} ) -c^{\ast }}{n\sigma( c^{\ast} ) -n+1-\alpha}, \\ \frac{dC^{\ast}}{dn} =&c^{\ast}-\frac{n ( \varTheta( c^{\ast } ) -c^{\ast} ) }{n\sigma( c^{\ast} ) -n+1-\alpha}. \end{aligned}$$

Let us show the first assertion. By the mean value theorem, given that Θ(0)=0, there exists c, with 0<c<c ^∗, such that

$$ \varTheta\bigl( c^{\ast} \bigr) =\sigma( c ) c^{\ast}. $$

Assume that σ(c)>1 for all c. Then, Θ(c ^∗)>c ^∗, implying that dc ^∗/dn<0. Likewise, one can prove that dc ^∗/dn=0 if σ(c)=1 for all c and dc ^∗/dn>0 if σ(c)<1 for all c.

As dC ^∗/dn=c ^∗+dc ^∗/dn, the second assertion directly follows from the first one.

Let us now prove the third assertion. Since

$$ n\varTheta\bigl( c^{\ast} \bigr) -nc^{\ast}=\delta x_{0}- ( 1-\alpha) c^{\ast}, $$

we can write, after substitution,

$$ \frac{dC^{\ast}}{dn}=c^{\ast}-\frac{\delta x_{0}}{n\sigma( c^{\ast } ) -n+1-\alpha}+\frac{ ( 1-\alpha) c^{\ast}}{n\sigma ( c^{\ast} ) -n+1-\alpha}. $$

Using anew Θ(c ^∗)=σ(c)c ^∗, we can show that

$$ c^{\ast}=\frac{\delta x_{0}}{n\sigma( c ) -n+1-\alpha}, $$

and, after substitution,

$$ \frac{dC^{\ast}}{dn}=\frac{n ( \sigma( c^{\ast} ) -\sigma ( c ) ) }{ ( n\sigma( c ) -n+1-\alpha ) ( n\sigma( c^{\ast} ) -n+1-\alpha) }\delta x_{0}+\frac{ ( 1-\alpha) c^{\ast}}{n\sigma( c^{\ast} ) -n+1-\alpha}. $$

It follows that dC ^∗/dn>0 whenever σ(c ^∗)≥σ(c), which proves the property.

1.6 A.6 Proof of Proposition 3

Consider any player i. Assume that all individuals j≠i play the Markovian strategy $s_{j}^{\ast}=f ( x )$, where f(x) satisfies Θ(f(x))=δx/n. Then, noting that y _i (t)=ϕ _i (f(x(t)),…,f(x(t)))=f(x(t)), z _i (t)=φ _i (f(x(t)),…,f(x(t)))=f(x(t)) and τ(z _i (t))≡(n−1+α)u′(f(x(t)))/((1−α)n) for all t, the problem of the remaining player i is to find a consumption path, c _i (⋅), maximizing

$$\begin{aligned} &\int_{0}^{\infty} \left [ \begin{array}{c} v ( c_{i} ( t ) ,f ( x ( t ) ) ) \\ -\tau( f ( x ) ) ( c_{i} ( t ) -f ( x ( t ) ) )\end{array} \right ] e^{-\delta t}\,dt, \\ &{}\text{where:} \\ &{}\dot{x} ( t ) =- \bigl( c_{i} ( t ) + ( n-1 ) f \bigl( x ( t ) \bigr) \bigr), \qquad x ( 0 ) =x_{0}, \\ &{}c_{i} ( t ) \geq0, \qquad x ( t ) \geq0. \end{aligned}$$

Define the current-value Hamiltonian

$$ H_{i} ( x,c_{i},\lambda_{i} ) =v \bigl( c_{i},f ( x ) \bigr) -\tau\bigl( f ( x ) \bigr) \bigl( c_{i}-f ( x ) \bigr) -\lambda_{i} \bigl( c_{i}+ ( n-1 ) f ( x ) \bigr), $$

where λ _i is a costate multiplier associated with the state x.

A feasible control path c _i (⋅), with corresponding state trajectory x(⋅), is optimal if there exists λ _i (⋅) such that

$$\begin{aligned} &v_{1} \bigl( c_{i} ( t ) ,f \bigl( x ( t ) \bigr) \bigr) - \tau\bigl( f \bigl( x ( t ) \bigr) \bigr) -\lambda_{i} ( t ) \leq0 \notag \\ &\quad \text{and}\quad \bigl( v_{1} \bigl( c_{i} ( t ) ,f \bigl( x ( t ) \bigr) \bigr) -\tau\bigl( f \bigl( x ( t ) \bigr) \bigr) - \lambda_{i} ( t ) \bigr) c_{i} ( t ) =0, \end{aligned}$$

(15)

$$\begin{aligned} &\dot{\lambda}_{i} ( t ) = \bigl( \delta+ ( n-1 ) f^{\prime } \bigl( x ( t ) \bigr) \bigr) \lambda_{i} ( t ) -v_{2} \bigl( c_{i} ( t ) ,f \bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) \notag \\ &\quad \qquad{} + \bigl( \tau^{\prime} \bigl( f \bigl( x ( t ) \bigr) \bigr) \bigl( c_{i} ( t ) -f \bigl( x ( t ) \bigr) \bigr) -\tau\bigl( f \bigl( x ( t ) \bigr) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr), \end{aligned}$$

(16)

$$\begin{aligned} &\lim_{t\rightarrow\infty}e^{-\delta t}\lambda_{i} ( t ) x ( t ) =0. \end{aligned}$$

(17)

Now, f(x) being such that Θ(f(x))=δx/n for all x, let c _i (t) and x(t) for all t satisfy

$$\begin{aligned} &c_{i} ( t ) =f \bigl( x ( t ) \bigr), \\ &\dot{x} ( t ) =-nf \bigl( x ( t ) \bigr), \qquad x ( 0 ) =x_{0}. \end{aligned}$$

We show below that the proposed consumption path c _i (⋅)=f(x(⋅)) is feasible, and we find λ _i (⋅) to satisfy conditions (15), (16), and (17). Proposition 3 follows.

(1) Feasibility The proposed path, being identical to the socially optimal consumption path, is clearly feasible (see Lemmas 1 and 2).

(2) Necessary conditions Let T represent the time of depletion of the resource stock (including the possibility that T=∞). We first construct λ _i (⋅) to simultaneously satisfy the maximum condition (15) and the adjoint condition (16). We then prove that the transversality condition (17) holds.

For all t<T, as x(t)>0 (by definition), c _i (t)=f(x(t))>0. Thus, the maximum condition (12) requires that (using v ₁(c _i (t),f(x(t)))=u′(f(x(t)))/(1−α) and τ(f(x(t)))=(n−1+α)u′(f(x(t)))/((1−α)n))

$$ \lambda_{i} ( t ) =v_{1} \bigl( c_{i} ( t ) ,f \bigl( x ( t ) \bigr) \bigr) -\tau\bigl( f \bigl( x ( t ) \bigr) \bigr ) = \frac{1}{n}u^{\prime} \bigl( f \bigl( x ( t ) \bigr) \bigr). $$

To show that the adjoint condition (16) holds, differentiate this expression to get

$$ \dot{\lambda}_{i} ( t ) =\frac{1}{n}u^{\prime\prime} \bigl( f \bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) \dot{x} ( t ). $$

Substitute $\dot{x} ( t ) =-nf ( x ( t ) )$ and use λ _i (t)=u′(f(x(t)))/n to obtain

$$ \dot{\lambda}_{i} ( t ) =n\sigma\bigl( f \bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) \lambda_{i} ( t ). $$

Now, by differentiation, Θ(f(x))=δx/n for all x implies that

$$ n\sigma\bigl( f ( x ) \bigr) f^{\prime} ( x ) =\delta\quad \text{for all }x. $$

Thus, after substitution, we can write

$$ \dot{\lambda}_{i} ( t ) =\delta\lambda_{i} ( t ). $$

Finally, we can verify that this precisely coincides with the adjoint equation (16). Indeed, using c _i (t)=f(x(t)), v ₁(c _i (t),f(x(t)))=u′(f(x(t)))/(1−α), v ₂(c _i (t),f(x(t)))=−αu′(f(x(t)))/(1−α), and λ _i (t)=u′(f(x(t)))/n, we can show that

$$\begin{aligned}[c] &( n-1 ) f^{\prime} \bigl( x ( t ) \bigr) \lambda_{i} ( t ) -v_{2} \bigl( c_{i} ( t ) ,f \bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) \\ &\quad {}+ \bigl( \tau^{\prime} \bigl( f \bigl( x ( t ) \bigr) \bigr) \bigl( c_{i} ( t ) -f \bigl( x ( t ) \bigr) \bigr) -\tau\bigl( x ( t ) \bigr) \bigr) f^{\prime} \bigl( x ( t ) \bigr) =0. \end{aligned} $$

For t≥T, as x(t)=0, c _i (t)=f(0)=0. Thus, the maximum condition (15) requires that

$$ v_{1} ( 0,0 ) -\tau( 0 ) =\frac{1}{n}u^{\prime } ( 0 ) \leq\lambda_{i} ( t ). $$

Assume that λ _i (t) satisfies the adjoint equation (16) with λ _i (T)=v ₁(0,0)−τ(0). Then, we can show that

$$ \lambda_{i} ( t ) = \biggl( \lambda_{i} ( T ) - \frac{( v_{2} ( 0,0 ) +\tau( 0 ) ) f^{\prime } ( 0 ) }{\delta+ ( n-1 ) f^{\prime} ( 0 ) }\biggr) e^{ ( \delta+ ( n-1 ) f^{\prime} ( 0 ) ) ( t-T ) }+\frac{ ( v_{2} ( 0,0 ) +\tau ( 0 ) ) f^{\prime} ( 0 ) }{\delta+ ( n-1 ) f^{\prime} ( 0 ) }. $$

Substituting v ₂(0,0)+τ(0)=(n−1)λ _i (T) and rearranging, we can write

$$ \lambda_{i} ( t ) =\frac{\delta e^{ ( \delta+ ( n-1 ) f^{\prime} ( 0 ) ) ( t-T ) }+ ( n-1 ) f^{\prime} ( 0 ) }{\delta+ ( n-1 ) f^{\prime } ( 0 ) }\lambda_{i} ( T ). $$

As e ^{(δ+(n−1)f′(0))(t−T)}≥1, this implies that λ _i (t)≥λ _i (T)=v ₁(0,0)−τ(0), which proves that the maximum condition (15) is true.

The transversality condition (17) is immediate since if T is finite, x(t)=0 for all t≥T, and if T is infinite, then $e^{-\delta t}\lambda_{i} ( t ) =\frac{1}{n}u^{\prime } ( f ( x_{0} ) )$ for all t, and lim _t→∞ x(t)=0.