<p>Let <span class="math-container">$c = 0$</span> and <span class="math-container">$Q_k = \sum_{i = 1}^k{q_i}$</span>.
For <span class="math-container">$j = n$</span> the best response is given by
<span class="math-container">\begin{align}
b_n(Q_{n-1}) = \arg\max_{q_n}(1 - Q_{n-1} - q_n)q_n = \frac{1- Q_{n-1}}{2}.
\end{align}</span>
For <span class="math-container">$j = n-1$</span> the best response is given by
<span class="math-container">\begin{align}
b_{n-1}(Q_{n-2}) =&amp; \arg\max_{q_{n-1}}(1 - Q_{n-2} - q_{n-1} - b_n(Q_{n-1}))q_{n-1} = \frac{1- Q_{n-2}}{2}.
\end{align}</span>
It can be shown that for any <span class="math-container">$k \in \{1,\ldots,n\}$</span> the best response is thus given by
<span class="math-container">\begin{align}
b_k(Q_{k-1}) = \arg\max_{q_k}\left(1 - Q_{k-1} - q_k - \sum_{\ell=k+1}^n{b_\ell}(Q_{\ell-1})\right)q_{n-1} = \frac{1- Q_{k-1}}{2}.
\end{align}</span>
Note that <span class="math-container">$q_1^* = b_1(0) = \frac{1}{2}$</span>.
It then follows
<span class="math-container">\begin{align}
q_n^* = b_n(Q^*_{n-1}) =&amp; \frac{1 - (q_1^* + q_2^* + q_3^* + \ldots + q_{n-1}^*)}{2}\\
=&amp; \frac{1 - \left(q_1^* + \frac{1 - q_1^*}{2} + \frac{1 - q_1^* - q_2^*}{2} + \ldots + \frac{1 - (q_1^* + q_2^* + q_3^* + \ldots + q_{n-2}^*)}{2}\right)}{2}\\
=&amp; \frac{1 - \left(q_1^* + \frac{1 - q_1^*}{2} + \frac{1 - q_1^* - \frac{1 - q_1^*}{2}}{2} + \ldots + \frac{1 - \left(q_1^* + \frac{1 - q_1^*}{2} + \frac{1 - q_1^* - q_2^*}{2} +\ldots + q_{n-2}^*\right)}{2}\right)}{2}\\
\vdots\\
=&amp; \frac{1-q_1^*}{2^{n-1}}\\
=&amp; \frac{1}{2^n}
\end{align}</span></p>

CC BY-SA 4.0
2019-05-15T21:08:24.270Z
2019-05-15T21:08:24.270Z
clueless
<p>Because each new player faces the residual demand curve of the prior player, therefore he always ends up producing half the quantity of the prior player. </p>

CC BY-SA 4.0
2018-12-16T17:16:45.697Z
OP asks for a "rigorous solution", not an intuition.
Herr K.
CC BY-SA 4.0
2018-12-16T16:00:01.677Z
2018-12-16T16:00:01.677Z
Becky
<p>In the classical two-player Stackelberg game (sequential Cournot) we have a linear demand function <span class="math-container">$P = 1 - Q$</span> where <span class="math-container">$Q = \sum_0^2 q_i $</span>, and we assume homogenous production cost <span class="math-container">$c$</span>. By starting from the "bottom" of the game where player 2 knows player 1's chosen output and best-response reacts to it, and then moving "upwards" assuming player 1 anticipates this response, the game can be shown to have equilibrium solutions:</p>

<p><span class="math-container">$$ q_1^\star = \frac{1-c}{2} $$</span>
<span class="math-container">$$ q_2^\star = \frac{1-c}{4} .$$</span></p>

<p>We can follow the same procedure with 3, 4 or more players and discover the pattern:
<span class="math-container">$$ q_n^\star = \frac{1-c}{2^n} .$$</span></p>

<p><strong>I have been trying to work out a rigorous mathematical proof, say by induction or other methods, to show that this has to be the case, but not managed to find any. Can anybody help please?</strong></p>

CC BY-SA 4.0
2018-12-16T17:14:46.493Z
A sketch of proof was given by Andy Skrzypacz on Quora: https://qr.ae/TUtWqN
Herr K.
<microeconomics><game-theory><extensive-games>
Rigorous solution of Stackelberg leader-follower game with N-players?
Jhonny
# Rigorous solution of Stackelberg leader-follower game with N-players?

(Asked by: username_0 on 2018-12-16)

In the classical two-player Stackelberg game (sequential Cournot) we have a linear demand function $P = 1 - Q$ where $Q = \sum\_0^2 q\_i $, and we assume homogenous production cost $c$. By starting from the "bottom" of the game where player 2 knows player 1's chosen output and best-response reacts to it, and then moving "upwards" assuming player 1 anticipates this response, the game can be shown to have equilibrium solutions:

$$ q\_1^\star = \frac{1-c}{2} $$
$$ q\_2^\star = \frac{1-c}{4} .$$

We can follow the same procedure with 3, 4 or more players and discover the pattern:
$$ q\_n^\star = \frac{1-c}{2^n} .$$

**I have been trying to work out a rigorous mathematical proof, say by induction or other methods, to show that this has to be the case, but not managed to find any. Can anybody help please?**

**Tags:** microeconomics, game-theory, extensive-games

> **Comments:**
> **username_1**: A sketch of proof was given by Andy Skrzypacz on Quora: https://qr.ae/TUtWqN  

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## 2 answers

### Answer by username_2 on 2019-05-15. Score: 1 | Accepted answer

Let $c = 0$ and $Q\_k = \sum\_{i = 1}^k{q\_i}$.
For $j = n$ the best response is given by
\begin{align}
b\_n(Q\_{n-1}) = \arg\max\_{q\_n}(1 - Q\_{n-1} - q\_n)q\_n = \frac{1- Q\_{n-1}}{2}.
\end{align}
For $j = n-1$ the best response is given by
\begin{align}
b\_{n-1}(Q\_{n-2}) =& \arg\max\_{q\_{n-1}}(1 - Q\_{n-2} - q\_{n-1} - b\_n(Q\_{n-1}))q\_{n-1} = \frac{1- Q\_{n-2}}{2}.
\end{align}
It can be shown that for any $k \in \{1,\ldots,n\}$ the best response is thus given by
\begin{align}
b\_k(Q\_{k-1}) = \arg\max\_{q\_k}\left(1 - Q\_{k-1} - q\_k - \sum\_{\ell=k+1}^n{b\_\ell}(Q\_{\ell-1})\right)q\_{n-1} = \frac{1- Q\_{k-1}}{2}.
\end{align}
Note that $q\_1^\* = b\_1(0) = \frac{1}{2}$.
It then follows
\begin{align}
q\_n^\* = b\_n(Q^\*\_{n-1}) =& \frac{1 - (q\_1^\* + q\_2^\* + q\_3^\* + \ldots + q\_{n-1}^\*)}{2}\\
=& \frac{1 - \left(q\_1^\* + \frac{1 - q\_1^\*}{2} + \frac{1 - q\_1^\* - q\_2^\*}{2} + \ldots + \frac{1 - (q\_1^\* + q\_2^\* + q\_3^\* + \ldots + q\_{n-2}^\*)}{2}\right)}{2}\\
=& \frac{1 - \left(q\_1^\* + \frac{1 - q\_1^\*}{2} + \frac{1 - q\_1^\* - \frac{1 - q\_1^\*}{2}}{2} + \ldots + \frac{1 - \left(q\_1^\* + \frac{1 - q\_1^\*}{2} + \frac{1 - q\_1^\* - q\_2^\*}{2} +\ldots + q\_{n-2}^\*\right)}{2}\right)}{2}\\
\vdots\\
=& \frac{1-q\_1^\*}{2^{n-1}}\\
=& \frac{1}{2^n}
\end{align}


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### Answer by username_3 on 2018-12-16. Score: 0

Because each new player faces the residual demand curve of the prior player, therefore he always ends up producing half the quantity of the prior player.

> **Comments:**
> **username_1**: OP asks for a "rigorous solution", not an intuition.  

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