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I am trying to solve the above problem. I know that the two Nash Equilibria are $(D,L)$ and $(C,R)$, so the first four pure strategy Nash equilibria would just involve playing either of these strategies in both periods. I understand that, for all other possible pure strategy Nash equilibria, the second/last game must still end with one of these Nash equilibria as well but could start in the first game with something that is not a Nash equilibria.

However, I am somewhat confused by what the other two Nash equilibria are.

One of them is:

If $(U,L)$ is played in the first period, then $(D,L)$ is played in the second period. Otherwise, $(C,R)$ is played in the second period.

The other one is:

If $(C,M)$ is played in the first period, then $(C,R)$ is played in the second period. Otherwise, $(D,L)$ is played in the second period.

I have noted the discount factor of $\frac{3}{4}$, so understand that if, for instance if $(U,L)$ is played in the first period and $(D,L)$ in the second, then the payoff for the row player would be $5+\frac{3}{4}\left(6\right)\:=\:9.5$, but I am not sure how to apply this here, although I know I have to.

Any help would be highly appreciated!

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1 Answer 1

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Let's analyze your first suggestion:

"If $(U,L)$ is played in the first period, then $(D,L)$ is played in the second period. Otherwise, $(C,R)$ is played in the second period."

The idea here is that Row player "sacrifices" something in the first period (he gets $5$ instead of $6$) so that his favorite NE will be played in the second period. The $(C,R)$ equilibrium is the way the column player "punishes" him for deviations.

So the payoff of the row player in this scenario is indeed $5+\tfrac{3}{4}\cdot 6=9.5$. If he deviates, however, and plays $D$ in the first round, he gets $6+\tfrac{3}{4}\cdot 4=9$. You can repeat this calculation for Player 2 and verify that this is indeed an equilibrium.

Same idea applies to other potential equilibrium profiles.

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