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I know that in the paper "Perfect Bayesian equilibrium and sequential equilibrium", the authors proved that for signaling games (considering the simple case that there are two periods and the sender has two types), PBE is equivalent to SE.

But the result is based on the assumption that the action space is finite (i.e. in signaling games that means the message space of the sender should be finite). My question is that in the games like the Spence's model where the message space could be continuous, whether every PBE is also SE?

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Good question.

At the moment, we still do not have a good grasp on what it means for an equilibrium to be a sequential equilibrium in the class of games you mention, with continuum of actions or types, because what does a pertubation of strategies mean here?

We think of a strategy pertubation (in SE) as one that puts positive probability in each possible node, but that is clearly impossible in games with infinite actions. Would it be one with full support? We still do not have a concensus.

As a reference, Myerson and Reny (2020) is the current "gold standard" of what a SE in such games could be, but their approach drops the whole existence of strategies, so I find it impractical.

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  • $\begingroup$ Thanks for your response. In many letures about spence's model, the authors often write that the consistency condition of SE imposes no constraints on the off-equilibrium path beliefs but without any further proof or explanation. If SE and PBE are not equivalent, why can they claim that result? That's really confusing to me. $\endgroup$
    – DevinY
    Jan 21 at 16:55
  • $\begingroup$ @IronMan I'm not exactly sure what you mean, but its important to note that every SE is automatically a PBE, while vice-versa it is not true. Perhaps ask a new question here, and I or some other expert might be able to help if you fill in more details. $\endgroup$ Jan 21 at 17:50
  • $\begingroup$ Thanks again. I am clear that "SE is automatically a PBE, while vice-versa it is not true". Here is my new question about my last comment, please see link $\endgroup$
    – DevinY
    Jan 21 at 18:16

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