Econ414 January 2010
Final Exam
Instructor: Jeff Borowitz ANSWERS
Wednesday, July 8 2009
1. (10 points)
In the course, we studied four types of games:
static games of complete information,
dynamic games of complete information, static games of incomplete information, and dynamic games of
incomplete information. In words, what makes static and dynamic games different, and what makes games
of complete and incomplete information different? Be very specific and do not just write down definitions.
Answer:
I am not looking for a regurgitation of definitions here. The essential difference between games of dynamic
and static timing is that in a static game, no player knows any other player’s strategy at any point before
the game is over.
In a dynamic game, on the other hand, one player sees the previous action of another
player at some point. An answer that only mentions that in a dynamic game, actions are taken sequentially
instead of simultaneously will not receive full credit.
1
In games of incomplete information, players do not know each other’s utility functions. Instead they have
beliefs about what each other’s utility functions might be. Each player’s utility function is determined from a
set of possible utility functions randomly right before the game is played. In games of complete information,
players know their opponent’s utility function, and it is always the same.
2. (30 points)
Find all Nash Equilibria in the following games
(a)
A
B
L
C
R
T
3
,
2
5
,
2
2
,
3
M
4
,
4
2
,
2
3
,
1
B
2
,
2
4
,
3
1
,
3
(b)
A
B
L
C
R
T
6
,
3
2
,
2
1
,
0
M
1
,
0
3
,
0
2
,
1
B
2
,
2
4
,
1
0
,
1
Answer:
(a)
This game can be solved either by iterated elimination of underlining. For brevity, I will use underlining
here.
1
To see why, think about the Prisoner’s Dilemma when the prisoners are interrogated sequentially by the same officer, but
are locked in separate interrogation rooms. As long as the second prisoner
doesn’t know
what the first has done, the game is
exactly the as when the prisoners are interrogated simultaneously.
1
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A
B
L
C
R
T
3
,
2
5
,
2
2
,
3
M
4
,
4
2
,
2
3
,
1
B
2
,
2
4
,
3
1
,
3
The only Nash Equilibrium here is (
M, L
).
(b)
Let’s start to solve this problem by iterated elimination. Using iterated elimination, note that
L
weakly
dominates
C
for player B. Therefore, the game is:
A
B
L
R
T
6
,
3
1
,
0
M
1
,
0
2
,
1
B
2
,
2
0
,
1
Noting that now,
T
dominates
B
for A, we have the game:
A
B
L
R
T
6
,
3
1
,
0
M
1
,
0
2
,
1
Iterated elimination can’t take us any further. Now, notice that there are two Nash Equilibria in pure
strategies. Underlining, we find:
A
B
L
R
T
6
,
3
1
,
0
M
1
,
0
2
,
1
So the pure strategy Nash Equilibria are (
T, L
) and (
M, R
).
There is also a mixed strategy Nash
Equilibrium here. To find it, allow Alice to play
T
with probability
p
and Bob to play
L
with probability
q
. Equating expected utilities, we find:
E
[
u
A
(
T
)] =
E
[
u
A
(
M
)]
6
q
+ 1(1

q
) =1
q
+ 2(1

q
)
5
q
=1

q
q
=1
/
6
E
[
u
B
(
L
)] =
E
[
u
B
(
R
)]
3
p
+ 0(1

p
) =0
p
+ 1(1

p
)
3
p
=1

p
p
=1
/
4
Therefore, all Nash Equilibria in this game are:
braceleftbigg
(
T, L
)
,
(
M, R
)
,
parenleftbigg
T
+ 3
M
4
,
L
+ 5
R
6
parenrightbiggbracerightbigg
3. (30 points)
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 Summer '08
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 Game Theory, Bob Alice C

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