Zero-Sum Game

Bruce C. Dieffenbach ()
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Bruce C. Dieffenbach: Independent author

Chapter 16 in Conjugate Duality in Economic Analysis, 2026, pp 123-126 from Springer

Abstract: Abstract Nash equilibrium is the cornerstone of game theory. For a two-player, zero-sum game, a Nash equilibrium is a saddle point of the payoff as a function of the strategies of the two players. This function is the Lagrangian of a primal/dual pair. In the primal one player maximizes the minimum payoff across his pure strategies, and in the dual the other player minimizes the maximum payoff across his pure strategies. The game has a Nash equilibrium, and one can calculate the optimum strategies by solving a convex optimization. A mixed strategy is a choice of probabilities for the different possible pure strategies. Let A denote the payoff matrix. Let x∈Pn denote a mixed strategy for player x, and let y*∈Pm denote a mixed strategy for player y. Then $$\left \langle \boldsymbol {y}^{\ast },{\textit {{\sf { {A}}}}}\boldsymbol {x}\right \rangle$$ is the payoff to x and the negative of the payoff to y. A Nash equilibrium $$\left ( \boldsymbol {x},\boldsymbol {y}^{\ast }\right )$$ is a saddle point of $$L\left ( \boldsymbol {x},\boldsymbol {y}^{\ast }\right ) =\left \langle \boldsymbol {y}^{\ast },\boldsymbol {{\textit {{\sf { {A}}}}}\boldsymbol {x}}\right \rangle -\delta _{P^{n}}\left ( \boldsymbol {x}\right ) + \delta _{P^{m}}\left ( \boldsymbol {y}^{\ast }\right ) \!.$$ At a Nash equilibrium, x maximizes the payoff given y*, and y* minimizes the payoff given x. Given the strategy of the other player, each player chooses his optimum strategy. Recognizing that the function $$L\left ( \boldsymbol {x},\boldsymbol {y}^{\ast }\right )$$ is the Lagrangian of an extended Fenchel duality primal/dual pair, we calculate the primal and the dual from the Lagrangian. Perturbation duality establishes that a Nash equilibrium exists. The objective function in the primal is $$\min _{\boldsymbol {y}^{\ast }\in P^{m}} {L\left ( \boldsymbol {x},\boldsymbol {y}^{\ast }\right ) },$$ so the primal is $$\max _{\boldsymbol {x}\in P^{n}}\left [ \min _{i}{\textit {{\sf { {A}}}}}\boldsymbol {x}\right ] .$$ Analogously, the dual is $$\min _{\boldsymbol {y}^{\ast }\in P^{m}}\left [ \max _{i}{\textit {{\sf { {A}}}}}^{\top }\boldsymbol {y}^{\ast }\right ] \!.$$

Date: 2026
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DOI: 10.1007/978-3-032-21396-9_16

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