Econ 171 Fall 2008: Extensive Form Games & Subgame Perfection - Practice Qs & Solutions - , Assignments of Economics

Practice questions and solutions for the topic of extensive form games and subgame perfection in the economics 171 course during the fall 2008 semester.

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Pre 2010

Uploaded on 08/30/2009

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Econ 171 Fall 2008
Practice Questions & Solutions
Part II: Extensive Form Games and Subgame Perfection
Practice questions with solutions below:
Watson 3.1 - 3.4, 15.1, 15.3 - 15.5
3.1 Exercise 1: SL={A, B}.SM={Rr, Rg, Gr, Gg}.SJ={Aa, Ab, Ba, B b}. Exercise 4:
Si={R, P, S }for i= 1,2.
3.2 No, “not hire” is not a strategy for the manager. A strategy must specify an action for
every contingency, but “not hire” does not specify what the manager does contingent
upon the worker being hired and exerting a particular level of effort.
3.3 See the next page.
3.4 Player 2 has four strategies: {(c, f ),(c, g),(d, f ),(d, g)}.
15.1 (a) (AF, C )
(b) (BH JK N, CE )
(c) (I, C , X)
15.3 (a) (AHI LN, C E)
(b) 6
15.4 For any given x,y
1(x) = y
2(x) = x; and x= 2.
15.5 (a) Not shown.
(b) We proceed by backwards induction. In the last round (5), player 1 will choose S.
Continuing in this fashion, we find that, in equilibrium, each player will choose S
any time it is her turn. Thus the BI solution is (SS S, SS).
(c) For any finite k, the backwards induction outcome is that player 1 chooses Sin
the first round and each player receives $1.
1
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Econ 171 Fall 2008 Practice Questions & Solutions Part II: Extensive Form Games and Subgame Perfection

Practice questions with solutions below:

Watson 3.1 - 3.4, 15.1, 15.3 - 15.

3.1 Exercise 1: SL = {A, B}. SM = {Rr, Rg, Gr, Gg}. SJ = {Aa, Ab, Ba, Bb}. Exercise 4: Si = {R, P, S} for i = 1, 2.

3.2 No, “not hire” is not a strategy for the manager. A strategy must specify an action for every contingency, but “not hire” does not specify what the manager does contingent upon the worker being hired and exerting a particular level of effort.

3.3 See the next page.

3.4 Player 2 has four strategies: {(c, f ), (c, g), (d, f ), (d, g)}.

15.1 (a) (AF, C) (b) (BHJKN, CE) (c) (I, C, X)

15.3 (a) (AHILN, CE) (b) 6

15.4 For any given x, y 1 ∗ (x) = y∗ 2 (x) = x; and x∗^ = 2.

15.5 (a) Not shown. (b) We proceed by backwards induction. In the last round (5), player 1 will choose S. Continuing in this fashion, we find that, in equilibrium, each player will choose S any time it is her turn. Thus the BI solution is (SSS, SS). (c) For any finite k, the backwards induction outcome is that player 1 chooses S in the first round and each player receives $1.

CE CF DE DF

A 0 , 0 0 , 0 1 , 1 1 , 1

B 2 , 2 3 , 4 2 , 2 3 , 4

(a)

I O

IU 4 , 0 − 1 , − 1

ID 3 , 2 − 1 , − 1

OU 1 , 1 1 , 1

OD 1 , 1 1 , 1

(b) AC AD BC BD UE 3 , 3 3 , 3 5 , 4 5 , 4 UF 3 , 3 3 , 3 5 , 4 5 , 4 DE 6 , 2 2 , 2 6 , 2 2 , 2 DF 2 , 6 2 , 2 2 , 6 2 , 2 (c) A B UXW 3 , 3 5 , 1 UXZ 3 , 3 5 , 1 UYW 3 , 3 3 , 6 UYZ 3 , 3 3 , 6 DXW 4 , 2 2 , 2 DXZ 9 , 0 2 , 2 DYW 4 , 2 2 , 2 DYZ 9 , 0 2 , 2 (d) U D A 2 , 1 1 , 2 B 6 , 8 4 , 3 C 2 , 1 8 , 7 (e) A B UXP 3 , 8 1 , 2 UXQ 3 , 8 1 , 2 UYP 8 , 1 2 , 1 UYQ 8 , 1 2 , 1 DXP 6 , 6 5 , 5 DXQ 6 , 6 0 , 0 DYP 6 , 6 5 , 5 DYQ 6 , 6 0 , 0 (f)