<p>No, there is not. Consider a game with two players, Ann and Bob. Both choose such vectors with entries <span class="math-container">$0$</span> or <span class="math-container">$1$</span> of the form <span class="math-container">$(a_1,a_2,\ldots,a_J)$</span> or <span class="math-container">$(b_1,b_2,\ldots,b_J)$</span>, respectively. If <span class="math-container">$\sum_{i=1}^J a_i+b_i$</span> is odd, Ann wins and Bob loses. If the number is even, Ann loses and Bob wins. Clearly, one of them loses in every profile of pure strategies, and the loser can win by changing only one entry of their chosen vector. So there is no pure strategy profile at which the revealed preference inequalities hold.</p>

CC BY-SA 4.0
2021-04-16T14:37:51.850Z
Thanks. Perhaps, I'm asking something very silly, but is there any learning dynamic concept (say, fictitious play) that would converge to something such that the revealed preference inequalities are satisfied?
Star
CC BY-SA 4.0
2021-04-16T19:15:00.093Z
@user3285148 In the example, one would have to go to mixed strategies.
Michael Greinecker
CC BY-SA 4.0
2021-04-08T18:30:07.380Z
2021-04-08T18:30:07.380Z
Michael Greinecker
<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 zero or one. The payoff of each player <span class="math-container">$i$</span> is <span class="math-container">$u_i(a_i, a_{-i})$</span>, where <span class="math-container">$a_{-i}$</span> denotes the actions of the other players.</p>
<p>As a first best, I would like to show that this game has a pure strategy Nash equilibrium (PSNE). I'm unable to do so and I also don't want to allow for mixed strategies. As a second best, I would be happy to consider a weaker notion of equilibrium (still, in pure strategy) and show its existence.</p>
<p>In particular, by revealed preference, if <span class="math-container">$(a_1,...,a_N)$</span> is a PSNE, then the following holds:</p>
<p><span class="math-container">$$
\forall i=1,...N \quad \forall j=1,...,J: \quad \text{if } a_{i,j}=1\text{, then } u_i(a_i, a_{-i})\geq  u_i(a_{i, \{-j\}}, a_{-i})\\
\hspace{6.5cm} \text{if } a_{i,j}=0\text{, then } u_i(a_{i}, a_{-i})\geq  u_i(a_{i,\{+j\}}, a_{-i})$$</span></p>
<p>where <span class="math-container">$a_{i, \{-j\}}$</span> denotes <span class="math-container">$a_i$</span> where <span class="math-container">$a_{i,j}=1$</span> is replaced by <span class="math-container">$a_{i,j}=0$</span>; <span class="math-container">$a_{i, \{+j\}}$</span> denotes <span class="math-container">$a_i$</span> where <span class="math-container">$a_{i,j}=0$</span> is replaced by <span class="math-container">$a_{i,j}=1$</span>.</p>
<p><strong>Question:</strong> Is there any equilibrium notion in pure strategy such that:</p>
<ul>
<li><p>it is weaker than PSNE</p>
</li>
<li><p>it implies the revealed preference inequalities reported above</p>
</li>
<li><p>its existence is typically easier to be shown (and, if you could give references on this)</p>
</li>
</ul>

CC BY-SA 4.0
2021-04-08T18:24:37.417Z
Perhaps a weaker notion would be rationalisable strategies? And although pure strategy nash would be impossible to guarantee in all games of your type, perhaps you could look at supermodular games (i.e. games with strategic complimentarities/substitutes) where people focus on these pure strategy existence, since you have a lattice structure on actions.
Walrasian Auctioneer
CC BY-SA 4.0
2021-04-08T18:26:38.773Z
Thanks. Unfortunately, my game is nor supermodular  or submodular.
Star
CC BY-SA 4.0
2021-04-08T18:27:27.800Z
How would you apply here the notion of rationalisable strategies and how do people prove existence of those typically?
Star
<game-theory><nash-equilibrium>
Relaxing the notion of Nash Equilibrium
Star
# Relaxing the notion of Nash Equilibrium

(Asked by: username_0 on 2021-04-08)

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. The payoff of each player $i$ is $u\_i(a\_i, a\_{-i})$, where $a\_{-i}$ denotes the actions of the other players.

As a first best, I would like to show that this game has a pure strategy Nash equilibrium (PSNE). I'm unable to do so and I also don't want to allow for mixed strategies. As a second best, I would be happy to consider a weaker notion of equilibrium (still, in pure strategy) and show its existence.

In particular, by revealed preference, if $(a\_1,...,a\_N)$ is a PSNE, then the following holds:

$$
\forall i=1,...N \quad \forall j=1,...,J: \quad \text{if } a\_{i,j}=1\text{, then } u\_i(a\_i, a\_{-i})\geq u\_i(a\_{i, \{-j\}}, a\_{-i})\\
\hspace{6.5cm} \text{if } a\_{i,j}=0\text{, then } u\_i(a\_{i}, a\_{-i})\geq u\_i(a\_{i,\{+j\}}, a\_{-i})$$

where $a\_{i, \{-j\}}$ denotes $a\_i$ where $a\_{i,j}=1$ is replaced by $a\_{i,j}=0$; $a\_{i, \{+j\}}$ denotes $a\_i$ where $a\_{i,j}=0$ is replaced by $a\_{i,j}=1$.

**Question:** Is there any equilibrium notion in pure strategy such that:

* it is weaker than PSNE
* it implies the revealed preference inequalities reported above
* its existence is typically easier to be shown (and, if you could give references on this)

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: Perhaps a weaker notion would be rationalisable strategies? And although pure strategy nash would be impossible to guarantee in all games of your type, perhaps you could look at supermodular games (i.e. games with strategic complimentarities/substitutes) where people focus on these pure strategy existence, since you have a lattice structure on actions.  
> **username_0**: Thanks. Unfortunately, my game is nor supermodular  or submodular.  
> **username_0**: How would you apply here the notion of rationalisable strategies and how do people prove existence of those typically?  

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## 1 answers

### Answer by username_2 on 2021-04-08. Score: 9 | Accepted answer

No, there is not. Consider a game with two players, Ann and Bob. Both choose such vectors with entries $0$ or $1$ of the form $(a\_1,a\_2,\ldots,a\_J)$ or $(b\_1,b\_2,\ldots,b\_J)$, respectively. If $\sum\_{i=1}^J a\_i+b\_i$ is odd, Ann wins and Bob loses. If the number is even, Ann loses and Bob wins. Clearly, one of them loses in every profile of pure strategies, and the loser can win by changing only one entry of their chosen vector. So there is no pure strategy profile at which the revealed preference inequalities hold.

> **Comments:**
> **username_0**: Thanks. Perhaps, I'm asking something very silly, but is there any learning dynamic concept (say, fictitious play) that would converge to something such that the revealed preference inequalities are satisfied?  
> **username_2**: @user3285148 In the example, one would have to go to mixed strategies.  

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