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I know that it is possible to have a Nash equilibrium which is not an equilibrium in dominant strategy, but is it also applicable for equilibrium in weakly dominant strategy (i.e. a Nash equilibrium which is NOT an equilibrium in weakly dominant strategy)? If yes, what would be an example?

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\begin{array}{|c|c|c|} \hline &L&R\\\hline T&1,1&0,0\\\hline B&0,0&0,0\\\hline \end{array}

In the game above, there are two pure strategy Nash equilibria:

  • $(T,L)$ is an equilibrium in weakly dominant strategies;
  • $(B,R)$ is an equilibrium in weakly dominated strategies.

Noting that "dominant" and "dominated" are two different words, I believe the above example answers both the question in your post as well as your question in the comment below.

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  • $\begingroup$ This is not the question I asked. I asked about Nash equilibrium which is NOT an equilibrium in weakly dominated strategies. $\endgroup$
    – Aqqqq
    Commented Nov 21, 2019 at 22:08
  • $\begingroup$ @Aqqqq Actually, this is the answer to the question you asked! You edited the question later. This is clear from the edit history. $\endgroup$
    – Giskard
    Commented Nov 21, 2019 at 22:09
  • $\begingroup$ @Giskard I edited the question to make it clearer for understanding. No meaning was changed. The only thing I added was the content in the bracket. $\endgroup$
    – Aqqqq
    Commented Nov 21, 2019 at 22:11
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    $\begingroup$ @Aqqqq: Your question asks for NE that's not in weakly dominant strategies. $(B,R)$ is NOT a Nash equilibrium in weakly dominant strategies, it's in weakly dominated strategies. $\endgroup$
    – Herr K.
    Commented Nov 21, 2019 at 22:12
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    $\begingroup$ @Aqqqq I believe you. Herr K's answer does answer both your questions though. $\endgroup$
    – Giskard
    Commented Nov 21, 2019 at 22:15

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