# A question about Nash Equilibrium

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 $$$$\max_{q_i} \pi_i=(200-q_i-\sum_{k\neq i,\\i\in I}q_k-\sum_{l\in II}q_l)q_i-c_1q_i\\ F.O.C\qquad \frac{\partial \pi_i}{\partial q_i}=200-2q_i-\sum_{k\neq i\\k\in I}q_k-\sum_{l\in II}q_l-c_1=0\\ BR_{q_i}(q_{-i})=q_i=100-\frac{1}{2}\sum_{k\neq i\\k\in I}q_k-\frac{1}{2}\sum_{l\in II}q_l-\frac{1}{2}c_1$$$$

Similarly, for $$\forall j\in II$$, we have best reaction function $$$$BR_{q_j}(q_{-j})=q_j=100-\frac{1}{2}\sum_{k\in I}q_k-\frac{1}{2}\sum_{l\neq j\\l\in II}q_l-\frac{1}{2}c_2$$$$

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.

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, $$q_{II}^* =\frac{200-c_2+N(c_1-c_2)}{2N+1}$$

• 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. Dec 11 '20 at 23:38
• 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. Dec 12 '20 at 1:13
• 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. Dec 12 '20 at 1:38
• @BongseoChoi yes that sounds absolutely correct Dec 12 '20 at 2:21
• Thank you for discussing this question. Dec 12 '20 at 2:49