<p>The gist/shortened and generalized version of the above answer:</p>
<p>In the context where <span class="math-container">$Q = \sum_i q_i$</span> the equation
<span class="math-container">$$ \frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q} = \frac{\partial \pi_i}{\partial q_i} + \frac{\partial Q}{\partial q_i}\frac{\partial \pi_i}{\partial Q} $$</span>
holds as
<span class="math-container">$$ \frac{\partial Q}{\partial q_i} = 1. $$</span></p>

CC BY-SA 4.0
2020-08-26T07:51:13.430Z
2020-08-26T07:51:13.430Z
Giskard
<p>I am reading an article that has this description of the first-order condition for a Cournot n-firm game:</p>
<p>Take <span class="math-container">$P(Q) = Q^{-1}$</span>, <span class="math-container">$\pi_i(q_i, Q) = (Q^{-1} - c_i)q_i$</span>.</p>
<p>Then the first-order condition for an interior profit-maximizing choice of <span class="math-container">$q_i$</span> requires that</p>
<p><span class="math-container">$$ \frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q} = Q^{-1} - c_i - q_iQ^{-2} = 0.$$</span></p>
<p>I am trying to understand why it is OK to simply take <span class="math-container">$\frac{\partial \pi_i}{\partial Q}$</span> ignoring the fact that <span class="math-container">$Q$</span> is actually a function of <span class="math-container">$q_i$</span>. If I expand the term so that <span class="math-container">$Q = q_i + q_{-i}$</span> and take the partial derivatives <span class="math-container">$\frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial q_{-i}}$</span>, the solution is not the same as the one that is written in the article. Would appreciate any explanation.</p>

<nash-equilibrium><oligopoly>
Cournot oligopoly - first-order condition
user219923
# Cournot oligopoly - first-order condition

(Asked by: username_0 on 2020-08-25)

I am reading an article that has this description of the first-order condition for a Cournot n-firm game:

Take $P(Q) = Q^{-1}$, $\pi\_i(q\_i, Q) = (Q^{-1} - c\_i)q\_i$.

Then the first-order condition for an interior profit-maximizing choice of $q\_i$ requires that

$$ \frac{\partial \pi\_i}{\partial q\_i} + \frac{\partial \pi\_i}{\partial Q} = Q^{-1} - c\_i - q\_iQ^{-2} = 0.$$

I am trying to understand why it is OK to simply take $\frac{\partial \pi\_i}{\partial Q}$ ignoring the fact that $Q$ is actually a function of $q\_i$. If I expand the term so that $Q = q\_i + q\_{-i}$ and take the partial derivatives $\frac{\partial \pi\_i}{\partial q\_i} + \frac{\partial \pi\_i}{\partial q\_{-i}}$, the solution is not the same as the one that is written in the article. Would appreciate any explanation.

**Tags:** nash-equilibrium, oligopoly

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## 1 answers

### Answer by username_1 on 2020-08-26. Score: 3

The gist/shortened and generalized version of the above answer:

In the context where $Q = \sum\_i q\_i$ the equation
$$ \frac{\partial \pi\_i}{\partial q\_i} + \frac{\partial \pi\_i}{\partial Q} = \frac{\partial \pi\_i}{\partial q\_i} + \frac{\partial Q}{\partial q\_i}\frac{\partial \pi\_i}{\partial Q} $$
holds as
$$ \frac{\partial Q}{\partial q\_i} = 1. $$


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