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There are (n+1) firms (Firm 0, Firm 1, ..., Firm n) in a market for a good where $n \ge 2$. The price in the market is given by the inverse demand equation $P = 100 - \sum_{i=0}^n q_i$, where $\sum_{i=0}^n q_i$ is the total output in the market and $q_i$; is the quantity produced by Firm i. For simplicity assume that the cost is 0 for each firm. The firms choose their quantities as follows: (1) Firm 0 chooses its output level $q_0$ (2) After observing Firm 0's choice, the remaining n firms simultaneously choose their outputs. Each firm wants to maximize its profits. What is the Stackelberg Equilibrium for this market?

I found SPNE={$q_0^*=50$, $q_i=50/(n-1)$} for all $i\ge 1$

But I don’t know exactly how I can solve this question.

Please help me to solve this question

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  • $\begingroup$ Are costs normalized to 0? $\endgroup$ Oct 17, 2022 at 17:49
  • $\begingroup$ @MatteoBulgarelli yes, true! $\endgroup$
    – ThePooh
    Oct 17, 2022 at 18:27
  • $\begingroup$ It seems you did everything correctly! $\endgroup$ Oct 18, 2022 at 6:31

1 Answer 1

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Edit: I didn't see that it's a two-stage game. I considered an $n$-stage game in my previous answer. Here's the new modified answer:

The leader firm is denoted by $F_0$ and the next $n$ firms by $F_i$ where $i = 1,2, \cdots, n$.

Let's say $F_0$ chooses quantity $q_0$. The second stage is basically a Cournot competition between the firms $F_1, \cdots, F_n$. We will solve this first.

The profit of $F_i$ $(i > 0)$ is $\pi_i = (100 - Q)q_i$ which is maximized at $q_i = \frac{1}{2} \left( 100 - Q_{-i} \right)$ where $Q_{-i}$ denotes $Q - q_{i}$.

Clearly, the optimal $q_i = 100 - Q$ which means they are all equal (for $i > 0$). From this, it can be deduced that $q_i = 100 - q_0 - nq_i \implies q_i = \frac{100 - q_0}{n+1}$ $\forall \ i > 0$.

The first stage game can now be solved as $$q_0 = \text{argmax}[\pi_1(q_0)] = \text{argmax}[P(Q)q_0] = \text{argmax}\left[\left(\frac{100 - q_0}{n+1}\right)q_0\right] = 50$$

Consequently, the remaining firms produce $\displaystyle q_i = \frac{50}{n+1}$ $(i > 0)$.

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  • $\begingroup$ I am confused at a point. You found $q_1^*=50$, which means firm 0’s output (the leader firm defined in the question). Or, This is two stage game. So, you calculated the output in the last stage (by backward induction) and you left the calculation of first stage for firm 0? $\endgroup$
    – ThePooh
    Oct 17, 2022 at 20:30
  • $\begingroup$ By the way, thank you for your great help:) $\endgroup$
    – ThePooh
    Oct 17, 2022 at 20:31
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    $\begingroup$ You're welcome! $\endgroup$
    – Rick_Morty
    Oct 17, 2022 at 20:46
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    $\begingroup$ That's wrong. Every firm has the same best response function, you sum up the FOCs and back out $q_i(q_1, ..., q_n)$. Then , and only then, you solve the first problem. You can't plug the BR of a player into the profits another player that plays simoultaneously and optimize... $\endgroup$ Oct 17, 2022 at 22:42
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    $\begingroup$ Thanks for updating! $\text{armgax}$ made my day lmao. $\endgroup$ Oct 18, 2022 at 6:31

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