<p>That's a generalized guessing game. I think the &quot;intended&quot; answer is that if <span class="math-container">$\theta&lt;1/100$</span> (&quot;guess some fraction <span class="math-container">$&lt;1$</span> of the average&quot;), the only rationalizable strategy is <span class="math-container">$0$</span>, if <span class="math-container">$\theta&gt;1/100$</span> (&quot;guess some multiple <span class="math-container">$&gt;1$</span> of the average&quot;), the only rationalizable strategy is <span class="math-container">$1$</span>, and if <span class="math-container">$\theta=1/100$</span> (&quot;guess the average&quot;), all strategies are rationalizable.</p>
<p>However, the way it is formulated actually yields a different answer. The &quot;most closed&quot; number should presumably be the one which is &quot;closest&quot; (of all chosen numbers) to the target number. Since no tie-breaking rule is stated, this implies that if all chosen numbers are equal, then all those numbers are closest to the target number and everyone gets a payoff of <span class="math-container">$1$</span>. So everyone choosing the same (arbitrary) number is a Nash equilibrium for every <span class="math-container">$\theta$</span>. Therefore, every number is rationalizable.</p>

CC BY-SA 4.0
2022-12-19T09:25:07.627Z
2022-12-19T09:25:07.627Z
VARulle
<p>For the question below, how can we solve it generally for every value of θ? As the θ is not discrete, I am not sure how to apply iterated elimination of dominated strategies in this question. And is there any Nash equilibrium? Any help is greatly appreciated.</p>
<p><a href="https://i.sstatic.net/yrHpv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yrHpv.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-12-18T22:43:26.407Z
What is the "official" definition of rationalizability you should use here?
Michael Greinecker
CC BY-SA 4.0
2022-12-19T09:03:50.410Z
@MichaelGreinecker, it's an exercise from a script that allows for correlated beliefs.
VARulle
<microeconomics><mathematical-economics><game-theory><nash-equilibrium>
Rationalizable strategies/ Nash equilibrium
KK econ
# Rationalizable strategies/ Nash equilibrium

(Asked by: username_0 on 2022-12-18)

For the question below, how can we solve it generally for every value of θ? As the θ is not discrete, I am not sure how to apply iterated elimination of dominated strategies in this question. And is there any Nash equilibrium? Any help is greatly appreciated.

[![enter image description here](https://i.sstatic.net/yrHpv.png)](https://i.sstatic.net/yrHpv.png)

**Tags:** microeconomics, mathematical-economics, game-theory, nash-equilibrium
> **Comments:**
> **username_1**: What is the "official" definition of rationalizability you should use here?  
> **username_2**: @MichaelGreinecker, it's an exercise from a script that allows for correlated beliefs.  

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## 1 answers

### Answer by username_2 on 2022-12-19. Score: 3

That's a generalized guessing game. I think the "intended" answer is that if $\theta<1/100$ ("guess some fraction $<1$ of the average"), the only rationalizable strategy is $0$, if $\theta>1/100$ ("guess some multiple $>1$ of the average"), the only rationalizable strategy is $1$, and if $\theta=1/100$ ("guess the average"), all strategies are rationalizable.

However, the way it is formulated actually yields a different answer. The "most closed" number should presumably be the one which is "closest" (of all chosen numbers) to the target number. Since no tie-breaking rule is stated, this implies that if all chosen numbers are equal, then all those numbers are closest to the target number and everyone gets a payoff of $1$. So everyone choosing the same (arbitrary) number is a Nash equilibrium for every $\theta$. Therefore, every number is rationalizable.


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