Agri-environmental Policies and Public Goods: An Assessment of Coalition Incentives and Minimum Participation Rules

Abstract

An increasing number of papers analyse the inclusion of collective/spatial conditionality constraints in agricultural policies dealing with natural resource management. In this article we theoretically assess the conditions in which employing collective conditionality constraints linked to incentives better reach the social preferences on PG provision by agriculture. We deal with this issue by using a coalition formation model to endogenize the size of the group of farmers cooperating, and investigate how it is affected by different policy schemes. We analyse and compare the following policy schemes: (1) a homogenous payment that target the whole population of farmers, (2) a coalition bonus, that incentivizes only the contributions by the coalition members, and (3) a coalition bonus associated to a MPR on the size of the coalition. The results show that formulating payments that discriminate between co-operators and free-riders, and associating to such a payment a MPR, is relatively more effective than the traditional homogenous payments. However this is true only under some (local) conditions that we theoretically derived.

Appendices

Appendix A

We derive the model used in the article from a simple model of land allocation between agriculture (x, with benefit y) and PG (l, with benefit g). Agriculture has a quadratic cost function with parameter k. Total available land is L. Thus:

$$ \begin{aligned} & \mathop {max}\limits_{l,x} \pi = gl + yx - \frac{1}{2}kx^{2} \\ & s.t. \\ & l + x \le L \\ \end{aligned} $$

In case if g = 0 we have that l = 0, no land is allocated to PG and the land allocated to agriculture is $x^{*} = \frac{y}{k}$. The allocation of land to PG is fully costly in terms of agricultural production as long as $L \le \frac{y}{k}$. Solving the land constraint for x and substituting it in the profit function yields: $\pi = gl + y(L - l) - \frac{1}{2}kL^{2} - \frac{1}{2}kl^{2} + klL$. If we substitute $L = \frac{y}{k}$ profit becomes $\pi = gl + \frac{1}{2}\frac{{y^{2} }}{k} - \frac{1}{2}kl^{2}$ which has a fixed component $\frac{1}{2}\frac{{y^{2} }}{k}$ and the rest is the function that we use in the article. By taking FOC with respect to l yields $l^{*} = g/k$, which must be lower than y/k, so that g≤ y. y however does not directly enter our problem.

Appendix B

2.1 Appendix B1

To find the stable coalition after the introduction of a payment, start from π_f(s − 1):

$$ \pi_{f} (s - 1) = g\left( {(s - 1)\left( {\frac{g(s - 1) + p + b}{k}} \right) + (n - s + 1)\frac{(g + p)}{k}} \right) + p\frac{(g + p)}{k} - \frac{1}{2}k\frac{(g + p)}{{k^{2} }}_{f}^{2} $$

which becomes:

$$ \pi_{f} (s - 1) = \frac{1}{k}\left( {g^{2} s^{2} + \frac{3}{2}g^{2} - 3sg^{2} + sbg - bg + ng^{2} + ngp + \frac{1}{2}p^{2} } \right) $$

We then put π_f(s − 1) and π_m(s) in the stability function

$$ \begin{aligned} Z & = \frac{1}{2}g^{2} s^{2} + gbs + ng^{2} - sg^{2} + npg + \frac{1}{2}p^{2} + pb + \frac{1}{2}b^{2} - g^{2} s^{2} - \frac{3}{2}g^{2} + 3sg^{2} - sbg + bg - ng^{2} - ngp - \frac{1}{2}p^{2} \\ Z & = - \frac{1}{2}g^{2} s^{2} + gbs - sg^{2} + pb + \frac{1}{2}b^{2} - \frac{3}{2}g^{2} + 3sg^{2} - sbg + bg \\ Z & = - \frac{1}{2}g^{2} s^{2} + pb + \frac{1}{2}b^{2} - \frac{3}{2}g^{2} + 2sg^{2} + bg \\ \end{aligned} $$

We set the stability function equal to 0, which after some steps becomes:

$$ \begin{aligned} & \frac{1}{2}g^{2} s^{2} + \frac{3}{2}g^{2} - 2sg^{2} - bg - pb - \frac{1}{2}b^{2} = 0 \\ & g^{2} s^{2} - 4sg^{2} + 3g^{2} - 2bg - 2pb - b^{2} = 0 \\ & s = \frac{{4g^{2} + \sqrt {16g^{4} - 4g^{2} (3g^{2} - 2bg - 2pb - b^{2} )} }}{{2g^{2} }} \\ & s = 2 \pm \frac{{\sqrt {g^{2} + 2bg + 2pb + b^{2} } }}{g} \\ \end{aligned} $$

We have two solutions. First it is observed that:

$$ 2 - \frac{1}{g}\sqrt {g^{2} + b^{2} + 2pb + 2gb} < 0 $$

holds when b > g and p > g:

$$ \begin{aligned} 4 & < \frac{1}{{g^{2} }}(g^{2} + b^{2} + 2pb + 2gb) \\ 4 & < \left( {1 + \frac{{b^{2} }}{{g^{2} }} + \frac{2pb}{{g^{2} }} + \frac{2b}{g}} \right) \\ \end{aligned} $$

Since a negative coalition is meaningless, we exclude the solution: $s = 2 - \frac{{\sqrt {g^{2} + 2bg + 2pb + b^{2} } }}{g}$.

We then evaluate the second solution by substituting it in the first derivative of Z with respect to s, and assess whether is negative: $Z_{s} \left( {2 + \frac{{\sqrt {g^{2} + 2bg + 2pb + b^{2} } }}{g}} \right) < 0$.

The derivative is: $\frac{dZ}{ds} = - g^{2} s + 2g^{2}$. We set $\frac{dZ}{ds} < 0$ obtaining: $- s + 2 < 0$. Substituting $s = 2 + \frac{1}{g}\sqrt {g^{2} + b^{2} + 2pb + 2gb}$ yields:

$$ - 2 - \frac{1}{g}\sqrt {g^{2} + b^{2} + 2pb + 2gb} + 2 < 0 $$

which is surely negative and thus that confirms:

$$ s^{*} = 2 + \frac{1}{g}\sqrt {g^{2} + b^{2} + 2pb + 2gb} $$

2.2 Appendix B2

Profits for coalition members in the CB scheme are:

$$ \pi_{m}^{CB} (s = t) = \frac{1}{k}\left( {\frac{1}{2}g^{2} s^{2} + s(gb - g^{2} ) + \frac{1}{2}b^{2} + ng^{2} } \right) $$

Profits for free riders are:

$$ \pi_{f} \left( {s - 1} \right) = g\left[ {\frac{g}{k}\left( {s - 1} \right)^{2} + \frac{g}{k}\left( {n - s + 1} \right)} \right] - \frac{1}{2}k\left( {\frac{g}{k}} \right)^{2} = \frac{1}{2}\frac{{g^{2} }}{k}(2s^{2} - 6s + 2n + 3) $$

The coalition of size s = t (with a MPR) is stable as long as $\pi_{m}^{CB} (s = t) \ge \pi_{f} (s = t - 1)$:

$$ \begin{aligned} & \frac{1}{2}g^{2} s^{2} + gbs + ng^{2} - sg^{2} + \frac{1}{2}b^{2} > g^{2} s^{2} + \frac{3}{2}g^{2} - 3sg^{2} + ng^{2} \\ & g^{2} s^{2} + 2gbs + 2ng^{2} - 2sg^{2} + b^{2} > 2g^{2} s^{2} + 3g^{2} - 6sg^{2} + 2ng^{2} \\ & g^{2} s^{2} - s(4g^{2} + 2gb) - b^{2} + 3g^{2} < 0 \\ & s = \frac{{(4g^{2} + 2gb) \pm \sqrt {(4g^{2} + 2gb) - 4g^{2} ( - b^{2} + 3g^{2} )} }}{{2g^{2} }} \\ & s = 2 + \frac{b}{g} \pm \frac{1}{g}\sqrt {g^{2} + 2b^{2} + 4gb} \\ \end{aligned} $$

note that:

$$ \begin{aligned} & 2 + \frac{b}{g} - \frac{1}{g}\sqrt {g^{2} + 2b^{2} + 4gb} < 3 + \frac{b}{g} \\ & - \frac{1}{g}\sqrt {g^{2} + 2b^{2} + 4gb} < 1 \\ & {\text{so}}\,t^{min} = 3 + \frac{b}{g} \\ & {\text{and}}\, - \frac{1}{g}\sqrt {g^{2} + 2b^{2} + 4gb} < 1 \\ \end{aligned} $$

2.3 Appendix B3

The total contribution from the whole population:

$$ \begin{aligned} L^{*} & = l_{m}^{*} \cdot s^{*} + (n - s^{*} ) \cdot l_{f}^{*} \\ L^{*} & = \frac{{gs^{*} + p + b}}{k}s^{*} + (n - s^{*} )\frac{g + p}{k} \\ L^{*} & = \frac{1}{k}(gs^{*2} + ps + bs^{*} + gn - gs^{*} + pn - ps^{*} ) \\ L^{*} & = \frac{1}{k}(gs^{*2} + s^{*} (b - g) + n(g + p)) \\ \end{aligned} $$

The stable coalition when p = 0 is:

$$ \begin{aligned} s^{*} & = 2 + \frac{1}{g}\sqrt {g^{2} + b^{2} + 2gb} \\ s^{*} & = 2 + \frac{1}{g}\sqrt {(g + b)^{2} } \\ s^{*} & = 2 + \frac{1}{g}(g + b) \\ s^{*} & = 3 + \frac{b}{g} \\ \end{aligned} $$

Thus L becomes:

$$ \begin{aligned} L^{*} & = \frac{1}{k}\left( {g\left( {3 + \frac{b}{g}} \right)^{2} + \left( {3 + \frac{b}{g}} \right)(b - g) + ng} \right) \\ L^{*} & = \frac{1}{k}\left( {g\left( {9 + \frac{{b^{2} }}{{g^{2} }} + 6\frac{b}{g}} \right) + 3b - 3g + \frac{{b^{2} }}{g} - b + ng} \right) \\ L^{*} & = \frac{1}{k}\left( {9g + \frac{{b^{2} }}{g} + 6b + 3b - 3g + \frac{{b^{2} }}{g} - b + ng} \right) \\ L^{*} & = \frac{1}{k}\left( {6g + 2\frac{{b^{2} }}{g} + 8b + ng} \right) \\ \end{aligned} $$

Setting a homogenous payment leads to a size of the stable coalition of 3; total protected land is given by:

$$ L^{*} = \frac{1}{k}(gs^{*2} - gs^{*} + n(g + p)) $$

and after substituting $s^{*}$ = 3 becomes:

$$ L^{*} = \frac{1}{k}(6g + n(g + p)) $$

When is discriminating convenient? Or when L^CB > L^HP:

$$ \frac{2}{g}b^{2} + 8b + 6g + gn > pn + 6g + gn $$

After substituting ρ:

$$ \begin{aligned} & \frac{2}{g}\rho^{2} + 8\rho + 6g + gn > \rho n + 6g + gn \\ & \frac{2}{g}\rho^{2} + 8\rho - \rho n > 0 \\ & 2\rho + 8g - gn > 0 \\ & 2\rho + 8g - gn > 0 \\ \end{aligned} $$

which can be substituted in either L^CB > L^HP to find $\hat{L}$.