Finding pure-strategy subgame-perfect Nash equilibria

Question

I'm interested in finding the pure-strategy subgame-perfect Nash equilibria of the game below. What is confusing me is that after player A chooses between reducing and not reducing his end payoffs, the game is not sequential anymore, but simultaneous. How do I go about finding the subgame-perfect Nash equilibria?

I thought of first finding the Nash equilibria in the two simultaneous-move subgames and I got 4 equilibria in total (written as the outcomes):

$(2,4)$ and $(4,2)$ for the left subgame, and

$(1,4)$ and $(3,2)$ for the right subgame.

From this, I deduced what player A would choose in the first stage (don't reduce or reduce), based on possible payoffs, so I got the following two subgame-perfect Nash equilibria:

$ \{Don't\space reduce, a_2, b_2\}$, and

$\{Reduce,a_2,b_2\}$.

I'm not sure if this is correct. After determining the Nash equilibria in the subgames, I get confused trying to determine what player A will do in the first stage. Am I doing this right and can someone explain how to approach solving this?

Answer 1

Because you are looking for subgame-perfect equilibria, it is the correct approach to solve this game backwards. There are 2 proper subgames, and you identified the correct Nash equilibria there. However, note that you specified the payoffs, not the strategies. The correct formulation would be $(a_1,b_1)$ instead of $(2,4)$.

Next, you substitute the subgame perfect payoffs for the subgame and check for the optimal selection of A in the beginning. It is important that an equilibrium is always a full strategy profile. That is, it needs to specify an action for every information set, regardless of whether it is reached on equilibrium path. A has three actions, one at the beginning, one in the left and one in the right subgame. B has two actions, one for each subgame. So a strategy profile would be of this form [(first,left,right),(left,right)].

You have to check for all of the combinations. If you consider payoff (2,4) on the left and (1,4) on the right, you have a SPNE [(don't reduce,a1,a1),(b1,b1)]. Next, consider (2,4) and (3,2), you have a SPNE [(reduce,a1,a2),(b1,b2)]. You also have two more SPNE: [(don't,a2,a1),(b2,b1)] and [(don't,a2,a2),(b2,b2)]. That is, there are four SPNE.

If you do not specify what happens off path, you cannot evaluate what happens when player A deviates in the first stage and thus cannot determine whether no profitable deviations exist.