On the Geometric Structure of the Cournot Equilibrium Set: The Case of Concave Industry Revenue and Convex Costs

Abstract

The recent results in von Mouche and Quartieri (Econ Bull 35(2):1299–1305, 2015) on equilibrium (semi-)uniqueness for homogeneous Cournot oligopolies with concave industry revenue and convex costs are refined and conceptualised. For this class of oligopolies also new results concerning the geometric structure of the equilibrium set E are provided. In particular, a subclass is identified for which E is a non-empty polytope on which the aggregator is constant and a subclass for which E is a 1-dimensional polytope on which the aggregator is injective.

Appendix

This appendix contains some results, which are not claimed to be new, but for which it is difficult to give a proper reference.

1.1 Effective Capacity Constraints

For equilibrium existence compact production level sets are very welcome. However, effective capacity constraints (see Sect. 2) also will do the job as Theorem 4 below shows.

Proposition 7

Sufficient for firm i to have an effective capacity constraint $\overline{x}_{i}$ is that $(r_{p} - c_{i})(x_{i}) <-c_{i}(0)$ for every x _i ∈ X _i with $x_{i}> \overline{x}_{i}$ and that p is decreasing on $\{y \in Y \;\vert \;y \geq \overline{x}_{i}\}$ . $\diamond$

Proof

Suppose x _i  ∈ X _i with $x_{i}> \overline{x}_{i}$. I show that x _i is strongly dominated by 0 ($\in [0,\overline{x}_{i}]$). So fix $\mathbf{z} \in \mathbf{X}_{\hat{\imath }}$. Now, as desired, $u_{i}^{(\mathbf{z})}(0) = -c_{i}(0)> (r_{p}-c_{i})(x_{i}) = p(x_{i})x_{i}-c_{i}(x_{i}) \geq p(x_{i}+z_{\hat{i}})x_{i}-c_{i}(x_{i}) = u_{i}^{(\mathbf{z})}(x_{i})$. □ 

The condition in Proposition 7 has the following economic interpretation: the variable monopoly profit becomes negative.

Theorem 4

Consider a Cournot oligopoly Γ where each firm i has an effective capacity constraint $\overline{x}_{i}$.

• 1. Let $K_{i}\;:=\; [0,\overline{x}_{i}]\;(i \in N)$ and let Γ′ be the game in strategic form with for firm i as production level set K _i and as payoff function $u_{i}\upharpoonright \mbox{ $K_{i}$}$ . Then the equilibrium sets of Γ and Γ′ are the same.

• 2. If, in Γ, each profit function is continuous and each conditional profit function is quasi-concave, then the equilibrium set E of Γ is a non-empty compact subset of $\mathbb{R}^{n}$ with $E \subseteq \mathsf{X}_{i=1}^{n}[0,\overline{x}_{i}]$ . $\diamond$

Proof

• 1. It is straightforward to verify that $\mathrm{argmax}\,u_{i}^{(\mathbf{z})}\upharpoonright \mbox{ $K_{i}$} =\mathrm{ argmax}\,u_{i}^{(\mathbf{z})}$ for every i ∈ N and $\mathbf{z} \in \mathbf{X}_{\hat{\imath }}$. This implies that Γ and Γ′ have the same equilibrium set.

• 2. The Nikaido-Isoda theorem guarantees that the equilibrium set of Γ′ is non-empty. So by part 1, Γ has an equilibrium and $E \subseteq \mathsf{X}_{i=1}^{n}[0,\overline{x}_{i}]$. As the profit functions of Γ′ are continuous, the equilibrium set of Γ′ is is closed in $\mathsf{X}_{i=1}^{n}[0,\overline{x}_{i}]$ and therefore also in $\mathbb{R}^{n}$. So E is a closed bounded subset of $\mathbb{R}^{n}$ and therefore compact. □ 

1.2 Some Convex Analysis Related to Price Functions

First, I introduce some useful notations. For a proper real interval with $J \subseteq \mathbb{R}_{+}$ and $k \in \mathbb{R}_{+}$, let

$$\displaystyle{J_{k}\;:=\; (J -\{ k\}) \cap \mathbb{R}_{+}.}$$

So J ₀ = J and also J _k is an interval; for k ∈ J ^⊖, the interval J _k is proper. And for a function $g: J \rightarrow \mathbb{R}$ with J and k as above, the function $g^{(k)}: J_{k} \rightarrow \mathbb{R}$ is defined by

$$\displaystyle{ g^{(k)}(x)\;:=\; g(x + k) }$$

(8)

and the function $r_{g}: J \rightarrow \mathbb{R}$ is defined by

$$\displaystyle{r_{g}\;:=\; g \cdot \mathrm{ Id}.}$$

In the rest of this appendix, K is a proper real interval with $0 \in K \subseteq \mathbb{R}_{+}$, $f: K \rightarrow \mathbb{R}$ is a function and

$$\displaystyle{\tilde{f}\;:=\; f\upharpoonright \mbox{ $K^{\oplus }$}.}$$

Lemma 5

Suppose r _f is (strictly) concave.

• 1. f is semi-differentiable at each interior point of its domain and in the case K = [0,m], the (left) derivative D ⁻ f(m) of f at m exists as element of $\mathbb{R} \cup \{-\infty \}$ ;

• 2. $D^{+}f(\,y) \leq D^{-}f(\,y)\;(\,y \in \mathrm{ Int}(K))$ ;

• 3. D ⁻ f (<) ≤  0;

• 4. $\tilde{f}$ is (strictly) decreasing. $\diamond$

Proof

1–3. As r _f is concave, r _f , and therefore f too, is semi-differentiable at each interior point of its domain. As r _f is (strictly) concave, it holds for all y ∈ Int(K) that $D^{+}r_{f}(\,y) \leq D^{-}r_{f}(\,y)\,(<) \leq \frac{r_{f}(\,y)-r_{f}(0)} {y-0}$. So $D^{+}f(\,y)y + f(\,y) \leq D^{-}f(\,y)y + f(\,y)\,(<) \leq f(\,y)$ and thus $D^{+}f(\,y) \leq D^{-}f(\,y)\,(<) \leq 0$. Thus the proof is complete if K ≠ [0, m].

Now consider the case where K = [0, m] and $D^{-}f(m)\neq -\infty$. So f is differentiable at m and therefore $Dr_{f}(m)\,(<) \leq \frac{r_{f}(m)-r_{f}(0)} {m-0}$ and Df(m) ( < ) ≤ 0 follows.

4. By parts 1 and 3. □ 

Lemma 6

If r _f is concave and on Int (K) decreasing, then V  := { y ∈ K ^⊕  | f( y) > 0} is a real interval with $]0,y] \subseteq V \;(\,y \in V )$ , D ⁻ f < 0 on V and f is strictly decreasing on V. $\diamond$

Proof

By Lemma 5(4), $\tilde{f}$ is decreasing. This implies that $V =\{ y \in K^{\oplus }\;\vert \;\tilde{f}(\,y)> 0\} =\{ y \in K^{\oplus }\;\vert \;f(\,y)> 0\}$ is a real interval with $]0,y] \subseteq V \;(\,y \in V )$. As $r_{\tilde{f}}$ is decreasing on Int(K) and $r_{\tilde{f}}$ is concave, even $r_{\tilde{f}}$ is decreasing. So, for y ∈ V, $0 \geq D^{-}r_{\tilde{f}}(\,y) = D^{-}\tilde{f}(\,y)y +\tilde{ f}(\,y)$. As $\tilde{f}> 0$ on V, it follows that $D^{-}f(\,y) = D^{-}\tilde{f}(\,y) <0\;(\,y \in V )$. As f is continuous on Int(K), it follows that f is strictly decreasing on V. □ 

A weaker variant of the next result can be found in Murphy et al. (1982); in particular I note that in this result there is no explicit assumption about the monotonicity properties of f.

Proposition 8

• 1. If r _f is (strictly) concave, then for every k ∈ K ^⊖ the function $r_{f^{(k)}}$ is (strictly) concave.

• 2. If r _f is concave and $\tilde{f}$ is strictly decreasing, then for every k ∈ Int (K) the function $r_{f^{(k)}}$ is strictly concave. $\diamond$

Proof

• 1. This is evident for k = 0. Now suppose k ∈ Int(K). Suppose r _f is strictly concave; the proof for the case of concave r _f is analogous. I give the proof by contradiction. So suppose $r_{f^{(k)}}$ is not strictly concave. This implies that there exist x, y ∈ K _k with x < y and t ∈ ]0, 1 [ such that for $z = tx + (1 - t)y$ the inequality $r_{f^{(k)}}(z) \leq tr_{f^{(k)}}(x) + (1 - t)r_{f^{(k)}}(\,y)$ holds. By Lemma 5(4), the function f is strictly decreasing on K ^⊕. So f ^(k) also is strictly decreasing.

  As z + k ≥ 0, the inequality

  $$\displaystyle{r_{f^{(k)}}(z)(z + k) \leq (tr_{f^{(k)}}(x) + (1 - t)r_{f^{(k)}}(\,y))(z + k)}$$

  follows. As $x + k,y + k \in K$, also $z + k = t(x + k) + (1 - t)(\,y + k) \in K$. As r _f is strictly concave and z > 0, I also have the strict inequality

  $$\displaystyle{r_{f}(z + k)z> (tr_{f}(x + k) + (1 - t)r_{f}(\,y + k))z.}$$

  These inequalities together with $r_{f}(z + k)z = r_{f^{(k)}}(z)(z + k)$ imply

  $$\displaystyle{(tr_{f^{(k)}}(x) + (1 - t)r_{f^{(k)}}(\,y))(z + k)> (tr_{f}(x + k) + (1 - t)r_{f}(\,y + k))z.}$$

  Using the definition of r _f and $r_{f^{(k)}}$ this inequality becomes

  $$\displaystyle{tx(z+k)f(x+k)+(1-t)y(z+k)f(\,y+k)> t(x+k)zf(x+k)+(1-t)(\,y+k)zf(\,y+k).}$$

  This implies, by using $y - z = t(\,y - x)$ and $z - x = (1 - t)(\,y - x)$, that $kt(1 - t)f(\,y + k)(\,y - x)> kt(1 - t)f(x + k)(\,y - x)$ and, as k > 0,

  $$\displaystyle{f(\,y + k)> f(x + k).}$$

  But this is a contradiction with the decreasingness of f ^(k).

• 2. This is proved by making obvious changes in the proof of part 1. □