Could free-riders promote cooperation in the commons?

Abstract

It is widely recognized that informal monitoring and sanctions foster cooperation in common resource dilemmas. Theoretical models usually assume that agents who punish are those who cooperate on efficient harvesting norms. However, observations from the field show that sanctions are also practiced by those who overexploit the resource. Emotional reasons such as revenge or spiteful motives are suggested to account for this behavior, but none of them insist on economic incentives. Using an evolutionary model, we provide an alternative explanation for the presence of punishing actions undertaken by overexploiters. By assuming that overexploiters have the opportunity to sanction other harvesters, we find that a dual state composed of both cooperators and punishing overexploiters can be stable. Despite the typically negative judgment about such behavior, punishments inflicted by overexploiters may increase cooperation on more efficient harvesting levels. We discuss the implication of this result for resource management.

Appendix

1.1 Proof of Propositions 1 and 2

The two-dimensional system (10) is in equilibrium when the shares of the different strategies do not change, formally $\dot{s_1}$ and $\dot{s_2}$ must equal zero. Let us rewrite (10) with the payoff functions (11–13),

$$ \left\{ \begin{array}{cl} \dot{s_1} =& s_1[-(1-s_1)(e_{\rm h}-e_{\rm l})(A(E)-c) + (1-s_1-s_2)(1-s_1-1/n)(\sigma + \gamma)]\\ \dot{s_2} =& s_2[s_1(e_{\rm h}-e_{\rm l})(A(E)-c) + (1-s_1-s_2)[\gamma-(s_1+ 1/n)(\sigma + \gamma) ] ] \end{array}\right. $$

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This system, consisting of autonomous differential equations, is nonlinear. However, stability of steady states can be checked by linearizing around the steady states. Hence, we apply the Jacobian analysis to determine the possible stable equilibria.

$$ J = \left(\begin{array}{ll} J_{11} & J_{12} \\ J_{21} & J_{22} \end{array}\right) = \left(\begin{array}{ll} \frac{\partial \dot{s_1}}{\partial s_1} & \frac{\partial \dot{s_1}}{\partial s_2} \\ \frac{\partial \dot{s_2}}{\partial s_1} & \frac{\partial \dot{s_2}}{\partial s_2} \end{array}\right), $$

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where

$$ \begin{aligned} J_{11} &= -(e_{\rm h}-e_{\rm l})[(1-2s_1)(A-c)+s_1(1-s_1)A_E E_{s_1}] \\&+ [(1-2s_1-s_2)(1-s_1-1/n)-s_1(1-s_1-s_2)](\sigma + \gamma) \\ J_{12} &= -s_1 (1-s_1-1/n)(\sigma + \gamma) \\ J_{21} &= s_2[(e_{\rm h}-e_{\rm l})(A-c+s_1 A_E E_{s_1}) -\gamma -(1-2s_1-s_2-1/n)(\sigma + \gamma)] \\ J_{22} &= s_1(e_{\rm h}-e_{\rm l})(A-c) + (1-s_1- 2s_2)[\gamma-(s_1+ 1/n)(\sigma + \gamma)] \\ \end{aligned} $$

with A = A(E), and $A_E E_{s_1}=n(e_{\rm h}-e_{\rm l}) \frac{f(E) - E f_E(E)}{E^2} >0$ by concavity of f(E). The positive sign describes the obvious fact that average returns increase when the proportion of cooperators rises, and conversely.

In a steady state, all remaining strategies must yield the same payoff, which is equivalent to the average payoff $\bar{\pi}.$ All steady states possibilities (in qualitative terms) and their stability are listed below:

• $S_{\rm C}=(1,0,0).$ For the state composed only of cooperators, the elements of J are:

  $$ \begin{aligned} J_{11} &= (e_{\rm h}-e_{\rm l})(A-c) + \frac{\sigma + \gamma}{n}>0 \\ J_{12} &= \frac{\sigma + \gamma}{n} \\ J_{21} &= 0 \\ J_{22} &= (e_{\rm h}-e_{\rm l})(A-c)>0 \end{aligned} $$

  J ₁₁ and J ₂₂ are positive, so S _C is always unstable.

• $S_{\rm D}=(0,1,0).$ For defectors, J becomes:

  $$ \begin{aligned} J_{11} &= -(e_{\rm h}-e_{\rm l})(A-c)<0 \\ J_{12} &= 0 \\ J_{21} &= (e_{\rm h}-e_{\rm l})(A-c) + \frac{\sigma + \gamma}{n}-\gamma \\ J_{22} &= \frac{\sigma + \gamma}{n} -\gamma \end{aligned} $$

  J ₂₂ < 0 if γ (n − 1) > σ. This condition is necessary and sufficient for $\det(J)>0$ and $\mathrm{tr}(J)<0.$ Then, S _D is locally asymptotically stable when γ (n − 1) > σ.

• $S_{\rm P}=(0,0,1).$ For punishers only, J is:

  $$ \begin{aligned} J_{11} &= -(e_{\rm h}-e_{\rm l})(A-c) + (1-1/n)(\sigma +\gamma) \\ J_{12} &= 0 \\ J_{21} &= 0 \\ J_{22} &= \gamma - \frac{\sigma + \gamma}{n} \end{aligned} $$

  (e _h − e _l)(A − c) > (1 − 1/n)(σ + γ) and γ (n − 1) < σ are necessary and sufficient conditions for asymptotic stability.

• $S_{\rm CD}=(s_1^*,s_2^*,0).$ Obviously, a state composed only of cooperators and defectors cannot be a stable state, since without punishment defectors perform better than cooperators. Formally, the steady state condition π₁ = π₂ would imply e _h(A − c) = e _l(A − c), which is rejected by definition.

• $S_{\rm CP}=(s_1^*,0,s_3^*).$ Setting s ₂ = 0 and (e _h − e _l)(A − c) = (1 − s ₁ − 1/n)(σ + γ) (for π₁ = π₃), the elements of the jacobian are:

  $$ \begin{aligned} J_{11} &= -s_1(1-s_1)(\sigma+\gamma+ (e_{\rm h}-e_{\rm l}) A_E E_{s_1})<0 \\ J_{12} &= -s_1 (1-s_1-1/n)(\sigma + \gamma)\\ J_{21} &= 0 \\ J_{22} &= (1-s_1-1/n)\gamma-\sigma/n \end{aligned} $$

  J ₁₁ is always negative, whereas J ₂₂ < 0 if σ > (n s ₃ − 1)γ with s ₃ = 1 − s ₁. Eliminating s ₃, this condition can be rewritten as $\frac{\sigma}{\gamma}>n\frac{(e_{\rm h}-e_{\rm l})(A(E)-c)}{(\sigma+\gamma)},$ and is sufficient for global asymptotic stability of S _CP.

  Of course, S _P and S _CP cannot be stable at the same time. We verify this because average returns A(E) evaluated at S _CP are higher than A(E) in S _P, hence conditions (e _h − e _l)(A − c) = (1 − s ₁ − 1/n)(σ + γ) for S _CP and (e _h − e _l)(A − c) > (1 − 1/n)(σ + γ) for S _P cannot be fulfilled simultaneously. Indeed, it would require s ₁ < 0 for (1 − s ₁ − 1/n)(σ + γ) > (1 − 1/n)(σ + γ) to hold.

• $S_{\rm DP}=(0,s_2^*,s_3^*).$ With the condition σ = (n − 1)γ for π₂ = π₃, the elements of the jacobian are

  $$ \begin{aligned} J_{11} &= -(e_h-e_l)(A-c) + (1-s_2)(n-1)\gamma \\ J_{12} &= 0 \\ J_{21} &= s_2[(e_h-e_l)(A-c) -(1-s_2)n\gamma] \\ J_{22} &= 0 \end{aligned} $$

  J ₁₁ < 0 if (e _h − e _l)(A − c) > s ₃(n − 1)γ. J ₂₂ = 0 means that we obtain a nonhyperbolic equilibrium in this case. Consequently, linearization cannot help us to ascertain stability. One must use techniques of nonlinear analysis to check whether the equilibrium is locally stable or not. Stability would require the coincidence that the punishment technology σ/γ exactly equals n − 1.

• $S_{\rm CDP}=(s_1^*,s_2^*,s_3^*).$ This steady state requires $\bar{\pi}=\pi_1=\pi_2=\pi_3,$ which implies σ = γ(n(1 − s ₁) − 1) and (e _h − e _l)(A − c) = (1 − s ₁ − s ₂)σ. Then, the elements of the Jacobian are now

  $$ \begin{aligned} J_{11} &= -s_1[(1-s_1-s_2)\gamma +(1-s_1)(\sigma +A_E E_{s_1})]<0 \\ J_{12} &= -s_1 n (1-s_1)(1-s_1-1/n)\gamma <0 \\ J_{21} &= s_1 s_2 (\sigma + A_E E_{s_1}) - (1-s_1-s_2)\gamma \\ J_{22} &= s_1 s_2 \sigma >0 \end{aligned} $$

  $\mathrm{tr}(J)= -s_1(1-s_1)[(1-s_1-s_2-s_3)n\gamma+A_E E_{s_1}] <0.$ However, after some calculation we find $\det(J)= -s_1(1-s_1)[(1-s_1-s_2)n \gamma+A_E E_{s_1}] <0.$ Thus, the state is unstable. All three types of agents never coexist in the long run.

$\square$

1.2 Proof of Proposition 3

Under payoff function (14–16), the system (10) becomes

$$ \left\{ \begin{array}{cl} \dot{s_1} = s_1[-(1-s_1)(e_{\rm h}-e_{\rm l})(A(E)-c)\\ + (1-s_1-s_2)(\sigma + \gamma)(s_2+\lambda(1-s_1-s_2-1/n))] \\ \dot{s_2} = s_2[s_1(e_{\rm h}-e_{\rm l})(A(E)-c) - (1-s_1-s_2)\gamma \\ +(1-s_1-s_2)(\sigma + \gamma)(s_2+\lambda(1-s_1-s_2-1/n))]\end{array}\right. $$

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Evaluated at $S_{\rm CP}=(s_1^*,0,s_3^*),$ the elements of the jacobian are:

$$ \begin{aligned} J_{11} &= -s_1(1-s_1)[(e_{\rm h}-e_{\rm l}) A_E E_{s_1}+\lambda(\sigma + \gamma)]<0 \\ J_{12} &= s_1 (\sigma + \gamma)[(1-2\lambda)(1-s_1)+ \lambda/n] \\ J_{21} &= 0 \\ J_{22} &= -(1-s_1)\sigma + \lambda(\sigma + \gamma)(1-s_1-1/n) \\ \end{aligned} $$

$\det(J)>0$ and $\mathrm{tr}(J)<0$ if J ₂₂ < 0, which is the case when [(1 − λ)s ₃ + λ/n]σ > λ(s ₃ − 1/n) γ. Substituting s ₃ gives the stability condition [(1 − λ)(e _h − e _l)(A(E) − c) + λ(σ + γ)/n]σ > λ (e _h − e _l)(A(E) − c) γ. When the number of agents tends to infinity, it reduces to (1 − λ)σ > λγ. $\square$

1.3 Proof of Claim 3

The implicit function theorem is used to perform comparative statics. The effects of changes in parameters (respectively c, σ, γ, λ and n) on $(s_1^*)$ are determined by the signs of derivatives:

$$ \frac{\partial s_1^*}{\partial c} = - \frac{\partial \dot{s_1}/ \partial c}{\partial \dot{s_1} / \partial s_1^*}= \frac{e_h-e_l}{(e_{\rm h}-e_{\rm l}) A_E E_{s_1}+\lambda (\sigma + \gamma)}>0 $$

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$$ \frac{\partial s_1^*}{\partial \sigma} = \frac{(1-s_1-1/n)\lambda}{(e_{\rm h}-e_{\rm l}) A_E E_{s_1}+\lambda (\sigma + \gamma)}>0 $$

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$$ \frac{\partial s_1^*}{\partial \gamma} = \frac{\partial s_1^*}{\partial \sigma} >0 $$

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$$ \frac{\partial s_1^*}{\partial \lambda} = \frac{(1-s_1-1/n)(\sigma + \gamma)}{(e_{\rm h}-e_{\rm l}) A_E E_{s_1}+\lambda (\sigma + \gamma)} >0 $$

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$$ \frac{\partial s_1^*}{\partial n} = \frac{(e_{\rm h}-e_{\rm l})A_E E_n - \lambda (\sigma + \gamma)/n^2}{(e_{\rm h}-e_{\rm l}) A_E E_{s_1}+\lambda (\sigma + \gamma)} \lesseqgtr 0 $$

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$\square$