Luận văn thạc sĩ Toán ứng dụng: A stronger criterion for the weak weak axiom of revealed preference

ĐẠI HỌC QUỐC GIA TP HỒ CHÍ MINHTRƯỜNG ĐẠI HỌC BÁCH KHOA

LUẬN VĂN THẠC SĨ TOÁN HỌC

TP HỒ CHÍ MINH, tháng 07 năm 2022

THIS THESIS IS COMPLETED AT

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGYAdvisor: Assoc Prof Dr Phan Thành An

Examiner 1: Dr Nguyễn Bá Thi

Examiner 2: Assoc Prof Dr Nguyễn Huy Tuấn

Master’s thesis is defended at Ho Chi Minh City University of Technology on15 July 2022.

The board of the Master’s Thesis Defense Council includes:

1 Chairman: Assoc Prof Dr Nguyễn Đình Huy.2 Secretary: Dr Huỳnh Thị Ngọc Diễm

3 Reviewer 1: Dr Nguyễn Bá Thi

4 Reviewer 2: Assoc Prof Dr Nguyễn Huy Tuấn5 Member: Dr Cao Thanh Tình

Verification of the Chairman of the Master’s Thesis Defense Council and theDean of the Faculty after the thesis being corrected (if any).

Assoc Prof Dr Nguyễn Đình Huy Assoc Prof Dr Trương Tích Thiện

THE SOCIALIST REPUBLIC OF VIET NAMIndependence - Freedom - Happiness

TASK SHEET OF MASTER’S THESIS

Name: Trần Trọng Hoàng TuấnStudent code: 2070243Date of birth: 31 March 1993Place of birth: Bạc LiêuMajor: Applied MathematicsMajor code: 8460112

- Study the stability introduced by An in "An, Phan Thanh, Stability of generalizedmonotone maps with respect to their characterizations, Optimization, Vol 55, 289–299(2006)" w.r.t the criteria introduced by Brighi.

III DATE OF TASK ASSIGNMENT: 14 February 2022IV DATE OF TASK COMPLETION: 6 June 2022V ADVISOR: Assoc Prof Dr Phan Thành An

Ho Chi Minh City, 15 July 2022

Assoc Prof Dr Phan Thành AnDr Nguyễn Tiến Dũng

DEAN OF FACULTY

Assoc Prof Dr Trương Tích Thiện

First of all, I would like to sincerely thank Ho Chi Minh City Universityof Technology for sponsoring my research in this thesis, and thank mythesis advisor, Assoc Prof Dr Phan Thành An for his careful and clearguidance.

In addition, I want to show my grateful attitude to my family membersfor the sympathy as they are always stand by me during my master thesisperiod.

Last but not least, I would like to deliver my gratitude to Divisionof Applied Mathematics, Faculty of Applied Science, Ho Chi Minh CityUniversity of Technology for giving me such valuable opportunity to applyall knowledge I have learned during my master of mathematics.

Ho Chi Minh City, 15 July 2022Author

Trần Trọng Hoàng Tuấn

i

In this thesis, the following stronger criteria (A), (B), (C), (A’) for theweak weak axiom of revealed preference introduced by Brighi in 2004 arepresented:

Criterion (A): For all x ∈ D and v ∈ Rn

ii

TÓM TẮT LUẬN VĂN

Trong bài luận văn, ta nghiên cứu các điều kiện (A), (B), (C), (A’) đượcgiới thiệu bởi Luigi Brighi trong năm 2004 cho tiên đề yếu yếu về lí thuyếtsở thích được bộc lộ như sau:

Điều kiện (A): Với mọi x ∈ D và v ∈ Rn

iii

DECLARATION OF AUTHORSHIP

I hereby declare that this thesis was carried out by myself under theguidance and supervision of Assoc Prof Dr Phan Thành An, and thatthe work contained and the results in it are true by author and have notviolated research ethics.

In addition, other comments, reviews and data used by other authors,and organizations have been acknowledged, and explicitly cited.

I will take full responsibility for any fraud detected in my thesis HoChi Minh City University of Technology is unrelated to any copyrightinfringement caused on my work (if any).

Ho Chi Minh City, 15 July 2022Author

Trần Trọng Hoàng Tuấn

iv

1.2 Demand maps and excess demand maps 4

2 THE CRITERIA FOR THE WEAK WEAK AXIOMOF REVEALED PREFERENCE 52.1 Pseudo-monotone maps 5

2.2 Quasi-monotone maps 11

2.3 Weak weak axiom of revealed preference 13

v

2.4 Weak axiom of revealed preference 17

3 THE STABILITY OF GENERALIZED MONOTONEMAPS WITH RESPECT TO SOME CRITERIA 213.1 S-quasimonotone maps 21

3.2 The stronger version of Wald’s Axiom 23

3.3 Stability with respect to criterion (A) 25

3.4 Stability with respect to criterion (A’) 27

3.5 Stability with respect to criterion (B) and (C) 28

vi

LIST OF ACRONYMS AND NOTATIONS

vii

1

The consumer or decision maker’s choices based on his preferences.

A  B will mean that A is at least as good as B.

A  B will mean that A is strictly preferred to B.

A ∼ B will mean that the decision maker is indifferent between A andB.

1.1.2Rational preference

Rational preference relation satisfies:

Completeness: we have A  B or B  A or both.Transitivity: if A  B and B  C, then A  C.

1.1.3Local nonsatiation

Preferences are locally nonsatiated onX if∀x ∈ X and∀ > 0,∃y ∈ X

such that ky − xk ≤  and y  x.

2

Applied MathematicsMaster’s Thesis1.1.4Utility maps

A map u : X → R is a utility map representing preference relation if∀x, y ∈ X, we have

x  y ⇔ u(x) ≥ u(y).

Proposition 1.1.1 Only rational preferences relations can be sented by a utility map Conversely, if X is finite, any rational preferencerelation can be represented by a utility map.

Property 1.1.2 Walras’ Law

If the preference is locally non-satiated, for any pair (p, w), and x ∈x(p, w), where, x(p, w) is optimal solution’s set of the following utilityoptimization problem:

max u(x) such that pTx ≤ w, x ∈ B(p, w) (1.1)We have

pTx = w.

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Proof We will prove by the contradiction Assuming that x ∈ x(p, w)

for pTx < w We have ∃ > 0 such that ∀y for kx − yk < , pTy < w.However, by locally non-satiation, ∃y such that p · y < w and y  x.Therefore x 6∈ x(p, w) This leads to the contradiction.

1.2 Demand maps and excess demand maps

Suppose that we have x∗ is the optimal solution of the optimizationproblem (1.1) We define the demand map at the price p is D(p) = x∗.We choose e such that pTe = w We have excess demand map is Z(p) =D(p)−e We have the following properties related to excess demand maps

Z(p), where p is the price vector:

Property 1.2.1 Homogeneity of degree zero

Z(λp) = Z(p), ∀λ > 0.

Property 1.2.2 Walras’ Law

pTZ(p) = 0.

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Chapter 2

THE CRITERIA FOR THEWEAK WEAK AXIOM OFREVEALED PREFERENCE

In this chapter, we study the concepts related to pseudo-monotonemaps, quasi-monotone maps, weak axiom, weak weak axiom, and the cri-teria for these concepts (see [3], [5] and [10]).

Since F (x) ≥ 0, ∀x ∈ R, we have

(y − x)F (x) ≤ 0 ⇒ y ≤ x ⇒ (y − x)F (y) ≤ 0.

Applied MathematicsMaster’s Thesis

This implies that F (x) is pseudo-monotone However, F (x) is not tone.

mono-Property 2.1.1 (see [3]) If pseudo-monotone maps are continuous withan open convex domain, the set of zeroes of such maps is convex.

Next, we will consider the following criteria for the equivalence ofpseudo-monotone maps.

Criterion (A): For all x ∈ D and v ∈ Rn

satisfies criterion (A).

However, if we choose x = 0, y = 1, we have

(y − x)F (x) = 0, (y − x)F (y) = 1 > 0.

Thus x2 is not a pseudo-monotone map.

Example 2.1.3 For x = (x1, x2)T, let F (x) = (x2, −x1)T We calculateJacobian matrix of F (x) as follows:

∂F (x) =

0 1−1 0

.

Applied MathematicsMaster’s Thesis

We have vT∂F (x)v = 0 Moreover, F (x) = 0 when x = 0 Thus, vTF (x+tv) = vTF (tv) = 0 This implies that F (x) satisfies criterion (B).

Example 2.1.4 Let F (x) = −x3 We have F (x) = 0 and F0(x) = 0

when x = 0 Hence,

vF (x + tv) = vF (tv) = −t3v4 ≤ 0 (2.1)This implies that F (x) satisfies criterion (C) Also, F (x) satisfies crite-rion (B).

We have the following theorem

Theorem 2.1.1 ([3]) Let F : D → R be a C1 map on the open convexset D ⊆ R Then, the followings are equivalent:

(i): F is pseudo-monotone

(ii): F satisfies the criteria (A) and (C)(iii): F satisfies the criteria (A) and (B).Proof.

We will prove that (i) ⇒ (ii) (see [3]).

Let vTF (x) = 0 and put y = x + tv for any t > 0 close to 0 Since

(y − x)TF (x) = tvTF (x) = 0, and F (x) is a pseudo-monotone maps, wehave:

(y − x)TF (x) = 0 ⇒ 0 ≥ (y − x)TF (y) = tvTF (x + tv)

⇒ (1/t)vT[F (x + tv) − F (x)] ≤ 0 (2.2)Taking the limit as t approaches 0, we obtain

Applied MathematicsMaster’s Thesis

We will prove by the contradiction Let us assume that criteria (A) and(C) hold, but criterion (B) is violated, i.e F (x) = 0, uT = vT∂F (x) 6= 0

and ∃t > 0 such that vTF (x + tv) > 0 for all t ∈ (0, t] Since uTu > 0,by continuity of the Jacobian, there exists an open ball Bx centered at x,such that vTF (z)Tu > 0,∀z ∈ Bx Let us define the map:

f (s, t) = vTF (x + su + vt).

in a sufficiently small neighborhood of the origin of R2, so that all thevectors x + su + vt are in Bx By assumption, f is C1, and its partialderivatives are, respectively,

s0(t) = −ft(s(t), t)fs(s(t), t).

Since vTF (x + s(t)u + vt) = 0, ∀t ∈ (−, ), criterion (A) requires that:

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and by the definition of implicit map, we have

We will prove that (iii) ⇒ (i) (see [5]).

We will prove by the contradiction, let us assume thatF is not a monotone map, i.e there exist x, y ∈ D such that

pseudo-(y − x)TF (x) ≤ 0, (y − x)TF (y) > 0.

Let v = y − x and consider the map f : I → R, defined on, I = {λ ∈ R :x + λv ∈ D} and

f (λ) = vTF (x + λv).

By definition of f, f (0) ≤ 0, F (1) > 0 Since f is continuous, ∃τ ∈ [0, 1[

such that f (τ ) = 0, and f (λ) > 0, ∀λ ∈]τ, 1] Without loss of generality,assume that τ = 0 By definition of f, we obtain vTF (x + λv) > 0, ∀λ ∈]0, 1] This implies that F (x) 6= 0 since otherwise it follows from

v · ∂F (x)v = limλ→0

TF (x + λv) ≥ 0.

This contradicts criterion (A).

Let xλ = x + λv and defined vλ := v − α(λ)F (x), where

α(λ) = f (λ)F (x)TF (xλ).

Since F (x) 6= 0, F (x)TF (xλ) > 0, for λ close to 0, i.e., α(λ) > 0, for all

λ > 0 close to 0.

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α(λ)2F (x)T∂F (xλ)F (x) − α(λ)[F (x)T∂F (xλ)v + vT∂F (xλ)F (x)]= α(λ)hv

T∂F (xλ)v

α(λ)F (x)T∂F (xλ)F (x) − F (x)T∂F (xλ)v − vT∂F (xλ)F (x)i (2.4)By definition of f, we have

vλT∂F (xλ)vλ > 0.

Applied MathematicsMaster’s Thesis

for some λ sufficiently close to 0.

This contradicts criterion (A) since vλ belongs to the space orthogonal to

Therefore, F (x) is not a pseudo-monotone map.

Next, we will consider the criteria of equivalence of quasi-monotonemaps The regularity of map F to be defined as follows:

Let F : X → Rn be continuously differentiable on the open convexdomain X ⊆Rn,∀x ∈ X:

F (x) = 0 ⇒ det ∂F (x) 6= 0.

We have the following theorem

Theorem 2.2.1 ([3]) Let F be regular, then the followings are lent:

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(i): F is pseudo-monotone(ii): F is quasi-monotone(iii): F satisfies criterion (A).Proof.

We will prove that (i) ⇒ (ii).

Pseudo-monotonicity trivially implies quasi-monotonicity.We will prove that (ii) ⇒ (iii).

We will prove by the contradiction Assuming that criterion (A) does nothold, i.e., ∃x ∈ X and v ∈ Rn such that vTF (x) = 0 và vT∂F (x)v > 0.It follows that

vT∂F (x)v = limλ→0

We obtain

x1 − x2 = (λ1 + λ2)v.

(x2 − x1)TF (x1) < 0,(x2 − x1)TF (x2) > 0.

By definition, F is not quasi-monotone.We will prove that (iii) ⇒ (i).

By Theorem 2.1.1, we only need to show that criterion (B) holds Let

Applied MathematicsMaster’s Thesisx ∈ X, v ∈ Rn, λ > 0 such that F (x) = 0, vT∂F (x)v = 0 We have toprove the existence of some λ ∈]0, λ] with vTF (x + λv) ≤ 0 Since F isregular, F (x) = 0 implies the existence of an open convex neighborhood

U of x such that F (y) 6= 0, ∀y ∈ U \{x} Consider some λ ∈]0, λ] with

xλ = x + λv ∈ U We will prove that vTF (xλ) ≤ 0 Assume the contrary,i.e vTF (xλ) > 0 We will prove the following claim that: vT∂F (x)w ≥0, ∀w ∈ Rn, w 6= 0 Since vT∂F (x)v = 0, the above claim is true for any

w that is collinear to v Consider w is not collinear to v From (xλ −x)TF (xλ) = λvTF (xλ) > 0 it follows that (xλ − yν)TF (xν) > 0, for

yν = x + νw ∈ U and ν > 0 sufficiently small The segment [xλ, yν] iscontained in an open convex subset of U on which F does not vanish Wehave

From vT∂F (x)w ≥ 0 for all w 6= 0, it follows that vT∂F (x) = 0 Since F

is regular, ∂F (x) is not regular Hence, v = 0, contradicts vTF (xλ) > 0.

2.3 Weak weak axiom of revealed preference

In this section, we study the criteria related to the weak weak axiomof revealed preference (see [3]).

Definition 2.3.1 (see [3]) The excess demand map Z satisfies the Weakweak axiom (WWA) iff for any pair of price vectors p and q, the following

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criterion holds

qTZ(p) ≤ 0 ⇒ pTZ(q) ≥ 0.

Example 2.3.1 ([3]) Consider the excess demand map Z(p) as follows:

Z(p) = 1p1

Firstly, we prove that Z(p) satisfies Walras’ Law.Indeed, we have

pTZ(p) = 0.Z(p) is homogeneous of degree zero because

Z(p) = Z(λp).

Next, we will prove that Z(p)satisfies WWA Supposing that p =

Let the vector q =

We have

qTZ(p) = q1(p3 − p2) + q2(p1 − p3) + q3(p2 − p1),pTZ(q) = −[q1(p3 − p2) + q2(p1 − p3) + q3(p2 − p1)].

qTZ(p) ≤ 0 ⇒ pTZ(q) ≥ 0.

It follows that Z(p) satisfies WWA

Example 2.3.2 Consider the map Z(p) : P ⊂ R2>0 → R2 to be definedas follows:

Z(p) = − f (p1p2),

p1p2f (

.

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where f (x) = 1 + x2 We have pTZ(p) = 0, i.e Z(p) satisfies Walras’Law Moreover Z(λp) = Z(p), i.e Z(p) satisfies homogeneous of degreezero Since f (x) > 0, ∀x, we get qTZ(p) ≤ 0 ⇒ pTZ(q) ≥ 0, i.e Z(p)

satisfies WWA.

Next, we will consider the criteria of equivalence of WWA Let Z be a

C1 excess demand map,p ∈ Rn>0, v ∈ Rn We have the following property:Property 2.3.1 ([3]) The excess demand map Z satisfies WWA ⇔ Theexcess demand map Z satisfies Walras’ Law and pseudo-monotonicity.The following criteria are defined as follows (see [3]):

We study the following theorem

Theorem 2.3.1 ([3]) The excess demand map Z satisfies WWA ⇔ Theexcess demand map Z satisfies criteria (A’) and (C’).

Proof Given Theorem 2.1.1 and Property 2.3.1, we need to show criteria(A’) and (C’) imply criteria (A) and (C).

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We will show that (A’) ⇒ (A).

Let vTZ(p) = 0 and notice that v can be written as v = u + αp with

Thus (A’) implies (A).

We will show that (C’) ⇒ (C).

Let Z(p) = 0 and vT∂Z(p) = 0 Vector v can be written as v = u + αp

withα = (vTp)/(pTp), where the vector uis orthogonal to p Since α canbe negative, we choose t > 0 be sufficiently small so that (1 + αt) > 0.Then, by homogeneity of degree zero, we have

Z(p + tv) = Z(p + t(u + αp)) = Z(p + su) (2.5)where s = t/(1 + tα),

and using twice Walras’ Law yields:

pTZ(p + tv) + tvTZ(p + tv) = pTZ(p + su) + suTZ(p + su) (2.6)First notice that, by (2.5), we have

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Since 0 = vT∂Z(p) = 0 = (u + αp)T∂Z(p) = uT∂Z(p), we have (C’)implies (C).

2.4 Weak axiom of revealed preference

In this section, we study the criteria related to the weak axiom ofrevealed preference (see [3] and [10]).

Definition 2.4.1 ([3]) The excess demand map Z satisfies weak axiom(WA) of revealed preference if and only if:

(q − p)TZ(p) ≤ 0(q − p)TZ(q) ≥ 0

⇒ Z(p) = Z(q).

Example 2.4.1 ([3]) We consider the map Z(p) as mentioned in 2.3.1.We will prove that Z(p) does not satisfies WA We choose p and q suchthatpis not collinear toq Thus Z(p) 6= Z(q), however,(q −p)TZ(p) = 0

and (q − p)TZ(q) = 0 Therefore, Z(p) does not satisfies WA.

Next, we will consider the criteria of equivalence of weak axiom Thecriteria are defined as follows (see [10]):

Theorem 2.4.1 ([3]) LetZ be a C1 excess demand map The followingsare equivalent:

(i): Z satisfies WA