Answers to Problem Set 3
Economics 703
Spring 2016
Comment: To find Nash equilibria, one can use either the normal form or reduced
normal form. In the former, some equilibria will be identical to each other and the reduced
normal form will eliminate the redundant equilibria. For subgame perfect equilibria, one
needs the full strategy profile.
a) The normal form is
ah
ai
bh
bi
c
d
2, 1, 1
0, 3, 2
2, 1, 1
1, 0, 0
3, 2, −1 3, 2, −1
3, 2, −1 3, 2, −1
e
ah
ai
bh
bi
c
2, 1, 1
2, 1, 1
1, 4, 3
1, 4, 3
f
d
0, 3, 2
1, 0, 0
1, 4, 3
1, 4, 3
ah
ai
bh
bi
c
2, 1, 1
2, 1, 1
5, 3, 2
5, 3, 2
g
d
0, 3, 2
1, 0, 0
5, 3, 2
5, 3, 2
The Nash equilibria are (ai, c, f ), (bh, d, f ), and (bi, d, f ). The backward induction
solution and hence the subgame perfect equilibrium is easily seen to be (ai, c, f ).
To carry out successive elimination of dominated strategies, we should consider the
reduced normal form to get rid of the redundancies in the normal form. The reduced
normal form is
c
d
c
d
c
d
ah 2, 1, 1
0, 3, 2
ah 2, 1, 1 0, 3, 2
ah 2, 1, 1 0, 3, 2
ai 2, 1, 1
1, 0, 0
ai 2, 1, 1 1, 0, 0
ai 2, 1, 1 1, 0, 0
b 3, 2, −1 3, 2, −1
b 1, 4, 3 1, 4, 3
b 5, 3, 2 5, 3, 2
e
f
g
For the first round of elimination, note that f dominates e and g for player 3. Also, ai
dominates ah for player 1. Nothing else is dominated on the first round. After eliminating
e, f , and ah, we have just
c
d
ai 2, 1, 1 1, 0, 0
b 1, 4, 3 1, 4, 3
f
1
Now ai dominates b for player 1 and c dominates d for player 2. After eliminating
the dominated strategies, then, we are left with (ai, c, f ), the unique subgame perfect
equilibrium.
b) The normal form is
agi
agj
ahi
ahj
bgi
bgj
bhi
bhj
ce
3, 1
3, 1
0, 0
0, 0
5, 5
2, 2
5, 5
2, 2
cf
3, 1
3, 1
0, 0
0, 0
1, 0
0, 4
1, 0
0, 4
de
0, 0
0, 0
1, 3
1, 3
5, 5
2, 2
5, 5
2, 2
df
0, 0
0, 0
1, 3
1, 3
1, 0
0, 4
1, 0
0, 4
The pure Nash equilibria are (bgi, ce), (bhi, ce), (agi, cf ), (agj, cf ), (bgi, de), (bhi, de),
(ahi, df ), and (ahj, df ).
Turning to dominance solvability, the reduced normal form is
ag
ah
bi
bj
ce
3, 1
0, 0
5, 5
2, 2
cf
3, 1
0, 0
1, 0
0, 4
de
0, 0
1, 3
5, 5
2, 2
df
0, 0
1, 3
1, 0
0, 4
For the first round of elimination, note that for player 1, bi dominates ah and bj. None
of 2’s strategies are dominated. After removing 1’s dominated strategies, we have
ce cf de df
ag 3, 1 3, 1 0, 0 0, 0
bi 5, 5 1, 0 5, 5 1, 0
Now ce dominates cf , de, and df for player 2. Nothing new is dominated for player 1 on
the second round. After eliminating 2’s dominated strategies, we have
ce
ag 3, 1
bi 5, 5
so the game is dominance solvable with solution (bi, ce).
To find subgame perfect equilibria, note that there are three subgames: the two
subgames starting at 2’s information sets and the game itself. The normal form for the
subgame starting at 2’s left–most information set (the one following a by player 1) is
c
d
g 3, 1 0, 0
h 0, 0 1, 3
2
This game has two pure Nash equilibria, (g, c) and (h, d). The normal form for the
subgame starting from 2’s other information set is
e
f
i 5, 5 1, 0
j 2, 2 0, 4
The unique Nash equilibrium of this subgame is (i, e).
To find the subgame perfect equilibria of the overall game, then, we have to consider
two cases: one where (g, c) is played in the left–hand subgame and (i, e) in the right–hand
subgame and one with (h, d) in the left and (i, e) in the right. Note that player 1’s payoff
in the right–hand subgame is 5, while his payoff in either of the two left–hand subgame
equilibria is smaller. Hence his optimal strategy either way is b. So there are two pure
strategy subgame perfect equilibria: (bgi, ce) and (bhi, de).
c) The normal form is
agi
agj
ahi
ahj
bgi
bgj
bhi
bhj
ce
3, 1
3, 1
0, 0
0, 0
2, 2
0, 0
2, 2
0, 0
cf
3, 1
3, 1
0, 0
0, 0
0, 0
4, 4
0, 0
4, 4
de
0, 0
0, 0
1, 3
1, 3
2, 2
0, 0
2, 2
0, 0
df
0, 0
0, 0
1, 3
1, 3
0, 0
4, 4
0, 0
4, 4
The Nash equilibria are (agi, ce), (agj, ce), (bgj, cf ), (bhj, cf ), (bgi, de), (bhi, de), (bgj, df ),
and (bhj, df ).
Turning to dominance solvability, the reduced normal form is
ag
ah
bi
bj
ce
3, 1
0, 0
2, 2
0, 0
cf
3, 1
0, 0
0, 0
4, 4
de
0, 0
1, 3
2, 2
0, 0
df
0, 0
1, 3
0, 0
4, 4
No strategies are dominated, so the game is not dominance solvable.
To find subgame perfect equilibria, note that the normal form of the subgame on the
left is the same as in (b) above. So there are two pure Nash equilibria in this subgame,
(g, c) and (h, d). The normal form for the other subgame is
e
f
i 2, 2 0, 0
j 0, 0 4, 4
3
There are two Nash equilibria in this subgame, (i, e) and (j, f ).
Because there are two pure equilibria in each subgame, there are four pure subgame
perfect equilibria corresponding to the four possibilities for play in the subgames. If (g, c)
and (i, e) are played in the subgames, 1’s best strategy at his initial information set is
a, so (agi, ce) is a subgame perfect equilibrium. If (g, c) and (j, f ) are played, 1’s best
strategy is b, so (bgj, cf ) is subgame perfect. Similar reasoning shows that the other
subgame perfect equilibria are (bhi, de) and (bhj, df ).
d) The normal form is
af
ag
bf
bg
cf
cg
dh
di
eh
ei
1, 2
1, 2 1, 2 1, 2
1, 2
1, 2 1, 2 1, 2
−1, 0 −1, 0 1, 1 1, 1
−1, 0 −1, 0 1, 1 1, 1
0, 1
0, 1 2, 2 0, 0
0, 1
0, 1 0, 0 −1, 1
The Nash equilibria are (af, dh), (af, di), (af, ei), (ag, dh), (ag, di), (ag, ei), (bf, ei),
(bg, ei), and (cf, eh).
The reduced normal form is
d
a 1, 2
b −1, 0
cf 0, 1
cg 0, 1
eh
ei
1, 2 1, 2
1, 1 1, 1
2, 2 0, 0
0, 0 −1, 1
On the first round, we can eliminate only b and cg for player 1 as both are dominated by
a. On the second round, we can eliminate d and ei for 2 since both are dominated by eh.
Finally, we eliminate a for 1, leaving only (cf, eh). So the game is dominance solvable.
To find out which of the Nash equilibria are subgame perfect, consider the subgame
starting where 1 picks between f and g. The normal form of this subgame is
h
i
f 2, 2 0, 0
g 0, 0 −1, 1
The unique equilibrium of this subgame is (f, h). Hence all the Nash that did not call for
these strategies in the subgame are not subgame perfect. So the only subgame perfect
equilibria are (af, dh) and (cf, eh).
e) We did most of this in class. To try to minimize confusion, I’ll refer to the actions
following 1’s choice of b with a b subscript and similarly for the actions following 1’s
choice of n. With this notation, the reduced normal form is
4
bUb
bDb
nUn
nDn
Lb Ln Lb Rn Rb Ln Rb Rn
2, 1
2, 1 −1, 0 −1, 0
−1, 0 −1, 0 0, 3
0, 3
3, 1
0, 0
3, 1
0, 0
0, 0
1, 3
0, 0
1, 3
The Nash equilibria are (nUn , Lb Ln ), (bUb , Lb Rn ), (nUn , Rb Ln ), and (nDn , Rb Rn ).
Turning to dominance solvability, 2 has no dominated strategies. For 1, bDb is dominated by nUn and nDn . This is the only strategy that gets eliminated on the first
round.
After eliminating bDb for 1, Lb Ln dominates Rb Ln and Lb Rn dominates Rb Rn for 2.
Intuitively, once we eliminate bDb , 2 knows that if 1 chooses b, he will follow it with U .
Hence 2 will want to choose L if 1 chooses b. Nothing new is dominated for 1, so the
only strategies we remove on the second round are Rb Ln and Rb Rn for 2.
Now that these strategies are eliminated, nDn is dominated by bUb for 1. Nothing
new is dominated for 2, so this is the only strategy eliminated on the third round.
Turning to the fourth round, we now see that Lb Ln dominates Lb Rn for 2. After eliminating this strategy (the only dominated strategy at the fourth round), nUn dominates
bUb for 1. Hence the game is dominance solvable with solution (nUn , Lb Ln ).
Finally, let’s compute the subgame perfect equilibria. There are again three subgames.
The normal form for the subgame after 1 plays b is
Lb
Rb
Ub 2, 1 −1, 0
Db −1, 0 0, 3
There are two pure equilibria in this subgame, (Ub , Lb ) and (Db , Rb ). Similarly the normal
form for the subgame after 1 plays n is
Ln Rn
Un 3, 1 0, 0
Dn 0, 0 1, 3
Again, there are two pure equilibria in this subgame, (Un , Ln ) and (Dn , Rn ). This gives
four combinations for finding pure subgame perfect equilibria. If we have the (U, L)
outcome in both subgames, 1 clearly prefers n, so (nUb Un , Lb Ln ) is a subgame perfect
equilibrium. Similarly, (nDb Dn , Rb , Rn ) is a subgame perfect equilibrium. If we have the
(U, L) outcome after b and the (D, R) after n, 1 prefers b. So (bUb Dn , Lb Rn ) is a subgame
perfect equilibrium. Finally, if we have the (U, L) outcome after n and the (D, R) after
b, 1 chooses n, so (nDb Un , Rb Ln ) is a subgame perfect equilibrium.
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