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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!

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    $\begingroup$ 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. $\endgroup$ – Ubiquitous Nov 17 '16 at 7:54
  • $\begingroup$ This question is crossposted on math.SE. And since that is where questions about adding equations belong, I am voting to close it here. $\endgroup$ – Giskard Nov 17 '16 at 8:36
  • $\begingroup$ I'm voting to close this question as off-topic because it is a duplicate of math.stackexchange.com/questions/2017877/… $\endgroup$ – Ubiquitous Nov 17 '16 at 9:57
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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

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

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    $\begingroup$ This solution method would be correct if the OP had calculated the correct FOC to begin with. $\endgroup$ – Ubiquitous Nov 17 '16 at 9:57
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    $\begingroup$ @KitsuneCavalry I think you might want to try again :) $\endgroup$ – Giskard Nov 17 '16 at 17:16
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    $\begingroup$ @KitsuneCavalry Or maybe not. After all, it is a crossposted off-topic question with no visible effort... $\endgroup$ – Giskard Nov 17 '16 at 17:26
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    $\begingroup$ FAM that's what I had at the end @denesp you charlatan $\endgroup$ – Kitsune Cavalry Nov 17 '16 at 22:34
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    $\begingroup$ 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). $\endgroup$ – Ubiquitous Nov 18 '16 at 7:54

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