ECOS3012 Strategic Behaviour
Quiz for Lecture 5 solution
1. Consider a repeated game in which the following stage game is played 3 times.
Player 2
L C R
Player 1
U 6, 2 5, 4 2, 3
M 8, 4 4, 5 1, 6
D 7, 4 4, 6 3, 5
In the last stage, players can play......?
A: (U, C) is the unique Nash equilibrium of the stage game. C strictly dominates L. After
deleting L, U strictly dominates M. After deleting L and M, C strictly dominates R.
After deleting L, M, and R, U strictly dominates D.
Therefore, (U, C) is the unique subgame perfect outcome of every stage.
2. (Continue)
In the first stage, players can play......?
A: (U, C). See Q1.
3. First, consider a Stackelberg game with 3 firms. Firm 1 chooses the quantity of its produc-
tion first, then firms 2 and 3 choose their quantities simultaneously after observing firm 1’s
quantities. Let S denote firm 1’s profit in the subgame perfect equilibrium of this Stackelberg
game.
Next, consider a Cournot game in which the same three firms choose their quantities simul-
taneously. Let C denote firm 1’s profit in the Nash equilibrium of the Cournot game.
Select the correct statement from the following.
A: C is never strictly greater than S.
Let qS1 be firm 1’s quantity in the Stackelberg equilibrium and q
C
1 be its quantity in the
Cournot equilibrium. S = C if and only if qS1 = q
C
1 , in which case each firm chooses the
same quantity in both equilibria. Let’s prove the answer by contradiation. Suppose that
C > S. Then, qS1 6= qC1 . However, this implies that firm 1 can profitably deviate to qC1 in the
Stackelberg game to get the higher profit C. Therefore, S cannot be an equilibrium profit.
This contradiction proves that C must be smaller than or equal to S.
4. Consider the same Stackelberg game with three firms as described by the previous question.
Firm 1 chooses the quantity of its production first, then firms 2 and 3 choose their quantit-
ies simultaneously after observing firm 1’s quantities. Suppose that they produce the same
product with different cost functions. Firm 1’s total cost is C1 (q1) = 10q1 + 10, firm 2’s
1
total cost isC2 (q2) = 8q2, and firm 3’s total cost isC3 (q3) = q23. The firms produce identical
goods and the market price is P(q1,q2,q3) = 300−q1−q2−q3.
What quantity does each firm produce in the subgame-perfect equilibrium? Please round
your answers to 3 decimal places (e.g., write 0.333 if your answer is 1/3).
A: q1 = 142.333. q∗2 = 63. q
∗
3 = 23.667.
We solve this game backwards.
First, taking q1 as given, firms 2 and 3 solve their maximisation problems.
max
q2
pi2 = (300−q1−q2−q3)q2−8q2
pi
′
2 (q2) = 0⇒ 300−q1−2q2−q3−8 = 0
q2 (q1,q3) =
292−q1−q3
2
(1)
max
q3
pi3 = (300−q1−q2−q3)q3−q23
pi
′
3 (q3) = 0⇒ 300−q1−q2−4q3 = 0
q3 (q1,q2) =
300−q1−q2
4
(2)
Plugging (2) into (1), we get
q∗2 (q1) = 124−
3
7
q1
q∗3 (q1) = 44−
1
7
q1
Next, given q∗2 (q1) and q
∗
3 (q1), firm 1 solves the following maximisation problem.
max
q1
pi1 =
[
300−q1−
(
124− 3
7
q1
)
−
(
44− 1
7
q1
)]
q1−10q1−10
pi
′
1 (q1) = 0⇒ 300−2q1−124+
6
7
q1−44+ 27q1−10 = 0
q∗1 =
427
3
≈ 142.333
q∗2 = 63
q∗3 =
71
3
≈ 23.667
2
5. Consider the following bank run game. (Note that, unlike the example in class, an investor’s
payoff in this game is 0 if he never withdraws.)
investor 2
withdraw don’t
investor 1
withdraw 20, 20 30, 0
don’t 0, 30 next stage
Stage 1
investor 2
withdraw don’t
investor 1
withdraw 25, 25 50, 0
don’t 0, 50 0, 0
Stage 2
Recall that if both investors withdraw in stage 1, it is called a “bank run”.
Question:
To prevent a bank run, suppose that the bank decides to reward an investor X dollars as long
as this investor does not withdraw in stage 1.
Assume that X has to be an integer. What is the smallest value of X to eliminate bank run as
a subgame perfect outcome?
A: 16. Following the bank’s policy, the updated stage 1 payoff matrix is
investor 2
withdraw don’t
investor 1
withdraw 20, 20 30, 0+X
don’t 0+X, 30 next stage+X
Stage 1
(W, W) is no longer a subgame perfect outcome when X > 20. Since X must be an
integer, the answer is 21.
6. (Continue)
Suppose, instead, that both investors need to pay a penalty of Y dollars if there is a bank run.
Assume that Y has to be an integer. What is the smallest value of Y to eliminate bank run as
a subgame perfect outcome?
A: 21.
Suppose that the bank decides to charge both investors Y dollars if and only if both of
them withdraw in stage 1.
investor 2
withdraw don’t
investor 1
withdraw 20-Y, 20-Y 30, 0
don’t 0, 30 next stage
Stage 1
(Withdraw, Withdraw) is no longer a NE in stage 1 as long as 20−Y < 0⇒ Y ≥ 21.
3
7. Consider the repeated game in which the following stage game is played twice.
Player 2
Hawk Dove
Player 1
Hawk 0, 0 3, 1
Dove 1, 3 2, 2
Select all pairs of payoffs that can be achieved in the second stage of a subgame-perfect
equilibrium.
A: Players must play a NE of the stage game in the second stage. The payoffs can be (1, 3),
(3, 1) if they play a pure NE, or (3/2, 3/2) if they play the mixed NE: Pr(Hawk) = 1/2
for both players.
8. (Continue) Select all pairs of actions that can be chosen in the first stage of a subgame-perfect
equilibrium.
Suppose that, in the second stage of a SPE, the players’ payoffs are:
(a) (A1, A2) if the outcome in the first stage is (Hawk, Hawk)
(b) (B1, B2) if the outcome in the first stage is (Hawk, Dove)
(c) (C1, C2) if the outcome in the first stage is (Dove, Hawk)
(d) (D1, D2) if the outcome in the first stage is (Dove, Dove)
Player 2
Hawk Dove
Player 1
Hawk 0 + A1, 0 + A2 3 + B1, 1 + B2
Dove 1 + C1, 3 + C2 2 + D1, 2 + D2
Based on the answer to Q7, the values of each pair (A1, A2), (B1, B2), (C1, C2), or (D1,
D2) must be equal to (1, 3), (3, 1), or (3/2, 3/2), and the total payoff is summarised by the
table above.
• Can players play (Hawk, Hawk) in stage 1?
No. (H, H) is a NE of the updated game above if and only if A1 ≥ 1+C1 and A2 ≥
1+B2. Recall that (A1,A2) = (1,3),(3,1), or (3/2,3/2). Also notice that C1≥ 1 and
B2≥ 1. If (A1,A2) = (1,3), then we can never satisfy A2≥ 1+B2. If (A1,A2) = (3,1),
then we can never satisfy A1 ≥ 1+C1. If (A1,A2) = (3/2,3/2), we can never satisfy
either inequality. Therefore, players cannot play (H, H) in stage 1 because it is never a
NE of the updated game matrix.
• Can players play (Hawk, Dove) in stage 1?
Yes. For example, let players play (Hawk, Dove) in stage 2 regardless of the outcome
in stage 1. Then, all pairs (A1, A2), ... (D1, D2) are equal to (3, 1) and the updated
game matrix is
Player 2
Hawk Dove
Player 1
Hawk 3, 1 6, 2
Dove 4, 4 5, 3
4
and (Hawk, Dove) is indeed a NE of this updated game matrix.
In general, there is always a SPE such that in the first stage, players play the NE of the
original stage game.
• Can players play (Dove, Hawk) in stage 1?
Yes. Logic similar to the last bullet point.
• Can players play (Dove, Dove) in stage 1?
No. (D, D) is a NE of the updated game above if and only if 2+D1 ≥ 3+B1 and
2+D2 ≥ 3+C2. This is equivalent to D1 ≥ 1+B1 and D2 ≥ 1+C2. We can use a
similar argument to the first bullet point to show that these two inequalities will never
be satisfied simultaneously. Therefore, the answer is “no”.
• Can players play mixed strategies in stage 1?
Yes. They can play the mixed NE of the original game: Pr(Hawk) = 1/2 for both
players. This is subgame perfect if, for example, they always play the same NE in
stage 2 regardless of the outcome in stage 1.
9. Consider the repeated game in which the following stage game is played twice.
Player 2
Stag Hare
Player 1 Stag 2, 2 0, 1Hare 1, 0 1, 1
In a pure-strategy SPE, suppose that players play (Stag, Hare) in stage 1. Then, in stage 2,
they will play
______ if the outcome in stage 1 is (Stag, Stag);
______ if the outcome in stage 1 is (Stag, Hare);
______ if the outcome in stage 1 is (Hare, Stag);
______ if the outcome in stage 1 is (Hare, Hare).
A: Suppose that in stage 2, the players
play a pair of actions with payoffs (x1,y1) if the outcome in stage 1 is (Stag, Stag);
play a pair of actions with payoffs (x2,y2) if the outcome in stage 1 is (Stag, Hare);
play a pair of actions with payoffs (x3,y3) if the outcome in stage 1 is (Hare, Stag);
play a pair of actions with payoffs (x4,y4) if the outcome in stage 1 is (Hare, Hare).
The total payoff matrix will be:
Player 2
Stag Hare
Player 1
Stag 2+x1, 2+y1 0+x2, 1+y2
Hare 1+x3, 0+y3 1+x4, 1+y4
In order for (S, H) to be a SPE strategy in stage 1, we must prevent profitable deviations for
both players, i.e., we must have
5
0+ x2 ≥ 1+ x4
1+ y2 ≥ 2+ y1
Moreover, in a pure-strategy SPE, the pair (xi,yi) must be a pair of pure-strategy NE payoffs
from the stage game for any i = 1,2,3,4. Because the stage game has only two pure NE,
(Stag, Stag) with payoff (2, 2) and (Hare, Hare) with payoff (1, 1), we must have (xi,yi) =
(2,2) or (1,1) for any i= 1,2,3,4. The only way to satisfy this condition and the inequalities
above is to let
x2 = y2 = 2
x4 = y4 = 1
y1 = x1 = 1
x3 = y3 = 1 or 2
The total payoff matrix is
Player 2
Stag Hare
Player 1 Stag 2 + 1, 2 + 1 0 + 2, 1 + 2Hare 1 + (1 or 2), 0 + (1 or 2) 1 + 1, 1 + 1
Therefore, the answer is
__(Hare, Hare)__ if the outcome in stage 1 is (Stag, Stag);
__(Stag, Stag)__ if the outcome in stage 1 is (Stag, Hare);
__(Hare, Hare) or (Stag, Stag)__ if the outcome in stage 1 is (Hare, Stag);
__(Hare, Hare)__ if the outcome in stage 1 is (Hare, Hare).