Bargaining Over Environmental Budgets: A Political Economy Model with Application to French Water Policy

Abstract

In decentralized water management with earmarked budgets financed by user taxes and distributed back in the form of subsidies, net gains are often heterogeneous across user categories. This paper explores the role of negotiation over budget allocation and coalition formation in water boards, to provide an explanation for such user-specific gaps between tax payments and subsidies. We propose a bargaining model to represent the sequential nature of the negotiation process in water districts, in which stakeholder representatives may bargain upon a fraction of the budget only. The structural model of budget shares estimated from the data on French Water Agencies performs well as compared with reduced-form estimation. Empirical results confirm the two-stage bargaining process and provide evidence for systematic net gains from the system for agricultural water users.

Appendices

Appendix 1: Proof of Proposition 1

Taking derivatives of V with respect to $\alpha$ one gets:

$$\begin{aligned} V_{i}^{\prime }\left( \alpha \right)= & {} -\widehat{p}_{i}u_{i}^{\prime }\left( \gamma _{i}+(1-\alpha )\sum _{j\in N\backslash (S_{i}\cup \left\{ i\right\} )}\gamma _{j}\right) \sum _{j\in N\backslash (S_{i}\cup \left\{ i\right\} )}\gamma _{j} \nonumber \\&+\left( 1-\widehat{p}_{i}-P_{i}\right) u_{i}^{\prime }(\alpha \gamma _{i})\gamma _{i}. \end{aligned}$$

(23)

and

$$\begin{aligned} V_{i}^{\prime \prime }\left( \alpha \right)= & {} \widehat{p}_{i}u_{i}^{\prime \prime }\left( \gamma _{i}+(1-\alpha )\sum _{j\in N\backslash (S_{i}\cup \left\{ i\right\} )}\gamma _{j}\right) \left( \sum _{j\in N\backslash (S_{i}\cup \left\{ i\right\} )}\gamma _{j}\right) ^{2} \nonumber \\&+\left( 1-\widehat{p}_{i}-P_{i}\right) u_{i}^{\prime \prime }(\alpha \gamma _{i})\left( \gamma _{i}\right) ^{2}. \end{aligned}$$

(24)

Since $u_{i}^{\prime \prime }(\cdot )<0$ it follows from (24) that $V_{i}^{\prime \prime }\left( \alpha \right) \le 0$.

From (23) it follows that:

$$\begin{aligned} V_{i}^{\prime }\left( 1\right) =u_{i}^{\prime }(\gamma _{i})\left[ \left( 1- \widehat{p}_{i}-P_{i}\right) \gamma _{i} - \widehat{p}_{i}\sum _{j\in N\backslash (S^{i}\cup \left\{ i\right\} ) }\gamma _{j}\right] . \end{aligned}$$

Therefore, for $\gamma _{i}\ge \overline{\gamma }_{i}=\left( \widehat{p}_{i}\sum _{j\in N\backslash (S_{i}\cup \left\{ i\right\} )}\gamma _{j}\right) /\left( 1- \widehat{p}_{i}-P_{i}\right)$, the function $V_{i}^{\prime }\left( 1\right) \ge 0$, and for $\gamma _{i}\le \overline{\gamma }_{i}$ the opposite inequality holds true.

The derivative of V at $\alpha =0$ is:

$$\begin{aligned} V_{i}^{\prime }\left( 0\right) =-\widehat{p}_{i}u_{i}^{\prime }\left( \sum _{j\in N\backslash S_{i}}\gamma _{j}\right) \sum _{j\in N\backslash (S_{i}\cup \left\{ i\right\} )} \gamma _{j} + (1-\widehat{p}_{i}-P_{i})u_{i}^{ \prime }(0)\gamma _{i}. \end{aligned}$$

One can check that: $V_{i}^{\prime }\left( 0\right) \le 0$ if and only if $\gamma _{i}\le \underline{\gamma }_{i}$, where $\underline{\gamma }_{i}$ satisfies (14).

Since $u_{i}^{\prime \prime }\le 0$ we can deduce that $u_{i}^{\prime }(0)\ge u_{i}^{\prime }\left( \sum _{j\in N\backslash S^{i}}\gamma _{j}\right)$. Substituting this into (14) we prove that $\underline{\gamma }_{i}\le \overline{\gamma }_{i}$.

Summing up, for $0\le \gamma _{i}\le \underline{\gamma }_{i}$ the function $V_{i}\left( \alpha \right)$ is decreasing on the whole interval $\left[ 0,1 \right]$, for $\gamma _{i}\ge \overline{\gamma }_{i}$ it is increasing on the whole interval, and for $\underline{\gamma }<\gamma _{i}<\overline{ \gamma }_{i}$ it has unique maximum on the interval $\left[ 0,1\right]$.

Appendix 2: Derivation of Optimal $\alpha ^*$

From Proposition 1, the thresholds for group 2 are:

$$\begin{aligned} \overline{\gamma }_{2}= & {} \frac{\widehat{p}_{2}\gamma _{3}}{\widehat{p}_{3}} \text { and} \\ \underline{\gamma }_{2}= & {} \frac{\widehat{p}_{2}\gamma _{3}u_{2}^{\prime }\left( \gamma _{2}+\gamma _{3}\right) }{\widehat{p}_{3}u_{2}^{\prime }\left( 0\right) }. \end{aligned}$$

Therefore, the behavior of player 2 can be described as follows:

• for $\displaystyle {\frac{\gamma _{3}}{\gamma _{2}} < \frac{\widehat{p}_{3}}{ \widehat{p}_{2}}}$, function $V_{2}\left( \alpha \right)$ increases on the whole interval $\left[ 0,1\right]$ and therefore $\alpha _{2}^{*}=1$;

• for $\displaystyle {\frac{\widehat{p}_{3}}{\widehat{p}_{2}} < \frac{\gamma _{3}}{\gamma _{2}}<\frac{\widehat{p}_{3}}{\widehat{p}_{2}}\frac{ u_{2}^{\prime }(0)}{u_{2}^{\prime }\left( \gamma _{2}+\gamma _{3}\right) }}$ , function $V_{2}\left( \alpha \right)$ has an inferior maximum $\alpha _{2}^{*}$ on $\left[ 0,1\right]$ which is defined from the equality $V_{2}^{\prime }\left( \alpha \right) =0$, that is,

  $$\begin{aligned} -\widehat{p}_{2}u_{2}^{\prime }\left( \alpha \gamma _{2}+(1-\alpha ) \left( 1-\gamma _{1}\right) \right) \gamma _{3}+\widehat{p}_{3}u_{2}^{\prime }(\alpha \gamma _{2})\gamma _{2}=0\text {;} \end{aligned}$$

  (25)

• for $\displaystyle {\frac{\gamma _{3}}{\gamma _{2}} > \frac{\widehat{p}_{3}}{ \widehat{p}_{2}}\frac{u_{2}^{\prime }(0)}{u_{2}^{\prime }\left( \gamma _{2}+\gamma _{3}\right) }}$, function $V_{2}\left( \alpha \right)$ is decreasing on the whole interval $\left[ 0,1\right]$ and therefore $\alpha _{2}^{*}=0$.

In a similar way, the thresholds on the tax share for player 3 can be expressed as follows:

$$\begin{aligned} \overline{\gamma }_{3}= & {} \frac{\widehat{p}_{3}\gamma _{2}}{1-\widehat{p}_{3}} \text { and} \\ \underline{\gamma }_{3}= & {} \frac{\widehat{p}_{3}\gamma _{2}u_{3}^{\prime }\left( \gamma _{2}+\gamma _{3}\right) }{\left( 1-\widehat{p}_{3}\right) u_{3}^{\prime }\left( 0\right) }. \end{aligned}$$

The behavior of player 3 can be summarized as:

• for $\displaystyle {\frac{\gamma _{3}}{\gamma _{2}} < \frac{\widehat{p}_{3}}{ 1-\widehat{p}_{3}}\frac{u_{3}^{\prime }\left( \gamma _{2}+\gamma _{3}\right) }{u_{3}^{\prime }(0)}}$ function $V_{3}\left( \alpha \right)$ is decreasing on the whole interval $\left[ 0,1\right]$ and therefore $\alpha _{3}^{*}=0$;

• for $\displaystyle {\frac{\widehat{p}_{3}}{1-\widehat{p}_{3}}\frac{ u_{3}^{\prime }\left( \gamma _{2}+\gamma _{3}\right) }{u_{3}^{\prime }(0)}< \frac{\gamma _{3}}{\gamma _{2}}<\frac{\widehat{p}_{3}}{1-\widehat{p}_{3}}}$, function $V_{3}\left( \alpha \right)$ has an inferior maximum $\alpha _{3}^{*}$ on $\left[ 0,1\right]$ and it is defined from $V_{3}^{\prime }\left( \alpha \right) =0$:

  $$\begin{aligned} -\widehat{p}_{3}u_{3}^{\prime }\left( \alpha \gamma _{3} + (1-\alpha ) \left( 1-\gamma _{1}\right) \right) \gamma _{2}+(1-\widehat{p}_{3})u_{3}^{\prime }(\alpha \gamma _{3})\gamma _{3}=0\text {;} \end{aligned}$$

  (26)

• for $\displaystyle {\frac{\gamma _{3}}{\gamma _{2}} > \frac{\widehat{p}_{3}}{ 1-\widehat{p}_{3}}}$, function $V_{3}\left( \alpha \right)$ is increasing on the whole interval $\left[ 0,1\right]$ and therefore $\alpha _{3}^{*}=1$.

We can identify the median voter: it is either player 2 if $\displaystyle { \frac{\gamma _{3}}{\gamma _{2}} > \frac{\widehat{p}_{3}}{\widehat{p}_{2}}}$ or player 3 if the opposite inequality holds.