I have a game with 4 players, each of whom must minimize a cost function. The strategic leverages of the 4 players are indicated by the variables: $s_i$, $s_j$, $c_i$, $c_j$.
I would like to find the Nash equilibrium of the $4$-player game using their best response functions. The best response functions aim at maximizing the players' utilities:
$$
\begin{cases}
\max _{s_i} & u_{1}(c_i, s_i, s_j) \\
\max _{c_i} & u_{2}(c_i, s_i)\\
\max _{s_j} & u_{3}(c_j, s_i, s_j) \\
\max _{c_j} & u_{4}(c_j, s_j)
\end{cases}
$$
We assume that $s_i,s_j\in\mathbb R^+$ and $c_i\in\left[d_i, s_i\right]\subset \mathbb R^+$, $c_j\in\left[d_j, s_j\right]\subset \mathbb R^+$. Given some positive numbers $d_i$, $d_j$, $\alpha$, $\beta$, $\gamma$, $\omega$, $\kappa$ and $\rho$, the best response functions are:
$$
\begin{cases}
s_i & = \alpha c_i+ \beta s_j \\
c_i & = \frac{\kappa}{s_i-d_i}+s_i\\
s_j & = \gamma c_j+ \omega s_i\\
c_j & = \frac{\rho}{s_j-d_j}+s_j
\end{cases}
$$
I would like to visualize these functions, but if I were to represent them all together, I would need a 4-dimensional space because I have 4 variables, one for each player. For example, if I only had the first two equations and $s_j$ disappeared from the first one, I would have a pair of curves that could be easily represented on a plane, but that's not the case here. Do you have any suggestions for visualizing the functions in a reduced space, perhaps leveraging the particular form of the system I have written, which I cannot see? Thank you in advance.
Best Answer
The following might help.
We can represent $c_i,s_j,c_j$ by $s_i$.
From the second equation, we have $$c_i = \frac{\kappa}{s_i-d_i}+s_i\tag1$$
From the first equation with $(1)$, we have $$\begin{align}s_j&=\frac{1}{\beta}(s_i -\alpha c_i) \\\\&=\frac{1}{\beta}\bigg(s_i -\alpha \bigg(\frac{\kappa}{s_i-d_i}+s_i\bigg)\bigg) \\\\&=\frac{1}{\beta}\bigg(s_i -\frac{\alpha\kappa}{s_i-d_i}-\alpha s_i\bigg)\tag2\end{align}$$
From the third equation with $(2)$, we have
$$\begin{align}c_j&=\frac{1}{\gamma}(s_j -\omega s_i) \\\\&=\frac{1}{\gamma}\bigg(\frac{1}{\beta}\bigg(s_i -\frac{\alpha\kappa}{s_i-d_i}-\alpha s_i\bigg)-\omega s_i\bigg) \\\\&=\frac{1}{\gamma}\bigg(\frac{s_i}{\beta} -\frac{\alpha\kappa}{\beta(s_i-d_i)}-\frac{\alpha s_i}{\beta}-\omega s_i\bigg)\tag3\end{align}$$
Finally, from the fourth equation with $(2)(3)$, $s_i$ has to satisfy $$\frac{1}{\gamma}\bigg(\frac{s_i}{\beta} -\frac{\alpha\kappa}{\beta(s_i-d_i)}-\frac{\alpha s_i}{\beta}-\omega s_i\bigg)= \frac{\rho}{\frac{1}{\beta}\bigg(s_i -\frac{\alpha\kappa}{s_i-d_i}-\alpha s_i\bigg)-d_j}+\frac{1}{\beta}\bigg(s_i -\frac{\alpha\kappa}{s_i-d_i}-\alpha s_i\bigg)$$
which can be written as
$$As_i^4+Bs_i^3+Cs_i^2+Ds_i+E=0\tag4$$ where $$A=- \alpha^2 \gamma + \alpha^2 + \alpha \beta \omega + 2 \alpha \gamma - 2 \alpha - \beta \omega - \gamma + 1,$$
$$B= 2 \alpha^2 d_i \gamma - 2 \alpha^2 d_i - 2 \alpha \beta d_i \omega - \alpha \beta d_j \gamma + \alpha \beta d_j - 4 \alpha d_i \gamma + 4 \alpha d_i + \beta^2 d_j \omega+ 2 \beta d_i \omega + \beta d_j \gamma - \beta d_j + 2 d_i \gamma - 2 d_i ,$$
$$C=-\alpha^2 d_i^2 \gamma + \alpha^2 d_i^2 - 2 \alpha^2 \kappa \gamma + 2 \alpha^2 \kappa + \alpha \beta d_i^2 \omega + 2 \alpha \beta d_i d_j \gamma- 2 \alpha \beta d_i d_j + \alpha \beta \kappa \omega + 2 \alpha d_i^2 \gamma - 2 \alpha d_i^2 + 2 \alpha \kappa \gamma - 2 \alpha \kappa - 2 \beta^2 d_i d_j \omega - \beta^2 \rho \gamma - \beta d_i^2 \omega - 2 \beta d_i d_j \gamma + 2 \beta d_i d_j - d_i^2 \gamma + d_i^2 , $$
$$D= 2 \alpha^2 d_i \kappa \gamma - 2 \alpha^2 d_i \kappa - \alpha \beta d_i^2 d_j \gamma + \alpha \beta d_i^2 d_j - \alpha \beta d_i \kappa \omega - \alpha \beta d_j \kappa \gamma + \alpha \beta d_j \kappa - 2 \alpha d_i \kappa \gamma + 2 \alpha d_i \kappa + \beta^2 d_i^2 d_j \omega + 2 \beta^2 d_i \rho \gamma + \beta d_i^2 d_j \gamma - \beta d_i^2 d_j ,$$
$$E=- \alpha^2 \kappa^2 \gamma + \alpha^2 \kappa^2 + \alpha \beta d_i d_j \kappa \gamma - \alpha \beta d_i d_j \kappa- \beta^2 d_i^2 \rho \gamma $$
So, when we are given $d_i,d_j,\alpha,\beta,\gamma,\omega,\kappa$ and $\rho$, we can get $s_i$ by solving $(4)$, and then get $c_i,s_j,c_j$ from $(1)(2)(3)$.