Advanced Microeconomics Lecture NotesAnders Munk-Nielsen

Fall, 2012

Disclaimer. These notes are my own personal notes from taking this course at the University ofCopenhagen in the fall of 2012. The lecturer is in no way responsible or related to this materialand I am merely putting these notes out to help fellow students of economics.

1

Contents1 2012-09-04 First lecture 5

1.1 Preparations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Lecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Plan (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.2 Commodity space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.3 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.4 Production (possibility) sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Assumptions on the producer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.1 Assumption P1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.3 More assumptions (P2 among other things) . . . . . . . . . . . . . . . . . . . . . 8

2 2012-09-05 82.1 General stuff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 More on the assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 The aggregate production set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Efficient productions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.2 Producer’s problem (PP) (MWG: Producer’s maximization problem) . . . . . . . 112.2.3 Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 2012-09-11 113.1 FB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Thm [4.2].1 — red proof (p. 29 NotesThPr) . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.3.1 Proof of (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3.2 Proof of (b) (on my own!!) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3.3 Proof of (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3.4 Proof of (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.4 Subspaces and linear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.5 A Decentralization Theorem (thm. 7.1, p 35 NotesThPr.) . . . . . . . . . . . . . . . . . 15

3.5.1 Proof of part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5.2 Proof of part (a) — LEFT FOR US TO DO BY OURSELVES . . . . . . . . . . 16

3.6 Representing production sets with production functions . . . . . . . . . . . . . . . . . . 16

4 2012-09-12 174.1 Representing production sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Thm. [8.2.].1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.2.1 Law of supply . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 Moving on to consumers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.4.1 F1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5 2012-09-18 205.1 Last time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.2 Preferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

5.2.1 Proof of continuity — sketch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2.2 Monotonicity assumptions F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2.3 Convex preferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.2.4 Strictly convex preferences (F3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.3 Towards calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.3.1 Assumption of differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2

5.3.2 Assumption F4 — bordered Hessian . . . . . . . . . . . . . . . . . . . . . . . . . 225.4 Consumer problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.4.1 Proposition 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6 2012-09-19 236.1 Stuff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6.1.1 Recapping the proof of proposition 1 . . . . . . . . . . . . . . . . . . . . . . . . . 246.2 Proposition 2 — solution to the CP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

6.2.1 Marshallian Demand Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.2.2 UMP and Expenditure Minimization . . . . . . . . . . . . . . . . . . . . . . . . . 276.2.3 Walrasian Demand Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7 2012-09-25 277.1 Last time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.2 Concluding the study of the consumer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

7.2.1 Prop 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.2.2 Proving proposition 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.2.3 Proposition 6 — differentiable demand . . . . . . . . . . . . . . . . . . . . . . . . 30

7.3 Economies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.3.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.3.2 Example: Edgeworth box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.3.3 Pareto optimal allocations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.3.4 Proposition concerning Assumption F2’ (variant of F2) . . . . . . . . . . . . . . 31

8 2012-09-26 318.1 Last time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

8.1.1 Proposition 5 — characterizing the solution . . . . . . . . . . . . . . . . . . . . . 318.1.2 Economies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

8.2 Economies cont’d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328.2.1 Koopman diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328.2.2 Assumption F2’ — Strong monotonicity. . . . . . . . . . . . . . . . . . . . . . . 328.2.3 Proposition [2.3.2].1 in NotesOpt. . . . . . . . . . . . . . . . . . . . . . . . . . . 328.2.4 Proposition [2.3.2].2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328.2.5 Section 2.3.3: Fairness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

8.3 Section 3.1: Social utility functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.4 Arrow’s impossibility theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

8.4.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.4.2 Assumptions on the Social Welfare Function (SWF) . . . . . . . . . . . . . . . . 338.4.3 Arrow’s Impossibility Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

8.5 Market Equilibrium (in MWG: Equilibrium with Transfers) . . . . . . . . . . . . . . . . 348.5.1 Theorem [4.2].1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

9 2012-10-02 359.1 Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

9.1.1 Proposition 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359.1.2 Proposition 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

9.2 Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.3 The first theorem of welfare economics. . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

9.3.1 Theorem [4.2].1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.4 The Second Theorem of Welfare Economics: Differentiable Approach . . . . . . . . . . . 37

9.4.1 Lemma [5.2].1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379.4.2 Theorem [5.2].1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3

9.4.3 Holy fuck in the structure... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389.4.4 Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

9.5 Sum of two “no-worse-than” sets — The Scitovsky contour . . . . . . . . . . . . . . . . 39

10 2012-10-03 4010.1 Towards the theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

10.1.1 Mathematical background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4010.2 Separating Hyperplane Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

10.2.1 Extension — boundary piont. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210.2.2 Addendum — Difference sets and separating hyperplanes . . . . . . . . . . . . . 43

10.3 The Second Theorem of Welfare Economics . . . . . . . . . . . . . . . . . . . . . . . . . 4310.3.1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.3.2 Lemma required to be proven first . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.3.3 Proof of the Second Welfare Theorem of Economics . . . . . . . . . . . . . . . . 44

11 2012-10-10 4511.1 Last time (I wasn’t there!!) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4511.2 Long proof — Theorem 10 — Existence of Walras EQ . . . . . . . . . . . . . . . . . . . 46

11.2.1 Theorem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4611.2.2 Extra stuff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

11.3 Uniqueness of Walras Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4911.3.1 Intro . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4911.3.2 New assumption — Gross Substitutes . . . . . . . . . . . . . . . . . . . . . . . . 5011.3.3 Theorem 17 — Uniqueness of W EQ . . . . . . . . . . . . . . . . . . . . . . . . . 50

12 2012-10-23 — Uncertainty and time 5012.1 Multi-period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

12.1.1 Proposition 2.3.A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5112.2 Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

13 2012-10-24 — Uncertainty cont’d 5313.1 Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5313.2 Farkas’ Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5513.3 Existence of Discount Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

13.3.1 Corollary — (V, q) arbitrage free . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4

1 2012-09-04 First lecture1.1 PreparationsQ. How do MWG use inequalities on vectors? Does y ≥ 0 mean y1 ≥ 0 ∧ ... ∧ yL ≥ 0?

Q. Set operations. Is A− B = A \ B?

A. No!Answer. Take A,B ⊂ RL. Then

A+B = x ∈ RL|x = a+ b, a ∈ A, b ∈ B.

”all vectors that can be created as the sum of a vector from A and one from B”. Similarly,

A−B = x ∈ RL|x = a− b, a ∈ A, b ∈ B.

1.2 Lecture1.2.1 Plan (?)

Production. PMP (Profit Maximization Problem)

Consumers. UMP (Utility Maximization Problem)

Also. Indirect utility functions and expenditure function.Interesting relationships (duality?)

Welfare thms.

Existence of Walras equilibria. And uniqueness and stability.

Time and uncertainty.

Why?

Production. We start with production because he finds that simpler...

1.2.2 Commodity space

(we’ll forget this after today (??))

Space. The set of commodities is L = 1, ..., L.

Commodity concept. We consider a finite number of (pysical characteristics, delivery times, deliverylocations, ...).

1.2.3 Production

Production. ConsiderF : Rq+ → R,

where

Rk+ = x ∈ Rk|xi ≥ 0, i = 1, ..., kRk++ = x ∈ Rk|xi > 0, i = 1, ..., k.

5

Normally. Normally, we think of a production function as a function that produces output Z as

Z = F (K,L).

This has three important characteristics,

1. Single output (Z ∈ R).(a) Rarely true (e.g. different vintages of wine from the same wineyard)

2. Know inputs and output.(a) A farmer may buy or sell young calves depending on prices (is it more profitable to sell

them quickly (meet very pricy) or to get the milk (milk pricy)).3. “Technological maximization” is implicit — we think of Z as the highest possible output...

i.e. on the production possibility frontier.4. Positive marginal product but diminishing returns

(a) ∂F∂K (K,L) ≥ 0 (positive MP) but ∂2F

∂K2 (K,L) < 0.(b) Turns out this will be captured by (something something convexity??) of the production

possibilities set...5. Non-increasing returns to scale.

(a) λ ∈ [0; 1], then λF (K,L) ≤ F (λK, λL).

1.2.4 Production (possibility) sets

Example. Suppose we have hay, livestock and fertilizer.

Input, a Output, b Net Output, b− aHay 10 0 -10

Livestock 5 10 5Fertilizer 0 0 0

Result. By considering net output, we get that inputs have a negative sign and outputs have apositive.

Decomposing the net output vector Suppose we have some y ∈ Rk. Then we can always write

y = y+ − y−, y+, y− ∈ Rk+.

This decomposition is unique when we couple it with the assumption that we know what is inputs andoutputs.

Further requirement. Furthermore, we require that any commodity is either an input or an output.Otherwise, you could have (y+ +a)−(y−+a) = y as well and the decomposition would no longerbe unique.

Production process. Is like y− → → y+.

Example. Consider(−1, 2, 3, 0,−5) = (0, 2, 3, 0, 0)︸ ︷︷ ︸

y+, outputs

− (1, 0, 0, 0, 5)︸ ︷︷ ︸y+, inputs

.

6

1.3 Assumptions on the producerConsider Y ⊂ RL and a specific y ∈ Y .

1.3.1 Assumption P1

Assume that

(a) 0 ∈ Y — inaction is a possibility.

(b) Y is a closed set — consider ynn∈N with yn → y and yn ∈ Y for all n. Then y ∈ Y .

⇒ Otherwise we’d have to work with suprema instead of maxima.

(c) Y is a convex set.

(d) y ∈ Y and −y ∈ Y implies that y = 0 ∈ RL — irreversibility.

(e) Take y ∈ Y and z ∈ RL+. Then y − z ∈ Y — free disposability.

Implications

(a) and (c). If 0 ∈ Y and Y is convex, then we have non-increasing returns to scale.

Proof. Let y ∈ Y and λ ∈ [0; 1]. Then λy ∈ Y because Y 3 λy + (1− λ)0 = λy.

(d). Consider a production where we take 1 horse and 1 chicken to make chicken patty. That is, thenet production vector is y = (−1,−1, 1). Reversing is −y = (1, 1,−1) would imply making onehorse and one chicken from a chicken patty, which is clearly impossible.

(e). Consider y − z and look at the `’th element, y` − z`. Since z ∈ RL+, z` ≥ 0, meaning that wedestroy some of the net output.

Actually what the assumption says is that we can do this freely.Moreover—pos MP. It’s also related to the assumption of positive MP. Suppose for contra-

diction that Y is the area below a −αx2 + βx + χ. Then there’s a “toppunkt” and to theleft of that we can’t produce. So taking z = (2, 0) would take us to an impossible set.Why? In this case, taking in more y1 will actually at one point, for y2 fixed, destroy the

output (shown by the fact that we suddenly end up outside of Y ).

Tell apart the following!

Non-decreasing returns to scale. Take y ∈ Y, λ ≥ 1, then λy ∈ Y .Non-increasing returns to scale. Take y ∈ Y, λ ∈ [0; 1], then λy ∈ Y (?? is this correct??)

(this is correct: λF (K,L) ≤ F (λK, λL))

1.3.2 Example

Consider the setY =

y ∈ R3∣∣y2, y3 ≤ 0, y1 ≤ F (−y2,−y3)

Example 2.3.2. Let

Y =y ∈ R4

∣∣∣ (y21 + y2

2)½ − (−y3)½(−y4)½ ≤ 0, y3 ≤ 0, y4 ≤ 0

.

Four variables, so impossible to draw. But of course, we could fix some and look at the remainingvariables.

7

Y. Look in the (y1, y2)-plane. The set is a circle, but only a quarter (pizza slize) of it.Problem. We don’t have free disposal because the Y set ends when y1, y2 falls below zero.Solution. Consider the set Y −R4

+. That would be Y ∪ all the stuff created by subtractingfrom it... i.e. the circle casts “shadows” down the negative plane...

Contour lines. In y4, y3 it’s like x 7→ 1/x mirrored in x 7→ −x.

1.3.3 More assumptions (P2 among other things)

Additivity. y ∈ Y ⇒ y + y ∈ Y . ALso, y, y′ ∈ Y ⇒ y + y′ ∈ Y .

Assumption P2. This assumption states that Y is a cone.

y ∈ Y, λ ≥ 0⇔ λy ∈ Y.

P2+Convexity ⇒ Y convex cone. If we also have the convexity assumption from P1, then Y is aconvex cone! That is

α, β ∈ R+, y, y′ ∈ Y =⇒ αy + βy′ ∈ Y.

So any non-negative linear combination will be in Y .

Sub-space? NO! Because we restrict α, β to be non-negative.

Result in MWG. Take any Y ⊂ RL, then you can define some Y ′ ⊂ RL+1 which is consistent withCRS.

2 2012-09-052.1 General stuffSetup. We have the goods L = 1, ..., L goods (with member `).

Spaces. Vectors will be inRL ≡ R× · · · × R,

and

RL+ ≡ x ∈ RL|x` ≥ 0, ` ∈ L,RL++ ≡ x ∈ RL|x` > 0, ` ∈ L.

Vector inequalities.x ∈ RL : x ≥ 0⇔ x` ≥ 0∀` ∈ L

x ∈ RL : x > 0⇔ x` ≥ 0 ∧ x 6= 0, 0 ∈ RL

x ∈ RL 0⇔ x` > 0∀` ∈ LFor x 0 we say that x is strongly greater than 0 (he just made that up :)

Intuition. x > 0 think — x is at least as great as but not equal to 0.

Operations on sets. Let A,B ⊂ RL.

A+B ≡ z ∈ RL|z = xA + xB , xA ∈ A, xB ∈ B.

Also, for λ ∈ R,λA ≡ z ∈ RL|z = λxA, xA ∈ A.

Since we can do this for λ < 0, we can do

A−B ≡ A+ (−1B).

8

Free disposability—Interpretation. Suppose you hav a point a ∈ R2 and you want the setthat is the rectangle that has upper-right corner in a and tends to −∞,−∞. Then the freedisposability assumption is that if a ∈ Y , then a − R2

+ ⊂ Y . Or in other words,

free disposability⇔ Y − R2+ ⊂ Y.

(isn’t it???)

Convexity. Take A ⊂ RL.

A is a convex set⇔ ∀a, a′ ∈ A∀α ∈ [0; 1] it holds that αa+ (1− α)a′ ∈ A.

(in his exposition he uses α, β ≥ 0 where α+ β = 1 and then αa+ βa′)

Definition—Cone. C ⊂ RL is a cone if

c ∈ C,α ≥ 0⇒ αc ∈ C.

Convex cone. C is a convex cone if it is a convex set and it is a cone.

Proposition.C is a convex cone⇐⇒ c, c′ ∈ C,α, β ≥ 0⇒ αc+ α′c ∈ C.

Proof.“⇒”. CASE: If α, β = 0, then take some c, c′ ∈ C αc+ βc′ = 0 + 0 = 0 and 0 ∈ C since we

know that C is a cone (just set α = 0 in the definition of a cone).CASE: Suppose either α 6= 0 or β 6= 0. Consider some c, c′ ∈ C

α

α+ βc+ β

α+ βc′

(which is fine because at least one of them is non-zero so we don’t divide by zero)α

α+ βc+ β

α+ βc′ = α

α+ βc+

(1− α

α+ β

)c′ ∈ C

because C is a convex set. Ok so we now have some number in C so because C is acone, we may multiply by any λ so let λ := α+ β, then we have

C 3 αc+ βc′,

which is precisely what we wanted. “⇐”. To prove a cone, take β = 0 and vary the α. To prove convexity, we have to restrict

α, β to α + β = 1. But this is a subset of the parameters for which it holds so inparticular it holds for α, β|α+ β = 1. In more detail. Todo: Prove that C is a cone and convex. We know: c, c′ ∈ C,α, β ≥

0⇒ αc+ α′c ∈ C.Cone: Just set β = 0.Convex: It holds for all α, β ≥ 0. Then restrict attention to α, β|α+ β = 1. Forthe larger set, the linear sum is in C but we only needs it to hold for the subset ofthe parameters.

Decomposition. Consider y ∈ RL. Let’s decompose

y = y+ − y−.

The decomposition is not unique unless we demand something more, let’s demand that

y+` y−` = 0∀` ∈ L.

Hence, for some y = (y1, ..., yL), we must have that either y+` or y−` is equal to zero (or both)!

That is the assumption that it is either an input or an output (or not relevant).

9

2.1.1 More on the assumptions

Assumptions (from last time).

Irreversibility. The intuitive argument is that production is a linear process in time — inputtoday, output tomorrow. Example: We can’t bring the dead cow meat back to life.

Free disposal.(free disposal ass.) : y ∈ Y, z ∈ RL+ =⇒ y − z ∈ Y.

He would also call it that it’s “Downward comprehensive” (in the sense that the set “expands”to −∞ in all directions... we can always put less stuff into the machine and just destroy it).MWG writes

Y − RL+ ⊂ Y.

2.2 The aggregate production setNow—Many firms. Now we have many firms — we’re interested in what they can do together and

look at the “aggregate production process”.

Turns out. The aggregate production set, Y , is

Y = Y1 + Y2 + ...+ YN ,

where Yi are the individual firm production sets.

Notation. Note how nicely the “+” notation for sets work for this.Implicit complexity. The notation hides some of the complexity — the first firm’s output can

be the input for the second and so forth.Howto? Sum over the indices separately. yout

1 = y11 + y12 + ...+ y1N .Nice? Yes, because with production functions, it’s very difficult to find the aggregate production

vector.

Inheritance of properties.

Q. If the individual production sets have all the nice properties, will the aggregate also havethem?

A. No, some of them are inherited but not necessarily all. Additional assumptions can be madeto rectify the situation.

2.2.1 Efficient productions

Goal. Single out production plans where we don’t “waste” resources.

Definition—Domination. We say that for y, y ∈ Y ,

y dominates y if y > y,

that is y ≥ y and y 6= y.

Example—Free lunch. If you can produce the same output without using any more input.

Definition—Eficiency. We say that y ∈ Y is an efficient production if

@y ∈ Y : y > y.

Example—Graphics. If you draw the rectangle that has the point y as the lower-left point(i.e. everything to the up-right of the point) and there is nothing there you could havechosen instead, then you’re good.

10

2.2.2 Producer’s problem (PP) (MWG: Producer’s maximization problem)

Interest: Existence of solutions and their properties.

The Producer’s problem (PP). Given prices p ∈ RL+, p 6= 0,

(PP) : maxy∈Y

py = p1y1 + ...+ pLyL.

Alternatively, we might write

py = p(y+ − y−) = py+ − py− = revenue− cost.

Drawing. Consider the (y1, y2)-plane and a vector p, say p = (1, 2). Then p defines a hyperplane,Hk = y|py = k for some k ≥ 0, then for different k, you get different “hyperplanes” (justlines in R2) that are parallel. We want to get to the hyperplane corresponding to the highestk-value.

(notation) He writes y → py for the notation y 7→ py.Multidimension. Here, the hyperplane is simply orthogonal to the gradient!!! The gradient is

orthogonal to the tangent of the set.

Example. If the production set has an asymptotic tangent that is orthogonal to the prices (i.e. theoptimal hyperplane, Hk, are asymptotically tangent to the hyperplane), then there might neverbe an optimal production plan, cause you’d always keep moving out along the frontier, all thetime improving.

Surprise—P1 ok. You can make up production sets that satisfy P1 and are strictly convexand still may end up in that weird asymptote case.

In MWG—exposition. They allow p ∈ R ∪ ∞... when they say p = ∞ is the solution, we willsay that no solution exists.

2.2.3 Something

Consider. The PP and assume that P1 and P2 are satisfied. And let’s draw on the simplest casethat y1 is input and y2 is output.

(graph) see the iPad from notes from today for a nice simple example with a piece-wise linear bound-ary.

READ ON YOUR OWN. There’s a simple proof (in his notes or MWG?)

3 2012-09-113.1 FBConsumption set. X ⊂ RL represents the non-economic restrictions on consumption.

Assumptions:

• X is non-empty.• X is closed.• X is convex.• X is a lower bounded set (in vector ordering).

11

– I.e. ∃b ∈ RL : x ∈ X ⇒ x ≥ b.(recall that x ≥ b⇔ x` ≥ b`∀`; x > b⇔ x` ≥ b ∧ x 6= b; x b⇔ x` > b`∀`)

• X is upward comprehensive.– x ∈ X, z ∈ RL+ ⇒ x+ z ∈ X.

Preference relation.

Continuous utility. The preference relation % can be represtented by a continuous utilityfunction if it holds that

∀x ∈ X : x ∈ X|u(x) ≥ u(x) and x ∈ X|u(x) ≤ u(x) are closed sets in X,

and that

∀x ∈ X : x ∈ X|u(x) > u(x) and x ∈ X|u(x) < u(x) are open sets in X.

Convexity. % is convex if the set if for all x ∈ X, the set x′ ∈ X|x′ % x is convex.Utility. If u represents %, then % is convex iff u is quasi-concave.

Strict convexity. If

x1, x2, x3 ∈ X, x1 % x2, x3 = λx1 + (1− λ)x2, λ ∈ (0; 1)⇒ x3 x2.

Utility.u(x1) ≥ u(x2), x3 = λx1 + (1− λ)x2, λ ∈ (0; 1)⇒ u(x3) > u(x2).

Differentiable. % is differentiable if there exists a continuous utility function, u, representing% such that1. Each partial derivative, ∂u(x)/∂x`, exists for x ∈ intX.2. The partial derivatives, ∂u(x)/∂x` : intX → R, ` ∈ L, are continuous.3. The gradient is not the zero vector, i.e.

∇u(x) =(∂u∂x1

(x), · · · ∂u∂xL

(x))6= 0∀x ∈ intX.

Differentiable demand fct? To obtain this, we need a strong assumption.

Required assumption: There exists a utility function, u, with continuous first and secondorder partial derivatives representing % such that det H 6= 0 where H is the borderedHessian,

H =

D2

11u(x) D212u(x) · · · D2

1Lu(x) D1u(x)D2

21u(x) D222u(x) · · · D2

2Lu(x) D2u(x)...

... . . . ......

D2L1u(x) D2

L2u(x) · · · D2LLu(x) DLu(x)

D1u(x) D2u(x) · · · DLu(x) 0

.

3.2 IntroductionNotation—again.

Number spaces. Recall that z ∈ RL++ ⇔ z >> 0.Mapping (function). He would write (y1, y2, y3) → y1 for the function I would write as

(y1, y2, y3) 7→ y1.

12

Convexity. He argues that if F is concave, then y1 − F (−y2,−y3) is convex because(y1, y2, y3)→ y1 is convex and (y1, y2, y3)→ −F (−y2,−y3).

Lower contour set. He looks at the lower contour set of a number c ∈ R as

lower contour set of c : Cc = y ∈ RL|y1 − F (−y2,−y3) ≤ ccontour set of c : Cc = y ∈ RL|y1 − F (−y2,−y3) = c

upper contour set of c : Cc = y ∈ RL|y1 − F (−y2,−y3) ≥ c.

Hold on. If the contour set is generated by a function (y1, y2, y3)→ y1−F (−y2,−y3) thatis convex, then ... ???But for utility functions, the upper contour set will most often be a convex set.

Q from student last time. R

Returns to scale. Consider y ∈ Y . If then λy ∈ Y for λ ∈ Λ, then XXX

1. Λ = [0; 1]; Y exhibits non-increasing returns to scale2. Λ = [1;∞); Y exhibits non-decreasing returns to scale3. Λ = [0;∞); Y exhibits constant returns to scale.

(see) the drawing I did in hand today...

Producer’s max problem (PMP).(PMP) : max

y∈Ypy,

i.e. find y0 ∈ Y s.t. py0 ≥ py∀y ∈ Y .

Cost-revenue.π = py = py+ − py−.

Isoprofit hyperplanes. Important to study. Hyperplanes are the contour sets for a linear function— hence they get almost full dimension. Say, 2-dimensional in R3.

Supply correspondance. The set of productions, y ∈ Y , giving maximal profit,

η(p) = y ∈ Y |py ≥ py ∀y ∈ Y .

Often contains more than one vector, may be empty.

MWG notation. MWG use y for the production and use y(p) for the supply curve.

3.3 Thm [4.2].1 — red proof (p. 29 NotesThPr)Let p ∈ RL+, p 6= 0 and let Y satisfy assumption P1. Then the following hold

(a) η(p) is a convex set, possibly empty

(b) η(αp) = η(p) for α > 0, α ∈ R++.

(c) py ≥ 0 for y ∈ η(p) and py is constant on η(p).

(d) if p ∈ RL++ then each y ∈ η(p) is an efficient production in Y .

further if P1 and P2 are satisfied, then

(e) py = 0 for each y ∈ η(p)

(f) if η(p) is non-empty, then η(p) is a convex cone which may be degenerate, that is, η(p) = 0.

13

3.3.1 Proof of (a)

Proposition. p ∈ RL+, p 6= 0 and let Y satisfy assumption P1. Then η(p) is a convex set, possiblyempty.

Proof. Assume that y, y ∈ η(p) and let α, β ∈ [0; 1] be given s.t. α + β = 1. Choose some arbitraryy ∈ Y . Then we know that

py ≥ py ∧ py ≥ py,

because of the definition of η(p). From the simple rules of inequalities and multiplying with ascalar,

αpy ≥ αpy ∧ βpy ≥ βpy.

Now add the inequalities,

αpy + βpy ≥ αpy + βpy

⇔ p(αy + βy) ≥ py,

which proves that αy + βy ∈ η(p).

3.3.2 Proof of (b) (on my own!!)

Proposition. η(αp) = η(p) for all α ∈ R++ (i.e. scalars α > 0.

Proof.

“⊂”. Take y ∈ η(αp). To show: y ∈ η(p). To whos this, take some y ∈ Y , it must then holdthat py ≥ py. But we know that

(αp)y ≥ (αp)y⇔ py ≥ py.

“⊃”. As above.

3.3.3 Proof of (c)

Proof. Let y, y ∈ η(p). By P1, we know that 0 ∈ Y , and py ≥ p0. Then py ≥ py∀y ∈ Y so inparticular for y = y, i.e. py ≥ py. Similarly, since y ∈ η(p), py ≥ py. Thus, py = py.

Intuition. The profit at maximum is unique... although the way of getting there need not be.

3.3.4 Proof of (d)

Proposition. If p ∈ RL++ then each y ∈ η(p) is an efficient production in Y .

Note. Note that now we’ve strengthened from p ∈ RL+ to p ∈ RL++.

To show. Show that y ∈ η(p) is efficient.

Method. Proof by contradiction.

Proof.

14

Assume contradiction. We assume that y is not efficient. Then, there is a y ∈ Y that domi-nates y, i.e. y > y. Elementwise, this means that

y` ≥ y`∀` and ∃l : yl > yl

with strict inequality for at least one `. Now multiply prices on both sides, maintaininginequality (note that prices are non-zero)

p`y` ≥ p`y`

and summing givespy > py

(due to the at least one surviving strict inequality). Hence, y /∈ η(p). `

Note. It was essential that p` 6= 0 for all ` (or at least for the ` where the inequality was strict).

3.4 Subspaces and linear functionsLinear functions. Are functions R2 → R. For a given p, we are in particular interested in y → py.

Homogenous hyperplane. A hyperplane that contains the zero vector. Hence, the family of hyper-planes,

Hα = y ∈ Y |py = α

have only one member that is homogenous and that is H0.

Subspace. Note that the homogenous hyperplanes are all linear subspaces (I think?).Affine subspaces. Any hyperplane is an affine subspace. However, these are not very studied

by mathematicians.

3.5 A Decentralization Theorem (thm. 7.1, p 35 NotesThPr.)Let Yjj∈J be J production sets and p ∈ RL+ , p 6= 0. Consider the J individual PMP,

maxyj∈Yj

pyj

and the aggregate PMPmax

y∈Y1+Y2+...+YJ

py.

Notation. Recall thatA+ B = c|∃a ∈ A, b ∈ B : c = a+ b.

Proposition.

(a) Ifyjj∈J solves the individual PMP, then y = y1 + ...+ yJ solves the aggregate PMP.

(b) If y ∈ Y1 + ... + YJ solves the aggregate PMP then for any yj ∈ Yj , j ∈ J such thaty = y1 + ...+ yJ it is the case that yj solves the individual PMP for j ∈ J.

15

3.5.1 Proof of part (b)

Setup. We are given y ∈∑j∈J Yj and we know that

py ≥ py∀y ∈∑

Yj .

We also know that since y ∈∑j∈J Yj that there exist y1, ..., yJ such that

y = y1 + ...+ yJ .

Assume the contradiction. Assume that one yj violates producer j’s problem (it could be thatmultiple did it but this will suffice). And let’s consider j = 1 for notation.

Then there isy1 ∈ Y1 : py1 > py1.

But. Then compare the aggregate production plan y = y1 +y2 +...+yJ with y =y1 +y2 +...+yJ .We know that y ∈

∑j Yj because all the elements are in the respective ones. Now consider

py =∑j

pyj < py1 +J∑j=2

pyj = py.

Hence,∃y ∈

∑j

Yj : py > py. `

contradiction!Hence. we conclude that yj must solve the PMP for each individual producer j.

3.5.2 Proof of part (a) — LEFT FOR US TO DO BY OURSELVES

Setup. Suppose that yj solves producer j’s problem for each j ∈ J. To show: y ≡∑j yj solves the

aggregate problem.

My idea. Well if you maximize each component of the sum then of course the sum itself must bemaximized.

3.6 Representing production sets with production functions(also: transformation functions)

First stupid attempt. Let Y be given, Y ⊂ RL. We can always describe Y with a function, forexample 1Y (y) (the indicator function for the set...).

Second attempt. We want to use calculus.

Production function. Y ⊂ RL can be described by a production function if there a functionF : RL → R (note, F must be defined on all of RL!) such that

(i) F (y) ≤ 0 whenever y ∈ Y .– That is, Y should be the lower contour set of F .

(ii) F (y) = 0 if and only if y is an efficient production in Y .(iii) F ∈ C1 (so F has continuous partial derivatives; continuously differentiable) on all of RL!

(no problem with interiors here!!)

16

The problem. There will often be kinks, for example at zero. And this can’t always be C1.But. Often, productions “after kinks” (e.g. on the second axis when we have a negativity

constraint) are often inefficientIdea. We only need sensible behaviour close to the optimal solution.

4 2012-09-124.1 Representing production setsRepresenting production. We can always represent Y by a function. We might just use the indi-

cator function.

Pro. Always feasibleCon. Not a nice function... we’d like to use differential calculus.

Alternatively.F : RL → R represents Y iff the following hold

(i) : F (y) ≤ 0 if y ∈ Y(ii) : F (y) = 0 iff y is an efficient production plan

(iii) : F ∈ C1(RL).

Pro. NicerCon. Often there will be kinks in the Y frontier, for example due to non-negativity sets.IMPORTANT: Suppose Y is the entire South-West (third quadrant) plus the part of the 4th

quadrant that is below a mirrored log function.Then in the 3rd quadrant, F (y) ≤ 0 (cause y ∈ Y ) and in the 2nd quadrant F (y) > 0

(because y /∈ Y ).BUT! What about the 2nd axis? It’s between them so for continuity we need F (0, y2) = 0

but it’s not efficient, which doesn’t work with (ii).

Solution—Vicinity. Consider y ∈ Y ⊂ RL and let U ⊂ RL be the open sphere centored on y.

Local representation.F : U → R locally represents Y at y iff

(i) : F (y) ≤ 0 for y ∈ Y ∩ U(ii) : F (y) = 0 iff y is an efficient production in Y ∩ U

(iii) : F ∈ C1(U).

Tangent hyperplane. To get the hyperplane, consider gradF (y). The tangent hyperplane is tangen-tial to the gradient!

Moving the coordinate system. You should move the coordinate system to have the origincentered at 0 := y and then look at the vector ∇F as eminating from that point.

Example. Consider Example [8.2].1 which we did in the tutorials.

17

4.2 Thm. [8.2.].1Assume. Assume that the prdocution possibility set Y can be locally represented by the production

function F at y and p ∈ RL++.

Proposition. The proposition has two parts, (I) and (II).

(I) If y is a solution to the PMP at prices p, then

F (y) = 0 and ∇F (y) = µp for some µ ∈ R and if ∇F (y) 6= 0, µ > 0.

Intuition. µ doesn’t mean anything — only relative prices matter....∇F sign. ∇F (y) ≥ 0 for y ∈ U \Y and ∇F (y) ≤ 0 for y ∈ U ∩Y because F (y) ≷ 0 respectively

of which side of the frontier y falls in.⇒ Hence, ∇F “points” in the direction of outside Y . That’s why it’ll always point in thesame direction as p. And therefore, µ ≥ 0 always. Hence, µ ∈ R+.

Think. y solves max py s.t. F (y) = 0.Problem. What if y is not a solution to the PMP?

Then. it might just be a local optimum...Graphs. Look at [8.2].1 and [8.2].2 depict examples (with a “wavy” frontier).

(II) If for some µ > 0, y ∈ Y satisfies

F (y) = 0 ∧∇F (y) = µp ∧ Y is convex,

then y is a solution to PMP.(if ∇F (y) = 0 (the zero vector) there is a problem... but he doesn’t have anything to say aboutit!)

Intuition. The differential calculus provides us with the candidate solutions,

candidates = y ∈ Y |F (y) = 0,∇F (y) = µp, µ > 0.

4.2.1 Law of supply

Suppose.y0 solves PMP at prices p0,

y1 solves PMP at pricse p1,

which of couse implies that p0y0 ≥ p0y1 and vice versa. Using these, we get

p0(y0 − y1) ≥ 0and 0 ≥ p1(y0 − y1)

⇒ p0(y0 − y1) ≥ p1(y0 − y1)⇔ (p0 − p1)(y0 − y1) ≥ 0,

or, ∆p∆y ≥ 0, i.e. sign∆p = sign∆p. That is, p` ↑⇒ y` ↑.(note the sign conventions, p` < 0 for an input where ↑ means closer to zero, so cheaper:)

Comparison to Slutsky. For the producer, we have nice and sensible substitution patterns inthe directions we’d assume

18

4.3 Moving on to consumersTime constraints ⇒ we skip talking about expenditure functions and indirect utility functions and

only use what is necessary to get onwards to the welfare stuff.

Notation—Wealth, w. He originally the latex w which is very close to ω. He wanted to change itto w.

Consumption (possibility) sets.

MWG. They just set X = RL+.Here. We want to allow more general X.What is it? Non-economic constraints: For example nutritional requirements, can’t work neg-

ative hours,

Demand function.

MWG. They use the demand function (p, w)→ x(p, w), where x(·, ·) is a function.Here. Since x could also be a variable x ∈ X, we instead use (p, w) → ξ(p, w), where x :

RL+ × R→ X.

Setup—Consumer stuff.

Commodities. L = 1, 2, ..., L commodities.

Sign stuff.

Recall for producers. There, we had y− → Y1 + ... + YJ → y+, where y− was inputs withpositive sign.(and recall how deliveries from firm j to k cancelled out in the aggregate...)

Consumers—Decomposition. Here, we also decompose

RL ⊃ X 3 x = x+ − x−

4.4 Assumptions4.4.1 F1

The consumption set X ⊂ RL satisfies

(a) X is non-empty.

(b) X is a closed set.

(c) X is a convex set.

(d) X is a lower bounded set in the vector ordering (∃b ∈ RL∀x ∈ X : x ≥ b)

(e) X is upward comprehensive (x ∈ X, z ∈ RL+ ⇒ x+ z ∈ X).

Note. These requirements are non-economic in the sense that there are no prices and so forth.

Notes.

(c) Convexity implies that all goods should be divisible!! (because if e` = (0, ..., 0, 1, 0, ..., 0) ∈ X,then λe` ∈ X)

19

(d) The lower bound requires that no consumer can supply ∞ of anything! Visually, in R2, itmeans that we can put down a 90-degree triangle (parallel to the axes) and then it’s linesnever touch X.

(e) Upward comprehensibility

Boundedness. A vector can be bounded in several senses:

Lower bounded. (in vector ordering)Upper bounded. (in vector ordering)Bounded. (in norm) A set A ⊂ RL is bounded if there exists an ε > 0 and a point (any point?)

a ∈ A such that the sphere centered on a with radius ε, S(a, ε) = x ∈ A|d(x, a) < ε,conttains A fully.(alternatively, have the sphere centored on the origin — i.e. there exists ε > 0 such that∀a ∈ A : ‖a‖ < ε.

Note: Lower bounded + upper bounded ⇒ bounded.

5 2012-09-185.1 Last timeDiscussed the consumer

Assumption F1.The consumption set X ⊂ RL satisfies

(a) X is non-empty.(b) X is a closed set (mathematical convenience).(c) X is a convex set (implies that commodities are divisible — counter-example is discrete

goods).(d) X is a lower bounded set in the vector ordering (∃b ∈ RL∀x ∈ X : x ≥ b) — means that

the consumer can’t deliver ∞ of any ressource to the firms.(e) X is upward comprehensive (x ∈ X, z ∈ RL+ ⇒ x+ z ∈ X) — MWG write X + RL+ ⊂ X.

5.2 PreferencesRational preference relation.

Rationality. Means that the preference relation is Total and Transitive (completeness is not apart of the definition).

Total. x, x′ ∈ X implies x % x′ or x′ % x.Transitive. x, x′, x′′ ∈ X and x % x′ and x′ % x′′ implies x % x′′.

Other properties.

Reflexive. Often we’lll assume this as well: x ∈ X then x % x.Monotonicity. A set of assumption F2 shown later in this note.

Intuition. Monotonicity ensures that we will get scarcity (⇒ positive prices, etc.)Satiation/bliss point. We want exclude this.

20

Local non-satiation. For any ∀x ∈ X, ε > 0∃x′ ∈ S(x, ε) ∩ X : x′ % x, where S(x, ε) is thesphere with center x and radius ε.Note—intersection. We need S(x, ε)∩X because x might for example be on the edge of

X.Convexity. See below

Relations.

Strictly preferred. Let x, x′ ∈ X. Then x x′ if x % x′ but it is not the case that x′ % x.Indifference. Let x, x′ ∈ X. Then x ∼ x′ if x % x′ and x′ % x.

5.2.1 Proof of continuity — sketch

Goal. Constructing a utility function for % and proving continuity.

Idea—Constructive. Consider the (x1, x2)-plane and consider the 45 line. Then think of all theindifference curves as Cobb-Douglas-style and they cut the 45-line further and further out. Thenlet the utility function, u(·), simply assign how far out of the 45-line you went before you gotto the intersection with the indifference curve on which (x1, x2).

Now. u : X → R represents %. But take any ϕ : R → R, ϕ increasing, and consider ϕ u — thisfunction will also represent % .

Requirement for existence. Assume that X is closed and non-empty. Then % can be representedby a continuous utility function if and only if ∀x ∈ X, both the upper and lower contour sets,x ∈ X|x % x and x ∈ X|x - x respectively, are closed sets.

Small note. For closedness, we require a metric, but recall that X ⊂ RL so the euclideandistance is used (I guess?).

Definition—Continuous preference relation. We actually say that % is continuous if itthe lower and upper contour sets of any x ∈ X are closed (i.e. it can be represented by acontinuous u).

Contour sets. Note that under

x ∈ X|x % x = x ∈ X|u(x) ≥ u(x),

and assuming that the preferences are continuous,

x ∈ X|x % x = u−1(

[u(x);∞))

which will be closed ...

Further on—Always assume continuity.

5.2.2 Monotonicity assumptions F2

F2. x1, x2 ∈ X and x1 ≥ x2 ⇒ x1 % x2.

F2 MWG Def 3.B.2. x1, x2 ∈ X and x1 >> x2 ⇒ x1 x2

Think. We want to exclude “thick” indifference curves — however Leontief (90 indiff curves)where if you get more of some goods but not all (and you start in an “optimal” point) thenyou are indifferent!!

F2 strong monotonicity MWG Def 3.B.2. x1, x2 ∈ X and x1 > x2 ⇒ x1 x2.

21

5.2.3 Convex preferences

Definition—Convexity. We say that a preference relation, %, is a convex preference relation iff forany x ∈ X, the set x′ ∈ X|x′ % x is a convex set. (MWG def 3.B.2)

Utility. If u represents %, then % is convex iff u is quasi-concave.

Quasi-concavity.f : R→ R increasing =⇒ f quasi-concave.

Example—Cobb-Douglas. Something something u = xα1xβ2 , when α + β = 1 lower contour

sets (??) are concave but α + β > 1 ruins concavity but something something (xα1xβ2 )2

retains quasi-concavity...

5.2.4 Strictly convex preferences (F3)

Setup. Let x1, x2, x3 ∈ X such that x3 = tx1 + (1 − t)x2 for some t ∈ (0; 1) (note that t 6= 0, t 6= 1)and x1 6= x2.

Preferences. % is a strictly convex preference relation if x1 % x2 implies x3 % x2.

Utility version. The utility function is strictly quasi-concave if u(x1) ≥ u(x2), then u(x3) ≥ u(x2).

Alternative. An alternative combination could be

u[tx1 + (1− t)x2] > min

[u(x1), u(x2)

].

Interpretation. A proper convex combination of two different points is better than either of the two.

Purpose. Prevents “flat” sets on the indifference curves.Note, it also prevents “fat” indifference curves because you can choose two points in the fatindifference and then the line between is also in the indifference set.

5.3 Towards calculus5.3.1 Assumption of differentiability

Definition. We say that % is diff if

1. Each partial deriv exists2. The partial derivatives, Dlu(x) = ∂u

∂xl(x), are continuous on intX.

3. For any x ∈ intX, the gradient, ∇u(x) = (D1u(x), ..., DLu(x)), is not the zero vector (i.e.not all the partial derivatives can be zero in the same point).

5.3.2 Assumption F4 — bordered Hessian

Assumption F4. There exists a utility function, u, with continuous first and second order partialderivatives representing % such that det H 6= 0 where H is the bordered Hessian,

H =

D2

11u(x) D212u(x) · · · D2

1Lu(x) D1u(x)D2

21u(x) D222u(x) · · · D2

2Lu(x) D2u(x)...

... . . . ......

D2L1u(x) D2

L2u(x) · · · D2LLu(x) DLu(x)

D1u(x) D2u(x) · · · DLu(x) 0

.

Mechanics. Prevents the Gaussian curvature from being zero.

22

5.4 Consumer problemSetup. We’re given (p, w), prices and wealth (in notes: he half-heartedly changed w to w to avoid

confusion with ω.

Budget constraint.∀x ∈ X : p · x ≤ w.

Zero prices. ALWAYS take note of whether we assume p ∈ RL+ or p ∈ RL++.

5.4.1 Proposition 1

Proposition 1. Let (X,u) be a consumer. Assume that

(i) X satisfies F1 (non-empty, closed, convex, lower bounded, upward comprehensive).(ii) p ∈ RL++ (positive strictly positive) and w ≥ minx∈X p ·x (MWG use infx∈X p ·x — but it’s

an irrelevant problem here).

Then the budget set,B(p, w) ≡ x ∈ X|p · x ≤ w,

is a non-empty, compact set.

Notes.

Non-empty. Follows from (ii).

Proof—Compactness.

Step 1: Closedness. Note that

B(p, w) = x ∈ RL|p · x ≤ w ∩X.

If we can show that the two sets are closed then the intersection is closed. The first of thetwo is the inverse image (“urbillede”) of the linear function x 7→ p ·x on the set (−∞;w] andthe inverse image of a linear function on a closed set is closed. X is closed by assumption.QED.

Step 2: Boundedness. Let x ∈ X. To show: There exists ε > 0 such that ‖x‖ < ε. First, notethat by F1, there is b ∈ −RL+ such that x ≥ b (if the lower bound has positive coordinates,we can always find an even lower bound that has all negative coordinates). Now note that

p1(x1 − b1) ≤ p1(x1 − b1) + ...+ pL(xL − bL) ≤ p · x− p · b ≤ w − p · b

x1 ≤ b1 + w − p · bp1

= d1.

so since xl ≤ dl for all l, ‖x‖ ≤ ‖d‖ =: ε.

6 2012-09-196.1 StuffRecall—Budget set. B(p, w) = x ∈ X|p · x ≤ w.

Proposition 1.

Assume. 1) X satisfies F1 and 2) p ∈ RL++ and w ≥ minx∈X p · x.Note. Part 2) ensures that B(p, w) is non-empty.

Then. B(p, w) is a non-empty, compact and convex set.

23

6.1.1 Recapping the proof of proposition 1

Proof of closedness. We have that

B(p, w) = x ∈ RL|p · x ≤ w ∩X = ϕ−1(

(−∞;w])∩X,

where ϕ(x) = p · x so the inverse image of a closed set by a continuous function is closed.Also X is closed by F1.Hence, the intersection is closed.

Proof of boundedness. To show: ∀x ∈ B(p, w)∃k ∈ R : ‖x‖ ≤ k.

Use lower boundedness of X. By F1 we know that ∃b ∈ RL : x ∈ X ⇒ x ≥ b. Now, choosec even smaller than the guarenteed b so that b ∈ −RL+ (i.e. all non-positive entries) and sothat p · x > p · c.WARNING! He uses b as my c!

Consider the setC = x ∈ RL|x ≥ b, p · x ≤ w.

Now note that.

pl · (xl − bl) ≤∑j∈L

pj · (xj − bj) = p · x− p · b ≤ w − p · b

⇔ bl ≤ xl ≤ bl + w − p · bpl

=: dl.

Warning—Sign! We do not know the sign and we want to take the square. So if it’snegative, x2

l > d2l . But then we’re saved by the other end point since then b2

l > x2l > d2

l .Hence, in a coordinate-wise fashion, we have

x2l ≤ max(b2

l , d2l )

Hence, the summation over l ∈ L will maintain the inequality,∑l∈L

x2l ≤

∑l∈L

max(b2l , d

2l )

⇔ ‖x‖ ≤ ‖e‖,

wheree =

(max(b2

1, d21), ...,max(b2

L, d2L)).

Proof that B(p, w) is non-empty. Let x ∈ X be given and let w = px. We know that B(p, w) iscompact so the function x→ p · x attains a minimum on B(p, w) because x→ p · x is linear andthus continuous.

Then. There is x ∈ B(p, w) such that it is cheaper than all other bundles, i.e.

∃x ∈ B(p, w)∀x ∈ B(p, w) : p · x ≤ p · x

and p · x ≤ w since x ∈ B(p, w). Then consider any x ∈ X \ B(p, w), it must hold thatp · x ≥ w because otherwise x ∈ B(p, w). Hence, in particular, p · x ≥ p · x.

Hence, p · x = minx∈X

p · x.

24

Proof that B(p, w) is a convex set. Note that

B(p, w) =x ∈ RL|p · x ≤ w

∩X = ϕ−1

((−∞;w]

)∩X,

which is again the intersection of two convex sets because X is convex by F1 and it is a mathe-matical fact that the first set is the half-space of RL (partitioned by a hyperplane) which is veryeasy to prove:

Proposition. A =x ∈ RL|p · x ≤ w

is convex.

Proof. Take x, x′ ∈ A and λ ∈ [0; 1]. Then

px ≤ w, px′ ≤ w

⇒ λpx ≤ λw, (1− λ)px′ ≤ (1− λ)w

⇒ λpx+ (1− λ)px′ ≤ λw + (1− λ)w

⇔ p · [λx+ (1− λ)x′] ≤ w

⇔ λx+ (1− λ)x′ ∈ A.

Intuition. The lower boundedness and the positive prices are what ensures a compact choice set.

6.2 Proposition 2 — solution to the CPSequences. Any infinite sequence on a compact set will have a convergent subsequence.

Assumptions. Let (X,u) be a consumer where X satisfies F1 and let p ∈ RL++ and w ≥ minx∈X p ·x.

Proposition 2 (with proofs?). The following hold.

(a) There is a solution to the CP.Proof. Consider the restriction of u to B(p, w). That function is continuous. Then there

is x ∈ B(p, w) such that∀x ∈ B(p, w) : u(x) ≥ u(x),

that is, there is a solution to the Consumer Problem.

(b) If u satisfies F2 then each solution to the CP belongs to the budget hyperplane.Proof by contradiction. Assume that there is a solution x ∈ B(p, w) with p · x < w

(“below” the budget hyperplane). To do: construct x such that x % x. Let

x = x+ (w − p · x) 1L

(1p1,

1p2, ...,

1pL

).

Claim. x ∈ B(p, w).Proof.

p · x = p · x+ (w − p · x) 1Lp ·(

1p1,

1p2, ...,

1pL

)= p · x+ (w − p · x) 1

L

L∑l=1

plpl

= p · x+ (w − p · x) 1LL

= w.

25

Claim. x % x.Proof. Since xl > xl in all coordinates (because w > p · x and p−1

l > 0 for all l) andsince F2 (“normal” monotonicity) gives that xl > x′l∀l⇒ x x′, we have x % x. Ifwe assume strong monotonicity, then because xl > xl for all l ∈ L, x x.

WARNING! I think we actually assumed strong monotonicity above.(c) If u satisfies F3 then there is a unique solution to the CP.

Proof. Assume that x, x with x 6= x are both solutions to the UMP, i.e.

u(x) = u(x).

Then consider the convex combination by λ = ½, x′ = ½x + ½x. By convexity ofB(p, w), x′ ∈ B(p, w). By strict convexity of u,

u(x′) = u (½x+ ½x) > min [u(x), u(x)] = u(x) = u(x).

But that contradicts that these could be solutions to UMP.

Comments.

(b) and F20. In this we have weak monotonicity. That allows for thick curves. But then theones on the budget hyper plane are equivalent to ones just close to them (still on the thickarea).

(c) and F3. If we don’t have convexity, there might be several solutions to the UMP (if theslope is the same over a region)

6.2.1 Marshallian Demand Functions

Setup. X satisfies F1, u is continuous and u satisfies F2 and F3.

Consider.D = (p, w) ∈ RL+1|p ∈ RL++, w ≥ min

x∈Xp · x.

(recall) then a unique solution exists in the budget hyperplane.

Definition. Then the Marshallian Demand function is

ξ(p, w), ξ : D → RL

(MWG use x(p, w)) and we can characterize it as

u [ξ(p, w)] ≥ u(x), ∀x ∈ B(p, w)

Continuity. ξ is continuous for (p, w) such that w > minx∈X p · x.

Proof—Not shown here!! Follows from Claude Berge’s theorem of maixmum or something...Minimum wealth situation. Is when w = minx∈X p · x.Illustration of the problem.

To show. An example where we have a sequence pn, wnn∈N where ξ(pn, wn) = 0 for alln but limn→∞ ξ(pn, wn) > 0.

How? Let X be on the 1st axis until (0,0) where it turns into the line x 7→ −x and after awhile it kinks and is parallel to the 2nd axis.Notes: See the consumption notes p. 17 — it’s drawn there!!

What happens? When the prices are p = (1, 1− 1n ) and w = 0, the optimum is x = (0, 0)

but if p = (1, 1) and w = 0, the optimum is at the point before it goes vertical. So itjumps across the part where the frontier of X is x 7→ −x.

26

6.2.2 UMP and Expenditure Minimization

Assumptions. Let (X,u) be a consumer satisfying F1 and let p ∈ RL++.

The following hold.

(a) If the consumer satisfies F2 and x solves the CP given (p, w), then x solves the expenditureminimization problem,

min p · x s.t. u(x) ≥ u(x).

(b) Let x be a solution to the expenditure minimization given p and u = u(x). Put w = px . Ifw = px > minx∈X p · x, then x is a solution to the Consumer Problem given (p, w)

Why F2 under (a)? Because you might have a thick indifference curve — if the solution is insidethe thick part of it, then some of them are solutions but waste money (it just wouldn’t help tosave money).

Local nonsatiation. Would work too. We essentially just need to disallow thick indifferencecurves.

6.2.3 Walrasian Demand Function

Setup. Consider (X,u, e), where e ∈ X is an initial endownment.

Wealth? Simply follows from the price system, p · e, p ∈ RL+.

Consider. The mapping p 7→ (p, p · e) 7→ ξ(p, p · e).

Typical assumption? e ∈ int(X) so as to avoid the minimum wealth situation.

MWG. They don’t make a distinction between the Walrasian and the Marshallian demand.

7 2012-09-257.1 Last timeExample with disc. He constructed an example where the demand fct. was discontinuous

Kinked cons. set. We condisdered a consumption set where the frontier was (x-axis until (0,0))+ (kinked line with slope -1 up until about (-1,-1)) + (line parallel to y-axis (at x = -1) upto ∞).

Prices. pn, wn = (1− n−1, 1, 0).Demand. ξ(pn, wn) = (0, 0) for all n ∈ N. But ξ(limn→∞(pn, wn)) = ξ(1, 1) = (−1,−1) —

because when prices become tangential to the “slope -1” part of the X-frontier, we followthe indifference curves upwards (at each jump increasing utility) until we reach the nextkink point at (-1,-1).

Properties of demand.

Homogenous of deg. 0.ξ(αp, αw) = α0ξ(p, w) = ξ(p, w).

Budget hyperplane.B(αp, aw) = B(p, w).

27

Walras’ law.pξ(p, w) = w

(we always choose something on the budget hyperplane)

Problem set 3—Revisions. In the CES function it said that ρ > 1 — it should be ρ ≤ 1.

CES fct.Leontief. ρ→ −∞ (small elasticity of subst) tends to 90 legs.Cobb-Douglas. ρ = 0. Here, the axes are asymptotes for the indifference curves.ρ ∈ (0; 1). Now indiff curves touch the axes.

7.2 Concluding the study of the consumer7.2.1 Prop 5

Let (X,u) be aconsumer satisfying F1, F2 and differentiability and (p, w) ∈ D.

Proposition 5.

(a) If x ∈ int(X) is a solution to the CP then there is λ > 0 s.t.gradu(x)− λp = 0w − px = 0.

(b) If furthermore F3 is satisfied and x ∈ int(X) is a consumption which for some λ > 0 satisfiesgradu(x)− λp = 0w − px = 0,

then x is the unique solution to the CP.

Thinking.

(a) Something something we could rescale u = λ−1u...(b) We already know that by F1–F3 there is a unique solution. So this thm. says whwat the

solution actually is.

7.2.2 Proving proposition 5

We know.

To show. x solves the expendituer minimization problem. That is

u(x) ≥ u(x)⇒ px ≥ px.

(recall there’s a proposition that the two are equivalent under certain restrictions)

We know. λp = grad(x).

Let x be such that u(x) ≥ u(x).

Now considerx+ t(x− x) = (1− t)x+ tx, t ∈ [0; 1]

which is the line segment from x to x.

28

By F3. We want to take the derivative of

u [x+ t(x− x)] ,

so we put forth the newtonian,

limt→0

u [x+ t(x− x)]− u(x)t

.

Digression. Think of this as f [x(t), y(t), z(t)]. To take this derivative, use the total derivative,

ddtf(x, y, z) = x′(t)f ′1 + y′(t)f ′2 + z′(t)f ′3 = grad(f) ·

(x′ y′ z′

),

where (·) is the scalar product.Applying this.

limt→0

u [x+ t(x− x)]− u(x)t

= grad(u) ·[x+ t(x− x)1 x+ t(x− x)2 · · · x+ t(x− x)L

]= ∂u

∂x1(x)(x1 − x1) + ...+ ∂u

∂xL(x)(xL − xL)

= grad(u) · (x− x),

because the inner vector, x+ t(x− x), has derivative ∇t[x+ t(x− x)] = x− x.Non-negativity. Because we have chosen x so that x ≥ x, we know that u(x) ≥ u(x). But

this means that since the convex combination is then at least as large as x as well, i.e.x+ t(x− x) ≥ x, then utility

u(x) ≥ u[x+ t(x− x)] ≥ u(x),

so that we know that utility is on the increase in that direction, i.e.

grad[x+ t(x− x)] ≥ 0.

Insert grad(u). By assumption, we know that

grad(u) = λp.

Then

λp(x− x) ≥ 0⇔ px ≥ px,

which is what we wanted to prove.

What about uniqueness? We already knew that F1–F3 ⇒ unique solution so it x solves the CPthen it must have been unique.

Understanding. Consider the optimization problem,

max u(x) s.t. p · x = w.

I THINK WHAT HE SAYS IS: Suppose that x solves the CP. Then there is λ such that (x, λ)is a critical point of the lagrangian, L(x, λ) = u(x) + λ(w− p · x). Furthermore, λ is going to bethe marginal utility of w (income).

Summing up. We got the demand function characterized as the solution to a set of equations.

29

7.2.3 Proposition 6 — differentiable demand

Intro.

F1(x, λ, p, w) = ∂u

∂x1(x)− λp1 = 0

F2(x, λ, p, w) = ∂u

∂x2(x)− λp2 = 0

...FL(x, λ, p, w) = ∂u

∂xL(x)− λpL = 0

FL+1(x, λ, p, w) = w − px = 0

Why this system? I’m pretty sure that this system is the system of

Linear algebra. If we have a linear system written in matrix algebra as Ax− b = 0.

Implicit fct. theorem. Gives us

SOMETHING????? Related to the Ax− b system but only a local thing...Solve we can solve for (x1, x2, ..., xL, λ) if the bordered Hessian (from ass. F4) has a non-zero

determinant.Bordered hessian. It has the Hessian of u inside because we have ∂u

∂x`−λp` so that ∂

∂xj( ∂u∂x`

−λp`) = D`ju but ∂

∂λ ( ∂u∂x`− λp`) = −p` but we know that marginal utility and prices are

proportional so −p` = D`u.Hence. Whenever the bordered Hessian has non-zero determinant at x, we’re golden (IN WHAT

SENSE??).

Required assumption. ξ(·, ·) is differentiable at (p, w).

7.3 Economies7.3.1 Notation

Economy : E = ((Xi, Si)i∈I, (Yj)j∈J) .Private ownership economy : E = ((Xi, Si, ωi)i∈I, (Yj)j∈J, (θij)i∈I,j∈J) ,

where θij is i’s share of j’s profits and ωi is i’s initial endownment.

Pure exchange private ownership economies : E = ((Xi, Si, ωi)i∈I) .

Diagram—Illustrating the sign convention. He has a picture where y+ (output from firms) be-comes x+ (income for consumers) and x− (work, etc.) from consumers becomes the input toproduction, y−.

Endownments. We may think of the ωi’s as inputs and outputs at the initial starting point ofthe economy.

Definition—Feasible allocations. A list of I + J vectors in RL, (x1, x2, ..., xL, y1, y2, ..., yL) givesan allocation in E . The allocation is feasible if:

feasible allocation :xi ∈ Xi, yj ∈ Yj ∀i, j (individually feasible)∑i∈I xi =

∑j∈J yj + ω (market balance).

We can write an allocation as ((xi)i∈I, (yj)j∈J).

30

7.3.2 Example: Edgeworth box

Case. WhenEdgeworth box : E =

((R2

+, Sa), (R2+, Sb), ω

).

Graph. The Edgeworth graph is nice because only points inside the box are feasible (in that x1a+x1b =ω1 and x2a + x2b = ω2.

7.3.3 Pareto optimal allocations

Let ((xi)i∈I, (yj)j∈J) be an allocation.

Definition The allocation ((xi)i∈I, (yj)j∈J) dominates ((xi)i∈I, (yj)j∈J) if

Si(xi) ≥ Si(xi)∀i ∈ I and ∃i′ ∈ I : Si′(xi′) > Si(xi).

Edgeworth box. If the indifference curves in the Edgeworth box are tangential to each other at somepoint then that point is pareto optimal (there are no points in “the cigar”.

Contract cuve. Connecting all the pareto optimal points, you get a curve through the Edgeworth box(which contains (0,0) and (ω1, ω2) so long as F2’ is satisfied; monotonicity) called the contractcurve.

7.3.4 Proposition concerning Assumption F2’ (variant of F2)

Recall—F2’. Let x1, x2 ∈ X with x1 ≥ x2 and x1 6= x2 then S(x1) > S(x2) (strict inequality).

Proposition. (section 5.2.3, NotesOpt) If the consumer (X,S) satisfies assumptions F1–F3, then(X,S) satisfies F2’.

8 2012-09-268.1 Last time8.1.1 Proposition 5 — characterizing the solution

Point. We wanted to show that the solution to the problem can be found as the solution to a systemof equations.

Considered. We considered gradu(x)− λp = 0w − px = 0.

as a system of equations in x and λ (i.e. given (p, w)), as F (x, λ) = 0.

Proof—idea. The idea for the proof was to take some x and consider the “line” between x and xdefined as x+ t(x− x), t ∈ [0; 1]. Then we found the newton derivative and let t→ 0 and foundthe derivative and showed that it was positive.

31

8.1.2 Economies

Definitions. We defined a lot of stuff.

Utility. Note that utility is Si(xi) rather than ui.

Feasible allocations. Two conditions:

(i) xi ∈ Xi and yj ∈ Yj(ii)

∑i xi =

∑j yj + ω (balanced)

Ad (ii) This is included as the definition of an allocation earlier on.

Edgeworth. Looked at the good ole box.

Pareto domination. Feasible allocation such that Si(xi) ≥ Si(xi) for all i ∈ I and Si(xi) ≥ Si(xi)for some i.

Not PO? Then we’re “wasting ressources” (given that monotonicity holds).

8.2 Economies cont’d8.2.1 Koopman diagram

Direction. Both consumer and production goes outwards (i.e. the diagram turns the right way fromthe point of view of both of them.

Consumer. Represented by indifference curves.

Firm. Represented by the set

Y + ω =x ∈ RL|x = y + ω, y ∈ Y

.

Note that the set Y usually starts in (0,0) so move out in the direction of the ω vector and fromthat point and downwards, it’s a line parallel to the 2nd axis. From that point upwards, it curvesconcavely to the right (but forming a convex set:)

8.2.2 Assumption F2’ — Strong monotonicity.

Ass F2’. Strengthens: x1 ≥ x2 and x1 6= x2 (greater than in some coordinate), then S(x1) > S(x2).

8.2.3 Proposition [2.3.2].1 in NotesOpt.

Proposition. If the consumer satisfies assumptions F1, F2 and F3, then he also satisfies F2’.

Note. If consumers a, b satisfy F2’, then the southwest and northeast corners are in the contractcurve.

8.2.4 Proposition [2.3.2].2

Let ((xi)i∈I, (yj)j∈J) be a pareto optimal allocation for E = (Xi, Si)i∈I, (Yj)j∈J, ω, where consumers(Xi, Si) satisfy F1 and F2’ (the strong variant).

Proposition. Then∑j∈J yj ∈

∑j∈J Yj is an efficient production in

∑j∈J Yj .

32

8.2.5 Section 2.3.3: Fairness

Definition—Fairness. An allocation (xi)i∈I (each in RL+) is said to be fair (or envy free) if allconsumers i ∈ I prefer no other bundle of any consumer j ∈ I, i.e.

(xi)i∈I is fair ⇐⇒ Si(xi) ≥ Si(xj)∀j ∈ I.

Example. Supposexi = 1

|I|ω,

then (xi)i∈I is fair. However, it might be pareto inefficient.

IMPORTANT! Suppose xj ∈ Xj but xj /∈ Xi. Then i can’t be envious of j because i could neverconsume what j is... for example, men can’t envy women supplying lesbian porn! Their utilityfunction isn’t defined on that output.

8.3 Section 3.1: Social utility functionsGoal. Define a function no (xi)i∈I that maps into R (utility for the social planner).

(xi)i∈I −→(S1(x1), ..., S|I|(x|I|)

)−→ U

(S1(x1), ..., S|I|(x|I|)

)−→ R.

Examples.

Weighted sum. U(· · · ) =∑i∈I αiSi(xi).

Min. U(· · · ) = mini∈I Si(xi).

Arrow’s impossibility thm. Explains that if you have a few basic requirements for U then it’simpossible to find one!

8.4 Arrow’s impossibility theorem8.4.1 The problem

Setup. Assume that there are only three individuals and three alternatives (need at least three).

Rankings: Below, we have the three persons (1,2,3) and how they would choose between the threealternatives that the society is faced with. Each has their own preferred scenario / choice / policy(a, b or c).

1 2 3 Social planner1st choice b c a ?

2nd a b c ?3rd c a b ?

8.4.2 Assumptions on the Social Welfare Function (SWF)

A1. Unrestricted domain. The SWF should have unrestricted domain.

A2. Pareto compatibility. If in a scheme (preference profiles; set of utility functions?) everyonehas x before y, then the SWF should map the scheme into a ranking where x is before y.

A3. Independence of irrelevant (or binarity). Consider the following three schemes:

33

Scheme I.

1 2 31st choice a b a

2nd b b3rd a

Scheme II.

1 2 31st choice a b a

2nd a3rd b b

Scheme III.

1 2 31st choice b b b

2nd a3rd a a

Suppose. Let I and II be two schemes and x, y two alternatives. x, y ∈ a, b, c and x 6= y.Assume that the x-y pattern in I and II are the same.

The assumption—IIA. Then scheme I is mapped to a ranking with x before y if and only ifscheme II is mapped to a ranking with x before y.

Intuition. We have removed the c option from the above matrices to focus on how a and b areplaced relative to each other alone.

A4. No dictator. For each i ∈ I = 1, 2, 3 there is some scheme where society’s ranking (the SWF)disagrees with i’s ranking.

Note. We can’t just say that we consider the scheme where Si = S∀i cause we want to considerall possible schemes.

8.4.3 Arrow’s Impossibility Theorem

Proposition. There is no Social Welfare Function (SWF) that satisfies A1–A4 if the number ofalternatives is greater than 2.

Alternative formulation. If we have a SWF that satisfies A1–A3, then it’s a dictator.

2 individuals. Then it is much simpler to show (?)

IMPORTANT!

Allocation vs own bundle. It only holds for these discrete situations where each individualmust have preferences over the allocation, not just over their own bundle!!

Discrete vs continuous. We have only finitely many here.

Note—Voting. If you are voting, then there might be an incentive to vote differently to tilt thechoice of society (unless there’s a dictator).

8.5 Market Equilibrium (in MWG: Equilibrium with Transfers)8.5.1 Theorem [4.2].1

Assumptions. Consider an economy E = ((Xi, Si)i∈I, (Yj)j∈J, ω) where the consumers each satisfyassumptions F1 and F2. Consider the allocation(

(xi)i∈I, (yj)j∈J, p, (Ri)i∈I),

where Ri is individual i’s income.

34

Prices. Prices are p ∈ RL+ \ 0, i.e. p 6= 0. This is because F2 gives us kind of non-satiation sowe could (?) get u =∞...WAIT! Don’t we then need F2’?No. We just say that prices can’t all be zero...Moreover—Balancing. Balancing will also cause something to break down if a price is

zero, I think?

Definition. We say that such an allocation is a market equilibrium if

1. (cons). For i ∈ I, xi solves the consumer’s problem given prices and income, (p,Ri).(i.e. all consumers have chosen the best consumption in B(p,Ri), their budget set)

2. (prod). For j ∈ J, yj solves producer j’s problem given prices, p.3. (comp).

((xi)i∈I, (yj)j∈J

)is a feasible allocation.

Theorem [4.2].1. Then the market equilibrium allocation,((xi)i∈I, (yj)j∈J

), is a pareto optimal

allocation.

Proof—Technique. Assume that we have a pareto dominating allocation — then it’s too expensive.

Proof. By F1 and F2, p ∈ RL+ \ 0 and

p · xi = Ri ∀i ∈ I.

2beCont’d... next time!

9 2012-10-02Last time. We proved proposition 2.

Assumes p ∈ RL++.Prop 2(b). This says that if u satisfies F2, then each solution to the CP belongs to the budget

hyperplane.

Wish: We want to extend the proof to hold for p ∈ RL+ \ 0.

IMPORTANT—Extending other propositions. It turns out that a lot of the propositions canbe extended to hold for p ∈ RL+ \ 0 and not just p ∈ RL++ as we actually proved them for.

Affected: Propositions 2(b), 4(a), 5(b).

WARNING! Note that if one good has a the price pl = 0, then there will often not exist a solutionto the CP!!!!! But all (or almost) of the theorems say if there is a solution to the CP thensomething...

9.1 Propositions9.1.1 Proposition 4

Assume. Let (X,u) be a consumer satisfying F1 and let p ∈ RL++.

Proposition.

(a) If ...

See slides...

35

9.1.2 Proposition 5

Assumptions. (X,u) satisfies F1, F2 and DIfferentiability and (p, w) ∈ D (positive prices and w ≥min pw).

Says. Then If x is a solution to the CP then there is λ > 0 such that it comes from Lagrangian stuff.Also if F3 is satisfied and x satisfies Lagrangians, then it’s a solution to the CP.

Extension. Also holds for p ∈ RL+ \ 0.

9.2 SomethingBasic economy. There is a difference between the notes and MWG — In the notes they didn’t care

about who had what of the ressources; In MWG, that should be specified as a part of theeconomy.

MWG. In MWG,E = ((Si, Xi)i∈I, (Yj)j∈J, (ωi)i∈I) .

Notes. In the notes,E = ((Si, Xi)i∈I, (Yj)j∈J, ω) .

Comparison. The notes just use ω =∑i ωi.

Equilibrium with transfers. Then you have (xi), (yj), p, (ti), where∑i∈I ti = 0 (transfers).

Wealth of cons. i. Is given as p · ωi + ti.

9.3 The first theorem of welfare economics.9.3.1 Theorem [4.2].1

Assume. Consider an economyE = ((Si, Xi)i∈I, (Yj)j∈J, ω) .

Assume that consumers satisfy F1 and F2 and let((x0i ), (y0

j ), p, (Ri))

be a market equilibrium.

Theorem. Then((x0i ), (y0

j ), p, (Ri))is a pareto optimal allocation.

Proof. By F1 and F2, p ∈ RL+ \ 0. For i ∈ I, px0i = Ri (by proposition 2(b)).

Assume contradiction. Now we assume that there exists an allocation,((x′i), (y′j)

), which is

(a) feasible and (b) pareto dominates((x0i ), (y0

j )).

WoLoG. Assume that it is consumer 1 that is better off, i.e.

S1(x′1) > S1(x01).

Then it must hold thatpx′1 > px0

1,

because otherwise it violates prop 2(b) (is it? the budget hyperplane one anyways)

36

Also. It must also hold thatSi(x′i) ≥ Si(x0

i ).

Well if it gives at least as much it must cost at least as much by expenditure minimization(prop 4(a) I think). That is,

px′i ≥ px0i , i ∈ I \ 1.

Now—Producers.py0j ≥ py′j

due to profit maximization. Rewrite this to get

−py0j ≤ −py′j , j ∈ J.

Add up inequalities.

px′1 +∑

i∈I\1

px′i +∑j∈J−py

′

j > px01 +

∑i∈I\1

px0i +

∑j∈J−py0

j∑i∈I

p · x′i −∑j∈J

p · y′j >∑i∈I

px0i −

∑j∈J

py0j

⇔ p

∑i∈I

x′i −∑j∈J

y′j

> p

∑i∈I

x0i −

∑j∈J

y0j

Now use the fact that ∑

i

x′i =∑j

y′j + ω,

because it was a feasible allocation by assumption.

⇔ pω′ > pω0,

but since they are from the same economy, the total amount of ressources was the same

ω′ = ω0 = ω,

so⇔ pω > pω,

which is a contradiction.

Note.

Assumptions needed. We need F2 to exclude thick indifference curves... otherwise an EQ notbe PO because one consumer might not’ve cared if we took something from him.

9.4 The Second Theorem of Welfare Economics: Differentiable Approach9.4.1 Lemma [5.2].1

Let((xi), (yj)

)be a feasible allocation for the economy

E = ((Si, Xi)i∈I, (Yj)j∈J, ω) ,

with utility imputation (si) = (Si(xi)).

Lemma.

37

(a)((xi), (yj)

)is a pareto optimal allocation for E if and only if ((xi), y) with y =

∑j∈J yj is a

pareto optimal allocation for the economy

E = ((Si, Xi)i∈I, Y, ω) , Y ≡∑j∈J

Yj .

(b) Let the consumers in E satisfy F1–F4 and differentiability and assume that Y satisfiesassumption P1 and that Y can be represented by a production function F : RL → R.(A).(B). ((xi), y) solves the problem

maxS1(x1), s.t. xi ∈ Xi, i ∈ I and(i) : Si(xi) = si, i ∈ I \ 1,(ii) : F (y) = 0,

(iii) :∑i∈I

xi = y + ω.

Cool. (A) implies (B). Furthermore, if for i ∈ I, xi ∈ intXi, then (B) implies (A).

Notes.

(a) Means that we can confine our study to economies with only one producer.

9.4.2 Theorem [5.2].1

Theorem. If for i ∈ I consumer i satisfies F1–F3, Differentiability and xi ∈ intXi and the total pro-duction set Y ≡

∑j Yj , satisfies assumption P1 and Y is represented by the production function

F : RL → R, then there are prices p ∈ RL++ and wealth (Ri)i∈I such that((xi), (yj), p, (Ri)

)is a

market EQ.

9.4.3 Holy fuck in the structure...

9.4.4 Lagrangian

Let’s go. We’re solving.

maxS1(x1), s.t. xi ∈ Xi, i ∈ I and(i) : Si(xi) = si, i ∈ I \ 1,(ii) : F (y) = 0,

(iii) :∑i∈I

xi = y + ω.

Lagrangian. Becomes

L(x1, ..., xI , y, β, γ, λ) = S1(x1) +I∑i=2

γi [Si(xi)− si] + βF (y)− λ(

I∑i=1

xi − y − ω

),

where λ ∈ RL so that we have a λl for each xil, yl, ωl touple.Note that whether you have minus or plus in front of the constraints doesn’t matter (absorbedinto mutipliers)

Suppose (xi), (y) is a solution to the problem.

38

Then there are β, (γi)Ii=2 and λ ≡ (λ1, ..., λL) such that all aprtial derivatives of L evaluated at the“bar” values are zero.

Assymetry. We don’t have a multiplier for consumer i = 1.

Remedy. We just consider γ1 := 1.

Now—Consider. Consider the γ-derivative for each consumer i and obtain the gradient (in RL-space)evaluated at x,

grad γiSi(xi)− λ = 0, i ∈ IβgradF (y)− λ = 0.

Differentiability says that at least one Si(xi) has strictly positive derivative and by F2 none arenegative (?) so λ must be positive (?)

Understanding... maybe he’s taking the gradient wrt. (x1, ..., xL)??

Choose. Now choose λ as prices and give consumer i the income λxi =: Ri (so that the consumercan buy the PO solution).

Use previous results. Since xi is the solution to a system of the type we looked on in proposition5 (??), we know that it is a solution to the CP.

Moreover. When y satisfies the marginal relations, then we can draw on the results about productionto say that y is a solution to the producer problem, i.e. λy ≥ λy for any y ∈ Y .

Decentralization. Now we can draw upon the result about decentralization that y =∑j yj .

Since y is a solution to the aggregate problem, the theorem of decentralization gives us thatyj must be solution to each of the small firm j problems.

END OF PROOF! We found prices for consumers and firms and showed that given those prices,everyone was optimizing.

Interpretation of λl? Well, consider

∂L∂ωl

= λl, and∂L∂ωh

= λh,

so λl/λh is the economy’s relative valuation of the two goods. Lagrange multipliers indicate thevalue of loosening their respective restrictions.

9.5 Sum of two “no-worse-than” sets — The Scitovsky contourMotivation—Graphical. It’s quite hard to grasp, but I think he’s moving around the (ω1, ω2)-point

of the graph (i.e. the origin for consumer 2). Then the same indifference curve for consumertwo will be tangential to the same indifference curve for consumer one (which has not moved atall) at a different point. Then we cet a curve of points, ω(h)

1 , ω(h)2 h, which moves the economy

around in a way such that utility is kept constant for each consumer.

Mathematically. Let the two utility-levels, s1, s2 ∈ R, be given and consider the upper contour setsfor each consumer,

A(s1) = x1 ∈ X1|S1(x1) ≥ s1 ,A(s2) = x2 ∈ X2|S2(x2) ≥ s2 .

39

Ifω ∈ A(s1) +A(s2),

then we can writeω = x1 + x2, x1 ∈ A(s1), x2 ∈ A(s2).

Moreover, if (x1, x2) is a pareto optimal allocation in E = ((Xi, Si)i=1,2, ω) giving utility levelsS1(x1) = s1, S2(x2) = s2, then their sum is a boundary point, i.e. ω is a boundary point of theset A(s1) +A(s2).

Production? We have to do something about production too.

Somethingsomething...

Convexity. Then we have a convex set with a point, ω, on the boundary of that set. Then we canfind a hyperplane supporting it by that point (i.e. tangential there). That is prices???

10 2012-10-03Next week. Lengthy theorem on the existence of Walras equilibria.

Last time—Prices and Lagrange. Last time we made a constructive argument where we foundthe prices as the lagrange multipliers.

Today: Welfare theorem.

10.1 Towards the theorem10.1.1 Mathematical background

First. Consider p 6= 0, p : RL → R defined by z 7→ p · z (scalar product.

Hyperplane? Recall that pz = w defines a hyperplane when z ∈ RL and w ∈ R. From that wecan define an upper and a lower half-space of the space.

Optimization. Will be the same whether it’s maximum or minimum, so let’s choose one.

Minimization. Consider the problemmin

c∈C1+C2+C3p · c,

where C1, C2, C3 ⊂ RL.

Suppose. Suppose that c is the minimizer, pc ≤ pc for all c ∈ C1 +C2 +C3. By definition, thismeans that

c = c1 + c2 + c3, ci ∈ Ci, i = 1, 2, 3.

Q. What can be said about ci, i = 1, 2, 3?A. Well, we know that

∀c1 ∈ C1 : p (c1 + c2 + c3) ≤ p (c1 + c2 + c3) .

(it holds for all c ∈∑Ci so in particular for that combination.

⇒ pc1 ≤ pc1.

Conclusion. When we’ve solved the minimization over the summation set,∑3i=1 Ci, we have

in particular solved in for any ci ∈ Ci considered by itself.

40

10.2 Separating Hyperplane TheoremLet. Let Z ⊂ RL be a non-empty convex set and let a /∈ Z (where Z is the closure of Z).

Proposition. Then there is p ∈ RL, p 6= 0, such that

pz ≥ pa, for all z ∈ Z.

Intuition.

Hyperplanes. For any convex set Z and any point, a, outside of it, there is a hyperplanethrough a that has the entire (closure of) Z above it.

Proof.

Let a ∈ RL be given so that a /∈ Z..Translation to different set. We want to instead consider the set Z − a and note that

Z − a ≡x ∈ RL|x = z − a, z ∈ Z

.

We will first prove that for any z ∈ Z−a, there exists p so that pz > 0 and then concludethat then in particular, for z ∈ Z, note that z−a ∈ Z−a and so p·(z−a) > 0⇔ p·z > p·a.First we need. Not that 0 ∈ Z − a ⇔ a ∈ Z so 0 /∈∈ Z − a ⇔ a /∈ Z.Then we can consider the question of whether or not 0 ∈ Z. We have so to speak translated

the problem to considering 0 instead of a.Secondly: Note that since any continuous function on a closed set has a maximum and a

minimum, we have that there exists a solution to

min ‖q‖ , s.t. q ∈ Z − a ,

because Z − a is a closed set.Now—idea. Since 0 /∈ Z, we can consider the closed point in Z to the origin, 0, and draw the

arrow from 0 to that point. That vector defines the hyperplane we are looking for.Let p be the solution,

p = arg minp∈Z−a

‖p‖ .

The vector p has the interpretation as the point in Z − a closest to the origin.To prove: p · z ≥ p · 0 = 0 for any z ∈ Z − a.Let z ∈ Z − a be given and consider the vector z which is an arrow from 0 to the point z in

the set.Then consider the vector from p to z, call it z′

z′ = p+ λ(z − p), λ ∈ [0; 1]= (1− λ)p+ λx

and note trivially thatz′ ∈ Z − a

because it is a convex set!!

41

Now: Use the fact that p solved minZ−a ‖q‖. Then note that

p · p = ‖p‖2

and also consider

‖z′‖2 = ‖p+ λ(z − p)‖2

= [p+ λ(z − p)] · [p+ λ(z − p)]≥ ‖p‖2

,

because p was defined as the minimizer.Expand.

‖z′‖ = p · p+ 2λp · (z − p) + λ2(z − p) · (z − p) ≥ p · p⇔ 2λp · (z − p) + λ2(z − p) · (z − p) ≥ 0,

for λ ∈ [0; 1]. Restrict to the particular intervalll λ ∈ (0; 1], where

2p · (z − p) + λ(z − p) · (z − p) ≥ 0.

Since this holds for all λ ∈ (0; 1], by letting λ → 0, it must hold that (you can prove it bycontradiction)

2p · (z − p) ≥ 0⇔ p · (z − p) ≥ 0

⇔ p · z ≥ p · p > 0, z ∈ Z − a,

which proves part of what we needed.Still missing to prove. p · z ≥ p · a for z ∈ Z. Let z ∈ Z. Then z − a ∈ Z − a and then use

the result we just proved to note that

p · (z − a) > 0⇔ p · z > p · a.

which is what we were required to prove.

Comment—Closure. We assumed z ∈ Z, we need z ∈ Z and in particular boundary points (??) .

10.2.1 Extension — boundary piont.

Let. Let Z be a non-empty convex set and a boundary point of Z.

Recall—Definition: boundary. A boundary point — for all r > 0 and all z ∈ Z \ Z, we have thatB(z, r) has points in common with Z and some points in common with ZC .

Proposition. Then there is p ∈ RL, p 6= 0 such that

p · z ≥ p · a, for z ∈ Z.

Intuition. a is a minimizer of the linear function z 7→ pz on Z.

Proof—In the notes.

Idea. Take a sequence outside of Z and let them converge to ??? nad prices increase and ???and then the inequality emerges in the limit case.

42

10.2.2 Addendum — Difference sets and separating hyperplanes

Suppose Z1 ∩ Z2 = ∅. Then 0 /∈ Z1 − Z2.

Proposition. Consider some z1 − z2 ∈ Z1 − Z2. Then there is p such that

p(z1 − z2) ≥ 0⇔ pz1 ≥ pz2.

10.3 The Second Theorem of Welfare Economics10.3.1 Theorem

Consider. Consider an economy E = ((Xi, Si)i∈I, (Yj)j∈J, ω) where the consumers satisfy assumptionsF1, F2, F3 and each producer satisfies P1.Let

((xi)i∈I, (yj)j∈J

)be a Pareto Optimal allocation for E with x0

i ∈ intXi all i.

Proposition. Then there is a price system p 6= 0, p ∈ RL+ and incomes (Ri)i∈I such that((xi)i∈I, (yj)j∈J, p, (Ri)i∈I

)is a market equilibrium.

Interpretation. If, in some magic way, we have found a pareto optimal allocation, then it can besupported as a market equilibrium.

Note—Non-negative prices. It is correct that p ∈ RL+ but it is only proved for p ∈ RL++ (is it??something about this) but we proved last time that F1–F3 ⇒ F2’ (strong; more of just one isbetter).

10.3.2 Lemma required to be proven first

Introduce stuff. Denote si = Si(xi) for all i ∈ I.Define

Ai(si) ≡ x ∈ Xi|Si(x) ≥ si ≡ upper contour set.

DefineZ ≡

∑i∈I

Ai(si)−∑j∈J

Yj .

Note—Convexity. Z is convex because all the summation-term-sets are.

To prove: ω is a boundary point for Z.

Q. Why is the Z set relevant to us?A.

Suppose we have z ∈ Z, i.e. z =∑i xi −

∑j yj , whre xi gives at least the utility of si for

i.⇔∑i

xi =∑j

yj + z.

Look. This looks like a market economy relation where the initial endownment is z.So, interpretation: z ∈ Z if an economy with the initial endownment z can achieve at

least the utilities (si)i∈I.We have. ω ∈ Z because it was a feasible allocation and hence satisfied the market balance

equation.

43

Assume contradiction. Assume that ω ∈ intZ.Definition of interior. Then for any ε > 0 there is r > 0, the sphere B(ω, r) ⊂ Z.Alternative def. Introduce ∆ ∈ RL++ very small and consider ω−∆, if ∆ has entries small

enough, ω −∆ ∈ Z.Then. But

ω −∆ ∈ Z ⇒ ω −∆ =∑i∈I

xi −∑j∈J

yj

⇔ (x1 + ∆) +∑

i∈I\1

xi −∑j∈J

yj − ω = 0.

Since ∆ ∈ RL++, x1 + ∆ ∈ X1 and there is nothing with a budget set here.Note. Now note that consumer 1 is strictly better off (he has strictly more) and all other

consumers and all firms are equally well off. Also, the ressource constraint (what wasthe name) still holds.

Conclusion (contradiction). We have found a new allocation that pareto dominates theoriginal one. This is a contradiction with the assumption that the original setup waspareto optimal.

10.3.3 Proof of the Second Welfare Theorem of Economics

Proof.

Step 1. Define a suitable convex set,

Z ≡∑i

Ai(si)−∑j

Yj ,

and then the lemma from above, showing that

ω ∈ boundary(Z).

Step 2. Invoke the Supporting Hyperplane Theorem. [result that we did not prove]

We know. ω ∈ boundary(Z).Thus. There is p ∈ RL, p 6= 0, such that p · z ≥ p · ω for z ∈ Z.Intuition. Look at the linear functional z 7→ p · z — ω is a minimizer for that.

Step 3. Show that the prices are non-negative

Proof—Contradiction. Assume the contradiction; that p1 < 0. But ω+(a, 0, ..., 0) with a > 0belongs to Z (more of the first good; suppose we give to consumer 1; since F1 it remains inX1 and by F2,F3⇒F2’ he likes it more). Hence, ω + ae1 ∈ Z.But from the above it must hold that p · (ω + ae1) ≥ p · ω.But rewriting this, p1a ≥ 0 but by the contradictory assumption, p1 < 0, whereby we arriveat the contradiction.

Step 4. Show that yj is a solution to producer j’s problem.

Proof.We know: p · ω ≤ p · z, for all z ∈ Z ≡

∑iAi(si) +

∑j(−Yj).

44

Recall—Small cute lemma. Suppose

c = arg minc∈C1+C2

p · c⇒ c1 = arg minc1∈C1

p · c1.

Then. Write ω =∑i xi +

∑j(−yj). But then it follows from the cute lemma that

p · (−yj) ≤ p · (−yj), ∀ − yj ∈ −Yj⇔ p · (yj) ≥ p · (yj), ∀yj ∈ Yj

⇔ yj is profit maximizing for firm j.

Step 5. Show that xi is a solution to consumer i’s problem given prices p and income pxi (he mustbe able to buy the preferred thing).

First—Proposition. xi is an expenditure minimizer.Proof. Consider, again ω =

∑i xi +

∑j(−yj). Since ω = arg min p · z for z ∈ Z, in

particularp · xi ≤ p · xi ∀xi ∈ Xi

⇔ xi is an expenditure minimizer to achieve utility si given prices p.Second—Proposition.

Motivation. What if xi was the very cheapest thing you could find? Then we wouldn’tbe done because you just couldn’t afford ... ??? WTF???

Give. Give consumer 1 income R1 := p · x1 .Then. xi ∈ intXi implies that x1 solves UMP at income (p,R1).

Notes.

The set Z. Very important to understand.Exam question. One possible exam question could be: “Prove that ω /∈ intZ”.MWG proof. They use a separating hyperplane theorem instead. Much easier to remember

and thus nicer but (???)(something with the diagram where the convex set Y + ω should be separated from A(s)(or whatever...????)... the Robinson Crusoe picture anyways.

11 2012-10-1011.1 Last time (I wasn’t there!!)Definition—Walras EQ.

Helpful: Consider producers first.⇒ then put that input into the consumers’ thingThen check for balanced.

(a) yj solves PP(b) ...

Walras in pure EQ.

Now: Initial endownment, ωi, that gives the income!

45

Normalization of prices—Unit simplex. Prices can be normalized to be on the unit simplex justas well as we could normalize prices with the unit circle (or unit sphere in RL).

Properties of total excess demand.

Proposition 6. There are (a)–(e) properties — (e) is very difficult to prove and not proven inMas-Colell.

Implication of Walras’ law. If you find a price the equilibrates the first L− 1 marketse, thenit also satisfies the L’th market. Holds in market economies as well.

(something) shows that almost any function can be an excess demand function.

Figure—Walras EQ L = 2. notice that −ω1 has subscript 1 — something something it doesn’t inthe notes.

Continuum of EQ. Non-typical.

Perturbing. If you have an economy with some endownment then if you change endownment a little,you can easily get to a case with an odd-number of “off-zeros”???

Brouwer’s Fixed Point Theorem.

Theorem 9. Let A ⊂ RL be non-empty convex and compact and g : A → A be continuous.Then f has a fixed point. That is, there is x ∈ A such that

f(x) = x.

Applying—Prices. We need to consider prices in the unit simplex,

∆ ≡p ∈ RL

∣∣∣∣∣∑`∈L

p` = 1.

Simplex—Hyperplane interpretation. We can interpret ∆ (the unit simplex in RL) asthe intersection between RL+ and a particular hyperplane.

Modified demand. We say; In EQ, the consumers with Xi = RL they can’t choose x >> ω...

Trying again. Consider the x = 2ω where ω is the total endownment of the economy.Restrict. Then restrict attention to the kunstige set X := 0 < x ≤ ω (in vector ordering).

Lemma 8. Let p ∈ RL+ \ 0.

11.2 Long proof — Theorem 10 — Existence of Walras EQ11.2.1 Theorem 10

Let Ecpr = (Xi, ui, ωi)i∈I.

Method. Find a suitable mapping so that Brouwer’s fixed point thm. can be applied. Then showthat the result is Walras EQ.⇒ Idea: Raise price to remove excess demand.

Proof.

Let g : ∆→ ∆ withζ(p) = ζ+(p)− ζ−(p)

where the sup + or - means that we split in positive and negative prices.

46

A failed attempt at g(p).gbad attempt(p) = p+ ζ+(p).

Then we take ∑l

pl +∑l

ζ+l (p) = 1 +

∑l

ζ+l (p),

where∑pl = 1 because p ∈ ∆. But then clearly gbad attempt(p) /∈ ∆ (it has too big norm).

Now it’s clear that we should define g component-wise to be

gl(p) = pl + ζ+l (p)

1 +∑l ζ

+l (p)

Property of g. Now g satisfies thatg : ∆→ ∆,

and clearly g is continuous because it’s entries, gl, are continuous (because excess demand iscontinuous (because demand is)).

Brouwer. By Brouwer,

Brouwer’s FP thm.⇒ ∃p∗ ∈ ∆ : gl(p∗) = p∗l , ∀l ∈ L.

Use def. From it’s very definition, we have

gl(p∗) ≡ g

p∗1...p∗L

= p∗l

⇔p∗l + ζ+

l (p∗)1 +

∑l ζ

+l (p∗)

= p∗l

⇔ p∗l + ζ+l (p∗) =

[1 +

∑l

ζ+l (p∗)

]p∗l .

Done. This was the first step. It will turn out that this mapping is very neat.

NEXT STEP—Prove p∗ ∈ RL++.

Create a bounded hyper rectangle. Let’s consider the rectangle that in R2 would be

x ∈ R2|x < 2ω.

Why 2? Could be arbitraryx < kω, k > 1.

Assume the contradiction p∗1 = 0. Then

ζ+1 (p∗) = |I| · (demand from the I consumers of good 1)− total initial of 1

= I · 2ω1 − ω1 > 0.

We know.

p∗l + ζ+l (p∗) =

[1 +

∑l

ζ+l (p∗)

]p∗l .

47

Here[1 +

∑l ζ

+l (p∗)

]p∗l = 0 because p∗l = 0.

BUT! We know that ζ+l (p∗) > 0 from I · 2ω1 − ω1 > 0.

Contradiction.Conclusion. p∗l > 0.

Done.

New proposition.∑l ζ

+l (p∗) = 0.

Assume contradiction that∑l ζ

+l (p∗) > 0.

Look at the result

p∗l + ζ+l (p∗) =

[1 +

∑l

ζ+l (p∗)

]p∗l .

Together, this means that [1 +

∑l

ζ+l (p∗)

]p∗l > p∗l

⇔ p∗l + ζ+l (p∗) > p∗l

⇔ ζ+l (p∗) > 0.

But this was for an arbitrary l. In other words,∑l

ζ+l (p∗) > 0⇒ ζ+

l (p∗) > 0∀l ∈ L.

Now think about where we got the p∗ from.

Recall that by conventionζ(p) = ζ+(p)− ζ−(p),

soζ+l (p) > 0∀p⇔ ζl(p) = ζ+

l (p)∀!!!

Now—Walras’ law. Walras’ law gives that

0 = p∗ζ(p∗).

But we know that0 = p∗ζ(p∗) = p∗ζ+(p∗)

but ζ+l (p∗) > 0∀l and p∗l > 0∀l implies that

0 = p∗ζ(p∗) = p∗ζ+(p∗) > 0⇒ CONTRADICTION!

Conclusion. Then it must be the case that∑l ζ

+l (p∗) = 0.

Implication. This implies that ζ+l (p∗) = 0 for all l.

(if∑i ai = 0 with ai ≥ 0∀i then ai = 0)

Done.

Proposition. ζ−(p∗) = 0.

48

Walras’ law. By Walras’ law,

0 W′s L= p∗ζ(p∗) = p∗[ζ+(p∗)− ζ−(p∗)

]= p∗ ζ+(p∗)︸ ︷︷ ︸

=0L×1 by above

− p∗ζ−(p∗)

= −p∗ζ−(p∗),

Implication

0 = −p∗ζ−(p∗) ∧ p∗ >> 0⇒ ζ(p∗) = 0.

Done.

Proof done.

Oh no—Proof was for modified demand. But because of some theorem of modified demand,

ξi(p∗, p∗ωi) = ξi(p∗, p∗ωi)

and (ξi(p∗, p∗ωi), p∗

)is a Walras EQ.

QED.

11.2.2 Extra stuff

Figure in MWG.

Walras’ law. Really states that the excess demand must belong to the homogenous hyperplane(pζ(p) = 0 (scalar product)).

(see drawing)

11.3 Uniqueness of Walras Equilibrium11.3.1 Intro

When Xi = R2, points outside of the box also have interpretations even though we can never bethere...

Moreover. Contract curve might go outside the box.

What to fix? We can’t really fix it so we fix the size of the box (??) ω.

Start to end. If you “start” outside of the contract curve, then you could imagine several budgethyperplanes that go through the initial endownment, and thus they could be intersecting thecontract curve many places.

Prices work well when: You are close to the contract curve.Conversely: They don’t work well when you are far away (there are many equilibria).

The theorem. THere is a unique Walras EQ if you are already directly on the contract curve.

49

11.3.2 New assumption — Gross Substitutes

Defitinition 18. Let ζ : RL++ → RL be the excess demand function for an exchange economy. Thenζ satisfies Gross Substitutes property if

(a) ζ is differentiable(b) for l ∈ L the derivatives

∂ζl∂pk

(p) > 0, k ∈ L, k 6= l.

Since for l ∈ L ζl() is homogenous of degree 0 in price, we have

∂ζl∂p1

(p) + ...+ pL∂ζl∂pL

(p) = 0,

implying that∂ζl∂pl

(p) < 0, l ∈ L.

11.3.3 Theorem 17 — Uniqueness of W EQ

12 2012-10-23 — Uncertainty and timeNotation: Let

T ≡ 0, 1, ..., T, T1 ≡ 1, 2, ..., T

Setup: In period 0, all market activity is carried out.

Time. We write x(t), ω(t) (ω =endownment) and P (t) for t = 0, 1, ..., T . P (t) is the price at date 0for obtaining a delivery of x(t) at time t.

Budget restriction w/ universal market at date 0. Sell all your endownments, ω(t)Tt=0 at timezero and buy some goods to be delivered at different points in time. Then what you get fromselling must not be smaller than what you buy of deliveries in the future.

T∑t=0

P (t)x(t) ≤T∑t=0

P (t)ω(t).

Can be written in terms of the net trades,

T∑t=0

P (t) [x(t)− ω(t)] ≤ 0. (SIN)

12.1 Multi-periodGoal: Compare with the situation where there is a spot market at each point in time.

Dating and prices? Then at date t, prices should be denominated in day t-prices.Market for date-prices. At date 0 for date t prices, t ∈ T1 (i.e. t ∈ 1, 2, ..., T).

Price of t-crowns?

β(t) = price of date t-crowns (in date 0-crowns),β(0) = 1.

50

Budget restrictions.p(0) [x(0)− ω(0)] ≤ r(0)p(1) [x(1)− ω(1)] ≤ r(1)

...p(T ) [x(T )− ω(T )] ≤ r(T )

T∑t=0

β(t)r(t) = 0

(SEQ)

Here, (??) is the budget restriction with one consumer but sequential periods. r(t) ≷ 0 dependingon whether he wants to consume more or less than his endownment in that period.

Last equation:

cost of your future must be zero :T∑t=0

β(t)r(t) = 0.

Hence, if∑Tt=1 β(t)r(t), then you must pay for it by having r(t) < 0.

Something-something. With just one consumer, we can disregard the problem of taking ex-pectations over future prices... with many, you might assume that people had rationalexpectations and solved the full model but then there’d be no guarentee that the EQ priceswould be unique!

12.1.1 Proposition 2.3.A

Motivation. With (SEQ) it seems that there are too many goods being traded all the time... can wemaybe somehow reduce to (SIN)?

Let p ∈ RL++ (spot prices) and the prices for delivery of t-crowns should be β(t)t∈T, β(0) = 1 andβ(t) > 0.

Define. What x ∈ RL that the consumer can buy with universal markets at date 0?

A =x ∈ RL|x satisfies (SIN) with P (t) = β(t)p(t)

.

Define. Another set...B =

x ∈ RL|∃r ∈ RT+1 : (x, r) satisfies (SEQ)

.

Proposition. A = B.

Proof part 1: B ⊂ A. Let x ∈ B. Hence, x satisfies the budget restrictions in (SEQ), i.e. p(t)[x(t)−ω(t)] ≤ r(t) and β satisfies

∑Tt=0 β(t)r(t) = 0.

Now multiply by β(t) in each of the (SEQ) equations. Then we get the system of equations

β(t)p(t)[x(t)− ω(t)] ≤ β(t)r(t), ∀t ∈ T.

Sum over time periodsT∑t=0

β(t)p(t)[x(t)− ω(t)] ≤T∑t=0

β(t)r(t) (∗)= 0,

where (∗) is given by assumption of what β satisfies.Done.

51

Proof part 2: A ⊂ B. Then we know that x satisfies the single budget restriction (SIN) with P (t) =β(t)p(t),

T∑t=0

β(t)p(t)x(t) ≤T∑t=0

β(t)p(t)ω(t).

Define as a first guess r as

r(t) ≡ β(t)p(t)[x(t)− ω(t)]β(t)

⇔ β(t)r(t) = β(t)p(t)[x(t)− ω(t)].

Sum over t to get

T∑t=0

β(t)r(t) =T∑t=0

β(t)p(t)[x(t)− ω(t)]

=T∑t=0

P (t)[x(t)− ω(t)]

≤ 0,

by assumption, meaning that∑Tt=0 β(t)r(t) ≤ 0.

Hmm this is not quite what we wanted.Define the first period r as

r(0) = r(0)−T∑t=0

β(t)r(t),

where we know from the inequality above that

r(0) ≥ r(0).

Future periods are defined asr(t) = r(t), t ∈ T \ 0.

Then by construction,T∑t=0

β(t)r(t) = 0

Done.

Intuition. There is an asset that can be bought at time 0 (an “Arrow-Debreu security”) which paysone t-crown at time t. It costs (???) crowns at period 0.

12.2 UncertaintyConsumption plans. In this setting, the consumer will have to decide on his plan (or “strategy”)

which is to say what he will do in any state of the world that he might find himself in.

Note: Difference between planned consumption (assigns a consumption to each state of theworld) and realized consumption (what he ends up consuming).

Assets. Assets may have different payoffs in different states of the world.

52

Something. Suppose there are J contracts, each paying a particular dividend in each period, asspecified by their dividen vector, vj ∈ RT . A portfolio means deciding on θ ∈ RJ such that

r =

r(1)...

r(T )

=

v1(1)...

v1(T )

θ1 + ...+

vJ(1)...

vJ(T )

θJ = V θ,

where V is a T × J matrix.

Signs. that θj ≷ 0 (short or long position) and vj(t) ≷ 0 depending on whether he has to payor not (for example, a loan would have vj(1) > 0 and then vj(t) < 0, t ≥ 2.

NOTE! If the columns of V are linearly independent, then we can get any payment stream rby simply choosing the θjs in the correct way (θ := V −1r).

Linear space. Consider the linear space spanned by the colums of V ,

〈V 〉 ≡ span(v1, ..., vJ).

Complete markets.markets are complete⇔ dim (〈V 〉) = T.

Otherwise, there are income stream patterns, r, which we cannot replicate using the contracts,v1, ..., vJ .

Price of assets at time 0. let q ∈ RJ denote the prices in 0-crowns of each asset.

Extended dividend matrix. Let

W ≡(−qV

)≡

−q1 · · · −qJv1(1) · · · vJ(1)... . . . ...

v1(T ) · · · vJ(T )

.

Definition—Arbitrage. θ ∈ RJ is an arbitrage portfolio if Wθ > 0.

Intuition. At some day you get something positive but at no date do you pay anything.

Definition—Arbitrage free. W is said to be arbitrage free if

W arbitrage free iff 〈W 〉 ∩ RT+1+ = 0.

Tomorrow: Farkas’ lemma. Is a separation theorem as well.

13 2012-10-24 — Uncertainty cont’d13.1 UncertaintyResult. If a certain connection was between the spot market and the (??? market? Market for future

crowns?), then the consumer is just as well off with the simple market as trading at each pointin time.

Refinements of sets. He explains how we get to the whole sigma algebra thing / the sequence ofpartitions.

53

Example: At each day 1 and 2 it can either rain or sun.Outcomes. There are 4 — Ω = (su, su), (su, ra), (ra, su), (ra, ra).

Day 0. We know nothing, so the partition is just the whole set, F = Ω.Day 1. Here, we know the event on the first coordinate, so either we’re in (su, su), (su, ra) or

(ra, su), (ra, ra). This is the new partition.Day 2. Here, the partition is just the set of all singleton sets (we’ve realized all uncertainty).

Dividend matrix. Consider future dates. If, for example, you are promised a payment stream (1, 1)then you receive something at each date so if you pay −q1 for that, then you need q1 > 0 toavoid an arbitrage.

W =(qV

),

where V contains the payments promised by the J contracts and q is the “prices” of them atperiod 0.

Define.

W =

−q1 −q21 −11 −1

, q1 > 0, q2 < 0, .

(and some restriction must hold on one of these? q1 = −q2?Portfolio: θ ∈ RJ , specifies how much of each of the J contracts to buy.??

r = Wθ,

where we also split up r = r+ − r− (unique by taking positive and negative parts).

Span.〈V 〉 ≡ span of column vectors of V.

Hence, 〈W 〉 is the set of payment streams that you can obtain.

Definition—Complete asset markets.

dim 〈V 〉 = T iff complete asset markets.

Anddim 〈V 〉 < T iff incomplete asset markets.

Recall def—Orthogonal complement. If dim 〈V 〉 < T , then dim(OrthCompl〈V 〉) > 1 (?? orzero??) and vectors in the orthogonal complement can’t be replicated using the asset market.

Definition—Arbitrage portfolio.

Wθ ∈ RT+1+ \ 0 ⇐⇒ θ is an arbitrage portfolio.

Definition—Arbitrage free. W is arbitrage free iff no arbitrage portfolios exist. This is equiv-alent to

W arbitrage free⇐⇒ 〈W 〉 ∩ RT+1+ = 0.

54

13.2 Farkas’ LemmaLet A be a k × l matrix.

Claim. Precisely one of the following alternatives is true.

(I) There is x ∈ Rl such that Ax > 0.(II) There is y ∈ Rk++ such that yA = 0.Note. Ax > 0 means that the k × 1 vector, Ax, has no negative elements but at least one

non-zero element.Ad (I) Then x corresponds to an arbitrage portfolio.Ad (II) Think of it as (recalling y = (y1, ..., yk), row vector)

yA = 0⇔ yA(:, 1) = 0 ∧ yA(:, 2) = 0 ∧ ... ∧ yA(:, l) = 0.

Intuition for proof. Think of the separating hyperplane.

Recall. We looked at a point a and a set C and said that there existed a hyperplane defined bysome price vector p (that existed) such that a was in the lower half space and all c ∈ C werein the upper.

Proof.

Proof of (II)⇒¬(I). Hence, there is y ∈ Rk++ where yA = 0. Now consider for any x ∈ Rl, then

yAx = 0

because yA = 0. Then by the zero rule,

yAx = 0⇔ y = 0 ∨Ax = 0.

But y = 0 can’t be cause y ∈ Rk++. ... I’M MISSING SOMETHING HERE!!!

Proof of ¬(I)⇒(II). Assume that (I) is false. To show: (II) is true.

If (I) is false, then 〈A〉 ∩ Rlk+ = 0 — i.e. there is no x ∈ Rl such that Ax > 0.WARNING! Dimensions fucked up... note that Ax takes x ∈ Rl but 〈A〉 ⊂ Rk because it’s

the output of z 7→ Az.Introduce ∆. The unit simplex

∆ ≡x ∈ Rk

∣∣∣∣∣k∑i=1

xi = 1.

Note that ∆ does not intersect the closed set 〈A〉.Consider ∆− 〈A〉. If 0 ∈ ∆− 〈A〉, then A would have a vector in ∆. Hence, 0 /∈ ∆− 〈A〉.

∆− 〈A〉 is closed. Clear, because both sets are closed and a sequence in ∆− 〈A〉 requiresa sequence xn = (dn, an) so there should be a convergent sequence in each of thesets and then the limit will be in them and then they’re in the difference-set.

Separate 0 and ∆− 〈A〉. There is y ∈ Rk and α such that

y · 0 < α < y · (z − x), for z ∈ ∆, x ∈ 〈A〉 so that z − x ∈ ∆− 〈A〉

⇔ 0 < α < y · (z − x).Hence, y defines a hyperplane that separates 0 (the zero vector in Rk) from the set ∆−〈A〉.

55

Specializer: Since x = 0 ∈ Rk is a possibility, we get in particular, that0 < α < yz z ∈ ∆.

Now considereh = (0, ..., 0, 1, 0, ..., 0)

(which has a one at the h’th coordinate). Theneh ∈ ∆,

and hence,y · eh = yh > 0.

Hence,⇒ yh > 0 for all h = 1, ..., l,

⇒ y ∈ Rk++

Now rearrange ineq.y · z > y · x+ α > y · x for x ∈ 〈A〉 .

Now reduce We can violate this inequality unless some restrictions hold on y · z. In particular,if x ∈ 〈A〉 then also −x ∈ 〈A〉 because it’s a linear subspace. But then we can always violateit.

⇒ y · x = 0 for x ∈ 〈A〉 ,and we know that if y · x∀x ∈ 〈A〉, then in particular y · A(:, i) = 0 for i = 1, ..., l (that is,y is in particular orthogonal to the columns of A (they are the bases for the linear span〈A〉)). But if y is orthogonal to all the column vectors of A, then in particular, yA = 0 ∈ Rl(that’s just saying that y is orthogonal in one sentence). That is

yA = 0,which was what we wanted.

Thus (II) is true.

13.3 Existence of Discount FactorsLet a (V, q) be a market and let

W =(−qV

).

Theorem. (V, q) is arbitrage free if and only if there is β ∈ RT+1++ (positive!) with β(0) = 1 (normal-

ization) such thatβW = 0.

(hence, β should be orthogonal to each of the column vectors)

Interpretation. At prices β, all assets have value zero.

Proof.

“If”. To be done at the tutorials.

“Only if”. By assumption here, 〈W 〉∩RT+1+ = 0. Then, by Farkas’ lemma, there is β ∈ RT+1

++ suchthat βW = 0.

Thinking. The statement 〈W 〉∩RT+1+ = 0 means that for all θ ∈ 〈W 〉, the only θ such thatWθ ≥ 0

is θ = 0 = (0, ..., 0)′.

Comment. Note that Farkas’ lemma says nothing about uniqueness and often there can indeed bemany prices (even when normalized).

56

13.3.1 Corollary — (V, q) arbitrage free

Corollary. If there is (at least one) β ∈ RT+1++ so that βW = 0, and in addition rankW = T , then the

following hold:

(a) β is unique.(b) 〈W 〉 =

r ∈ RT+1|βr = 0

, that is, what you can achieve in the asset markets is exactly the

same as you could from trading in the money market.

Implication. Has all we’ve done been in vain?

My thinking. I guess (b) is the same as saying 〈W 〉 = 〈diag(r)〉 (where diag(r) is a matrix thathas (r(0), r(1), ..., r(T )) in the diagonal and zeros everywhere else...). Or alternatively that〈e1r(0), e2r(1), ..., eT+1r(T )〉.

More thinking.

Recall—Complete. A market is complete if dim 〈V 〉 = T .Complete markets. Suppose markets are complete. Then, for example the stream (1, 0, ..., 0)

costs β(1) at time zero, so

Wθ =

−β(1)

10...0

.

Incomplete markets. With incomplete markets there is no unique discount factor. He thenmentioned suppose you want a payment stream (1, 2, 0, ..., 0) and wrote

Wθ =

−β(1)− 2β(2)120...0

.

Here, we can’t create

An important note. rankW = rankV (or something??? Something with the q row being a linearcombination of the other.....)

57