<p>From symmetry we can say that for each player <span class="math-container">$i$</span> in <span class="math-container">$I$</span>, <span class="math-container">$q_r=q_k$</span> <span class="math-container">$\forall r,k \in I$</span>. Similarly for <span class="math-container">$II$</span>. Let this be <span class="math-container">$q_I$</span> for producers in <span class="math-container">$I$</span> and <span class="math-container">$q_{II}$</span> for <span class="math-container">$II$</span>.</p>
<p>So profit function for individual <span class="math-container">$i$</span> in <span class="math-container">$I$</span>:
<span class="math-container">\begin{align}
\max_{q_i} \pi_i&amp;=(200-(N-1)q_I-Nq_{II}-q_i))q_i-c_1q_i\\
\text{FOC: }\,\,\frac{\partial \pi_i}{\partial q_i}&amp;=200-(N-1)q_I-Nq_{II}-q_i-q_i-c_1=0 \\
\implies q_i^*(q_I,q_{II})&amp;=\frac{200-Nq_{II}-c_1-(N-1)q_I}{2}
\end{align}</span></p>
<p>Similarly, we will get for individual <span class="math-container">$j$</span> in <span class="math-container">$II$</span>:</p>
<p><span class="math-container">$$q_j^*(q_I,q_{II})=\frac{200-Nq_I-c_2 - (N-1)q_{II}}{2N}$$</span></p>
<p>At NE, we know that <span class="math-container">$q_i^*=q_I^*$</span> and <span class="math-container">$q_j^*=q_{II}$</span>. Using this:</p>
<p><span class="math-container">\begin{align}
q_I^*&amp;=\frac{200-Nq_{II}^*-c_1-(N-1)q_I^*}{2} \\
&amp;=\frac{200-Nq_{II}^*-c_1}{N+1}
\end{align}</span></p>
<p>And so, <span class="math-container">$$q_{II}^*=\frac{200-Nq_{I}^*-c_2}{N+1}$$</span></p>
<p>Solving for <span class="math-container">$q_I^*$</span>:</p>
<p><span class="math-container">\begin{align}
(N+1)q_I^*&amp;=200-N\frac{200-Nq_I^*-c_2}{N+1}-c_1 \\
\implies q_I^* &amp;=\frac{200-c_1+N(c_2-c_1)}{2N+1}
\end{align}</span>
Similarly, <span class="math-container">$$q_{II}^* =\frac{200-c_2+N(c_1-c_2)}{2N+1}$$</span></p>

CC BY-SA 4.0
2020-12-11T23:38:15.540Z
I just don’t understand why $n(q_1+q_2)$ can be directly substituted into the reaction function. Isn’t it calculated under the conditions of other people? If you substitute $n(q_1+q_2)$ at the beginning, it seems unreasonable.
Bongseo Choi
CC BY-SA 4.0
2020-12-12T01:13:06.120Z
Conceptually it may seem unreasonable but think of this as a mathematical trick to simplify calculation. We know that it's going to be true so we might as well use it.
Dayne
CC BY-SA 4.0
2020-12-12T01:38:00.120Z
From the final result you have derived, I think it is equivalent to considering only one person in each of the two districts of the village at the beginning, and get their corresponding Nash Equilibrium, and then divide the total number of each district evenly among the $N$ people in the same district. In this way, we can see the $N$ is on the denominator.
Bongseo Choi
CC BY-SA 4.0
2020-12-12T02:21:07.193Z
@BongseoChoi yes that sounds absolutely correct
Dayne
CC BY-SA 4.0
2020-12-12T02:49:58.037Z
Thank you for discussing this question.
Bongseo Choi
CC BY-SA 4.0
2020-12-12T03:02:39.653Z
@BongseoChoi  Did you confirm that this answer is correct? It would be rewarding I think to verify that this answer matches with that we get from your approach.
Dayne
CC BY-SA 4.0
2020-12-12T04:23:37.267Z
Actually, I don't have an official answer to this question. If I follow my method to solve this question, the Nash Equilibrium for $q_1^*=\frac{200-c_1+N(c_2-c_1)}{2N+1}$,which is quite different with your answer-$q_1^*=\frac{200+c_2-2c_1}{3N}$.
Bongseo Choi
CC BY-SA 4.0
2020-12-12T05:09:51.547Z
Then my answer is incorrect. I was wondering that it may be wrong as when I am taking derivative, the $q_i$ for same village should have been taken differently. I will try to edit it.
Dayne
CC BY-SA 4.0
2020-12-12T05:36:34.787Z
@BongseoChoi So the idea is that instead of summation, we take production by other producers in the same village as same variable. Now we get same answer as you did.
Dayne
CC BY-SA 4.0
2020-12-12T05:57:51.200Z
yes, the former method you used may cause some errors to obtain the FOC.
Bongseo Choi
CC BY-SA 4.0
2020-12-11T17:43:35.180Z
2020-12-12T05:35:25.210Z
2020-12-12T05:35:25.210Z
Dayne
<p>I have some trouble with Nash Equilibrium. The specific question as follows.</p>
<blockquote>
<p>Suppose that there are <span class="math-container">$2N$</span> people in the village, of which <span class="math-container">$N$</span> residents live in the first district, and each person chooses to raise <span class="math-container">$q_1$</span> sheep, and the cost of each sheep is <span class="math-container">$c_1$</span>. In addition, n people live in the second district, and each person chooses to raise <span class="math-container">$q_2$</span> sheep. The cost of a sheep is <span class="math-container">$c_2$</span>. The income brought by each sheep is <span class="math-container">$200-q$</span>, where <span class="math-container">$q$</span> is the total number of sheep in the village.
<br />
Q: Find the number of sheep raised by each resident in the two regions under the Nash equilibrium of this game.</p>
</blockquote>
<p>Here is my thought. For simplicity, let <span class="math-container">$I$</span> denotes the index set of individuals in the first district, similarly, <span class="math-container">$II$</span> for the second district.</p>
<p>For <span class="math-container">$\forall i\in I$</span>, the profit maximization problem is
<span class="math-container">\begin{equation}
\max_{q_i} \pi_i=(200-q_i-\sum_{k\neq i,\\i\in I}q_k-\sum_{l\in II}q_l)q_i-c_1q_i\\
F.O.C\qquad \frac{\partial \pi_i}{\partial q_i}=200-2q_i-\sum_{k\neq i\\k\in I}q_k-\sum_{l\in II}q_l-c_1=0\\
BR_{q_i}(q_{-i})=q_i=100-\frac{1}{2}\sum_{k\neq i\\k\in I}q_k-\frac{1}{2}\sum_{l\in II}q_l-\frac{1}{2}c_1
\end{equation}</span></p>
<p>Similarly, for <span class="math-container">$\forall j\in II$</span>, we have best reaction function
<span class="math-container">\begin{equation}
BR_{q_j}(q_{-j})=q_j=100-\frac{1}{2}\sum_{k\in I}q_k-\frac{1}{2}\sum_{l\neq j\\l\in II}q_l-\frac{1}{2}c_2
\end{equation}</span></p>
<p>Since every individual belonging to the same area is in the same situation, we can assume that
<span class="math-container">\begin{align}
\begin{cases}
q_i^*=a, &amp;\forall i\in I\\
q_j^*=b, &amp;\forall j\in II
\end{cases}
\end{align}</span>
and then substitute it into the best reaction function, we have
<span class="math-container">\begin{align}
\begin{cases}
(N+1)a+Nb=200-c_1\\
Na+(N+1)b=200-c_2
\end{cases}
\end{align}</span>
<span class="math-container">\begin{align}
\begin{cases}
a=\frac{
\begin{vmatrix}
200-c_1 &amp; N\\
200-c_2 &amp; N+1
\end{vmatrix}}{\begin{vmatrix}
N+1 &amp; N\\
N &amp; N+1
\end{vmatrix}}\\
b=\frac{
\begin{vmatrix}
N+1 &amp; 200-c_1\\
N &amp; 200-c_2
\end{vmatrix}}{\begin{vmatrix}
N+1 &amp; N\\
N &amp; N+1
\end{vmatrix}}
\end{cases}
\end{align}</span></p>
<p>I don’t know if I’m doing it right, and I’m going to solve it under the assumption that there is an interior point solution. I think if it is strictly proved, the cost <span class="math-container">$c_1,c_2$</span> may be considered.</p>

<game-theory><production-function><nash-equilibrium>
A question about Nash Equilibrium
Bongseo Choi
# A question about Nash Equilibrium

(Asked by: username_0 on 2020-12-11)

I have some trouble with Nash Equilibrium. The specific question as follows.

> Suppose that there are $2N$ people in the village, of which $N$ residents live in the first district, and each person chooses to raise $q\_1$ sheep, and the cost of each sheep is $c\_1$. In addition, n people live in the second district, and each person chooses to raise $q\_2$ sheep. The cost of a sheep is $c\_2$. The income brought by each sheep is $200-q$, where $q$ is the total number of sheep in the village.
>   
> Q: Find the number of sheep raised by each resident in the two regions under the Nash equilibrium of this game.

Here is my thought. For simplicity, let $I$ denotes the index set of individuals in the first district, similarly, $II$ for the second district.

For $\forall i\in I$, the profit maximization problem is
\begin{equation}
\max\_{q\_i} \pi\_i=(200-q\_i-\sum\_{k\neq i,\\i\in I}q\_k-\sum\_{l\in II}q\_l)q\_i-c\_1q\_i\\
F.O.C\qquad \frac{\partial \pi\_i}{\partial q\_i}=200-2q\_i-\sum\_{k\neq i\\k\in I}q\_k-\sum\_{l\in II}q\_l-c\_1=0\\
BR\_{q\_i}(q\_{-i})=q\_i=100-\frac{1}{2}\sum\_{k\neq i\\k\in I}q\_k-\frac{1}{2}\sum\_{l\in II}q\_l-\frac{1}{2}c\_1
\end{equation}

Similarly, for $\forall j\in II$, we have best reaction function
\begin{equation}
BR\_{q\_j}(q\_{-j})=q\_j=100-\frac{1}{2}\sum\_{k\in I}q\_k-\frac{1}{2}\sum\_{l\neq j\\l\in II}q\_l-\frac{1}{2}c\_2
\end{equation}

Since every individual belonging to the same area is in the same situation, we can assume that
\begin{align}
\begin{cases}
q\_i^\*=a, &\forall i\in I\\
q\_j^\*=b, &\forall j\in II
\end{cases}
\end{align}
and then substitute it into the best reaction function, we have
\begin{align}
\begin{cases}
(N+1)a+Nb=200-c\_1\\
Na+(N+1)b=200-c\_2
\end{cases}
\end{align}
\begin{align}
\begin{cases}
a=\frac{
\begin{vmatrix}
200-c\_1 & N\\
200-c\_2 & N+1
\end{vmatrix}}{\begin{vmatrix}
N+1 & N\\
N & N+1
\end{vmatrix}}\\
b=\frac{
\begin{vmatrix}
N+1 & 200-c\_1\\
N & 200-c\_2
\end{vmatrix}}{\begin{vmatrix}
N+1 & N\\
N & N+1
\end{vmatrix}}
\end{cases}
\end{align}

I don’t know if I’m doing it right, and I’m going to solve it under the assumption that there is an interior point solution. I think if it is strictly proved, the cost $c\_1,c\_2$ may be considered.

**Tags:** game-theory, production-function, nash-equilibrium


---

## 1 answers

### Answer by username_1 on 2020-12-11. Score: 0 | Accepted answer

From symmetry we can say that for each player $i$ in $I$, $q\_r=q\_k$ $\forall r,k \in I$. Similarly for $II$. Let this be $q\_I$ for producers in $I$ and $q\_{II}$ for $II$.

So profit function for individual $i$ in $I$:
\begin{align}
\max\_{q\_i} \pi\_i&=(200-(N-1)q\_I-Nq\_{II}-q\_i))q\_i-c\_1q\_i\\
\text{FOC: }\,\,\frac{\partial \pi\_i}{\partial q\_i}&=200-(N-1)q\_I-Nq\_{II}-q\_i-q\_i-c\_1=0 \\
\implies q\_i^\*(q\_I,q\_{II})&=\frac{200-Nq\_{II}-c\_1-(N-1)q\_I}{2}
\end{align}

Similarly, we will get for individual $j$ in $II$:

$$q\_j^\*(q\_I,q\_{II})=\frac{200-Nq\_I-c\_2 - (N-1)q\_{II}}{2N}$$

At NE, we know that $q\_i^\*=q\_I^\*$ and $q\_j^\*=q\_{II}$. Using this:

\begin{align}
q\_I^\*&=\frac{200-Nq\_{II}^\*-c\_1-(N-1)q\_I^\*}{2} \\
&=\frac{200-Nq\_{II}^\*-c\_1}{N+1}
\end{align}

And so, $$q\_{II}^\*=\frac{200-Nq\_{I}^\*-c\_2}{N+1}$$

Solving for $q\_I^\*$:

\begin{align}
(N+1)q\_I^\*&=200-N\frac{200-Nq\_I^\*-c\_2}{N+1}-c\_1 \\
\implies q\_I^\* &=\frac{200-c\_1+N(c\_2-c\_1)}{2N+1}
\end{align}
Similarly, $$q\_{II}^\* =\frac{200-c\_2+N(c\_1-c\_2)}{2N+1}$$

> **Comments:**
> **username_0**: I just don’t understand why $n(q_1+q_2)$ can be directly substituted into the reaction function. Isn’t it calculated under the conditions of other people? If you substitute $n(q_1+q_2)$ at the beginning, it seems unreasonable.  
> **username_1**: Conceptually it may seem unreasonable but think of this as a mathematical trick to simplify calculation. We know that it's going to be true so we might as well use it.  
> **username_0**: From the final result you have derived, I think it is equivalent to considering only one person in each of the two districts of the village at the beginning, and get their corresponding Nash Equilibrium, and then divide the total number of each district evenly among the $N$ people in the same district. In this way, we can see the $N$ is on the denominator.  
> **username_1**: @username_0 yes that sounds absolutely correct  
> **username_0**: Thank you for discussing this question.  
> **username_1**: @username_0  Did you confirm that this answer is correct? It would be rewarding I think to verify that this answer matches with that we get from your approach.  
> **username_0**: Actually, I don't have an official answer to this question. If I follow my method to solve this question, the Nash Equilibrium for $q_1^*=\frac{200-c_1+N(c_2-c_1)}{2N+1}$,which is quite different with your answer-$q_1^*=\frac{200+c_2-2c_1}{3N}$.  
> **username_1**: Then my answer is incorrect. I was wondering that it may be wrong as when I am taking derivative, the $q_i$ for same village should have been taken differently. I will try to edit it.  
> **username_1**: @username_0 So the idea is that instead of summation, we take production by other producers in the same village as same variable. Now we get same answer as you did.  
> **username_0**: yes, the former method you used may cause some errors to obtain the FOC.  

---

