# Stackelberg Social Equilibrium in Water Markets

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## Abstract

**:**

## 1. Introduction

## 2. Duopoly Water Markets

## 3. Motivating Example

## 4. Stackelberg Social Equilibrium

**Theorem**

**1.**

**Proof.**

#### The Stackelberg Social Equilibrium of Example 4.3 of Ansink and Houba (2012)

- The profit-maximizing quantity for downstream coincides with the upper bound on water extraction, i.e., ${R}_{D}^{*}\left({y}_{U}\right)={e}_{D}+{e}_{U}-{y}_{U}$. Then, upstream’s profit function ${\pi}_{U}\left({y}_{U},{R}_{D}^{*}\left({y}_{U}\right)\right)=2\left(1-{e}_{D}-{e}_{U}\right){y}_{U}$ is linear and increasing, and full extraction by upstream is profit maximizing.
- The profit-maximizing quantity for downstream lies below the upper bound on water extraction, i.e., ${R}_{D}^{*}\left({y}_{U}\right)={\textstyle \frac{1}{2}}\left(1-{y}_{U}\right)<{e}_{D}+{e}_{U}-{y}_{U}$. Then, upstream’s profit function is ${\pi}_{U}\left({y}_{U},{R}_{D}^{*}\left({y}_{U}\right)\right)=\left(1-{y}_{U}\right){y}_{U}$. Then, either $\frac{1}{2}$ extraction by upstream is profit maximizing whenever feasible, i.e., ${e}_{U}\ge \frac{1}{2}$, or full extraction is profit maximizing if constrained, i.e., ${e}_{U}\le \frac{1}{2}$.
- Downstream will adopt the best response below the upper bound for low levels of extraction by upstream, while downstream chooses this upper bound for high levels of extraction by upstream. In this case, upstream’s profit-maximizing extraction is a mix of the first two cases, either full extraction or $\frac{1}{2}$ extraction.

## 5. Conclusions

## Author Contributions

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

## Appendix A. Derivation of the SSE of Example 4.3 of Ansink and Houba (2012)

- $min\left\{2{e}_{D}+2{e}_{U}-1,{e}_{U}\right\}={e}_{U}$ ⟺ $2{e}_{D}+{e}_{U}\ge 1$ ⟺ ${R}_{D}^{*}\left({y}_{U}\right)={\textstyle \frac{1}{2}}\left(1-{y}_{U}\right)$.
- $max\left\{0,2{e}_{D}+2{e}_{U}-1\right\}=0$ ⟺ ${e}_{D}+{e}_{U}\le \frac{1}{2}$ ⟺ ${R}_{D}^{*}\left({y}_{U}\right)={e}_{D}+{e}_{U}-{y}_{U}$.
- Otherwise, we have to deal with two subintervals of $[0,{e}_{U}]$.

- ${e}_{D}+{e}_{U}\le \frac{1}{2}:$
- In this case, upstream solves ${max}_{{y}_{U}\in \left[0,{e}_{U}\right]}2\left(1-{e}_{D}-{e}_{U}\right){y}_{U}$. The linear function increases because of water scarcity, i.e., ${e}_{D}+{e}_{U}<1$. Hence, we obtain boundary solution ${y}_{U}={e}_{U}$ and upstream’s profit $2\left(1-{e}_{D}-{e}_{U}\right){e}_{U}>0$. Furthermore, ${R}_{D}\left({e}_{U}\right)={e}_{D}$ and $2\left(1-{e}_{D}-{e}_{U}\right){e}_{D}\ge 0$ is downstream’s profit.
- $2{e}_{D}+{e}_{U}\ge 1:$
- Here, upstream solves ${max}_{{y}_{U}\in \left[0,{e}_{U}\right]}\left(1-{y}_{U}\right){y}_{U}$. The maximum is given by ${y}_{U}=min\left\{{e}_{U},\frac{1}{2}\right\}$, which implies two subcases. If ${e}_{U}\le \frac{1}{2}$, then ${y}_{U}={e}_{U}$ yields upstream’s profit $\left(1-{e}_{U}\right){e}_{U}$. Furthermore, ${R}_{D}\left({e}_{U}\right)=\frac{1}{2}\left(1-{e}_{U}\right)$, and $\left(1-{e}_{U}\right){e}_{D}\ge 0$ is downstream’s profit. Similarly, if ${e}_{U}\ge \frac{1}{2}$ then ${y}_{U}=\frac{1}{2}$, upstream’s profit is $\frac{1}{4}$, ${R}_{D}\left(\frac{1}{2}\right)=\frac{1}{4}$, and downstream’s profit is $\frac{1}{8}$.
- ${e}_{D}+{e}_{U}>\frac{1}{2}:$
- and $2{e}_{D}+{e}_{U}<1:$ Then, $2{e}_{D}+2{e}_{U}-1\in \left(0,{e}_{U}\right)$, and it partitions the interval $\left[0,{e}_{U}\right]$ into two subintervals. Both the linear and quadratic optimization under the maximum of upstream’s optimization problem should be considered. One difference with the previous cases is that ${max}_{{y}_{U}\in \left[0,2{e}_{D}+2{e}_{U}-1\right]}$ $\left(1-{y}_{U}\right){y}_{U}$ has ${y}_{U}=min\left\{2{e}_{D}+2{e}_{U}-1,\frac{1}{2}\right\}$ as its maximum. Note that $2{e}_{D}+2{e}_{U}-1<\frac{1}{2}$ ⟺ ${e}_{D}+{e}_{U}<\frac{3}{4}$. After modifying previous arguments, we obtain: If ${e}_{D}+{e}_{U}<\frac{3}{4}$, then ${y}_{U}=2{e}_{D}+2{e}_{U}-1\in \left(0,{e}_{U}\right)$ generates profit $2\left(1-{e}_{D}-{e}_{U}\right)\left(2{e}_{D}+2{e}_{U}-1\right)>0$ for upstream and ${y}_{D}={R}_{D}\left(2{e}_{D}+2{e}_{U}-1\right)=1-{e}_{D}-{e}_{U}>0$ (because of water scarcity) with positive profit $2{\left(1-{e}_{D}-{e}_{U}\right)}^{2}$ for downstream. If ${e}_{D}+{e}_{U}\ge \frac{3}{4}$, then previous results imply ${y}_{U}=\frac{1}{2}$ generates profit $\frac{1}{4}$ for upstream and ${y}_{D}={R}_{D}\left(\frac{1}{2}\right)=\frac{1}{4}$ generates profit $\frac{1}{8}$ for downstream. Finally, the substitution of previous results yields the following upstream’s maximal profit$$\left\{\begin{array}{cc}max\left\{2\left(1-{e}_{D}-{e}_{U}\right)\left(2{e}_{D}+2{e}_{U}-1\right),\phantom{\rule{4pt}{0ex}}2\left(1-{e}_{D}-{e}_{U}\right){e}_{U}\right\},\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{e}_{D}+{e}_{U}<\frac{3}{4},\hfill \\ max\left\{\frac{1}{4},\phantom{\rule{4pt}{0ex}}2\left(1-{e}_{D}-{e}_{U}\right){e}_{U}\right\},\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{e}_{D}+{e}_{U}\ge \frac{3}{4}.\hfill \end{array}\right.$$Because $2{e}_{D}+{e}_{U}<1$ in this third main case and $1-{e}_{D}-{e}_{U}{e}_{U}\le \left(1-{e}_{U}\right){e}_{U}\le \frac{1}{4}$, we obtain the maximal profit$$\left\{\begin{array}{cc}2\left(1-{e}_{D}-{e}_{U}\right){e}_{U},\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{e}_{D}+{e}_{U}<\frac{3}{4},\hfill \\ \frac{1}{4},\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{e}_{D}+{e}_{U}\ge \frac{3}{4},\hfill \end{array}\right.$$

## Notes

1 | |

2 | For background and further details of social equilibrium, see, e.g., Dasgupta and Maskin (2015) [7]. |

3 | There is an interesting analogy with the classic Stackelberg model. In that model, the leader setting the Cournot equilibrium quantity and the follower’s (constant) strategy to produce the Cournot equilibrium quantity irrespective of the leader’s quantity forms a Nash equilibrium. Obviously, this Nash equilibrium lacks backward induction and involves non-credible threats made by the follower. And it is the reason that the economic literature analyzes Stackelberg equilibrium instead. |

4 | The below-represented equations correspond to the equations (A)–(D) in [1]. |

5 | This reference also discusses in detail how these conditions restrict both benefit and cost functions. |

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**MDPI and ACS Style**

Houba, H.; Tomori, F.
Stackelberg Social Equilibrium in Water Markets. *Games* **2023**, *14*, 54.
https://doi.org/10.3390/g14040054

**AMA Style**

Houba H, Tomori F.
Stackelberg Social Equilibrium in Water Markets. *Games*. 2023; 14(4):54.
https://doi.org/10.3390/g14040054

**Chicago/Turabian Style**

Houba, Harold, and Françeska Tomori.
2023. "Stackelberg Social Equilibrium in Water Markets" *Games* 14, no. 4: 54.
https://doi.org/10.3390/g14040054