Convergence Conditions for the Dynamics of Reflexive Collective Behavior in a Cournot Oligopoly Model under Incomplete Information

Abstract

This paper considers a Cournot oligopoly model with an arbitrary number of rational agents under incomplete information in the classical case (linear cost and demand functions). Within the dynamic reflexive collective behavior model, at each time instant each agent adjusts his output, taking a step towards the maximum profit under the expected choice of the competitors. Convergence conditions to a Cournot–Nash equilibrium are analyzed using the errors transition matrices of the dynamics. Restrictions on the ranges of agents’ steps are imposed and their effect on the convergence properties of the dynamics is demonstrated. Finally, a method is proposed to determine the maximum step ranges ensuring the convergent dynamics of collective behavior for an arbitrary number of agents.

APPENDIX

Proof of Proposition 3. This result is established by transforming the right-hand side of (12):

$$\begin{gathered} \sum\limits_{i \in N}^{} {\left\{ {\left( {1 - \beta _{i}^{{t + 1}}} \right){{{\left[ {{{\varepsilon }_{i}} - \frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} \right]}}^{2}} + \beta _{i}^{{t + 1}}{{{\left[ {{{\varepsilon }_{i}} - \frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} \right]}}^{2}}} \right\}} \\ = \sum\limits_{i \in N}^{} {{{{({{\varepsilon }_{i}})}}^{2}} - \sum\limits_{i \in N}^{} {{{\varepsilon }_{i}}\vec {\gamma }_{i}^{{t + 1}}\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right) + {{{\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}}^{2}}} } {{\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)}^{2}} \\ - \,\,\sum\limits_{i \in N}^{} {\beta _{i}^{{t + 1}}\left\{ {{{{\left[ {{{\varepsilon }_{i}} - \frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} \right]}}^{2}} - {{{\left[ {{{\varepsilon }_{i}} - \frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} \right]}}^{2}}} \right\}} \\ \end{gathered} $$

$$\begin{gathered} = 1 - \sum\limits_{i \in N}^{} {{{\varepsilon }_{i}}\vec {\gamma }_{i}^{{t + 1}}} \left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right) + {{\left( {\frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}} \right)}^{2}}\sum\limits_{i \in N}^{} {{{{\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)}}^{2}}} \\ - \,\,\sum\limits_{i \in N}^{} {\beta _{i}^{{t + 1}}\left[ {2{{\varepsilon }_{i}} - \left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2} + \frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}} \right)\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} \right]} \left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2} - \frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}} \right)\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right) \\ = 1 - \sum\limits_{i \in N}^{} {2{{\varepsilon }_{i}}} \left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)\left[ {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2} + \beta _{i}^{{t + 1}}\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2} - \frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}} \right)} \right] \\ \end{gathered} $$

$$\begin{gathered} + \,\,\sum\limits_{i \in N}^{} {{{{\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)}}^{2}}\left[ {{{{\left( {\frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}} \right)}}^{2}} + \beta _{i}^{{t + 1}}\left( {{{{\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}}^{2}} - {{{\left( {\frac{{\vec {\gamma }_{i}^{{t + 1}}}}{2}} \right)}}^{2}}} \right)} \right]} \\ = 1 - \sum\limits_{i \in N}^{} {2{{\varepsilon }_{i}}\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} \mu _{i}^{{t + 1}} + {{\sum\limits_{i \in N}^{} {\left( {{{\varepsilon }_{i}} + \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)} }^{2}}\eta _{i}^{{t + 1}} \\ = 1 - 2\sum\limits_{i \in N}^{} {{{{({{\varepsilon }_{i}})}}^{2}}\mu _{i}^{{t + 1}}} - \sum\limits_{i \in N}^{} {2{{\varepsilon }_{i}}\mu _{i}^{{t + 1}}} \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}} + \sum\limits_{i \in N}^{} {{{{({{\varepsilon }_{i}})}}^{2}}\eta _{i}^{{t + 1}}} } + \sum\limits_{i \in N}^{} {2{{\varepsilon }_{i}}\eta _{i}^{{t + 1}}} \sum\limits_{j \in N}^{} {{{\varepsilon }_{j}} + \sum\limits_{i \in N}^{} {{{{\left( {\sum\limits_{j \in N}^{} {{{\varepsilon }_{j}}} } \right)}}^{2}}\eta _{i}^{{t + 1}}} } \\ \end{gathered} $$

$$\begin{gathered} = 1 - \sum\limits_{i \in N}^{} {{{{({{\varepsilon }_{i}})}}^{2}}} \left( {4\mu _{i}^{{t + 1}} - 3\eta _{i}^{{t + 1}} - \sum\limits_{k \in N}^{} {\eta _{k}^{{t + 1}}} } \right) \\ - \,\,\sum\limits_{i \in N}^{} {\sum\limits_{k \in N{{\backslash }}\{ i\} }^{} {{{\varepsilon }_{i}}{{\varepsilon }_{j}}\left[ {\left( {\mu _{i}^{{t + 1}} - \eta _{i}^{{t + 1}}} \right) + \left( {\mu _{j}^{{t + 1}} - \eta _{j}^{{t + 1}}} \right) - \sum\limits_{k \in N}^{} {\eta _{k}^{{t + 1}}} } \right]} } . \\ \end{gathered} $$

If the principal minors of the matrix (15) have positive determinants, then the quadratic form

$$\sum\limits_{i \in N}^{} {{{{({{\varepsilon }_{i}})}}^{2}}\left( {4\mu _{i}^{{t + 1}} - 3\eta _{i}^{{t + 1}} - \sum\limits_{k \in N}^{} {\eta _{k}^{{t + 1}}} } \right)} + \sum\limits_{i \in N}^{} {\sum\limits_{i \in N{{\backslash }}\{ i\} }^{} {{{\varepsilon }_{i}}{{\varepsilon }_{j}}\left[ {(\mu _{i}^{{t + 1}} - \eta _{i}^{{t + 1}}) + (\mu _{j}^{{t + 1}} - \eta _{j}^{{t + 1}}) - \sum\limits_{k \in N}^{} {\eta _{k}^{{t + 1}}} } \right]} } $$

is positive definite and ||B ^t+1|| < 1 due to (7) and (10). In other words, the process (4), (5) converges for the parameter values $\vec {\gamma }_{i}^{{t + 1}}$, $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$, $\beta _{i}^{{t + 1}}$.

The proof of Proposition 3 is complete.

Proof of Lemma 2. We take advantage of the following properties of determinants.

(1) The determinant of a matrix does not change when multiplying any row by an arbitrary number and adding the result to any other row.

(2) The determinant of a triangular square matrix is equal to the product of its diagonal elements. Adding to each row the next row multiplied by (–1), we obtain

$$\det (A) = \left| {\begin{array}{*{20}{c}} a&b&b& \cdots &b \\ b&a&b& \cdots &b \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ b&b&b& \cdots &a \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {a - b}&{b - a}&0& \cdots &0 \\ b&a&b& \cdots &b \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ b&b&b& \cdots &a \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {a - b}&{b - a}&0& \cdots &0 \\ 0&{a - b}&{b - a}& \cdots &0 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ b&b&b& \cdots &a \end{array}} \right|.$$

Using the row expansion for the last row and the triangular determinant formula, we finally arrive at det(A) = a(a – b)^m–1 + (m – 1)b(a – b)^m–1 = (a – b)^m–1 [a + (m – 1)b]. The proof of Lemma 2 is complete.

Proof of Lemma 3. We have $\gamma _{i}^{{t + 1}}$ = $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$; in view of (15), for $\beta _{i}^{{t + 1}}$ = 1,

$$\mu _{i}^{{t + 1}} = \frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2},\quad \eta _{i}^{{t + 1}} = {{\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}^{2}},$$

$$f_{{ii}}^{{t + 1}} = 2\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}} - (3 + n){{\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}^{2}},\quad f_{{ij}}^{{t + 1}} = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}} - (2 + n){{\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}^{2}},\quad i,j \in N,\quad i \ne j.$$

According to Lemma 2, det(F ^t+1) = (1 + n)${{\left( {\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}^{n}}{{\left( {2 - \frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right)}^{{n - 1}}}\left[ {2 - (1 + n)\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}}}{2}} \right]$. The determinant is positive for $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$ < $\frac{4}{{1 + n}}$. Under this inequality, the kth principal minor of the matrix F  ^{t + 1} (k < n) has a positive determinant as well. Therefore, a + (k – 1)b = $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$[2(1 + k) – (1 + 2k + nk)$\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$/2]/2.

For $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$ = $\frac{4}{{1 + n}}$, it follows that a + (k – 1)b = $\frac{{4(n - k)}}{{{{{(1 + n)}}^{2}}}}$ > 0. For $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma } _{i}^{{t + 1}}$ < $\frac{4}{{1 + n}}$, the positivity of the determinant of the kth principal minor is obvious.

The proof of Lemma 3 is complete.

Proof of Proposition 4. When restricting the step ranges, it is desirable to have their right bounds as close to 1 as possible. Let all agents have the same step range at all time instants. Therefore, we omit the superscript (t + 1) for f and F.

Based on Lemma 2, for the Cournot duopoly, the right bound of $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma }$ of the range should be taken equal to 1 and  f(1, 1) < 1.

Consider  f($\vec {\gamma }$, 1). The matrix (15) corresponding to this quadratic form is defined for β₁ = 0 and β₂ = 1: F = $\left( {\begin{array}{*{20}{c}} {2\vec {\gamma } - {{{\vec {\gamma }}}^{2}} - 0.25}&{\vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2} \\ {\vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{1 - {{{\vec {\gamma }}}^{2}}{\text{/}}4} \end{array}} \right)$. For the convenience of calculations, the determinant of this matrix can be simplified to det(F) = $\left| {\begin{array}{*{20}{c}} {3.75}&{\vec {\gamma }{\text{/}}2 - 2} \\ {\vec {\gamma }{\text{/}}2 - 2}&{1 - {{{\vec {\gamma }}}^{2}}{\text{/}}4} \end{array}} \right|$. The matrix is positive definite if $\vec {\gamma }$ ≥ 0.136. Therefore,  f(0.136, 1) < 1 $\forall$ε₁, ε₂.

Naturally,  f(0.136, 1) = f(1, 0.136) < 1.

Consider  f($\vec {\gamma }$, $\vec {\gamma }$). The matrix (15) corresponding to this quadratic form is defined for β₁ = β₂ = 0: F = $\left( {\begin{array}{*{20}{c}} {2\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{\vec {\gamma } - {{{\vec {\gamma }}}^{2}}} \\ {\vec {\gamma } - {{{\vec {\gamma }}}^{2}}}&{2\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4} \end{array}} \right)$. The matrix is positive definite if  $\vec {\gamma }$ ≠ 0. Therefore,  f(0.136, 0.136) < 1 $\forall$ε₁, ε₂.

The desired result follows from (11), and the proof of Proposition 4 is complete.

Proof of Proposition 5.

Consider  f(1, 1, $\vec {\gamma }$). The matrix (15) corresponding to this quadratic form is defined for β₁ = β₂ = 1 and β₃ = 0: F = $\left( {\begin{array}{*{20}{c}} {0.75 - {{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - {{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2} \\ { - {{{\vec {\gamma }}}^{2}}{\text{/}}4}&{0.75 - {{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2} \\ { - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{ - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{ - 0.5 + 2\vec {\gamma } - {{{\vec {\gamma }}}^{2}}} \end{array}} \right)$. For the convenience of calculations, the determinant of this matrix can be simplified to det(F) = $\left| {\begin{array}{*{20}{c}} {0.75}&{ - {{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2} \\ 0&{0.75 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{ - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}} \\ 0&{ - 0.25 + \vec {\gamma }{\text{/}}2 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{ - 0.5 + 2\vec {\gamma } - {{{\vec {\gamma }}}^{2}}} \end{array}} \right|$.

The minimum value $\vec {\gamma }$ under which the matrix F remains positive definite is 0.334. Hence,  f(1, 1, 0.334) < 1 $\forall$ε₁, ε₂.

In addition, we have f(1, 1, 0.334) = f(1, 0.334, 1) = f(0.334, 1, 1) < 1 $\forall$ε₁, ε₂.

Consider  f(1, $\vec {\gamma }$, $\vec {\gamma }$). The matrix (15) corresponding to this quadratic form is defined for β₁ = 1 and β₂ = β₃ = 0: F = $\left( {\begin{array}{*{20}{c}} {1 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{\vec {\gamma }{\text{/}}2 - 3{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{\vec {\gamma }{\text{/}}2 - 3{{{\vec {\gamma }}}^{2}}{\text{/}}4} \\ {\vec {\gamma }{\text{/}}2 - 3{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + 2\vec {\gamma } - 5{\kern 1pt} {\kern 1pt} {{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + \vec {\gamma } - {{{\vec {\gamma }}}^{2}}} \\ {\vec {\gamma }{\text{/}}2 - 3{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + \vec {\gamma } - {{{\vec {\gamma }}}^{2}}}&{ - 0.25 + 2\vec {\gamma } - 5{\kern 1pt} {\kern 1pt} {{{\vec {\gamma }}}^{2}}{\text{/}}4} \end{array}} \right)$. For the convenience of calculations, the determinant of this matrix can be simplified to det(F) = $\left\| {\begin{array}{*{20}{c}} {1 - {{{\vec {\gamma }}}^{2}}{\text{/}}2}&{\vec {\gamma } - 3{{{\vec {\gamma }}}^{2}}{\text{/}}2}&{1 - 3{{{\vec {\gamma }}}^{2}}{\text{/}}4} \\ {\vec {\gamma }{\text{/}}2 - 3{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.5 + 3\vec {\gamma } - 9{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{ - 0.25 + \vec {\gamma } - {{{\vec {\gamma }}}^{2}}} \\ 0&0&{\vec {\gamma } - {{{\vec {\gamma }}}^{2}}{\text{/}}4} \end{array}} \right\|$.

The minimum value $\vec {\gamma }$ under which the matrix F remains positive definite is 0.22. Hence,  f(1, 0.22, 0.22) < 1 $\forall$ε₁, ε₂.

Also,  f(0.22, 1, 0.22) = f(0.22, 0.22, 1) = f(1, 0.22, 0.22) < 1 $\forall$ε₁, ε₂.

Consider f($\vec {\gamma }$, $\vec {\gamma }$, $\vec {\gamma }$). The matrix (15) corresponding to this quadratic form is defined for β₁ = β₂ = β₃ = 0 : F = $\left( {\begin{array}{*{20}{c}} {2\vec {\gamma } - 3{{{\vec {\gamma }}}^{2}}{\text{/4}}}&{\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4} \\ {\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{2\vec {\gamma } - 3{{{\vec {\gamma }}}^{2}}{\text{/}}\vec {\gamma }}&{\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4} \\ {\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{\vec {\gamma } - 5{{{\vec {\gamma }}}^{2}}{\text{/}}4}&{2\vec {\gamma } - 3{{{\vec {\gamma }}}^{2}}{\text{/}}\vec {\gamma }} \end{array}} \right)$ .

By Lemma 2, det(F) = ${{\vec {\gamma }}^{3}}$(2 – $\vec {\gamma }$/2)²(1 – $\vec {\gamma }$). The matrix is positive definite if $\vec {\gamma }$ ≠ 0.1. From these lower bounds we choose the maximum value, i.e., 0.334.

Based on Lemma 2, for the Cournot oligopoly model with three agents, the right bound $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\gamma }$ of the step range should be chosen less than 1.

The desired result follows from (11), and the proof of Proposition 5 is complete.