Utility function and location in the Hotelling game

Abstract

It is demonstrated herein that a slight expansion of the utility function in the duopolistic Hotelling game enables any symmetric location pair with respect to the center to be in equilibrium, which implies that any level of locational differentiation between the minimum Hotelling (Econ. J.39:41–57, 1929) and maximum D’Aspremont (Econom. 47:1145–1150, 1979) is obtained in one model. The location equilibrium is monotone with respect to the introduced parameter (k), while the equilibrium price and profits are not monotone (they are U-shaped). That is, the nearer the two firms are located, the higher their prices are set (with an upper limit) when k is sufficiently large. This counterintuitive phenomenon is interpreted as an example of strategic complementarity that is inherent in the Hotelling games.

Appendices

Appendix

On Proposition 2

We have $\partial \pi _1/\partial x_1=\,g_0\cdot g_1$ and $\partial \pi _2/\partial x_2=\,g_0\cdot g_2$, where

$$\begin{aligned} g_0=\, & {} \left\{ 2k(k+2)(x_2-x_1)\right\} ^{-1}\{t(x_2-x_1)((2-x_1-x_2)+2)\}^{1/k}k^{-1/k}(k+2)^{-1/k}\ne 0\\ g_1=\, & {} k^2(x_2-x_1)+2k x_1+2\\ g_2=\, & {} k^2(x_2-x_1)+2k (1-x_2)+2. \end{aligned}$$

Therefore, the first-order conditions $\partial \pi _i/\partial x_i=0$ degenerate into $g_1=0$ and $g_2=0$. Solving the two equations yield (13) in the proposition. Evaluating the second derivative $\partial ^2 \pi _i/\partial x_i^2$ at Eq. (13), we have:

$$\begin{aligned} \frac{\partial ^2 \pi _i}{\partial x_i^2}_{(x_1,x_2)=(x_1^*,x_2^*)}=-\frac{t}{2k}\left( \frac{t(k+2)}{k^2(k+1)}\right) ^{-1+1/k}<0, \end{aligned}$$

which implies that the second-order conditions are satisfied.

On ${\tilde{k}}$

Let f(k) be the right-hand side of Eq. (16). We can easily have $df(k)/dk>0$ with $\lim _{k\rightarrow 0} f(k)=-\infty$ and $\lim _{k\rightarrow \infty } f(k)=\infty$. This implies that there exists a unique solution for Eq. (16) for any $t\in (0,\infty )$, and we also find $d{\tilde{k}}/dt=(2k^4+14k^3+29k^2+24k+8)/k(k^2+3k+2)^2>0$.