micro prelim solutions

7/27/2019 Micro Prelim Solutions

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Micro Prelim Solutions

Joseph Steinberg

May 5, 2009

Contents

1 Fall 2008, I.1 (Werner, Pratt theorem) 2

2 Fall 2008, I.2 (Werner, Profit rationalization) 3

3 Spring 2008, I.1 (Werner, Risk) 4

4 Spring 2008, I.1 (Werner, Reavealed preference) 5

5 Fall 2007, I.1 (Werner, Homothetic preferences & Homogeneous utility functions) 6

6 Fall 2007, I.2 (Werner, Risk) 7

7 Fall 2007, II.1 (B. Allen, FWT) 8

8 Fall 2007, II.2 (B. Allen, Sonnenscheins conjecture) 9

9 Fall 2007, III.1 (Aldo, Nash equilibrium, dominance 11

10 Spring 2007, I.1 (Werner, Continuity of preferences) 14

11 Spring 2007, I.1 (Werner, Pratts theorem) 15

12 Spring 2007, II.1 (B. Allen, Generic economies) 16

13 Spring 2007, II.2 (B. Allen, FWT/Continuum of agents) 18

14 Spring 2007, III.1 (Aldo, Extensive form games) 20

15 Fall 2006, I.1 (Werner, Profit function and supply correspondence) 21

16 Fall 2006, I.2 (Werner, Expected utility) 23

17 Fall 2006, II.1 (B. Allen, Nonconvexities) 24

18 Fall 2006, II.1 (B. Allen, Generic economies) 26

19 Fall 2006, III.1 (Aldo, Perfect equilibrium) 28

20 Fall 2005, I.1 (Werner, Risk) 28

21 Fall 2005, III.1 (Aldo, Nash equilibrium) 31

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1 Fall 2008, I.1 (Werner, Pratt theorem)

Definition. Let v be a twice-differentiable and strictly increasing von Neumann-Morgenstern utility func-tion. The Arrow-Pratt measure of (absolute) risk aversion is

A(x) = v(x)

v(x)

.

Definition. Let v be a von Neumann-Morgenstern utility function. Given fixed deterministic level ofinitial consumptrion w, the risk compensation for additional risky consumption plan z with E[z] = 0 is(w, z, defined by

v(w (w, z)) = E[v(w + z)].

Part (a)

Theorem (Pratts theorem). Let v1, v2 be two strictly increasing, C2, von Neumann-Morgenstern utility

functions with risk compensation 1, 2 and absolute risk aversion measures A1, A2, respectively. Then thefollowing three conditions are equivalent:

(A) A1(x) A2(x) for every x R.

(B) 1(w, z) 2(w, z)) for every w R and every risky consumption plan z with E[z] = 0.

(C) v1 is a concave transformation of v2, i.e., v1(x) = f(v2(x)) for every x R, where f is concave andstrictly increasing.

Part (b)

(i) Proof that (A) implies (C). Suppose (A) holds. Then A1(x) A2(x), x R. Define f by f(t) =v1(v

12 (t)) for evert t. Then

f(t) =v1(t)(v

12 (t))

v2(v12 (t))

and

f(t) = v1 (x) (v2 (x)v1(x))/v2(x)[v2(x)]2

where x = v12 (t). We can rewrite this as

f(t) = [A2(x) A1(x)]v1(x)

[v2(x)]2.

Since (A) holds, f(t) 0, t R. So f is concave, and thus (A) implies (C).

(ii) Proof that (C) implies (B). Suppose (C) holds. Let w R and let z be a risky consumption plan withE[z] = 0. By definition of 1,

v1(w 1(w, z)) = E[v1(w + z)].

Since v1 = f(v2) and f is concave, Jensens inequality implies that

v1(w 1(w, z)) = E[v1(w + z)] = E[f(v2(w + z))] f(E[v2(w + z)]).

By definition of2,f(E[v2(w + z)]) = f(v2(w 2(w, z))) = v1(w 2(z)).

Combining the last two inequalities, we have

v1(w 1(w, z)) v1(w 2(z)).

Since v1 is strictly increasing, it must be that 1(w, z) 2(w, z). Thus (C) implies (B).

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Part (c)

Let v1(x) =x1

1 for (0, 1) and let v2(x) = ex for > 0. Then

A1(x) = v1 (x)v1(x)

= x1

x=

x

x1+=

x.

andA2(x) =

v2 (x)v2(x)

= 2ex

ex=

2ex

ex= .

Then for x >

, A1(x) < A2(x), and for x A2(x). Thus neither v1 nor v2 is more risky thanthe other in the sense of Pratts theorem.

2 Fall 2008, I.2 (Werner, Profit rationalization)

Part (a)

Weak axiom of profit maximization (WAPM): The set of observations {pt, yt}Tt=1 satisfies the WAPMif

pt

yt

pt

ys

, s,t.

Part (b)

Suppose that there exists a closed, convex production set Y that profit-rationalizes the observations. Thenyt Y for all t and pt yt = maxyY pty for all t = 1, . . . , T . Take s, t {1, . . . , T }. Then since yt, ys Yand pt yt pt y, y Y, we have pt yt pt ys. Thus {pt, yt}Tt=1 satisfies the WAPM. How does Y beingclosed and convex matter for this direction?

Now suppose that these observations satisfy the WAPM. Then pt yt pt ys, s, t. Let Y be the convexhull of{y1, . . . , yT}, i.e.,

Y =

y Rn : y =

T

t=1tyt, (1, . . . , T) T

where

T =

(1, . . . , T) : t 0, t,

Tt=1

t = 1

.

Then Y is convex by definition. Take {yn} Y such that yn y. Then yn is a convex combination of{y1, . . . , yT} for all n, i.e.,

yn =Tt=1

tnyt

for (1n, . . . , Tn )

T. Note that T is closed. Then (1n, . . . , Tn ) (

1, . . . , T) T, which impliesthat

yn y =T

t=1

tyt.

Then y Y, so Y is closed. We know pt yt pt ys, s, t. Then

pt yt T

s=1

s(pt ys), t, (alpha1, . . . , T) T.

We can rewrite this as

pt yt pt

Ts=1

s ys

, t, (alpha1, . . . , T) T.

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This implies thatpt yt pt y, t, y Y.

In other words, pt yt = maxyY pt y. Thus Y profit-rationalizes {pt, yt}Tt=1.

3 Spring 2008, I.1 (Werner, Risk)

Part (a)

Let v : R+ R be a von Neumann-Morgenstern utility function defined as v(x) = ( x)2 where isgreater than the largest possible value than either y or z can take. Then v(x) is concave and nondecreasingfor all possible values ofy and z. Suppose y is more risky than z. We want to show that var(y) var(z), i.e,

E[y2] (E[y])2 E[z2] (E[z])2.

Since E[z] = E[y], we just need to show that

E[z2] E[y2].

Since y is more risky than z and v is continuous, nondecreasing and concave, it must be that E[v(z)]

E[v(y)], i.e.,E[( z)2] E[( y)2].

We can rewrite this asE[2 + 2z z2] E[2 + 2y y2].

Since E(y) = E(z), several terms cancel out and we are left with

E[z2] E[y2].

Multiplying by 1 and reversing the inequality gives us

E[y2] E[z2].

Thus var(y) var(z).

Part (b)

I assume that the support of both y and z is the closed interval [a, b]. Suppose that y is more risky than zand z is more risky than y. Since y is more risky than z, then by definition z second-order stochasticallydominates y, i.e., w

a

Fz(t) dt

wa

Fy(t) dt, w [a, b].

Similarly, since z is more risky than y,

w

a

Fy(t) dt w

a

Fz(t) dt, w [a, b].

Then wa

Fz(t) dt =

wa

Fy(t) dt, w [a, b].

Does this imply that Fz(t) = Fy(t) for all t [a, b]?

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Part (c)

Define the random variables y and z as

y =

2 p = 1/2

3 p = 1/4

7 p = 1/4

and

z =

1 p = 1/4

3 p = 1/4

4 p = 1/4

6 p = 1/4

Note that E[y] = E[z] = 3.5.Let v1(x) = ln(x). Then v1 is continuous, nondecreasing and concave, and we have

E[v1(y)] = 1/4 ln(84) > 1/4 ln(72) = E[v1(z)].

Then y is not more risky than z.Now let v2(x) = (10x)2, which is also continuous, nondecreasing and concave (at least on the relevant

range). ThenE[v2(y)] = 46.5 < 45.5E[v2(z)].

Thus z is not more risky than y.

4 Spring 2008, I.1 (Werner, Reavealed preference)

Part (a)

Generalized weak axiom of revealed preference (GWARP): The set of observations {pt, xt}Tt=1satisfies the GWARP if

pt xs pt xt ps xs ps xt, s,t.

Part (b)

Suppose that locally non-satiated utility function u rationalizes {pt, xt}Tt=1. Since u rationalizes these obser-vations, it must be that (i) for all t = 1, . . . , T , u(xt) u(x) for all x such that pt x pt xt.

Pick t {1, . . . , T } and take x such that pt x < pt xt. Then we can find > 0 small enough so thatfor all y with y x < , pt y < pt xt. By local non-satiation of u, for each such there exists y withy x < and u(y) > u(x). By (1), u(xt) u(y), so it must be that u(xt) > u(x). Thus we also have (ii)for all t = 1, . . . , T , u(xt) > u(x) for all x such that pt x < pt xt.

Take s, t such that pt xs pt xt. Then by (i), u(xt) u(xs). Suppose that ps xt < ps xs. Then (ii)implies that u(xs) > u(xt) which is a contradiction. Therefore it must be that ps xt ps xs. So we have

pt xs pt xt ps xs ps xt, s,t.

Thus the GWARP holds.

Part (c)

Let u(x) = 1 for all x Rn+, i.e., u is constant. Then for any x, y Rn+ such that u(y) > u(x), so u is

globally satiated and thus also locally satiatied. Let n = 2 and T = 2. Let the observations of prices andconsumption bundles be as follows:

p1 = (1, 2), x1 = (1, 2) p2 = (2, 1), x2 = (2, 1).

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Then we have

p1 x1 = 5

p1 x2 = 4

p2 x1 = 4

p2

x2

= 5

So p1 x2 < p1 x1 but p2 x2 > p2 x1. Then the GWARP does not hold.

Part (d)

First, define the relations R (weak) and P (strict) between observation xt and commodity bundle x as

pt x pt xt xtRx

pt x < pt xt xtP x

I will write (xtRx) and (xtP x) to denote the negations of these relations.

Theorem (Afriat). Observations{pt, xt}Tt=1 are rationalized by a locally non-satiatied utility function if andonly if these observations satisfy the Generalized axion of revealed preference (GARP):

xt1Rxt2, xt2Rxt3, . . . , xtn1Rxtn (xtnP xt1), {t1, . . . , t n} {1, . . . , T }.

5 Fall 2007, I.1 (Werner, Homothetic preferences & Homogeneousutility functions)

Definition. u() is a utility representation of if for all x, x, x x u(x) u(x).

This question could be answered in two ways. First, the statement we are asked to prove is false.Lexicographic preferences satisfy all assumptions of the question but have no utility representation at all.The other possible answer would be to add the assumption of continuity and prove existence of a utility

representation (and then show that it is homogeneous of degree one).

First answer. Let be a lexicographic preference relation on R2+, i.e., x y if x1 y1 or (x1 = y1 andx2 y2). This preference relation is clearly reflexive, transitive and complete. Let x, y R2 such that x yand x = y. Then either x1 > y2 or (x1 = y2 and x2 > y2). In either case, x y, so is strongly monotone.Now let x, y R2+ such that x y. Then x1 = y1 and x2 = y2. Let > 0. Then x1 = y1 and x2 = y2,so x y. Therefore is homothetic.

Suppose for contradiction that u : X R such that u is a utility representation of. For each x R+,we can find a rational number r(x) such that u(x, 2) > r(x) > u(x, 1). Let x, x R+ such that x > x.Then

u(x, 2) > r(x) > u(x, 1) > u(x, 2) > r(x) > u(x, 1).

Thus r(x) > r(x), so we have a one-to-one map r : R+ Q. This is a contradiction since R+ is uncountable

andQ

is countable. Therefore has no utility representation.Second answer (additional assumption of continuity). Let be a reflexive, transitive, complete, strictlymonotone, and continuous preference relation on R2+. Suppose that there exists a utility representationu : R2+ R of that is homogeneous of degree 1. Let x, y R

2+ such that x y and let > 0. Since u

is a utility representation of , u(x) = u(y). Then u(x) = u(y). Since u is homogeneous of degree one,u(x) = u(y). This implies that x y, so is homothetic.

Now suppose that is homothetic. Let x Rl+. Let e = (1, 1, . . . , 1) R++. Then e Rl+ R+.

Choose such that e x. Since is monotone, e x. Also note that 0e x, so x 0e.

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Define A+ and A as

A+ = { R+ : e x}

A = { R+ : x e}

(1)

Since is continuous, A+

and A are closed. A+

and 0 A, so both sets are also nonempty. Bycompleteness of ,

R+ = A+ A.

Since R+ is connected, it cannot be separated into two disjoint closed sets, so A+ A = . Let (x) A+ A. Then (x)e x.

Now suppose that A+ A such that = (x). Then either > (x) or (x) > . If theformer case is true, then since is monotone, e (x)e x. If the latter, then e (x)e x. Ineither case, it cannot be true that A+ A. Therefore (x) is unique. Let y Rl+, and suppose thaty x. Then by definition of (x), we have

(y)e y x (x)e.

By transitivity of , (y)e (x)e, and so by monotonicity of, (y) (x).

Now suppose that (y) (x). Then by monotonicity of , (y)e (x)e. By definition of(x), wehave

y (y)e (x)e x.

By transitivity of, y x. Thus we have y x (y) (x), so u(x) = (x) is a utility representationof . Note that u : Rl+ R+. For any x R

l+, we can write u

1([0, (x)]) and u1([(x), )) as

u1([0, (x)]) = {y Rl+ : (x) u(y)} = {y Rl+ : (x)e y}

u1([(x), )) = {y Rl+ : u(y) (x)} = {y Rl+ : y (x)e}

By continuity of , both of these sets are closed. So we have for every x Rl+, [0, (x)] and u1([0, (x)])

are closed, and [(x), ) and u1([(x), )) are closed. Thus by the topological definition of continuity,u(x) = (x) is continuous. Thus we have shown that has a continuous utility representation.

Let > 0. By definition of u, u(x) = (x) such that (x)e x. Since is homothetic, (x)e x.Using the definition of (x) from the proof of existence of u, this implies that (x) = (x). Using thedefinition of u again, this gives us u(x) = u(x), so u is homothetic.

Thus we have shown that is homothetic if and only if it has a utility representation that is homogeneousof degree 1.

6 Fall 2007, I.2 (Werner, Risk)

Definition. Random variable Z is more risky than random variable Y if E(Y) = E(Z) and E[v(Y)] E[v(Z)] for every continuous, nondecreasing and concave function v. In other words, every risk-averse agentprefers Y to Z.

Part (a)Note that

E(Y + Z) = 1E(Y + Z|Y = y1) + 2E(Y + Z|Y = y2) = 1y1 + 2y2 = E(Y).

Let v be concave. We can write E[v(Y + z)] as

E[v(Y + Z)] = 1E[v(Y + Z)|Y = y1] + 2E[v(Y + Z)|Y = y2].

By Jensens inequality,E[v(Y + Z)|Y = y1] v[E(Y + Z|Y = y1)]

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andE[v(Y + Z)|Y = y2] v[E(Y + Z|Y = y2)].

ThusE[v(Y + Z)] 1v[E(Y + Z|Y = y1)] + 2v[E(Y + Z|Y = y2)].

Since E[Z|Y = y1] = E[Z|Y y2] = 0, this implies that

E[v(Y + Z)] 1v(y1) + 2v(y2) = E[v(Y)].

Therefore Y + Z is more risky than Y.

Part (b)

Given the specification of the problem, E(Y + 2Z) = E(Y + Z) = E(Y). Note that Y + Z = 12Y +12

Y + 2Z.Again, let v be concave. Then

E[v(Y + Z)] = E

v

1

2Y +

1

2Y + 2Z

E12

v(Y) +1

2

v(Y + 2Z)=

1

2E[v(Y)] +

1

2E[v(Y + 2Z)]

1

2E[v(Y + Z)] +

1

2E[v(Y + 2Z)]

where the last line comes from the result in part (a). Rearranging, we see that

1

2E[v(Y + Z)]

1

2E[v(Y + 2Z)]

soE[v(Y + Z)] E[v(Y + 2Z)].

Therefore Y + 2Z is more risky that Y + Z.

7 Fall 2007, II.1 (B. Allen, FWT)

Part (a)

We need the following definitions before proceeding with the theorem:

Definition. An allocation is a vector x Rln.

Definition. Trader is budget set is Bi(p, ei) = {xi Xi : p xi p ei}.

Definition. Trader is demand is xi(p, ei) = {xi Bi(p, ei) : xi i xi xi Bi(p, ei)}.

Definition. An allocation x is feasible ifxi Xi for all i {1, . . . , n} andn

i=1 xi n

i=1 ei.

Definition. An allocation x is (strongly) Pareto optimal if it is feasible and there is no other feasibleallocation x such that xi xi for all i {1, . . . , n} and x

j xj for some j {1, . . . , n}.

Definition. A competitive equilibrium is an allocation x and a price vector p Rl+ such that:

(i) xi Xi for all i {1, . . . , n}.

(ii) xi xi(p, ei) for all i {1, . . . , n}.

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(iii) x is feasible.

Definition. Preference relation i is locally nonsatiated if for all xi and all > 0, xi such that xxxi <

and xi i xi.

Theorem (First welfare theorem). Consider a pure exchange economy with l goods and n traders indexedby i {1, . . . , n}, each of which having endowment ei Rl and preferences i which are complete and

continuous preorders on their respective consumption sets Xi. Let(x, p) be a competitive equilibrium inthis economy. Then if i are locally nonsatiated for all i, x is Pareto optimal.

Part (b)

Proof. Let (x, p) be a competitive equilibrium in this economy and suppose for contradiction that x is notPareto optimal. Then there exists another feasible allocation y Rln such that yi i xi for all i {1, . . . , n}and yj j xj for at least one j. Since x

j xj(p

, ej) it must be that yj B(p, ej), i.e., p yj > p ej.Now suppose for contradiction that for some i, p yi 0 y

i such that y

i yi <

and yi i yi. Pick small enough so that p yi p ej , we can see that ni=1 p yi > i=1 p ei. Since

p = 0, this implies that

ni=1 yi >

i=1 ei, so y is not feasible. This is a contradiction, so x is Paretooptimal.

Part (c)

(i) Convexity of preferences is not one of the assumptions of the theorem so this complication doesntaffect the proof. Thus the theorem still holds.

(ii) As long as the LNS assumption still holds, the answer for (i) applies here as well. Note that weakconvexity is compatible with thick indifference curves (an example of preferences that are not locallynonsatiated).

(iii) If preferences are nonsatiated but not locally nonsatiated we cannot prove that p yi p ei for all i.More specifically, given the supposition (for contradiction) that p yi < p ei, we may not be able tofind an small enough that there exists yi such that yi yi < , yi i yi, and p yi < p ei. Thusthe theorem may not hold with this complication.

(iv) ??? Not entirely sure ???

Part (d)

The FWT tells us that under fairly weak assumptions the set of competitive equilibrium allocations is asubset of the set of Pareto optimal allocations. In other words, competitive equilibrium is a mechanism forarriving at a Pareto optimal allocation. This implies that CE allocations are non-wasteful. However, itis important to note that the FWT does not say anything about the existence of competitive equilibria. Itis not logically inconsistent to say that the FWT holds when there are zero competitive equilibria. In fact,this is really a trival case of the theorem; if CE = , then it must be that CE P O since is a subset of

every set.

8 Fall 2007, II.2 (B. Allen, Sonnenscheins conjecture)

Part (a)

Theorem (Sonnenscheins conjecture). Define as

=

p :

pjpk

, j, k

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Let Z : Rn be a continuous function that is homogeneous of degree zero and satisfies Walras law.Then there is a pure exchange economy of n consumers whose aggregate demand is represented by Z on thedomain P.

Part (b)

(Z is a continuous function)Lemma. The budget set Bi(, ei) : R+ is a continuous (u.h.c. and l.h.c.) correspondence.

Proof. Fix ei R+. By observation, B(p, ei) is compact for all p , i.e., B(, ei) is compact-valued on .Let {pn} such that pn p and let {xn} X such that xn B(pn, ei), n. Since pn R++and pn p , > 0 such that pn,i , n. Then every B(pn, ei) contains only commodity bundles inwhich the quantity of each commodity is finite. Thus {xn} is bounded, so there exists a subsequence {xnk}such that xnk x. We know that xnk B(pnk , ei), nk. Then pnk xnk pnk ei. By continuity of thedot product, p x p ei. Then x B(p, ei). Therefore B(, ei) is u.h.c.

Again, let {pn} such that pn p . Let x B(p). We want to show that {xn} such thatxn x and xn B(pn, ei), n.

Case 1: Suppose p x 0 such that for all n N, pn (x ei) < 0. Then for n N, pn x < pn ei. Sincexn = x, n, we have pn xn < pn ei. Thus xn B(pn, ei), n N. Since xn x B(p, ei), B(, ei) isl.h.c. in this case.

Case 2: Suppose p x = p ei. Construct {xn} so that xn =pneipnx

x, n. Then

pn xn = pn

pn eipn x

x =

pn eipn x

pn x = pn ei.

Thus xn B(pn, ei), n. Since pn p and p x = p ei,

xn

p eip x

x = x.

Thus B(, ei) is l.h.c. in this case as well.Since B(, ei) is u.h.c. and l.h.c., B(, ei) is continuous.

Lemma. Let i be a complete, continuous and monotone preorder and let ei R+. Then xi(, ei) is anonempty, compact-valued and u.h.c. correspondence on .

Proof. Given the assumptions on preferences, for all i I, i has a continuous utility representation ui :X R. Then we can express demand xi(p, ei) as

xi(p, ei) = argmaxxB(p,ei)

ui(x).

Since ei R+, B(, ei) is nonempty and compact-valued on . By the previous lemma, B(, ei) is continuouson . Thus the theorem of the maximum implies that xi(, ei) is nonempty, compact-valued and u.h.c. on

.

Lemma. If i is strictly convex, then xi(p, ei) is single-valued.

Proof. Let x xi(p, ei) and suppose for contradiction that y xi(p, ei) such that x = y. Since x, y xi(p, ei), x i y. Let (0, 1). Since p x p ei and p y p ei,

p ei (p x) + (1 )p y = p [x + (1 )y].

So x + (1 )y B(p, ei). By strict convexity of i, x + (1 )y i x. This contradicts x xi(p, ei), soit must be that xi(p, ei) is single-valued.

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With these helpful results out of the way, we can move on to what we want to prove.

Proposition. Continuity of Z requires that be complete, continuous, monotone and strictly convex forall i I.

Proof. Given the assumptions on preferences, the lemmas above imply that each consumers demand xi(p, ei)is u.h.c. and single-valued. These two properties together are equivalent to continuity, so each consumers

demand is a continuous function. Since Z is just the sum of the consumers demand functions (minus aconstant for the endowments), Z is also a continous function.

(Z is homogeneous of degree zero)

Proposition. Z is homogeneous of degree zero

Proof. Consumer is budget set is

Bi(p, ei) = {x R+ : p x p ei}.

Since any x that satisfies the constraint p x p ei also satisfies (p) x (p) ei, Bi(p, ei) = Bi(p,ei).Thus every consumers budget set is homogeneous of degree zero in prices. This implies that each consumersdemand

xi(p, ei) = {x B(p, ei) : x i y, y B(p, ei)}is also homogeneous of degree zero in prices. Since Z is the sum of the consumers demands minus a constant,Z is also homogeneous of degree zero.

(Z satisfies Walras law)

Lemma. If a preorder is monotone then it is locally nonsatiated.

Proof. Let x X and let > 0. let e denote the unit vector in R. Let y = x + 2

e. Then y x. Since

is monotone, y x. By construction, y x = 2 < . Thus is locally nonsatiated.

Lemma. If i is locally nonsatiated then p x = p ei, p , x xi(p, ei).

Proof. Suppose not. Then p and x xi(p, ei) such that p x = p ei. By definition of xi(p, ei),

p x p ei, so it must be that p x 0 such that for all y B(x) (the open ball of radius around x), p y < p ei. Then for all y B(x), y B(p, ei). By local nonsatiation ofi, x B(x) suchthat x i x. This contradicts x xi(p, ei), so it must be that p x = p ei x xi(p, ei).

Proposition. The result follows from the two lemmas above.

Part (c)

to come later

9 Fall 2007, III.1 (Aldo, Nash equilibrium, dominance

Part (a)

I will state some definitions and prove some lemmas to do this part.

Definition. The vector = (1(ai)aiAi)iI) is a perturbation vector if i(ai) > 0, i I, ai Ai andaiAi

i(ai) < 1, i I.

Definition. Given a NFG G and a perturbation vector , the perturbed game G has the same playersutilities and action sets, but the players mixed strategies are constrained to the sets

Si = {si Si : si(ai) i(ai), ai Ai}.

The constrained set of mixed strategy profiles is S = iISi.

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Definition. The set of Nash equilibria in the perturbed game G is

N E(G) = {s S : ui(si, si) ui(ti, si), i I, ti Si}.

Definition. A mixed strategy profile s is a perfect equilibrium if there exists a sequence of perturbationvectors n such that i(ai) 0, i I, ai Ai, and there exists a sequence sn N E(Gn) such thatsn s. In other words, a perfect equilibrium is the limit of equilibria in some sequence of perturbed games.

Lemma. In a finite game, for every perturbation vector , N E(G) = .

Proof. Clearly, S is nonempty, compact and convex. The constrained best response correspondence is

BRi(s) = {si Si : u

i(si, si) ui(ti, si), ti Si}.

We just need to verify that BRi(s) is nonempty, closed, convex and u.h.c. for all s S. fill this in,same as for normal BR correspondence. Then BR = BRi also has those properties, so Kakutani impliesexistence of a fixed point. This is basically exactly the same as proving existence of a Nash equilibrium inan unconstrained game.

Lemma. In any finite game, the set of perfect equilibria is nonempty.

Proof. For any sequence of perturbation vectors n 0, we can use the previous theorem to get sn

N E(Gn). Then sn is a sequence in S. And since S is compact, there exists a convergent subsequencesnk s S. Then s is a perfect equilibrium by definition.

Lemma. A perfect equilibrium is a Nash equilibrium.

Proof. Let s be a perfect equilibrium. By definition, there exists a sequence of perturbation vectors n 0and a sequence sn N E(Gn) such that sn s. We want to show that s N E(G), i.e.,

ui(si, si) ui(ti, si), i I, ti Si.

We haveui(sin, s

in ) u

i(ti, sin ), i I, ti Sin.

For any ti Si we can find a sequence tin Sin

such that tin ti. Then we have

ui(sin, sin ) ui(tin, sin )

so the conclusion follows from continuity of ui.

So we now know that the set of perfect equilibria is nonempty, and that every perfect equilibrium is aHash equilibrium.

Proposition. In a perfect equilibrium, no players assigns a positive probability to a weakly dominated purestrategy.

Proof. Suppose for contradiction that there exists a perfect equilibrium s such that si(aik) > 0 for someweakly dominated strategy aik A

i. Since s is a perfect equilibrium, {n} such that

1. in(ai) > 0, i I, ai Ai, n,

2.

aiAi in(ai) < 1, i I, n,

3. in(ai) 0, i I, ai Ai,

and n, sn N E(Gn) such that sn s. Then s

in

(aik) si(aik) > 0. Since

in(a

ik) 0, s

in

(aik) > in(a

ik)

for n sufficiently large. Let n = sin(aik)

in(a

ik). Then n > 0 for n sufficiently large.

Let ti Si weakly dominate aik. Define the mixed strategy rin

by

rin(aik) =

in(a

ik) + nt

i(aik)

rin(aij) = s

in

(aij) + nti(aij), a

ij = a

ik

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Note that

aiAi rin

(ai) = 1. Then

ui(rin , sin

) ui(sn) =aA

P rsin(ai)rin(a

i)ui(a) aA

P rsin(ai)sin(a

i)ui(a)

= aA

P rsin(ai) rin(ai) sin(ai)ui(a)

=aA

P rsin

(ai)

aijAi,ai

j=ai

k

nti(aij , a

i)

+ (in(aik) + nti(aik) sin(aik))ui(aik, ai)

=aA

P rsin

(ai)

n

aijAi,ai

j=ai

k

ti(aij , ai)

+ n(ti(aik) 1)ui(aik, ai)

= naA

P rsin

(ai)

aijAi,ai

j=ai

k

ti(aij , ai)

+ (ti(aik) 1)ui(aik, ai)

= naA P rs

in (a

i

)

aiAi ti

(a

i

, ai

)

u

i

(a

i

k, ai

)

= naA

P rsin (ai)[ui(ti, ai) ui(aik, a

i)]

Since ti weakly dominates aik, ui(ti, ai) ui(aik, a

i) for all ai Ai, and bi Ai such thatui(ti, bi) > ui(aik, b

i). Since sn Sn, sn is completely mixed. Then P rsin (ai) > 0 for all ai Ai.

Since n > 0, this implies that ui(rin , sin

) > ui(sn). But this contradicts the fact that sn N E(Gn).

Therefore it must be that no player plays a weakly dominated strategy with positive probability.

Thus we have shown that the set of perfect equilibria is nonempty, a perfect equilibrium is a Nashequilibrium, and in every perfect equilibrium no player plays a weakly dominated strategy with positiveprobability. QED.

Part (b)

Lemma. The following three statements are equivalent:

(A) s is a perfect equilibrium (as in definition in part (a)).

(B) There exists a sequence of completely mixed strategy profiles sn s such that si is a best response tosin for all i I and all n sufficiently large.

(C) There exists a sequence n 0 such that n > 0 and a corresponding sequence of n-perfect strategiessn such that sn s.

Proof. We will show that A C B A. First, suppose that s is a perfect equilibrium. Then there exists

a sequence of perturbation vectors n 0 and a corresponding sequence of perturbed-game Nash equilibriasn N E(Gn) such that sn s. Since I and A are finite, we can define n = maxiI,aiAi{

in(a

i)}. Byconstruction, sin BR

in

(sn), so any pure strategy ai that is not a best response to sn will be played with

probability in(ai). Thus each sn is n-perfect, so A C.

Now suppose that t satisfies C. Then n 0 and a corresponding sequence tn ofn-perfect strategiessuch that tn t. Since I and A are finite, we can find d > 0 such that for all a

i support(ti), tin(ai) d.

So for n large enough, d > n, which implies that any ai support(ti) is a best response to tin . This in turn

implies that ti is a best response to tin . Thus C B.Finally, suppose r satisfies B. Then a sequence rn of completely mixed strategies such that rn r

and ri is a best response to rin for all i I and all n sufficiently large. We want to show that r is a perfect

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equilibrium, i.e., a sequence of perturbation vectors n 0 and a sequence rn N E(Gn) such that

rn r. Define n by

in(ai) =

1/n ai support(ri)

rin(bi) bi support(ri)

Then in(ai) > 0, i I, ai Ai, n, and for n sufficiently large, aiAi in(ai) < 1, i I, and most

importantly, in(a

i

) 0, i I, ai

Ai

. We then have rin(a

i

) ri

(ai

) > 0, ai

support(ri

) andrin(b

i) = in(bi), bi support(ri). By assumption, ri is a best response to rin for all i I and all n

sufficiently large. Then for all i I, rin BRin

(rin ), the n-constrained best response set. Then we haverin N E(G

n) for n sufficiently large. Therefore B A.

Proposition. If s is perfect, then s is undominated.

Proof. Suppose not. Then for some i I, ti Si such that si is weakly dominated by ti. Then

ui(si, ai) ui(si, ai), ai Ai

and bi Ai such thatui(si, bi) < ui(si, bi).

Since s is a perfect equilibrium, condition (B) in the lemma above implies that there exists a sequence sn of

fully mixed strategies such that sn s and for n sufficiently large, si BRi(sn) for all i I. In particular,this holds for the player i who plays a weakly dominated strategy. This means that for n sufficiently large,

ui(si, sin ) ui(ti, sin ), t

i Si.

Thus for n sufficiently large,ui(si, sin ) u

i(si, sin )

where si is the strategy that weakly dominates si. But since sn are fully mixed, P rsin (bi) > 0 for bi

defined above, so we should have

ui(si, sin ) =

aiAiP r

sin

(ai)ui(si, ai) 0 such that B1(y), the open ball of radius 1 aroundy, is a subset of P(x)c. Since yn y, N1 > 0 such that n N1, yn B1(y) P(x)

c. Then n N1,yn x. Similarly, C2 implies L(y) is closed, so

L(y)c = {x X : x y}

is open. Again, x L(y)c, so there exists an 2 > 0 such that B2(x) L(x)c. Since xn x, N2 > 0 suchthat n N2, xn B2(x) L(x)

c. Then n N2, xn y.Let N = max(N1, N2). Then we have shown that x yN and xN y. Then x L(yN)c and

y P(xN)c. Both of these sets are open. By an open ball argument similar to the one used above, there

exists an M N such that for all n M, xn L(yN)c and yn P(xN)c. In other words, xn yN andxN yn for all n M. Recall that by construction of{xn} and {yn}, yn xn n. Then yN xN. Butthen transitivity implies that xn yn which is a contradiction. Therefore it must be that x y, so C2implies C1.

11 Spring 2007, I.1 (Werner, Pratts theorem)

Definition. The risk compensation for a risky claim z such that E[z] = 0 and deterministic initial

consumption w is (w, z) that solves E[v(w + z)] = v(w (w, z)).

Definition. If v is twice-differentiable and strictly increasing, the Arrow-Pratt measure of risk aversion is

A(w) v(w)v(w) .

Theorem (Pratts theorem). Let v1, v2 be two C2, strictly increasing vN-M utility functions with 1, 2,and A1, A2 respectively. Then the following conditions are equivalent:

(i) A1(w) A2(w) for every w R.

(ii) 1(w, z) 2(w, z) for every w R and every risky plan z with E[z] = 0.

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(iii) v1 is a concave transformation of v2, i.e., v1(w) = f(v2(w)) for every w R, where f is concave andstrictly increasing.

Part (i)

Define the function v1 : R R by v1(w) = v(w + w) where w 0 for all w R. Then v1 is also a

strictly increasing, twice-differentiable vN-M utility function. Risk compensation 1(w, z) is defined by

E[v1(w + z)] = v1(w 1(w, z))

which is the same asE[v(w + w + z)] = v(w + w 1(w, z))

so 1(w, z) = (w + w, z). We also have

A1(w) = v1 (w)v1(w)

= v(w + wv(w + w

= A(w + w).

By the Pratts theorem, A1(w) A(w) for all w R if and only if 1(w, z) (w, z) for every w Rand every risky claim z with E[z] = 0. By the above results concerning A1(w) and 1(w, z), this implies

that A(w + w) A(w) for every w R if and only if (w + w, z) (w, z) for every w R and everyrisky claim z with E[z] = 0. Thus A is weakly decreasing in w if and only if is weakly decreasing in w forevery z with E[z] = 0.

Part (ii)

If we let w 0 instead of w 0 in the definition ofv1 in part (i), the same proof used above implies thatA is weakly increasing in w if and only if is weakly increasing in w for every z with E[z] = 0. Combinedwith the original result in part (i), this implies that A is constant for all w R if and only if (w, z) isindependent ofw for every z with E[z] = 0. Thus it suffices to show that v(w) = ew is, up to an increasinglinear transformation, the only strictly-increasing, twice-differentiable vN-M utility function for which A(w)is constant.

Let v be a strictly increasing, twice-differentiable vN-M utility function with A(w) = > 0 for all w R.

Then A(w) = v

(w)v(w) = . We can solve this differential equation:

v(w)v(w)

= d

dw[ln v(w)] =

ln v(w) = w + C1

v(w) = ew+C1

v(w) = 1

ew+C1 + C2

v(w) = 1

eC1ew + C2

So v is an increasing linear transformation ofex. Thus any strictly increasing, twice-differentiable vN-M

utility function with constant risk aversion must be an increasing linear transformation of ex

. Thiscompletes the proof.

12 Spring 2007, II.1 (B. Allen, Generic economies)

Part (a)

Sufficient conditions for Z to be a smooth function are: For all i I,

ui is C2

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ui is strictly differentiably monotone (Dui(x) 0, x)

ui is strictly differentiably concave (D2ui(x) is negative definite x)

ui satisfies the boundary condition

cl{y R++ : ui(y) ui(x)} R++ = , x.

In order to guarantee that each individuals prefefences are represented by a continuous utility function, werequire that each individuals preference relation be complete, continuous, and monotone.

What about endowments?

Part (b)

The generic approach uses techniques from analysis and differential topology to study the frequency of badlybehaved economies (those with unstable numbers of equilibria, a continuum of equilibria, etc.). It beginswith the assumption that Z is smooth (and thus implicity assumes the conditions on part (a)). We consideraggregate excess demand as a function of both prices and endowments. To avoid dimensionality issues, weconsider aggregate excess demand in the first 1 commodities. Let Z(p, e) denote this function. By Walraslaw, if Z(p, e) = 0, then aggregate excess demand in the last commodity is zero as well.

Let E P denote the set of pairs of prices and endowments. Note that E P has dimension n + 1.It can be shown that 0 is a regular value of Z. Define M = Z1({0}), the preimage of zero under Z, as theequilibrium manifold. Since 0 is a regular value of Z, the regular value theorem implies that M is a smoothmanifold of dimension m. Define the map T : M E as the pro jection ofM onto the space of endowments,i.e., T(p, e) = e. We will define a regular economy as a regular value of T. Similarly, a critical economyis a critical value of T. It is straightforward to show that T is smooth, and thus Sards theorem impliesthat the set of critical economies has Lebesgue measure zero. In other words, the set of regular economies isgeneric (this result is where the name of the generic approach comes from).

After showing that the set of regular economies is generic, we can use the inverse function theorem andthe implicit function theorem to show that a regular economy has a finite number of equilibria, and on aneighborhood of each equilibrium, the equilibrium price vector depends smoothly on the endowments. Thusregular economies are well-behaved, in that they do not have continuous sets of equilibria, nor is the number

of equilibria unstable. In other words, the set of badly-behaved economies has measure zero.The conclusions of the generic approach are important for several reasons. First, these results act as asort of rebuttal to Sonnenscheins conjecture, which states that any continuous function defined on a compactsubset of that is homogeneous of degree zero and satisfies Walras law represents aggregate excess demandfor some pure exchange economy. So while Sonnenscheins conjecture tells us that there exist economiesthat are extraordinarily badly behaved, the generic approach tells us that the set of such economies hasmeasure zero, i.e., virtually all economies are well-behaved. Second, the generic approach tells us that mosteconomies are stable in the sense that a small perturbation of the endowment vector will have little effecton the equilibrium price(s). Anything else?

Part (c)

Im not exactly sure what she means by the equilibrium price correspondence. Is it the projection of T1

onto the space of prices? If so, we could say that it is a local diffeomorphism around regular values, allowingus to use the inverse function theorem. . .

Part (d)

As stated in part (b), a generic economy (here an economy is defined by the initial endowment vector)has a finite number of equilibrium price vectors, and each one depends smoothly on the endowment vector.Again, the economic implications of this are that virtually all economies are well-behaved, so Sonnenscheinsconjecture has very little bite.

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Part (e)

If we just want Z to be continuous, we only require that individual preferences be complete, continuous,monotone, and strictly convex. To ensure that Z satisfies Walras law, we just need individual preferencesto be locally nonsatiated. Monotonicity takes care of this.

Part (f)If Z, and thus Z is just C0, the regular value theorem no longer applies, so we cannot be sure that M is asmooth manifold. This implies that we cannot be sure the the projection of M onto the space of prices issmooth either.Is there something else to this? This cant be all she wants.

Part (g)

In addition to the answer for (f), Sards theorem no longer applies, so we cannot be sure that the set ofwell-behaved economies is generic. In other words, we cannot say much about the properties of an arbitrarygeneric economy. Again, not sure what she wants here.

13 Spring 2007, II.2 (B. Allen, FWT/Continuum of agents)

Part (a)

Theorem (First welfare theorem). Let E = {(i, ei)iI} be a pure exchange economy with goods and ntraders indexed by I = {1, . . . , n}, each of which having endowment ei R+ and preferences i with arecomplete, continuous, and locally nonsatiatied preorders onR+. Then if(x

, p) is a competitive equilibriumin this economy, x is Pareto optimal.

Part (b)

Proof of FWT. Let (x, p) be a competitive equilibrium in this economy and suppose for contradiction thatx is not Pareto optimal. Then there exists another feasible allocation y Rn such that yi i xi for alli I and yj j xj for at least one j I. First, note that since x

is a competitive equilibrium, it must be

that xi xi(p, ei) (trader is demand correspondence at price p), defined asxi(p

, ei) = {xi B(p, ei) : xi i xi, xi B(p

, ei)}

where B(p, ei) is trader is budget set

B(p, ei) = {xi R+ : p xi p ei}.

Since xj xj(p, ej) (trader js demand correspondence at price p), it must be that yj B(p, ej)

(trader js budget set), i.e., p yj > p ej .Now suppose for contradiction that for some i I, i = j, that p yi 0, y

i such

that yi yi < and yi i yi. Since the dot product is a continuous function, we can pick small enough

so that p yi p ej, we have

iIp yi >

iI

p ei.

Since p = 0, this implies that iI

yi >iI

ei.

Thus y is not feasible. This is a contradiction, so it must be that x is Pareto optimal.

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Part (c)

A complete preference relation is one that can compare all possible commodity bundles. This propertyimplies that the agent is able to determine which of two commodity bundles he prefers, regardless of whatthose two bundles are.

A continuous preference relation is one that maintains its ranking between two commodity bundles whenthose bundles are slightly perturbed. This property implies that the benefit from each commodity does notchange drasticly when the amount of the commodity is changed slightly.

Lastly, local nonsatiation means that in each neighborhood of a commodity bundle, no matter how smallthe radius, there exists another commodity bundle that is strictly better. This property implies that theagent can always make himself better off given slightly more resources; when his choice set expands (nomatter how slightly) he will always find a newly affordable alternative that he strictly prefers to his choicegiven the smaller choice set.

Part (d)

Let the set of traders be I = [0, 1]. In this context, an allocation is a vector-valued (Lebesgue) integrablefunction mapping from I to R+. Then the initial endowments are now represented by a function e : I R

+,

i.e., e(i) is trader is endowment. For the sake of simple notation, I will write x to denote

x(i) di. Eachindividual traders preference relation is the same as in the discrete case. However, since we have a continuumof agents, I will let preferences be defined as a map from I to R, which I will use to denote the space ofcomplete, continuous preorders on R+. Then trader is preference relation is (i). To ensure measurabilitywhere we need it, we will require that for all allocations x, y, the set {i : x(i) (i) y(i)} I is measurable.Budget sets and demand correspondences are also the same as before: trader is budget set is

B(p, i) = {x R+ : p x p e(i)}

and his demand isx(p, i) = {x B(p, i) : x (i) y, y B(p, i)}.

Definition. A feasible allocation is an allocation x : I R+ such that

x =

e.

Definition. An allocation x : I R+ is Pareto optimal if another feasible allocation y : I R+ such

that: y(i) (i) x(i), i, and

S I such that y(j) (j) x(j), j S and S has positive measure.

Definition. A competitive equilibrium is an allocation/price vector pair (x, p) such that x is feasibleand x(i) x(p, i), i I.

Proposition. Suppose that (i) is monotone for all i I (in addition to being complete and continuousas given by the problem). If (x, p) is a competitive equilibrium, then x is Pareto optimal.

Sketch of proof. Suppose not. Then there exists a feasible allocation y such that y(i) (i) x(i), i andy(j) (j) x(j), j S, where S I has positive measure. By the exact same logic as in the proof ofthe FTW in the context of a discrete set of traders, p y(i) p e(i), i and p y(j) > p e(j), j S.

Then since all allocations are Lebesgue integrable and S has positive measure, we should be able to writep

y > p

e. This would imply that y is not feasible - a contradiction. Thus x should be Paretooptimal.

Note that if the set S had measure zero and p y(i) = p e(i), i I \ S, then p

y = p

e eventhough p y(j) > p e(j), j S. Thus we need S to have positive measure in order to show that y is notfeasible.

Part (e)

Dont know much about OLG models, at least in the GE context.

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14 Spring 2007, III.1 (Aldo, Extensive form games)

Part (a

An linear extensive form game is:

A set I = {1, . . . , n} of players.

Nature.

A set X of nodes.

A set Z X of final nodes.

An initial node o X.

Player utilities ui : Z R, i I.

A partial order over X where:

x y means x comes before y.

is reflexive, assymetric, and transitive.

For all x X, o x.

For all x X, the set of predecessor nodes p(x) {y : y x} is linearly ordered by .

For all x X, the set of immediate predecessor nodes is

P(x) {y p(x) : y s.th. y = y, y = x, z y x}.

For all x X, the set of immediate successor nodes is

S(x) {y X : y x, y s.th. y y x}.

Z = {x X : S(x) = }.

Player partition P = (P1, . . . , P n).

Information partitions Ui of Pi, i I.

Choice sets Cu, u Ui, i I.

such that #[p(z) u] 1 for all z Z, for all i I, and all i = u Ui.

Part (b)

Let z Z be a final node. Since the game is linear, for all i I and all u Ui, #(p(z) u) 1. Note thatp(z) is unique and thus there exists a unique set of choices in iI(uUiCu) for the players that leads toz. Let (si, si) represent the pure strategy profile that plays these choices.

Let bi Bi be a behavioral strategy for player i. Given a pure strategy s1 of player is opponents, the

probability induced on final node z Z is

P r(bi,si)(z) =uUi

biu(siu)

uUisiu (s

iu ).

where siu (c) = 1 ifsiu = c and s

iu (c) = 0 ifs

iu = c. Define the mixed strategy

i as

i(si) = i

(siu)uUi

=uUi

biu(siu), s

i Si

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where si Si is a pure strategy of player i and siu Cu is the choice associated with that pure strategy atinformation set u. Note that i(si) 0 by construction, and

siSii(si) =

siSi

uUi

biu(siu) = 1

why??? So i (Si), i.e., 1 is indeed a mixed strategy for player i.

The probability induced by (i, si) on z is

P r(bi,si)(z) = i(si)

uUi

siu (siu )

=uUi

biu(siu)

uUisiu (s

iu )

= P r(bi,si)(z)

So for any si Si, i induces the same probability on final nodes as bi. Thus i is equivalent to bi.

Part (c)

Consider the game in figure 2. There are 2 players plus nature, and player 1 has an information set u in

which he does not recall what happened previously. Let

1 = (p1(L, ), p2(L, r), p3(R, ), p4(R, r)) = (0, 1/2, 1/2, 0)

and let s2 = A. Then the probability distribution induced on final nodes by (1, s2) is

P r(1,s2)(z1, z2, z3, z4, z5, z6) = (1/4, 1/4, 0, 1/4, 1/4, 0).

Suppose there exists a behavioral strategy b1 that is equivalent to 1. Then b1 and 1 must induce thesame probability on all final nodes when the opponent plays s2. Since P r(1 ,s2)(z1) = 1/4, it must bethat b1(L) = 1/2, which means that b1(R) = 1/2 as well. Then since P r(1,s2)(z2) = 1/4, it must be thatb1() = 1, which means that b1(r) = 0. But then P r(b1,s2)(z5) = 0 which contradicts the fact that b

1 isequivalent to 1.

15 Fall 2006, I.1 (Werner, Profit function and supply correspon-dence)

Definition. The profit function is defined as

(p) = supyY

p y.

Definition. The supply correspondence s is

s(p) = {y Y : p y p y, y Y}.

Part (a)

I will make use of the theorem of the maximum so I will state it here:Theorem. Let X Rn and Y Rm. Let : Y X be a nonempty, compact-valued, continuouscorrespondence and let f : X Y R be a continuous function. Then

h(x) = maxy(x)

f(x, y)

is a continuous function andg(x) = argmax

y(x)f(x, y)

is a nonempty, compact-valued, u.h.c. correspondence.

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Nature

1

1/2

z1

L

1

R

z2

z3

r

2

1/2

1

A

z4

z5

r

z6

B

u

Figure 2: A linear game that is not of perfect recall

Note that in this case, Y is closed and bounded (i.e., compact). Then since the dot product is acontinuous function it must achieve a maximum on Y, so we can replace the sup with max in thedefinition of the profit function, i.e,

(p) = maxyY p y.

We can think of Y as a nonempty, compact-valued, continuous (since its constant) correspondence of p.Then by the theorem of the maximum, is a continuous function of p.

To see that is convex, let p, p Rn and let [0, 1]. Let p = p + (1 )p and let y s(p).

Then

(p) = p y= [p + (1 )p] y= [p y] + (1 )[p

y]

[maxyY

p y] + (1 )[maxyY

p y]

= (p) + (1 )(p)

Thus is convex.

Part (b)

By the theorem of the maximum (see part (a)), s is a u.h.c. correspondence. Since Y (the range of s)is compact, this implies that s has a closed graph. We can also prove the result directly. Let pn Rn

such that pn p Rn and let yn s(pn) such that yn y. We want to show that y s(p). Sinceyn s(pn), n,

pn yn pn y, y Y.

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Since pn p and yn y, continuity of the dot product implies that

limn

[pn yn] = [ limn

pn] [ limn

yn] = p y p y = [ limn

pn] y = limn

[pn y], y Y.

Then y s(p), so s has a closed graph.

Part (c)This is a direct result of the envelope theorem but I will prove it here. Suppose is differentiable at p.This implies that s(p) is a singleton. Let f(p, y) = p y. Then

(p) = maxyY

f(p, y) = f(p, s(p)).

By the chain rule,

D(p) = Dp(f(p, y(p))) = Dp(f(p, y))|y=s(p) + Dyf(p, s(p))Dps(p).

But Dy(f(y, p) = 0 is a FOC of the maximization problem, so we have

D(p) = Dp(f(p, y))|y=s(p).

By construction, Dpf(p, y) = y, soD(p) = s(p).

Part (d)

Assume and s are differentiable at p. By part (c), D(p) = s(p). Differentiating again, we have

D2(p) = Ds(p).

Since is a convex function, it must be that D2(p) is positive semi-definite, i.e., all leading principalminors ofD2(p) must be non-negative. Then all leading principal minors ofDs(p) must be non-negative.This implies that

si (p)pi

0.

16 Fall 2006, I.2 (Werner, Expected utility)

Definition. Utility function U has a state-separable representation if there exists functions vs : R R, s = 1, . . . , S such that

U(c) U(c) =S

s=1

sv(cs) S

s=1

svs(cs).

Definition. Utility function U has a concave expected utility representation if there exists a concavefunction v : R R such that

U(c) U(c) E[v(c)] E[v(c)].

Part (a)

If S 3, U has a state-separable representation if and only if it satisfies the sure-thing axiom, i.e.,

U(cs, x) U(ds, x) U(cs, y) U(ds, y)

for all c, d RS, all x, y R, and all s S, where cs denote the consumption plan c with cs replaced byx. See http://econ.ucsb.edu/~mkapicka/RG/Werner-Axiomatization.pdf.

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http://econ.ucsb.edu/~mkapicka/RG/Werner-Axiomatization.pdfhttp://econ.ucsb.edu/~mkapicka/RG/Werner-Axiomatization.pdfhttp://econ.ucsb.edu/~mkapicka/RG/Werner-Axiomatization.pdfhttp://econ.ucsb.edu/~mkapicka/RG/Werner-Axiomatization.pdf

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Part (b)

Suppose U has a state-separable representation and is risk averse with respect to . Consider...Finish this!

17 Fall 2006, II.1 (B. Allen, Nonconvexities)

Part (a)

Definition. Trader is budget set is

B(p, ei) = {x Xi : p x p ei}.

Definition. Trader is demand is

xi(p, ei) = {x B(p, ei) : x i y, y B(p, ei)}.

Definition. A competitive equilibrium in this economy is an allocation and price vector pair (x, p)such that p 0,

iIxi

iIei, and xi xi(p

, ei), i.

Part (b)

Definition. Trader is preference relation i is locally nonsatiated if for all x Xi and all > 0, y Xisuch that y = x, y x < , and y i x.

Part (c)

(i) Some (or all) traders preference relations can be nonconvex. This can be interpreted as. . .

(ii) Some (or all) traders consumption sets can be nonconvex. The economic interpretation for this wouldbe that some goods are indivisible, e.g., cars, houses, etc.

(iii) Anything else?

Part (d)

(i) TRUE: Nonconvexity of preferences can cause a discontinuity that may prevent existence of competitiveequilibrium.

Nonconvex preferences can cause individual demand (and thus aggregate excess demand) to be acorrespondence with non-convex values rather than a continuous (single-valued) function. In this case,equilibrium may fail to exist.

Note that nonconvex preferences cause the assumptions of all 4 of the existence theorems (very easy,extended very easy, easy, and extended easy) we covered in class to fail. The very easy and easyexistence theorems require that aggregate excess demand be a continuous function (which requires strictconvexity of all consumers preference relations). The extended VEET and extended EET require thataggregate excess demand be a convex-valued correspondence (which requires all consumers preferencesto be weakly convex).

The following example illustrates these concepts.

Example. Consider a pure exchange economy with two commodities, x and y, with prices p and 1 prespectively. Suppose there are two consumers, each with endowment (1 , 1) and preferences representedby the following utility functions:

u1(x, y) = min(x, y)

andu2(x, y) = x

2 + y2.

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Note that this implies that consumer 2s preference relation is nonconvex.

Consumer 1s budget set is

B1(p, e1) = {x R2+ : px + (1 p)y 1}.

Her demand is

x1(p, e1) = argmax(x,y)B1(p,e1) min(x, y) = (1, 1).

Consumer 2s budget set is

B2(p, e2) = {x R2+ : px + (1 p)y 1}

and her demand is

x2(p, e2) = argmax(x,y)B2(p,e2)

x2 + y2

=

(1/p, 0) p < 1/2

{(2, 0), (0, 2)} p = 1/2

(0, 1/(1 p)) p > 1/2

Note that consumer 2s demand is a correspondence and x2(1/2, e2) is nonconvex.

Looking at x1(p, e1) and x2(p, e2), we can see this economy has no competitive equilibrium. If p < 1/2,

then x1(p, e1) = (1, 1) and x2(p, e1) = (1/p, 0). Thus aggregate demand for commodity x is 1+1/p > 2,

and aggregate demand for commodity y is 1 < 2. Thus Z(p) = 0, i.e., p < 1/2 is not a CE price vector.

Similarly, if p > 1/2, then x1(p, e1) = (1, 1) and x2(p, e2) = (0, 1/(1 p)). Then aggregate demand for

x is 1 < 2 and aggregate demand for y is 1 + 1/(1 p) > 2, so p > 1/2 is not a CE price vector.

Finally, if p = 1/2, then x1(p, e1) = (1, 1) and x2(p, e2) = {(2, 0), (0, 2)}. Then for both elements of

x2(p, e2), aggregate excess demand will be nonzero since aggregate demand for one commodity will begreater than 3 and aggregate demand for the other commodity will be less than 1. Thus p = 1/2 isnot a CE price vector. So for all p [0, 1], p is not a CE price vector, i.e., the economy has no CE.

(ii) TRUE: Nonconvexity of consumption sets can cause discontinuities that prevents existence of a compet-itive equilibrium. In particular, nonconvex consumption sets lead to a discontinuous individual demandfunction or nonconvex demand correspondence, depending on whether or not demand is single-valued.These properties carry over to aggregate excess demand, which causes all four of the usual theoremsof existence of equilibrium to fail. For the VEET or the EET, discontinuous aggregate excess demandprevents the use of Brouwers fixed point theorem. For the extended versions of those theorems, non-convexity of the aggregate excess demand correspondence prevents the application of Kakutanis fixedpoint theorem. The next example illustrates this problem.

Example. Consider a pure exchange economy with two commodities and two consumers, each ofwhich with consumption set Xi = Z2+ (both commodities must be consumed in non-negative, integerquantities). Let both consumers have endowment og (1, 1). Let both consumers have strictly convexpreferences represented by separable utility functions of the form

ui(x, y) = vi(x) + wi(y)

where vi and wi are strictly convex. In particular, suppose that vi(0) = wi(0), i = 1, 2,

0 < wi(x/2) < vi(x) < wi(x), x > 0.

If we normalize prices as usual, both consumers have the budget set

Bi(p, ei) = {(x, y) Z2+ : px + (1 p)y px}.

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For p < 1/2, both consumers will demand 2 units ofx since neither can afford 2 units ofy and they bothprefer 2 units ofx to 1 unit ofy. For p 1/2, both consumers will demand 2 units of y since they prefery to x for equal amounts of both commodities. Thus both consumers have the same discontinuousdemand function

xi (p, ei) =

(2, 0) p < 1/2

(0, 2) p 1/2

So for p < 1/2, aggregate excess demand for x is 2 and aggregate excess demand for y is 2. Forp 1/2, aggregate excess demand for x is 2 and aggregate excess demand for y is 2. Thus there isno competitive equilibrium.

Part (e)

(i) The problems caused by nonconvex preferences for existence of competitive equilibria do in fact dis-appear when the economy is large. To see this, first consider the example in (d)(i), and supposethat the economy now has two identical consumers of type 2 plus one consumer of type 1 (whichmeans that aggregate endowment of each commodity is 3). At p = 1/2, x1(p, e1) = ( 1, 1) andx2(p, e2) = {(2, 0), (0, 2)}. Then if we give one consumer of type 2 the allocation (2, 0) and we give theother (0, 2), then aggregate demand for each commodity is 1 + 2 = 3. Thus aggregate excess demand

for both commodities is zero, so p = 1/2 is a CE price vector in this modified economy.More generally, suppose we have an exchange economy with I types of consumers. Suppose type ihas nonconvex preferences. Then excess demand for this type zi(p) is a correspondence. In a replicaeconomy with R consumers of each type, the average excess demand correspondence for type i is

zRi (p) =1

F

Rr=1

zi(p) =1

R{zi1 + . . . + ziR : zir zi(p), r = 1, . . . , R}.

Consider the economy of figure 3 below and let p = p (the price vector shown in the figure). We cansee that as R , zR1 (p + e1 fills the entire segment between the points x1 and x

1. In particular, for

any [0, 1] and any R, we can find aR [0, R] such that

aRR

1

R.

By putting aR consumers x1 and R aR consumers at x1, we can get average consumption for type 1of

xR1 =aRR

x1 +

1 aRR

x1.

By making R large enough we can get arbitrarily close to x1 + (1 )x1. We can see that forany p 0, as R , zRi (p) converges to the convex hull of zi(p). Therefore at the limit, theexcess aggregate demand correspondence will be convex-valued and we will be able to apply one of theextended existence theorems to get existence of an equilibrium. In this sense, as R , the economymust have an allocation and price vector the constitute an approximate equilibrium.

If we go even further and allow for a continuum of traders rather than a countable set, existence ofcompetitive equilibrium can be proven without any assumption about convexity of preferences. Theintuition for this is more or less the same as the above argument, taken to its logical conclusion whenthe set of traders gets even larger than a countably infinite set. See Aumann (19660) for the proof.

(ii) Nonconvexity of consumption sets can cause discontinuities that prevent existence of equilibrium re-gardless of the size of the economy. Looking at the example in (d)(ii), we can see that regardless ofthe number of agents of each type, there will never be an equilibrium.

18 Fall 2006, II.1 (B. Allen, Generic economies)

Part (a)

See the answer for Spring 2007, II.1, part (b).

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O1

O2

e

x1

x1

2

1

Figure 3: An economy with nonconvex preferences

Part (b)

(i)

Definition. A continuous function ui : R++ R is concave if for every x, y R++ and every

[0, 1],ui(x + (1 )y) ui(x) + (1 )ui(y).

(ii) For i to be represented by a continuous utility function, it is necessary that i be complete, transitive,continuous, and monotone. If i has these properties, we can find a continuous utility representationas follows.

For every x R++, let A+ = { R+ : e i x} and let A = { R+ : e i x} (where

e = (1, 1, . . . , 1) R). Completeness of i implies that A+ A = R+. Continuity of i impliesthat A+ and A are closed. Monotonicity of i implies that both sets are nonempty: x e A

and x e A+. Connectedness ofR+ implies that A+ A = . Denote A+ A as ui(x).Transitivity of i ensures that ui(x) is unique, i.e., ui is a function. Finally, continuity ofi implies

that ui is continuous by using a topological continuity argument (continuity of ensures that thepreimage sets u1([0, ui(x)]) and u1([ui(x), ]) are closed for all x R++).

Note that it is possible for a discontinuous utility function to represent a continuous preference relation.However, continuity of i guarantees existence of a continuous utility representation.

For ui to be concave, we require that. . .

(iii)

Definition. Preference relation i is weakly convex if for all x, y R++ and all [0, 1],

y i x y + (1 )x i x.

Definition. Preference relation i is convex if for all x, y R++ and all (0, 1),

y i x y + (1 )x i x.

Definition. Preference relation i is strictly convex if for all x, y R++ and all (0, 1),

(y i x AND y = x) y + (1 )x i x.

(iv) Finish this!

(v) Finish this!

(vi) Finish this!

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19 Fall 2006, III.1 (Aldo, Perfect equilibrium)

Definition. Let G = (I, (Ai, ui)iI) be a normal form game. A mixed strategy profile s is a perfectequilibrium if there exists a sequence of perturbation vectors n such that ein(a

i) 0, i I, ai Ai,and there exists a sequence sn N E(Gn) such that sn s. In other words, a perfect equilibrium is thelimit of equilibria in a sequence of perturbed games Gn that converges to the game G.

Part (a)

Let s be a perfect equilibrium. Then there exists a sequence of perturbation vectors n such that in(a

i) 0, i I, ai Ai and a sequence sn N E(Gn) such that sn s. We want to show that s N E(G),i.e.,


Since sn N E(Gn), n, we have

ui(sin, sin ) u

i(ti, sin ), i I, ti Sin

whereSin = {s

i Si : si(ai) in(ai), ai Ai}, i I.

Take i I and ti Si. We can find a sequence tin Sin

such that tin ti. Then we have

ui(sin, sin ) u

i(tin, sin ).

Since ui is continuous, sn s, and tin t

i, we have

ui(si, si) ui(ti, si).

Since i I and ti Si were picked arbitrarily, we have shown that


Thus s N E(G).

Part (b)

Let sn be a sequence of perfect equilibria such that sn s S. We want to show that s is also a perfectequilibrium. Since sn is a perfect equilibrium for all n, there are sequences of perturbation vectors kn suchthat kn 0 and sequences skn N E(G

kn ) of perturbed game equilibria such that skn sn, n. Take > 0. Then for each n we can find sKn such that sKn sn < /2. Since sn s, N > 0 such that forn N, sn s < /2. Then the triangle inequality implies that for n N,

sKn s sKn sn + sn s < .

Then sKn s. Since sKn is a sequence of pertubed game equilibria converging to s, we have shown that sis a perfect equilibrium. Therefore the set of perfect equilibria is closed.

20 Fall 2005, I.1 (Werner, Risk)

Definition. Random variable Z is more risky than random variable X if E(X) = E(Z) and X second-order stochastically dominates Z. Note that we have two equivalent definitions of second-order stochasticdominance:

1. X SSD Z ifwa

FX(t) dt wa

FZ(t) dt for all w [a, b].

2. X SSD Y is E[v(X)] E[v(Y)] for every continuous, nondecreasing, concave function v.

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Part (a)

Let Z be a random variable with E(Z) = 0 and let X be a random variable such that X = w Z for somedeterministic real number w. Then X + Z = w Z + Z = w, i.e., X + Z is deterministic. Then for anyw < w,

w

a

FX+Z(t)dt = 0 < w

a

FX(t)dt.

Then Y does not second-order stochastically dominate X+ Z, so X+ Z is not more risky than X.Alternative solution: Suppose there are two states, s = 1, 2, with respective probabilities 1 = 1/4 and

2 = 3/4. Define the random variables (here we will think of them as risky consumption plans) X and Z as

X =

3 if s = 1

1 if s = 2Z =

3 if s = 1

1 if s = 2

In other words, Z = X. Note that E(X) = E(Z) = 0, so E(X + Z) = 0. Then E(X) = E(X + Z). Alsonote that X+ Z = (0, 0), i.e., X+ Z gives 0 deterministically. Let v be a continuous, nondecreasing, strictlyconcave function. Then

E[v(X)] = (1/4)v(3) + (3/4)v(1) < v(0) = E[v(X+ Z)].

Then X does not second-order stochastically dominate X+ Z, so X+ Z is not more risky than X.

Part (b)

Proposition. If X and Z are independently distributed random variables and E(Z) = 0, then X + Z ismore risky than Z.

Part (c)

Proof. Let X and Z be independently distributed random variables such that E(Z) = 0. Then E(X) =E(X+ Z). Let v be a continuous, nondecreasing, concave function. Then

E[v(X+ Z)] =X

Zv(x + z) dF

X(x) dFZ(z) (since X and Z are independently distributed)

=

X

Z

v(x + z) dFZ(z)

dFX(x)

X

v

Z

(x + z) dFZ(z)

dFX(x) (Jensens inequality since v is concave)

=

X

v (x + E(Z)) dFX(x)

=

X

v(x)dFX (x) (since E(Z) = 0)

= E[v(X)]

Thus E[v(X+ Z)] E[v(X)], so X second-order stochastically dominates X+ Z. Therefore X+ Z is morerisky than X.

Fall 2005, I.1 (Werner, utility representations)

Definition. Preference relation is continuous if it satisfies the following equivalent conditions:

(i) For all convergent sequences {xn}, {yn}, xn yn, n lim xn lim yn.

(ii) For all x Rn+, the sets {y Rn+ : y x} and {z R

n+ : x x} are closed.

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Definition. Preference relation is strictly increasing (strongly monotone) if for all x, y Rn+ suchthat x y and x = y, x y.

Definition. Function u is a utility representation for preference relation if for all x, y Rn+,

u(x) u(y) x y.

Part (a)

Let x Rn+ and define the sets A+ and A as

A+ = {u R+ : ue x}

andA = {u R+ : x ue}

where e = (1, 1, . . . , 1) Rn+. Since x R+, x 0e = (0, 0, . . . , 0). By strict monotonicity, x 0e.Then 0e A. Similarly, take M = max(x1, . . . , xn). Then Me x, so by strict monotonicity, M e x.Then M e A+. Thus A+ and A are both nonempty. By continuity of , A+ and A are closed. Bycompleteness of , A+ A = R+. Since R+ is connected, it cannot be the union of two nonempty, closed,

disjoint sets. Thus A+

A = . Then u(x) A+

A, i.e., u(x)e x.Suppose that u(x) is not unique. Then v(x) R+ such that v(x) = u(x) and v(x)e x. Sinceu(x)e x and v(x)e x, transitivity of implies that u(x)e v(x)e. But since v(x) = u(x), eitheru(x) > v(x) or u(x) < v(x). Ifu(x) > v(x), then u(x)e v(x)e, so strict monotonicity of implies thatu(x)e v(x)e. This is a contradiction. Similarly, if u(x) < v(x), then by the same reasoning we haveu(x)e v(x)e which is also a contradiction. Therefore it must be that u(x) is the unique element of R+such that u(x)e x. This implies that u(x) is a well-defined function on Rn+.

To see that u(x) is a utility representation of , take x, y Rn+. Suppose that x y. Then u(x)e x y u(y)e, so transitivity of implies that u(x)e u(y)e. By strict monotonicity of , u(x) u(y). Nowsuppose that u(x) u(y). Then by strict monotonicity, u(x)e u(y)e. Then we have x u(x)e u(y)e y,so transitivity of implies that x y. Therefore x y u(x) u(y), so u is a utility representationof .

Part (b)Let n = 2 and define the preference relation as

x y x1 = y1

andx y x1 < y1.

Take two sequences {xn} and {yn} such that xn x, yn y, and xn yn, n. Then xn1 yn1, n. Sincexn x and yn y, xn1 x1 and yn1 yn. This implies that x1 y1, so x y. Then is continuous.However, take x = (2, 2) and y = (1, 1). Clearly, x > y but x y. Then is not (strictly) monotone.

Let x = (1, 1) and let y = (2, 2). Then x y. Note that e = (1, 1), the unit vector in R2+. Then1e = (1, 1) x u(x) = 1 and 2e = (2, 2) y u(y) = 2. But then we have x y and u(x) < u(y), so u

is not a utility representation of .

Part (c)

Let n = 2 and let be the lexicographic preference relation on R2+ defined as

x y x = y

andx y x1 > y1 OR (x1 = y1 AND x2 > y2).

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Note that for all n > 0, xn = (1 1n

, 1) (1, 0) = y but x = (1, 1) (1, 0) = y. Therefore is notcontinuous.

Suppose that has a utility representation u. Then for each x R+, we can find a rational numberr(x) such that u(x, 2) > r(x) > u(x, 1). Let x, x R+ such that x > x. Then

u(x, 2) > r(x) > u(x, 1) > u(x, 2) > r(x) > u(x, 1)

so r(x) > r(x). But this implies that r : Q+ R+ is one-to-one. This is a contradiction since Q+is countable while R+ is uncountable. Therefore has no utility representation. This implies that u(x)specified in the problem cannot be a utility representation for .

21 Fall 2005, III.1 (Aldo, Nash equilibrium)

Part (a)

Note that for a correspondence f : S S with compact range and closed values, the closed graph conditionis equivalent to upper hemicontinuity.

Theorem (Kakutanis fixed point theorem). Suppose S Rn is nonempty, compact and convex, and the

correspondencef : S S is nonempty-valued, closed-valued, convex-valued and upper hemicontinuous. Thenf has a fixed point in S.

Part (b)

Definition (Nash equilibrium). Let Si denote the set of player is mixed strategies, i.e., Si = (Ai). ANash equilibrium is a mixed strategy profile s S = iISi such that si is a best response to si for alli I, i.e.,


Part (c)

First, note that by definition, Si = (Ai), so Si is clearly nonempty, compact and convex. Define player

is best response correspondence BRi

as

BRi(s) = {ti Si : ui(ti, si) ui(ri, si), ri Si}.

Note that BRi : S Si. The first step of the proof is to show that BRi is nonempty-valued, closed-valued, compact-valued and u.h.c.

To see that BRi is nonempty-valued, note that we can write it as

BRi(s) = argmaxtiSi

ui(ti, si).

Since Si is compact and ui is continuous, BRi(s) = , i.e., BRi is nonempty-valued.To see that it is closed-valued, let tin BR

i(s) such that tin ti Si. We want to show that

ti BRi(s). Since tin BRi(s),

ui(tin, si) ui(ri, si), ri Si, n.

Continuity of ui implies thatui(ti, si) ui(ri, si), ri Si.

Then ti BRi(s), so BRi is closed-valued.To see that it is convex-valued, let ti, ri BRi(s). Then it must be that

ui(ti, si) = ui(ri, si) ui(qi, si), qi Si.

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By definition of a mixed strategy, we can write this asaiAi

ti(ai)ui(ai, si) =aiAi

ri(ai)ui(ai, si)

aiAiqi(ai)ui(ai, si), qi Si.

Let [0, 1]. ThenaiAi

ti(ai)ui(ai, si) +

aiAi(1 )ri(ai)ui(ai, si)

aiAi

qi(ai)ui(ai, si), qi Si

or simply aiAi

[ti(ai) + (1 )ri(ai)]ui(ai, si) aiAi

qi(ai)ui(ai, si), qi Si

Thenui(ti + (1 )ri, si) ui(qi, si), qi Si.

This implies that ti + (1 )ri BRi(s), so BRi is convex-valued.Finally, to see that BRi is u.h.c., note that since BRi is closed-valued and has compact range (S), it

suffices to show that BRi has a closed graph. Let sn S such that sjn s S. Let tin Si such that

tin BRi(sn), n and t

in t

i Si. We just need to show that ti BRi(s). Since tin BRi(s), n,

ui(tin, sin ) u

i(ri, sin ), ri Si, n.

By continuity of ui, we have

ui(ti, si) = ui

limn

tin, limn

sin

= limn

[ui(tin, sin )] lim

n[ui(ri, sin )] = u

i

ri, limn

sin

= ui(ri, si), ri Si

Then ti BRi(s), so BRi has a closed graph and is therefore u.h.c.Define the correspondence BR : S S as

BR(s) = iIBRi(s).

Since BRi is nonempty-valued, closed-valued, convex-valued and u.h.c. for all i I, BR also has

those properties since it is just the Cartesian product of the #I correspondences BRi. Since S is nonempty,compact and convex, S and BR satisfy the assumptions of Kakutanis fixed point theorem. Therefore BRhas a fixed point s S. By definition of BR,


Thus s is a Nash equilibrium. Since s exists, G has a Nash equilibrium.

micro prelim solutions

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