14.12 Game Theory
Muhamet Yildiz
Fall 2005
Solution to Homework 3
Problem One
a) This game has two subgames. One subgame is the game as a whole and the
other is the subgame after both players say yes. In this last subgame we have 3 Nash
equilibria (NE) which are
I
,
I
,
C
,
C
and a mixed strategy. The mixed strategy NE is
computed as follows:
Let’s say player 1 plays
I
with probability
µ
and player 2 plays
I
with probability
.
Then the mixed equilibrium should satisfy the following equations:
Player 2 is indifferent between
I
and
C
µ
∗
100
1
−
µ
∗
0
µ
∗
60
1
−
µ
∗
60
u
0.6
Player 1 is indifferent between
I
and
C
∗
100
1
−
∗
0
∗
60
1
−
∗
60
0.6
Therefore the solution is
µ
0.6.
This implies that the mixed Nash equilibrium is
0.6
∗
I
0.4
C
,0.6
I
0.4
∗
C
.
To compute the subgameperfect equilibrium (SPE) we have to check if the
strategy profile is a NE in every subgame. Thus, to obtain the SPE we have to assume
that outcome of the smaller subgame will be one of those NE.
When both players choose
I
in the smaller subgame, the reduced form of the game
is:
Y
N
Y
100, 100
80, 80
N
80, 80
80, 80
There are 2 types of NE in this case
Y
,
Y
and
N
,
N
. Therefore the SPE are
YI
,
YI
and
NI
,
NI
When both players choose
C
in the smaller subgame. The reduced form of the
game is
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N
Y
60,60
80,80
N
80,80
80,80
There are 3 pure NE in this case
Y
,
N
,
NY
and
N
,
N
and 2 mixed NE
N
,
pY
1
−
p
N
and
pY
1
−
p
N
,
N
,
0
≤
p
≤
1.
Therefore the SPE are
YC
,
NC
,
NC
,
YC
,
NC
,
NC
,
NC
,
pY
1
−
p
N
C
and
pY
1
−
p
N
C
,
NC
.
When both players choose to play the mixed NE in the smaller subgame the
reduced form is the same as in the former case and therefore the NE are also the
same. Thus, SPE of the game are
YC
,
NC
.
NC
,
YC
,
NC
,
NC
,
NC
,
pY
1
−
p
N
C
and
pY
1
−
p
N
C
,
NC
.
Problem Two
Denote
V
t
i
the continuation value to player
i
when the game goes to the period
t
.
Clearly, in the last period,
t
3
n
, player 2 will offer
0,1
thus
V
3
n
2
1
.
1
0
and
V
3
n
In period
t
3
n
−
3
k
3,
k
1,2,.
..
n
−
1
,
player 2 makes an offer. He will offer
1
1
V
3
n
−
3
k
4
,1
−
V
3
n
−
3
k
4
that will be accepted by player 1.
In period
t
3
n
−
3
k
2
, player 1 is the one to make an offer and will offer
2
2
1
−
V
3
n
−
3
k
3
,
V
3
n
−
3
k
3
.
2
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 Spring '05
 MuhammadYildiz
 Game Theory, Period, player

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