Consider two people have a mutually advantageous relationship. That is, if both dedicate more effort to the relationship both improve. Specifically, each individual chooses a level of effort $x_{i}\in [0,1]$ and the payoff of the person $i$ is

$\pi_{i}(x_{i},x_{j})=cx_{i}+x_{i}x_{j}-x^2_{i}$. where $c$ is constan $c\in [0,1]$

Verify conditions that guarantee the existence of Nash equilibrium in pure strategies and determine such equilibrium.


Q1: for the equilibrium.


$\max_{\mathbf{x}_{i}\in [0,1]} \pi_{i}(x_{j})=\max_{\mathbf{x}_{i}\in [0,1]}(cx_{i}+x_{i}x_{j}-x^2_{i})$

First order conditions $c+x_{j}-2x_{i}=0$ $$x_{i}=\dfrac{c+x_{j}}{2}$$ $\max_{\mathbf{x}_{j}\in [0,1]} \pi_{j}(x_{i})=\max_{\mathbf{x}_{j}\in [0,1]}(cx_{j}+x_{j}x_{i}-x^2_{j})$

First order conditions $c+x_{i}-2x_{j}=0$ $$x_{j}=\dfrac{c+x_{i}}{2}$$ So then I can say $x_{i}=x_{j}=c$ is nash equilibrium.

Is correct?

Q2: (Conditions for existence). I think that I must verify the conditions of the theorem(Debreu. 1952;Fan, 1952;Glicksberg, 1952): Let $G$ be a game in strategic form such that, for each $i=1,2,...,n, S_{i}\subset \mathbb{R^{m}}$ is compact and convex, and the function $\pi_{i}:S_{1}\times...\times S_{n}\rightarrow \mathbb{R}$ is continuous in $s=(s_{1},s_{2},...,s_{n})$ and quasi-concave in $s_{i}$.Then, the game $G$ has a Nash equilirbium in pure strategies.

$\rho:[0,1]\rightrightarrows [0,1]$ where $\rho_{i}(x_{j})=arg \max_{x_{i}\in [0,1]}\pi_{i}(x_{j})$

*$S\neq \phi$

*$S=[0,1]$ is compact.

*$\rho_{i}(x_{j}) \neq \phi$ and convex...



But The teacher indicates that it is not the theorem that must be analyzed. But I think he can be wrong. Do you recommend that I continue with this?

Or who will be asking me to apply?

Thanks for your help.

  • 1
    $\begingroup$ I'm voting to close this question because it is not clear what you are asking. Do you have to show that a Nash equilibrium exists? You just did in Q1. Do you have to analyze whether the conditions of a given theorem (Glicksberg/Nash/Brouwer) apply? Then analyze it for the given theorem, and not for another. You have to talk to your professor about this, not strangers on the net. $\endgroup$ – Giskard May 19 '19 at 8:34
  • $\begingroup$ I did not know if I should use the theorem, as I said the professor told me no. (I think he could be wrong). And that theorem I think could solve the question. That's why I came to consult the question, maybe there was another method, that was my doubt. But if you tell me that I should continue with the theorem, I will move forward, thank you very much. $\endgroup$ – Franciscolli May 19 '19 at 22:42

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