General distribution of consumers in pure Hotelling games

Abstract

A pure Hotelling game is a spatial competition between a finite number of players who simultaneously select a location in order to attract as many consumers as possible. In this paper, we study the case of a general distribution of consumers on a network generated by a metric graph. Because players do not compete on price, the continuum of consumers shop at the closest player’s location. If the number of sellers is large enough, we prove the existence of an approximate equilibrium in pure strategies, and we construct it.

Appendix

Lemma 19

Let $\epsilon _1 \in (0,\frac{\min \lambda _e K}{2})$, $\hat{e} \in E$ and $\hat{i} \in \{0,\ldots ,I_e-1\}$, then $\Omega (\epsilon _1) \ge h(\frac{\ell _{\hat{e}}g_{\hat{e}}^{\hat{i}}}{10},\epsilon _1)$ .

Proof

It follows from the definition of $\ell _e$ and the inequality $m-\epsilon _1 \le g_e \le M + \epsilon _1$ that:

$$\begin{aligned} h\left( \frac{\ell _{\hat{e}}g_{\hat{e}}^{\hat{i}}}{8},\epsilon _1\right)&= 2 {{\mathrm{Card}}}(E_L) + \sum _e \sum _i \left\lceil \frac{10}{2} \frac{\ell _e(\epsilon _1) g_e^i}{\ell _{\hat{e}(\epsilon _1)}g_{\hat{e}}^{\hat{i}}} \right\rceil + 2 \sum _e \left\lceil \frac{\lambda _e K}{2 \epsilon _1} \right\rceil \\&\le 2 {{\mathrm{Card}}}(E_L) + \sum _e \left\lceil \frac{\lambda _e K}{2 \epsilon _1} \right\rceil \left\lceil 5 \frac{\lambda _e \left\lceil \frac{\lambda _{\hat{e}}K}{2 \epsilon _1}\right\rceil (M+\epsilon _1)}{\lambda _{\hat{e}} \left\lceil \frac{\lambda _{e}K}{2 \epsilon _1}\right\rceil (m-\epsilon _1)} \right\rceil + 2 \sum _e \left\lceil \frac{\lambda _e K}{2 \epsilon _1} \right\rceil . \end{aligned}$$

Moreover, because $x \le \left\lceil x \right\rceil \le x+1$ we also have:

$$\begin{aligned} h\left( \frac{\ell _{\hat{e}}g_{\hat{e}}^{\hat{i}}}{8},\epsilon _1\right)&\le 2 {{\mathrm{Card}}}(E_L) + \sum _e \left( 5 \frac{\lambda _e (\frac{\lambda _{\hat{e}}K}{2 \epsilon _1}+1) (M+\epsilon _1)}{\lambda _{\hat{e}} (m-\epsilon _1)} \right) +3 \sum _e \left\lceil \frac{\lambda _e K}{2 \epsilon _1} \right\rceil \\&\le 2 {{\mathrm{Card}}}(E_L) + \frac{5}{2} \sum _e \frac{\lambda _e K (M+\epsilon _1)}{\epsilon _1 (m-\epsilon _1)} +\sum _e 5 \frac{\lambda _e}{\lambda _{\hat{e}}}\frac{M+\epsilon _1}{m-\epsilon _1} \\&\quad +3 \sum _e \left\lceil \frac{\lambda _e K}{2 \epsilon _1} \right\rceil . \\ \end{aligned}$$

Finally, we use the notation $L:= \sum _e \lambda _e$ and the fact that $\sum _e \left\lceil \frac{\lambda _e K}{2 \epsilon _1} \right\rceil \le {{\mathrm{Card}}}(E)+\frac{L K}{2 \epsilon _1}$.

$$\begin{aligned} h\left( \frac{\ell _{\hat{e}}g_{\hat{e}}^{\hat{i}}}{8},\epsilon _1\right)&\le 2 {{\mathrm{Card}}}(E_L) + \frac{ 5 L K (M+\epsilon _1)}{2 \epsilon _1 (m-\epsilon _1)} + \frac{5 L}{\min \lambda _e}\frac{M+\epsilon _1}{m-\epsilon _1} +3 {{\mathrm{Card}}}(E) \\&\quad + \frac{3 L K}{2 \epsilon _1} \\&\le 5 {{\mathrm{Card}}}(E) + \frac{5 L (M+\epsilon _1)}{(m-\epsilon _1)}\left( \frac{K}{2 \epsilon _1} + \frac{1}{\min \lambda _e}\right) + \frac{3 L K}{2 \epsilon _1} \le \Omega (\epsilon _1). \\ \end{aligned}$$

$\square$

Lemma 20

If $n \ge \Omega (\epsilon _1)$ then $\overline{\theta } \le \frac{\epsilon _1 (M+ \epsilon _1)}{4K}$

Proof

Remember that $\overline{\theta }$ was defined in Eq. (8) as a real number such that $n \le h(\overline{\theta }) \le n+ \sum _{e \in E} I_e$. We suppose here that $n \ge \Omega (\epsilon _1)$ and we have, thanks to Lemma 19, $\Omega (\epsilon _1) \ge h(\frac{\ell _e g_e^i}{10},\epsilon _1)$. Putting all these inequalities together gives:

$$\begin{aligned} h(\overline{\theta },\epsilon _1) \ge h\left( \frac{\ell _e g_e^i}{10},\epsilon _1\right) \end{aligned}$$

Because h is a decreasing function, and replacing $\ell$ by its definition in Definition 10, we get:

$$\begin{aligned} \overline{\theta } \le \frac{\lambda _e g_e^i}{10 \left\lceil \frac{\lambda _e K}{2 \epsilon _1}\right\rceil } \le \frac{\epsilon _1 g_e^i}{5K} \end{aligned}$$

Using the inequality $g_e^i \le M+\epsilon$ we obtain:

$$\begin{aligned} \overline{\theta } \le \frac{\epsilon _1 (M+ \epsilon _1)}{5K}. \end{aligned}$$

$\square$