<p>Let <span class="math-container">$N=2$</span> and for <span class="math-container">$(x,y)$</span> and <span class="math-container">$(p,q)$</span> in <span class="math-container">$[0,1]^2$</span> let <span class="math-container">$d_{p,q}(x,y)$</span> be the Euclidean distance between <span class="math-container">$(p,q)$</span> and <span class="math-container">$(x,y)$</span>, i.e. <span class="math-container">$d_{p,q}(x,y)=[(p-x)^2+(q-y)^2]^{1/2}$</span>.</p>
<p>Choose <span class="math-container">$k&gt;0$</span> such that <span class="math-container">$k&lt;a-b$</span> and <span class="math-container">$k&lt;b$</span>.</p>
<p>Now define
<span class="math-container">$$u(p,q):=\max\{ -d_{p,q}(a,a),-d_{p,q}(b,b),k-d_{p,q}(b,0),k-d_{p,q}(0,b)  \}.$$</span>
This function is continuous as a maximum of continuous functions.</p>
<p>The construction makes <span class="math-container">$u(p,q)$</span> negative everywhere except at <span class="math-container">$(a,a)$</span> and at <span class="math-container">$(b,b)$</span>, where it is <span class="math-container">$0$</span>, and in the (small) <span class="math-container">$k$</span>-discs around <span class="math-container">$(b,0)$</span> and <span class="math-container">$(0,b)$</span>, where it reaches a maximum value of <span class="math-container">$k$</span> at these points.</p>
<p>In the unrestricted case, <span class="math-container">$(a,a)$</span> is a symmetric, pure NE, but <span class="math-container">$(b,b)$</span> is not, since each player would want to unilaterally deviate to playing <span class="math-container">$0$</span>. In all other symmetric strategy profiles players would want to unilaterally deviate to either <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, or <span class="math-container">$0$</span>.</p>
<p>In the restricted case, <span class="math-container">$(b,b)$</span> is also a pure, symmetric NE, since a downward deviation is no longer possible.</p>
<p>This example can straightforwardly be generalized to <span class="math-container">$N\ge 3$</span> players. I leave the refinement part unanswered here but I think this could also be constructed in a similar way.</p>

CC BY-SA 4.0
2021-07-03T08:55:51.273Z
I've added some context to the question on the particular circumstance I had in mind, which you may find interesting.
cfp
CC BY-SA 4.0
2021-07-02T18:59:40.317Z
2021-07-02T18:59:40.317Z
VARulle
<p>Suppose <span class="math-container">$N$</span> players play a game, where each player's action space is <span class="math-container">$[0,1]$</span>.</p>
<p>Each player has an identical continuous utility function <span class="math-container">$u:[0,1]\times [0,1]^{N-1}\rightarrow\mathbb{R}$</span>, where the first argument is their own action, and the other arguments are the actions of other players, with <span class="math-container">$u$</span> being invariant under re-orderings of the other players.</p>
<p>Suppose that the game has a unique pure, symmetric, Nash equilibrium with each player taking action <span class="math-container">$a\in (0,1)$</span>.</p>
<p>Now suppose that the game's rules are changed, and players can now only choose actions in the interval <span class="math-container">$[b,1]$</span>, where <span class="math-container">$b\in (0,a)$</span>. By revealed preference, each player taking action <span class="math-container">$a$</span> is still a Nash equilibrium (if they did not have a profitable deviation before, they do not now).</p>
<p>Is it possible (for some <span class="math-container">$u$</span>, hopefully not too weird!) that each player taking action <span class="math-container">$b$</span> is now an additional Nash equilibrium, even though it was not before? Is it possible that the <span class="math-container">$b$</span> equilibrium survives some refinement (e.g. trembling hand) that the <span class="math-container">$a$</span> equilibrium does not?</p>
<hr />
<p>Now this has been answered, I'll briefly note the particular circumstance I had in mind. The <span class="math-container">$N$</span> players are countries, choosing the level of their corporation tax rate. <span class="math-container">$a$</span> is the moderate level of corporation taxes (most) countries have set in the absence of international agreement. <span class="math-container">$b$</span> is the new minimum level now internationally agreed. Whether the <span class="math-container">$u$</span> in the answer looks anything like countries' pay-offs in setting their corporation tax rate is another question!</p>

CC BY-SA 4.0
2021-07-02T16:58:32.430Z
Before the change, players cannot choose 1, after the change they can. Is this a typo?
VARulle
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2021-07-02T17:13:16.613Z
Before the change the action space is $[0,1]$, so they can choose $1$. Maybe we're misunderstanding each other?
cfp
CC BY-SA 4.0
2021-07-02T17:59:57.523Z
Yeah, sorry, I misread the $a\in(0,1)$ in the third paragraph as the action space...
VARulle
CC BY-SA 4.0
2021-07-02T18:34:00.053Z
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?
VARulle
CC BY-SA 4.0
2021-07-02T18:37:08.270Z
@VARulle I meant that there was only one symmetric, pure NE. There may be other NE that are either non pure, or non symmetric.
cfp
CC BY-SA 4.0
2021-07-03T11:12:07.830Z
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.
Giskard
CC BY-SA 4.0
2021-07-03T11:46:15.663Z
Dixit-Stiglitz converges to Bertrand but has nice continuous pay offs so it may not make so much difference? But in any case, wouldn't that make the b equilibrium more easy to sustain?
cfp
<game-theory><nash-equilibrium><theory>
Effect of bounding action space on the set of equilibria
cfp
# Effect of bounding action space on the set of equilibria

(Asked by: username_0 on 2021-07-02)

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!

**Tags:** game-theory, nash-equilibrium, theory
> **Comments:**
> **username_1**: Before the change, players cannot choose 1, after the change they can. Is this a typo?  
> **username_0**: Before the change the action space is $[0,1]$, so they can choose $1$. Maybe we're misunderstanding each other?  
> **username_1**: Yeah, sorry, I misread the $a\in(0,1)$ in the third paragraph as the action space...  
> **username_1**: 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?  
> **username_0**: @VARulle I meant that there was only one symmetric, pure NE. There may be other NE that are either non pure, or non symmetric.  
> **username_2**: 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.  
> **username_0**: Dixit-Stiglitz converges to Bertrand but has nice continuous pay offs so it may not make so much difference? But in any case, wouldn't that make the b equilibrium more easy to sustain?  

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## 1 answers

### Answer by username_1 on 2021-07-02. Score: 8 | Accepted answer

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 and $k.

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.

> **Comments:**
> **username_0**: I've added some context to the question on the particular circumstance I had in mind, which you may find interesting.  

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