# Nash equilibria in games with infinitely many strategies

As a simple example, suppose two players A and B play a game wherein each picks a positive integer, and if they both pick the same integer $N$ then B pays $f(N)$ dollars to A, for some given payoff function $f$. This game should be particularly simple since the payoffs are symmetric, i.e., if A and B choose different strategies then no money is exchanged at all. What general conditions can one place on $f$ to ensure this game has a strict Nash equilibrium in mixed strategies?

For instance, we might suppose $f$ to assume nonnegative integer values with a unique minimum of $0$ at some integer, say $M$. Clearly then B should always play $M$, and then it doesn't matter what A plays. This isn't a strict equilibrium, though, because A has no preference at all for what number to pick given that B picks $M$.

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If there are positive and negative payoffs, both players will only pick payoffs in their favour, and the value of the game is $0$. You've already stated that the value is also $0$ if one of the payoffs is $0$. Thus the only interesting form of the game is when all payoffs are non-zero with the same sign, say, positive.
In equilibrium all pure responses by $A$ to $B$'s mixed strategy $\{p_N\}$ must yield the same expected payoff $p_Nf(N)$ and vice versa, so an equilibrium strategy would need to have the form
$$p_N=f(N)^{-1}\left(\sum_{k=1}^\infty f(k)^{-1}\right)^{-1}\;.$$
Thus an equilibrium strategy exists iff $f(N)^{-1}$ is summable. For instance, for $f(N)=N^2$, the equilibrium strategy (for both players) is $p_N=6/(\pi N)^2$, and the value of the game is $6/\pi^2$.
If $f(N)^{-1}$ is not summable, $B$ can make the expected payoff as low as she wants by spreading her strategy arbitrarily far out to infinity. For instance for $f(N)=N$, the strategy $p_N=N^\alpha/\sum N^\alpha$ with $\alpha\lt-1$ allows $A$ to pick $N=1$ for a payoff of $\zeta(-\alpha)^{-1}$, which goes to zero as $\alpha\nearrow-1$.