<p>Edit: I didn't see that it's a two-stage game. I considered an <span class="math-container">$n$</span>-stage game in my previous answer. Here's the new modified answer:</p>
<p>The leader firm is denoted by <span class="math-container">$F_0$</span> and the next <span class="math-container">$n$</span> firms by <span class="math-container">$F_i$</span> where <span class="math-container">$i = 1,2, \cdots, n$</span>.</p>
<p>Let's say <span class="math-container">$F_0$</span> chooses quantity <span class="math-container">$q_0$</span>. The second stage is basically a Cournot competition between the firms <span class="math-container">$F_1, \cdots, F_n$</span>. We will solve this first.</p>
<p>The profit of <span class="math-container">$F_i$</span> <span class="math-container">$(i &gt; 0)$</span> is <span class="math-container">$\pi_i = (100 - Q)q_i$</span> which is maximized at <span class="math-container">$q_i = \frac{1}{2} \left( 100 - Q_{-i} \right)$</span> where <span class="math-container">$Q_{-i}$</span> denotes <span class="math-container">$Q - q_{i}$</span>.</p>
<p>Clearly, the optimal <span class="math-container">$q_i = 100 - Q$</span> which means they are all equal (for <span class="math-container">$i &gt; 0$</span>). From this, it can be deduced that <span class="math-container">$q_i = 100 - q_0 - nq_i \implies q_i = \frac{100 - q_0}{n+1}$</span> <span class="math-container">$\forall \ i &gt; 0$</span>.</p>
<p>The first stage game can now be solved as <span class="math-container">$$q_0 = \text{argmax}[\pi_1(q_0)] = \text{argmax}[P(Q)q_0] = \text{argmax}\left[\left(\frac{100 - q_0}{n+1}\right)q_0\right] = 50$$</span></p>
<p>Consequently, the remaining firms produce <span class="math-container">$\displaystyle q_i = \frac{50}{n+1}$</span> <span class="math-container">$(i &gt; 0)$</span>.</p>

CC BY-SA 4.0
2022-10-17T20:30:54.900Z
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?
ThePooh
CC BY-SA 4.0
2022-10-17T20:31:19.837Z
By the way, thank you for your great help:)
ThePooh
CC BY-SA 4.0
2022-10-17T20:39:43.287Z
If you see, I mentioned that I will use notations $F_1, F_2, \cdots, F_n$. Instead of $n+1$ firms, I am considering $n$ firms (as it doesn't make a difference). The leader is $F_1$, then $F_2$ chooses and then $F_3$ and so on. The leader $(F_1)$ will produce $50$ units, followed by $25$ units by $F_2$, $12.5$ units by $F_3$ and so on.
Rick_Morty
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2022-10-17T20:41:41.927Z
I understand now. Thanks a lot.
ThePooh
CC BY-SA 4.0
2022-10-17T20:46:54.373Z
You're welcome!
Rick_Morty
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2022-10-17T22:42:56.350Z
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...
Matteo Bulgarelli
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2022-10-17T22:49:39.667Z
Only one firm has a first mover advantage
Matteo Bulgarelli
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2022-10-17T23:31:41.107Z
@MatteoBulgarelli Thanks for the catch! I thought it's an $n$-stage game.
Rick_Morty
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2022-10-18T06:31:37.580Z
Thanks for updating! $\text{armgax}$ made my day lmao.
Matteo Bulgarelli
CC BY-SA 4.0
2022-10-18T19:13:11.067Z
@MatteoBulgarelli Haha
Rick_Morty
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2022-10-17T19:56:49.413Z
2022-10-18T12:44:00.953Z
2022-10-18T12:44:00.953Z
Rick_Morty
<p>There are (n+1) firms (Firm 0, Firm 1, ..., Firm n) in a market for a good where <span class="math-container">$n \ge 2$</span>. The price in the market is given by the inverse demand equation <span class="math-container">$P = 100 - \sum_{i=0}^n q_i$</span>, where <span class="math-container">$\sum_{i=0}^n q_i$</span> is the total output in the market and <span class="math-container">$q_i$</span>; 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 <span class="math-container">$q_0$</span> (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?</p>
<p>I found SPNE={<span class="math-container">$q_0^*=50$</span>, <span class="math-container">$q_i=50/(n-1)$</span>} for all <span class="math-container">$i\ge 1$</span></p>
<p>But I don’t know exactly how I can solve this question.</p>
<p>Please help me to solve this question</p>

CC BY-SA 4.0
2022-10-17T17:49:52.830Z
Are costs normalized to 0?
Matteo Bulgarelli
CC BY-SA 4.0
2022-10-17T18:27:36.227Z
@MatteoBulgarelli yes, true!
ThePooh
CC BY-SA 4.0
2022-10-18T06:31:29.713Z
It seems you did everything correctly!
Matteo Bulgarelli
<microeconomics><game-theory><self-study>
Stackelberg equilibrium with n+1 firms
ThePooh
# Stackelberg equilibrium with n+1 firms

(Asked by: username_0 on 2022-10-17)

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

**Tags:** microeconomics, game-theory, self-study

> **Comments:**
> **username_1**: Are costs normalized to 0?  
> **username_0**: @MatteoBulgarelli yes, true!  
> **username_1**: It seems you did everything correctly!  

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## 1 answers

### Answer by username_2 on 2022-10-17. Score: 5 | Accepted answer

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 $\displaystyle q\_i = \frac{50}{n+1}$ $(i > 0)$.

> **Comments:**
> **username_0**: 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?  
> **username_0**: By the way, thank you for your great help:)  
> **username_2**: If you see, I mentioned that I will use notations $F_1, F_2, \cdots, F_n$. Instead of $n+1$ firms, I am considering $n$ firms (as it doesn't make a difference). The leader is $F_1$, then $F_2$ chooses and then $F_3$ and so on. The leader $(F_1)$ will produce $50$ units, followed by $25$ units by $F_2$, $12.5$ units by $F_3$ and so on.  
> **username_0**: I understand now. Thanks a lot.  
> **username_2**: You're welcome!  
> **username_1**: 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...  
> **username_1**: Only one firm has a first mover advantage  
> **username_2**: @MatteoBulgarelli Thanks for the catch! I thought it's an $n$-stage game.  
> **username_1**: Thanks for updating! $\text{armgax}$ made my day lmao.  
> **username_2**: @MatteoBulgarelli Haha  

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