<p>Your FOC is wrong.</p>

<p>So $$\max_{q_i} \quad q_i(a − \sum_{j=1}^nq_j) - q_i^2$$</p>

<p>$$\implies \max_{q_i} \quad aq_i − q_i\sum_{j=1}^nq_j - q_i^2$$</p>

<p>The real FOC is</p>

<p>$$a - \sum_{j\neq i}q_j - 2q_i - 2q_i = 0.$$</p>

CC BY-SA 3.0
2016-11-17T09:57:58.170Z
This solution method would be correct if the OP had calculated the correct FOC to begin with.
Ubiquitous
CC BY-SA 3.0
2016-11-17T15:03:07.660Z
Oh dammit you're right.
Kitsune Cavalry
CC BY-SA 3.0
2016-11-17T15:09:06.560Z
It should be fixed.
Kitsune Cavalry
CC BY-SA 3.0
2016-11-17T17:16:18.727Z
@KitsuneCavalry I think you might want to try again :)
Giskard
CC BY-SA 3.0
2016-11-17T17:26:46.210Z
@KitsuneCavalry Or maybe not. After all, it is a crossposted off-topic question with no visible effort...
Giskard
CC BY-SA 3.0
2016-11-17T20:52:52.927Z
@denesp I'm traaaaaash *headdesk* (If this is wrong again please just delete this answer out of my misery ayy lmao)
Kitsune Cavalry
CC BY-SA 3.0
2016-11-17T22:14:07.510Z
Soooo torn between editing and voting to delete.
Giskard
CC BY-SA 3.0
2016-11-17T22:16:57.880Z
oh editing works too
Kitsune Cavalry
CC BY-SA 3.0
2016-11-17T22:34:18.970Z
FAM that's what I had at the end @denesp you charlatan
Kitsune Cavalry
CC BY-SA 3.0
2016-11-18T07:34:32.867Z
:) I rolled it back. But look, you are correct at the end of the answer but incorrect at the beginning. I mean you do not even give the same solution via the two methods.
Giskard
CC BY-SA 3.0
2016-11-18T07:54:10.950Z
The problem with this new version is that you impose symmetry before differentiating. That means that you are allowing $i$ to change not only his own quantity, but also everyone else's! My advice would be to write profit at $qi[a-Q(q_i)]-q_i^2$ and differentiate with the product rule (we know $Q'(q_i)=1$). You should get the FOC $a-Q(q_i)-q_iQ'(q_i)-2q_i=a-Nq_i-3q_i=0$ (where the first equality follows by imposing symmetry).
Ubiquitous
CC BY-SA 3.0
2016-11-18T08:12:11.023Z
I am really out of it today. I'll roll it back to denesp's edit >o< Thanks for your patience.
Kitsune Cavalry
CC BY-SA 3.0
2016-11-17T06:24:51.737Z
2016-11-18T08:12:25.077Z
2016-11-18T08:12:25.077Z
Kitsune Cavalry
<p>This is the full question: <strong>Find all pure strategy Nash equilibria of the Cournot oligopoly game with $N$ firms facing linear demand $P = a − Q$ when the total cost for each firm is $c_i(q_i) = q_i^2$.</strong></p>

<p>My understanding is that:</p>

<p>$$\pi_i = q_i(a − Q) - q_i^2$$</p>

<p>$$\max_{q_i} \quad q_i(a − Q) - q_i^2$$</p>

<p>Therefore, the FOC: </p>

<p>$a - Q - 2q_i = 0$</p>

<p>This FOC can be derived for each firm. </p>

<p>This is the part that I get stuck at. How do you add all the FOCs of $N$ firms together? Thank you! </p>

CC BY-SA 3.0
2016-11-17T07:54:14.200Z
I think your FOC is wrong. Isn't $Q=q_1+q_2+q_3+\ldots+q_N$? But then this term should be differentiated WRT $q_i$ too.
Ubiquitous
CC BY-SA 3.0
2016-11-17T08:36:31.290Z
This question is crossposted on [math.SE](http://math.stackexchange.com/questions/2017877/cournot-oligopoly-game-with-n-firms). And since that is where questions about adding equations belong, I am voting to close it here.
Giskard
CC BY-SA 3.0
2016-11-17T09:57:01.887Z
I'm voting to close this question as off-topic because it is a duplicate of http://math.stackexchange.com/questions/2017877/cournot-oligopoly-game-with-n-firms
Ubiquitous
<microeconomics><game-theory>
Solving a Cournot oligopoly game with N firms
D. Daniels
# Solving a Cournot oligopoly game with N firms

(Asked by: username_0 on 2016-11-17)

This is the full question: **Find all pure strategy Nash equilibria of the Cournot oligopoly game with $N$ firms facing linear demand $P = a − Q$ when the total cost for each firm is $c\_i(q\_i) = q\_i^2$.**

My understanding is that:

$$\pi\_i = q\_i(a − Q) - q\_i^2$$

$$\max\_{q\_i} \quad q\_i(a − Q) - q\_i^2$$

Therefore, the FOC:

$a - Q - 2q\_i = 0$

This FOC can be derived for each firm.

This is the part that I get stuck at. How do you add all the FOCs of $N$ firms together? Thank you!

**Tags:** microeconomics, game-theory
> **Comments:**
> **username_1**: I think your FOC is wrong. Isn't $Q=q_1+q_2+q_3+\ldots+q_N$? But then this term should be differentiated WRT $q_i$ too.  
> **username_2**: This question is crossposted on [math.SE](http://math.stackexchange.com/questions/2017877/cournot-oligopoly-game-with-n-firms). And since that is where questions about adding equations belong, I am voting to close it here.  
> **username_1**: I'm voting to close this question as off-topic because it is a duplicate of http://math.stackexchange.com/questions/2017877/cournot-oligopoly-game-with-n-firms  

---

## 1 answers

### Answer by username_3 on 2016-11-17. Score: 1

Your FOC is wrong.

So $$\max\_{q\_i} \quad q\_i(a − \sum\_{j=1}^nq\_j) - q\_i^2$$

$$\implies \max\_{q\_i} \quad aq\_i − q\_i\sum\_{j=1}^nq\_j - q\_i^2$$

The real FOC is

$$a - \sum\_{j\neq i}q\_j - 2q\_i - 2q\_i = 0.$$

> **Comments:**
> **username_1**: This solution method would be correct if the OP had calculated the correct FOC to begin with.  
> **username_3**: Oh dammit you're right.  
> **username_3**: It should be fixed.  
> **username_2**: @KitsuneCavalry I think you might want to try again :)  
> **username_2**: @KitsuneCavalry Or maybe not. After all, it is a crossposted off-topic question with no visible effort...  
> **username_3**: @denesp I'm traaaaaash *headdesk* (If this is wrong again please just delete this answer out of my misery ayy lmao)  
> **username_2**: Soooo torn between editing and voting to delete.  
> **username_3**: oh editing works too  
> **username_3**: FAM that's what I had at the end @denesp you charlatan  
> **username_2**: :) I rolled it back. But look, you are correct at the end of the answer but incorrect at the beginning. I mean you do not even give the same solution via the two methods.  
> **username_1**: The problem with this new version is that you impose symmetry before differentiating. That means that you are allowing $i$ to change not only his own quantity, but also everyone else's! My advice would be to write profit at $qi[a-Q(q_i)]-q_i^2$ and differentiate with the product rule (we know $Q'(q_i)=1$). You should get the FOC $a-Q(q_i)-q_iQ'(q_i)-2q_i=a-Nq_i-3q_i=0$ (where the first equality follows by imposing symmetry).  
> **username_3**: I am really out of it today. I'll roll it back to denesp's edit >o< Thanks for your patience.  

---

