# Symmetric Cournot equilibrium: suffciency without second order conditon

Let $$q_i \in Q = \mathbb R_+$$ denote the quantity produced by firm $$i \in \{1,2\}$$. Further let $$\pi_i(q_1,q_2) = (1-q_1-q_2)q_i$$ denote the profits of $$i$$. A Nash equilibrium $$(q_1^*,q_2^*) \in Q^2$$ satisfies \begin{align} &\pi_1(q_1^*,q_2^*) \geq \pi_1(q_1,q_2^*) \quad \forall q_1 \in Q\\ &\pi_2(q_1^*,q_2^*) \geq \pi_2(q_1^*,q_2) \quad \forall q_2 \in Q. \end{align} We are considering symmetric equilibria of the form $$q^* = q_1^* = q_2^*$$ and therefore apply the symmetric opponents form approach. Define $$\pi(q,q^*) = \pi_1(q,q^*)$$. There exists a unique symmetric root to the first order condition $$\pi_q(q^*,q^*) = 0$$ given by $$q^* = \frac{1}{3}$$.

Claim The candidate $$q = \frac{1}{3}$$ is the unique symmetric maximizer of $$\pi(q,q^*)$$.

Problem: The candidate might be a minimum or saddle.

The idea: In economic settings equilibrium quantities are basically restricted by individual rationality, i.e. $$\pi(q^*,q^*) = (1-2q^*)q^*$$ implies $$q^* \in [0,\frac{1}{2}]$$. Since $$\pi(\frac{1}{3},\frac{1}{3}) = \frac{1}{9} > 0$$, the claim follows.

Edit I edit the question to further clarify the issue. Suppose I don't have any information about concavity of $$\pi(q,q^*)$$ w.r.t. $$q$$.

A general argument: We need to distinguish 4 cases.

1. $$q^*$$ is a saddle and $$\pi(q^*,q^*) > 0$$ and $$\pi(\infty,q^*) = \infty$$.
2. $$q^*$$ is a saddle and $$\pi(q^*,q^*) < 0$$ and $$\pi(\infty,q^*) = -\infty$$.
3. $$q^*$$ is a minimum and $$\pi(q^*,q^*) < 0$$ and $$\pi(\infty,q^*) = \infty$$.
4. $$q^*$$ is a maximum and $$\pi(q^*,q^*) > 0$$ and $$\pi(\infty,q^*) = -\infty$$.

Since case 4 is considered here $$\frac{1}{3} = \arg\max_q\pi(q,q^*)$$.

• So, what is your question? – Herr K. Apr 24 '19 at 16:25
• Is my reasoning sound? – clueless Apr 29 '19 at 9:14
• Since I do not rely on concavity in the sense of $\pi_{qq}(q^*,q^*)<0$. – clueless Apr 29 '19 at 9:31
• Since it is not this particular Cournot example you are after, why don't you give the problem in which you cannot determine the second-order condition? In your example, you have a function that is clearly concave. – Bayesian Apr 30 '19 at 15:40

How can it be a minimum or saddle since for every $$q_2$$ the profit function is strictly concave? If anything, the implicit non-negativity constraints might be binding, for example, if $$q_2\geq 1/2$$.
• You do use the second order derivative, but this is a classic problem so many people already know that $\frac{\partial^2\pi_i(q_1,q_2)}{\partial q_i^2}=-2$ so the function is every-where concave, for all $q_{-i}$, and because there is no marginal cost, the first order condition will give the unique maximum. Your argument about individual rationality doesn't really make sense to me about how that deals with the concern of having a local minimum or saddle, instead of a maximum. – Regio Apr 29 '19 at 17:09
• I don't really think there is a way around arguing at least local concavity in order to justify that a critical point is a local maximum. However, concavity arguments not necessarily need to use derivatives. Concavity is defined by inequalities: $F$ is concave if for any two points $x,y$ and $\lambda\in[0,1]$, $\lambda F(x)+(1-\lambda)F(y)\leq F(\lambda x+(1-\lambda) y)$, and there are many operations that preserve concavity, but limit arguments, like the ones you are using in your four cases, not necessarily suffice in general. It would be useful to see the particular case you are considering – Regio Apr 30 '19 at 17:57