Reflexion Processes and Equilibrium in an Oligopoly Model with a Leader

Abstract

An oligopoly model with a leader in the class of linear demand and cost functions of agents is considered, and the dynamic processes of reflexive behavior in this model are analytically studied. Dynamic decision-making processes with the inaccurate beliefs of agents about the choice of competitors are implemented not through the optimal responses to their expected actions, but as repeated static games on a range of admissible responses. Such an approach to decision-making is demonstrated to be justified. Observing the current state of the market and considering current economic restrictions (competitiveness and profit), agents refine their outputs in game-to-game dynamics and take steps towards the current position of their goal. A Stackelberg leader and other agents with the Cournot response choose step sizes independently of each other. Sufficient conditions on the step sizes under which the dynamics converge to an equilibrium are established.

APPENDIX

Proof of Proposition 1. For the agents with the Cournot response, from (5), (11), and (12) it follows that

$${q}_{i}^{t+1}=\frac{1}{2}\left({h}_{i}-{Q}^{t}+{q}_{i}^{t}\right).$$

(A.1)

The difference for the two successive time instants has the form

$${q}_{i}^{t+1}-{q}_{i}^{t}=\frac{1}{2}\left({q}_{i}^{t}-{q}_{i}^{t-1}\right)-\frac{1}{2}\left({Q}^{t}-{Q}^{t-1}\right).$$

Summing over i = 2, …, n, arrive at the equality

$${Q}_{-1}^{t+1}-{Q}_{-1}^{t}=-\frac{n-2}{2}\left({Q}_{-1}^{t}-{Q}_{-1}^{t-1}\right)-\frac{n-1}{2}\left({q}_{1}^{t}-{q}_{1}^{t-1}\right).$$

(A.2)

For the Stackelberg leader, from (10) and (12) it follows that

$${q}_{1}^{t+1}=\frac{n}{1+n}\left({h}_{1}-{Q}_{-1}^{t}\right).$$

(A.3)

Then (A.3) implies the equalities

$${Q}_{-1}^{t}-{Q}_{-1}^{t-1}=-\frac{1+n}{2}({q}_{1}^{t+1}-{q}_{1}^{t}),\quad {Q}_{-1}^{t+1}-{Q}_{-1}^{t}=-\frac{1+n}{2}({q}_{1}^{t+2}-{q}_{1}^{t+1}).$$

(A.4)

Substituting these expressions into (A.2) gives

$${q}_{1}^{t+2}-{q}_{1}^{t+1}=\frac{2-n}{2}\left({q}_{1}^{t+1}-{q}_{1}^{t}\right)+\frac{n(n-1)}{2(1+n)}\left({q}_{1}^{t}-{q}_{1}^{t-1}\right).$$

(A.5)

Introduce the notations

$$p=\frac{2-n}{2},$$

(A.6)

$$g=\frac{n(n-1)}{2(1+n)},$$

(A.7)

$${y}_{t}={q}_{1}^{t+1}-{q}_{1}^{t}.$$

(A.8)

Hence, (A.5) is a linear recursive sequence with constant coefficients and initial values ${y}_{0}={q}_{1}^{1}-{q}_{1}^{0}$, ${y}_{1}={q}_{1}^{2}-{q}_{1}^{1}$, that has the form

$${y}_{t+2}=p{y}_{t+1}+g{y}_{t}\quad (t\ge 0).$$

(A.9)

As is known [22], a linear recursive sequence with constant coefficients satisfies the relation

$${y}_{t}={s}_{1}{({x}_{1})}^{t}+{s}_{2}{({x}_{2})}^{t}.$$

(A.10)

The subscript t in (A.10), indicating time instant, is the exponent for the roots x₁ and x₂ of the characteristic equation x² − px − g = 0. The values s₁ and s₂ are obtained by solving the system of linear equations

$$\left\{\begin{array}{l}{y}_{0}={s}_{1}{x}_{1}+{s}_{2}{x}_{2}\\ {y}_{1}={s}_{1}{\left({x}_{1}\right)}^{2}+{s}_{2}{\left({x}_{2}\right)}^{2}.\end{array}\right.$$

(A.11)

These roots are

$${x}_{1}=\frac{p}{2}+\sqrt{\frac{{p}^{2}}{4}+g},\quad {x}_{2}=\frac{p}{2}-\sqrt{\frac{{p}^{2}}{4}+g}.$$

(A.12)

Lemma A.1.

The roots x₁ and x₂ of the characteristic equation x² − px − g = 0 have the following properties: a) differ from one another; b) have real values; c) are not identically equal to 0; d) have different signs; e) for $n=2,{x}_{1}=\frac{1}{\sqrt{3}}$ (the greater root) and ${x}_{2}=-\frac{1}{\sqrt{3}}$ (the smaller root); f) for n ≥ 3, the negative root is greater by magnitude and its magnitude exceeds 1; g) the positive root is smaller than 1.

Proof of Lemma A.1. The radicand $\frac{{(2-n)}^{2}}{16}+\frac{n(n-1)}{2(1+n)}$ in (A.12) is positive, i.e., the equation has simple real roots. Assume on the contrary that there exist zero roots. In this case, from x² − px − g = 0 it follows that g = 0. However, by (A.7) this situation is possible only for n = 1. Properties a)–c) are proved.

Property d) is immediate due to ${x}_{1}{x}_{2}=-g=-\frac{n(n-1)}{2(1+n)}$. For n = 2, $x=\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right);$ if n = 3, then $x=\left(\frac{\sqrt{11}-1}{4},\frac{-\sqrt{11}-1}{4}\right)$. For n ≥ 4, the inequality x₁x₂ < − 1 obviously holds. Therefore, properties e) and f) are true. Now establish property g). By (A.12), ${x}_{1}=\frac{2-n}{4}+\sqrt{\frac{{(2-n)}^{2}}{16}+\frac{n(n-1)}{2(1+n)}},$ and x₁ < 1 since $\sqrt{\frac{{(2-n)}^{2}}{16}+\frac{n(n-1)}{2(1+n)}}<1-\frac{2-n}{4}$ or $\frac{n(n-1)}{2(1+n)}<1-\frac{2-n}{2}=\frac{n}{2}.$

The proof of Lemma A.1 is complete.

Lemma A.2.

If the greater simple root x₁ of the characteristic equation of the sequence y_t+2 = py_t+1 + gy_t is $\frac{{y}_{1}}{{y}_{0}}({y}_{0}\ne 0),$ then the sequence converges to 0.

Proof of Lemma A.2. Let ${x}_{1}=\frac{{y}_{1}}{{y}_{0}}({y}_{0}\ne 0)$. In accordance with Lemma 1, x₁ ≠ 0. From the system of Eqs. (A.11) it follows that s₂x₂(x₁ − x₂) = 0. Because the roots are different and x₂ ≠ 0, s₂ = 0. Then, by (A.10), ${y}_{t}={s}_{1}{({x}_{1})}^{t}$. Due to Lemma (A.1), x₁ < 1, and hence $\left\{{y}_{t}\right\}$ converges to 0.

The proof of Lemma A.2 is complete.

The superscript (s) below indicates the parameter values in the static Stackelberg equilibrium of the basic model (1), (2).

Lemma A.3.

If ${y}_{t}={q}_{1}^{t+1}-{q}_{1}^{t}\to 0$ as t → ∞, then: a) ${q}_{1}^{t}\to {q}_{1}^{(s)},$ b) ${q}_{i}^{t}\to {q}_{i}^{(s)}\forall i\in N\backslash \left\{1\right\}$.

Proof of Lemma A.3. From (A.4) it follows that ${Q}_{-1}^{t}-{Q}_{-1}^{t-1}\to 0$. Next, show that ${q}_{1}^{t}\to {q}_{1}^{(s)}$. Transform (A.1) to ${q}_{i}^{t+1}-{q}_{i}^{t}=\frac{1}{2}\left({h}_{i}-{Q}^{t}-{q}_{i}^{t}\right)$. Summing this expression over i = 2, …, n yields

$${Q}_{-1}^{t+1}-{Q}_{-1}^{t}=\frac{1}{2}\left(\mathop{\sum }\limits_{i=2}^{n}{h}_{i}-n{Q}^{t}+{q}_{1}^{t}\right)=\frac{1}{2}\left(n{h}_{1}-n{Q}^{t}-{q}_{1}^{t}+\mathop{\sum }\limits_{i=2}^{n}{h}_{i}-n{h}_{1}+2{q}_{1}^{t}\right).$$

By (A.3), ${q}_{1}^{t+1}-{q}_{1}^{t}=\frac{1}{n+1}(n{h}_{1}-n{Q}^{t}-{q}_{1}^{t})$. Hence, $n{h}_{1}-n{Q}^{t}-{q}_{1}^{t}\to 0$ and ${q}_{1}^{t}\to \frac{1}{2}\left(n{h}_{1}-\mathop{\sum }\limits_{i=2}^{n}{h}_{i}\right)={q}_{1}^{(s)}$. Thus, ${q}_{1}^{t}\to {q}_{1}^{(s)}$ and ${Q}^{t}\to {h}_{1}-\frac{{q}_{1}^{(s)}}{n}={Q}^{(s)}$. Item a) is established.

From (A.1) and Q^t → Q^(s) it follows that ${q}_{i}^{t+1}-\frac{1}{2}{q}_{i}^{t}=\frac{1}{2}({h}_{i}-{Q}^{t})$ and ${q}_{i}^{t+1}-\frac{1}{2}{q}_{i}^{t}\to \frac{1}{2}\left({h}_{i}-{Q}^{(s)}\right)$. Therefore, the sequences $\left\{\frac{1}{2}{q}_{i}^{t}-\frac{1}{4}{q}_{i}^{t-1}\right\}$ are convergent, and the sequences $\left\{{q}_{i}^{t+1}-\frac{1}{4}{q}_{i}^{t-1}\right\}$ as the sums of convergent sequences are also convergent. The convergence of $\left\{\frac{1}{4}{q}_{i}^{t-1}-\frac{1}{8}{q}_{i}^{t-2}\right\}$ and $\left\{{q}_{i}^{t+1}-\frac{1}{8}{q}_{i}^{t-2}\right\},$ etc., can be established by analogy. In final analysis, the sequences $\left\{{q}_{i}^{t+1}-\frac{1}{{2}^{t+1}}{q}_{i}^{0}\right\}$ and $\left\{{q}_{i}^{t+1}\right\}$ are also convergent. Then from (A.1) it follows that ${q}_{i}^{t}\to {q}_{i}^{(s)}={h}_{i}-{Q}^{(s)}(i\in N\backslash \left\{1\right\})$.

The proof of Lemma A.3 is complete.

Now get back to the proof of Proposition 1.

For n = 2, from (A.10) and item e) of Lemma A.1 it follows that ${y}_{t}={q}_{1}^{t+1}-{q}_{1}^{t}\to 0$. Then by Lemma A.3, ${q}_{1}^{t}\to {q}_{1}^{(s)}$ and ${q}_{2}^{t}\to {q}_{2}^{(s)}$. In the case n ≥ 3 with ${x}_{1}=\frac{{y}_{1}}{{y}_{0}}$ and ${y}_{0}\ne 0,{y}_{t}={q}_{1}^{t+1}-{q}_{1}^{t}\to 0$ by Lemma A.2 and ${q}_{i}^{t}\to {q}_{i}^{(s)}\forall i\in N$ by Lemma A.3. Items a) and b) of Proposition 1 are established.

In the case n ≥ 3 with ${x}_{1}\ne \frac{{y}_{1}}{{y}_{0}}$, s₂ ≠ 0 in (A.10). In accordance with Lemma A.1, 0 < x₁ < 1 and x₂ < − 1. Therefore, the sequence ${y}_{t}={q}_{1}^{t+1}-{q}_{1}^{t}$ does not converge to 0.

The proof of Proposition 1 is complete.

Remark. If y₀ = 0 and s₂ = 0, from the system of Eqs. (A.11) it follows that s₁ = 0 and y_t ≡ 0. The process will stand still, without any convergence.

Proof of Proposition 2. Introduce the indicator functions [11] that characterize the deviations of the current outputs from the current optimums, ${\alpha }_{i}^{t}=2({x}_{i}^{t}-{q}_{i}^{t})$ for the agents with the Cournot response and ${\alpha }_{1}^{t}=\frac{1+n}{n}({x}_{1}^{t}-{q}_{1}^{t})$ for the Stackelberg leader. Here the coefficients 2 and $\frac{1+n}{n}$ are introduced for convenience. Using (7) and also the relations ${h}_{i}={Q}^{(s)}+{q}_{i}^{(s)}(i\in N\backslash \left\{1\right\})$ and ${h}_{1}={Q}^{(s)}+\frac{{q}_{1}^{(s)}}{n}$ (which are due to (10) and (11)), write

$${\alpha }_{i}^{t}={Q}^{(s)}+{q}_{i}^{(s)}-{Q}^{t}-{q}_{i}^{t},\quad i\in N\backslash \left\{1\right\},$$

(A.13)

$${\alpha }_{1}^{t}={Q}^{(s)}+\frac{1}{n}{q}_{1}^{(s)}-{Q}^{t}-\frac{1}{n}{q}_{1}^{t}.$$

(A.14)

The solution of the homogeneous system of Eqs. (A.13), (A.14) is an equilibrium output ${q}_{i}^{t}={q}_{i}^{(s)}(i\in N)$. As it will be demonstrated below, besides the indicator functions, the analysis and proof of the convergence of process (10), (11), (13) involves the expression $\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}$.

Introduce the notations ${N}_{1}^{t}=\left\{i\left|{x}_{i}^{t}>0,i\in N\right.\right\}$ and ${N}_{2}^{t}=\left\{\left.i\right|{x}_{i}^{t}\le 0,i\in N\right\}$. Then ${N}_{1}^{t}\bigcap {N}_{2}^{t}=\otimes$ and ${N}_{1}^{t}\bigcup {N}_{2}^{t}=N$.

In view of these notations and (7), (10), (11), (A.13), and (A.14), write (14) as

$${q}_{i}^{t+1}=\left\{\begin{array}{ll}{q}_{i}^{t}+\frac{{\gamma }_{i}^{t+1}}{2}{\alpha }_{i}^{t},&i\in {N}_{1}^{t}\backslash \left\{1\right\},\quad {\gamma }_{i}^{t+1}\in \left[0,1\right]\\ 0,&i\in {N}_{2}^{t}\backslash \left\{1\right\};\end{array}\right.$$

(A.15)

$${q}_{1}^{t+1}=\left\{\begin{array}{ll}{q}_{1}^{t}+\frac{{\gamma }_{1}^{t+1}n}{1+n}{\alpha }_{1}^{t},&1\in {N}_{1}^{t},\quad {\gamma }_{1}^{t+1}\in \left[0,\ 1\right]\\ 0,&1\in {N}_{2}^{t}.\end{array}\right.$$

(A.16)

For the convenience of further transformations, redefine the parameters γ in the following way: ${\lambda }_{i}^{t+1}={\gamma }_{i}^{t+1}$, $i\in N\backslash \left\{1\right\}$; ${\lambda }_{1}^{t+1}=\frac{{\gamma }_{1}^{t+1}2n}{1+n}(t=0,1,2,\ldots )$. Then:

$${Q}^{t+1}={Q}^{t}+\sum _{j\in {N}_{1}^{t}}\frac{{\lambda }_{j}^{t+1}}{2}{\alpha }_{j}^{t}-\sum _{j\in {N}_{2}^{t}}{q}_{j}^{t};$$

(A.17)

$$\begin{array}{c}{Q}^{(s)}+{q}_{i}^{(s)}-{Q}^{t+1}-{q}_{i}^{t+1}\\ ={Q}^{(s)}+{q}_{i}^{(s)}-{Q}^{t}-{q}_{i}^{t}-\frac{{\lambda }_{i}^{t+1}}{2}{\alpha }_{i}^{t}-{Q}^{t+1}+{Q}^{t},\quad i\in {N}_{1}^{t}\backslash \left\{1\right\};\\ {Q}^{(s)}+\frac{1}{n}{q}_{1}^{(s)}-{Q}^{t+1}-\frac{1}{n}{q}_{1}^{t+1}={Q}^{(s)}+\frac{1}{n}{q}_{1}^{(s)}-{Q}^{t}-\frac{1}{n}{q}_{1}^{t}-\frac{1}{n}\frac{{\lambda }_{1}^{t+1}}{2}{\alpha }_{1}^{t}-{Q}^{t+1}+{Q}^{t};\\ {\alpha }_{i}^{t+1}=\left(1-\frac{{\lambda }_{i}^{t+1}}{2}\right){\alpha }_{i}^{t}-{Q}^{t+1}+{Q}^{t}\\ \end{array}$$

(A.18)

$$\begin{array}{c}=\left(1-\frac{{\lambda }_{i}^{t+1}}{2}\right){\alpha }_{i}^{t}-{\displaystyle {\sum_{j\in {N}_{1}^{t}}}}\frac{{\lambda }_{j}^{t+1}}{2}{\alpha }_{j}^{t}+{\displaystyle \sum _{j\in {N}_{2}^{t}}}{q}_{j}^{t},\quad i\in {N}_{1}^{t}\ \backslash \left\{1\right\};\\ {\alpha }_{1}^{t+1}=\left(1-\frac{{\lambda }_{1}^{t+1}}{2n}\right){\alpha }_{1}^{t}-{Q}^{t+1}+{Q}^{t}\end{array}$$

(A.19)

$$\begin{array}{c}=\left(1-\frac{{\lambda }_{1}^{t+1}}{2n}\right){\alpha }_{1}^{t}-{\displaystyle \sum _{j\in {N}_{1}^{t}}}\frac{{\lambda }_{j}^{t+1}}{2}{\alpha }_{j}^{t}+{\displaystyle \sum _{j\in {N}_{2}^{t}}{q}_{j}^{t}},\quad 1\in {N}_{1}^{t};\\ {Q}^{(s)}+{q}_{i}^{(s)}-{Q}^{t+1}-{q}_{i}^{t+1}={Q}^{(s)}+{q}_{i}^{(s)}-{Q}^{t}-{q}_{i}^{t}+{q}_{i}^{t}-{Q}^{t+1}+{Q}^{t},\quad i\in {N}_{2}^{t}\backslash \left\{1\right\};\\ {Q}^{(s)}+\frac{1}{n}{q}_{1}^{(s)}-{Q}^{t+1}-\frac{1}{n}{q}_{1}^{t+1}={Q}^{(s)}+\frac{1}{n}{q}_{1}^{(s)}-{Q}^{t}-\frac{1}{n}{q}_{1}^{t}+\frac{1}{n}{q}_{1}^{t}-{Q}^{t+1}+{Q}^{t};\\ {\alpha }_{i}^{t+1}={\alpha }_{i}^{t}+{q}_{i}^{t}-{Q}^{t+1}+{Q}^{t}={\alpha }_{i}^{t}+{q}_{i}^{t}-{\displaystyle \sum _{j\in {N}_{1}^{t}}}\frac{{\lambda }_{j}^{t+1}}{2}\ {\alpha }_{j}^{t}+{\displaystyle \sum _{j\in {N}_{2}^{t}}}{q}_{j}^{t},\quad i\in {N}_{2}^{t}\ \backslash \left\{1\right\};\end{array}$$

(A.20)

$${\alpha }_{1}^{t+1}={\alpha }_{1}^{t}+\frac{1}{n}{q}_{1}^{t}-{Q}^{t+1}+{Q}^{t}={\alpha }_{1}^{t}+\frac{1}{n}{q}_{1}^{t}-{\displaystyle \sum _{j\in {N}_{1}^{t}}}\frac{{\lambda }_{j}^{t+1}}{2}\ {\alpha }_{j}^{t}+{\displaystyle \sum _{j\in {N}_{2}^{t}}}{q}_{j}^{t},\quad 1\in {N}_{2}^{t}.$$

(A.21)

Lemma A.4.

If the sequence $\left\{{\alpha }_{i}^{t},\ i\in N\right\}$ for process (10), (11), (13) contains not only positive terms, then

$$\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}<\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}.$$

Proof of Lemma A.4. There are 4 possible cases for agents i and j, $\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\},$ as follows: 1) $i,j\in {N}_{1}^{t}$; 2) $i\in {N}_{1}^{t}$ and $j\in {N}_{2}^{t}$; 3) $i\in {N}_{2}^{t}$ and $j\in {N}_{1}^{t};$ 4) $i,j\in {N}_{2}^{t}$.

Consider the first case, in which $i,j\in {N}_{1}^{t}$. For the sake of definiteness, let $i,j\in {N}_{1}^{t}\backslash \left\{1\right\}$. If $i\in {N}_{1}^{t}\backslash \left\{1\right\}$ and j = 1, or if i = 1 and $j\in {N}_{1}^{t}\backslash \left\{1\right\},$ the proof is similar. Denoting ${\alpha }_{{M}_{1}^{t+1}}^{t+1}=\mathop{\max }\limits_{i}\left\{{\alpha }_{i}^{t+1},i\in {N}_{1}^{t}\right\}$ and ${\alpha }_{{m}_{1}^{t+1}}^{t+1}=\mathop{\min }\limits_{i}\left\{{\alpha }_{i}^{t+1},i\in {N}_{1}^{t}\right\},$ by (A.18) write

$${\alpha }_{{M}_{1}^{t+1}}^{t+1}-{\alpha }_{{m}_{1}^{t+1}}^{t+1}=\left(1-\frac{{\lambda }_{{M}_{1}^{t+1}}^{t+1}}{2}\right){\alpha }_{{M}_{1}^{t+1}}^{t}-\left(1-\frac{{\lambda }_{{m}_{1}^{t+1}}^{t+1}}{2}\right){\alpha }_{{m}_{1}^{t+1}}^{t}.$$

But ${\alpha }_{{M}_{1}^{t+1}}^{t}\le {\alpha }_{{M}^{t}}^{t}>0$ and ${\alpha }_{{m}_{1}^{t+1}}^{t}\ge {\alpha }_{{m}^{t}}^{t}\le 0$. Therefore,

$$\mathop{\max }\limits_{i,j\in {N}_{1}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}<{\alpha }_{{M}_{1}^{t+1}}^{t}-{\alpha }_{{m}^{t}}^{t}<{\alpha }_{{M}^{t}}^{t}-{\alpha }_{{m}^{t}}^{t}=\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}.$$

Consider the second case in which $i\in {N}_{1}^{t}$, $j\in {N}_{2}^{t}$. For the sake of definiteness, let $i\in {N}_{1}^{t}\backslash \left\{1\right\}$ and j = 1. If i = 1 and $j\in {N}_{2}^{t}\backslash \left\{1\right\},$ or if $i\in {N}_{1}^{t}\backslash \left\{1\right\}$ and $j\in {N}_{2}^{t}\backslash \left\{1\right\},$ the proof is similar. Due to (A.18) and (A.21),

$${\alpha }_{{M}_{1}^{t+1}}^{t+1}-{\alpha }_{{m}_{2}^{t+1}}^{t+1}=\left(1-\frac{{\lambda }_{{M}_{1}^{t+1}}^{t+1}}{2}\right){\alpha }_{{M}_{1}^{t+1}}^{t}-\left({\alpha }_{{m}_{2}^{t+1}}^{t}+\frac{1}{n}{q}_{{m}_{2}^{t+1}}^{t}\right).$$

Here ${\alpha }_{{m}_{2}^{t+1}}^{t+1}=\mathop{\min }\limits_{i}\left\{{\alpha }_{i}^{t+1},i\in {N}_{2}^{t}\right\}$, ${m}_{2}^{t+1}=1$ and ${\alpha }_{{M}_{1}^{t+1}}^{t}\le {\alpha }_{{M}^{t}}^{t}>0$. Then

$$\mathop{\max }\limits_{i\in {N}_{1}^{t},j\in {N}_{2}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}<{\alpha }_{{M}^{t}}^{t}-{\alpha }_{{m}_{2}^{t+1}}^{t}\le {\alpha }_{{M}^{t}}^{t}-{\alpha }_{{m}^{t}}^{t}=\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}.$$

Consider the case $i\in {N}_{2}^{t}$, $j\in {N}_{1}^{t}$. For the sake of definiteness, let i = 1 and $j\in {N}_{1}^{t}\backslash \left\{1\right\}$. If $i\in {N}_{2}^{t}\backslash \left\{1\right\}$ and j = 1, or if $i\in {N}_{2}^{t}\backslash \left\{1\right\}$ and $j\in {N}_{1}^{t}\backslash \left\{1\right\},$ the proof is similar. Then ${M}_{2}^{t+1}=1.$ Due to (10), ${x}_{1}^{t}\le 0$, (A.14), and ${h}_{1}={Q}^{(s)}+\frac{{q}_{1}^{(s)}}{n},$ it follows that ${\alpha }_{1}^{t}+\frac{1}{n}{q}_{1}^{t}={h}_{1}-{Q}_{-1}^{t}-{q}_{1}^{t}<0<{\alpha }_{{M}^{t}}^{t}$. Consequently, by (A.21) and (A.18),

$$\begin{array}{l}{\alpha }_{{M}_{2}^{t+1}}^{t+1}-{\alpha }_{{m}_{1}^{t+1}}^{t+1}=\left({\alpha }_{{M}_{2}^{t+1}}^{t}+\frac{1}{n}{q}_{{M}_{2}^{t+1}}^{t}\right)-\left(1-\frac{{\lambda }_{{m}_{1}^{t+1}}^{t+1}}{2}\right){\alpha }_{{m}_{1}^{t+1}}^{t}\\ <{\alpha }_{{M}^{t}}^{t}-\left(1-\frac{{\lambda }_{{m}_{1}^{t+1}}^{t+1}}{2}\right){\alpha }_{{m}^{t}}^{t}<{\alpha }_{{M}^{t}}^{t}-{\alpha }_{{m}^{t}}^{t}=\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}.\end{array}$$

Now consider the case $i,j\in {N}_{2}^{t}$. For the sake of definiteness, let i = 1 and $j\in {N}_{2}^{t}\backslash \left\{1\right\}$. If $i\in {N}_{2}^{t}\backslash \left\{1\right\}$ and j = 1, or if $i\in {N}_{2}^{t}\backslash \left\{1\right\}$ and $j\in {N}_{2}^{t}\backslash \left\{1\right\},$ the proof is similar. Due to (A.20) and (A.21), ${\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}=({\alpha }_{i}^{t}+\frac{1}{n}{q}_{i}^{t})-({\alpha }_{j}^{t}+{q}_{j}^{t})$. Next,

$$\mathop{\max }\limits_{i,j\in {N}_{2}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}<\mathop{\max }\limits_{i\in {N}_{1}^{t},j\in {N}_{2}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\},$$

since $i={M}_{2}^{t+1}=1$ and ${\alpha }_{1}^{t}+\frac{1}{n}{q}_{1}^{t}<{\alpha }_{{M}^{t}}^{t}.$ Generalizing all the cases considered, write

$$\begin{array}{c}\begin{array}{l}\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}={\rm max}\left\{\mathop{\max }\limits_{i,j\in {N}_{1}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\},\mathop{\max }\limits_{i\in {N}_{1}^{t},j\in {N}_{2}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\},\right.\\ \left.\mathop{\max }\limits_{i\in {N}_{2}^{t},j\in {N}_{1}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\},\mathop{\max }\limits_{i,j\in {N}_{2}^{t}}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}\right\}<\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}.\end{array}\end{array}$$

The proof of Lemma A.4 is complete.

Lemma A.5.

Consider process (10), (11), (13), and assume that ${\gamma }_{i}^{t+1}\in \left(0,\frac{2}{1+n}\right]$ for the agents with the Cournot response and ${\gamma }_{1}^{t+1}\in \left(0,\frac{1}{n}\right]$ for the Stackelberg leader. The following assertions are true: a) if the sequence $\left\{{\alpha }_{i}^{t},i\in N\right\}$ contains positive terms, then the sequence $\left\{{\alpha }_{i}^{t+1},i\in N\right\}$ also contains positive terms; b) if the sequence $\left\{{\alpha }_{i}^{t},i\in N\right\}$ contains negative or zero terms and ${N}_{1}^{t}=N,$ then the sequence $\left\{{\alpha }_{i}^{t+1},i\in N\right\}$ contains negative terms.

Proof of Lemma A.5. Establish item a). Under the assumptions above, ${\alpha }_{{M}^{t}}^{t}>0$ and ${M}^{t}\in {N}_{1}^{t}$. Due to (A.18) and (A.19),

$${\alpha }_{{M}^{t}}^{t+1}>\left(1-\frac{{\lambda }_{{M}^{t}}^{t+1}}{2}\right){\alpha }_{{M}^{t}}^{t}-{\alpha }_{{M}^{t}}^{t}\sum _{j\in {N}_{1}^{t}}\frac{{\lambda }_{j}^{t+1}}{2}={\alpha }_{{M}^{t}}^{t}\left(1-\frac{{\lambda }_{{M}^{t}}^{t+1}}{2}-\sum _{j\in {N}_{1}^{t}}\frac{{\lambda }_{j}^{t+1}}{2}\right)$$

if ${M}^{t}\in {N}_{1}^{t}\backslash \left\{1\right\}$, or

$${\alpha }_{{M}^{t}}^{t+1}>\left(1-\frac{{\lambda }_{{M}^{t}}^{t+1}}{2n}\right){\alpha }_{{M}^{t}}^{t}-{\alpha }_{{M}^{t}}^{t}\sum _{j\in {N}_{1}^{t}}\frac{{\lambda }_{j}^{t+1}}{2}={\alpha }_{{M}^{t}}^{t}\left(1-\frac{{\lambda }_{{M}^{t}}^{t+1}}{2n}-\sum _{j\in {N}_{1}^{t}}\frac{{\lambda }_{j}^{t+1}}{2}\right)$$

if ${M}^{t}=1\in {N}_{1}^{t}$. If $1-\frac{{\lambda }_{{M}^{t}}^{t+1}}{2}-{\displaystyle \sum _{j\in {N}_{1}^{t}}}\frac{{\lambda }_{j}^{t+1}}{2}>0,$ which holds by the redefinition of the parameters (see above), then ${\alpha }_{{M}^{t}}^{t+1}>0$ for ${\lambda }_{i}^{t+1}={\gamma }_{i}^{t+1}\in \left(0,\frac{2}{1+n}\right](i\in N\backslash \left\{1\right\})$ and ${\lambda }_{1}^{t+1}=\frac{{\gamma }_{1}^{t+1}2n}{1+n}\in \left(0,\frac{2}{1+n}\right]$.

Item b) of this lemma can be established by analogy, using formulas (A.18) and (A.19).

The proof of Lemma A.5 is complete.

Lemma A.6.

Process (10), (11), (13) satisfies the following inequalities: a) Q^(s) − Q^t > Q^(s) − Q^t+1 > 0 if ${\alpha }_{i}^{t},{\alpha }_{i}^{t+1}\ge 0(\forall i\in N)$ and there are nonzero terms ${\alpha }_{i}^{t}$ and ${\alpha }_{i}^{t+1};$ b) Q^t − Q^(s) > Q^t+1 − Q^(s) > 0 if ${\alpha }_{i}^{t},{\alpha }_{i}^{t+1}\le 0(\forall i\in N)$ and there are nonzero terms ${\alpha }_{i}^{t}$ and ${\alpha }_{i}^{t+1}$.

Proof of Lemma A.6. By (A.17), ${Q}^{(s)}-{Q}^{t+1}={Q}^{(s)}-{Q}^{t}-{\displaystyle \sum _{j\in {N}_{1}^{t}}}\frac{{\lambda }_{j}^{t+1}}{2}{\alpha }_{j}^{t}+{\displaystyle \sum _{j\in {N}_{2}^{t}}}{q}_{j}^{t}$. Under the conditions on ${\alpha }_{i}^{t}$ in item a) of this lemma, ${N}_{2}^{t}$ will be non-empty and Q^(s) − Q^t > Q^(s) − Q^t+1. Moreover, if ${\alpha }_{i}^{t+1}\ge 0$ and some of these terms are nonzero, then

$$\sum _{i\in N\backslash \left\{1\right\}}{\alpha }_{i}^{t+1}+n{\alpha }_{1}^{t+1}=2n({Q}^{(s)}-{Q}^{t+1})>0$$

due to (A.13) and (A.14); as a result, Q^(s) − Q^t > Q^(s) − Q^t+1 > 0. Item b) of this lemma can be established by analogy.

The proof of Lemma A.6 is complete.

The next lemma describes the change of signs in the sequence $\left\{{\alpha }_{i}^{t},i\in N\right\}$ while passing between time instants t and (t + 1).

Lemma A.7.

Consider process (10), (11), (13). a) If some negative term of the sequence $\{{\alpha }_{i}^{t},i\in N\}$ becomes positive in the sequence $\{{\alpha }_{i}^{t+1},i\in N\}$, then all positive terms of $\{{\alpha }_{i}^{t},i\in N\}$ will keep their signs in $\{{\alpha }_{i}^{t+1},i\in N\}$. b) If some positive term of the sequence $\{{\alpha }_{i}^{t},i\in N\}$ becomes negative in the sequence $\{{\alpha }_{i}^{t+1},i\in N\}$, then all negative terms of $\{{\alpha }_{i}^{t},i\in N\}$ will keep their signs in $\{{\alpha }_{i}^{t+1},i\in N\}.$

Proof of Lemma A.7. Establish item a). Let k be the index of a negative term in the former sequence that becomes positive in the latter, $k\in {N}_{1}^{t}$. By (A.18) and (A.19), the signs of the positive terms ${\alpha }_{i}^{t}$ will not change. Let $k\in {N}_{2}^{t}$. Taking into account that $2{x}_{k}^{t}={\alpha }_{k}^{t}+2{q}_{k}^{t}\le 0(k\ne 1)$ or $\frac{1+n}{n}{x}_{1}^{t}={\alpha }_{1}^{t}+\frac{1+n}{n}{q}_{1}^{t}\le 0$(k = 1), due to (A.20) and (A.21) the signs of the positive terms ${\alpha }_{i}^{t}(i\in {N}_{2}^{t})$ (due to (A.18) and (A.19), also the signs of the positive terms ${\alpha }_{i}^{t}$$(i\in {N}_{1}^{t})$) will not change. Item a) is established. Item b) can be established by analogy.

The proof of Lemma A.7 is complete.

With these auxiliary lemmas at hand, now prove Proposition 2.

First of all, consider the sequences with negative and zero terms only. At a next time instant, such a sequence may pass into one of the following sequences: 1) a sequence containing positive terms, 2) a sequence without positive terms.

In the first case, a sequence containing negative and zero terms only will not be observed at successive time instants: in accordance with Lemma A.5, a sequence with at least one positive term cannot pass into a sequence with negative and zero terms only if ${\gamma }_{i}^{t+1}\in \left(0,\frac{2}{1+n}\right](i\in N\backslash \left\{1\right\})$ and ${\gamma }_{1}^{t+1}\in \left(0,\frac{1}{n}\right]$. Therefore, all successive sequences will have positive terms only.

In the second case, Lemma A.6 implies 0 < Q^t+1 − Q^(s) < Q^t − Q^(s). It is again possible that only negative and zero terms will occur at a time instant (t + 2). Thus, the sequences with negative and zero terms only can occur either at the initial stage of the process or during the entire process. Consider the second option in detail. A successive application of Lemma A.6 gives the chain of inequalities Q⁰ − Q^(s) > Q¹ − Q^(s) > … > Q^t − Q^(s) > Q^t+1 − Q^(s) > … > 0(t > 1), from which it follows that Q^t → Q^(s) and ${\displaystyle {\sum_{i\in N\backslash \left\{1\right\}}}}{\alpha }_{i}^{t}+n{\alpha }_{1}^{t}=2n({Q}^{(s)}-{Q}^{t})\to 0$. Therefore, ${\alpha }_{i}^{t}\to 0,$ and by (A.13) and (A.14) ${q}_{i}^{t}\to {q}_{i}^{(s)}$(i ∈ N). Thus, the process is convergent.

Now let ${\alpha }_{i}^{t}>0(\forall i\in N)$. Since ${\alpha }_{i}^{t}=2({x}_{i}^{t}-{q}_{i}^{t}),$ it follows that ${x}_{i}^{t}>0\forall (i\in N)$, and all agents calculate their current output by formula (13). Hence, Lemma A.6 leads to the inequality Q^(s) − Q^t+1 < Q^(s) − Q^t. If the signs of all terms at the successive time instants remain positive, then due to the chain of inequalities Q^(s) − Q^t > Q^(s) − Q^t+1 > … > Q^(s) − Q^t+k > Q^(s) − Q^t+k+1 > … > 0(k > 1) the total output will converge to the equilibrium, i.e., Q^t → Q^(s). As a result, ${\alpha }_{i}^{t}\to 0$ and ${q}_{i}^{t}\to {q}_{i}^{(s)}$(i ∈ N).

Assume that the sequence $\left\{{\alpha }_{i}^{t},i\in N\right\}$ contains not only positive terms. By Lemma A.4 the process will make a sequential approximation to the equilibrium, because $\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}<\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}$. Due to Lemma A.5, the sequence $\left\{{\alpha }_{i}^{t+1},i\in N\right\}$ will contain positive terms. If it also has negative or zero terms, then the process will make a next approximation to the equilibrium $\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+2}-{\alpha }_{j}^{t+2}\right\}<\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}$. If such a situation occurs during the entire process, $\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t+1}-{\alpha }_{j}^{t+1}\right\}<\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}<\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t-1}-{\alpha }_{j}^{t-1}\right\}<\ldots <\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{0}-{\alpha }_{j}^{0}\right\}$. Thus, $\mathop{\max }\limits_{i,j\in N}\left\{{\alpha }_{i}^{t}-{\alpha }_{j}^{t}\right\}\to 0$ as t → ∞. Since the signs of ${\alpha }_{{m}_{t}}^{t}$ and ${\alpha }_{{M}_{t}}^{t}$ do not coincide, then $\forall i\in N\,{\alpha }_{i}^{t}\to 0$ as t → ∞, and consequently Q^t → Q^(s) as well as ${q}_{i}^{t}\to {q}_{i}^{(s)}$. The process is convergent. In addition, note a series of fruitful properties connected with the change (or retention) of signs of the terms of the sequences $\left\{{\alpha }_{i}^{t},i\in N\right\},$ which have been described by Lemma A.7.

Well, process (10), (11), (13) is convergent for any initial outputs $\left\{{q}_{i}^{0},i\in N\right\}$ of agents.

The proof of Proposition 2 is complete.