Game Theory

Game Theory

Question 1. Describing individual’s behaviors

Each individual plays A 80% of the time and B 20% of the time.

Question 2. Difference between a pure strategy and a pure ESS.

A pure strategy is a composition of behaviors that one will instill in a given set of circumstances while a pure ESS is a single strategy that cannot be invaded by any other known strategy.

Question 3. Difference between a mixed strategy and a mixed ESS.

A mixed strategy is one which is made up of many different strategy components while a mixed ESS which has several behaviors which unite to make up an equilibrium.

Question 4. Frequency of strategy B.

The frequency of strategy B: B= 1.0-a =1.0-0.9999 =0.0001

Question 5. Frequency of A vs. A interactions.

Overall in the entire population, = a^2 = 0.9999^2 = 0.9998

From the point of view of an A strategist =a= 0.9999

Question 6. Frequency of B vs. B interactions.

Overall in the entire pop=b^2 =0.0001^2= 0.00000001(1e-8)

From the point of view of a B strategist =b = 0.0001

Question 7. Frequency of A vs. B interactions.

Where the payoff is to A – that is, E(A,B) =a*b =0.9999*0.0001 = 0.0000999

Question 8. Frequency of B vs A interactions.

Where the payoff is to B - - that is, E(B,A) = b * a = 0.9999 * 0.0001 = 0.0000999

Question 9. Will proportion of the total number of payoffs to A when vs. B be any different than the proportion of the total number of payoffs to B when vs. A?

As it is in the calculations in question 7 and 8, it can be noted that the frequency in the whole population is all the equal

Question 10. Expression of determining whether or not strategy B is a pure ESS against A.

Use the opponent’s strategy

Focal strategy             A                     B

A                                 E(A,A) =0       E(A,B) =1

B                                 E(B,A) = -0.5  E(B,B) = 0.5

Question 11. Using the expression for B vs. A that you just wrote whether or not B is stable against invasion by A

First review the expressions above. From the payoff matrix, E(B,B) = 0.5 and E(A,B)= 1.0. Therefore, B is not stable to invasion A.

Question 12. Low frequencies A can be considered more fit than B, what does this mean?

This means that if there is more than one B invader, probably there may be some rare interactions switch payoff E(B,B) (Ichiishi, 2014). Also, in this same situation, the payoff E(A,B) starts to matter, even though it is extremely rare. Again, this can be referred the stability property.

Question 13. General form of equations.

 Payoff (to strat., when vs. astrat.) =

[(chance of win) * (resource value – cost of win)]

                                    +

[(chance of loss) 8 cost of loss]

Question 14. Calculate the payoff of the aggressor vs. aggressor.

From the equation in question 13 E(H,H) = (0.5*50) + (0.5*-100) = 25-50

                                                                                    = -25

Question 15. Calculation of the payoff to aggressor when vs. Non-Aggressor.

E(H,D) = 1.0 * 50 -0 = +50

Question 16. Calculation of the payoff to non-aggressor when vs Aggressor.

E(D,H) = 0 * 50 + 1.0 * 0 = 0

Question 17. Calculation of the payoff to non-aggressor when vs. non-aggressor.

E(D,D) = (0.5) * (50 -10) – (0.5) * (-10) = +15

Question 18. So defied by these payoffs (equations), the pay-off matrix is:

                                               

Opponent

Focal Strategy            Aggressor                               Non-Aggressor

Aggressor                   -25                                           +50

Non-Aggressor           0                                              +15

Question 19. Does our example meet the criteria for a pure ESS, why or why not?

Yes, it meets the criteria for a pure ESS.

Define: Non-aggressive A and Aggressive B.

E(A,A) is E(H,H) = -25,

E(B,A) is E(D,H) = 0

E(H,H) < E(D,H), therefore H is not an ESS

for D

E(D,D) = +15 and E(H,D) = +50

therefore D is not an ESS since E(D,D) < E(H,D)

NO PURE ESS

Question 20. What happens at the equilibrium point? What if you add individuals of either strategy?

At the equilibrium point, there will be a mixed ESS which is known as the frequencies of the strategies where the two of them have same fitness. Adding individuals of either strategy lowers relative fitness of all members of the strategy.

Question 21. What does addition of more aggressors do?

Addition of aggressors lowers their fitness relative to Non-Aggressors.

Question 21. How does the aggressor strategy do when in high frequency compared to when its rare? What does this mean?

Aggressor does very poorly when in high frequencies compared to when it is rare (Ichiishi, 2014). This means that it is easily invaded by what might seem to be most improbable of invaders, the pacific “non-aggressor”

Question 22. Is the Non-Aggressive strategy stable? Why or why not?

The non-aggressive strategy is not stable since the aggressor does extremely well when entering the population.

Question 23. What assumptions would you have to make to turn this into a pure ESS?

Doing the problem above, you will realize that neither, aggressor or non-aggressor is pure ESSs given the payoffs calculated from the equations and values for benefits and costs presented in the above related question (Ichiishi, 2014). To turn this into a pure you need to use simulation.

Question 24. What can we generally say about the evolutionary strategy of the two sable island horse populations?

The result is evolutionary stasis with respect to the behaviors being considered -- there is no change in relative frequency of strategies over time.

References

Ichiishi, T. (2014). Game Theory for Economic Analysis. London: Elsevier.