# Relaxing the notion of Nash Equilibrium

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)

• 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. Apr 8, 2021 at 18:24
• Thanks. Unfortunately, my game is nor supermodular or submodular. Apr 8, 2021 at 18:26
• How would you apply here the notion of rationalisable strategies and how do people prove existence of those typically? Apr 8, 2021 at 18:27

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.