<p>Firm 3 will choose <span class="math-container">$q_3$</span> after observing <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> in the following way. It solves the following problem:</p>
<p><span class="math-container">\begin{eqnarray*}\max_{q_3\geq 0} &amp; \ (16-q_1-q_2-q_3)q_3\end{eqnarray*}</span></p>
<p>and we get
<span class="math-container">$q_3=\dfrac{16-q_1-q_2}{2}$</span></p>
<p>Firm 2 will anticipate firm 3's response as above and choose <span class="math-container">$q_2$</span> after observing <span class="math-container">$q_1$</span> in the following way. It solves the following problem:</p>
<p><span class="math-container">\begin{eqnarray*}\max_{q_2\geq 0} &amp; \ (16-q_1-q_2-q_3)q_2 \\ \text{s.t.} &amp; q_3=\dfrac{16-q_1-q_2}{2}\end{eqnarray*}</span></p>
<p>which can be re-written as
<span class="math-container">\begin{eqnarray*}\max_{q_2\geq 0} &amp; \ \frac{(16-q_1-q_2)q_2}{2} \end{eqnarray*}</span></p>
<p>and we get
<span class="math-container">$q_2=\dfrac{16-q_1}{2}$</span>.</p>
<p>Firm 1 will anticipate firm 2 and 3 to respond in the manner above and choose <span class="math-container">$q_1$</span> by solving
<span class="math-container">\begin{eqnarray*}\max_{q_1\geq 0} &amp; \ (16-q_1-q_2-q_3)q_1  \\ \text{s.t.} &amp; q_3=\dfrac{16-q_1-q_2}{2} \\ \text{and} &amp;  q_2=\dfrac{16-q_1}{2}\end{eqnarray*}</span></p>
<p>which can be re-written as</p>
<p><span class="math-container">\begin{eqnarray*}\max_{q_1\geq 0} &amp; \ \frac{(16-q_1)q_1}{4} \end{eqnarray*}</span></p>
<p>and we get <span class="math-container">$q_1=8$</span>.</p>

CC BY-SA 4.0
2024-06-20T01:29:07.960Z
2024-06-20T01:29:07.960Z
Amit
<p>My friend and I were solving the following game for assignment, and came up with different solutions. Could any one help us know whose solution is right?</p>
<p>Suppose there are 3 firms in a homogeneous good market, which set their quantities sequentially (firm <span class="math-container">$i$</span> in period <span class="math-container">$i$</span>). Suppose firms have zero marginal cost of production.The industry faces an inverse demand <span class="math-container">$P(q) = 16-q$</span> where <span class="math-container">$q=\sum_{i=1}^3q_i$</span> is the aggregate quantity. Solve for <span class="math-container">$q_1$</span>.</p>
<p>Both agreed to the following:(See <a href="https://economics.stackexchange.com/questions/26054/rigorous-solution-of-stackelberg-leader-follower-game-with-n-players">Rigorous solution of Stackelberg leader-follower game with N-players?</a>)
<span class="math-container">$$q_i = \frac{1}{2}(16-\sum_{j=1}^{i-1}q_j)$$</span>
<span class="math-container">$$q_i = \frac{1}{2}q_{i-1}$$</span></p>
<p>My Solution:
<span class="math-container">$$max_{q_1}Pq_1=max(16-q_1-\frac{1}{2}q_1-(\frac{1}{2})^2q_1)q_1 $$</span>
<span class="math-container">$$=(16-\frac{7}{4}q_1)q_1$$</span>
Applying the FOC results in <span class="math-container">$q_1=32/7$</span> with the profit level <span class="math-container">$\frac{32\cdot8}{7}$</span>.</p>
<p>Her Solution:
Applying i=1 to the first equation provides <span class="math-container">$q_1=8$</span>. Hence, the profit level is <span class="math-container">$(16-(8+4+2))\times 8 = 16$</span>.</p>
<p>Could anyone tell us which solution is right?</p>

<dynamic-games>
Stackelberg Game with three players
Wooyong Park
# Stackelberg Game with three players

(Asked by: username_0 on 2024-06-20)

My friend and I were solving the following game for assignment, and came up with different solutions. Could any one help us know whose solution is right?

Suppose there are 3 firms in a homogeneous good market, which set their quantities sequentially (firm $i$ in period $i$). Suppose firms have zero marginal cost of production.The industry faces an inverse demand $P(q) = 16-q$ where $q=\sum\_{i=1}^3q\_i$ is the aggregate quantity. Solve for $q\_1$.

Both agreed to the following:(See [Rigorous solution of Stackelberg leader-follower game with N-players?](https://economics.stackexchange.com/questions/26054/rigorous-solution-of-stackelberg-leader-follower-game-with-n-players))
$$q\_i = \frac{1}{2}(16-\sum\_{j=1}^{i-1}q\_j)$$
$$q\_i = \frac{1}{2}q\_{i-1}$$

My Solution:
$$max\_{q\_1}Pq\_1=max(16-q\_1-\frac{1}{2}q\_1-(\frac{1}{2})^2q\_1)q\_1 $$
$$=(16-\frac{7}{4}q\_1)q\_1$$
Applying the FOC results in $q\_1=32/7$ with the profit level $\frac{32\cdot8}{7}$.

Her Solution:
Applying i=1 to the first equation provides $q\_1=8$. Hence, the profit level is $(16-(8+4+2))\times 8 = 16$.

Could anyone tell us which solution is right?

**Tags:** dynamic-games


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## 1 answers

### Answer by username_1 on 2024-06-20. Score: 2 | Accepted answer

Firm 3 will choose $q\_3$ after observing $q\_1$ and $q\_2$ in the following way. It solves the following problem:

\begin{eqnarray\*}\max\_{q\_3\geq 0} & \ (16-q\_1-q\_2-q\_3)q\_3\end{eqnarray\*}

and we get
$q\_3=\dfrac{16-q\_1-q\_2}{2}$

Firm 2 will anticipate firm 3's response as above and choose $q\_2$ after observing $q\_1$ in the following way. It solves the following problem:

\begin{eqnarray\*}\max\_{q\_2\geq 0} & \ (16-q\_1-q\_2-q\_3)q\_2 \\ \text{s.t.} & q\_3=\dfrac{16-q\_1-q\_2}{2}\end{eqnarray\*}

which can be re-written as
\begin{eqnarray\*}\max\_{q\_2\geq 0} & \ \frac{(16-q\_1-q\_2)q\_2}{2} \end{eqnarray\*}

and we get
$q\_2=\dfrac{16-q\_1}{2}$.

Firm 1 will anticipate firm 2 and 3 to respond in the manner above and choose $q\_1$ by solving
\begin{eqnarray\*}\max\_{q\_1\geq 0} & \ (16-q\_1-q\_2-q\_3)q\_1 \\ \text{s.t.} & q\_3=\dfrac{16-q\_1-q\_2}{2} \\ \text{and} & q\_2=\dfrac{16-q\_1}{2}\end{eqnarray\*}

which can be re-written as

\begin{eqnarray\*}\max\_{q\_1\geq 0} & \ \frac{(16-q\_1)q\_1}{4} \end{eqnarray\*}

and we get $q\_1=8$.


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