Y
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