Suppose $N$ players play a game, where each player's action space is $[0,1]$.

Each player has an identical continuous utility function $u:[0,1]\times [0,1]^{N-1}\rightarrow\mathbb{R}$, where the first argument is their own action, and the other arguments are the actions of other players, with $u$ being invariant under re-orderings of the other players.

Suppose that the game has a unique pure, symmetric, Nash equilibrium with each player taking action $a\in (0,1)$.

Now suppose that the game's rules are changed, and players can now only choose actions in the interval $[b,1]$, where $b\in (0,a)$. By revealed preference, each player taking action $a$ is still a Nash equilibrium (if they did not have a profitable deviation before, they do not now).

Is it possible (for some $u$, hopefully not too weird!) that each player taking action $b$ is now an additional Nash equilibrium, even though it was not before? Is it possible that the $b$ equilibrium survives some refinement (e.g. trembling hand) that the $a$ equilibrium does not?

Now this has been answered, I'll briefly note the particular circumstance I had in mind. The $N$ players are countries, choosing the level of their corporation tax rate. $a$ is the moderate level of corporation taxes (most) countries have set in the absence of international agreement. $b$ is the new minimum level now internationally agreed. Whether the $u$ in the answer looks anything like countries' pay-offs in setting their corporation tax rate is another question!

  • $\begingroup$ Before the change, players cannot choose 1, after the change they can. Is this a typo? $\endgroup$
    – VARulle
    Jul 2, 2021 at 16:58
  • $\begingroup$ Before the change the action space is $[0,1]$, so they can choose $1$. Maybe we're misunderstanding each other? $\endgroup$
    – cfp
    Jul 2, 2021 at 17:13
  • $\begingroup$ Yeah, sorry, I misread the $a\in(0,1)$ in the third paragraph as the action space... $\endgroup$
    – VARulle
    Jul 2, 2021 at 17:59
  • $\begingroup$ With "a unique pure, symmetric, Nash equilibrium" do you mean a unique symmetric NE among all pure ones, or a unique pure NE among all symmetric ones, or both? $\endgroup$
    – VARulle
    Jul 2, 2021 at 18:34
  • 2
    $\begingroup$ In response to the edit: $u$ is unlikely to be continuous. A simple model of the problem of corporate/capital taxation would be a type of Bertrand competition, where the state setting the lowest tax rate gets all business. In reality there is undoubtedly some friction, but jumps are still likely when the tax rate is low enough to entice a megacorporation. $\endgroup$
    – Giskard
    Jul 3, 2021 at 11:12

1 Answer 1


Let $N=2$ and for $(x,y)$ and $(p,q)$ in $[0,1]^2$ let $d_{p,q}(x,y)$ be the Euclidean distance between $(p,q)$ and $(x,y)$, i.e. $d_{p,q}(x,y)=[(p-x)^2+(q-y)^2]^{1/2}$.

Choose $k>0$ such that $k<a-b$ and $k<b$.

Now define $$u(p,q):=\max\{ -d_{p,q}(a,a),-d_{p,q}(b,b),k-d_{p,q}(b,0),k-d_{p,q}(0,b) \}.$$ This function is continuous as a maximum of continuous functions.

The construction makes $u(p,q)$ negative everywhere except at $(a,a)$ and at $(b,b)$, where it is $0$, and in the (small) $k$-discs around $(b,0)$ and $(0,b)$, where it reaches a maximum value of $k$ at these points.

In the unrestricted case, $(a,a)$ is a symmetric, pure NE, but $(b,b)$ is not, since each player would want to unilaterally deviate to playing $0$. In all other symmetric strategy profiles players would want to unilaterally deviate to either $a$, $b$, or $0$.

In the restricted case, $(b,b)$ is also a pure, symmetric NE, since a downward deviation is no longer possible.

This example can straightforwardly be generalized to $N\ge 3$ players. I leave the refinement part unanswered here but I think this could also be constructed in a similar way.

  • $\begingroup$ I've added some context to the question on the particular circumstance I had in mind, which you may find interesting. $\endgroup$
    – cfp
    Jul 3, 2021 at 8:55

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