Hurwicz’s criterion and the equilibria of duopoly models

Abstract

In this paper we investigate a model of duopolistic competition in an uncertain environment where the attitudes of the firms towards uncertainty are incorporated. In particular, we analyze an extension of a Cournot duopoly in which the firms face a different market demand in each of two scenarios, and make their output decisions before uncertainty is resolved. The way in which firms value the possible outcomes is critical when deciding their strategies. In real-life situations the attitudes that agents exhibit can vary from extreme pessimism to extreme optimism, and it is possible to characterize their behavior according to their degrees of optimism. In this context, we identify the sets of equilibria for the full range of degrees of optimism, and illustrate the results with the analysis of some cases in which the demand functions are linear.

Appendix

Proof Proposition 1

Recall that for each $\gamma ,$$0\le \gamma \le 1$, the Hurwicz function is

$$\begin{aligned} v^i_\gamma (q)=\left\{ \begin{array}{lcl} \gamma u^i_1(q)+(1- \gamma ) u^i_2(q) &{} \text{ if } \, \, q^1+q^2\le \bar{Q}\\ \gamma u^i_2(q)+(1- \gamma ) u^i_1(q)&{} \text{ if } \, \, q^1+q^2 \ge \bar{Q}.\, \\ \end{array} \right. \end{aligned}$$

Equivalently,

$$\begin{aligned} v^i_\gamma (q)=\left\{ \begin{array}{lcl} q^i(\gamma P_1(Q)+(1- \gamma ) P_2(Q)) &{} \text{ if } \, \, q^1+q^2\le \bar{Q}\\ q^i(\gamma P_2(Q)+(1- \gamma ) P_1(Q))&{} \text{ if } \, \, q^1+q^2 \ge \bar{Q}\, \\ \end{array} \right. \end{aligned}$$

a) If $\gamma \ge \frac{1}{2}$, we will prove that when $q^1+q^2 \le \bar{Q}$, then $\gamma u^i_1(q)+(1- \gamma ) u^i_2(q)\ge \gamma u_2^i(q)+(1- {{\gamma }}) u_1^i(q)$, and when $q^1+q^2 \ge \bar{Q}$, $\gamma u^i_1(q)+(1- \gamma ) u^i_2(q)\le \gamma u_2^i(q)+(1- {{\gamma }}) u_1^i(q)$. Thus, $v^i_\gamma (q)=max\{\gamma u^i_1(q)+(1- \gamma ) u^i_2(q), \gamma u_2^i(q)+(1- \gamma ) u_1^i(q)\}$ for any q.

First, note that we can write

$\gamma P_1(Q)+(1- \gamma ) P_2(Q)= (2 \gamma -1) (P_1(Q)-P_2(Q))+\gamma P_2(Q)+(1- \gamma )P_1(Q).$

For $q^1+q^2 \le \bar{Q}$, $P_1(Q)\ge P_2(Q)$. Since $\gamma \ge \frac{1}{2}$, $2\gamma -1\ge 0$, then $(2\gamma -1)(P_1(Q)-P_2(Q)) \ge 0$, and therefore $\gamma P_1(Q)+(1- \gamma ) P_2(Q)\ge \gamma P_2(Q)+(1- \gamma )P_1(Q)$. Thus $\gamma u^i_1(q)+(1- \gamma ) u^i_2(q)\ge \gamma u_2^i(q)+(1- {{\gamma }}) u_1^i(q)$.

On the other hand, for $q^1+q^2 \ge \bar{Q}$, $P_1(Q)\le P_2(Q)$, and since $(2\gamma -1)(P_1(Q)-P_2(Q)) \le 0,$ then $\gamma P_1(Q)+(1- \gamma ) P_2(Q) \le \gamma P_2(Q)+(1- \gamma )P_1(Q).$

Thus, $\gamma u^i_1(q)+(1- \gamma ) u^i_2(q)\le \gamma u_2^i(q)+(1- {{\gamma }}) u_1^i(q)$, and the result follows.

b) Anagolously, if $\gamma \le \frac{1}{2}$, then

$v^i_\gamma (q)=min\{\gamma u^i_1(q)+(1- \gamma ) u^i_2(q), \gamma u_2^i(q)+(1- \gamma ) u_1^i(q)\}$ for any q. $\square$

Proof Theorem 1

It follows from the definition that $q^*$ is an $H_{\gamma }$-equilibrium for the game G if and only if $q^*$ is a Nash equilibrium for the game $\{(A^i, v^i_\gamma )\}_{i =1,2}$. By applying Proposition 1, for $\gamma \ge \frac{1}{2}$, $v^i_\gamma (q)= max\{w^{i\gamma }_1(q),w^{i\gamma }_2(q)\}$, and for $\gamma \le \frac{1}{2}$, $v^i_\gamma (q)= min\{w^{i\gamma }_1(q),w^{i\gamma }_2(q)\}$, where $w^{i\gamma }_1(q)=\gamma u^i_1(q)+(1- \gamma ) u^i_2(q)$, and $w^{i\gamma }_2(q)= \gamma u_2^i(q)+(1- \gamma ) u_1^i(q)$.

Therefore, it follows that for $\gamma \ge \frac{1}{2}$, $q^*$ is an optimistic equilibrium for the game $G^\gamma =\{(A^i, w^{i\gamma }_1,w^{i\gamma }_2)\}_{i =1,2},$ and for $\gamma \le \frac{1}{2}, \, q^*$ is a conservative equilibrium for the game $G^{\gamma }.$$\square$