Solution for MIEG (Part I) Makeup Test, Winter Term, 2015

Solution for MIEG (Part I) Makeup Test, Winter Term, 2015

Ans1

Lets say firms from 1 to n are followers and (n+ 1)^th firm is a leader, index byL.

This a two stage game. First L will chooses q_L then given that n firms will choose their quantities q1 toqn simultaneously. To find SPNE we will solve it backwardly.

Given q_L, the profit function of each firm i from 1 to n is (1−q_L−Q−c).q_i where Q=q₁+q₂+...+q_n.

By differentiating it one can find best response which isqi = [1−qL−c−Q] for all i. Solving these equations we get q_i = 1−q_L−c

n+ 1

Substituting it in the profit function ofL to find his optimum choice.

So SPNE is qL = 1−c

2 and qi = 1−q_L−c

n+ 1 for every possible history/subgame starts with q_L for all i= 1 to n.

(b) Increase inn will not affectq_L but will reduce price. Hence profit of the leader will fall.

(c) Yes such NE exists. Intuitively, Cournot equilibrium of this game is a NE (but not SPNE). Formally,

q_L= 1−c n+ 2 and

∀i={1,2, ..., n} q_i =

( 1−c

n+ 2 whenever q_L = 1−c n+ 2

1 otherwise

You can check that no firm (including L) has any incentive to deviate from above strategy profile given the strategies of all other firms.

Ans2

S₁ =S₂ = {0,1, ...,10}. We will argue for player 1 same holds for player 2.

Step1: s₁ = 10 is strictly dominated with s₁ = 1 or 2 or ... 9. So both will eliminate 10.

Note at the beginning of the game no strategy other than 10 is strictly dominated, because each number 0 to 9 is a unique best response for some strategy of 2, being precise s1 =k is unique best response to s₂ =k+ 1 for k = 1,2, ...,9 and 0 is unique best response of 0.

Step2: Given 10 has been deleted, now 9 is strictly dominated (say) by 8.

Spet3: Now 8 is strictly dominated. and so on...

Iteratively applying it till we left with 0 and 1. And now no strategy is strictly dominated.

So IESDS gives the following four equilibria (0,0) (0,1) (1,0) (1,1) Ans3

Lets say (x,1−x) and (1−y, y) are the offers proposed by player 1 and 2 respectively. For it to be SPNE it must satisfy the following two conditions

1

(1−y)^α1 = δx^α1 for player 1 when 2 makes the offer and (1−x)^α2 = δy^α2 for player 2 when 1 makes the offer Solving them gives you x= 1−δ2

1−δ₁δ₂ and y= 1−δ1

1−δ₁δ₂ where δ_i =δ^αi1 .

2