Extended game mixed SPNE

I have doubts about whether my solution is right. Let $$x>z>y>1$$. Given the extended game above, we have two subgames and will find the Nash equilibria of these.

Observe that player 2 will play $$b$$ in the right subgame. In the left subgame player 2 will be indifferent between playing $$a$$ and $$b$$. Let $$0 \le r \le 1$$ such that $$(r,1-r)$$ is a mixed strategy for player 2 and such that the player plays this strategy in the left subgame.

Player 1 has expected payoffs $$E u(s) = (rx+(1-r)y) \cdot 1_{s=A} + z \cdot 1_{s=B}$$ Observe that player 1's best response is $$A$$ if $$r \ge (z-1)/(x-y)$$ and $$B$$ if $$r \le (z-1)/(x-y)$$, hence SPNE will be $$\{(A,(r,1-r))|r \ge (z-1)/(x-y) \} \cup \{(B,(r,1-r))|r \le (z-1)/(x-y)\}$$ Is this correct?

There are a couple of imprecisions. First, there are 3 subgames: two that start with player 2 moving, and the complete game is also a subgame.

Second I think that A is a best response for player 1 iff $$r\geq \frac{z-y}{x-y}$$ otherwise, B is the best response (perhaps this was just a typo).

Third, note that the fraction is guaranteed to be positive and smaller than 1 given the restrictions on the parameters, but it is not necessarily equal to $$0.5$$, For example, let $$x=5$$, $$z=3$$ and $$y=2$$, then the fraction is equal to 1/3.

Fourth, you are missing other equilibria, when $$r=\frac{z-y}{x-y}$$, then player 1 is indifferent between A and B, and any randomization of A and B also constitutes an equilibrium.

Lastly, when writing the equilibria, you must specify what player 2 does in each information set (or node in this game)

Thus: $$NE=\{(A, (ra+(1-r)b)b)|r\geq \frac{z-y}{x-y}\}\cup\{(B, (ra+(1-r)b)b)|r\leq \frac{z-y}{x-y}\}$$ $$\cup \{((pA+(1-p)B), (ra+(1-r)b)b)|r=\frac{z-y}{x-y}, 0\leq p\leq 1\}$$

Where $$(ra+(1-r)b)b$$ specifies that player 2 randomizes in the first node and chooses b in the second node.

• I don't see how the Third point is relevant. Other than that, the other points are valid.
– user28372
Jun 13 '20 at 9:01