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講義 1：最適化理論と消費者問題

上級ミクロ経済学^—財務省理論研修

安田 洋祐

政策研究大学院大学（GRIPS）

2012_年6_月4_日

Announcement | お知らせ

Course website You can find my corse websites from the link below: https://sites.google.com/site/yosukeyasuda/jp/lecture/mof12micro

Lecture slides Uploaded on the website (before) the lecture.

Textbooks There is a main textbook and three related books:

JR Jehle and Reny, Advanced Microeconomic Theory, 3rd. O 奥野正寛,『ミクロ経済学』東京大学出版会, 2008.

G _ギボンズ,『経済学のためのゲーム理論入門』創文社^{, 1995.} MWG Mas-Colell, Whinston and Green, Microeconomic Theory, 1995.

Symbols that we use in lectures

✄

✂Ex : Example,✁ ^✞

✝

☎

Fg : Figure,✆ ^✄

✂Rm : Remark,✁ ^✄

✂Q : Question,✁ Def : Definition, Thm : Theorem.

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What and How to Optimize? | 何をどうやって最適化するのか？

Optimization is a set of mathematical procedures to find the optimal value of some function.

We frequently adopt the assumption that an economic agent seeks to maximize (_最大化) or minimize (_最小化) some function, for example:

◮ _{効用最大化：}Consumer maximizes her utility function.

◮ _{費用最小化：}Firm minimizes its cost function.

◮ _{収入最大化：}Seller at the auction maximizes (expected) revenue function.

◮ _{社会厚生最大化：}Government tries to maximize the social welfare function.

We study and apply static optimization (静学的最適化).

→ Dynamic optimization (動学的最適化), a common tool in modern macroeconomics, will not be covered in our lectures...

Equality Constraints | 等号制約

Consider choosing x1and x2to maximize f (x1, x2) , when x1and x2must satisfy some particular relation to each other that we write implicit form as

g(x1, x2) = 0 .

Formally, we write this problem as follows:

xmax₁,x₂^{f(x1, x2}) subject to g(x1, x2) = 0.

◮ _{f(x1, x2}): objective function (目的関数).

◮ _{x1and x2}: choice variables (_選択変数).

◮ _{g(x1, x2}): constraint (制約).

◮ The set of all (x1, x2) that satisfy the constraint: feasible set.

✄

✂Q Does solution always exist?✁

→_{補論参照．}

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Lagrange’s Method | ラグランジュ（の未定乗数）法 (1)

There are two approaches to solve this type of optimization problems with equality constraints: substitution (_代入) and Lagrange’s method. Lagrange’s method is a powerful way to solve constrained optimization problems, which essentially translates them into unconstrained problems.

◮ _{等号制約付き}(constrained)_問題⇒_制約無し(unconstrained)_問題

Again, consider the following problem:

xmax₁,x₂^{f(x1, x2}) subject to g(x1, x2) = 0.

Let us construct a new function L(x1, x2, λ), called the Lagrangian function (_{ラグランジュ関数}) as follows:

L(x1, x2, λ) = f (x1, x2) + λg(x1, x2).

Lagrange’s Method | ラグランジュ（の未定乗数）法 (2)

Then maximize this Lagrangian function, that is, derive the first order conditions (一階条件):

∂L

∂x1

=^∂f(x

∗ 1^{, x}

∗ 2⁾

∂x1

+ λ^∗∂g(x

∗ 1^{, x}

∗ 2⁾

∂x1

= 0

∂L

∂x2

=^∂f(x

∗ 1, x^∗2⁾

∂x2

+ λ^∗∂g(x

∗ 1, x^∗2⁾

∂x2

= 0

∂L

∂λ ^{= g(x}

∗

1, x^∗2) = 0.

◮ Lagrange’s method asserts that if we find values x^∗1, x^∗2, and λ^∗that solves these three equations simultaneously, then we will have a critical point of along the constraint g(x1, x2) = 0.

◮ 制約条件を“ あたかも ”忘れて解くことができる！

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Practice of Lagrange’s Method | ラグランジュ法の練習

✄

✂Ex Example A2.11 (see JR, pp.508)✁

xmax₁,x₂^x1x2 subject to a − 2x1− 4x2= 0

Forming the Lagrangian, we get L = x1x2+ λ(a − 2x1− 4x2), with first order conditions:

∂L

∂x1

= x2− 2λ = 0, ^∂L

∂x2

= x1− 4λ = 0.

∂L

∂λ ^{= a − 2x1− 4x2= 0.} These can be solved to find

x1= ^a 4^{, x2=}

a 8^{, λ=}

a 16^.

Note that the solution of the problem is a function of parameter a.

Envelope Theorem | 包絡線定理 (1)

Consider the following constrained optimization problem P 1: P1 : max

x f(x, a) s.t. g(x, a) = 0.

where x is a vector of choice variables, and a := (a1, ..., am) is a vector of parameters (パラメータ) that may enter the objective function and constraint. Suppose that for each vector a, the solution is unique and denoted by x(a).

◮ A maximum-value function, denoted by M (a), is defined as follows: M(a) := max

x f(x, a) s.t. g(x, a) = 0, or equivalently, M (a) := f (x(a), a).

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Envelope Theorem | 包絡線定理 (2)

If the objective function, constraint, and the solutions are differentiable in the parameters, there is a very powerful theorem that shows how the solutions vary with the parameters.

Thm Envelope Theorem (_{包絡線定理})

Consider P 1 and suppose the objective function and constraint are

continuously differentiable in a. For each a, let x(a) ≫ 0 uniquely solve P 1 and assume that it is also continuously differentiable in the parameters a. Then, the Envelope theorem states that

∂M(a)

∂aj

= ^∂L

∂aj

|x(a),λ(a) j= 1, ..., m,

where the right hand side denotes the partial derivative of the Lagrangian function with respect to the parameter aj evaluated at the point (x(a),λ(a)).

Proof See JR, pp.506-507.

Practice of Envelope Theorem | 包絡線定理の練習

✄

✂Ex Example A2.11 (again)✁

xmax₁,x₂^x1x2s.t. a − 2x1− 4x2= 0.

We form the maximum-value function by substituting the solutions for x1 and x2into the objective function. Thus,

M(a) = x1(a)x2(a) = ^a 4^·

a 8⁼

a² 32^. Differentiating M (a) with respect to a, we get

dM(a)

da ⁼

a 16^,

which tells us how the maximized value varies with a. Applying the Envelope theorem, we directly obtain

dM(a)

da ⁼

∂L

∂a^{|x(a),λ(a)= λ(a).}

Since λ(a) =₁₆^a, we verified that the Envelope theorem works.

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What is Consumer Problem? | 消費者問題とはなにか？

Assume % is a consumer’s preference relation (_選好関係) on the consumption set (_{消費集合) X = R}ⁿ₊ where

Rⁿ+^{:= {(x}1, ..., xn)|xi≥ 0, i = 1, ..., n} ⊂ R^n.

◮ For any x, y ∈ X, x % y means x is at least as preferred as y.

◮ Consumption set contains all conceivable alternatives.

◮ A budget set (_予算集合) is a set of feasible consumption bundles, represented as B(p, ω) = {x ∈ X|px ≤ ω}, where p is an n-dimensional positive vector interpreted as prices, and ω is a positive number interpreted as the consumer’s wealth.

◮ We assume that the consumer is motivated to choose the most preferred feasible alternative according to her preference relation. That is, she seeks

x^∗∈ B(p, ω) such that x^∗%x for all x ∈ B(p, ω).

Utility Maximization | 効用最大化

Def We refer to the problem of finding the % best bundle in B(p, ω) as the consumer problem (_{消費者問題}).

◮ Function U : X → R represents (表現する) the preference % if for all x and y ∈ X, x % y if and only if U (x) ≥ U (y).

◮ If U represents a preference relation %, we call it a utility function (効用 関数), and we say that % has a utility representation (効用表現).

◮ Utility functions are useful since it is often more convenient to talk about the maximization of a numerical function than of a preference relation.

Given utility representation, consumer problem becomes:

x∈B(p,ω)max ^{U(x) or max}x∈X s.t. x ∈ B(p, ω).

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Monotonicity | 単調性

Monotonicity says that “more is better (than less).”

Def A preference relation % satisfies monotonicity at the bundle y if for all x ∈ X,

If xk≥ ykfor all k, then x % y, and If xk> ykfor all k, then x ≻ y.

Monotonicity can be expressed by a increasing utility function U , which is assumed in most of consumer problems.

Thm U (x) is strictly increasing if and only if % is monotonic.

Monotonicity says that if one bundle contains at least as much of every commodity as another bundle, then the one is at least as good as the other.

Inequality Constraints | 不等号制約

Consider a two-variable optimization problem in which the only constraint is given by the inequality g(x1, x2) ≥ 0. Formally, our problem is

xmax₁,x₂^{f(x1, x2) s.t. g(x1, x2) ≥ 0.}

Let us define Lagrangian function as if the constraint holds with equality. L = f (x1, x2) + λg(x1, x2).

The optimal solutions must satisfy the following Kuhn-Tucker conditions (クーン-タッカー条件):

∂L

∂x1

= ^∂f

∂x1

+ λ^∂g

∂x1

= 0, ^∂L

∂x2

= ^∂f

∂x2

+ λ^∂g

∂x2

= 0. λg(x1, x2) = 0, λ ≥ 0, g(x1, x2) ≥ 0.

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Solving Consumer Problem | 消費者問題を解く (1)

✄

✂Ex Practice: A consumer problem with n goods.✁

x∈Rmaxⁿ+

u(x) s.t. px ≤ ω.

The corresponding Lagrangian function is: L = u(x) + λ(ω − px) + λ1x1+ · · · + λnxn.

The Kuhn-Tucker conditions are as follows:

∂L

∂x1

= ^∂u

∂x1

− λp1+ λ1= 0. ...

∂L

∂xn

= ^∂u

∂xn

− λpn+ λn= 0. λ(ω − px) = 0, λ ≥ 0, ω − px ≥ 0,

λ1x1= 0, ..., λnxn= 0, λi≥ 0, xi≥ 0 for all i.

Solving Consumer Problem | 消費者問題を解く (2)

Suppose that solution x^∗_i and ^∂u(x_∂x^∗)

i is strictly positive for all i. Then, the corresponding Lagrangian multipliers (_{ラグランジュ乗数}) λimust be 0 for all i, which implies λ, pi>0 for all i. Therefore, for any two goods j and k, we can combine the conditions to conclude that

∂u(x^∗)

∂x_j

∂u(x^∗)

∂x_k

= ^λpj λpk

= ^pj pk

.

This says that at the optimum, the marginal rate of substitution (MRS,限界 代替率) between any two goods must be equal to the ratio of the goods’ prices. Note that, for two goods case along the indifference curve (無差別曲線):

0 = du = ^∂u(x

∗)

∂x1

dx1+^∂u(x

∗)

∂x2

dx2 ⇐⇒

∂u(x^∗)

∂x₁

∂u(x^∗)

∂x₂

= −^dx2 dx1

.

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Indirect Utility Function | 間接効用関数

To construct the indirect utility function, we fix market prices and “initial wealth,” and seek the maximum level of utility the consumer could achieve.

Def The indirect utility function is the maximum-value function

corresponding to the consumer’s utility maximization problem (UMP,効用最 大化問題), and it is denoted by v(p, ω). That is,

v(p, ω) = max

x∈Rⁿ+

u(x) s.t. px ≤ ω,

or, equivalently

v(p, ω) = u(x(p, ω))

where x(p, ω) is the solution of the UMP, known as Marshallian demand functions (_{マーシャルの需要関数}).

Example of UMP | 効用最大化問題の例

✄

✂Ex Cobb-douglas (✁ _コブ-_ダグラス) utility function with two goods max

x∈R²+

x^α1x^1−α₂ s.t. px (= p1x1+ p2x2) ≤ ω.

where α ∈ (0, 1).

✄

✂Rm u(x) = x✁ ^β₁x^γ₂(β, γ > 0) can express the identical preference. Note that x^α₁x^1−α₂ is a monotone transformation of u(x) = x^β₁x^γ₂.

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Expenditure Function | ( 最小 ) 支出関数

To construct the expenditure function, we fix prices and a “level of utility,” and seek the minimum level of money expenditure the consumer must make to achieve this particular level of utility.

Def The expenditure function is the minimum-value function corresponding to the consumer’s expenditure minimization problem (EMP,支出最小化問 題), and it is denoted by e(p, u). That is,

e(p, u) = min

x∈Rⁿ₊px s.t. u(x) ≥ u, or, equivalently

e(p, u) = px^h(p, u)

where x^h(p, u) is the solution of the EMP, known as Hicksian (Compensated) demand functions (ヒックスの(補償)需要関数).

Example of EMP | EMP の例

✄

✂Ex Cobb-douglas utility function with two goods✁ min

x∈R²+

p1x1+ p2x2s.t. x^α₁x^1−α₂ ≥ u

where α ∈ (0, 1).

✄

✂Rm EMP is the mirror image of UMP. This property is formally established as✁ duality (_双対性) (_講義2_を参照).

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Indirect Utility Function | 間接効用関数

There are several properties that the indirect utility function possesses.

Thm If u(x) is continuous and strictly increasing on Rⁿ+, then v(p, ω) is

1. Continuous in p and ω.

2. Homogeneous of degree zero in (p, ω). 3. Strictly increasing in ω.

4. Decreasing in p. 5. Quasiconvex in (p, ω).

6. Roy’s identity (_{ロアの恒等式}): If v(p, ω) is differentiable at (p⁰, ω⁰), then

xi(p⁰, ω⁰) = −

∂v(p⁰,ω⁰)

∂p_i

∂v(p⁰,ω⁰)

∂ω

, i = 1, ..., n.

More on Roy’s Identity | もっとロアの恒等式

Roy’s identity says that the consumer’s Marshallian demand for good i is simply the ratio of the partial derivatives of indirect utility with respect to pi

and ω after a sign change.

By the envelope theorem and Lagrangian method,

∂v(p⁰, ω⁰)

∂pi

= −λxi(p⁰, ω⁰) and

∂v(p⁰, ω⁰)

∂ω ^{= λ,}

which implies

−

∂v(p⁰,ω⁰)

∂p_i

∂v(p⁰,ω⁰)

∂ω

= xi(p⁰, ω⁰).

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Expenditure Function | 支出関数

There are several properties that the expenditure function possesses.

Thm If u(x) is continuous and strictly increasing on Rⁿ+, then e(p, u) is

1. Continuous in p and ω. 2. Homogeneous of degree 1 in p. 3. Strictly increasing in u, for all p ≫ 0. 4. Increasing in p.

5. Concave in p.

If, u(·) is strictly quasi-concave, we have

6. Shephard’s lemma (_{シェパードの補題}): e(p, u) is differentiable in p at (p⁰, u⁰) with p⁰≫ 0, and

∂e(p⁰, u⁰)

∂pi

= x^hi(p⁰, u⁰), i = 1, ..., n.

More on Shephard’s Lemma | もっとシェパードの補題

To prove Shephard’s Lemma, we appeal to the envelope theorem:

∂e(p⁰, u⁰)

∂pi

= ^∂L

∂pi

= x^hi(p⁰, u⁰).

where L is the corresponding Lagrangian function: L = px − λ(u(x) − u).

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【補論】 Existence of Solutions | 解の存在

◮ A set S in Rⁿis called bounded (_有界) if there exists some ε > 0 such that S ⊂ Bε(x) for some x ∈ R^n.

◮ A set S in Rⁿis called compact if it is closed (_閉) and bounded.

Thm A1.10 (Weierstrass) Existence of Extreme Values

Let f : S → R be a continuous real-valued function where S is a non-empty compact subset of Rⁿ. Then f has its maximum and minimum values. That is, there exists vectors x and x such that

f(x) ≤ f (x) ≤ f (x) for all x ∈ S.

◮

✞

✝

☎

Fg Figure A1.18 (see JR, pp.522)✆

◮ Most problems in economics have compact domains (_定義域) and continuous objective functions. ⇒ Solutions guaranteed!

【補論】 Intuitions of Lagrange’s method | ラグランジュ法の直観 (1)

✄

✂Q Why does Lagrange’s method work?✁ Take the total differential of the Lagrangian:

dL = ^∂L

∂x1^dx1

+ ^∂L

∂x2^dx2

+^∂L

∂λ^dλ.

When (x1, x2, λ) = (x^∗₁, x^∗₂, λ^∗), it can be re-written as follows: dL = (^∂f(x

∗ 1^{, x}

∗ 2⁾

∂x1

+ λ^∗∂g(x

∗ 1^{, x}

∗ 2⁾

∂x1

)dx1

+ (^∂f(x

∗ 1, x^∗2⁾

∂x2

+ λ^∗∂g(x

∗ 1, x^∗2⁾

∂x2

)dx2+ g(x^∗₁, x^∗₂)dλ = 0. Since g(x^∗₁, x^∗₂) = 0,

0 = dL =^∂f(x

∗ 1^{, x}

∗ 2⁾

∂x1

dx1+^∂f(x

∗ 1^{, x}

∗ 2⁾

∂x2

dx2

+ λ^∗(^∂g(x

∗ 1, x^∗2⁾

∂x1

dx1+^∂g(x

∗ 1, x^∗2⁾

∂x2

dx2) for all dx1 and dx2that satisfy the constraint g.

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【補論】 Intuitions of Lagrange’s method (2) | ラグランジュ法の直観 (2)

Note that, dg= ^∂g(x

∗ 1, x^∗2⁾

∂x1 ^dx1

+^∂g(x

∗ 1, x^∗2⁾

∂x2 ^dx2

= 0. So, we can show that

dL = ^∂f(x

∗ 1, x^∗2⁾

∂x1

dx1+^∂f(x

∗ 1, x^∗2⁾

∂x2

dx2= 0

for all dx1 and dx2that satisfy the constraint g. Thus, (x^∗₁, x^∗₂) is indeed a critical point of f given that the variables must satisfy the constraint.

Lagrange’s method is very clever and useful. In effect, it offers us an algorithm for identifying the constrained optima in a wide class of practical problems.

【補論】 Remarks on Utility Function | 効用関数に関する注意

✄

✂Rm Utility function has NO meaning other than that of representing a✁ preference relation %.

Thm If U represents %, then for any strictly increasing function f : R → R, the function V (x) = f (U (x)) represents % as well.

Proof Note that increasing function implies the second step. a% b⇔ U (a) ≥ U (b) ⇔ f (U (a)) ≥ f (U (b)) ⇔ V (a) ≥ V (b).

✄

✂Q Under what conditions do utility representations exist?✁

→_講義3_で確認．

Thm If % is represented by continuous utility function U , then any consumer problem has a solution.

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【補論】 Convex Sets and Concave Functions | 凸集合と凹関数

✄

✂^{Def S ⊂ R}✁

nis a convex set if for all x¹,x²∈ S, tx¹+ (1 − t)x²∈ S ∀t ∈ [0, 1].

✄

✂Def Let D be a convex set and x✁ ^t= tx¹+ (1 − t)x². f: D → R is a concave function if for all x¹,x²∈ D,

f(x^t) ≥ tf (x¹) + (1 − t)f (x²) ∀t ∈ [0, 1].

✞

✝

☎

Fg Figures A1.5 and A1.27 (see JR, pp.502 and pp.534)✆

The points below the graph of all concave regions appear to be convex. Formally, the next theorem holds.

Thm Let A := (x, y)|x ∈ D, f (x) ≥ y be the set of points “on and below” the graph of f : D → R, where D ⊂ Rⁿis a convex set. Then,

f is a concave function (_凹関数) ⇐⇒ A is a convex set (_凸集合).

【補論】 Convex Preference | 凸選好 (1)

There are three equivalent definitions of convex preferences. Def The preference relation % satisfies convexity (_凸性) if

1. x % y and α ∈ (0, 1) implies that αx + (1 − α)y % y.

2. For all x, y and z such that z = αx + (1 − α)y for some α ∈ (0, 1), either z % x or z % y.

3. For all y, the set AsGoodAs(y) := {z ∈ X|z % y} is convex.

The notion of convex preferences captures the following intuitions:

1. If x is preferred to y, then going part of the way from y to x is also an improvement upon y.

2. If z is between x and y then it is impossible that both x and y are better than z.

3. If both x1and x2are better than y, then the average of x1and x2is definitely better than y.

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【補論】 Convex Preference | 凸選好 (2)

Convexity has a stronger version.

Def The preference relation % satisfies strict convexity if a % y, b % y, a 6= b and λ ∈ (0, 1) imply that λa + (1 − λ)b ≻ y.

✄

✂Q How is the convexity of preferences translated into properties of the utility✁ function?

Def A function u(·) : X → R is quasi-concave if for all x, y ∈ X, u(αx + (1 − α)y) ≥ min[u(x), u(y)]

holds for all α ∈ (0, 1). u(·) is strictly quasi-concave if for all x 6= y in X, u(αx + (1 − α)y) > min[u(x), u(y)]

holds for all α ∈ (0, 1).

【補論】 Properties of Convex Preference | 凸選好の性質

Convex preferences expressed by utility functions.

Thm Suppose % is represented by utility function u(·). Then,

1. u(x) is quasiconcave if and only if % is convex.

2. u(x) is strictly quasiconcave if and only if % is strictly convex.

Convexity induces a simple solution structure.

Thm Convex preference satisfies the following properties.

1. If % is convex, then the set of solutions for a choice from B(p, ω) is convex.

2. If % is strictly convex, then every consumer problem has at most one solution.

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【補論】 Proof | 証明

We give the proof for each property in the second theorem.

1. Assume that both x and y maximize % given B(p, ω). By the convexity of the budget set B(p, ω) we have αx + (1 − α)y ∈ B(p, ω) and, by the convexity of %, αx + (1 − α)y % x % z for all α ∈ [0, 1] and z ∈ B(p, ω). Thus, αx + (1 − α)y is also a solution to the consumer problem. 2. Assume that both x and y (where x 6= y) are solutions to the consumer

problem B(p, ω). By the convexity of the budget set B(p, ω) we have αx + (1 − α)y ∈ B(p, ω) and, by the strict convexity of %,

αx + (1 − α)y ≻ z for all α ∈ (0, 1) and z ∈ B(p, ω), which is a contradiction of x being % optimal in B(p, ω).

【補論】 (Marshallian) Demand Function | ( マーシャルの ) 需要関数

1. Homogeneous of degree zero (0_次同次性): x(p, ω) = x(λp, λω) for any λ > 0.

2. Walras’s Law (_{ワルラス法則}): If the preferences are monotonic, then any solution x to the consumer problem B(p, ω) is located on its budget line, i.e., px(p, ω) = ω.

3. Continuity (_連続性): If % is a continuous preference, then the demand function is continuous in p and in ω.

Proof We give the sketch of the proof for 1 and 2.

1. The budget sets are identical, i.e, B(p, ω) = B(λp, λω).

2. If px(p, ω) < ω, there must exist some consumption bundle x^′ with x ≪ x^′ and px(p, ω) ≤ ω. By monotonicity, x^′must be strictly preferred to x, which contradicts x being a solution of the consumer problem.

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