# Strong sequential equilibria and the existence of others

I am working on the following game

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?

Thanks!

• 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. Apr 16 at 20:09
• @HerrK, well, strong is standart sequential equilibria, not weak. And I don't see why they are not fully mixed Apr 16 at 20:12
• Aren't $y$ and $r$ pure strategies? Apr 16 at 20:15
• @HerrK, yes they are Apr 16 at 20:16

There are three classes of equilibria of this game.

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

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

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 $$$$\mu_2(a\mid y)=1-\mu_2(b\mid y)=\frac58,$$$$ so that P2 can play a mixed strategy that puts sufficiently high probability on $$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.