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Consider a $n$-player continuous game $G=(P,S,U)$ where:

  • $P=\{1,2,\dots,n\}$ is the set of $n$ players.
  • $S=\{S_1,S_2,\dots,S_n\}$ where $S_i=\mathbb{R}^n_+$ is the $i$- th player's set of pure strategies, and $\mathbb{R}^n_+$ is the set of non-negative $n$-tuples.
  • $U=\{u_1,u_2,\dots,u_n\}$ where $u_i:S\rightarrow\mathbb{R}$ is the utility function of player $i$.

Let $\sigma_i\in S_i=\mathbb{R}^n_+$ denote a single strategy for player $i$, $\sigma=\{\sigma_1,\sigma_2,\dots,\sigma _n\}\in S$ denote a strategy profile, and $\sigma_{-i}$ denote a strategy profile of all players except for player $i$.

A strategy profile $\sigma^*=\{\sigma_1^*,\sigma_2^*,\dots,\sigma_n^*\}\in S$ is said to be a Nash equilibrium if the strategy $\sigma_i^*$ is a local maximum for the utility function $u_i(\sigma_i; \sigma_{-i}^*)$ for all the players $i$.

Question:

Is there any property for the utility functions $u_i$ that could guarantee the existence of at least one Nash equilibrium? for example, concavity or quasi-concavity.

I know that if all the utility functions are continuous and the sets $S_i$ are compact, then, the existence of a Nash equilibrium is guaranteed, but $\mathbb{R}^n_+$ is not bounded.

AlephZero
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1 Answers1

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Concavity or quasi-concavity is not enough, since it can be linear (for example take $u_i(\sigma)=\sigma_i\ \forall i$).

If $\{-u_i\}$ are continuous and coercive (meaning $\lim_{|\sigma|\rightarrow\infty} u_i(\sigma)=-\infty\ \forall i$) then you can restrict the problem to a compact rectangle subset of $S$ on which the functions $u_i$ are not to small and then use the result with compact strategy sets.

In particular, if $\{u_i\}$ are continuous and strongly concave, one can show $\{-u_i\}$ are also coercive.

MOMO
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