Consumer theory
Preference relation
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.
Rational preference
Completeness: we have A B or B A or both.
Local nonsatiation
Preferences are locally nonsatiated onX if∀x ∈ X and∀ >0,∃y ∈ X such that ky −xk ≤ and y x.
Utility maps
A map u : X → R is a utility map representing preference relation if
Proposition 1.1.1 Only rational preferences relations can be repre- sented by a utility map Conversely, if X is finite, any rational preference relation can be represented by a utility map.
Budget sets
Given that p is price vector and w is wealth, we define the budget set as follows:
Property 1.1.1 Homogeneity of degree zero
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 utility optimization problem: maxu(x) such that p T x ≤ w, x ∈ B(p, w) (1.1)
Proof We will prove by the contradiction Assuming that x ∈ x(p, w) for p T x < w We have ∃ > 0 such that ∀y for kx−yk < , p T y < w.However, by locally non-satiation, ∃y such that p ã y < w and y x.Therefore x6∈ x(p, w) This leads to the contradiction.
Demand maps and excess demand maps
Suppose that we have x ∗ is the optimal solution of the optimization problem (1.1) We define the demand map at the price p is D(p) = x ∗
We choose e such that p T e = 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
THE CRITERIA FOR THE WEAK WEAK AXIOM
Pseudo-monotone maps
In this section, we study the criteria related to pseudo-monotone maps (see [3] and [5]).
Definition 2.1.1 (see [3]) A map F : D ⊆ R n → R n is said to be pseudo-monotone iff ∀x, y ∈ D,
This implies that F(x) is pseudo-monotone However, F(x) is not mono- tone.
Property 2.1.1 (see [3]) If pseudo-monotone maps are continuous with an open convex domain, the set of zeroes of such maps is convex.
Next, we will consider the following criteria for the equivalence of pseudo-monotone maps.
Criterion (B): For all x ∈ D and v ∈ R n , if F(x) = 0 and v T ∂F(x)v = 0, then
Criterion (C): For all x ∈ D and v ∈ R n , if F(x) = 0 and v T ∂F(x) = 0, then
However, if we choose x = 0, y = 1, we have
Thus x 2 is not a pseudo-monotone map.
Example 2.1.3 For x = (x 1 , x 2 ) T , let F(x) = (x 2 ,−x 1 ) T We calculate Jacobian matrix of F(x) as follows:
We have v T ∂F(x)v = 0 Moreover, F(x) = 0 when x = 0 Thus, v T F(x+ tv) = v T F(tv) = 0 This implies that F(x) satisfies criterion (B).
Example 2.1.4 Let F(x) = −x 3 We have F(x) = 0 and F 0 (x) = 0 when x = 0 Hence, vF(x+tv) =vF(tv) =−t 3 v 4 ≤ 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 C 1 map on the open convex set D ⊆ R Then, the followings are equivalent:
(ii): F satisfies the criteria (A) and (C)
(iii): F satisfies the criteria (A) and (B).
We will prove that (i) ⇒ (ii) (see [3]).
Let v T F(x) = 0 and put y = x + tv for any t > 0 close to 0 Since (y−x) T F(x) = tv T F(x) = 0, and F(x) is a pseudo-monotone maps, we have:
⇒(1/t)v T [F(x+tv)−F(x)] ≤ 0 (2.2) Taking the limit as t approaches 0, we obtain limt→0(1/t)v T [F(x+ tv)−F(x)] = v T ∂F(x)v ≤ 0.
It implies that criterion (A) holds As shown in (2.2), we have v T F(x+ tv) ≤0, it implies that criterion (C) holds.
We will prove that (ii) ⇒ (iii) (see [3]).
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, u T = v T ∂F(x) 6= 0 and ∃t > 0 such that v T F(x+ tv) > 0 for all t ∈ (0, t] Since u T u > 0, by continuity of the Jacobian, there exists an open ball Bx centered at x, such that v T F(z) T u > 0,∀z ∈ Bx Let us define the map: f(s, t) =v T F(x+su+vt). in a sufficiently small neighborhood of the origin of R 2 , so that all the vectors x+ su+ vt are in B x By assumption, f is C 1 , and its partial derivatives are, respectively, fs(s, t) = v T ∂F(x+su+vt)u, (2.3) f t (s, t) = v T ∂F(x+su+vt)v.
Since x + su + vt ∈ B x , f s (s, t) > 0 Since f(0,0) = v T F(x) = 0 and f s (0,0) = v∂F(x)u = u T u > 0, the equation f(s, t) = 0 define an implicit map, i.e there exists > 0 and a map s(t) defined in the open interval (−, ), such that s(0) = 0 and f(s(t), t) = 0 for all t The implicit map is differentiable and: s 0 (t) = −ft(s(t), t) f s (s(t), t). Since v T F(x+s(t)u+vt) = 0,∀t ∈ (−, ), criterion (A) requires that:
Moreover,f s (s(t), t) > 0and we obtain s 0 (t) ≥0 Therefore s(t) ≥ 0and non decreasing in the interval [0, ).
Now, let us take 0 < t ∗ < min (t, ) From the violation of criterion (B) we have f(0, t ∗ ) = v T F(x+t ∗ v) > 0.
Applied Mathematics Master’s Thesis and by the definition of implicit map, we have
Thus, s(t ∗ ) 6= 0 and since s(t) ≥ 0, it must be s ∗ = s(t ∗ ) > 0 By the Mean Value Theorem, there exists s, 0 < s < s ∗ , such that
Therefore, fs(s, t ∗ ) < 0, and this contradiction completes the proof. Therefore, we have criterion (B) holds.
We will prove that (iii) ⇒ (i) (see [5]).
We will prove by the contradiction, let us assume thatF is not a pseudo- monotone map, i.e there exist x, y ∈ D such that
Let v = y−x and consider the map f : I →R, defined on, I = {λ ∈ R : x+λv ∈ D} and f(λ) =v T F(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 v T F(x+λv) > 0,∀λ ∈ ]0,1] This implies that F(x) 6= 0 since otherwise it follows from v ã∂F(x)v = lim λ→0
F(x) T F(x λ ). Since F(x) 6= 0, F(x) T F(x λ ) > 0, for λ close to 0, i.e., α(λ) > 0, for all λ >0 close to 0.
Since f(λ) = v T F(x λ ), we have v λ T F(x λ ) = v T F(x λ )−α(λ)F(x) T F(x λ ) = 0. this implies that vλ belongs to the space orthogonal to F(xλ).
By definition of v λ , for λ >0 close to 0, we obtain v λ T ∂F(x λ )v λ v T ∂F(x λ )v + α(λ) 2 F(x) T ∂F(x λ )F(x)−α(λ)[F(x) T ∂F(x λ )v +v T ∂F(x λ )F(x)]
By definition of f, we have v T ∂F(x λ )v = f 0 (λ).
We have λ→0lim + f(λ) = 0 λ→0lim + ln (λ) =−∞.
The derivative of ln (f(λ)) with respect to λ là f 0 (λ)/f(λ) has no upper bound when λ approaches 0 We have λ→0lim + F(x) T F(x λ ) = F(x) T F(x) > 0.
Thus, the first term in the square blackets of (2.4) is not upper-bounded when λ approaches 0 Moreover, since the other terms inside the square brackets of (2.4) are bounded when λ approaches 0, we obtain v λ T ∂F(x λ )v λ > 0.
Applied Mathematics Master’s Thesis for some λ sufficiently close to 0.
This contradicts criterion (A) since v λ belongs to the space orthogonal to
Quasi-monotone maps
In this section, we study the criteria related to quasi-monotone maps (see [5]).
Definition 2.2.1 (see [5]) LetF : D ⊆ R →R n is quasi-monotone map if ∀x, y ∈ D:
Example 2.2.1 We consider F(x) = x 2 We have
Consequently, F(x) is a quasi-monotone map However, if we choose x 0, y = 1, we have
Therefore, F(x) is not a pseudo-monotone map.
Next, we will consider the criteria of equivalence of quasi-monotone maps The regularity of map F to be defined as follows:
Let F : X → R n be continuously differentiable on the open convex domain X ⊆R n ,∀x ∈ X:
We have the following theorem
Theorem 2.2.1 ([3]) Let F be regular, then the followings are equiva- lent:
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 not hold, i.e., ∃x ∈ X and v ∈ R n such that v T F(x) = 0 và v T ∂F(x)v > 0.
Hence, there exist λ 1 , λ 2 > 0 such that v T F(x−λ 1 v) < 0< v T F(x+λ 2 v).
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 Mathematics Master’s Thesis x ∈ X, v ∈ R n , λ > 0 such that F(x) = 0, v T ∂F(x)v = 0 We have to prove the existence of some λ ∈]0, λ] with v T F(x+λv) ≤ 0 Since F is regular, 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 x λ = x+λv ∈ U We will prove that v T F(x λ ) ≤ 0 Assume the contrary, i.e v T F(x λ ) > 0 We will prove the following claim that: v T ∂F(x)w ≥
0,∀w ∈ R n , w 6= 0 Since v T ∂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) T F(x λ ) = λv T F(x λ ) > 0 it follows that (x λ − y ν ) T F(x ν ) > 0, for y ν = x + νw ∈ U and ν > 0 sufficiently small The segment [x λ , y ν ] is contained in an open convex subset of U on which F does not vanish We have
From v T ∂F(x)w ≥ 0 for all w 6= 0, it follows that v T ∂F(x) = 0 Since F
Weak weak axiom of revealed preference
In this section, we study the criteria related to the weak weak axiom of revealed preference (see [3]).
Definition 2.3.1 (see [3]) The excess demand map Z satisfies the Weak weak axiom (WWA) iff for any pair of price vectors p and q, the following
Applied Mathematics Master’s Thesis criterion holds q T Z(p) ≤ 0⇒ p T Z(q) ≥ 0.
Example 2.3.1 ([3]) Consider the excess demand map Z(p) as follows:
Firstly, we prove that Z(p) satisfies Walras’ Law.
Z(p) is homogeneous of degree zero because
Next, we will prove that Z(p) satisfies WWA Supposing that p
It follows that Z(p) satisfies WWA
Example 2.3.2 Consider the map Z(p) : P ⊂ R 2 >0 → R 2 to be defined as follows:
Applied Mathematics Master’s Thesis where f(x) = 1 + x 2 We have p T Z(p) = 0, i.e Z(p) satisfies Walras’ Law Moreover Z(λp) = Z(p), i.e Z(p) satisfies homogeneous of degree zero Since f(x) > 0,∀x, we get q T Z(p) ≤ 0 ⇒ p T Z(q) ≥ 0, i.e Z(p) satisfies WWA.
Next, we will consider the criteria of equivalence of WWA Let Z be a
C 1 excess demand map,p ∈ R n >0 , v ∈ R n We have the following property: Property 2.3.1 ([3]) The excess demand map Z satisfies WWA ⇔ The excess demand map Z satisfies Walras’ Law and pseudo-monotonicity.
The following criteria are defined as follows (see [3]):
Example 2.3.3 ([4]) Let us consider the simple case of a pure exchange economy with only two goods The excess demand map is given by
Z(p) = (h(p 1 /p 2 ),−p 1 h(p 1 /p 2 )/p 2 ). whereh : (0,∞) →R is defined byh(x) = (x−1) 3 /(x+1) 3 SinceZ(p)is not a pseudo-monotone map,Z(p) does not satisfy WWA However,Z(p) satisfies criterion (A’) because Jacobian matrix ofZ(p)at the equilibrium price vector as 0.
We study the following theorem
Theorem 2.3.1 ([3]) The excess demand map Z satisfies WWA ⇔ The excess 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).
Let v T Z(p) = 0 and notice that v can be written as v = u+ αp with α = (v T p)/(p T p).
Clearly, the vector u is orthogonal to both p and Z(p).
By Walras’ Law, we have p T ∂Z(p) =−Z(p) T
By homogeneity of degree zero, we have
Let Z(p) = 0 and v T ∂Z(p) = 0 Vector v can be written as v = u+ αp withα = (v T p)/(p T p), where the vector uis orthogonal to p Since α can be 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: p T Z(p+tv) +tv T Z(p+tv) =p T Z(p+su) +su T Z(p+su) (2.6) First notice that, by (2.5), we have p T Z(p+tv) =p T Z(p+su).
Thus, for t > 0 sufficiently small, also s > 0 approaches 0, and by (2.6), we have
Weak axiom of revealed preference
In this section, we study the criteria related to the weak axiom of revealed 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:
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 such thatpis not collinear toq Thus Z(p) 6= Z(q), however,(q−p) T Z(p) = 0 and (q −p) T Z(q) = 0 Therefore, Z(p) does not satisfies WA.
Next, we will consider the criteria of equivalence of weak axiom The criteria are defined as follows (see [10]):
Let v = q −p If v T Z(p) = 0 and Z(p) 6= Z(q), then ∃t ∈ [0,1) such that v T ∂Z(p+tv)v < 0.
Let v = q−p If v T p = 0, v T Z(p) = 0 and Z(p) 6= Z(q), then ∃t ∈ [0,1) such that v T ∂Z(p+tv)v < 0.
We study the following theorem
Theorem 2.4.1 ([3]) LetZ be a C 1 excess demand map The followings are equivalent:
(ii): Z satisfies WWA and criterion (LZ)
(iii): Z satisfies criteria (A’), (C’), and (LZ’).
We will show that (i) ⇒ (ii).
We only need to show that WA implies criterion (LZ) because WA implies WWA Let (q −p) T Z(p) = 0 and Z(p) 6= Z(q) By the WA, we have
Set g(t) =v T Z(p+tv) with v = q −p such that g(0) = 0, g(1) < 0.
By the Mean Value Theorem,∃t ∈ (0,1) such that0> g 0 (t) = v T ∂Z(p+ tv)v, which implies criterion (LZ) holds.
We will prove that (ii) ⇒ (iii).
By Theorem 2.3.1, WWA implies (A’) and (C’) Moreover (LZ) implies (LZ’), Q.E.D.
We will show that (iii) ⇒ (i).
We will show by the contradiction, let us assume that the WA is violated , and criteria (A’), (C’) and (LZ’) hold By Theorem 2.3.1, we have Z satisfies WWA However, by assumption, WA does not hold, therefore (q−p) T Z(p) = (q−p) T Z(q) = 0andZ(p) 6= Z(q) Setq 0 = (p T p)/(p T q)q and v = q 0 −p, such that v T p = 0.
By homogeneity of degree zero and Walras’ Law, we have v T Z(p) = v T Z(q 0 ) = 0,
Define the map g(t) = v T Z(p+tv) We have
By pseudo-monotonicity of Z, we have
Thus, g is also pseudo-monotone By Property 2.1.1, the set of zeroes of a pseudo-monotone map is convex, moreover g(1) = g(0) = 0, the map g is constant on [0,1] and criterion (LZ’) cannot hold This contradiction completes the proof.
See [3] and other theorem proved above, we study the characteristics of demand map and excess demand map as follows: For demand map, cri- teria (A) and (LZ) characterize the WA For excess demand map, criteria (A), (LZ) and (C) characterize the WA In other words, the difference between characteristics of WA for demand map and excess demand map is pseudo-monotonicity.
We consider the following example of a map satisfying criteria (A) and (LZ) but not satisfying pseudo-monotonicity:
Example 2.4.2 ([3]) Let F(x) = x 3 (2 + sin(1/x)) and F(0) = 0 As- suming that x < 0 < y, we have
It follows that F(x) is not pseudo-monotone.
Since F(0) = 0 when x = 0 and F 0 (0) = 0, criterion (A) holds.
To show that (LZ) holds, since x 6= 0,(y −x)F(x) = 0 iff y = x, we need to consider the case x = 0 We have ∀y 6= 0, F(y) 6= F(0) and
F 0 (y) < 0 wheny = 1/2πk or y = −1/(2k+ 1)π, where k is any positive integer Thus, in any neighbourhood of of 0, the derivative of F(x) takes on negative values, i.e criterion (LZ) holds.
Next, we will consider the excess demand map satisfying criteria (A’) and (LZ’), but not satisfying WA.
Example 2.4.3 ([3]) We consider the excess demand map as follows:
Applied Mathematics Master’s Thesis where h(s) = F(s − 1), with F(s) = s 3 /(2 + sin(1/s)), F(0) = 0 At the unique equilibrium p ∗ = (1,1), the Jacobian matrix of Z(p) is null, therefore criterion (A’) holds We consider criterion (LZ’) If Z(p) 6= 0, then we have v = 0 because v T Z(p) = 0, criterion (LZ’) holds Next, we consider v = q −p ∗ such that v T p ∗ Excluding the trivial case v = 0, we obtain v 2 = −v 1 with v 1 6= 0 because v T p ∗ = 0 We consider g(t) =v T Z(p ∗ +tv) = v 1 h(s(t))(1 + s(t)). where s(t) = 1 +tv 1
We calculate the derivative of g(t) as follows: g 0 (t) =v 1 s 0 (t)[h 0 (s(t))(1 +s(t)) +h(s(t)] = v T ∂Z(p ∗ +tv)v.
If v 1 > 0 we choose t = 1/(v 1 (4πk + 1)) for k sufficiently large such that t ∈ [0,1], we have g 0 (t) < 0 If v 1 < 0 we choose t such that 1/(s(t)− 1) = −(2k + 1)π, we have g 0 (t) < 0 It follows that criterion (LZ’) holds.
However, the map Z(p) does not satisfies WA We choose p ∗ = (1,1) and q = (2,1), we have (q −p ∗ ) T Z(p ∗ ) = 0,(q −p ∗ ) T Z(q) = 2 > 0.
THE STABILITY OF GENERALIZED MONOTONE
S-quasimonotone maps
In this section, we study the criteria related to s-quasimonotone maps (see [1]).
Definition 3.1.1 ([1]) A mapZ is said to be s-quasimonotone iff∃σ > 0 such that
Theorem 3.1.1 ([1]) The followings are equivalent:
(ii): Z is stable w.r.t s-quasimonotonicity property
(iii): Z is stable w.r.t pseudo-monotonicity property.
We will show that (i) ⇒ (iii)
Assuming that Z is s-quasimonotone, and there exists σ Z satisfies defini- tion 3.1.1 of s-quasimonotone Set := σ We assume that q 6= p and for all a ∈ R n ,kak < such that (q−p) T (F(p) +q) ≥ 0 Therefore, we have
(q −p) T Z(p) kq −pk ≥ −(q−p) T a kq −pk Clearly, we have
Since Z is s-quasimonotone, we have
It follows that (q −p) T (Z(q) +a) ≥ 0 , i.e Z +a is pseudo-monotone.
We will show that (iii) ⇒ (i)
Assuming that there exists such that Z +a is pseudo-monotone for all a ∈ R n satisfies kak< Set σ Z = , and assume that:
(q −p) T Z(p) kq −pk ≥ δ for all |δ| < σ Z Choose a ∈ R n such that kak = |δ| and
(q −p) T Z(p) kq −pk ≥ −(q−p) T a kq −pk Therefore (q −p) T (Z(p) +a) ≥ 0 Since Z +a is pseudo-monotone, we have (q −p) T (Z(q) + a) ≥ 0 It follows that
We will show that (i) ⇒ (ii)
As have been shown above, there exists > 0 such that Z +a is pseudo- monotone for all a ∈ R n satisfies kak < Choose a ∈ R n such that kak< /2 For each a 1 ∈ R n satisfies ka 1 k < /2, we have ka 1 +ak < Therefore,Z+a+a1 is pseudo-monotone withka 1 | < /2 Thus, as proof above, Z +a is s-quasimonotone for all a ∈ R n satisfies kak < /2.
We will show that (ii) ⇒ (i)
Choose a = 0 we have the proof completed.
The stronger version of Wald’s Axiom
In this section, we study criteria related to the stronger version of Wald’s Axiom (see [2]).
Definition 3.2.1 (see [2]) The excess demand map Z satisfies Wald’s Axiom iff: p T Z(q) ≤0 q T Z(p) ≤0
Definition 3.2.2 ([2]) A stronger version of Wald’s Axiom (SWA) is defined as follows : ∃σ > 0 such that ∀δ,|δ| < σ, Z(p) 6= Z(q), q T Z(p)− δ ≤ 0, implies p T Z(q) +δ >0.
Clearly, SWA implies Wald’s Axiom.
The criterion is defined as follows:
Theorem 3.2.1 ([2]) Assuming that diamP < ∞ SWA of Z implies s-quasimonotone property of −Z.
Proof Assuming thatZ satisfies SWA withσ > 0 We setσ 1 = σ/diamP
Applied Mathematics Master’s Thesis and assuming that
By Walras’ Law, since p T Z(p) = 0, it follows that q T Z(p)−δ1kq −pk ≤ 0.
Set δ = δ 1 kq −pk, we have |δ| < σ By SWA, we conclude that p T Z(q) + δ >0 By Walras’ Law, we have
(q −p) T Z(p) kq −pk −δ 1 = −p T Z(q) +δ kq −pk < 0. which implies that −Z is s-quasimonotone.
Theorem 3.2.2 ([2]) Assuming that kq −pk ≥ 1 and Z satisfies crite- rion (D) We have the property of s-quasimonotone of −Z implies SWA of Z.
Proof Assuming that −Z is s-quasimonotone with σ > 0 and q T Z(p)−δ ≤ 0,|δ| < σ, Z(p) 6= Z(q).
By Walras’ Law, we have
Set δ 1 = δ/kq −pk By condition kq −pk ≥ 1, we have |δ 1 | ≤ |δ| < σ.
By the property of s-quasimonotone of −Z, we have
Applying Walras’ Law, we obtain p T Z(q) +δ ≥ 0 We need to show that p T Z(q) + δ > 0 We assuming that p T Z(q) + δ = 0 Applying Walras’ Law, we obtain
Since −Z is s-quasimonotone, we have
By Walras’ Law, we obtain −q T Z(p) +δ ≤ 0 Combining with q T Z(p)− δ ≤ 0, it follows that q T Z(p) = δ Therefore, q T Z(p) = −p T Z(q)δ. Moreover, by Walras’ Law, we have (q − p) T Z(p) = (q − p) T Z(q) = δ. For any p 0 ∈]p, q[, we have
(p 0 −p) T Z(p) kp 0 −pk = (q −p) T Z(p) kq −pk = δ1. Since −Z is s-quasimonotone, we have
On the other hand, we have
(p 0 −q) T Z(q) kp 0 −qk = (p−q) T Z(q) kp−qk = −δ 1 Since −Z is s-quasimonotone, we have
Therefore, we conclude that (q − p) T Z(p 0 ) = δ Thus (q − p) T Z(p) (q − p) T Z(p 0 ) = δ It follows that (p − p 0 ) T (Z(p) − Z(p 0 )) = 0, this contradicts the criterion (D).
Stability with respect to criterion (A)
Property 3.3.1 (see [1]) Pseudo-monotone maps are not stable w.r.t the criterion (A).
Example 3.3.1 (see [1]) We consider the map F(x) to be defined as follows:
Consider the case x < 0, we have
Consider the case x = 0 we have
(y −x)F(y) = yF(y) ≤0 Consider the case x > 0, we have
For k is integer and k ≥ 1, we set xk = − 1
We calculate the derivative of F(x) as follows:
F 0 (x k ) = x 2 k (1− 4 kπ) > 0 Thus, the criterion (A) is not satisfied Since lim k→+∞ ak = 0, there not exists >0 such that F +a satisfies (A) with |a| < It follows that (A) is not stable w.r.t pseudo-monotonicity.
Stability with respect to criterion (A’)
Consider the map Z(p) as follows:
Z(p) = −F(p 1 p 2 ),p 1 p 2 F(p 1 p 2 ) Taking v such that v T p= 0, we choose v = (−p 2 , p 1 ) T We have v T Z(p) = 0 ⇔F(p 1 p 2 )(1 + p 2 1 p 2 ) = 0. and v T ∂Z(p)v = −(p 2 1 +p 2 2 ) 2 F 0 ( p p 1
We choose F(x) as in example 3.3.1 and p 2 ≤ 0, we have the map is not stable w.r.t (A’) for case 1.
Stability with respect to criterion (B) and (C)
Property 3.5.1 The pseudo-monotone map is not stable w.r.t criterion (B) and (C).
Example 3.5.1 (see [1]) We consider the map F(x) to be defined as follows:
As proved in the example 3.3.1, we have F(x)is pseudo-monotone More- over, we have
Since F(x) < 0,∀x > 0, F(x) satisfies criterion (B) (also criterion (C)). For x > 0, we have
We consider the map: g(x) = cos( x 1 )
2 +k2π. Since g(a k ) > 0> g(b k ), by Intermediate Value Theorem, there exists c k where b k < c k < a k satisfying g(c k ) = 0.
Assuming that the root c k of F 0 (x) = 0 can be written as c k = u 1 k +k2π with π 4 < u k < π 2
We have sin(u k ) > 0 since π 4 < u k < π 2 Since c k is the root of F 0 (x) = 0, g(ck) = 0 We have g(ck) = 0 ⇔ k 4(2+sin(u k )) cos(u k ) −u k 2π >
Thus, F 00 (c k ) > 0, which implies c k is a local minimum of the map F(x). For ck chosen as above, we have
It follows that for any , we can take k sufficiently large such that
We set x = c k is the root of the equation F(x) − F(c k ) = (F(x) −
F(ck)) 0 = 0 Since ck is the local minimum, we cannot find t > 0 in the neighborhood of c k such that (F +a k )(x+t) =F(c k +t)−F(c k ) ≤ 0.
It follows that pseudo-monotone map is not stable w.r.t criterion (B) (also criterion (C)).
We have studied criteria (A), (B), (C), (A’) of pseudo-monotone maps, quasi-monotone maps, weak weak axiom of revealed preference and weak axiom of revealed preference Also, we have studied the stability of gen- eralized monotone maps w.r.t these criteria.
Our future tasks are to study the stability index and algorithm for finding the stability index.
[1] P T An "Stability of generalized monotone maps with respect to their characterizations" Optimization, vol 55, 289–299 (2006)
[2] P T An and V T T Binh "Stability of excess demand functions with respect to a strong version of Wald’s axiom".Asia-Pacific Jour- nal of Operational Research, vol 26, no 4, 523–532 (2006)
[3] L Brighi "A stronger criterion for the Weak Weak Axiom" Journal of Mathematical Economics vol 40, 93-103 (2004)
[4] L Brighi and R John "Characterizations of pseudomonotone maps and economic equilibrium" Journal of Statistics and Management Systems vol 5, issue 1-3 (2002)
[5] R John "A first order characterization of generalized monotonicity". Mathematical Programming vol 88, 147–155 (2000)
[6] A Mas-Colell, M D Whinston, and J R Green Microeconomic Theory Oxford University Press (1995)
[7] W E Nicholson and C M Snyder Microeconomic Theory 10th edition, Thomson/South-Western (2007)
[8] J E Roger and D M Zin Revealed preference theory Encyclopedia Britannica (2013)
[9] H R Varian.Microeconomic Analysis 3rd edition New York, Norton (1992)
[10] L Zhou "A characterization of demand functions that satisfy the Weak Axiom of Revealed Preference" Economics Letters vol 49, 403–406 (1995)