Uniqueness Versus Indeterminacy in the Tragedy of the Commons: A ‘Geometric’ Approach

Abstract

This paper characterizes continuous Markov perfect equilibria as smooth connections between an ‘initial’, i.e., at the origin of the state space, and an ‘end’ manifold that result from patching with the boundary solution. The major result is that multiple equilibria require a non-monotonic initial manifold. This necessary condition for multiple equilibria can be tested without (or prior to) solving the Hamilton-Jacobi-Bellman equation. Application to a familiar dynamic tragedy of the commons with nonlinear instead of linear-quadratic utilities shows that the elasticity of marginal utility is the crucial property: If this elasticity is (everywhere) greater than $\frac{n-1}{n}$, n=number of polluters, then the Nash equilibrium is unique. Assuming the opposite inequality (globally) implies that no saddle-point equilibrium exists. Therefore, the ‘focal’ point equilibrium is gone and all conceivable boundary conditions determine a corresponding equilibrium, e.g. ‘anything goes’ for power utility functions.

Appendix

1.1 7.1 Proof of Proposition 1

First, value matching is obvious to rule out arbitrage from increasing or reducing X marginally so that the proof is confined to establish smooth pasting.

After moving for some time along the boundary (thus assuming the normalization (B1) and also symmetry), the optimal strategy of an individual player to start emitting again at T is

$$\begin{aligned} &\max_{T}\int _{0}^{T}e^{-rt} \bigl( -C \bigl( X ( t ) \bigr) \bigr) +e^{-rT}V \bigl( X ( T ) \bigr) , \\ &\quad\dot{X} =-\delta X, \end{aligned}$$

in which the interior value function V describes the scrap value. Defining the Hamiltonian, H=−C−λδX, the condition for the optimal choice of T is

$$H ( T ) =-C \bigl( X ( T ) \bigr) -\lambda ( T ) \delta X ( T ) =rV \bigl( X ( T ) \bigr) . $$

Using $V_{b}^{\prime}=\lambda$ from continuous dynamic programming on the left hand side, (7) and continuity of the strategies (x(T)=0) on the right hand side, yields:

$$-C \bigl( X ( T ) \bigr) -V_{b}^{\prime} \bigl( X ( T ) \bigr) \delta X ( T ) =-C \bigl( X ( T ) \bigr) -\delta X ( T ) V^{\prime} \bigl( X ( T ) \bigr) , $$

which requires $V_{b}^{\prime}=V^{\prime}$. □

1.2 7.2 Proof of Proposition 2

The following sketch of a proof follows Rowat (2000, 2006). Due to (12),

$$V_{b}^{\prime}=-\frac{Ar}{\delta}X^{-\frac{r+\delta}{\delta}}- \frac {cX}{(r+2\delta)}=-u^{\prime} ( 0 ) $$

characterizes the border between interior and boundary solutions, which establishes a relation between the state X and the constant of integration A as shown in Fig. 6. If A>0 then for x=0: V=V _b →∞ for X→0, contradicting the in (A1) assumed upper bound. The boundary strategy is impossible for A<0 in a surrounding of X=0 (see Fig. 1) and the knife-edge case, A=0, can be excluded by showing that a deviating strategy, x>0 at least for X close to 0, is profitable (see Rowat 2006).  □

Fig. 6

Interior and boundary domain in terms of the state X and integration constant A, similar to Rowat (2006) for the linear-quadratic case

1.3 7.3 Proof of Lemma 1

We show below that the inequality (15) is necessary and sufficient for the existence of Q=0 for any $X<\frac{n\widehat {x}}{\delta}$. From this follows immediately the first part for satiating utilities, $\widehat{x}<\infty$, since then σ→∞ for $x\rightarrow\widehat{x}$ and this inequality (15) is trivially fulfilled. Assuming instead non-satiating utilities, inequality (16) follows readily from the second Inada condition, $\widehat {x}=\infty$.

Sufficiency: We have to show that inequality (15) ensures the existence of Q=0. Rearranging (14) and holding $X<\frac {n\widehat{x}}{\delta}$ fixed yields

$$ \sigma \bigl( f \bigl( V^{\prime} \bigr) \bigr) =\frac{n-1}{n-\frac {\delta X}{f ( V^{\prime} ) }}, $$

(28)

i.e., two functions—one on the left hand side and the other on the right hand side—defined for V′≤0. The elasticity of marginal utility, σ, is positive and non-decreasing in x (due to Assumption (B2)) and thus also in V′ (due to (6)). Hence, the left hand side is positive, non-decreasing, σ′f′≥0, and reaches its maximum $\sigma ( f ( 0 ) ) =\sigma ( \widehat{x} )$ at V′=0, since V′→0 implies u′→0 and thus $x\rightarrow \widehat{x}$. The right hand side declines monotonically for all V′∈(s(X),0] and assumes all positive real numbers from $[ \frac {n-1}{n-\frac{\delta X}{\widehat{x}}},\infty )$. The upper bound results from V′→s(X), the lower from the above implication of V′=0. Summarizing: (a) the right hand side of (28) is declining and covering $[ \frac{n-1}{n-\frac{\delta X}{\widehat {x}}},\infty )$; (b) the left hand side is positive, increasing (or constant) and surpassing the minimal level of the right hand side if inequality (15) holds. Therefore, there exists a unique intersection that determines the value of V′ uniquely for each value of $X<\frac{n\widehat{x}}{\delta}$ along Q=0 and thus the function, V′=q(X).

Necessity follows because the opposite of the inequality of (15) rules out an intersection of the left hand and right hand sides in (28) so that {Q=0}=∅ ∀X≥0 in contradiction to the assumed existence.

The claim that q>s holds in fact in general and without Assumption (B2). For this purpose, rewrite (14) moving the term defining s to the right hand side

$$Q=0\quad\Longleftrightarrow\quad\frac{ ( n-1 ) V^{\prime}}{u^{\prime \prime }}= ( nf-\delta X ) =\dot{X}. $$

This implies that V′ from Q=0 is in the domain of $\dot{X}>0$ (for any $X<\frac{n\widehat{x}}{\delta}$ since V′=q<0). Hence q(X)>s(X) since it takes a larger negative value of V′ to induce $\dot{X}=0$.

Finally, the last claim, q′>0, is proven. The choice of X has no effect on the left hand side of (28), but a larger value of X reduces the support on the right hand side (s(X),0], because s(X) is increasing. This implies in turn a larger value of V′ (i.e. a smaller negative number), shifts the curve defined by the right hand side of (28) upwards and, as a consequence, the point of intersection at V′ must increase (from applying Assumption (B2) to the left hand side). Hence the set {Q=0} defines (if existing) an increasing function in the (X,V′) plane.  □

1.4 7.4 Proof of Theorem 1

The first part of the Theorem for $\sigma>\frac{n-1}{n}\ \forall x>0$ follows directly from Lemma 1 and Proposition 5 since only passing through the then unique intersection between P=0 and Q=0 allows for a globally defined strategy.

If in contrast $\sigma<\frac{n-1}{n}\ \forall x>0$ and thus in particular for all V′∈(−∞,0], then the left hand side of (28) is always below the right hand side for V′∈(s(X),0], and always above for V′∈(−∞,s(X)) (irrespective of the sign of σ′). Therefore, the set Q=0 is empty for all X. P=0 is linearly declining and thus crossing the entire relevant quadrant X≥0 and V′<0. The set s is monotonically increasing and also covering the entire relevant domain if u satisfies the two Inada conditions: s(X)→−∞ for X→0, s(X)→0 for X→∞. As a consequence, any solution curve V′(X) starting at an arbitrary initial value V′(0)<0 (and V(0) determined from the initial manifold (17)) must stay within the quadrant X≥0 and V′<0 since it must start above s, must increase until it intersects P=0 at a feasible value of V′<0 and must decline after this intersection. Therefore this solution has global, interior support.

It remains to show that each strategy originating from an arbitrary V′(0)<0 ensures a unique and stable steady state. Stability of the associated Markov strategy requires that such a solution curve V′ crosses s from above (and only from above). And clearly given the shapes of P=0 and s any of these strategies must cross s, which ensures stability of this steady state. Therefore further and in particular unstable steady states must be excluded. Indirectly, assume such an unstable steady state at X ₂, (i.e., intersections from below yet clearly in the domain V′′>0⟺P<0):

$$V^{\prime\prime}=\frac{-u^{\prime\prime} [ cX+ ( r+\delta ) V^{\prime} ] }{ ( n-1 ) V^{\prime}}>s^{\prime }=-\frac{\delta u^{\prime\prime}}{n} \quad\Leftrightarrow\quad V^{\prime}<-\frac {ncX}{nr+\delta}\text{ at }X=X_{2} . $$

Since V′(0)>lim _X→0 s=−∞ it must follow an intersection from above at X ₁<X ₂: $V^{\prime \prime}<s^{\prime}\Leftrightarrow V^{\prime} ( X_{1} ) >-\frac{ncX_{1}}{nr+\delta}$. Therefore, V′(X ₁)>V′(X ₂), which is impossible in the domain P<0 and thus V′′>0. Contradiction.  □

1.5 7.5 Proof of Proposition 6

The proof falls into three parts. First, the socially optimal policy is derived. Second a lower bound on steady states attainable by Nash behavior is derived. Finally, this lower bound is compared with the efficient stationary pollution.

1.5.1 7.5.1 Derivation of the Social Optimum for Utility (24)

The Hamilton-Jacobi-Bellman equation for the corresponding value function W is

$$ rW(X)=\max_{x\geq0} \biggl\{ n \biggl( \frac{x^{a}}{a}- \frac {c}{2}X^{2} \biggr) +W^{\prime}(X) ( nx-\delta X ) \biggr\} . $$

(29)

Maximization on the right hand side yields for interior policies:

$$x^{\ast}= \bigl( -W^{\prime} \bigr) ^{\frac{1}{a-1}}, $$

which implies for (29):

$$ rW=n \biggl( \frac{ ( -W^{\prime} ) ^{\frac{a}{a-1}}}{a}-\frac {c}{2}X^{2} \biggr) +W^{\prime} \bigl( n \bigl( -W^{\prime} \bigr) ^{\frac{1}{a-1}}-\delta X \bigr) . $$

(30)

Differentiating (30), simplifying and representing as a ratio yields,

$$W^{\prime\prime}=\frac{P(X,W^{\prime})}{Q(X,W^{\prime})}=\frac {cnX+ ( r+\delta ) W^{\prime}}{n ( -W^{\prime} ) ^{\frac{1}{a-1}}-\delta X}. $$

Therefore

$$\begin{aligned} P\bigl(X,W^{\prime}\bigr) =&0\quad\Longleftrightarrow\quad W^{\prime}=- \frac {ncX}{r+\delta}, \\ Q\bigl(X,W^{\prime}\bigr) =&0\quad\Longleftrightarrow\quad W^{\prime}=- \biggl( \frac {\delta X}{n} \biggr) ^{a-1}. \end{aligned}$$

Moreover, {Q=0} coincides with s. Therefore, the intersection of P=0 with Q=0 determines simultaneously a point of the saddle-point path and the steady state. Hence, solving the equation

$$-\frac{ncX}{r+\delta}=- \biggl( \frac{\delta X}{n} \biggr) ^{a-1} $$

yields

$$X^{\ast}= \biggl( \frac{nc}{r+\delta} \biggr) ^{\frac{1}{a-2}} \biggl( \frac{n}{\delta} \biggr) ^{\frac{a-1}{a-2}}. $$

 □

1.5.2 7.5.2 Lower Bound for $\bar{X}$

Figure 5 suggests that the solution curves cut the steady state line s only after the intersection of P=0 with s. This intersection between P and s occurs at

$$\widetilde{X}= \biggl( \frac{c}{r+\delta} \biggr) ^{\frac{1}{a-2}} \biggl( \frac{n}{\delta} \biggr) ^{\frac{a-1}{a-2}}. $$

This bound clearly exceeds the social optimum, $\widetilde{X}>X^{\ast}$, but one cannot exclude in principle the possibility of an intersection at a lower level, i.e. when V′′>0. A stable intersection (i.e. one from above) requires

$$V^{\prime}>-\frac{ncX}{nr+\delta}, $$

which is independent from the specification of u. Intersecting now this general (lower) bound of V′ with the specific s from (26) yields

$$-\frac{ncX}{nr+\delta}=- \biggl( \frac{\delta X}{n} \biggr) ^{a-1}. $$

The solution of this equation determines the lower bound on stable steady states

$$\underline{X}= \biggl( \frac{cn}{nr+\delta} \biggr) ^{\frac{1}{a-2}} \biggl( \frac{n}{\delta} \biggr) ^{\frac{a-1}{a-2}}. $$

Therefore it remains to compare the social optimum with the above lower bound,

$$ \frac{X^{\ast}}{\underline{X}}=\frac{ ( \frac{nc}{r+\delta} ) ^{\frac{1}{a-2}} ( \frac{n}{\delta} ) ^{\frac {a-1}{a-2}}}{ ( \frac{cn}{nr+\delta} ) ^{\frac{1}{a-2}} ( \frac{n}{\delta } ) ^{\frac{a-1}{a-2}}}= \biggl( \frac{nr+\delta}{r+\delta} \biggr) ^{\frac {1}{a-2}}= \biggl( \frac{r+\delta}{nr+\delta} \biggr) ^{\frac {1}{2-a}}<1. $$

(31)

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