Here are some things to know.
The box on isoclines remains hidden. When finding the equilibria of the dynamical system associated with the replicator equations, it is sufficient to confirm that the “obvious” equilibria are indeed equilibria. Since the the rate of change of the frequencies of the strategy types are proportional to the frequencies of the strategy types themselves, we would expect that any point where there is only a single strategy type present would be an equilibrium. Hence in the example in the practice final you would check that (0,0), (0,1), and (1,0) are indeed equilibria of your dynamical system by plugging those points into your replicator equations and checking that they are indeed zero. The other obvious equilibrium to check is the one corresponding to your mixed strategy nash equilibrium. In the case of the practice final that would be (1/2,1/2).
I’ll be at the exam tomorrow and you can just hand in your bonus assignment before you go into your exam.
Someone asked some good questions and I answered them:
1. We talked about dominated strategies, and how they could be “crossed out” or ignored for strong nash. But can a dominated strategy enter into a mixed Nash? Or is it always irrelevant to the game analysis entirely?
A dominated strategy will not be part of mixed strategy Nash equilibrium, it is essentially irrelevant.
2. For mixed strategy nash, we sometimes set payoffs equal for each strategy. Besides assuming the population is large, this also assumes a symmetric game. Is that right?
No, this technique works for asymmetric games as well. Setting the payoffs of player one’s actions equal to eachother allows you to find the probabiliites with which player two should choose their strategies in order to make player one indifferent to their own choice of action. And vice versa, Setting the payoffs of player two’s actions equal to eachother allows you to find the probabiliites with which player one should choose their strategies in order to make player two indifferent to their own choice of action
3. For the technique of setting payoffs equal to find mixed nashes, if there are 3+ strategies, do all three have to be equal, or just any two and both larger than the third?
Just any two and both that are equal larger than the other strategies.
4. In an evolutionary game, we have symmetry, but there can still be off-diagonal nash equilibria. They will not correspond to ESS since they are impossible strategies for a population. They also will not be fixed points for the replicator equations for the same reason. Is that true? Nash <==> fixed point otherwise though, right?
Yes this is true.
5. Is it true that ESS <==> stable fixed point? Or is that just a one way (ESS==> stable fixed point)?
Just one way ESS==> stable fixed point. (the double implication holds in most cases but not all)
6. For section 3 of question 2 in this year’s practice final, the online solutions says that the only nash is a (0,0), but in class when we drew it, there were clearly at least 3 points of intersection. Can you please clarify on which one is the correct solution?
There were actually only two points of intersection of the best response function (x=-3/sq3, y = 3) and
(x=2, y=1). The three intersections I believe you are referring to are from when we were comparing the payoff to player one at their two distinct local maxima to determine the global maxima.
7. Also, if the the one with 3 equilibria is correct, how do we know which one is the best for either player? just plug in an estimated value and check?
You plug in each equilibrium point into the original payoff functions and see which equilibrium is best for each player.