Download Answer Key: Midterm (some typos corrrected and clarifications added )

Answer Key: Midterm
(some typos corrrected and clari…cations added )
March 11, 2016
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Question 1
a. The maximum pro…t the two …rms could earn together through explicit collusion is monopoly pro…t. A monopolist
solves
max (48
2P ) P
P
The FOC is
48
4P = 0
m
so P = 12;
= 288.
b. Note that in this game …rms are competing by choosing prices, not quantities! To …nd the smallest
consistent with sustaining monopoly pro…t, we want to consider trigger strategies with the most severe punishment
that can be part of a subgame perfect equilibrium (“SPE”). Thus, we consider punishment strategies that set price
to marginal cost, yielding each …rm a per-period pro…t during punishment of P = 0. An optimal deviation for a
…rm is to set price 12
for arbitrarily small , yielding one-period pro…t D = m = 288. Finally, when both …rms
set the monopoly price, each …rm has per-period pro…t C = 288
2 = 144 from “cooperating.”For these strategies to
form a SPE, we must have
C
D
1
P
+
1
,
i.e.,
144
1
288:
1
Thus, we need
2:
c. With n …rms, the per-period pro…t from cooperation becomes C = 288
n . The pro…ts obtained during
punishment and from optimal deviation are unchanged. So sustaining the monopoly price in a SPE now requires
288
n (1
)
288
n 1
n 1
1
which means
n . For n > 2, n > 2 , i.e., sustaining collusion is “more di¢ cult.” This follows intuitively
C
because (a) the one-period gain from deviation, D
, is larger than before while (b) the loss from punishment
C
triggered by deviation, 1 , is smaller.
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d. In this case, a deviating …rm will obtain monopoly pro…ts for two periods before punishment is triggered. So
sustaining the monopoly price in SPE requires
144
1
288 + 288:
Simplifying, we obtain
1
2 (1 + )
1
1
2
2
1
1
p
2
Since p12 > 12 , now it is harder to sustain collusion.
punishment is delayed.
2
This is because the payo¤ from deviation lasts longer and
Question 2
a. The …rm solves the problem:
max
q1 ;T1 ;q2 ;T2
T1
cq1 + T2
cq2
s:t:
1 v (q1 )
T1
0
(V P 1)
2 v (q2 )
T2
0
(V P 2)
1 v (q1 )
T1
1 v (q2 )
T2
(IC1)
2 v (q2 )
T2
2 v (q1 )
T1
(IC2)
However, notice that (V P 1) together with (IC2) imply that (V P 2) will be slack:
2 v (q2 )
T2
>
2 v (q1 )
T1
1 v (q1 )
T1
0:
We will also conjecture that (IC1) is slack and ignore it, and then verify that the solution found satis…es this
conjecture. Finally, we observe that as long as the remaining constraints (V P 1) and (IC2) are not violated, the
…rm prefers to set T1 and T2 as large a possible. Thus the …rm wants these two constraints to bind (hold with
equality). Thus
T1 = 1 v (q1 )
(1)
and
2 v (q2 )
2 v (q1 )
+ T1 = T2 :
(2)
Substituting (1) into (2) gives
T2 =
2 v (q2 )
(
2
2
1 ) v (q1 ) :
(3)
Substituting (1) and (3) into the maximization problem (and ignoring (IC1) and (VP2) as we suggested already),
the …rm’s problem becomes:
max
q1 ;q2
1 v (q1 )
+
2 v (q2 )
(
1 ) v (q1 )
2
c (q1 + q2 ) :
The FOCs are
(2
1
2) v
0
(q1 )
=
c
(4)
2v
0
(q2 )
=
c:
(5)
These equations characterize the optimal tari¤ as long as (IC1) is actually slack. We will con…rm this after discussing
points b. and c.
b. The e¢ cient quality qiE for consumer i is the one for which marginal utility of consumption equals marginal
cost of production, i.e.,
0
E
= c:
i v qi
If q1 satis…es (4), then we have
(2
2) v
1
0
(q1 ) =
1v
0
q1E :
E
Since 2 1
2 > 1 and v ( ) is strictly concave, this requires q1 < q1 .
E
c. From (5) we immediately have q2 = q2 .
Notice that from (4) and (5),
c
v 0 (q2 ) =
<
2
c
2
1
= v 0 (q1 ) ;
2
which implies that q2 > q1 : This fact helps to con…rm that (IC1) is slack. We know that the LHS of (IC1) is zero
by (1). The RHS of (IC1) is (plugging in (IC2)):
1 v (q2 )
T2
=
=
1 v (q2 )
(
[ 2 v (q2 )
2 ) [v (q2 )
1
(
2
1 ) v (q1 )]
v (q1 )]
< 0;
which follows from the facts that ( 1
2 ) < 0; q2 > q1 and v increasing.
d. Given any q2 sold to type 2, the monopolist can extract all surplus available by setting
a=
2 v (q2 )
bq2
So the monopolist chooses q2 to maximize is pro…t
2 v (q2 )
cq2 :
The FOC is 2 v 0 (q2 ) = c, which implies the e¢ cient quality for type 2. More simply, because the monopolist
extracts all surplus created by selling a given qaulity q2 , the monopolist chooses the q2 that maximizes total
surplus.
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Question 3
a. Consider the following strategies for the three …rms:
(
a c
2b
qm =
if q1 = 0 & q2 = 0
otherwise
…rm 3
:
q3 =
…rm 2
:
q2 = 0 regardless of q1
…rm 1
:
q1 = 0
a
b
Note that, for …rms 2 and 3, specifying strategies (complete contingent plans) requires specifying the quantity to
set after every feasible history. The strategies above form a Nash equilibrium. To show this, we need to show that
no player has a pro…table deviation given the strategies of other …rms. Firm 3 makes the monopoly pro…t when
…rms use these strategies, so it cannot pro…tably deviate (although there are many di¤erent values of the “threat”
quantity, a=b here, that would also yield an equilibrium.) Given strategies of …rm 3 and …rm 1, …rm 2 doesn’t have
a pro…table deviation either because producing at any non-zero quantity causes …rm 3 (according to its strategy)
to produce a level of output ensuring that no …rm has positive pro…t. The same argument applies to …rm 1. Note
that to specify …rms’strategies, you need to descri
b. We want to work backward starting with stage 3. At that point q1 and q2 are given and …rm 3 maximizes:
max
3
q3
= (a
bq1
bq2
a
bq1
bq3 ) q3
cq3
Solving for q3 , the best response function is
q3 (q1 ; q2 ) =
bq2
c
2b
In Stage 2, …rm 2 maximizes its pro…ts, expecting …rm 3 to behave as we just derived:
max
2
q2
= (a
bq1
which is
max
q2
2
b
q1
2
a
2
=
bq2
bq3 (q1 ; q2 )) q2
b
c
q2 +
2
2
q2
cq2
cq2 :
The FOC gives us the best response function of …rm 2:
bq1 c
2b
In Stage 1, …rm 1 knows that …rm 2 will play as above. It also knows that …rm 3 will go on to react according to
q2 (q1 ) =
a
q3 (q1 ; q2 (q1 )) =
a
bq1
4b
c
:
Thus …rm 1 solves
max
q1
1
= [a
bq1
max
1
which is
q1
=
bq2 (q1 )
a
4
bq3 (q1 ; q2 (q1 ))] q1
b
3
q1 + c q1
4
4
4
cq1 :
cq1
From the FOC (and plugging in the solution to the best response functions above), we obtain
q1 =
a
c
2b
a
; q2 =
c
4b
; q3 =
a
c
8b
:
This yields pro…ts
2
2
2
(a c)
(a c)
c)
; 2=
; 3=
:
1
16b
32b
64b
So 1 > 2 > 3 : the …rms that move earlier have an advantage. This is due to the fact that setting quantity
…rst (or second) allows the …rm to commit to producing more output than would be optimal given the quantities
ultimately chosen by other …rms. (This illustrates the important distinction between best responses to others’
actions and best responses to others’strategies.)
c. Solving the Stackelberg game for two …rms using backwards induction, we have …rm 2 solving
=
(a
max
=
2
q2
(a
bq1
=) q2 (q1 ) =
bq2 ) q2
a
cq2
bq1
2b
c
Firm 1 solves
max
q1
1
=
(a
bq1
=) q1 =
a
bq2 (q1 )) q1
c
2b
; q2 =
a
cq1
c
4b
The quantities produced by …rms 1 and 2 are identical to those found in 3-…rm Stackelberg game! So …rm 3’s entry
has no e¤ect on their SPE quantities.
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