<p>Isn't the mixed extension of matching pennies a good example of this? The strategies <span class="math-container">$p,q$</span> are both elements of <span class="math-container">$[0,1]$</span>, and
<span class="math-container">$$
\begin{align*}
U_1(p,q) &amp; = pq + (1-p)(1-q) \\
\\
U_2(p,q) &amp; = p(1-q) + (1-p)q.
\end{align*}
$$</span>
There is a unique Nash-equilibrium at <span class="math-container">$p = q = 1/2$</span> but all <span class="math-container">$p$</span> and all <span class="math-container">$q$</span> are rationalizable, because in the equilibrium all strategies are best responses.</p>

<hr>

<p>Even if you were to edit your definition of symmetry to include symmetric utility functions, a game with
<span class="math-container">$$
U(p,q) = -(1/2 - p)^2 \cdot (1/2 - q)^2
$$</span>
yields the same result.</p>

<hr>

<p>If you insist on some shrinkage one could easily edit the utility function to
<span class="math-container">$$
U(p,q) = - 1 \text{ if } p+q \leq 6/10. 
$$</span>
Then some strategies - the ones under <span class="math-container">$1/10$</span> - would not be rationalizable.</p>

CC BY-SA 4.0
2019-03-20T19:02:27.757Z
2019-03-21T18:32:01.023Z
2019-03-21T18:32:01.023Z
Giskard
<p>Suppose we have a nice symmetric game with <span class="math-container">$n$</span> players, i.e. each player's action space is the same compact interval of the real line. I am tasked with identifying all of the rationalizable action profiles, and I am wondering if this set of action profiles will include profiles other than the unique symmetric Nash equilibrium. I know that there are often more rationalizable action profiles than there are Nash equilibria, but I'm specifically interested in nice symmetric games.</p>

<p>Starting with the initial action set for player <span class="math-container">$i$</span>, denoted <span class="math-container">$A_i$</span>, I have iterated the rationalization operator twice so far, and as expected, the set of available actions shrinks, and always includes the Nash equilibrium action. Is there some point when iterating the rationalization operator will cease to reduce the set of available actions?</p>

CC BY-SA 4.0
2019-03-20T16:55:35.693Z
"I am wondering if this set of action profiles will include profiles other than the unique Nash equilibrium" - How do you know that there is a unique NE?
Bayesian
CC BY-SA 4.0
2019-03-20T18:20:57.243Z
Other than the unique *symmetric* Nash equilibrium. I will edit.
David
<game-theory><nash-equilibrium><dominant-strategy>
Rationalizable action profiles in nice symmetric games
David
# Rationalizable action profiles in nice symmetric games

(Asked by: username_0 on 2019-03-20)

Suppose we have a nice symmetric game with $n$ players, i.e. each player's action space is the same compact interval of the real line. I am tasked with identifying all of the rationalizable action profiles, and I am wondering if this set of action profiles will include profiles other than the unique symmetric Nash equilibrium. I know that there are often more rationalizable action profiles than there are Nash equilibria, but I'm specifically interested in nice symmetric games.

Starting with the initial action set for player $i$, denoted $A\_i$, I have iterated the rationalization operator twice so far, and as expected, the set of available actions shrinks, and always includes the Nash equilibrium action. Is there some point when iterating the rationalization operator will cease to reduce the set of available actions?

**Tags:** game-theory, nash-equilibrium, dominant-strategy
> **Comments:**
> **username_1**: "I am wondering if this set of action profiles will include profiles other than the unique Nash equilibrium" - How do you know that there is a unique NE?  
> **username_0**: Other than the unique *symmetric* Nash equilibrium. I will edit.  

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## 1 answers

### Answer by username_2 on 2019-03-20. Score: 4 | Accepted answer

Isn't the mixed extension of matching pennies a good example of this? The strategies $p,q$ are both elements of $[0,1]$, and
$$
\begin{align\*}
U\_1(p,q) & = pq + (1-p)(1-q) \\
\\
U\_2(p,q) & = p(1-q) + (1-p)q.
\end{align\*}
$$
There is a unique Nash-equilibrium at $p = q = 1/2$ but all $p$ and all $q$ are rationalizable, because in the equilibrium all strategies are best responses.

---

Even if you were to edit your definition of symmetry to include symmetric utility functions, a game with
$$
U(p,q) = -(1/2 - p)^2 \cdot (1/2 - q)^2
$$
yields the same result.

---

If you insist on some shrinkage one could easily edit the utility function to
$$
U(p,q) = - 1 \text{ if } p+q \leq 6/10.
$$
Then some strategies - the ones under $1/10$ - would not be rationalizable.


---

