联系我们： 手动添加方式: 微信>添加朋友>企业微信联系人>13262280223 或者 QQ: 1483266981

EC9011
UNIVERSITY OF WARWICK
January Examinations 2019/20
Economic Analysis: Microeconomics – SOLUTIONS
Time Allowed: 3 Hours plus 15 minutes reading time during which notes may be made (on the
question paper only) BUT NO ANSWERS MAY BE BEGUN.
Answer FOUR questions: TWO questions must be from Section A and TWO questions must be
from Section B. Answer Section A questions in one booklet and Section B questions in a separate
booklet. All questions carry equal weight.
Approved pocket calculators are allowed.
Read carefully the instructions on the answer book provided and make sure that the particulars
required are entered on each answer book. If you answer more questions than are required and do
not indicate which answers should be ignored, we will mark the requisite number of answers in the
order in which they appear in the answer book(s): answers beyond that number will not be
considered.
Section A: Answer TWO questions
1. (a) The yi are minimum necessary levels of consumption. The assumption z 0
so the goods are Hicksian substitutes. (5 marks)
2.(a)(i) –c + x is ordinary utility, v(c ≠ rc) + v(x ≠ rx) is the gain-loss part of the utility
function. (4 marks)
(ii) A personal equilibrium is a (cú, xú) where (1) (cú, xú) maximize u(c, x; rc, rx) subject to
pc + x = m; (2) rc = cú, rx = xú. (4 marks)
(iii) A PE with car purchase satisfies
– + m ≠ p m ≠ + (m ≠ (m ≠ p))
– + 2p p pmax
– +
2
(4 marks)
(iv) PE without car purchase satisfies
m – + m ≠ p + 1 ≠ p
– + 1 (1 + )p p pmin
–
1 +
+ 1

(4 marks)
3
(Question 2 continued overleaf)
EC9011
(v) Multiple equilibria exist iff
– +
2 p
– + 1
1 +
The size of this interval increases with – and . As 1, it shrinks to p –+1
2 .
(4 marks)
(vi) There would be path-dependence if –+
2 p –
1+
+1
. Specifically, if p is in this interval,
he would buy the car (not buy the car) if he had bought the car (not bought the car) in
the previous period. (5 marks)
3. (a) Generally, the investor will invest a positive share in the risky asset iff E[R] > 0. Here,
E[R] = qpr + (1 ≠ q)0 + q(1 ≠ p)(≠r) = q(2p ≠ 1)r
So, s > 0 iff p > 0.5. (5 marks)
(b) For R = r, 0, ≠r, final wealth is w(1 + rs), w, w(1 ≠ rs). (5 marks)
(c) Expected utility of final wealth is
Eu = qp ln(w(1 + sr)) + (1 ≠ q) ln(w) + q(1 ≠ p) ln(w(1 ≠ sr))
(5 marks)
(d) With additional assumptions, the maximand is
Eu = p ln(1 + sr) + (1 ≠ p) ln(1 ≠ sr)
Assume an interior solution first. This will satisfy the FOC
Eu
s = 1 +
pr
sr ≠ (1
1
≠ p)r
≠ sr
= 0
p
1 + sr = (1
1
≠ p)
≠ sr

p(1 ≠ sr) = (1 ≠ p)(1 + sr)
sú = 2p ≠ 1
r
As p > 0.5, we know that a corner with sú = 0 is impossible. The other possibility is
that

Eu
s |s=1 > 0. But this requires
Eu
s |s=1 = pr
1 + r ≠ (1
1
≠
≠
p
r
)r
> 0
p
1 ≠ p
>
1 + r
1 ≠ r
4
(Question 3 continued overleaf)
EC9011
So, the solution is
sú =
I 2p≠1
r , p
1≠p 1+r
1≠r
1, p
1≠p
> 1+r
1≠r
(5 marks)
(e) By inspection, from part (c), we can write
Eu = q[p ln(w(1 + sr)) + (1 ≠ p) ln(w(1 ≠ sr))] + (1 ≠ q) ln(w)
As the last term is independent of s, the optimal s must maximize the term in square
brackets. So, the answer is unchanged if q < 1. (5 marks) 4.(i)(a) We expect to see the derivations here i.e. Lagrangean for Amy, substituting xB 2 = —xB 1 into the budget constraint for Bob. This is standard and gives xA 1 = – p1 p1 = –, xA 2 = (1 ≠ –) p1 p2 xB 1 = p2 p1 + —p2 , xB 2 = —p2 p1 + —p2 (5 marks) (b) Set p1 = p, p2 = 1. The excess demand for good 1 is then – + 1 p + — ≠ 1 Setting this equal to zero, and solving for p, we get pú = 1 1≠– ≠ —, so p1 = 1 1 ≠– ≠ —, p2 = 1. (5 marks) (c) Allocations are xA 1 = – = –, xA 2 = (1 ≠ –) 3 1 1 ≠ – ≠ — 4 = 1 ≠ —(1 ≠ –) xB 1 = 1 1 1≠– ≠ — + — = 1 ≠ –, xB 2 = —(1 ≠ –) (5 marks) 5 (Question 4 continued overleaf) EC9011 (ii)(a) Following the previous case, xA 1 = – 2 p1 + p2 p1 = –, xA 2 = (1 ≠ –) 2 p1 + p2 p2 xB 1 = p2 p1 + —p2 , xB 2 = —p2 p1 + —p2 (5 marks) (b) Set p1 = p, p2 = 1. The excess demand for good 1 is then zA = – 2 p + 1 p + 1 p + — ≠ 1 2 Setting this equal to zero, and rearranging, we get (1 –(p + 1)(p + —)+2p ≠ p(p + —)=0 ≠ –)p2 + p(— ≠ –(1 + —) ≠ 2) ≠ –— = 0 So, any roots of this that are positive are equilibrium prices. Generally, if B = 2 ≠ — + –(1 + —): p = B + (B2 + 4(1 ≠ –)–—)0.5 2(1 ≠ –) p = B ≠ (B2 + 4(1 ≠ –)–—)0.5 2(1 ≠ –) By inspection, the second root is negative and so cannot be an equilibrium. So, the equilibrium is unique. (5 marks) Section B: Answer TWO questions Please use a separate booklet 5. (a) A 50/50 randomization over L and C strictly dominates R for the column player. (2 marks) Once R is eliminated, a 50/50 randomization over M and B strictly dominates T for the row player. (2 marks) 6 (Question 5 continued overleaf) EC9011 Once T is eliminated, C weakly dominates L for the column player. (1 mark) Once C is eliminated, M dominates B for the row player. (1 mark) This leaves (M,L) as the solution. (1 mark) (Award part marks if student hesitates on the weak dominance step on the grounds that weakly dominated strategies can be part of a N.E. The question is not asking about N.E., and so this is not a legitimate concern here.) (b) Let q denote the probability with which players select F in a symmetric, mixed-strategy equilibrium. Indifference between playing F and playing R then requires q · 1 + (1 ≠ q) · 0 = q · c + (1 ≠ q) · 1, (2 marks) which gives q = 1/(2 ≠ c). (2 marks) For c > 0, a mixed-strategy equilibrium with q (0, 1) then requires c (0, 1).
(2 marks)
(An equivalent answer is found if q denotes the probability of choosing R.)
(c)(i)) Differentiating i’s payoff re. ai, (2 marks)
setting ai = aN ’i, (1 mark)
equating the resulting expression to zero and solving for aNE gives aNE = 1 ≠ c/N. (2
marks)
(ii) Differentiating the expression provided re. aú, (2 marks)
and solving for aú gives aNE = 1 ≠ c/N2. (1 mark)
(iii) Proceeding as in (i), we get aNE = 1 ≠ (c ≠ t)/N ≠ t/N2, (2 marks)
which equals aú for t = c. (2 marks)
6. (a) Both are sealed-bid auctions (buyers submit their bids independently and without seeing
each other’s bids). In a SP action the good is awarded to the highest bidder and the
bidder pays a price equal to the highest next bid. In a FP action the good is awarded to
the highest bidder and the bidder pays a price equal to her bid. (1 mark)
The optimal bidding strategy in a SP auction is to bid an amount equal to one’s
valuation. Bidding above the valuation has no effect if there is there is a higher bid or if
the highest bid by other players is below the bidder’s valuation, but results in a negative
surplus if the highest bid by other players is above the valuation and below the bid.
Bidding below the valuation has no effect if there is a higher bid that is above the
valuation or if the highest bid by other players is below the bidder’s valuation, but
results in the loss of a positive surplus if the highest bid by other players is below the
valuation and above the bid. (2 marks)
7
(Question 6 continued overleaf)
EC9011
The equilibrium bidding strategy in a FP auction consists of bidding an amount equal to
the expected highest valuation of players that have a valuation lower than the bidder’s.
If all players do the same, this results in a win for the highest valuation bidder.
(2 marks)
It has been shown in lectures that the two auction format yield the same revenue.
(1 mark)
(b) (i) There are two pure-strategy NEs (L1 = L, L2 = L) and (L1 = R, L2 = R).
(2 marks)
(ii) If the rational types of both players select L, they both get 50 in expectations.
However, if one of them deviates from playing L and selects R instead, she gets 60
for sure, so (L, L) is not a NE. (3 marks)
If the rational types of both players select R, they both get 60 for sure. This is a
NE (no player can gain from deviating from it). (2 marks)
(c)(i) Differentiating A’s profits, qA(pA ≠ 1/2), with respect to pA, we obtain
pA = 3/4 + pB/4. Doing the same for B, we get pB = 3/4 + pA/4. Letting
pA = pB = p, gives and solving for p gives p = 1. (3 marks)
(ii) Differentiating profits for each type of firm B (B: high marginal cost, B: low
marginal cost) gives pB =1+ pA/4, pB = 1/2 + pA/4. (1 mark)
A cannot vary its price depending on B’s type. Differentiating A’s expected profits
gives pA = 3/4+(pB + pB)/8. (2 marks)
Solving for pA, pB and pB gives pA = 1, pB = 5/4 and pB = 3/4. (1 mark)
(iii) Proceeding as in (i) to solve the game between A and B, we get pA = 16/15 and
pB = 19/15. Doing the same for the game between A and B, we get pA = 14/15
and pB = 11/15. Using these values, we get expected profits of 229/900 ￥ 0.254
for A and 137/450 ￥ 0.3 for B. (2 marks)
Substituting the equilibrium prices found in (ii) into the expressions for profits we
get expected profits of 1/4 = 0.25 and 5/16 ￥ 0.31. So, expected profits are higher
for A under (iii) than under (ii), and the opposite is true for B. (2 marks)
7. (a) (i) Starting from the last stage, A would always choose not to try to undermine B (as
doing so gives A a lower payoff). So, if the game continues to the last stage, payoffs
will be 20 for both players. (2 marks)
Anticipating this, B chooses to enter in stage II (assuming the outside option of
stating out has zero value). (2 marks)
Anticipating this, A chooses to enter in stage I (assuming the outside option of
stating out has zero value). Equilibrium payoffs are thus 20 for both players.
(2 marks)
8
(Question 7 continued overleaf)
EC9011
(ii) Now if B enters in stage II, both sellers will obtain payoffs of 50 ≠ 25 ≠ 30 = ≠5 < 0; so B will choose not to enter in stage II and A will receive a payoff of 100 ≠ 50 ≠ 30 = 20. (3 marks) Anticipating this, A will choose to enter in stage I. (2 marks) (iii) For c = 0, A is indifferent (as it receives the same payoff under (i) and (ii)). For any c > 0, A’s payoff, net of C, would be higher under (ii), and so A would choose not to
adopt the new process. (2 marks)
(b) Using backward induction, fix w. We solve maxL fi = 6L1/2 ≠ wL. The FOC is
w = 3/

L. Hence L = 9/w2, the firm’s labour demand curve. (3 marks)
Now proceeding backwards, we solve maxw u = (w ≠ —)L. This gives wú = 2 —, and
hence Lú = 9/
1
4—2
2
. (3 marks)
(c) If one country deviated from low tariffs in a given period, it would experience a one-shot
deviation gain of 6 ≠ 5=1; however, from the next period onwards its payoff would be 4
instead of 5, i.e. a per-period indefinite loss of 1. (3 marks)
The present value of this indefinite stream, discounted at the discount factor ” (0, 1), is
≠”/(1 ≠ ”). So, deviation is not gainful if 1 ≠ ”/(1 ≠ ”) 0, i.e. if ” > 1/2. (3 marks)
8. (a) (i) In a separating equilibrium, we have w( ) = = 2 and w( ) = = 1. Let e( ) be
the high-productivity type’s equilibrium level of education, and assume that
off-equilibrium beliefs are such that any level of education not equal to e( ) signals
= 1. Then, unless e = e( ) is chosen, the only possible alternative choice that is
rational is e = 0. (2 marks)
Conditions for a separating equilibrium are then:
w( ) ≠ e
1

2
/
ò
w( ) 1 ≠ e( )/

2 0,
w( ) ≠ e
1

2
/
ò
w( ) 1 ≠ e( ) 0.
(3 marks)
The minimum level of e( ) satisfying these conditions is the one for which the second
condition holds with equality, i.e. e( )=1. (2 marks)
(ii) In a pooling equilibrium where e = 0 for both types, both types obtain a payoff of
(w( ) + w( ))/2=3/2. (1 mark)
In the lowest e( ) separating equilibrium, the low-ability type obtains w( )=1 and
the high-ability type obtains w( ) ≠ 1/
2 ￥ 1.3 < 1.5=3/2. (2 marks) 9 (Question 8 continued overleaf) EC9011 (b) (i) If the worker always chooses e = 1, she is always paid w(e = 1), and so E(w)=1 implies w(e = 1) = 1, which gives the worker a level of disposable income of 1. If the worker chose e = 0, she would get w(e = 0) = 0 and her disposable income, inclusive of the value of using the machine for private consumption, would be 1/2 < 1. (3 marks) So the worker will always chooses e = 1. (1 mark) (ii) In order for firms to break even, wage levels must satisfy (wH + wL)/2 = E(w)=1, which implies wH = 2 ≠ wL. (2 marks) The worker’s expected utility for e = 1 is (u(wH) + u(wL))/2=(u(2 ≠ wL) + u(wL))/2. (1 mark) The worker’s expected utility for e = 0 is u(wL + 1/2). (1 mark) The second condition identifying incentive-compatible contracts for which firms break even, is thus 1 u(2 ≠ wL) + u(wL) 2 /2 u(wL + 1/2). (3 marks) (iii) Solving the above condition when it holds with equality, for u(y) = y ≠ y2/2, gives wL = 3/4 (which implies wH = 5/4). (4 marks) (Only award full marks for (iii) if full derivations are provided.) 10 (End)