I am working on the following game

enter image description here

and I have to find all strong sequential equilibria here.

I determined that here any belief derived from a fully mixed strategy gives a distribution (1/2, 1/2) over the nodes in Player 2’s information set. Given this, player 2 will chose r and player 1 will chose y. I believe that this is the unique sequential equilibrium. But is it strong?

And are there any other Nash equilibria and if so why they are not sequential?


  • $\begingroup$ What do you mean by "strong" SE? Also the belief you derive for P2 in a fully mixed equilibrium seems incorrect. The resulting strategies ($y$ for P1 and $r$ for P2) are not "fully mixed", and do not induce the $(\frac12,\frac12)$ belief for P2. $\endgroup$ – Herr K. Apr 16 at 20:09
  • $\begingroup$ @HerrK, well, strong is standart sequential equilibria, not weak. And I don't see why they are not fully mixed $\endgroup$ – Rodger Haggings Apr 16 at 20:12
  • $\begingroup$ Aren't $y$ and $r$ pure strategies? $\endgroup$ – Herr K. Apr 16 at 20:15
  • $\begingroup$ @HerrK, yes they are $\endgroup$ – Rodger Haggings Apr 16 at 20:16

There are three classes of equilibria of this game.

The first class is sequential: \begin{equation} (s_1,s_2)=(y,r) \end{equation} and the beliefs are \begin{equation} \mu_1(a)=\mu_1(b)=\mu_2(a\mid y)=\mu_2(b\mid y)=\frac12. \end{equation}

enter image description here

The second class is not sequential, but weak perfect Bayesian: \begin{equation} (s_1,s_2)=(x,l) \end{equation} and the beliefs are \begin{equation} \mu_1(a)=\mu_1(b)=\frac12,\quad\text{but }\mu_2(a\mid y)=1-\mu_2(b\mid y)=p>\frac58. \end{equation}

enter image description here

The third class is actually a boundary case of the second class, which permits P2 to use mixed strategies. This equilibrium requires P2's off equilibrium belief to be \begin{equation} \mu_2(a\mid y)=1-\mu_2(b\mid y)=\frac58, \end{equation} so that P2 can play a mixed strategy that puts sufficiently high probability on $l$ ($\sigma_2(l)\ge\frac35$). It is also necessary that P1 does not play $y$ with positive probability, because if they do, Bayes rule would kick in, requiring that P2's belief be $(\frac12,\frac12)$ instead of $(\frac58,\frac38)$ which rationalizes P2's mixed strategy. enter image description here


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