Let $x \in [0,1]$ denote some state (e.g. market share). Let $i \in \{1,2\}$ denote an agent (e.g. firm). Im considering a model where payoffs $F_i(x)$ are perfectly invertible in the sense that the payoff functions can be mirrored along the line $x = \frac{1}{2}$ with $F_1(x) = F_2(1-x)$. Take $F_1(x) = x$ and $F_2(x) = 1 - x$ for instance. Is there a fixed term for these kind of games. It has somewhat a zero sum flavor, because preferences are perfectly opposing. Firm one wants increase its market share, and firm 2 wants to decrease the market share of firm 1 in the same manner. I can think of Hotelling or a linear city. But is this the most generic characterization?
Or, even better, is there a mathematical term for this kind of functions which are mirrored along a fixed line?

Edit: Example Consider the following advertising differential game defined by $(x,u_1,u_2) \in [0,1] \times \mathbb R_+ \times \mathbb R_+$ \begin{align} &v_1(x_0) = \max_{u_1}\int_0^\infty{e^{-t}(x - u_1^2/2)dt}\\ &v_2(x_0) = \max_{u_2}\int_0^\infty{e^{-t}(1 - x - u_2^2/2)dt}\\ \text{s.t.}\quad & \dot x = u_1 - u_2 \end{align} $x$ is the market share of firm 1, and $u_i$ is the respective advertising effort. A stationary feedback Nash equilibrium $(\phi_1(x), \phi_2(x))$ solves a coupled system of Hamilton-Jacobi-Bellman equations \begin{align} &v_1(x) = \max_{u_1}\{x - u_1^2/2 + v'_1(x)(u_1-\phi_2(x))\}\\ &v_2(x) = \max_{u_2}\{1 - x - u_2^2/2 + v'_2(x)(\phi_1(x)-u_2)\} \end{align}

It turns out that $(\phi_1(x), \phi_2(x)) = (1,-1)$ is the unique Nash equilibrium with associated values \begin{align} &v_1(x) = x - \frac{1}{2}\\ &v_2(x) = \frac{1}{2}-x\\ \Longrightarrow \quad &v_1(x) = v_2(1-x). \end{align}

  • How would one characterize the class of games in general? Symmetric antagonistic?

I may propose the following definition:

Let $(\phi_1(x), \phi_2(x))$ solve \begin{align} &v_1(x) = \max_{u_1}\{F_1(x,u_1,\phi_2(x)) + v'_1(x)f(x,u_1,\phi_2(x))\}\\ &v_2(x) = \max_{u_2}\{F_2(x,\phi_1(x),u_2) + v'_2(x)f(x,\phi_1(x),u_2)\}. \end{align} A differential game is symetrically antagonistic if \begin{align} v_1(x) = v_2(\overline x - x) \quad \forall x \in X \end{align} where $\overline x := \max X$.


1 Answer 1


A game where the payoff function $F_1,F_2$ are such that for any two strategy profiles $s,s'$ $$ F_1(s) > F_1(s') \Leftrightarrow F_2(s) < F_2(s') $$ is called an antagonistic game. Sounds like what you want is a symmetric antagonistic game?

  • $\begingroup$ Thanks for hinting. Antagonistic sounds about right. I added an example to clarify matters. For differential games it sound more approbiate to refer to the state instead of strategies such that $v_1(x) \geq v_1(x') \Leftrightarrow v_2(x) \leq v_2(x')$ for all $x \in X$ . $\endgroup$
    – clueless
    Mar 11, 2018 at 21:01

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