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Problems of Calculation Equilibria in Stackelberg Nonlinear Duopoly

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Frontiers of Dynamic Games

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Abstract

A duopoly model with a linear demand function and nonlinear cost functions of agents is considered. The game with the multilevel Stackelberg leadership is investigated. We analyze conjectural variations, i.e., the agent’s assumption about changes in the counterparty’s actions, which optimize the latter’s utility function. For an arbitrary Stackelberg leadership level, the formula for calculating the conjectural variations of agents is derived. The main insights are as follows: (1) the variations depend not only on the leadership level, but also on the product of the cost functions concavity/convexity indicators; (2) if at least one of agents has the concave cost function, then the variations can be not only negative, but also positive, and are not limited in absolute value, i.e., the bifurcations can occur.

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Correspondence to Mikhail I. Geraskin .

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Appendix

Appendix

Proof (Of Proposition 1)

We transform the expansion (6a) of continued fraction (6) in convergent fractions as follows:

$$\displaystyle \begin{aligned} p_{1(r)}= \left\{ \begin{array}{llllr} u_1^{r-\tau}u_2^{\tau-1} \quad for \quad r=1,2,\\ u_1^{r-\tau}u_2^{\tau-1}\left[ 1-(r-2)u_1^{-1}u_2^{-1}\right] \quad for \quad r=3,4,\\ u_1^{r-\tau}u_2^{\tau-1}\left[ 1-(r-2)u_1^{-1}u_2^{-1}+\frac{1}{2}(r-3)(r-4)u_1^{-2}u_2^{-2} \right] \quad for \quad r=5,6,\\ u_1^{r-\tau}u_2^{\tau-1}[ 1-(r-2)u_1^{-1}u_2^{-1}+\frac{1}{2}(r-3)(r-4)u_1^{-2}u_2^{-2}-\\ \qquad -\frac{1}{6}(r-6)(r-5)(r-4)u_1^{-3}u_2^{-3}] \quad for \quad r=7,8, \end{array} \right. \end{aligned}$$
$$\displaystyle \begin{aligned} q_{1(r)}= \left\{ \begin{array}{lllll} u_1^{r- \tau} u_2^{\tau} \quad for \quad r=1, \\ u_1^{r- \tau} u_2^{\tau} \left[ 1- (r-1)u_1^{-1} u_2^{-1} \right] \quad for \quad r=2,3, \\ u_1^{r- \tau} u_2^{\tau} \left[1 - (r-1)u_1^{-1} u_2^{-1} + \frac {1}{2} (r-2)(r-3) u_1^{-2} u_2^{-2} \right] \quad for \quad r=4,5, \\ u_1^{r- \tau} u_2^{\tau} [ 1 -(r-1)u_1^{-1} u_2^{-1} + \frac {1}{2} (r-2)(r-3) u_1^{-2} u_2^{-2} - \\ \qquad -\frac{1}{6} (r-5)(r-4)(r-3)u_1^{ -3} u_2^{-3} ] \quad for \quad r=6,7 \end{array} \right. \end{aligned}$$

Substituting these formulas in (6a), and, taking into account the fact that \(\frac {u_1^{r-\tau }u_2^{\tau -1}}{u_1^{r-\tau }u_2^{\tau }}=u_2^{-1}\) and the notation \(y=u_1^{-1}u_2^{-1}\), we obtain the following expression for the variation of the first agent:

$$\displaystyle \begin{aligned} x_{1(r)}=u_2^{-1} \frac{P_{(r)}}{Q_{(r)}}, \end{aligned}$$
$$\displaystyle \begin{aligned} P_{(r)}= \left\{ \begin{array}{llll} 1 \quad for \quad r=1,2,\\ 1-(r-2)y \quad for \quad r=3,4,\\ 1-(r-2)y+\frac{1}{2}(r-3)(r-4)y^2 \quad for \quad r=5,6,\\ 1-(r-2)y+\frac{1}{2}(r-3)(r-4)y^2-\frac{1}{6}(r-6)(r-5)(r-4)y^3 \quad for \quad r=7,8, \end{array} \right. \end{aligned}$$
$$\displaystyle \begin{aligned} Q_{(r)}= \left\{ \begin{array}{llll} 1 \quad for \quad r=1,\\ 1-(r-1)y \quad for \quad r=2,3,\\ 1-(r-1)y+\frac{1}{2}(r-2)(r-3)y^2 \quad for \quad r=4,5,\\ 1-(r-1)y+\frac{1}{2}(r-2)(r-3)y^2-\frac{1}{6}(r-5)(r-4)(r-3)y^3 \quad for \quad r=6,7. \end{array} \right. \end{aligned}$$

Similar reasoning for the second agent leads to the following formula for its variation:

$$\displaystyle \begin{aligned} x_{2(r)}=u_1^{-1} \frac{P_{(r)}}{Q_{(r)}}. \end{aligned}$$

We write the formulas P (r), Q (r) in the following generalized form:

$$\displaystyle \begin{aligned} P_{(r)}=\sum_{t=0}^{\lceil \frac{r}{2}\rceil-1} \frac{(1-)^t}{t!} y^t \prod_{\gamma=t+1}^{2t} (r-\gamma), Q_{(r)}=\sum_{t=0}^{\lceil \frac{r+1}{2}\rceil-1} \frac{(1-)^t}{t!} y^t \prod_{\gamma=t}^{2t-1} (r-\gamma). \end{aligned}$$

Because \(\prod _{\gamma =t+1}^{2t} (r-\gamma ) = \frac {(r-t-1)!}{(r-2t-1)!}\), \(\prod _{\gamma =t}^{2t-1} (r-\gamma ) =\frac {(r-t)!}{(r-2t)!}\), then these formulas have the following form [10]:

$$\displaystyle \begin{aligned} P_{(r)}=\sum_{t=0}^{\lceil \frac{r}{2}\rceil-1} \frac{(-1)^t}{t!} y^t \frac{(r-t-1)!}{(r-2t-1)!}, Q_{(r)}=\sum_{t=0}^{\lceil \frac{r+1}{2}\rceil-1} \frac{(-1)^t}{t!} y^t \frac{(r-t)!}{(r-2t)!}. \end{aligned}$$

Comparison of these formulas proves that Q (r) = P (r+1); therefore, \(x_{1(r)}=u_2^{-1} \frac {P_{(r)}}{P_{(r+1)}}\), \(x_{2(r)}=u_1^{-1} \frac {P_{(r)}}{P_{(r+1)}}\) and, in general, these expressions are written as (7).

Proof (Of Proposition 2)

We introduce the function f j(r−1) = u j − x j(r−1), then formula (6) has the following form:

$$\displaystyle \begin{aligned} x_{i(r)}=f_{j(r-1)}^{-1}. \end{aligned} $$
(9)

An analysis of the function f 2(r−1) demonstrate that f 2(0) = u 2, \(f_{2(1)}=u_2-\frac {1}{u_1}=u_2-f_{1(0)}^{-1}\), \(f_{2(2)}=u_2-\frac {1}{u_1-\frac {1}{u_2}}=u_2-f_{1(1)}^{-1}\), etc., therefore \(f_{j(r-1)}=u_j-f_{i(r-2)}^{-1}\). For variation x 1(r), we consider the function

$$\displaystyle \begin{aligned} f_{2(r-1)}=u_2-f_{1(r-2)}^{-1}=u_2-(u_1-x_{1(r-2)})^{-1}. \end{aligned} $$
(10)

The following cases are possible:

  1. (i)

    if f 1(r−2) > 0, i.e.x 1(r−2) < u 1, then f 2(r−1) < 0, therefore, according to (9),

    $$\displaystyle \begin{aligned} x_{1(r)}<0, \quad |x_{1(r)}|<1,\quad \lim_{f_{1(r-2)} \to \infty} |x_{1(r)}| = |u_2|{}^{-1};{} \end{aligned} $$
    (11)
  2. (ii)

    if f 1(r−2) < 0, i.e. x 1(r−2) > u 1, then, according to (10), two options are possible:

  3. (iii)

    for |f 1(r−2)| < |u 2|−1, the inequality f 2(r−1) > 0 holds; therefore,

    $$\displaystyle \begin{aligned} x_{1(r)}>0, \quad x_{1(r)}=\left(|f_{1(r-2)}|{}^{-1}-|u_2| \right)^{-1},{} \end{aligned} $$
    (12)
    $$\displaystyle \begin{aligned} |x_{1(r)}|>1, \lim_{f_{1(r-2)} \to |u_2|{}^{-1}} |x_{1(r)}|=\infty; \end{aligned}$$
  4. (iv)

    for |f 1(r−2)| > |u 2|−1, the inequality f 2(r−1) < 0 holds; therefore,

    $$\displaystyle \begin{aligned} x_{1(r)}<0, \quad x_{1(r)}=\left(|f_{1(r-2)}|{}^{-1}-|u_2| \right)^{-1},{} \end{aligned} $$
    (13)
    $$\displaystyle \begin{aligned} |x_{1(r)}|>1, \quad \lim_{f_{1(r-2)} \to |u_2|{}^{-1}} |x_{1(r)}| = \infty. \end{aligned}$$

We introduce the function α 1(r) = |f 1(r−2)|−1 −|u 2| = |u 1 − x 1(r−2)|−1 −|u 2|, and assume that the minimum value of this function is equal to \(A_{1(r)}=\min _{x_{1(r-2)} \in \varOmega _1(u_1, u_2)} |\alpha _{1(r)}|\), where Ω 1(u 1, u 2) is a set of admissible values of the variation x 1(r−2) for given values u 1, u 2. Then conditions (9)–(12) can be written as follows:

$$\displaystyle \begin{aligned} x_{1(r)} \left\{ \begin{array}{ll} <0 \quad if \quad f_{1(r-2)}>0 \vee (f_{1(r-2)}<0 \wedge \alpha_{1(r)}<0), \\ >0 \quad if \quad f_{1(r-2)}<0 \wedge \alpha_{1(r)}>0; \end{array} \right. \end{aligned}$$
$$\displaystyle \begin{aligned} |x_{1(r)}| \left\{ \begin{array}{ll} <|u_2|{}^{-1} \quad if \quad f_{1(r-2)}>0 , \\ <A_{1(r)}^{-1} \quad if \quad f_{1(r-2)}<0. \end{array} \right. \end{aligned}$$

Similar reasoning for the second agent leads to general notation (8).

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Geraskin, M.I. (2021). Problems of Calculation Equilibria in Stackelberg Nonlinear Duopoly. In: Petrosyan, L.A., Mazalov, V.V., Zenkevich, N.A. (eds) Frontiers of Dynamic Games. Trends in Mathematics. Birkhäuser, Cham. https://doi.org/10.1007/978-3-030-93616-7_6

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