Solution for MIEG (Part I) Midterm-1, Winter Term, 2015
Ans1
(a) It has unique pure strategy Nash equilibrium (S,S).
(b) Let p,q,r and a,b,c are probabilities by which H, T and S are played by 1 and 2 respec- tively. Strategy profile {(0.5,0.5,0)(0.5,0.5,0)} is a Nash equilibrium. Note that strategies H and T by 1 and 2 form the game of matching of pennies. When H and T both are played with positive probabilities two things follows:
1. H and T must be played with equal probabilities by other player because only then H and T can give equal payoff and that is 0.
2. Then S can’t be played with positive probabilities because if other player plays S then the expected payoff of a player from playing S is positive.
(c) We will show no other mixed strategy exists by taking every possible case. One important observation, each pure strategy played by other player has uniquebest response which is also a pure strategy and other than (S,S) no pure strategy profile is Nash equilibrium.
Case1: a,b 6= 0 and c = 0. Under this case the only possible Nash is as in part (b).
Case2: a,c 6= 0 and b = 0. It makes S strictly dominate T for 1. So q = 0, which makes S dominate H for 2. So equilibrium doesn’t exist.
Case3: b,c 6= 0 and a = 0. It makes S strictly dominates H for 1. So p = 0, which makes S dominates T for 2. So equilibrium doesn’t exist.
Case4: a,b,c 6= 0.
4.1 If a6= b then H and T can’t give equal payoff to player1. So either p = 0 or q = 0 then follow the case 2 or 3 above and again no equilibrium exists.
4.2 If a = b, since c 6= 0, S strictly dominates H and T for 1 which makes p = q = 0, but then 2 doesn’t find optimal playing either H or T, again no equilibrium.
Ans2
(a) SE2 = {(a1, a2, a3) | ai ∈ {In, Out}} where the first action (a1) corresponds to the information set ofE2 after the history whenE1 opts out, second action (a2) when E1 enters and M fights and third action (a3) when E1 enters andM accommodates respectively.
(b) The strategy profile sE1 = (Out), sE2 = (Out, Out, Out) and sM = (F, F, F, F) forms a nash equilibrium. There can be four possible histories where M can take an action, so his strategy should specify four actions for every possible history. GivenM going to fight every time no entrant can gain by entering. And if no entrant going to enter than M is indifferent between Fight and accommodate, so no incentive to gain.
(c) In the above equilibriumM enjoys monopoly profits in every period, best possible payoff in this game. But note that this is Nash but not SPNE because it doesn’t induce Nash in every subgame. Strategy of always fight is Non-credible threat, and makes the commitment problem for him. The only way to keep E2 out if M fights whenever he enters. But this action lacks commitment because when entrant actually enters, M would accommodates because now Fighting is not optimal.
Ans3
In this answer we going to assume whenever a player is indifferent from accepting and rejection, he is going to accept.
1
We going to solve it backwardly. Since time periods are always even, 2 will always going to make the last offer.
So in the last period 2 will make an offer (0,1), 0 to player 1 and 1 to himself, which 1 will accept because rejection makes the game end and each gets 0.
Given this in the second last period 1 will offer (c,1−c) i.e. 1−c to 2 and c to himself, which 2 will accept, because if he rejects the best he can get in the next period is 1 but after delay cost which makes it equal to 1−c.
Now given this the period before it 2 will make an offer (0,1), which 1 will accept because if he rejects the best he can get in the next period is c (because of the previous argument) but after delay cost which makes it equal to c−c= 0.
Now we can apply this argument iteratively, we get the following strategies at SPNE
s1 =
(c,1−c) whenever 1 makes a offer i.e. in all odd time periods
Accept any offer, when he is the respondent i.e. in all even time periods
s2 =
(0,1) whenever 2 makes a offer i.e. in all even time periods
Accept if he gets atleast 1−cas his share, when he is the respondent i.e. at all odd times
2
