<p>How can it be a minimum or saddle since for every <span class="math-container">$q_2$</span> the profit function is strictly concave? If anything, the implicit non-negativity constraints might be binding, for example, if <span class="math-container">$q_2\geq 1/2$</span>. </p>

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

<p>Not sure if I'm answering your question, but hopefully it helps.</p>

CC BY-SA 4.0
2019-04-29T09:19:49.970Z
How do you know that the profit function is concave without invoking second order partial derivatives?
clueless
CC BY-SA 4.0
2019-04-29T17:09:40.087Z
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
CC BY-SA 4.0
2019-04-30T07:47:08.940Z
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?
clueless
CC BY-SA 4.0
2019-04-30T15:30:57.747Z
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.
Bayesian
CC BY-SA 4.0
2019-04-30T17:57:38.117Z
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
CC BY-SA 4.0
2019-04-25T17:02:14.563Z
2019-04-25T17:02:14.563Z
Regio
<p>Let <span class="math-container">$q_i \in Q = \mathbb R_+$</span> denote the quantity produced by firm <span class="math-container">$i \in \{1,2\}$</span>. Further let <span class="math-container">$\pi_i(q_1,q_2)  = (1-q_1-q_2)q_i$</span> denote the profits of <span class="math-container">$i$</span>. A Nash equilibrium <span class="math-container">$(q_1^*,q_2^*) \in Q^2$</span> satisfies
<span class="math-container">\begin{align}
&amp;\pi_1(q_1^*,q_2^*) \geq \pi_1(q_1,q_2^*) \quad \forall q_1 \in Q\\
&amp;\pi_2(q_1^*,q_2^*) \geq \pi_2(q_1^*,q_2) \quad \forall q_2 \in Q. 
\end{align}</span>
We are considering symmetric equilibria of the form <span class="math-container">$q^* = q_1^* = q_2^*$</span> and therefore apply the symmetric opponents form approach.
Define <span class="math-container">$\pi(q,q^*) = \pi_1(q,q^*)$</span>.
There exists a unique symmetric root to the first order condition <span class="math-container">$\pi_q(q^*,q^*) = 0$</span> given by <span class="math-container">$q^* = \frac{1}{3}$</span>.</p>

<p><strong>Claim</strong> The candidate <span class="math-container">$q = \frac{1}{3}$</span> is the unique symmetric maximizer of <span class="math-container">$\pi(q,q^*)$</span>.</p>

<p>Problem: The candidate might be a minimum or saddle.</p>

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

<p><strong>Edit</strong> I edit the question to further clarify the issue. <em>Suppose</em> I don't have any information about concavity of <span class="math-container">$\pi(q,q^*)$</span> w.r.t. <span class="math-container">$q$</span>. </p>

<p>A general argument:
We need to distinguish 4 cases. </p>

<ol>
<li><span class="math-container">$q^*$</span> is a saddle and <span class="math-container">$\pi(q^*,q^*) &gt; 0$</span> and <span class="math-container">$\pi(\infty,q^*) = \infty$</span>.</li>
<li><span class="math-container">$q^*$</span> is a saddle and <span class="math-container">$\pi(q^*,q^*) &lt; 0$</span> and <span class="math-container">$\pi(\infty,q^*) = -\infty$</span>. </li>
<li><span class="math-container">$q^*$</span> is a minimum and <span class="math-container">$\pi(q^*,q^*) &lt; 0$</span> and <span class="math-container">$\pi(\infty,q^*) = \infty$</span>.</li>
<li><span class="math-container">$q^*$</span> is a maximum and <span class="math-container">$\pi(q^*,q^*) &gt; 0$</span> and <span class="math-container">$\pi(\infty,q^*) = -\infty$</span>.</li>
</ol>

<p>Since case 4 is considered here <span class="math-container">$\frac{1}{3} = \arg\max_q\pi(q,q^*)$</span>. </p>

CC BY-SA 4.0
2019-04-24T16:25:52.200Z
So, what is your question?
Herr K.
CC BY-SA 4.0
2019-04-29T09:14:19.697Z
Is my reasoning sound?
clueless
CC BY-SA 4.0
2019-04-29T09:31:18.477Z
Since I do not rely on concavity in the sense of $\pi_{qq}(q^*,q^*)<0$.
clueless
CC BY-SA 4.0
2019-04-30T15:40:57.980Z
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
<mathematical-economics><game-theory><optimization>
Symmetric Cournot equilibrium: suffciency without second order conditon
clueless
# Symmetric Cournot equilibrium: suffciency without second order conditon

(Asked by: username_0 on 2019-04-24)

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

**Tags:** mathematical-economics, game-theory, optimization

> **Comments:**
> **username_1**: So, what is your question?  
> **username_0**: Is my reasoning sound?  
> **username_0**: Since I do not rely on concavity in the sense of $\pi_{qq}(q^*,q^*)<0$.  
> **username_2**: 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.  

---

## 1 answers

### Answer by username_3 on 2019-04-25. Score: 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.

> **Comments:**
> **username_0**: How do you know that the profit function is concave without invoking second order partial derivatives?  
> **username_3**: 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.  
> **username_0**: 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?  
> **username_2**: 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.  
> **username_3**: 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  

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