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Using chemical reaction network theory to show stability of distributional dynamics in game theory

  • * Corresponding author: Vlastimil Křivan

    * Corresponding author: Vlastimil Křivan
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  • This article shows how to apply results of chemical reaction network theory (CRNT) to prove uniqueness and stability of a positive equilibrium for pairs/groups distributional dynamics that arise in game theoretic models. Evolutionary game theory assumes that individuals accrue their fitness through interactions with other individuals. When there are two or more different strategies in the population, this theory assumes that pairs (groups) are formed instantaneously and randomly so that the corresponding pairs (groups) distribution is described by the Hardy–Weinberg (binomial) distribution. If interactions times are phenotype dependent the Hardy-Weinberg distribution does not apply. Even if it becomes impossible to calculate the pairs/groups distribution analytically we show that CRNT is a general tool that is very useful to prove not only existence of the equilibrium, but also its stability. In this article, we apply CRNT to pair formation model that arises in two player games (e.g., Hawk-Dove, Prisoner's Dilemma game), to group formation that arises, e.g., in Public Goods Game, and to distribution of a single population in patchy environments. We also show by generalizing the Battle of the Sexes game that the methodology does not always apply.

    Mathematics Subject Classification: Primary: 91A22, 91A05, 91A06.


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  • Figure 1.  Distributional dynamics (10) of pairs (panel A) and groups of size four (panel B) for PGG (20). Both panels assume that initially there are only singles and initial conditions are $ n_1(0) = n_2(0) = 5 $ ($ n_C(0) = n_D(0) = 5 $) in panel A (panel B). Parameters used in Panel A: $ \lambda_{11} = \lambda_{12} = \lambda_{21} = \lambda_{22} = \lambda = 0.1 $, $ \tau_{11} = 5 $, $ \tau_{12} = \tau_{21} = 3 $, $ \tau_{22} = 1 $. Parameters used in Panel B: $ \lambda_i = \frac{\lambda}{4}\binom{4}{i} $ with $ \lambda = 0.05 $, $ \tau_{0} = 1 $, $ \tau_{1} = 2 $, $ \tau_{2} = 4 $, $ \tau_{3} = 6 $, $ \tau_4 = 15 $

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