Here is a practice final. It is dangerously similar to your real final. If you only do one practice final do this one.
I’m aiming to have solutions up for this by Monday.
I was just going through the Resource notes on the website and “Recipe 7.5 Finding the Fixed Points In 2-D systems using Isoclines” on page 7 of “Crash Course in Dynamical Systems” is cut off. I was wondering if you could post what that box is supposed to say.
Question about the first question on this practice final … in the first example ( four people are playing, and one player chooses 1, two players choose 2 and 1 player chooses 3), wouldn’t the person who played 1 win? He has the lowest integer and fewer than n (=1) other people chose 1 (in fact 0 other people chose it). Why would the players who chose 2 win?
Why would the players who chose 2 win? Because there is a typo which is now fixed. The player who chose one would win.
About the isocline box. I’m reworking those notes, I’ll try and get those up by Tuesday.
Would it be possible for you to post solutions to the assignments? Specifically, the non-analysis assignments, such as the “How Long to Wait” and the “ESS” assignment.
I will aim to have solutions for those assignments up by Thursday.
For the first question… I think I must be off somewhere….
For everyone to win, everyone must pick the same number and there is only one winning number (since it needs to be the lowest such number to satisfy the winning condition) so how is there two Nash Equilibria?
All will be revealed in the late but immenent solutions pdf.
Hi Dan, just wondering if a time and place has been decided for the review session?
There will be two. One tomorrow, Wednesday from 2 until 5. And another on Saturday starting at 5:30. I will be answering questions and if there are no questions I will be working through assignments and the practice final.
Will it be in the math help centre?
It’ll be in that room behind the math help centre and if too many people show up for that spot we’ll migrate elsewhere in jefferey and leave a note.
When are you planning on posting solutions to this year’s practice final?
are the isocline notes up, yet?
Nope! We’re screwed!
You are not as screwed as you think. As discussed in both help sessions, when dealing with replicator equations there are several obvious equilibria. Any point where there is only one strategy type, will certainly be equilibria. Similarly the mixed strategy Nash equilibrium should correspond to an equilibrium point in the replicator equations dynamical system. All you need to do to find the equilibria is to check that these “obvious” equilibria are indeed equilibria by confirming that both replicator equations are zero when you evaluate them at these candidate equilibria.
If I find an ESS (pure or mixed) and the analysis with Jacobian tells me the ESS is not stable, then I must have the replicator equation and/or the MSNE wrong?
Yes, an ESS will always be a stable equilibrium point!
Hopefully you see this in time.
When you’re drawing diagrams to summarize mixed strategy equilibria, and you have your isoclines drawn out, how do you determine the directions of the trajectories in each region?
I think you use could just plug in values on that isocline into your replicator equations and see how p and q are changing.
could you tell me how you found the isoclines though?
Just set the replicator equations equal to zero. You’ll get one isocline for p=0 and one for q=0, plus others that are equations with p in terms of q or vice versa
If there are easy to draw isoclines, then you should draw them.
p = 1-2q is an easy to draw isocline.
p= (1+2q-5q^2)/(1-5q) is a hard to draw isocline.
On one side of a p isocline the direction of the trajectories in the p dimension will be positive and on the other side of that isocline it will be negative. Similarly for a q isocline the direction of the trajectories in the q dimension will positive on one side of the isocline and negative on the other.
Setting the p and q replicator equations equal to zero is indeed how you find the isoclines for p and q respectively.
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