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 $\frac{1}{3} = \arg\max_q\pi(q,q^*)$.

  • $\begingroup$ So, what is your question? $\endgroup$
    – Herr K.
    Apr 24, 2019 at 16:25
  • $\begingroup$ Is my reasoning sound? $\endgroup$
    – clueless
    Apr 29, 2019 at 9:14
  • $\begingroup$ Since I do not rely on concavity in the sense of $\pi_{qq}(q^*,q^*)<0$. $\endgroup$
    – clueless
    Apr 29, 2019 at 9:31
  • $\begingroup$ 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. $\endgroup$
    – Bayesian
    Apr 30, 2019 at 15:40

1 Answer 1


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 $q_2\geq 1/2$.

When the objective function is strictly concave, the first order conditions are necessary and sufficient, and the maximum is unique.

Not sure if I'm answering your question, but hopefully it helps.

  • $\begingroup$ How do you know that the profit function is concave without invoking second order partial derivatives? $\endgroup$
    – clueless
    Apr 29, 2019 at 9:19
  • 2
    $\begingroup$ 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. $\endgroup$
    – Regio
    Apr 29, 2019 at 17:09
  • $\begingroup$ I'm aware. That's exactly the point. I don't want to invoke second order conditions. The Cournot example is just for illustrational reasons. The bigger questions is: how can we verify that a candidate is a maximum, if we have no information about concavity of the objective function? $\endgroup$
    – clueless
    Apr 30, 2019 at 7:47
  • 1
    $\begingroup$ So your question has nothing to do with your Cournot example? If you have the profit function as you gave it, you know that it is concave and it follows that you found a maximum. Maybe you should use an example in which the curvature of the objective is not clear. $\endgroup$
    – Bayesian
    Apr 30, 2019 at 15:30
  • 1
    $\begingroup$ 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 $\endgroup$
    – Regio
    Apr 30, 2019 at 17:57

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