Answers to Problem Set 4
Economics 703
Spring 2016
1. a) The monopolist facing no threat of entry will pick the first cost function. To see
this, calculate profits with each one. With the first cost function, the monopolist chooses
output, q, to maximize
π1 = (p − 3)q = (10 − q − 3)q = (7 − q)q
Differentiate with respect to q, set the derivative to zero and solve to get q = 7/2. Profits
are
π1 = 49/4.
With the second cost function, q maximizes
π2 = (p − 1)q − 10 = (9 − q)q − 10
Again, differentiate with respect to q and solve to get q = 9/2. Profits are
π2 = 81/4 − 10 = 41/4 < π1 .
Thus profits are higher with the first cost function.
b) The second firm will only enter the market if he can earn profits of at least 4. First,
suppose the monopolist picks the first cost function. Then if the second firm enters,
we have a Cournot problem, like the one we analyzed in class earlier. To compute the
equilibrium, let qi∗ be the equilibrium output of firm i. Then for each i, qi∗ must solve
max
(10 − q − qj∗ )q − ci (q)
q
is firm i’s cost function. The first order conditions for firm i are
10 − qj∗ − 3 − 2q = 0
or
q1∗ =
7 1 ∗
− q
2 2 2
1
and
7 1 ∗
− q .
2 2 1
Solving these two equations together, you get q1∗ = q2∗ = 7/3. The price will be 10 − 7/3 −
7/3 = 16/3. Profits for each firm will be (16/3 − 3)(7/3) = 49/9. Notice that 49/9 > 4
so that the second firm will enter. Thus if the monopolist picks the first cost function,
the second firm will enter and the first firm will earn profits of 49/9.
q2∗ =
Suppose then that the monopolist chooses the second cost function. Will the second
firm enter? Suppose he does. The first firm chooses output, q1 , to maximize
(p − 1)q1 − 10 = (10 − q1 − q2∗ − 1)q1 − 10
where q2∗ is firm 2’s output. Differentiate with respect to q1 and set the derivative to zero
to get
9 − q2∗
q1∗ =
2
Firm 2 will choose his output to maximize
(p − 3)q2 = (7 − q1∗ − q2 )q2
Differentiate with respect to q2 and set the derivative to zero to get
q2∗ =
7 − q1∗
2
Solve these two equations for q1∗ and q2∗ . Substituting the first equation into the second,
q2∗ =
7 − (9 − q2∗ )/2
14 − 9 + q2∗
=
2
4
or
3q2∗ = 5.
So q2∗ = 5/3. Substituting this into the equation above to get q1∗ , we see that
q1∗ =
9 − 5/3
27 − 5
=
= 22/6 = 11/3
2
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The price would be
10 − q1 − q2 = 10 − 5/3 − 11/3 = 10 − 16/3 = 14/3
Firm 2’s profits would be
(14/3 − 3)q2 = (5/3)(5/3) = 25/9
which is less than the entry cost of 4. Thus firm 2 would not enter. Therefore, if firm 1
picks the second cost function, there will be no entry and the profits firm 1 earns would
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be the same as the profits earned with this cost function as we calculated in (a) or 41/4.
Thus we see that the firm 1 earns 49/9 if he chooses the first cost function and 41/4 if
he chooses the second one. Since 41/4 > 49/9, he chooses the second one.
2. a) Since the game has perfect information, backward induction corresponds to sequential rationality at every information set. Hence the unique weak PBE and the unique
sequential equilibrium is (ai, c, f ). (Note that there is no issue of beliefs because of perfect
information.)
b) For weak perfect Bayesian equilibria, let’s work backwards. Sequential rationality
for 1 at his information set in the left–hand subgame doesn’t pin down his action: depending on his beliefs, he could play either g or h here. However, for any beliefs, the
only sequentially rational strategy at his information in the subgame on the right is i. So
first suppose 1 plays g in the left–hand subgame and i in the right. Then for sequential
rationality, 2 must play c in the game on the left and e in the game on the right. Given
this, 1 must play b at his initial information set. Since 1’s information set on the left is
not reached, we can assign any beliefs we like there, so, for example, we can assign 1 at
the node on the left and 0 at the node on the right in that information set. Since 1’s
information set on the right is reached, we have to assign 1 to the node on the left there
and 0 to the node on the right. With these beliefs, we have sequential rationality for all
players. Hence (bgi, ce) is a weak perfect Bayesian equilibrium.
Next, suppose 1 plays h in the left–hand subgame (and i, as he must, in the right–
hand subgame). For sequential rationality, 2 must play d and e. Given this, at his initial
information set, 1 must play b. Again, we can assign beliefs at the unreached information
set to make 1’s strategy sequentially rational. Hence (bhi, de) is a weak perfect Bayesian
equilibrium.
Which, if either, of these weak PBE’s is sequential? In both cases, 1’s information
set in the subgame on the right is reached and so weak consistency and consistency say
the same thing about the beliefs here. The only question is whether the beliefs at the
information set in the left subgame are consistent. In both cases, this is straightforward:
any totally mixed strategy for 2 will give beliefs at 1’s information set converging to
probability 1 on 1 correctly inferring 2’s action. In both cases, 1’s action is a best
response to 2, so we satisfy sequential rationality and consistency. Hence both weak
PBE’s are sequential.
c) For weak perfect Bayesian equilibria, let’s work backwards again. Sequential rationality for 1 does not pin down his strategy at either of those final information sets.
So we have four cases to check. First, suppose 1’s strategy is gi at those information
sets. Then sequential rationality for 2 implies that his strategy must be ce. Given this,
sequential rationality for 1 at his initial information set implies that he plays a. Weak
consistency then requires 1’s beliefs at his left–hand information set to put probability 1
on the node on the left, a belief which makes g sequentially rational. We can pick any
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belief for the information set in the right–hand subgame and so can pick one making i
sequentially rational. Hence (agi, ce) is a weak perfect Bayesian equilibrium.
Now suppose 1 plays gj. Now sequential rationality implies that 2’s strategy must
be cf and 1’s initial move must be b. Again, the beliefs implied by weak consistency are
the ones making 1’s strategy sequentially rational at the information set with positive
probability. Again, we can assign anything at the other information set. Hence (bgj, cf )
is a weak perfect Bayesian equilibrium.
The reasoning for the other two cases is similar: both (bhi, de) and (bhj, df ) are weak
perfect Bayesian equilibria. The reasoning for sequential for all three cases is similar to
part (b) — all three equilibria are sequential.
d) To compute weak perfect Bayesian equilibria, let’s work backward starting with
2’s second information set. Suppose 2 plays h here. Sequential rationality for 1 at his
immediately preceding information set would require that he play f . Once we have this,
we see that no matter what beliefs 2 has at his first information set, it is optimal for
him to play e. Given this, sequential rationality implies that 1 must play c at his first
information set. Since this equilibrium reaches every information set, it must be weak
perfect Bayesian. So (cf, eh) is weak perfect Bayesian. Recall from the answers to the
last problem set that (af, dh) is subgame perfect; we see here that it is not weak perfect
Bayesian.
So now suppose 2 plays i at that last information set. Again, sequential rationality
requires 1 to play f . Using this, 2’s strategy at his preceding information set can be either
d or e depending on his beliefs. So we have two subcases here. First, let’s suppose 2 plays
d here. Then a is the best thing for 1. Since this implies that neither of 2’s information
sets are reached, this must be a weak perfect Bayesian equilibrium. So (af, di) is weak
perfect Bayesian. (Again, comparing to the answers to the last problem set, we see that
this weak PBE is not subgame perfect.)
Turning to the second subcase, suppose 2 plays i at his last information set, 1 plays
f at his last one, and 2 plays e at his other information set. In this case, 1 is indifferent
between b and a. So we have two more subcases: where he plays a and where he plays
b. If he plays a, we know we get a weak perfect Bayesian equilibrium. Hence (af, ei) is
a weak perfect Bayesian equilibrium. (Again, it is not subgame perfect.)
So suppose 1 plays b. Then 2 must put probability 1 on the left–hand node in his
information set, making e his best strategy there. Again, the lower information set is not
reached, so weak perfect Bayesian allows any beliefs there. Hence (bf, ei) is also weak
perfect Bayesian. (Again, it is not subgame perfect.)
Turning to sequential equilibria, we know that every sequential equilibrium is both
weak PBE and subgame perfect. Hence if there is a pure sequential equilibrium, it must
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be (cf, eh). Note that this equilibrium reaches every information set, so it must be sequential. More specifically, 2’s beliefs at his first information set must put probability 1
on the right–hand node to satisfy weak consistency and his beliefs at his second information set must put probability 1 on the left–hand node. Taking totally mixed strategies
that put probability 1/n on each “mistake” (i.e., each deviation from these strategies),
we see that the probability on 2’s right–hand node in the first information set is
1−
1−
2
n
1
n
which does go to 1 as n → ∞. Similarly, 2’s belief on the left–hand node in the second
information set is
h
i3
1 − n1
1
i2 = 1 − ,
h
n
1 − n1
which also goes to 1 as n → ∞.
3. a) The normal form is
c
d
a 2, 2, 2 4, 3, 1
b 3, 0, 1 3, 1, 0
e
c
d
a 2, 2, 2 0, 0, 0
b 0, 0, 0 1, 0, 1
f
The pure strategy Nash equilibria are (a, d, e), (b, d, f ), and (a, c, f ). Since the only
subgame is the game itself, all are subgame perfect. Note that (a, d, e) and (b, d, f ) reach
every information set, so both must be weak perfect Bayesian. (It’s worth verifying this
for yourself directly!)
So consider the equilibrium (a, c, f ). This equilibrium is a weak perfect Bayesian
equilibrium. To see this, note that 2’s beliefs must put probability 1 on the node where
1 played a and 3’s beliefs are not pinned down by the strategies. So let 3’s beliefs put
probability 1 on the right–most node in his information set — the one where 1 played b
and 2 played d. Given these beliefs, 3’s strategy is sequentially rational, as is 2’s. Hence
this is a weak perfect Bayesian equilibrium.
Turning to sequential equilibrium, we know that every sequential equilibrium is a
weak perfect Bayesian equilibrium. Hence we only need to check the three weak perfect
Bayesian equilibria: (a, d, e), (b, d, f ), and (a, c, f ). The first two reach every information
set, so they must be sequential. To see this, simply note that any totally mixed strategies
that converge to these must generate beliefs converging to the unique weakly consistent
beliefs. Hence both satisfy consistency, not just weak consistency.
However, (a, c, f ) is not sequential. To see this, let µ be 3’s probability on that right–
most node, where 1 played b and 2 having played d. Clearly, the strategy f is sequentially
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rational iff µ ≥ 1/2. Is this consistent? Suppose it is. Let pn be 1’s probability on a and
qn be 2’s probability on c in our sequence of totally mixed strategies converging to this
equilibrium. We must have pn → 1 and qn → 1. Let µn be the probability put on the
node in question by Bayes’ Rule with these strategies. Then
µn =
(1 − pn )(1 − qn )
(1 − pn )(1 − qn )
<
= 1−pn .
(1 − pn )(1 − qn ) + qn (1 − pn ) + pn (1 − qn )
(1 − pn )(1 − qn ) + pn (1 − qn )
Since pn → 1, we must have µn → 0. Hence we cannot converge to these beliefs, so the
beliefs are not consistent and the equilibrium is not sequential.
b) The normal form is
c
d
a 1, 2, 0 −1, 1, 0
b 2, 1, 3 0, 3, 2
e
c
d
a 1, 2, 0 0, 3, 2
b 2, 1, 3 1, − 1, 0
f
Note that 1 has a strictly dominant strategy of b. Hence he must play this in every
Nash equilibrium. There are two Nash equilibria in pure strategies: (b, d, e) and (b, c, f ).
Because there is only one subgame, all are subgame perfect.
Both are weak perfect Bayesian equilibria. To see this, note that 2’s beliefs must put
probability 1 on the right–hand node in his information set in either equilibrium. For
(b, d, e), have 3’s beliefs put probability 1 on the right–hand node in his information set.
Given these beliefs, sequential rationality requires 3 to play e. Given this strategy by 3
and 2’s beliefs, sequential rationality requires 2 to play d. Hence this is a weak perfect
Bayesian equilibrium.
For (b, c, f ), let 3’s beliefs put probability 1 on the left–hand node in his information
set. Given these beliefs, f is sequentially rational. Given this and 2’s beliefs, 2 faces a
choice between 1 and −1, so c is sequentially rational for him. Hence, again, this is a
weak perfect Bayesian equilibrium.
Are the weak PBE’s sequential? Since (b, d, e) reaches every information set, it must
be sequential. Is (b, c, f )? No — there are no totally mixed strategies converging to
(b, c, f ) which generate beliefs converging to the ones we need to make f sequentially
rational. To see this, let pn be the probability 1 makes a mistake (plays a) and let qn be
the probability 2 makes a mistake in this equilibrium (plays d). Then by Bayes’ rule, the
probability 3 is at the left–hand node of his information set given that the information
is reached is
pn qn
pn qn
=
= pn .
pn qn + (1 − pn )qn
qn
But this must go to zero as n → ∞. Hence 3 must put probability zero on the left–hand
node, meaning that f cannot be sequentially rational given any consistent belief. So this
equilibrium is not sequential.
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