# Nash Equilibrium and Limits in Game Theory

I am sure the solution to this is easy, but I can't work it out.

Suppose we have an extremely simple game: There is only 1 player, who announces a number in the set [0,1]. His payoff is equal to his announcement, unless he announces 1, when his payoff is zero.

What is the Nash? It seems like it should be 1-$\epsilon$, for some arbitrarily small positive $\epsilon$, but then the player can surely improve by increasing his announcement by $\frac{\epsilon}{2}$.

If you think there must be 2 players for there to be a Nash, the following game has the equivalent problem: 2 players, who announce a number in the set [0,1], and their payoffs are equal to their own announcement unless they both announce 1, when both their payoffs are zero.

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Indeed, a game theoretic problem requires strategic interaction, i.e., two players. The problem you pose is a decision theoretic problem ("player against nature").

If we were to apply the Nash equilibrium concept anyway, then your reasoning is exactly right: there is no Nash equilibrium in this "game", because there is no action that maximizes the payoff (i.e., you can always find a deviation that increases the payoff). The payoff function is $$p(x)=\begin{cases} x & \text{if }x\in[0,1) \\ 0 & \text{else}, \end{cases}$$ and the discontinuity implies there is no maximum.

If the action space where discrete rather than continuous in $[0,1]$, then there would be a "Nash-equilibrium", by choosing the smallest increment below 1.

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It is ok to consider decision-theoretic settings as 1-player games

This game has no Nash equilibrium, not because it is a 1-player game, but rather because it is discontinuous

Nash existence theorem only applies for finite games. It can easily be extended to games with compact action spaces and upper hemi-continuous payoff functions, but it is easy to build examples of games with non-compact action spaces or discontinuous utility functions that don't admit any Nash equilibrium in pure or mixed strategies (like the one you propose)

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if I understand the exercise right, there must be two nash-equilibriums. At the strategies where both do not bit one, the arbitrarily small positive values can be different. That´s why I use different letters.

Pay-Off-Matrix

greetings,

calculus

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There plainly is no Nash equilibrium in this game, for exactly the reason given. $1$ can never be a best response, since $x$ does better for $x\in(0,1)$. Also, no $x\in[0,1)$ is a best response, since $x+(1-x)/2$ is strictly better. So no element of $[0,1]$ is a best response. Since mixed best responses are mixtures of pure best responses, there is no mixed best response either.

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