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What happens if we assume that there is incomplete information to the prisoner's dilemma game?

For example, suppose we have the following matrix with the utilities $T>R>P>S$ and $2R>S+T$ with only $2$ players and $2$ actions.

enter image description here

With $\Theta_i=\{\underline{\theta},\bar{\theta}\}$, does this mean that we have $2$ subgames (ex-interim) where we fix player $1$'s type and assign probabilities to player $2$'s types? Is this is the correct way to analyze the game?

If there is a paper studying the one-shot Prisoner's dilemma as a Bayesian game, I'd appreciate it very much if you share it with me!

Thank you.

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    $\begingroup$ If the types don't affect the payoffs, then there's no change in the predicted outcome. $\endgroup$ – Herr K. Feb 5 at 16:11
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    $\begingroup$ I found this blog related to one-shot Prisoner's dilemma online. schneier.com/blog/archives/2013/05/one-shot_vs_ite.html $\endgroup$ – Mike J Feb 5 at 19:13
  • $\begingroup$ @HerrK. But usually in a Bayesian game, we define the utility functions as $u:\Theta\times A\to\mathbb{R}$, where $\Theta$ is the type space and $A$ is the action space. My point is that the types will necessarily affect the payoffs, right? $\endgroup$ – johnny09 Feb 5 at 20:54
  • $\begingroup$ @johnny09: Yes. But how do types affect payoffs in your case? How do $P,R,S,T$ change when $\theta$ changes? $\endgroup$ – Herr K. Feb 6 at 1:20
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    $\begingroup$ @johnny09: Sorry I didn't get pinged for your last comment. The graduate micro text by Mas-Colell, Whinston and Green is a good reference. Bayesian games and the associated Bayesian Nash equilibrium is introduced on page 255. $\endgroup$ – Herr K. Feb 7 at 4:52

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