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

CC BY-SA 3.0
2018-03-11T21:01:30.697Z
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 $x \in X$ .
clueless
CC BY-SA 3.0
2018-03-11T20:11:42.990Z
2018-03-11T20:11:42.990Z
Giskard
<p>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?<br>
Or, even better, is there a mathematical term for this kind of functions which are mirrored along a fixed line?</p>

<p><strong>Edit: Example</strong> Consider the following advertising differential game defined by $(x,u_1,u_2) \in [0,1] \times \mathbb R_+ \times \mathbb R_+$
\begin{align}
&amp;v_1(x_0) = \max_{u_1}\int_0^\infty{e^{-t}(x - u_1^2/2)dt}\\
&amp;v_2(x_0) = \max_{u_2}\int_0^\infty{e^{-t}(1 - x - u_2^2/2)dt}\\
\text{s.t.}\quad &amp; \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}
&amp;v_1(x) = \max_{u_1}\{x - u_1^2/2 + v'_1(x)(u_1-\phi_2(x))\}\\
&amp;v_2(x) = \max_{u_2}\{1 - x - u_2^2/2 + v'_2(x)(\phi_1(x)-u_2)\}
\end{align}</p>

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

<ul>
<li>How would one characterize the class of games in general? Symmetric antagonistic?</li>
</ul>

<p>I may propose the following definition:</p>

<blockquote>
  <p>Let $(\phi_1(x), \phi_2(x))$ solve
  \begin{align}
&amp;v_1(x) = \max_{u_1}\{F_1(x,u_1,\phi_2(x)) + v'_1(x)f(x,u_1,\phi_2(x))\}\\
&amp;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 $\overline x := \max X$.</p>
</blockquote>

<microeconomics><mathematical-economics><game-theory>
Term for perfect symmetric asymmetry
clueless
# Term for perfect symmetric asymmetry

(Asked by: username_0 on 2018-03-11)

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 $\overline x := \max X$.

**Tags:** microeconomics, mathematical-economics, game-theory

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## 1 answers

### Answer by username_1 on 2018-03-11. Score: 0 | Accepted answer

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?

> **Comments:**
> **username_0**: 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 $x \in X$ .  

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