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ProblemSet04--solutions1.pdf
Problem Set 4 (Solutions)
ECO 444
Konrad Grabiszewski
Exercise 1
We need to find the best-response correspondence for Ann.
First, suppose that tB < vA. Then, Ann’s utility (as a function of tA) is as depicted in Figure 1.
Note that vA 2 − tB can be positive or negative; in the figure below, we assume that the value is
positive. However, the specific assumption about vA−tB 2
does not matter because it is always true
(as log as vA > tB) that vA − tB > vA−tB2 .
𝑡𝑡𝐴𝐴
𝑢𝑢𝐴𝐴
𝑡𝑡𝐵𝐵
𝑣𝑣𝐴𝐴 − 𝑡𝑡𝐵𝐵 𝑣𝑣𝐴𝐴 2 − 𝑡𝑡𝐵𝐵
Figure 1: Ann’s utility when tB < vA.
Hence, we conclude that if tB < vA, then Ann should choose tA > vB; i.e., any tA > vB is her
best-response.
Second, suppose that tB = vA. Then, Ann’s utility (as a function of tA) is as depicted in Figure
2. Note that vA 2 − tB < 0.
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𝑡𝑡𝐴𝐴
𝑢𝑢𝐴𝐴
𝑡𝑡𝐵𝐵𝑣𝑣𝐴𝐴 2 − 𝑡𝑡𝐵𝐵
𝑣𝑣𝐴𝐴 − 𝑡𝑡𝐵𝐵 = 0
Figure 2: Ann’s utility when tB = vA.
Hence, we conclude that if tB = vA, then Ann should choose either tA = 0 or tA > vB.
Third, suppose that tB > vA. Then, Ann’s utility (as a function of tA) is as depicted in Figure 3.
𝑡𝑡𝐴𝐴
𝑢𝑢𝐴𝐴
𝑡𝑡𝐵𝐵
𝑣𝑣𝐴𝐴 2 − 𝑡𝑡𝐵𝐵
𝑣𝑣𝐴𝐴 − 𝑡𝑡𝐵𝐵
Figure 3: Ann’s utility when tB > vA.
Hence, we conclude that if tB = vA, then Ann should choose tA = 0.
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The above analysis allows us to depict Ann’s best-response correspondence.
BRA(tB) =
(tB,∞) if tB < vA
(tB,∞) ∪{0} if tB = vA
{0} if tB > vA
(1)
We also derive Bob’s best-response correspondence.
BRB(tA) =
(tA,∞) if tA < vB
(tA,∞) ∪{0} if tA = vB
{0} if tA > vB
(2)
In this game, the simplest way to find the Nash equilibria is to conduct a graphical analysis of the
best-response correspondences. We start with Ann’s best-response correspondence (Figure 4).
𝑡𝑡𝐵𝐵
𝑡𝑡𝐴𝐴
𝑣𝑣𝐴𝐴
45° 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑣𝑣𝐴𝐴
Figure 4: Ann’s best-response correspondence (in blue).
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Next, we obtain Bob’s best-response correspondence (Figure 5).
𝑡𝑡𝐴𝐴
𝑡𝑡𝐵𝐵
𝑣𝑣𝐵𝐵
45° 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑣𝑣𝐵𝐵
Figure 5: Bob’s best-response correspondence (in red).
Finally, we combine Ann’s and Bob’s best-response correspondence in one graph (Figure 6). Note
that, in order to make a graph, we assume that vA > vB but this assumption; however, this
assumption is irrelevant.
𝑡𝑡𝐵𝐵
𝑡𝑡𝐴𝐴
𝑣𝑣𝐴𝐴
45° 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑣𝑣𝐴𝐴
𝑣𝑣𝐵𝐵
𝑣𝑣𝐵𝐵
Figure 6: Ann’s (in red) Bob’s (in blue) best-response correspondences.
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Where do Ann’s and Bob’s best-response correspondences intersect in Figure 6? There are two
places which are circled in Figure 7.
𝑡𝑡𝐵𝐵
𝑡𝑡𝐴𝐴
𝑣𝑣𝐴𝐴
45° 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑣𝑣𝐴𝐴
𝑣𝑣𝐵𝐵
𝑣𝑣𝐵𝐵
Figure 7: Ann’s (in red) Bob’s (in blue) best-response correspondences.
We conclude that there infinitely many Nash equilibria.
We can divide Nash equilibria into two types.
(t∗A, t ∗ B) such that t
∗ A = 0 and t
∗ B ≥ vA (3)
(t∗A, t ∗ B) such that t
∗ B = 0 and t
∗ A ≥ vB (4)
Exercise 2
(a) Normal-form representation.
• N = {A, B} is the set of players;
• Si = [0,∞) is the set of strategies of player i; a strategy si is a level of effort;
• ui is the utility function of player i defined as
ui(si, sj) = (k + sj)si −s2i (5)
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where (k + sj)si is the revenue, s 2 i is the cost, and k > 0.
(b) We solve the optimization problem of Firm i: maximize ui with respect to si.
∂ui ∂si
= k + sj − 2si (6)
∂2ui ∂s2i
= −2 (7)
Since ∂ 2ui ∂s2i
< 0 for each si, we know that FOC defines the maximum s̃i (best-response of i).
s̃i = 1
2 (k + sj) (8)
From equation (8) we obtain A and B best-responses.
s̃A = 1
2 (k + sB) (9)
s̃B = 1
2 (k + sA) (10)
(c) From (9) and (10), we derive two equations which describe A and B equilibrium strategies
denoted as s∗A and s ∗ B, respectively.
s∗A = 1
2 (k + s∗B) (11)
s∗B = 1
2 (k + s∗A) (12)
We solve that system of two equalities in order to obtain s∗A and s ∗ B.
s∗A = k (13)
s∗B = k (14)
(d) Comparative statics. First, we compute equilibrium values.
s∗A = k (15)
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s∗B = k (16)
uA (s ∗ A, s
∗ B) = k
2 (17)
uB (s ∗ A, s
∗ B) = k
2 (18)
Second, we analyze the impact of k on equilibrium values.
ds∗A dk
> 0 (19)
ds∗B dk
> 0 (20)
duA (s ∗ A, s
∗ B)
dk > 0 (21)
duB (s ∗ A, s
∗ B)
dk > 0 (22)
(e) We solve the optimization problem of Firm i: maximize ui with respect to si.
∂ui ∂si
= k + sj − 2cisi (23)
∂2ui ∂s2i
= −2ci (24)
Since ∂ 2ui ∂s2i
< 0 for each si, we know that FOC defines the maximum s̃i (best-response of i).
s̃i = 1
2ci (k + sj) (25)
From equation (25) we obtain A and B best-responses.
s̃A = 1
2cA (k + sB) (26)
s̃B = 1
2cB (k + sA) (27)
(f) From (28) and (29), we derive two equations which describe A and B equilibrium strategies
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denoted as s∗A and s ∗ B, respectively.
s∗A = 1
2cA (k + s∗B) (28)
s∗B = 1
2cB (k + s∗A) (29)
We solve that system of two equalities in order to obtain s∗A and s ∗ B.
s∗A = k(2cB + 1)
4cAcB − 1 (30)
s∗B = k(2cA + 1)
4cAcB − 1 (31)
(Now you can see why we needed our “technical” assumption 4cAcB > 1.)
(g) Comparative statics. First, we compute equilibrium values.
s∗A = k(2cB + 1)
4cAcB − 1 (32)
s∗B = k(2cA + 1)
4cAcB − 1 (33)
uA (s ∗ A, s
∗ B) =
k2cA(2cB + 1) 2
(4cAcB − 1)2 (34)
uB (s ∗ A, s
∗ B) =
k2cB(2cA + 1) 2
(4cAcB − 1)2 (35)
Second, we analyze the impact of k, cA, and cB on equilibrium values.
∂s∗A ∂k
> 0 (36)
∂s∗A ∂cA
< 0 (37)
∂s∗A ∂cB
= − 2k(2cA + 1)
(4cAcB − 1)2 < 0 (38)
∂s∗B ∂k
> 0 (39)
∂s∗B ∂cA
< 0 (40)
∂s∗B ∂cB
< 0 (41)
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∂uA (s ∗ A, s
∗ B)
∂k > 0 (42)
∂uA (s ∗ A, s
∗ B)
∂cA =
k2cA(2cB + 1) 2
(4cAcB − 1)4 (1 − 16c2Ac
2 B) < 0 (43)
∂uA (s ∗ A, s
∗ B)
∂cB =
k2cA(4cAcB − 1) (4cAcB − 1)4
(−4 − 8cA − 8cB − 16cAcB − 16cAc2B) < 0 (44)
∂uB (s ∗ A, s
∗ B)
∂k > 0 (45)
∂uB (s ∗ A, s
∗ B)
∂cA < 0 (46)
∂uB (s ∗ A, s
∗ B)
∂cB < 0 (47)
Exercise 3
(a) We impose two assumptions.
a > 2cA − cB (48)
cA > cB > 0 (49)
First, we show that a > cA. Note that 2cA−cB = cA + (cA−cB). By the assumption (49), cA > cB.
Hence, 2cA − cB > cA. By assumption (48), a > 2cA − cB. Hence, a > cA. Second, we show that
a > cA. By the assumption (49), cA > cB. Hence, 2cA−cB > 2cB −cB = cB. By assumption (48),
a > 2cA − cB. Hence,a > cB.
(b) Normal-form representation.
• N = {A, B} is the set of players;
• Si = [0,∞) is the set of strategies of player i; a strategy qi is the quantity produced;
• ui is the utility function of player i defined as
ui(qA, qB) = qiP (Q) − ciqi = qi(a− qA − qB) − ciqi (50)
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(c) We solve the optimization problem of Firm i: maximize ui with respect to qi.
∂ui ∂qi
= a− 2qi − qj − ci (51)
∂2ui ∂q2i
= −2 (52)
From (51), we drive the first-order condition (FOC). From (52) we know that we will have a
maximum. Let q̃i be the best-response function of player i to player j strategy qj.
q̃i = 1
2 (a− ci − qj) (53)
From equation (53) we obtain A and B best-responses.
q̃A = 1
2 (a− cA − qB) (54)
q̃B = 1
2 (a− cB − qA) (55)
(d) From (54) and (55), we derive two equations which describe A and B equilibrium strategies
denoted as q∗A and q ∗ B, respectively.
q∗A = 1
2 (a− cA − q∗B) (56)
q∗B = 1
2 (a− cB − q∗A) (57)
We solve that system of two equalities in order to obtain q∗A and q ∗ B.
q∗A = 1
3 (a− 2cA + cB) (58)
q∗B = 1
3 (a + cA − 2cB) (59)
We need to make sure that our solution makes sense. In particular, we need to check that both
equilibrium values are at least zero. By assumption, a > 2cA − cB. Hence, q∗A > 0. Next, note
that 2cA−cB > 2cB −cA; this is because cA > cB. Hence, a > 2cB −cA. This implies that q∗B > 0.
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(e) Comparative statics. First, we compute equilibrium values.
q∗A = 1
3 (a− 2cA + cB) (60)
q∗B = 1
3 (a + cA − 2cB) (61)
Q∗ = q∗A + q ∗ B =
1
3 (2a− cA − cB) (62)
P ∗ = a−Q∗ = 1
3 (a + cA + cB) (63)
uA (q ∗ A, q
∗ B) = q
∗ AP
∗ − cAq∗A = 1
3 (a− 2cA + cB)
[ 1
3 (a + cA + cB) − cA
] (64)
= 1
3 (a− 2cA + cB)(a− 2cA + cB) =
1
3 (a− 2cA + cB)2 (65)
uB (q ∗ A, q
∗ B) =
1
3 (a + cA − 2cB)2 (66)
Second, we analyze the impact of a, cA, and cB on equilibrium values.
∂q∗A ∂a
> 0 (67)
∂q∗A ∂cA
< 0 (68)
∂q∗A ∂cB
> 0 (69)
∂q∗B ∂a
> 0 (70)
∂q∗B ∂cA
> 0 (71)
∂q∗B ∂cB
< 0 (72)
∂Q∗
∂a > 0 (73)
∂Q∗
∂cA < 0 (74)
∂Q∗
∂cB < 0 (75)
∂P ∗
∂a > 0 (76)
∂P ∗
∂cA > 0 (77)
∂P ∗
∂cB > 0 (78)
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∂uA (q ∗ A, q
∗ B)
∂a > 0 (79)
∂uA (q ∗ A, q
∗ B)
∂cA < 0 (80)
∂uA (q ∗ A, q
∗ B)
∂cB > 0 (81)
∂uB (q ∗ A, q
∗ B)
∂a > 0 (82)
∂uB (q ∗ A, q
∗ B)
∂cA > 0 (83)
∂uB (q ∗ A, q
∗ B)
∂cB < 0 (84)
Exercise 4
(a) Best-response correspondences.
FOD and SOD for Ann.
∂uA ∂sA
= v sB
(sA + sB)2 − cA (85)
∂2uA ∂s2A
< 0 (86)
Note that ∂ 2uA ∂s2
A < 0 for every value of sA. Consequently, FOC defines the maximum. From FOC,
we derive the equation describing Ann’s best-response function s̃A.
v sB
(s̃A + sB)2 − cA = 0 (87)
By symmetry, we derive the equation describing Bob’s best-response function s̃B.
v sA
(sA + s̃B)2 − cB = 0 (88)
(b) Nash equilibria.
The following system describes Nash equilibrium (s∗A, s ∗ B).
v s∗B
(s∗A + s ∗ B)
2 − cA = 0 (89)
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v s∗A
(s∗A + s ∗ B)
2 − cB = 0 (90)
Hence, it must be true that s∗ A
cB =
s∗ B
cA which implies that s∗B =
cA cB s∗A. This allows us to derive the
equilibrium values: we replace s∗B by cA cB s∗A.
v s∗A
(s∗A + s ∗ B)
2 − cB = 0 (91)
v s∗A
(s∗A + cA cB s∗A)
2 − cB = 0 (92)
v s∗A
(s∗A) 2
(1 + cA cB
)2 − cB = 0 (93)
v c2B
s∗A(cA + cB) 2 − cB = 0 (94)
v cB
s∗A(cA + cB) 2
= 1 (95)
s∗A = v cB
(cA + cB)2 (96)
We also find s∗B = v cA
(cA+cB) 2 . Hence,
( v cB (cA+cB)
2 , v cA
(cA+cB) 2
) is the (only) Nash equilibrium in this
game.
(c) Comparative statics.
First, we compute equilibrium values. Let p∗A and p ∗ B denote equilibrium probability of Ann
winning and Bob winning, respectively.
s∗A = v cB
(cA + cB)2 (97)
s∗B = v cA
(cA + cB)2 (98)
p∗A = cB
cA + cB (99)
p∗B = cA
cA + cB (100)
Second, we compute derivatives with respect to exogenous parameters cA and cB.
∂s∗A ∂cA
< 0 (101)
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∂s∗A ∂cB
= v cA − cB
(cA + cB)3 > 0 (102)
∂s∗B ∂cA
= v cB − cA
(cA + cB)3 < 0 (103)
∂s∗B ∂cB
< 0 (104)
∂p∗A ∂cA
< 0 (105)
∂p∗A ∂cB
> 0 (106)
∂p∗B ∂cA
> 0 (107)
∂p∗B ∂cB
< 0 (108)
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