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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.

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  • $\begingroup$ What is the "official" definition of rationalizability you should use here? $\endgroup$
    – Michael Greinecker
    Commented Dec 18, 2022 at 22:43
  • $\begingroup$ @MichaelGreinecker, it's an exercise from a script that allows for correlated beliefs. $\endgroup$
    – VARulle
    Commented Dec 19, 2022 at 9:03

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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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