<p>Seems like this class of games is very general. (Or I don't get the definition.) Note that <span class="math-container">$\theta$</span> is not even used, it is just some parameter that has value <span class="math-container">$\theta_0$</span> throughout.</p>
<p>As long as there exist two games <span class="math-container">$G_{\delta}$</span> and <span class="math-container">$G_{\delta_0}$</span>, were both have the same number of players <span class="math-container">$J$</span>, each player has two strategies in both <span class="math-container">$G_{\delta}$</span> and <span class="math-container">$G_{\delta_0}$</span>, and <span class="math-container">$G_{\delta}$</span> has a Nash-equilibrium but <span class="math-container">$G_{\delta_0}$</span> does not, then one can get negation by defining
<span class="math-container">$$
v(a; \delta_0) := u_{\delta_0}(a) - u_{\delta}(a)
$$</span>
where <span class="math-container">$u_{\delta}$</span> is the vector of payoffs in <span class="math-container">$G_{\delta}$</span>, and <span class="math-container">$u_{\delta_0}$</span> is the vector of payoffs in <span class="math-container">$G_{\delta_0}$</span> given strategy profile <span class="math-container">$a$</span>.</p>
<p>There is still the possibility that either all or no games have a Nash-equilibria, but this is not true on the class of these two strategies per player games, as shown by Walrasian Auctioneer's <a href="https://economics.stackexchange.com/a/43907/1601">answer</a>.</p>

CC BY-SA 4.0
2021-05-12T20:27:56.360Z
2021-05-12T20:27:56.360Z
Giskard
<p>No. Consider the matching pennies game
<span class="math-container">$$ \begin{array}{c |c}
&amp; H &amp; T  &amp; \\ \hline
H &amp; (1,-1) &amp; (-1,1)\\ 
T &amp; (-1,1) &amp; (1,-1)\\
\end{array} $$</span>
which we know no PSNE exists.</p>
<p>Define the function
<span class="math-container">$v_i(\cdot,\cdot)$</span> by
<span class="math-container">$$
v_i(H,0) = 5 \\
v_i(x,y) = 0 \quad \forall (x,y) \not = (H,0)
$$</span></p>
<p>At <span class="math-container">$\delta = 0$</span>, we have the game
<span class="math-container">$$ \begin{array}{c |c}
&amp; H &amp; T  &amp; \\ \hline
H &amp; (5,4) &amp; (-4,1)\\ 
T &amp; (-1,5) &amp; (1,-1)\\
\end{array} $$</span>
which has a PSNE of <span class="math-container">$(H,H)$</span>.</p>
<p>For any other value of <span class="math-container">$\delta$</span>, we're back in the original game so no PSNE exists.</p>
<p>Redefine <span class="math-container">$H = 0, T = 1$</span> and you have your monotone decreasing requirement.</p>

CC BY-SA 4.0
2021-05-12T19:48:41.537Z
2021-05-12T19:48:41.537Z
Walrasian Auctioneer
<p>Consider a game with <span class="math-container">$N$</span> players, each indexed by <span class="math-container">$i=1,...,N$</span>. Every player <span class="math-container">$i$</span> has to choose a <span class="math-container">$J\times 1$</span> <strong>vector</strong> of actions <span class="math-container">$a_i\equiv (a_{i,1},...,a_{i,J})$</span> where each <span class="math-container">$a_{i,j}$</span> can be <strong>zero or one</strong>. For each player <span class="math-container">$i$</span>, let <span class="math-container">$a_{-i}$</span> denote the actions of the other players.</p>
<p>The payoff of each player <span class="math-container">$i$</span> is
<span class="math-container">$$
u_i(a_i, a_{-i}; \theta)+v_i(a_i; \delta)
$$</span>
where <span class="math-container">$u_i$</span> and <span class="math-container">$v_i$</span> are some parametric functions of the <strong>parameters</strong> <span class="math-container">$\theta$</span> and <span class="math-container">$\delta$</span>. Moreover <span class="math-container">$v_i$</span> is monotone decreasing in <span class="math-container">$\sum_{j=1}^J a_{i,j}$</span>.</p>
<p>A pure strategy Nash equilibrium (PSNE) of the game is <span class="math-container">$a^*\equiv (a_1^*,...,a_N^*)$</span> solving
<span class="math-container">$$
 a_i^*\in \arg\max_{a_i\in \{0,1\}^J} u_i(a_i, a^*_{-i}; \theta)+v_i(a_i; \delta) \quad \forall i=1,...,N
$$</span></p>
<p><strong>Question:</strong> Suppose that I'm able to show that a PSNE exists for <span class="math-container">$\theta=\theta_0$</span> and <span class="math-container">$\delta=\delta_0$</span> where <span class="math-container">$\theta_0$</span> and <span class="math-container">$\delta_0$</span> are some specific real values of the parameters <span class="math-container">$\theta$</span> and <span class="math-container">$\delta$</span>. Can I conclude that a PSNE exists for <span class="math-container">$\theta=\theta_0$</span> and <strong>any</strong> value of <span class="math-container">$\delta$</span>?</p>
<p>In particular, here, I'm wondering whether the claim may follow from the fact that <span class="math-container">$v_i(a_i; \delta)$</span> enters additively, does not depend on <span class="math-container">$a_{-i}$</span>, and it is monotone.</p>

CC BY-SA 4.0
2021-05-12T20:20:59.813Z
What does it mean that $v$ is monotone decreasing in $\sum_{j=1}^J a_{i,j}$? This is not a variable of $v$.
Giskard
<game-theory><nash-equilibrium>
Showing existence of a Nash equilibrium in pure strategy
Star
# Showing existence of a Nash equilibrium in pure strategy

(Asked by: username_0 on 2021-05-12)

Consider a game with $N$ players, each indexed by $i=1,...,N$. Every player $i$ has to choose a $J\times 1$ **vector** of actions $a\_i\equiv (a\_{i,1},...,a\_{i,J})$ where each $a\_{i,j}$ can be **zero or one**. For each player $i$, let $a\_{-i}$ denote the actions of the other players.

The payoff of each player $i$ is
$$
u\_i(a\_i, a\_{-i}; \theta)+v\_i(a\_i; \delta)
$$
where $u\_i$ and $v\_i$ are some parametric functions of the **parameters** $\theta$ and $\delta$. Moreover $v\_i$ is monotone decreasing in $\sum\_{j=1}^J a\_{i,j}$.

A pure strategy Nash equilibrium (PSNE) of the game is $a^\*\equiv (a\_1^\*,...,a\_N^\*)$ solving
$$
a\_i^\*\in \arg\max\_{a\_i\in \{0,1\}^J} u\_i(a\_i, a^\*\_{-i}; \theta)+v\_i(a\_i; \delta) \quad \forall i=1,...,N
$$

**Question:** Suppose that I'm able to show that a PSNE exists for $\theta=\theta\_0$ and $\delta=\delta\_0$ where $\theta\_0$ and $\delta\_0$ are some specific real values of the parameters $\theta$ and $\delta$. Can I conclude that a PSNE exists for $\theta=\theta\_0$ and **any** value of $\delta$?

In particular, here, I'm wondering whether the claim may follow from the fact that $v\_i(a\_i; \delta)$ enters additively, does not depend on $a\_{-i}$, and it is monotone.

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: What does it mean that $v$ is monotone decreasing in $\sum_{j=1}^J a_{i,j}$? This is not a variable of $v$.  

---

## 2 answers

### Answer by username_2 on 2021-05-12. Score: 4 | Accepted answer

No. Consider the matching pennies game
$$ \begin{array}{c |c}
& H & T & \\ \hline
H & (1,-1) & (-1,1)\\
T & (-1,1) & (1,-1)\\
\end{array} $$
which we know no PSNE exists.

Define the function
$v\_i(\cdot,\cdot)$ by
$$
v\_i(H,0) = 5 \\
v\_i(x,y) = 0 \quad \forall (x,y) \not = (H,0)
$$

At $\delta = 0$, we have the game
$$ \begin{array}{c |c}
& H & T & \\ \hline
H & (5,4) & (-4,1)\\
T & (-1,5) & (1,-1)\\
\end{array} $$
which has a PSNE of $(H,H)$.

For any other value of $\delta$, we're back in the original game so no PSNE exists.

Redefine $H = 0, T = 1$ and you have your monotone decreasing requirement.


---

### Answer by username_1 on 2021-05-12. Score: 2

Seems like this class of games is very general. (Or I don't get the definition.) Note that $\theta$ is not even used, it is just some parameter that has value $\theta\_0$ throughout.

As long as there exist two games $G\_{\delta}$ and $G\_{\delta\_0}$, were both have the same number of players $J$, each player has two strategies in both $G\_{\delta}$ and $G\_{\delta\_0}$, and $G\_{\delta}$ has a Nash-equilibrium but $G\_{\delta\_0}$ does not, then one can get negation by defining
$$
v(a; \delta\_0) := u\_{\delta\_0}(a) - u\_{\delta}(a)
$$
where $u\_{\delta}$ is the vector of payoffs in $G\_{\delta}$, and $u\_{\delta\_0}$ is the vector of payoffs in $G\_{\delta\_0}$ given strategy profile $a$.

There is still the possibility that either all or no games have a Nash-equilibria, but this is not true on the class of these two strategies per player games, as shown by Walrasian Auctioneer's [answer](https://economics.stackexchange.com/a/43907/1601).


---

