Mathematical Economics

Lecture Notes on Mathematics for Economists

Chien-Fu CHOU

September 2006

Contents
Lecture 1 Lecture 2 Lecture 3 Lecture 4 Lecture 5 Lecture 6 Lecture 7 Lecture 8 Lecture 9 Lecture 10 Static Economic Models and The Concept of Equilibrium Matrix Algebra Vector Space and Linear Transformation Determinant, Inverse Matrix, and Cramer’s rule Diﬀerential Calculus and Comparative Statics Comparative Statics – Economic applications Optimization Optimization–multivariate case Optimization with equality constraints and Nonlinear Programming General Equilibrium and Game Theory 1 5 10 16 25 36 44 61 74 89

1

1

Static Economic Models and The Concept of Equilibrium

Here we use three elementary examples to illustrate the general structure of an economic model. 1.1 Partial market equilibrium model

A partial market equilibrium model is constructed to explain the determination of the price of a certain commodity. The abstract form of the model is as follows. Qd = D(P ; a) Qd : Qs : P: a: Qs = S(P ; a) Qd = Qs ,

quantity demanded of the commodity quantity supplied to the market market price of the commodity a factor that aﬀects demand and supply

D(P ; a): demand function S(P ; a): supply function

Equilibrium: A particular state that can be maintained. Equilibrium conditions: Balance of forces prevailing in the model. Substituting the demand and supply functions, we have D(P ; a) = S(P ; a). For a given a, we can solve this last equation to obtain the equilibrium price P ∗ as a function of a. Then we can study how a aﬀects the market equilibrium price by inspecting the function. Example: D(P ; a) = a2 /P , S(P ) = 0.25P . a2 /P ∗ = 0.25P ∗ ⇒ P ∗ = 2a, Q∗ = Q∗ = s d 0.5a. 1.2 General equilibrium model

Usually, markets for diﬀerent commodities are interrelated. For example, the price of personal computers is strongly inﬂuenced by the situation in the market of microprocessors, the price of chicken meat is related to the supply of pork, etc. Therefore, we have to analyze interrelated markets within the same model to be able to capture such interrelationship and the partial equilibrium model is extended to the general equilibrium model. In microeconomics, we even attempt to include every commodity (including money) in a general equilibrium model. Qd1 = D1 (P1 , . . . , Pn ; a) Qd2 = D2 (P1 , . . . , Pn ; a) . . . Qdn = Dn (P1 , . . . , Pn ; a) Qsn = Sn (P1 , . . . , Pn ; a) Qs2 = S2 (P1 , . . . , Pn ; a) Qs1 = S1 (P1 , . . . , Pn ; a) Qdn = Qsn Qd2 = Qs2 Qd1 = Qs1 Qdi : quantity demanded of commodity i Qsi : quantity supplied of commodity i Pi : market price of commodity i a: a factor that aﬀects the economy Di (P1 , . . . , Pn ; a): demand function of commodity i Si (P1 , . . . , Pn ; a): supply function of commodity i We have three variables and three equations for each commodity/market.

2 Substituting the demand and supply functions, we have D1 (P1 , . . . , Pn ; a) − S1 (P1 , . . . , Pn ; a) ≡ E1 (P1 , . . . , Pn ; a) = 0 D2 (P1 , . . . , Pn ; a) − S2 (P1 , . . . , Pn ; a) ≡ E2 (P1 , . . . , Pn ; a) = 0 . . . . . . Dn (P1 , . . . , Pn ; a) − Sn (P1 , . . . , Pn ; a) ≡ En (P1 , . . . , Pn ; a) = 0. For a given a, it is a simultaneous equation in (P1 , . . . , Pn ). There are n equations and n unknown. In principle, we can solve the simultaneous equation to ﬁnd the ∗ ∗ equilibrium prices (P1 , . . . , Pn ). A 2-market linear model: D1 = a0 + a1 P1 + a2 P2 , S1 = b0 + b1 P1 + b2 P2 , D2 = α0 + α1 P1 + α2 P2 , S2 = β0 + β1 P1 + β2 P2 . (a0 − b0 ) + (a1 − b1 )P1 + (a2 − b2 )P2 = 0 (α0 − β0 ) + (α1 − β1 )P1 + (α2 − β2 )P2 = 0. 1.3 National income model

The most fundamental issue in macroeconomics is the determination of the national income of a country. C I Y S = = = = a + bY (a > 0, 0 < b < 1) I(r) ¯ C +I +G Y − C.

C: Consumption Y : National income I: Investment S: Savings ¯ G: government expenditure r: interest rate a, b: coeﬃcients of the consumption function. To solve the model, we substitute the ﬁrst two equations into the third to obtain ¯ ¯ Y = a + bY + I0 + G ⇒ Y ∗ = (a + I(r) + G)/(1 − b). 1.4 The ingredients of a model

We set up economic models to study economic phenomena (cause-eﬀect relationships), or how certain economic variables aﬀect other variables. A model consists of equations, which are relationships among variables. Variables can be divided into three categories: Endogenous variables: variables we choose to represent diﬀerent states of a model. Exogenous variables: variables assumed to aﬀect the endogenous variables but are not aﬀected by them. Causes (Changes in exogenous var.) ⇒ Eﬀects (Changes in endogenous var.) Parameters: Coeﬃcients of the equations.

3 End. Var. P , Qd , Qs Pi , Qdi , Qsi C, Y , I, S Ex. Var. a a ¯ r, G Parameters Coeﬃcients of D(P ; a), S(P ; a) a, b

Partial equilibrium model: General equilibrium model: Income model:

Equations can be divided into three types: Behavioral equations: representing the decisions of economic agents in the model. Equilibrium conditions: the condition such that the state can be maintained (when diﬀerent forces/motivations are in balance). Deﬁnitions: to introduce new variables into the model. Behavioral equations Equilibrium cond. Deﬁnitions Partial equilibrium model: Qd = D(P ; a), Qs = S(P ; a) Qd = Qs General equilibrium model: Qdi = Di (P1 , . . . , Pn ), Qdi = Qsi Qsi = Si (P1 , . . . , Pn ) ¯ Income model: C = a + bY , I = I(r) Y =C +I +G S =Y −C 1.5 The general economic model

Assume that there are n endogenous variables and m exogenous variables. Endogenous variables: x1 , x2 , . . . , xn Exogenous variables: y1 , y2 , . . . , ym. There should be n equations so that the model can be solved. F1 (x1 , x2 , . . . , xn ; y1, y2 , . . . , ym ) = 0 F2 (x1 , x2 , . . . , xn ; y1, y2 , . . . , ym ) = 0 . . . Fn (x1 , x2 , . . . , xn ; y1 , y2 , . . . , ym) = 0. Some of the equations are behavioral, some are equilibrium conditions, and some are deﬁnitions. In principle, given the values of the exogenous variables, we solve to ﬁnd the endogenous variables as functions of the exogenous variables: x1 = x1 (y1 , y2 , . . . , ym ) x2 = x1 (y1 , y2 , . . . , ym ) . . . xn = x1 (y1 , y2 , . . . , ym ). If the equations are all linear in (x1 , x2 , . . . , xn ), then we can use Cramer’s rule (to be discussed in the next part) to solve the equations. However, if some equations are nonlinear, it is usually very diﬃcult to solve the model. In general, we use comparative statics method (to be discussed in part 3) to ﬁnd the diﬀerential relationships between ∂x xi and yj : ∂yji .

4 1.6 Problems Qd = 3 − P 2 ,

1. Find the equilibrium solution of the following model: Qs = 6P − 4, Qs = Qd .

2. The demand and supply functions of a two-commodity model are as follows: Qd1 = 18 − 3P1 + P2 Qs1 = −2 + 4P1 Qd2 = 12 + P1 − 2P2 Qs2 = −2 + 3P2

Find the equilibrium of the model.

3. (The eﬀect of a sales tax) Suppose that the government imposes a sales tax of t dollars per unit on product 1. The partial market model becomes Qd = D(P1 + t), 1 Qs = S(P1 ), 1 Qd = Qs . 1 1 Eliminating Qd and Qs , the equilibrium price is determined by D(P1 + t) = 1 1 S(P1 ). (a) Identify the endogenous variables and exogenous variable(s). (b) Let D(p) = 120 − P and S(p) = 2P . Calculate P1 and Q1 both as function of t. (c) If t increases, will P1 and Q1 increase or decrease? 4. Let the national-income Y = C = G = (a) (b) (c) (d) model be: C + I0 + G a + b(Y − T0 ) gY

(a > 0, 0 < b < 1) (0 < g < 1)

Identify the endogenous variables. Give the economic meaning of the parameter g. Find the equilibrium national income. What restriction(s) on the parameters is needed for an economically reasonable solution to exist? C = 25 + 6Y 1/2 ,

5. Find the equilibrium Y and C from the following: Y = C + I0 + G0 , I0 = 16, G0 = 14.

6. In a 2-good market equilibrium model, the inverse demand functions are given by −2 −2 1 1 3 3 P1 = Q13 Q2 , P2 = Q1 Q23 . (a) Find the demand functions Q1 = D 1 (P1 , P2 ) and Q2 = D 2 (P1 , P2 ). (b) Suppose that the supply functions are Q1 = a−1 P1 , Q2 = P2 .
∗ ∗ Find the equilibrium prices (P1 , P2 ) and quantities (Q∗ , Q∗ ) as functions 1 2 of a.

5

2

Matrix Algebra

There are n columns each with m elements or m rows each with n elements. We say that the size of A is m × n. If m = n, then A is a square matrix. A m × 1 matrix is called a column vector and a 1 × n matrix is called a row vector. A 1 × 1 matrix is just a number, called a scalar number. 2.1 Matrix operations

A matrix is a two dimensional rectangular array of numbers:   a11 a12 . . . a1n  a21 a22 . . . a2n    A≡ . . .  .. . . .   . . . . am1 am2 . . . amn

Equality: A = B ⇒ (1) size(A) = size(B), (2) aij = bij for all ij. Addition/subtraction: A + B and A − B can be deﬁned only when size(A) = size(B), in that case, size(A + B) = size(A − B) = size(A) = size(B) and (A + B)ij = aij + bij , (A − B)ij = aij − bij . For example, A= 1 2 3 4 , B= 1 0 0 1 ⇒A+B = 2 2 3 5 , A−B = 0 2 3 3 .

Scalar multiplication: The multiplication of a scalar number α and a matrix A, denoted by αA, is always deﬁned with size (αA) = size(A) and (αA)ij = αaij . For 1 2 4 8 example, A = , ⇒ 4A = . 3 4 12 16 Multiplication of two matrices: Let size(A) = m × n and size(B) = o × p, the multiplication of A and B, C = AB, is more complicated. (1) it is not always deﬁned. (2) AB = BA even when both are deﬁned. The condition for AB to be meaningful is that the number of columns of A should be equal to the number of rows of B, i.e., n = o. In that case, size (AB) = size (C) = m × p.      a11 a12 . . . a1n b11 b12 . . . b1p c11 c12 . . . c1p  a21 a22 . . . a2n   b21 b22 . . . a2p   c21 c22 . . . c2p      A= . . . . .. .  ⇒C =  . . .. . .. . . , B =  . . .  .   .  . . . .  . . . . . . . . . . . am1 am2 . . . amn bn1 bn2 . . . anp cm1 cm2 . . . cmp where cij = Examples: 1 2 3 4 n k=1 aik bkj .



  , 

1 2 3 3+8 11 = = , 0 5 4 0 + 20 20 5 6 5 + 14 6 + 16 19 22 = = 7 8 15 + 28 18 + 32 43 50

,

6 1 2 = 3 4 1 2 Notice that 3 4 2.2 5 6 7 8 5 + 18 10 + 24 7 + 24 14 + 32 5 6 5 6 = 7 8 7 8 23 34 31 46 1 2 . 3 4

=

.

Matrix representation of a linear simultaneous equation system

A linear simultaneous equation system: a11 x1 + . . . + a1n xn = b1 . . . . . . an1 x1 + . . . + ann xn = bn      a1n x1 b1 . , x ≡  .  and b ≡  . . Then the equation . . .   .   .  . ann xn bn the simultaneous equation system.

a11 . . .  . .. Deﬁne A ≡  . . . an1 . . . Ax = b is equivalent to Linear 2-market model:



E1 = (a1 − b1 )p1 + (a2 − b2 )p2 + (a0 − b0 ) = 0 E2 = (α1 − β1 )p1 + (α2 − β2 )p2 + (α0 − β0 ) = 0 ⇒ a1 − b1 a2 − b2 α1 − β1 α2 − β2 p1 p2 + a0 − b0 α0 − β0 = 0 0 .

Income determination model:

In the algebra of real numbers, the solution to the equation ax = b is x = a−1 b. In matrix algebra, we wish to deﬁne a concept of A−1 for a n × n matrix A so that x = A−1 b is the solution to the equation Ax = b. 2.3 Commutative, association, and distributive laws

     C = a + bY 1 0 −b C a I = I(r) ⇒  0 1 0  I  =  I(r) . Y =C +I 1 1 −1 Y 0

The notations for some important sets are given by the following table. N = nature numbers 1, 2, 3, . . . I = integers . . . , −2, −1, 0, 1, 2, . . . m R = real numbers Q = rational numbers n Rn = n-dimensional column vectors M(m, n) = m × n matrices M(n) = n × n matrices A binary operation is a law of composition of two elements from a set to form a third element of the same set. For example, + and × are binary operations of real numbers R. Another important example: addition and multiplication are binary operations of matrices.

7 Commutative law of + and × in R: a + b = b + a and a × b = b × a for all a, b ∈ R. Association law of + and × in R: (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c) for all a, b, c ∈ R. Distributive law of + and × in R: (a+b)×c = a×c+b×c and c×(a+b) = c×a+c×b for all a, b, c ∈ R. The addition of matrices satisﬁes both commutative and associative laws: A + B = B + A and (A + B) + C = A + (B + C) for all A, B, C ∈ M(m, n). The proof is trivial. In an example, we already showed that the matrix multiplication does not satisfy the commutative law AB = BA even when both are meaningful. Nevertheless the matrix multiplication satisﬁes the associative law (AB)C = A(BC) when the sizes are such that the multiplications are meaningful. However, this deserves a proof! It is also true that matrix addition and multiplication satisfy the distributive law: (A + B)C = AC + BC and C(A + B) = CA + CB. You should try to prove these statements as exercises. 2.4 Special matrices

In the space of real numbers, 0 and 1 are very special. 0 is the unit element of + and 1 is the unit element of ×: 0 + a = a + 0 = a, 0 × a = a × 0 = 0, and 1 × a = a × 1 = a. In matrix algebra, we deﬁne zero matrices and identity matrices as     1 0 ... 0 0 ... 0  0 1 ... 0     . .. .  Om,n ≡  . . .  In ≡  . . . . . . . . .  . .  . . . . 0 ... 0 0 0 ... 1 Clearly, O+A = A+O = A, OA = AO = O, and IA = AI = A. In the multiplication 1 0 0 0 0 0 of real numbers if a, b = 0 then a×b = 0. However, = = 0 0 0 1 0 0 O2,2 . Idempotent matrix: If AA = A (A must be square), then A is an idempotent matrix. 0.5 0.5 Both On,n and In are idempotent. Another example is A = . 0.5 0.5 Transpose of a matrix: For a matrix A with size m × n, we deﬁne its transpose A′ as a matrix with size n × m such that the ij-th element of A′ is equal to the ji-th element of A, a′ij = aji .   1 4 1 2 3 A= then A′ =  2 5 . 4 5 6 3 6 Properties of matrix transposition: (1) (A′ )′ = A, (2) (A + B)′ = A′ + B ′ , (3) (AB)′ = B ′ A′ . Symmetrical matrix: If A = A′ (A must be square), then A is symmetrical. The

8 condition for A to be symmetrical is that aij = aji . Both On,n and In are symmetrical. 1 2 Another example is A = . 2 3 Projection matrix: A symmetrical idempotent matrix is a projection matrix. Diagonal matrix: A symmetrical matrix A is diagonal if aij = 0 for all i = j. Both   λ1 0 0 In and On,n are diagonal. Another example is A =  0 λ2 0  0 0 λ3 2.5 Inverse of a square matrix We are going to deﬁne the inverse of a square matrix A ∈ M(n). Scalar: aa−1 = a−1 a = 1 ⇒ if b satisﬁes ab = ba = 1 then b = a−1 . Deﬁnition of A−1 : If there exists a B ∈ M(n) such that AB = BA = In , then we deﬁne A−1 = B. −1 Examples: (1) Since II = I, I −1 = I. (2) On,n B = On,n ⇒ On,n does not exist. (3) a1 0 a−1 0 1 If A = , a1 , a2 = 0, then A−1 = . (4) If a1 = 0 or a2 = 0, 0 a2 0 a−1 2 then A−1 does not exist. Singular matrix: A square matrix whose inverse matrix does not exist. Non-singular matrix: A is non-singular if A−1 exists. Properties of matrix inversion: Let A, B ∈ M(n), (1) (A−1 )−1 = A, (2) (AB)−1 = B −1 A−1 , (3) (A′ )−1 = (A−1 )′ . 2.6 Problems

1. Let A = I − X(X ′X)−1 X ′ . (a) If the dimension of X is m × n, what must be the dimension of I and A. (b) Show that matrix A is idempotent. 2. Let A and B be n × n matrices and I be the identity matrix. (a) (A + B)3 = ? (b) (A + I)3 = ? 3. Let B = 0.5 0.5 , U = (1, 1)′, V = (1, −1)′ , and W = aU + bV , where 0.5 0.5 a and b are real numbers. Find BU, BV , and BW . Is B idempotent?

4. Suppose A is a n × n nonsingular matrix and P is a n × n idempotent matrix. Show that AP A−1 is idempotent. 5. Suppose that A and B are n × n symmetric idempotent matrices and AB = B. Show that A − B is idempotent. 6. Calculate (x1 , x2 ) 3 2 2 5 x1 x2 .

9 1 0 0 1 0 1 −1 0

7. Let I =

and J =

.

(b) Make use of the above result to calculate J 3 , J 4 , and J −1 . (c) Show that (aI + bJ)(cI + dJ) = (ac − bd)I + (ad + bc)J. 1 (d) Show that (aI + bJ)−1 = 2 (aI − bJ) and [(cos θ)I + (sin θ)J]−1 = a + b2 (cos θ)I − (sin θ)J.

(a) Show that J 2 = −I.

10

3

Vector Space and Linear Transformation

In the last section, we regard a matrix simply as an array of numbers. Now we are going to provide some geometrical meanings to a matrix. (1) A matrix as a collection of column (row) vectors (2) A matrix as a linear transformation from a vector space to another vector space

3.1

Vector space, linear combination, and linear independence

Each point in the m-dimensional Euclidean space can be represented as a m-dimensional   v1  .  column vector v =  . , where each vi represents the i-th coordinate. Two points . vm in the m-dimensional Euclidean space can be added according to the rule of matrix addition. A point can be multiplied by a scalar according to the rule of scalar multiplication.       v1 w1 v1 + w1  .   .    . . Vector addition:  .  +  .  =  . . . . vm w vm + wm  m  v1 αv1  .   .  Scalar multiplication: α .  =  . . . . vm αvm

x2

T

v2 0 v 1 Q

1 2 v +v 0

x2

T

E x1

v E x1

2v

With such a structure, we say that the m-dimensional Euclidean space is a vector space.

11     m m-dimensional column vector space: R =       v1  . , v ∈ R . .  i .   vm

We use superscripts to represent individual vectors.

A m × n matrix: a collection of n m-dimensional column vectors:         a11 a11 a12 . . . a1n a12 a1n   a21 a22 . . . a2n   a21   a22    a2n         . . . .. . . .  =  . ,  . , . . . ,  . .   .   .  . . . .   . . .   a am1 am2 . . . amn am2 amn m1 Linear combination of a collection of vectors {v , . . . , v }: w = (α1 , . . . , αn ) = (0, . . . , 0).
1 n

n

           αi v i , where

i=1

Linear dependence of {v 1 , . . . , v n }: If one of the vectors is a linear combination of others, then the collection is said to be linear dependent. Alternatively, the collection is linearly dependent if (0, . . . , 0) is a linear combination of it. Linear independence of {v 1 , . . . , v n }: If the collection is not linear dependent, then it is linear independent.      a1 0 0 Example 1: v 1 =  0 , v 2 =  a2 , v 3 =  0 , a1 a2 a3 = 0. 0 0 a3 If α1 v 1 + α2 v 2 + α3 v 3 = 0 then (α1 , α2 , α3 ) = (0, 0, 0). Therefore, {v 1 , v 2 , v 3 } must be linear independent.       1 4 7 1  2 , v 2 =  5 , v 3 =  8 . Example 2: v = 3 6 9 2 1 3 1 2 3 , is 2v = v + v . Therefore, {v , v v } linear dependent.   1 4 Example 3: v 1 =  2 , v 2 =  5 .  6   3 0 α1 + 4α2 α1 v 1 + α2 v 2 =  2α1 + 5α2  =  0  ⇒ α1 = α2 = 0. Therefore, {v 1 , v 2 } is 0 3α1 + 6α2 linear independent. Span of {v 1 , . . . , v n }: The space of linear combinations. If a vector is a linear combination of other vectors, then it can be removed without changing the span. 

12  v11 . . . v1n  . .  ≡ Dimension(Span{v 1 , . . . , v n }) = Maximum # of indepen.. .  Rank . . . . vm1 . . . vmn dent vectors. 

3.2

Linear transformation

Consider a m × n matrix A. Given x ∈ Rn , Ax ∈ Rm . Therefore, we can deﬁne a linear transformation from Rn to Rm as f (x) = Ax or    y1 a11 . . . a1n  .   . . .. .  f : Rn →Rm , f (x) =  .  =  . . . . .  ym am1 . . . amn   x1 . . .  . xn

It is linear because f (αx + βw) = A(αx + βw) = αAx + βAw = αf (x) + βf (w).       0 0 1  0   1   0        Standard basis vectors of Rn : e1 ≡  . , e2 ≡  . , . . . , en ≡  . . .  .   .   .  . . 1 0 0   a1i  .  i i Let v be the i-th column of A, v =  . . . ami      a11 . . . a1n 1 a11 .  .  ⇒ v 1 = f (e1 ) = Ae1 , etc. .. i i  ...  =  . .  .  v = f (e ):  . . . . . am1 0 am1 . . . amn Therefore, v i is the image of the i-th standard basis vector ei under f . Span{v 1 , . . . , v n } = Range space of f (x) = Ax ≡ R(A). Rank(A) ≡ dim(R(A)). Null space of f (x) = Ax: N(A) ≡ {x ∈ Rn , f (x) = Ax = 0}. dim(R(A)) + dim(N(A)) = n. Example 1: A = Example 2: B = Rank(B) = 1. 1 0 1 1 0 . N(A) = 2 1 . N(B) = 1 0 0 k −k , R(A) = R2 , Rank(A) = 2. , k ∈ R , R(B) = k k ,k ∈ R ,

The multiplication of two matrices can be interpreted as the composition of two linear transformations. f : Rn →Rm , f (x) = Ax, g : Rp →Rn , g(y) = By, ⇒ f (g(x)) = A(By), f ◦g : Rp →Rm .

13 The composition is meaningful only when the dimension of the range space of g(y) is equal to the dimension of the domain of f (x), which is the same condition for the validity of the matrix multiplication. Every linear transformation f : Rn →Rm can be represented by f (x) = Ax for some m × n matrix. 3.3 Inverse transformation and inverse of a square matrix

Remark: On,n represents the mapping that maps every point to the origin. In represents the identity mapping that maps a point to itself. A projection matrix repre0.5 0.5 sents a mapping that projects points onto a linear subspace of Rn , eg., 0.5 0.5 projects points onto the 45 degree line. d x d d d Ax d d d d d d E x1 s d d d d d d d d T

Consider now the special case of square matrices. Each A ∈ M(n) represents a linear transformation f : Rn →Rn . The deﬁnition of the inverse matrix A−1 is such that AA−1 = A−1 A = I. If we regard A as a linear transformation from Rn →Rn and I as the identity transformation that maps every vector (point) into itself, then A−1 is the inverse mapping of A. If dim(N(A)) = 0, then f (x) is one to one. If dim(R(A)) = n, then R(A) = Rn and f (x) is onto. ⇒ if Rank(A) = n, then f (x) is one to one and onto and there exits an inverse mapping f −1 : Rn →Rn represented by a n×n square matrix A−1 . f −1 f (x) = x ⇒ A−1 Ax = x. ⇒ if Rank(A) = n, then A is non-singular. if Rank(A) < n,  then f (x) is not onto, no inverse mapping  exists, and A is singular.   a1 0 0 1 4 7 Examples: Rank 0 a2 0  = 3 and Rank 2 5 8  = 2. 0 0 a3 3 6 9

x2

x= Ax = x′ =

1 2 .5 .5 1 .5 .5 2 k , Ax′ = −k = 0 0 1.5 1.5

14 3.4 Problems 

 0 1 0 1. Let B =  0 0 1  0 0 0

and TB the corresponding linear transformation TB : R3 → R3 , TB (x) = Bx,  x1 where x =  x2  ∈ R3 . x3   a (a) Is v 1 =  0 , a = 0, in the null space of TB ? Why or why not? 0   0 (b) Is v 2 =  0 , b = 0, in the range space of TB ? Why or why not? How b  c 3  d ? about v = 0 (c) Find Rank(B).

2. Let A be an idempotent matrix. (a) Show that I − A is also idempotent.

(b) Suppose that x = 0 is in the null space of A, i.e., Ax = 0. Show that x must be in the range space of I − A, i.e., show that there exists a vector y such that (I − A)y = x. (Hint: Try y = x.) (c) Suppose that y is in the range space of A. Show that y must be in the null space of I − A.

(d) Suppose that A is n × n and Rank[A] = n − k, n > k > 0. What is the rank of I − A?    1 1 1      1 0 0 1 1 3 3 3  0 1 0 , A =  1 1 1 , x =  a , y =  α , and 3. Let I = 3 3 3 1 1 1 0 0 1 b β 3 3 3 B = I − A. (a) Calculate AA and BB. (b) If y is in the range space of A, what are the values of α and β? (c) What is the dimension of the range space of A? (d) Determine the rank of A. (e) Suppose now that x is in the null space of B. What should be the values of a and b? (f) What is the dimension of the null space of B?

15 (g) Determine the rank of B? 4. Let A = 1/5 2/5 2/5 4/5  1 1 1 and B =  0 1 1 . 0 0 1 

(a) Determine the ranks of A and B.

(b) Determine the null space and range space of each of A and B and explain why. (c) Determine whether they are idempotent.

16

4

Determinant, Inverse Matrix, and Cramer’s rule

In this section we are going to derive a general method to calculate the inverse of a square matrix. First, we deﬁne the determinant of a square matrix. Using the properties of determinants, we ﬁnd a procedure to compute the inverse matrix. Then we derive a general procedure to solve a simultaneous equation. 4.1 Permutation group

A permutation of {1, 2, . . . , n} is a 1-1 mapping of {1, 2, . . . , n} onto itself, written as 1 2 ... n π= meaning that 1 is mapped to i1 , 2 is mapped to i2 , . . ., and i1 i2 . . . in n is mapped to in . We also write π = (i1 , i2 , . . . , ın ) when no confusing. Permutation set of {1, 2, . . . , n}: Pn ≡ {π = (i1 , i2 , . . . , in ) : π is a permutation}. P2 = {(1, 2), (2, 1)}. P3 = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}. P4 : 4! = 24 permutations. Inversions in a permutation π = (i1 , i2 , . . . , in ): If there exist k and l such that k < l and ik > il , then we say that an inversion occurs. N(i1 , i2 , . . . , ın ): Total number of inversions in (i1 , i2 , . . . , in ). Examples: 1. N(1, 2) = 0, N(2, 1) = 1. 2. N(1, 2, 3) = 0, N(1, 3, 2) = 1, N(2, 1, 3) = 1, N(2, 3, 1) = 2, N(3, 1, 2) = 2, N(3, 2, 1) = 3. 4.2 Determinant  a11 a12 a21 a22 . . . . . . an1 an2  . . . a1n . . . a2n   . : .. .  . . . . . ann (−1)N (i1 ,i2 ,...,in ) a1i1 a2i2 . . . anin .
(i1 ,i2 ,...,in )∈Pn

  Determinant of A =  

|A| ≡ a11 a12 a21 a22

n = 2:

= (−1)N (1,2) a11 a22 + (−1)N (2,1) a12 a21 = a11 a22 − a12 a21 .

a11 a12 a13 n = 3: a21 a22 a23 = a31 a32 a33 N (1,2,3) a11 a22 a33 + (−1)N (1,3,2) a11 a23 a32 + (−1)N (2,1,3) a12 a21 a33 (−1) (−1)N (2,3,1) a12 a23 a31 + (−1)N (3,1,2) a13 a21 a32 + (−1)N (3,2,1) a13 a22 a31 = a11 a22 a33 − a11 a23 a32 − a12 a21 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 .

+

17 a11 a11 n = 2: a21 a12 d d

a12 d d

a13
d d

a11
d d d d

a12
d d

n = 3:

a21 a31

a22 a32

a23 a33

a21 a31

a22 a32

a22

d d

4.3

Properties of determinant

Property 1: |A′ | = |A|. Proof: Each term of |A′ | corresponds to a term of |A| of the same sign. By property 1, we can replace “column vectors” in the properties below by “row vectors”. Since a n × n matrix can be regarded as n column vectors A = {v 1 , v 2, . . . , v n }, we can regard determinants as a function of n column vectors |A| = D(v 1 , v 2 , . . . , v n ), D : Rn×n →R. By property 1, we can replace “column vectors” in the properties below by “row vectors”. Property 2: If two column vectors are interchanged, the determinant changes sign. Proof: Each term of the new determinant corresponds to a term of |A| of opposite sign because the number of inversion increases or decreases by 1. 2 1 1 2 = 2 × 3 − 1 × 4 = 2, = 1 × 4 − 2 × 3 = −2, Example: 4 3 3 4 Property 3: If two column vectors are identical, then the determinant is 0. Proof: By property 2, the determinant is equal to the negative of itself, which is possible only when the determinant is 0. Property 4: If you add a linear combination of other column vectors to a column vector, the determinant does not change. Proof: Given other column vectors, the determinant function is a linear function of v i : D(αv i + βw i; other vectors ) = αD(v i ; other vectors ) + βD(w i; other vectors ). 1+5×2 2 1 2 5×2 2 1 2 2 2 Example: = + = +5× = 3+5×4 4 3 4 5×4 4 3 4 4 4 1 2 + 5 × 0. 3 4 Submatrix: We denote by Aij the submatrix of A obtained by deleting the i-th row and j-th column from A. Minors: The determinant |Aij | is called the minor of the element aij . Cofactors: Cij ≡ (−1)i+j |Aij | is called the cofactor of aij .

18 n Property 5 (Laplace theorem): Given i = ¯ |A| = i,

a¯ C¯ . ij ij j=1 Given j = ¯ |A| = n ai¯Ci¯. j, j i=1 j Proof: In the deﬁnition of the determinant of |A|, all terms with aij can be put togethere to become aij Cij . 1 2 3 5 6 4 6 4 5 Example: 4 5 6 = 1 × −2× +3× . 8 0 7 0 7 8 7 8 0 n Property 6: Given i = ¯ i,
′ j=1

ai′ j C¯ = 0. ij

Given j ′ = ¯ = n aij ′ Ci¯ = 0. j, j i=1 Therefore, if you multiply cofactors by the elements from a diﬀerent row or column, you get 0 instead of the determinant. Proof: The sum becomes the determinant of a matrix with two identical rows (columns). 5 6 4 6 4 5 Example: 0 = 4 × −5× +6× . 8 0 7 0 7 8 4.4 Computation of the inverse matrix

Using properties 5 and 6, can calculate the inverse of A as follows. we  C11 C12 . . . C1n  C21 C22 . . . C2n    1. Cofactor matrix: C ≡  . . . . .. . .  .  . . . . Cn1 Cn2 . . . Cnn   C11 C21 . . . Cn1  C12 C22 . . . Cn2    ′ 2. Adjoint of A: Adj A ≡ C =  . . . .. . . . .  . . . .  C1n C2n . . . Cnn   |A| 0 . . . 0  0 |A| . . . 0    1 ′ −1 3. ⇒ AC ′ = C ′ A =  . . .. .  ⇒ if |A| = 0 then |A| C = A . .  . . .  . . . 0 0 . . . |A| a11 a12 a22 −a21 Example 1: A = then C = . a21 a22 −a12 a11 1 ′ 1 a22 −a12 C = ; if |A| = a11 a22 − a12 a21 = 0. |A| a11 a22 − a12 a21 −a21 a11   1 2 3 Example 2: A =  4 5 6  ⇒ |A| = 27 = 0 and 7 8 0 5 6 4 6 4 5 C11 = = −48, C12 = − = 42, C13 = = −3, 8 0 7 0 7 8 A−1 =

19 2 3 = 24, C22 = 8 0 2 3 C31 = = −3, C32 = − 5 6   −48 42 −3 C =  24 −21 6 , C ′ −3 6 −3 C21 = − 3 1 2 = −21, C23 = − = 6, 0 7 8 3 1 2 = 6, C33 = = −3, 6 4 5     −48 24 −3 −48 24 −3 1 =  42 −21 6 , A−1 =  42 −21 6 . 27 −3 6 −3 −3 6 −3 1 7 1 4

If |A| = 0, then A is singular and A−1 does not  The reason is that |A| = exist.    a11 a1n 0  .   .   .  0 ⇒ AC ′ = 0n×n ⇒ C11  .  + · · · + Cn1  .  =  . . The column vectors . . . an1 ann 0 of A are linear dependent, the linear transformation TA is not onto and therefore an inverse transformation does not exist. 4.5 Cramer’s rule C′ . The solution to the simultaneous |A|

C ′b . |A| n |Ai | j=1 Cj bj = , where Ai is a matrix obtained by replacing Cramer’s rule: xi = |A| |A| the i-th column of A by b, Ai = {v 1 , . . . , v i−1, b, v i+1 , . . . , v n }. equation Ax = b is x = A−1 b = 4.6 Economic applications a1 − b1 a2 − b2 α1 − β1 α2 − β2 p1 p2 = b0 − a0 β0 − α0 .

If |A| = 0 then A is non-singular and A−1 =

Linear 2-market model:

b0 − a0 a2 − b2 β0 − α0 α2 − β2 p1 = a1 − b1 a2 − b2 α1 − β1 α2 − β2  1 Income determination model:  0 1 a 0 −b I(r) 1 0 0 1 −1 , 1 0 −b 0 1 0 1 1 −1

C=

I=

1 a −b 0 I(r) 0 1 0 −1 , 1 0 −b 0 1 0 1 1 −1

a1 − b1 b0 − a0 α1 − β1 β0 − α0 , p2 = . a1 − b1 a2 − b2 α1 − β1 α2 − β2     0 −b C a 1 0  I  =  I(r) . 1 −1 Y 0 1 0 1 1 0 1

Y =

0 a 1 I(r) 1 0 . 0 −b 1 0 1 −1

20 IS-LM model: In the income determination model, we regard interest rate as given and consider only the product market. Now we enlarge the model to include the money market and regard interest rate as the price (an endogenous variable) determined in the money market. good market: money market: C = a + bY L = kY − lR ¯ I = I0 − iR M =M ¯=Y C +I +G M =L end. var: C, I, Y , R (interest rate), L(demand for money), M (money supply) ¯ ¯ ex. var: G, M (quantity of money). parameters: a, b, i, k, l. Substitute into equilibrium conditions: good market: money market endogenous variables: ¯ ¯ a + bY + I0 − iR + G = Y , kY − lR = M, Y, R 1−b i k −l Y R = ¯ a + I0 + G ¯ M ¯ 1 − b a + I0 + G ¯ k M . 1−b i k −l

Y =

¯ a + I0 + G i ¯ M −l , 1−b i k −l

R=

Two-country income determination model: Another extension of the income determination model is to consider the interaction between domestic country and the rest of the world (foreign country). domestic good market: foreign good market: endogenous variables: C = a + bY C ′ = a′ + b′ Y ′ C, I, Y , ′ ′ I = I0 I = I0 M (import), ′ M = M0 + mY M ′ = M0 + m′ Y ′ X (export), ′ ′ ′ ′ ′ ′ C +I +X −M =Y C +I +X −M =Y C , I ′, Y ′, M ′, X ′. By deﬁnition, X = M ′ and X ′ = M. Substituting into the equilibrium conditions,
′ ′ ′ (1 − b + m)Y − m′ Y ′ = a + I0 + M0 − M0 (1 − b′ + m′ )Y ′ − mY = a′ + I0 + M0 − M0 .

1−b+m −m′ −m 1 − b′ + m′ Y =
′ a + I0 + M0 − M0 −m′ ′ ′ ′ a + I0 + M0 − M0 1 − b′ + m′ 1−b+m −m′ −m 1 − b′ + m′

Y Y′

=

′ a + I0 + M0 − M0 ′ ′ a′ + I0 + M0 − M0

.

Y′ =

′ 1 − b + m a + I0 + M0 − M0 ′ ′ ′ −m a + I0 + M0 − M0 . 1−b+m −m′ −m 1 − b′ + m′

4.7

Input-output table

Assumption: Technologies are all ﬁxed proportional, that is, to produce one unit of product Xi , you need aji units of Xj .

21  a11 a12 . . . a1n  a21 a22 . . . a2n    IO table: A =  . . .. . . . . .   . . . . an1 an2 . . . ann Column i represents the coeﬃcients of inputs needed to produce one unit of Xi .   x1  x2    Suppose we want to produce a list of outputs x =  . , we will need a list of inputs .   . xn   a11 x1 + a12 x2 + . . . + a1n xn  a21 x2 + a22 x2 + . . . + a2n xn    Ax =  . The net output is x − Ax = (I − A)x. . .   . an1 x1 + an2 x2 + . . . + ann xn   d1  d2    If we want to produce a net amount of d =  . , then since d = (I − A)x, .   . dn −1 x = (I − A) d.  4.8 A geometric interpretation of determinants

Because of properties 2 and 4, the determinant function D(v 1 , . . . , v n ) is called an alternative linear n-form of Rn . It is equal to the volume of the parallelepiped formed by the vectors {v 1 , . . . , v n }. For n = 2, |A| is the area of the parallelogram formed by a11 a21 , . See the diagram: a12 a22 x2
T

|A| = Area of D
¢ ¢ ¢ v 2 ¢ ¢ ¢ ¢ D ¢ ¢ ¢ ¢ ¢ ¢ 1 ¢ ¢ v I ¢ ¢ ¢

E x1

If the determinant is 0, then the volume is 0 and the vectors are linearly dependent, one of them must be a linear combination of others. Therefore, an inverse mapping does not exist, A−1 does not exist, and A is singular.

22 4.9 Rank of a matrix and solutions of Ax = d when |A| = 0

Rank(A) = the maximum # of independent vectors in A = {v 1 , . . . , v n } = dim(Range Space of TA ). Rank(A) = the size of the largest non-singular square submatrices of A.   1 2 3 1 2 1 2 Examples: Rank = 2. Rank 4 5 6  = 2 because is non3 4 4 5 7 8 9 singular. Property 1: Rank(AB) ≤ min{Rank(A), Rank(B)}. Property 2: dim(Null Space of TA ) + dim(Range Space of TA ) = n. Consider the simultaneous equation Ax = d. When |A| = 0, there exists a row of A that is a linear combination of other rows . and Rank(A) < n. First, form the augmented matrix M ≡ [A. and calculate .d] the rank of M. There are two cases. Case 1: Rank(M) = Rank (A). In this case, some equations are linear combinations of others (the equations are dependent) and can be removed without changing the solution space. There will be more variables than equations after removing these equations. Hence, there will be inﬁnite number of solutions. 1 2 x1 3 1 2 Example: = . Rank(A) = Rank = 1 = Rank(M) = 2 4 x2 6 2 4 1 2 3 Rank . 2 4 6 The second equation is just twice the ﬁrst equation and can be discarded. The sox1 3 − 2k lutions are = for any k. On x1 -x2 space, the two equations are x2 k represented by the same line and every point on the line is a solution. Case 2: Rank(M) = Rank(A) + 1. In this case, there exists an equation whose LHS is a linear combination of the LHS of other equations but whose RHS is diﬀerent from the same linear combination of the RHS of other equations. Therefore, the equation system is contraditory and there will be no solutions. 1 2 x1 3 1 2 Example: = . Rank(A) = Rank = 1 < Rank(M) = x2 2 4 7 2 4 1 2 3 Rank = 2. 2 4 7 Multiplying the ﬁrst equation by 2, 2x1 + 4x2 = 6, whereas the second equation x1 says 2x1 + 4x2 = 7. Therefore, it is impossible to have any satisfying both x2 equations simultaneously. On x1 -x2 space, the two equations are represented by two parallel lines and cannot have any intersection points.

23 4.10 Problems

1. Suppose v1 = (1, 2, 3)′, v2 = (2, 3, 4)′, and v3 = (3, 4, 5)′. Is {v1 , v2 , v3 } linearly independent? Why or why not? 6 5 . 2.. Find the inverse of A = 8 7   2 1 6 3. Given the 3 × 3 matrix A =  5 3 4 , 8 9 7 (a) calculate the cofactors C11 , C21 , C31 , (b) use Laplace expansion theorem to ﬁnd |A|, (c) and use Cramer’s rule to ﬁnd X1 of the following equation system:      2 1 6 X1 1  5 3 4   X2  =  2  . 8 9 7 X3 3 (Hint: Make use of the results of (a).) 4. Use Cramer’s rule to solve the national-income model C = a + b(Y − T ) T = −t0 + t1 Y Y = C + I0 + G 

(1) (2) (3)

 0 1 0 5. Let A =  0 0 1 . 0 0 0 (a) Find AA and AAA. (b) Let x = (1, 2, 3)′, compute Ax, AAx, and AAAx. (c) Find Rank[A], Rank[AA], and Rank[AAA].   1 −1 6. Let X =  1 0 . 1 1 ′ (a) Find X X and (X ′ X)−1 . (b) Compute X(X ′ X)−1 X ′ and I − X(X ′ X)−1 X ′ . (c) Find Rank[X(X ′ X)−1 X ′ ] and Rank[I − X(X ′X)−1 X ′ ]. 1 2 1 2 1 1 2 4 7. A = ,B= , and C = . 3 6 3 6 1 3 6 12 (a) Find the ranks of A, B, and C. (b) Use the results of (a) to determine whether the following system has any solution: 1 2 3 6 X1 X2 = 1 1 .

(c) Do the same for the following system: 1 2 3 6 X1 X2 = 4 12 .

24 3 2 , I the 2 × 2 identity matrix, and λ a scalar number. 1 2 (a) Find |A − λI|. (Hint: It is a quadratic function of λ.) (b) Determine Rank(A − I) and Rank(A − 4I). (Remark: λ = 1 and λ = 4 are the eigenvalues of A, that is, they are the roots of the equation |A − λI| = 0, called the characteristic equation of A.) (c) Solve the simultaneous equation system (A − I)x = 0 assuming that x1 = 1. (Remark: The solution is called an eigenvector of A associated with the eigenvalue λ = 1.) (d) Solve the simultaneous equation system (A − 4I)y = 0 assuming that y1 = 1. (e) Determine whether the solutions x and y are linearly independent. 8. Let A =

25

5

Diﬀerential Calculus and Comparative Statics

As seen in the last chapter, a linear economic model can be represented by a matrix equation Ax = d(y) and solved using Cramer’s rule, x = A−1 d(y). On the other hand, a closed form solution x = x(y) for a nonlinear economic model is, in most applications, impossible to obtain. For general nonlinear economic models, we use diﬀerential calculus (implicit function theorem) to obtain the derivatives of endogenous variables ∂xi with respect to exogenous variables : ∂yj f1 (x1 , . . . , xn ; y1 , . . . , ym ) = 0 . . . fn (x1 , . . . , xn ; y1, . . . , ym ) = 0      ∂x1 ∂x1 ∂f1 ∂f1 ∂f1 ∂f1  ∂x1 . . . ∂xn  ∂y1 . . . ∂ym   ∂y1 . . . ∂ym   .  .  .  . .  = − . . . .. .. ..  . .  . .  .  ⇒ . . . . . . . . .     ∂fn  ∂xn   ∂fn ∂fn ∂xn ∂fn  ... ... ... ∂x1 ∂xn ∂y1 ∂ym ∂y1 ∂ym    −1   ∂x1 ∂x1 ∂f1 ∂f1 ∂f1 ∂f1  ∂x1 . . . ∂xn   ∂y1 . . . ∂ym   ∂y1 . . . ∂ym   .  .  = − . .   . . . .. .. ..  . .  .   . .  ⇒ . . . . . . .  . .   .    ∂fn  ∂xn ∂xn  ∂fn   ∂fn ∂fn  ... ... ... ∂y1 ∂ym ∂x1 ∂xn ∂y1 ∂ym ∂xi ∂xi represents a cause-eﬀect relationship. If > 0 (< 0), then xi will increase Each ∂yj ∂yj (decrease) when yj increases. Therefore, instead of computing xi = xi (y), we want to ∂xi determine the sign of for each i-j pair. In the following, we will explain how it ∂yj works. 5.1 Diﬀerential Calculus x = f (y) ⇒ f ′ (y ∗) = dx dy ≡ lim f (y ∗ + ∆y) − f (y ∗) . ∆y→0 ∆y

y=y ∗

On y-x space, x = f (y) is represented by a curve and f ′ (y ∗ ) represents the slope of the tangent line of the curve at the point (y, x) = (y ∗, f (y ∗)). Basic rules: dx 1. x = f (y) = k, = f ′ (y) = 0. dy dx 2. x = f (y) = y n , = f ′ (y) = ny n−1 . dy dx 3. x = cf (y), = cf ′ (y). dy

26 4. x = f (y) + g(y), 5. x = f (y)g(y), dx = f ′ (y) + g ′ (y). dy

dx = f ′ (y)g(y) + f (y)g ′(y). dy dx f ′ (y)g(y) − f (y)g ′(y) 6. x = f (y)/g(y), = . dy (g(y))2 dx dx 1 7. x = eay , = aeay . x = ln y, = . dy dy y dx dx 8. x = sin y, = cos y. x = cos y, = − sin y. dy dy Higher order derivatives: d d d2 d d2 f ′′ (y) ≡ f (y) = 2 f (y), f ′′′ (y) ≡ f (y) dy dy dy dy dy 2 5.2 Partial derivatives

=

d3 f (y). dy 3

In many cases, x is a function of several y’s: x = f (y1 , y2 , . . . , yn ). The partial ∗ ∗ ∗ derivative of x with respect to yi evaluated at (y1 , y2 , . . . , yn ) = (y1 , y2 , . . . , yn ) is ∂x ∂yi
∗ ∗ ∗ (y1 ,y2 ,...,yn )

∗ ∗ ∗ ∗ ∗ ∗ f (y1 , . . . , yi + ∆yi , . . . , yn ) − f (y1 , . . . , yi , . . . , yn ) ≡ lim , ∆yi →0 ∆yi

that is, we regard all other independent variables as constant (f as a function of yi only) and take derivative. ∂xn xm 1 2 9. = nxn−1 xm . 1 2 ∂x1 Higher order derivatives: We can deﬁne higher order derivatives as before. For the case with two independent variables, there are 4 second order derivatives: ∂ ∂x ∂2x ∂ ∂x ∂2x ∂ ∂x ∂2x ∂ ∂x ∂2x = 2, = , = , = 2. ∂y1 ∂y1 ∂y1 ∂y2 ∂y1 ∂y2 ∂y1 ∂y1 ∂y2 ∂y1 ∂y2 ∂y2 ∂y2 ∂y2 Notations: f1 , f2 , f11 , f12 , f21 , f22 .   f1  .  ∇f ≡  . : Gradient vector of f . . f n  f11 . . . f1n  . .  H(f ) ≡  . . . . . : second order derivative matrix, called Hessian of f . . . fn1 . . . fnn Equality of cross-derivatives: If f is twice continously diﬀerentiable, then fij = fji and H(f ) is symmetric. 5.3 Economic concepts similar to derivatives

Elasticity of Xi w.r.t. Yj : EXi ,Yj ≡

Yj ∂Xi , the percentage change of Xi when Yj Xi ∂Yj P dQd increases by 1 %. Example: Qd = D(P ), EQd ,P = Qd dP

27 Basic rules: 1. EX1 X2 ,Y = EX1 ,Y + EX2 ,Y , 3. EY,X = 1/EX,Y . 2. EX1 /X2 ,Y = EX1 ,Y − EX2 ,Y ,

1 dX Growth rate of X = X(t): GX ≡ , the percentage change of X per unit of X dt time. 5.4 Mean value and Taylor’s Theorems

Continuity theorem: If f (y) is continuous on the interval [a, b] and f (a) ≤ 0, f (b) ≥ 0, then there exists a c ∈ [a, b] such that f (c) = 0. Rolle’s theorem: If f (y) is continuous on the interval [a, b] and f (a) = f (b) = 0, then there exists a c ∈ (a, b) such that f ′ (c) = 0. Mean value theorem: If f (y) is continously diﬀerentiable on [a, b], then there exists a c ∈ (a, b) such that f (b) − f (a) = f ′ (c)(b − a) or f (b) − f (a) = f ′ (c). b−a

Taylor’s Theorem: If f (y) is k + 1 times continously diﬀerentiable on [a, b], then for each y ∈ [a, b], there exists a c ∈ (a, y) such that f (y) = f (a) + f ′(a)(y − a) + 5.5 f (k) (a) f (k+1) (c) f ′′ (a) (y − a)2 + . . .+ (y − a)k + (y − a)k+1 . 2! k! (k + 1)!

Concepts of diﬀerentials and applications

Let x = f (y). Deﬁne ∆x ≡ f (y + ∆y) − f (y), called the ﬁnite diﬀerence of x. ∆x f (y + ∆y) − f (y) ∆x Finite quotient: = ⇒ ∆x = ∆y. ∆y ∆y ∆y dx, dy: Inﬁnitesimal changes of x and y, dx, dy > 0 (so that we can divid something by dx or by dy) but dx, dy < a for any positive real number a (so that ∆y→dy). Diﬀerential of x = f (y): dx = df = f ′ (y)dy. Chain rule: x = f (y), y = g(z) ⇒ x = f (g(z)),

dx = f ′ (y)g ′(z) = f ′ (g(z))g ′(z). dz dx = 3y 2 2z = 6z(z 2 + 1)2 . Example: x = (z 2 + 1)3 ⇒ x = y 3, y = z 2 + 1 ⇒ dz Inverse function rule: x = f (y), ⇒ y = f −1 (x) ≡ g(x), dy 1 dx = f ′ (y)dy, dy = g ′(x)dx ⇒ dx = f ′ (y)g ′(x)dx. = g ′(x) = ′ . dx f (y) dx 1 1 Example: x = ln y ⇒ y = ex ⇒ = x = . dy e y dx = f ′ (y)dy, dy = g ′(z)dz ⇒ dx = f ′ (y)g ′(z)dz.

28 5.6 Concepts of total diﬀerentials and applications

Let x = f (y1 , y2). Deﬁne ∆x ≡ f (y1 + ∆y1 , y2 + ∆y2 ) − f (y1, y2 ), called the ﬁnite diﬀerence of x. ∆x = f (y1 + ∆y1 , y2 + ∆y2 ) − f (y1, y2 ) = f (y1 + ∆y1 , y2 + ∆y2 ) − f (y1, y2 + ∆y2 ) + f (y1 , y2 + ∆y2 ) − f (y1 , y2) f (y1 + ∆y1 , y2 + ∆y2 ) − f (y1, y2 + ∆y2 ) f (y1, y2 + ∆y2 ) − f (y1 , y2 ) = ∆y1 + ∆y2 ∆y1 ∆y2 dx = f1 (y1 , y2 )dy1 + f2 (y1 , y2 )dy2.   dy1  .  dx, dy =  . : Inﬁnitesimal changes of x (endogenous), y1 , . . . , yn (exogenous). . dyn Total diﬀerential of x = f (y1 , . . . , yn ):   dy1  .  dx = df = f1 (y1 , . . . , yn )dy1 +. . .+fn (y1 , . . . , yn )dyn = (f1 , . . . , fn ) .  = (∇f )′ dy. . dyn

Implicit function rule: In many cases, the relationship between two variables are deﬁned implicitly. For ¯ example, the indiﬀerence curve U(x1 , x2 ) = U deﬁnes a relationship between x1 and dx2 , we use implicit function rule. x2 . To ﬁnd the slope of the curve dx1 U1 (x1 , x2 ) dx2 ¯ =− . dU = U1 (x1 , x2 )dx1 + U2 (x1 , x2 )dx2 = dU = 0 ⇒; dx1 U2 (x1 , x2 )
1 1 3 3 Example: U(x1 , x2 ) = 3x1 + 3x2 = 6 deﬁnes an indiﬀerence curve passing through the point (x1 , x2 ) = (1, 1). The slope (Marginal Rate of Substitution) at (1, 1) can be calculated using implicit function rule.

dx2 U1 x 3 1 =− = − 1 2 = − = −1. − dx1 U2 1 x 3
2

−2

Multivariate chain rule: x = f (y1, y2 ), y1 = g 1(z1 , z2 ), y2 = g 2 (z1 , z2 ), ⇒ x = f (g 1(z1 , z2 ), g 2(z1 , z2 )) ≡ H(z1 , z2 ). We can use the total diﬀerentials dx, dy1, dy2 to ﬁnd the derivative dx = (f1 , f2 ) dy1 dy2 = (f1 , f2 )
1 1 g1 g2 2 2 g1 g2

∂x . ∂z1 dz1 dz2 .

dz1 dz2

1 2 1 2 = (f1 g1 +f2 g1 , f1 g2 +f2 g2 )

29 ⇒ ∂H ∂x ∂H ∂x 1 2 1 2 = = f1 g1 + f2 g1 , = = f1 g2 + f2 g2 . ∂z1 ∂z1 ∂z2 ∂z2 ∂x 5 7 6 6 = 6y1 y2 (2) + 7y1 y2 (4). ∂z1

6 7 Example: x = y1 y2 , y1 = 2z1 + 3z2 , y2 = 4z1 + 5z2 ,

Total derivative: x = f (y1 , y2 ), y2 = h(y1 ), x = f (y1 , h(y1 )) ≡ g(y1), dx dy1 = f1 + f2 h′ . y2 =h(y1 )

⇒ dx = f1 dy1 + f2 dy2 = f1 dy1 + f2 h′ dy1 = (f1 + f2 h′ )dy1. Total derivative:

Partial derivative (direct eﬀect of y1 on x): Indirect eﬀect through y2 : Example: (x1 , x2 ) is ∂x dy2 = f2 h′ . ∂y2 dy1

∂x ∂f = = f1 (y1 , y2 ). ∂y1 ∂y1
1 3 3x2 ,

w.r.t. x1 along the indiﬀerence curve passing through (1, 1) is a total derivative dm dx1 5.7 d2 x2 = dx2 1 ∂m ∂m dx2 ∂m ∂m = + = + ∂x1 ∂x2 dx1 ∂x1 ∂x2 − x1
−2 3

1 3 Given the utility function U(x1 , x2 ) = 3x1 −2 dx2 U1 (x1 , x2 ) x 3 m(x1 , x2 ) = = − = − 12. − dx1 U2 (x1 , x2 ) x2 3

+

the MRS at a point

The rate of change of MRS

3x1

1/3

+3x2 =6

1/3

3x1

1/3

+3x2 =6

1/3

−2 x2 3

.

Inverse function theorem

In Lecture 3, we discussed a linear mapping x = Ay and its inverse mapping y = A−1 x when |A| = 0. x1 x2 = a11 a12 a21 a22 y1 y2 y1 y2 = a22 |A| −a21 |A| −a12 |A| a11 |A|

x1 x2

.

Therefore, for a linear mapping with |A| = 0, an 1-1 inverse mapping exists and the partial derivatives are given by the inverse matrix of A. For example, ∂x1 /∂y1 = a11 where∂y1 /∂x1 = a22 etc. The idea can be generalized to nonlinear mappings. |A|     y1 x1  .   .  A general nonlinear mapping from Rn to Rn , y =  .  → x =  . , is . . yn xn represented by a vector function     x1 f 1 (y1 , . . . , yn )  .    . . x= . =  ≡ F (y). . . n xn f (y1 , . . . , yn )

30  ∂x1 ∂x1 ... ∂y1 ∂yn . . .. . . . . . ∂xn ∂xn ... ∂y1 ∂yn 

Jacobian:

  Jacobian matrix: JF (y) ≡   

∂(x1 , . . . , xn ) ≡ |JF (y)|. ∂(y1 , . . . , yn )

    =  

 1 1 f1 . . . fn . .. .  . . . . . . n n f1 . . . fn

Inverse function theorem: If x∗ = F (y ∗ ) and |JF (y ∗ )| = 0 (JF (y ∗ ) is non-singular), then F (y) is invertible nearby x∗ ,   g 1(x1 , . . . , xn )   . . that is, there exists a function G(x) ≡   such that y = G(x) if . n g (x1 , . . . , xn ) ∗ ∗ −1 x = F (y). In that case, JG (x ) = (JF (y )) . Reasoning:      1 1 dx1 f1 . . . fn dy1  .   . .. . .  .   . = . . .  .  ⇒ . . . n n dxn f1 . . . fn dyn       −1   1 1 1 1 dy1 g1 . . . gn dx1 f1 . . . fn dx1  .   . .. .  .   . . . .   .   . = . . .  .  =  . . .   .  . . . . . . . n n n n dyn g1 . . . gn dxn f1 . . . fn dxn Example: x2 and x1 JG = (JF )−1 . When r = 0, J = 0 and the mapping is degenerate, i.e., the whole set {r = 0, −π ≤ θ < π} is mapped to the origin (0, 0), just like the case in Lecture 3 when the Null space is a line. J = |JF | = r(cos2 θ + sin2 θ) = r > 0, ⇒ r = x2 + x2 , θ = tan−1 1 2
1 1 Notice that g1 = 1/(f1 ) in general.

x1 x2

= F (r, θ) =

r cos θ r sin θ

. JF (r, θ) =

cos θ −r sin θ sin θ r cos θ

.

5.8

Implicit function theorem and comparative statics

If |A| = 0, then the solution is given by x = −A−1 (By + c). The derivative matrix [∂xi /∂yj ]ij = A−1 B. Using total diﬀerentials of the equations, we can derive a similar derivative matrix for general nonlinear cases.

Linear model: If all the equations are linear, the model can be represtned in matrix form as           a11 · · · a1n x1 b11 · · · b1m y1 c1 0  . .  . − .  =  . . . .  .   . .  .   .   .  Ax+By = c ⇔  . . . . .  . + . . . . . . . . . . . an1 · · · ann xn bn1 · · · anm ym cn 0

31 We can regard the LHS of a nonlinear economic model as a mapping from Rn+m to Rn :   f1 (x1 , . . . , xn ; y1 , . . . , ym ) = 0   . .   ⇔ F (x; y) = 0. . fn (x1 , . . . , xn ; y1, . . . , ym ) = 0   1 1 f1 . . . fn  . .  Jacobian matrix: Jx ≡  . . . . . . . . n n f1 . . . fn

Implicit function theorem: If F (x∗ ; y ∗) = 0 and |Jx (x∗ ; y ∗ )| = 0 (Jx (x∗ ; y ∗) is non-singular),  then F (x; y) = 0 is  solvable nearby (x∗ ; y ∗), that is, there exists a   x1 x1 (y1 , . . . , ym )  .    . . function x =  .  = x(y) =   such that x∗ = x(y ∗ ) and . . xn xn (y1 , . . . , ym ) F (x(y); y) = 0. In that case,    1 1  ∂x1 ∂x1  −1 ∂f . . . ∂f ... 1 1  ∂y1  ∂y1 f1 . . . fn ∂ym  ∂ym    . .  = − . . . .   . . . .. ..  . .  .   . . . .   . . . . . .   .  .   ∂f n  ∂xn n n ∂xn ∂f n  f1 . . . fn ... ... ∂y1 ∂ym ∂y1 ∂ym Reasoning:   ∂f 1 ∂f 1          1 1  ∂y1 . . . ∂ym  dy1 df 1 0 f1 . . . fn dx1  .   .   . .  .   . . .  .  = 0 .. .  .   .  =  .  ⇒  . . . . .  . + . . . . . . . .   .  ∂f n n n ∂f n  dym df n 0 f1 . . . fn dxn ... ∂y1 ∂ym   1 1    −1 ∂f . . . ∂f   1 1  ∂y1 dx1 f1 . . . fn ∂ym  dy1   .   . .  . . .. .   . .  .  ⇒  .  = − . . . . .   . . . . . . . .  n n   ∂f n n ∂f dxn f1 . . . fn dym ... ∂y1 ∂ym 1 2 2 Example: f = x1 x2 − y = 0, f = 2x1 − x2 − 1 = 0, When y = 1, (x1 , x2 ) = (1, 1) dx1 dx2 is an equilibrium. To calculate and at the equilibrium we use the implicit dy dy function theorem:     dx1 ∂f 1 −1 1 1  ∂y  f1 f2  dy     dx  = − 2 2  ∂f 2  2 f1 f2 dy ∂y = − 2x1 x2 x2 1 2 −1
−1

−1 0

=−

2 1 2 −1

−1

−1 0

=

1/4 1/2

.

32 5.9 Problems

1. Given the demand function Qd = (100/P ) − 10, ﬁnd the demand elasticity η.
2 2 2 2. Given Y = X1 X2 + 2X1 X2 , ﬁnd ∂Y /∂X1 , ∂ 2 Y /∂X1 , and ∂ 2 Y /∂X1 ∂X2 , and the total diﬀerential DY . 2 3. Given Y = F (X1 , X2 )+f (X1 )+g(X2), ﬁnd ∂Y /∂X1 , ∂ 2 Y /∂X1 , and ∂ 2 Y /∂X1 ∂X2 .

4. Given the consumption function C = C(Y − T (Y )), ﬁnd dC/dY . 5. Given that Q = D(q ∗ e/P ), ﬁnd dQ/dP .
2 6. Y = X1 X2 , Z = Y 2 + 2Y − 2, use chain rule to derive ∂Z/∂X1 and ∂Z/∂X2 .

7. Y1 = X1 + 2X2 , Y2 = 2X1 + X2 , and Z = Y1 Y2 , use chain rule to derive ∂Z/∂X1 and ∂Z/∂X2 .
2 2 8. Let U(X1 , X2 ) = X1 X2 + X1 X2 and X2 = 2X1 + 1, ﬁnd the partial derivative ∂U/∂X1 and the total derivative dU/dX1 .

9. X 2 + Y 3 = 1, use implicit function rule to ﬁnd dY /dX.
2 2 10. X1 + 2X2 + Y 2 = 1, use implicit function rule to derive ∂Y /∂X1 and ∂Y /∂X2 .

11. F (Y1 , Y2, X) = Y1 − Y2 + X − 1 = 0 and G(Y1 , Y2, X) = Y12 + Y22 + X 2 − 1 = 0. use implicit function theorem to derive dY1 /dX and dY2 /dX. 12. In a Cournot quantity competition duopoly model with heterogeneous products, the demand functions are given by Q1 = a − P1 − cP2 , Q2 = a − cP1 − P2 ; 1 ≥ c > 0.

(a) For what value of c can we invert the demand functions to obtain P1 and P2 as functions of Q1 and Q2 ? (b) Calculate the inverse demand functions P1 = P1 (Q1 , Q2 ) and P2 = P2 (Q1 , Q2 ). (c) Derive the total revenue functions T R1 (Q1 , Q2 ) = P1 (Q1 , Q2 )Q1 and T R2 (Q1 , Q2 ) = P2 (Q1 , Q2 )Q2 . 13. In a 2-good market equilibrium model, the inverse demand functions are given by P1 = A1 Qα−1 Qβ , P2 = A2 Qα Qβ−1 ; α, β > 0. 1 2 1 2 (a) Calculate the Jacobian matrix ∂(P1 , P2 ) ∂P1 /∂Q1 ∂P1 /∂Q2 and Jacobian . ∂P2 /∂Q1 ∂P2 /∂Q2 ∂(Q1 , Q2 ) What condition(s) should the parameters satisfy so that we can invert the functions to obtain the demand functions?

(b) Derive the Jacobian matrix of the derivatives of (Q1 , Q2 ) with respect to ∂Q1 /∂P1 ∂Q1 /∂P2 (P1 , P2 ), . ∂Q2 /∂P1 ∂Q2 /∂P2

33 5.10 Proofs of important theorems of diﬀerentiation

Rolle’s theorem: If f (x) ∈ C[a, b], f ′ (x) exists for all x ∈ (a, b), and f (a) = f (b) = 0, then there exists a c ∈ (a, b) such that f ′ (c) = 0. Proof: Case 1: f (x) ≡ 0 ∀x ∈ [a, b] ⇒ f ′ (x) = 0 ∀x ∈ (a, b)./ Case 2: f (x) ≡ 0 ∈ [a, b] ⇒ ∃e, c such that f (e) = m ≤ f (x) ≤ M = f (c) and M > m. Assume that M = 0 (otherwise m = 0 and the proof is similar). It is easy ′ ′ to see that f− (c) ≥ 0 and f+ (c) ≤ 0. Therefore, f ′ (c) = 0. Q.E.D. Mean Value theorem: If f (x) ∈ C[a, b] and f ′ (x) exists for all x ∈ (a, b). Then there exists a c ∈ (a, b) such that f (b) − f (a) = f ′ (c)(b − a). Proof: Consider the function φ(x) ≡ f (x) − f (a) + f (b) − f (a) (x − a) . b−a

It is clear that φ(x) ∈ C[a, b] and φ′ (x) exists for all x ∈ (a, b). Also, φ(a) = φ(b) = 0 so that the conditions of Rolle’s Theorem are satisﬁed for φ(x). Hence, there exists a c ∈ (a, b) such that φ′ (c) = 0, or φ′ (c) = f ′ (c) − f (b) − f (a) f (b) − f (a) = 0 ⇒ f ′ (c) = = 0. b−a b−a r (r+1)

Taylor’s Theorem: If f (x) ∈ C [a, b] and f there exists a c ∈ (a, b) such that

Q.E.D. (x) exists for all x ∈ (a, b). Then

1 1 1 f (r+1) (c)(b−a)r+1 . f (b) = f (a)+f ′ (a)(b−a)+ f ′′ (a)(b−a)2 +. . .+ f (r) (a)(b−a)r + 2 r! (r + 1)! Proof: Deﬁne ξ ∈ R (b − a)r+1 1 1 ξ ≡ f (b)− f (a) + f ′ (a)(b − a) + f ′′ (a)(b − a)2 + . . . + f (r) (a)(b − a)r . (r + 1)! 2 r! Consider the function 1 φ(x) ≡ f (b) − f (x) + f ′ (x)(b − x) + f ′′ (x)(b − x)2 + . . . + 2 1 (r) ξ f (x)(b − x)r + (b − x)r+1 . r! (r + 1)!

34 It is clear that φ(x) ∈ C[a, b] and φ′ (x) exists for all x ∈ (a, b). Also, φ(a) = φ(b) = 0 so that the conditions of Rolle’s Theorem are satisﬁed for φ(x). Hence, there exists a c ∈ (a, b) such that φ′ (c) = 0, or φ′ (c) = ξ − f (r+1) (c) = 0 ⇒ f (r+1) (c) = ξ. r! n Q.E.D. Inverse Function Theorem: Let E ⊆ R be an open set. Suppose f : E → Rn is C 1 (E), a ∈ E, f (a) = b, and A = J(f (a)), |A| = 0. Then there exist open sets U, V ⊂ Rn such that a ∈ U, b ∈ V , f is one to one on U, f (U) = V , and f −1 : V → U is C 1 (U). Proof: (1. Find U.) Choose λ ≡ |A|/2. Since f ∈ C 1 (E), there exists a neighborhood U ⊆ E with a ∈ U such that J(f (x)) − A < λ. (2. Show that f (x) is one to one in U.) For each y ∈ Rn deﬁne φy on E by φy (x) ≡ x + A−1 (y − f (x)). Notice that f (x) = y if and only if x is a ﬁxed point of 1 φy . Since J(φy (x)) = I − A−1 J(f (x)) = A−1 [A − J(f (x))] ⇒ J(φy (x)) < on U. 2 Therefore φy (x) is a contraction mapping and there exists at most one ﬁxed point in U. Therefore, f is one to one in U. (3. V = f (U) is open so that f −1 is continuous.) Let V = f (U) and y0 = f (x0 ) ∈ V for x0 ∈ U. Choose an open ball B about x0 with radius ρ such that the closure [B] ⊆ U. To prove that V is open, it is enough to show that y ∈ V whenever y − y0 < λρ. So ﬁx y such that y − y0 < λρ. With φy deﬁned above, ρ φy (x0 ) − x0 = A−1 (y − y0 ) < A−1 λρ = . 2

If x ∈ [B] ⊆ U, then φy (x) − x0 ≤ φy (x) − φy (x0 ) + φy (x0 ) − x0 < 1 ρ x − x0 + ≤ ρ. 2 2

That is, φy (x) ∈ [B]. Thus, φy (x) is a contraction of the complete space [B] into itself. Hence, φy (x) has a unique ﬁxed point x ∈ [B] and y = f (x) ∈ f ([B]) ⊂ f (U) = V . (4. f −1 ∈ C −1 .) Choose y1 , y2 ∈ V , there exist x1 , x2 ∈ U such that f (x1 ) = y1 , f (x2 ) = y2 . φy (x2 ) − φy (x1 ) = x2 − x1 + A−1 (f (x1 ) − f (x2 )) = (x2 − x1 ) − A−1 (y2 − y1 ). ⇒ (x2 −x1 )−A−1 (y2 −y1 ) ≤ or x2 − x1 ≤ 1 1 1 x2 −x1 ⇒ x2 −x1 ≤ A−1 (y2 −y1 ) ≤ y2 −y1 2 2 2λ

1 y2 − y1 . It follows that (f ′ )−1 exists locally about a. Since λ

f −1 (y2 ) − f −1 (y1 ) − (f ′ )−1 (y1 )(y2 − y1 ) = (x2 − x1 ) − (f ′ )−1 (y1 )(y2 − y1 ) = −(f ′ )−1 (y1 )[−f ′ (x1 )(x2 − x1 ) + f (x2 ) − f (x1 )],

35 We have f −1 (y2 ) − f −1 (y1) − (f ′ )−1 (y1 )(y2 − y1 ) (f ′ )−1 ≤ y2 − y1 λ f (x2 ) − f (x1 ) − f ′ (x1 )(x2 − x1 ) . x2 − x1

As y2 →y1 , x2 →x1 . Hence (f 1− )′ (y) = {f ′[f −1 (y)]} for y ∈ V . Since f −1 is diﬀerentiable, it is continuous. Also, f ′ is continuous and its inversion, where it exists, is continuous. Therefore (f −1 )′ is continuous or f −1 ∈ C 1 (V ). Q.E.D. Implicit Function Theorem: Let E ⊆ R(n+m) be an open set and a ∈ Rn , b ∈ Rm , (a, b) ∈ E. Suppose f : E → Rn is C 1 (E) and f (a, b) = 0, and J(f (a, b)) = 0. Then there exist open sets A ⊂ Rn and B ⊂ Rm a ∈ A and b ∈ B, such that for each x ∈ B, there exists a unique g(x) ∈ A such that f (g(x), x) = 0 and g : B → A is C 1 (B). Proof: Deﬁning F : Rn+m →Rn+m by F (x, y) ≡ (x, f (x, y)). Note that since  J(F (a, b)) =    ∂xi ∂xj 1≤i,j≤n ∂fi ∂xj 1≤i≤m,1≤j≤n  ∂xi ∂xn+j 1≤i≤n,1≤j≤m  =  ∂fi ∂xn+j 1≤i,j≤m I O N M ,

|J(F (a, b))| = |M| = 0. By the Inverse Function Theorem there exists an open set V ⊆ Rn+m containing F (a, b) = (a, 0) and an open set of the form A × B ⊆ E containing (a, b), such that F : A × B→V has a C 1 inverse F −1 : V →A × B. F −1 is of the form F −1 (x, y) = (x, φ(x, y)) for some C 1 function φ. Deﬁne the projection π : Rn+m →Rm by π(x, y) = y. Then π ◦ F (x, y) = f (x, y). Therefore f (x, φ(x, y)) = f ◦ F −1 (x, y) = (π ◦ F ) ◦ F −1 (x, y) = π ◦ (F ◦ F −1 )(x, y) = π(x, y) = y and f (x, φ(x, 0)) = 0. So, deﬁne g : A→B by g(x) = φ(x, 0). Q.E.D.

36

6
6.1

Comparative Statics – Economic applications
Partial equilibrium model Q = D(P, Y ) Q = S(P ) ∂D ∂D < 0, >0 ∂P ∂Y ′ S (P ) > 0 end. var: Q, P. ex. var: Y..

f 1 (P, Q; Y ) = Q − D(P, Y ) = 0

f 2 (P, Q; Y ) = Q − S(P ) = 0   dQ ∂D ∂D  dY  1 − , |J| =  dP  = ∂P ∂Y 1 −S ′ (P ) 0 dY ∂D ∂D − ∂D ′ ∂Y ∂P S (P ) − 0 −S ′ (P ) dQ = = ∂Y > 0, dY |J| |J| 6.2 Income determination model C = C(Y ) I = I(r) ¯ Y =C +I +G

dQ ∂D dP ∂D df 1 = − − =0 dY2 dY ∂P dY ∂Y . df dQ dP = − S ′ (P ) =0 dY dY dY ∂D ∂P 1 −S ′ (P ) 1 − 1 dP = dY = −S ′ (P ) + ∂D < 0. ∂P

∂D ∂Y 1 0 |J|

∂D = ∂Y > 0. |J| −

0 < C ′ (Y ) < 1. I ′ (r) < 0

end. var. C, Y, I ¯ ex. var. G, r.

¯ I ′ (r)dr + dG ¯ ¯ Y = C(Y ) + I(r) + G ⇒ dY = C ′ (Y )dY + I ′ (r)dr + dG ⇒ dY = . 1 − C ′ (Y ) ∂Y I ′ (r) ∂Y 1 = < 0, ′ (Y ) ¯ = 1 − C ′ (Y ) > 0. ∂r 1−C ∂G

6.2.1

Income determination and trade

Consider an income determination model with import and export: C = C(Y ) 1 > Cy > 0, M = M(Y, e) My > 0, Me < 0 C + I + X − M = Y, ¯ I = I, X = X(Y ∗ , e), Xy∗ > 0, Xe > 0

where import M is a function of domestic income and exchange rate e and export X is a function of exchange rate and foreign income Y ∗ , both are assumed here as exogenous variables. Substituting consumption, import, and export functions into the equilibrium condition, we have ¯ C(Y )+I+X(Y ∗ , e)−M(Y, e) = Y, ¯ ⇒F (Y, e, Y ∗ ) ≡ C(Y )+I+X(Y ∗ , e)−M(Y, e)−Y = 0..

37 Use implicit function rule to derive ∂Y ¯ and determine its sign: ∂I

1 ∂Y FI =− = . ¯ Fy 1 − C ′ (Y ) + My ∂I Use implicit function rule to derive ∂Y and determine its sign: ∂e

Fe Xe − Me ∂Y =− = . ∂e Fy 1 − C ′ (Y ) + My Use implicit function rule to derive ∂Y and determine its sign: ∂Y ∗

Fy∗ Xy ∗ ∂Y =− = . ∗ ′ (Y ) + M ∂Y Fy 1−C y 6.2.2 Interdependence of domestic and foreign income

Now extend the above income determination model to analyze the joint dependence of domestic income and foreign income: ¯ C(Y ) + I + X(Y ∗ , e) − M(Y, e) = Y ¯ C ∗ (Y ∗ ) + I ∗ + X ∗ (Y, e) − M ∗ (Y ∗ , e) = Y ∗ ,

∗ with a similar assumption on the foreigner’s consumption function: 1 > Cy∗ > 0. Since domestic import is the same as foreigner’s export and domestic export is foreigner’s import, X ∗ (Y, e) = M(Y, e) and M ∗ (Y ∗ , e) = X(Y ∗ , e) and the system becomes:

¯ C(Y ) + I + X(Y ∗ , e) − M(Y, e) = Y

¯ C ∗ (Y ∗ ) + I ∗ + M(Y, e) − X(Y ∗ , e) = Y ∗ , ¯ (Xe − Me )de + dI ¯ (Me − Xe )de + dI ∗

Calculate the total diﬀerential of the system (Now Y ∗ becomes endogenous): 1 − C ′ + My −Xy∗ −My 1 − C ∗ ′ + Xy∗ |J| = 1 − C ′ + My −Xy∗ −My 1 − C ∗ ′ + Xy∗ dY dY ∗ = ,

= (1 − C ′ + My )(1 − C ∗ ′ + My∗ ) − My Xy∗ > 0.

Use Cramer’s rule to derive

∂Y ∂Y ∗ and and determine their signs: ∂e ∂e

dY =

¯ (Xe − Me )de + dI −Xy∗ ¯∗ 1 − C ∗ ′ + Xy∗ (Me − Xe )de + dI |J|

¯ ¯ (Xe − Me )(1 − C ∗ ′ + Xy∗ − Xy∗ )de + (1 − C ∗ ′ + Xy∗ )dI + Xy∗ dI ∗ = , |J|

38 ¯ 1 − C ′ + My (Xe − Me )de + dI ¯∗ (Me − Xe )de + dI −My |J|

dY ∗ = =

Derive

∂Y ∗ ∂Y and ¯ and determine their signs: ¯ ∂I ∂I

∂Y (Xe − Me )(1 − C ∗ ′ ) = > 0, ∂e |J|

¯ ¯ −(Xe − Me )(1 − C ′ + My − My )de + (1 − C ′ + My )dI ∗ + Xy dI . |J|

∂Y ∗ −(Xe − Me )(1 − C ′ ) = < 0. ∂e |J|

∂Y ∗ ∂Y 1 − C ∗ ′ + My∗ My =− > 0, ¯ ¯ = |J| < 0. |J| ∂I ∂I 6.3 IS-LM model C = C(Y ) 0 < C ′ (Y ) < 1 ∂L ∂L M d = L(Y, r) > 0, 0, = > 0, ¯ ¯ |J| |J| ∂G ∂M 6.4

¯ ¯ Lr dG + I ′ dM , dr = |J|

∂r LY ∂r (1 − C ′ ) =− > 0, = < 0. ¯ ¯ |J| |J| ∂G ∂M

¯ 1 − C ′ dG ¯ LY dM |J|

=

Two-market general equilibrium model
1 2 2 Q1d = D 1 (P1 , P2 ) D1 < 0, Q2d = D 2 (P1 , P2 ) D2 < 0, D1 > 0. ′ ¯ Q2s = S2 (P2 ) S2 (P2 ) > 0 Q1s = S1 Q1d = Q1s Q2d = Q2s

end. var: Q1 , Q2 , P1 , P2 .

¯ ex. var: S1 . ⇒
1 1 ¯ D1 dP1 + D2 dP2 = dS1 2 2 ′ D1 dP1 + D2 dP2 − S2 dP2 = 0.

¯ D 1 (P1 , P2 ) = S1 2 D (P1 , P2 ) − S2 (P2 ) = 0

39
1 1 D1 D2 2 2 ′ D1 D2 − S 2

dP1 dP2

=

¯ d S1 0

, |J| =

1 1 2 2 Assumption: |D1 | > |D2 | and |D2 | > |D1 | (own-eﬀects dominate), ⇒ |J| > 0. ′ 2 dP1 dP2 D 2 − S2 −D1 = 2 < 0, = 0, where Qn is the quantity of new cars and Pn (Pu ) the price of a new (used) car. The supply function of new cars is Qn = S(Pn ), S ′ (Pn ) > 0. end. var: Pn , Qn . n ex. var: Pu . ∂D n ∂D n dPn + dPu = S ′ (Pn )dPn . ⇒ ∂Pn ∂Pu

D (Pn ; Pu ) = S(Pn );

dPn ∂D n /∂Pu dQn dPn S ′ (Pn )∂D n /∂Pu = ′ > 0, = S ′ (Pn ) = ′ > 0. dPu S (Pn ) − ∂D n /∂Pn dPu dPu S (Pn ) − ∂D n /∂Pn The markets for used cars and for new cars are actually interrelated. The demand for used cars is Qu = D u (Pu , Pn ), ∂D n /∂Pu < 0, ∂D n /∂Pn > 0. In each period, ¯ the quantity of used cars supplied is ﬁxed, denoted by Qu . Instead of analyzing the eﬀects of a change in Pu on the new car market, we want to know how a change in ¯ Qu aﬀects both markets. end. var: Pn , Qn , Pu , Qu . Qn = D n (Pn , Pu ), ⇒ D n (Pn , Pu ) = S(Pn ), n n Dn − S ′ Du u u Dn Du

¯ ex. var: Qu . Qn = S(Pn ); ¯ D u (Pu , Pn ) = Qu ; = 0 ¯ dQu , |J| = Qu = D u (Pu , Pn ), n n Dn − S ′ Du u u Dn Du

¯ Qu = Qu

n n Dn dPn +Du dPu = S ′ dPn ,

u u ¯ Dn dPn +Du dPu = Q

dPn dPu

n u n u = (Dn −S ′ )Du −Du Dn .

n n u u Assumption: |Dn | > |Du | and |Du | > |Dn | (own-eﬀects dominate), ⇒ |J| > 0. n dPn −Du dPu Dn − S ′ = < 0, = n < 0. ¯ ¯ |J| |J| dQu dQu

∂Qn ∂Qn ¯ ∂Qu From Qu = Qu , ¯ = 1. To calculate ¯ , we have to use chain rule: ¯ = ∂ Qu ∂ Qu ∂ Qu dPn S ′ ¯ < 0. dQn

40 6.5 Classic labor market model L = h(w) h′ > 0

labor supply function ∂Q w = MPPL = = FL (K, L) FLK > 0, FLL < 0 labor demand function. ∂L endogenous variables: L, w. exogenous variable: K. L − h(w) = 0 w − FL = 0 1 −h′ (w) −FLL 1 0 dL = dK 6.6 FLK dL dw −h′ (w) 1 |J| =
′

⇒

dL − h′ (w)dw = 0 . dw − FLL dL − FLK dK = 0 , |J| = 1 −h′ (w) −FLL 1 = = 1−h′ (w)FLL > 0. FLK > 0. |J|

0 FLK dK

=

h FLK dw > 0, = |J| dK

1 0 −FLL FLK |J|

Problem Q = D(P ) D ′ (P ) < 0 Q = S(P, t) ∂S/∂P > 0, ∂S/∂t < 0, where t is the tax rate on the commodity. (a) Derive the total diﬀerentail of each equation. (b) Use Cramer’s rule to compute dQ/dt and dP/dt. (c) Determine the sign of dQ/dt and dP/dt. (d) Use the Q − P diagram to explain your results.

1. Let the demand and supply functions for a commodity be

2. Suppose consumption C depends on total wealth W , well as on income Y . The IS-LM model becomes C = C(Y, W ) 0 < CY < 1 CW > 0 I = I(r) I ′ (r) < 0 M D = L(Y, r) LY > 0 Lr < 0

which is predetermind, as MS = M Y =C +I MS = MD

(a) Which variables are endogenous? Which are exogenous? (b) Which equations are behavioral/institutional, which are equilibrium conditions? The model can be reduced to Y − C(Y, W ) − I(r) = 0, L(Y, r) = M.

(c) Derive the total diﬀerential for each of the two equations. (d) Use Cramer’s rule to derive the eﬀects of an increase in W on Y and r, ie., derive ∂Y /∂W and ∂r/∂W . (e) Determine the signs of ∂Y /∂W and ∂r/∂W .

41 3. Consider a 2-industry (e.g. manufacturing and agriculture) general equilibrium model. The demand for manufacturing product consists of two components: private demand D 1 and goverment demand G. The agricultural products have only private demand D 2 . Both D 1 and D 2 depend only on their own prices. Because each industry requires in its production process outputs of the other, the supply of each commodity depends on the price of the other commodity as well as on its own price. Therefore, the model may be written as follows: Qd 1 Qd 2 Qs 1 Qs 2 Qd 1 Qd 2 = = = = = = D1 (P1 ) + G D2 (P2 ) S 1 (P1 , P2 ) S 2 (P1 , P2 ) Qs , 1 Qs . 2
′ D1 (P1 ) < 0, ′ D2 (P2 ) < 0, 1 1 1 1 S1 > 0, S2 < 0, S1 > |S2 |, 2 2 2 2 S1 < 0, S2 > 0, S2 > |S1 |,

(a) Which variables are endogenous? Which are exogenous? (b) Which equations are behavioral? Which are deﬁnitional? Which are equilibrium conditions? 4. The model can be reduced to S 1 (P1 , P2 ) = D1 (P1 ) + G S 2 (P1 , P2 ) = D2 (P2 )

(c) Compute the total diﬀerential of each equation. (d) Use Cramer’s rule to derive ∂P1 /∂G and ∂P2 /∂G. (e) Determine the signs of ∂P1 /∂G and ∂P2 /∂G. (f) Compute ∂Q1 /∂G and ∂Q2 /∂G. (Hint: Use chain rule.) (g) Give an economic interpretation of the results. 5. The demand functions of a 2-commodity market model are: Qd = D 1 (P1 , P2 ) 1 Qd = D 2 (P1 , P2 ). 2

The supply of the ﬁrst commodity is given exogenously, ie., Qs = S1 . The 1 supply of the second commodity depends on its own price, Qs = S2 (P2 ). The 2 equilibrium conditions are: Qd = Qs , 1 1 Qd = Qs . 2 2

(a) Which variables are endogenous? Which are exogenous? (b) Which equations are behavioral? Which are deﬁnitional? Which are equilibrium conditions? The model above can be reduced to : D 1 (P1 , P2 ) − S1 = 0 D 2 (P1 , P2 ) − S2 (P2 ) = 0

42 i Suppose that both commodities are not Giﬀen good (hence, Di < 0, i = 1, 2), i that each one is a gross substitute for the other (ie., Dj > 0, i = j), that i i ′ |Di | > |Dj |, and that S2 (P2 ) > 0. (c) Calculate the total diﬀerential of each equation of the reduced model. (d) Use Cramer’s rule to derive ∂P1 /∂S1 and ∂P2 /∂S1 . (e) Determine the signs of ∂P1 /∂S1 and ∂P2 /∂S1 . (f) Compute ∂Q1 /∂S1 and ∂Q2 /∂S1 and determine their signs.

6. The demand functions for ﬁsh and chicken are as follows:
′ Qd = DF (PF − PC ), DF < 0 F ′ Qd = DC (PC − PF ), DC < 0 C

where PF , PC are price of ﬁsh and price of chicken respectively. The supply of ﬁsh depends on the number of ﬁshermen (N) as well as its price PF : Qs = F F (PF , N), FPF > 0, FN > 0. The supply of chicken depends only on its price PC : Qs = C(PC ), C ′ > 0. The model can be reduced to C DF (PF − PC ) = F (PF , N) DC (PC − PF ) = C(PC ) (a) Find the total diﬀerential of the reduced system. (b) Use Cramer’s rule to ﬁnd dPF /dN and dPC /dN. (c) Determine the signs of dPF /dN and dPC /dN. What is the economic meaning of your results? (d) Find dQC /dN. 7. In a 2-good market equilibrium model, the inverse demand functions are given by P1 = U1 (Q1 , Q2 ), P2 = U2 (Q1 , Q2 ); where U1 (Q1 , Q2 ) and U2 (Q1 , Q2 ) are the partial derivatives of a utility function U(Q1 , Q2 ) with respect to Q1 and Q2 , respectively. (a) Calculate the Jacobian matrix ∂(P1 , P2 ) ∂P1 /∂Q1 ∂P1 /∂Q2 and Jacobian . ∂P2 /∂Q1 ∂P2 /∂Q2 ∂(Q1 , Q2 ) What condition(s) should the parameters satisfy so that we can invert the functions to obtain the demand functions?

(b) Derive the Jacobian matrix of the derivatives of (Q1 , Q2 ) with respect to ∂Q1 /∂P1 ∂Q1 /∂P2 (P1 , P2 ), . ∂Q2 /∂P1 ∂Q2 /∂P2 (c) Suppose that the supply functions are Q1 = a−1 P1 , Q2 = P2 ,

and Q∗ and Q∗ are market equilibrium quantities. Find the comparative 1 2 dQ∗ dQ∗ 1 2 statics and . (Hint: Eliminate P1 and P2 .) da da

43 8. In a 2-good market equilibrium model with a sales tax of t dollars per unit on product 1, the model becomes D 1 (P1 + t, P2 ) = Q1 = S1 (P1 ), D 2 (P1 + t, P2 ) = Q2 = S2 (P2 ).

i i i Suppose that Di < 0 and Si′ > 0, |Di | > |Dj |, i = j, i, j = 1, 2.

(a) Calculate dP2 /dt. (b) Calculate dQ2 /dt. i (c) Suppose that Dj > 0. Determine the signs of dP2 /dt and dQ2 /dt. i (d) Suppose that Dj < 0. Determine the signs of dP2 /dt and dQ2 /dt.

(e) Explain your results in economics.

44

7

Optimization

A behavioral equation is a summary of the decisions of a group of economic agents in a model. A demand (supply) function summarizes the consumption (production) decisions of consumers (producers) under diﬀerent market prices, etc. The derivative of a behavioral function represents how agents react when an independent variable changes. In the last chapter, when doing comparative static analysis, we always assumed that the signs of derivatives of a behavioral equation in a model are known. For example, D ′ (P ) < 0 and S ′ (P ) > 0 in the partial market equilibrium model, C ′ (Y ) > 0, I ′ (r) < 0, Ly > 0, and Lr < 0 in the IS-LM model. In this chapter, we are going to provide a theoretical foundation for the determination of these signs. 7.1 Neoclassic methodology

Neoclassic assumption: An agent, when making decisions, has an objective function in mind (or has well deﬁned preferences). The agent will choose a feasible decision such that the objective function is maximized. A consumer will choose the quantity of each commodity within his/her budget constraints such that his/her utility function is maximized. A producer will choose to supply the quantity such that his proﬁt is maximizaed. Remarks: (1) Biological behavior is an alternative assumption, sometimes more appropriate, (2) Sometimes an agent is actually a group of people with diﬀerent personalities like a company and we have to use game theoretic equilibrium concepts to characterize the collective behavior. Maximization ⇒ Behavioral equations Game equilibrium ⇒ Equilibrium conditions x1 , . . . , xn : variables determined by the agent (endogenous variables). y1 , . . . , ym : variables given to the agent (exogenous variables). Objective function: f (x1 , . . . , xn ; y1 , . . . , ym). Opportunity set: the agent can choose only (x1 , . . . , xn ) such that (x1 , . . . , xn ; y1 , . . . , ym) ∈ A ⊂ Rn+m . A is usually deﬁned by inequality constriants.   g 1 (x1 , . . . , xn ; y1 , . . . , ym ) ≥ 0  . . max f (x1 , . . . , xn ; y1 , . . . , ym ) subject to . x1 ,...,xn  k  g (x , . . . , x ; y , . . . , y ) ≥ 0. 1 n 1 m Solution (behavioral equations): xi = xi (y1 , . . . , ym ), i = 1, . . . , n (derived from FOC). ∂xi /∂yj : derived by the comparative static method (sometimes its sign can be determined from SOC).

Example 1: A consumer maximizes his utility function U(q1 , q2 ) subject to the budget constraint p1 q1 + p2 q2 = m ⇒ demand functions q1 = D 1 (p1 , p2 , m) and q2 = D 2 (p1 , p2 , m).

45 Example 2: A producer maximizes its proﬁt Π(Q; P ) = P Q − C(Q) where C(Q) is the cost of producing Q units of output ⇒ the supply function Q = S(P ). one endogenous variable: this chapter. n endogenous variables without constraints: next n endogenous variables with equality constraints: Nonlinear programming: n endogenous variables with inequality constraints Linear programming: Linear objective function with linear inequality constraints Game theory: more than one agents with diﬀerent objective functions 7.2 Diﬀerent concepts of maximization

Suppose that a producer has to choose a Q to maximize its proﬁt π = F (Q): maxQ F (Q). Assume that F ′ (Q) and F ′′ (Q) exist. A local maximum Ql : there exists ǫ > 0 such that F (Ql ) ≥ F (Q) for all Q ∈ (Ql − ǫ, Ql + ǫ). A global maximum Qg : F (Qg ) ≥ F (Q) for all Q. A unique global maximum Qu : F (Qu ) > F (Q) for all Q = Qu . F F F T l : local max. Q T T Qg : a global max. Qu : unique global max not global max but not unique

Ql

EQ

Qg

EQ

Qu

EQ

The agent will choose only a global maximum as the quantity supplied to the market. However, it is possible that there are more than one global maximum. In that case, the supply quantity is not unique. Therefore, we prefer that the maximization problem has a unique global maximum. A unique global maximum must be a global maximum and a global maximum must be a local maximum. ⇒ to ﬁnd a global maximum, we ﬁrst ﬁnd all the local maxima. One of them must be a global maximum, otherwise the problem does not have a solution (the maximum occurs at ∞.) We will ﬁnd conditions (eg., increasing MC or decreasing MRS) so that there is only one local maximum which is also the unique global maximum. 7.3 FOC and SOC for a local maximum

At a local maximum Ql , the slope of the graph of F (Q) must be horizontal F ′ (Ql ) = 0. This is called the ﬁrst order condition (FOC) for a local maximum. A critical point Qc : F ′ (Qc ) = 0. A local maximum must be a critical point but a critical point does not have to be a local maximum.

46 F F F

T

F ′′ (Qc ) < 0 local max.

T

T

F ′′ (Qc ) = 0 degenerate

F ′′ (Qc ) > 0 local min. Qc
EQ

Qc

EQ

Qc

EQ

A degenerate critical point: F ′ (Q) = F ′′ (Q) = 0. A non-degenerate critical point: F ′ (Q) = 0, F ′′ (Q) = 0. A non-degenerate critical point is a local maximum (minimum) if F ′′ (Q) < 0 (F ′′ (Q) > 0). FOC: F ′ (Ql ) = 0 SOC: F ′′ (Ql ) < 0

Example: F (Q) = −15Q + 9Q2 − Q3 , F ′ (Q) = −15 + 18Q − 3Q2 = −3(Q − 1)(Q − 5). There are two critical points: Q = 1, 5. F ′′ (Q) = 18 − 6Q, F ′′ (1) = 12 > 0, and F ′′ (5) = −12 < 0. Therefore, Q = 5 is a local maximum. It is a global maximum for 0 ≤ Q < ∞. Remark 1 (Degeneracy): For a degenerate critical point, we have to check higher order derivatives. If the lowest order non-zero derivative is of odd order, then it is a reﬂect point; eg., F (Q) = (Q − 5)3 , F ′ (5) = F ′′ (5) = 0 and F ′′′ (5) = 6 = 0 and Q = 5 is not a local maximum. If the lowest order non-zero derivative is of even order and negative (positive), then it is a local maximum (minimum); eg., F (Q) = −(Q − 5)4 , F ′ (5) = F ′′ (5) = F ′′′ (5) = 0, F (4) (5) = −24 < 0 and Q = 5 is a local maximum. Remark 2 (Unboundedness): If limQ→∞ F (Q) = ∞, then a global maximum does not exist. Remark 3 (Non-diﬀerentiability): If F (Q) is not diﬀerentiable, then we have to use other methods to ﬁnd a global maximum. Remark 4 (Boundary or corner solution): When there is non-negative restriction Q ≥ 0 (or an upper limit Q ≤ a), it is possible that the solution occurs at Q = 0 (or at Q = a). To take care of such possibilities, FOC is modiﬁed to become F ′ (Q) ≤ 0, QF ′ (Q) = 0 (or F ′ (Q) ≥ 0, (a − Q)F ′ (Q) = 0). 7.4 Supply function of a competitive producer

Consider ﬁrst the proﬁt maximization problem of a competitive producer: max Π = P Q − C(Q),
Q

FOC ⇒

∂Π = P − C ′ (Q) = 0. ∂Q

The FOC is the inverse supply function (a behavioral equation) of the producer: P = C ′ (Q) = MC. Remember that Q is endogenous and P is exogenous here. To ﬁnd

47 dQ the comparative statics , we use the total diﬀerential method discussed in the last dP chapter: dQ 1 dP = C ′′ (Q)dQ, ⇒ = ′′ . dP C (Q) To determine the sign of fore, dQs > 0. dP dQ ∂2Π , we need the SOC, which is = −C ′′ (Q) < 0. TheredP ∂Q2

Remark: The result is true no matter what the cost function C(Q) is. MC = C ′ (Q) can be non-monotonic, but the supply curve is only part of the increasing sections of the MC curve and can be discontinuous. P
T

P MC

MC is the supply curve.

T

MC

Pc

S(Pc ) = {Q1 , Q2 }, supply curve has two component.

EQ

Q1

Q2

EQ

7.5

Maximization and comparative statics: general procedure
X

Maximization problem of an agent: max F (X; Y ). FOC: FX (X ∗ ; Y ) = 0, ⇒ X ∗ = X(Y ) · · · · · · Behavioral Equation dX FXY Comparative statics: FXX dX + FXY dY = 0 ⇒ =− . SOC: FXX < 0 dY FXX dX FXY Case 1: FXY > 0 ⇒ =− > 0. dY FXX dX FXY Case 2: FXY < 0 ⇒ =− < 0. dY FXX dX Therefore, the sign of depends only on the sign of FXY . dY 7.6 Utility Function

A consumer wants to maximize his/her utility function U = u(Q) + M = u(Q) + (Y − P Q). ∂U FOC: = u′ (Q) − P = 0, ∂Q ⇒ u′ (Qd ) = P (inverse demand function) ⇒ Qd = D(P ) (demand function, a behavioral equation)

48 dQd ∂2U = UP Q = −1 ⇒ = D ′ (P ) < 0, the demand function is a decreasing ∂Q∂P dP function of price.

7.7

Input Demand Function

The production function of a producer is given by Q = f (x), where x is the quantity of an input employed. Its proﬁt is Π = pf (x) − wx, where p (w) is the price of the output (input). The FOC of proﬁt maximization problem is pf ′ (x) − w = 0 ⇒ f ′ (x) = w/p (inverse input demand function) ⇒ x = h(w/p) (input demand function, a behavioral equation) ∂2Π dx = −1 ⇒ = h′ (w/p) < 0, the input demand is a decreasing func∂x∂(w/p) d(w/p) w tion of the real input price . p 7.8 Envelope theorem

Deﬁne the maximum function M(Y ) ≡ maxX F (X, Y ) = F (X(Y ), Y ) then the total derivative dM ∂F (X, Y ) = M ′ (Y ) = . dY ∂Y X=X(Y ) dX Proof: M ′ (Y ) = FX + FY . At the maximization point X = X(Y ), FOC implies dY that the indirect eﬀect of Y on M is zero. In the consumer utility maximization problem, V (P ) ≡ U(D(P )) + Y − P D(P ) is called the indirect utility function. The envelope theorem implies that V ′ (P ) = dU ≡ −D(P ), this is a simpliﬁed version of Roy’s identity. dP Qd =D(P ) In the input demand function problem, π(w, p) ≡ pf (h(w/p)) − wh(w/p) is the proﬁt function. Let p = 1 and still write π(w) ≡ f (h(w)) − wh(w). The envelope theorem dΠ implies that π ′ (w) = = −h(w), a simpliﬁed version of Hotelling’s lemma. dw x=h(w) Example: The relationships between LR and SR cost curves STC(Q; K) = C(Q, K), K: ﬁrm size. Each K corresponds to a STC. LTC(Q) = minK C(Q, K), ⇒ K = K(Q) is the optimal ﬁrm size. LTC is the envelope of STC’s. Each STC tangents to the LTC (STC = LTC) at the quantity Q such that K = K(Q). Notice that the endogenous variable is K and the exogenous is Q here. By envelope theorem, LMC(Q) = dLTC(Q)/dQ = ∂C(Q, K(Q))/∂Q =SMC(Q; K(Q)). That is, when K = K(Q) is optimal for producing Q, SMC=LMC.

49 Since LAC(Q) =LTC(Q)/Q and SAC(Q) =STC(Q)/Q, LAC is the envelope of SAC’s and each SAC tangents to the LAC (SAC = LAC) at the quantity Q such that K = K(Q). 7.9 Eﬀect of a Unit Tax on Monopoly Output (Samuelson)

Assumptions: a monopoly ﬁrm, q = D(P ) ⇐⇒ P = f (q), (inverse functions), C = C(q), t = unit tax max π(q) = P q − C(q) − tq = qf (q) − C(q) − tq q: endogenous; t: exogenous FOC: ∂π/∂q = f (q) + qf ′ (q) − C ′ (q) − t = 0. The FOC deﬁnes a relationship between the monopoly output q ∗ and the tax rate t as q ∗ = q(t) (a behavioral equation). The derivative dq ∗ /dt can be determined by the sign of the cross derivative: ∂2π = −1 < 0 ∂q∂t Therefore, we have dq ∗ /dt < 0. The result can be obtained using the q–pdiagram. FOC ⇐⇒ MR = MC + t. Therefore, on q–p space, an equilibrium is determined by the intersection point of MR and MC + t curves. Case 1: MR is downward sloping and MC is upward sloping: When t increases, q ∗ decreases as seen from the left diagram below. P P
T T

MC+t MC MRE MC+t MC < q < MRE q

Case 2: Both MR and MC are downward sloping and MR is steeper. MC decreasing; MR decreasing more ⇒ t ↑ q ↓. Case 3: Both MR and MC are downward sloping, but MC is steeper. The diagram shows that dq ∗ /dt > 0. It is opposite to our comparative statics result. Why? MR = MC + t violates SOC < 0, therefore, the intersection of MR and MC + t is not a proﬁt maximizing point.

50 P

T

MR MC+t > 7.10 MCE q

A Price taker vs a price setter

A producer employs an input X to produce an output Q. The production function is Q = rX. The inverse demand function for Q is P = a − Q. The inverse supply function of X is W = b + X, where W is the price of X. The producer is the only seller of Q and only buyer of X. There are two markets, Q (output) and X (input). We want to ﬁnd the 2-market equilibrium, i.e., the equilibrium values of W , X, Q, and P . It depends on the producer’s power in each market. Case 1: The ﬁrm is a monopolist in Q and a price taker in X. To the producer, P is endogenous and W is exogenous. Given W , its object is max π = P Q−W X = (a−Q)Q−W X = (a−rX)(rX)−W X, x ⇒ FOC ar−2r 2 X−W = 0.

The input demand function is X = X(W ) = ar−W . 2r 2 ar−b Equating the demand and supply of X, the input market equilibrium is X = 2r2 +1 2b and W = b + X = ar+2r . 2r 2 +1 Substituting back into the production and output demand functions, the output market equilibrium is Q = rX = f ar 2 − br2r 2 + 1 and P = a − Q = f a + ar 2 + b2r 2 + 1. Case 2: The ﬁrm is a price taker in Q and a monopsony in X. To the producer, P is exogenous and W is endogenous. Given P , its object is b 2Q max π = P Q − (b + X)X = P Q − (b + (Q/r))(Q/r), ⇒ FOC P − − 2 = 0. Q r r The output supply function is Q = Q(P ) = r P2−br . 2 −br Equating the demand and supply of Q, the output market equilibrium is Q = ar2 −2 r 2a+br and P = a − Q = 2r2 +1 . Substituting back into the production and output demand functions, the output market equilibrium is X = Q/r = f ar − br 2 − 2 and W = b + X = f ar + br 2 − 3br 2 − 2. Case 3: The ﬁrm is a monopolist in Q and a monopsony in X. To the producer, both P and W are endogenous, its object is max π = (a − Q)Q − (b + X)X = (a − (rX))(rX) − (b + X)X. x 2

51 (We can also eliminate X instead of Q. The two procedures are the same.) (Show that π is strictly concave in X.) Find the proﬁt maximizing X as a function of a and b, X = X(a, b). ∂X ∂X Determine the sign of the comparative statics and and explain your results ∂a ∂b in economics. Derive the price and the wage rate set by the ﬁrm P and W and compare the results with that of cases 1 and 2. 7.11 Labor Supply Function

Consider a consumer/worker trying to maximize his utility function subject to the time constraint that he has only 168 hours a week to spend between work (N) and leisure (L), N + L = 168, and the budget constraint which equates his consumption (C) to his wage income (wN), C = wN, as follows: max U = U(C, L) = U(wN, 168 − N) ≡ f (N, w) Here N is endogenous and w is exogenous. The FOC requires that the total derivative of U w.r.t. N be equal to 0. FOC: dU = fN = UC w + UL (−1) = 0. dN

FOC deﬁnes a relationship between the labor supply of the consumer/worker, N, and the wage rate w, which is exactly the labor supply function of the individual N ∗ = N(w). The slope the supply function N ′ (w) is determined by the sign of the cross-derivative fN w C ← N ւ U, Uc , UL ւ տ տ L w fN w = UC + wNUCC − NULC The sign of fN w is indeterminate, therefore, the slope of N ∗ = N(w) is also indeterminate. Numerical Examples: √ √ Example 1: U = 2 C + 2 L elasticity of substitution σ > 1 √ √ √ √ U = 2 C+2 L = 2 wN+2 168 − N , dU w 1 w 1 = √ −√ = √ −√ = 0. dN 168 − N C L wN

52 Therefore, the inverse labor supply function is w = N/(168 − N), which is positively sloped. w w
T

σ>1

T

EN

σ 0, then we deﬁne y ≡ loga x = . ln a 1 d Using inverse function rule: dx = ey dy, ln x = ln x = 1/x. dx e y y = ex T y = ln x Ex x ax = eln a = e(ln a)x ⇒ d lna x = dx 1 ln a 1 x

d x a = (ln a)e(ln a)x = (ln a)ax . dx

growth rate of a function of time y = f (t): growth rate ≡ 1 dy f′ d = = [ln f (t)]. y dt f dt

Example: f (t) = g(t)h(t). ln f (t) = ln g(t) + ln h(t), therefore, growth rate of f (t) is equal to the sum of the growth rates of g(t) and h(t). Interest compounding A: principal (PV), V = Future Value, r = interest rate, n = number of periods V = A(1 + r)n If we compound interest m times per period, then the interest rate each time becomes r/m, the number of times of compounding becomes mn, and V = A[(1 + r/m)m ]n m→∞ lim (1 + r/m)m = 1 + r + r 2 /2! + · · · + r n /n! + . . . = er

Therefore, V → Aern , this is the formula for instantaneous compounding.

54 7.13 Timing: (when to cut a tree)

t: number √ years to wait, A(t) present value after t years of V (t) = Ke√ t : the market value of the tree after t years A(t) = Ke t e−rt is the present value We want to ﬁnd the optimal t such that the present value is maximized. max A(t) = Ke t e−rt . t √

The FOC is A′ = Because A(t) = 0, FOC implies:

1 √ − r A(t) = 0 2 t

1 √ − r = 0, t∗ = 1/(4r 2) 2 t For example, if r = 10%, t = 25, then to wait 25 years before cutting the tree is the optimum. Suppose that A(t) = ef (t) FOC becomes: A′ (t)/A(t) = f ′ (t) = r, ⇒ f ′ (t) is the instantaneous growth rate (or the marginal growth rate) at t, ⇒ at t = 25, growth rate = 10% = r, at t = 26, growth rate < 10%. Therefore, it is better to cut and sell the tree at t = 25 and put the proceed in the bank than waiting longer. SOC: It can be shown that A′′ < 0. 7.14 Problems

1. Suppose the total cost function of a competitive ﬁrm is C(Q) = eaQ+b . Derive the supply function. 2. The input demand function of a competitive producer, X = X(W ), can be derived by maximizing the proﬁt function Π = F (X) − W X with respect to X, where X is the quantity of input X and W is the price of X. Derive the comparative statics dX/dW and determine its sign. 3. The utility function of a consumer is given by U = U(X) + M, where X is the quantity of commodity X consumed and M is money. Suppose that the total income of the consumer is $ 100 and that the price of X is P . Then the utility function become U(X) + (100 − XP ). (a) Find the ﬁrst order condition of the utility maximization problem. (b) What is the behavior equation implied by the ﬁrst order condition? (c) Derive dX/dP and determine its sign. What is the economic meaning of your result? 4. The consumption function of a consumer, C = C(Y ), can be derived by maximizing the utility function U(C, Y ) = u1 (C) + u2 (Y − C), where u′1 (C) > 0, u′2 (Y − C) > 0 and u1 ”(C) < 0, u2 ”(Y − C) < 0. Derive the comparative statics dC/dY and determine its sign.

55 5. Consider a duopoly market with two ﬁrms, A and B. The inverse demand function is P = f (QA + QB ), f ′ < 0, f ” < 0. The cost function of ﬁrm A is T CA = C(QA ), C ′ > 0, C” > 0. The proﬁt of ﬁrm A is ΠA = P QA − T CA = QA f (QA + QB ) − C(QA ). For a given output of ﬁrm B, QB , there is a QA which maximizes ﬁrm A’s proﬁt. This relationship between QB and QA is called the reaction function of ﬁrm A, QA = RA (QB ).
′ (a) Find the slope of the reaction function RA = dQA . dQB

(b) When QB increases, will ﬁrm A’s output QA increase or decrease? 6. The proﬁt of a monopolistic ﬁrm is given by Π = R(x) − C(x, b), where x is output, b is the price of oil, R(x) is total revenue, and C(x, b) is the total cost function. For any given oil price b, there is an optimal output which maximizes proﬁt, that is, the optimal output is a function of oil price, x = x(b). Assume that Cbx = ∂ 2 C/∂b∂x > 0, that is, an increase in oil price will increase marginal cost. Will an increase in oil price increase output, that is, is dx/db > 0? 7. Consider a monopsony who uses a single input, labor (L), for the production of a commodity (Q), which he sells in a perfect competition market. His production function is Q = F (L), (f ′ (L) > 0). The labor supply function is L = L(w), or more convenient for this problem, w = L−1 (L) = W (L). Given the commodity price p, there is an optimal labor input which maximizes the monopsonist’s total proﬁt Π = pf (L) − W (L)L. In this problem, you are asked to derive the relation between L and p. (a) State the FOC and the SOC of the proﬁt maximization problem. (b) Derive the comparative statics dL/dp and determine its sign. 8. Suppose that a union has a ﬁxed supply of labor (L) to sell, that unemployed workers are paid unemployment insurance at a rate of $u per worker, and that the union wishes to maximize the sum of the wage bill plus the unemployment compensation S = wD(w) + u(L − D(w)), where w is wage per worker, D(w) is labor demand function, and D ′ (w) < 0. Show that if u increases, then the union should set a higher w. (Hint: w is endogenous and u is exogenous.)

9. The production function of a competitive ﬁrm is given by Q = F (L, K). where L is variable input and K is ﬁxed input. The short run proﬁt function is given by Π = pQ−wL−rK, where p is output price, w is wage rate, and r is the rental rate on capital. In the short run, given the quantity of ﬁxed input K,there is a L which maximizes Π. Hence, the short run demand for L can be regarded as a function of K. Assume that FLK > 0. (a) State the FOC and SOC of the proﬁt maximization problem. (b) Derive the comparative statics dL/dK and determine its sign.

56 7.15 Concavity and Convexity

The derivation of a behavioral equation X = X(Y ) from the maximization problem maxx F (X; Y ) is valid only if there exists a unique global maximum for every Y . If there are multiple global maximum, then X = X(Y ) has multiple values and the comparative static analysis is not valid. Here we are going to discuss a condition on F (X; Y ) so that a critical point is always a unique global maximum and the comparative static analysis is always valid. Convex sets A is a convex set if ∀X 0 , X 1 ∈ A and 0 ≤ θ ≤ 1, X θ ≡ (1 − θ)X 0 + θX 1 ∈ A. (If X 0 , X 1 ∈ A then the whole line connecting X 0 and X 1 is in A.) x2 x2
T 1( 0) d d T 52 43 E x1 some non-convex sets

some convex sets

E x1

(1) If A1 and A2 are convex, then A1 ∩ A2 is convex but A1 ∪ A2 is not necessarily convex. Also, the empty set itself is a convex set. (2) The convex hull of A is the smallest convex set that contains A. For example, the convex hull of {X 0 } ∪ {X 1 } is the straight line connecting X 0 and X 1 . Convex and concave functions Given a function F (X), we deﬁne the sets G+ ≡ {(x, y)| y ≥ F (x), x ∈ R}, F G− ≡ {(x, y)| y ≤ F (x), x ∈ R}, F
− G+ , GF ⊂ R 2 . F

If G+ (G− ) is a convex set, then we say F (X) is a convex function (a concave funcF F tion). If F (X) is deﬁned only for nonnegative values X ≥ 0, the deﬁnition is similar.
T T

G+ F

F (X)
+ GF

F (X)

G− F
EX

G− F
EX

G− is a convex set ⇒ F (X) is concave F

G+ is a convex set ⇒ F (X) is convex F

Equivalent Deﬁnition: Given X 0 < X 1 , 0 ≤ θ ≤ 1, denote F 0 = F (X 0 ), F 1 = F (X 1 ). Deﬁne X θ ≡ (1 − θ)X 0 + θX 1 , F (X θ ) = F ((1 − θ)X 0 + θX 1 ). Also

57 deﬁne F θ ≡ (1 − θ)F (X 0 ) + θF (X 1 ) = (1 − θ)F 0 + θF 1 . F 1 = F (X 1 ) T F (X θ ) F (X) Fθ F0

F1

T

F 0 = F (X 0 ) X
0

X

θ

X

1

EX

X

0

X

θ

X

1

EX

Therefore, (X θ , F θ ) is located on the straight line connecting (X 0 , F 0 ) and (X 1 , F 1 ) and when θ shifts from 0 to 1, (X θ , F θ ) shifts from (X 0 , F 0 ) to (X 1 , F 1 ) (the right ﬁgure). On the other hand, (X θ , F (X θ )) shifts along the curve representing the graph of F (X) (the left ﬁgure). Put the two ﬁgures together: F (X ) θ F θ Xθ − X0 θ(X 1 − X 0 ) θ = = , 1 − Xθ 1 − X 0) X (1 − θ)(X 1−θ

Fθ − F0 θ(F 1 − F 0 ) θ = = . 1 − Fθ 1 − F 0) F (1 − θ)(F 1−θ

T

F (X)

X0

Xθ

X1

EX

F (X) is strictly concave ⇒ if for all X 0 , X 1 and θ ∈ (0, 1), F (X θ ) > F θ . F (X) is concave ⇒ if for all X 0 , X 1 and θ ∈ [0, 1], F (X θ ) ≥ F θ .
T T

F (X) F (X) is concave F (X) (not strictly) E EX X Notice that these concepts are global concepts. (They have something to do with the whole graph of F , not just the behavior of F nearby a point.) The graph of a concave function can have a ﬂat part. For a strictly concave function, the graph should be curved everywhere except at kink points. F (X) is strictly convex ⇒ if for all X 0 , X 1 and θ ∈ (0, 1), F (X θ ) < F θ . F (X) is convex ⇒ if for all X 0 , X 1 and θ ∈ [0, 1], F (X θ ) ≤ F θ . F (X) is strictly concave

58 F F

T

T

F (X) is strictly convex
EX

F (X) is convex (not strictly)
H EX

Remark 1: A linear function is both concave and convex since F θ ≡ F (X θ ). F F F
T T $$ $$ T

¨ ¨¨ ¨¨ ¨ ¨ ¨¨

EX

¢ ¢

¢

¢

¢ ¢

a convex ¢ piecewise-linear ¢ function ¢
¢ ¢

¢

a concave piecewise-linear function
EX

EX

Remark 2: A piecewise-linear function consists of linear components; for example, the income tax schedule T = f (Y ) is a piecewise-linear function. Other examples are concave F (X) = 2X X≤1 1+X X >1 convex F (X) = X X ≤1 2X − 1 X > 1

In the following theorems, we assume that F ′′ (X) exists for all X. Theorem 1: F (X) is concave, ⇔ F ′′ (X) ≤ 0 for all X. F ′′ (X) < 0 for all X ⇒ F (X) is strictly concave. ¯ ¯ Proof: By Taylor’s theorem, there exist X 0 ∈ [X 0 , X θ ] and X 1 ∈ [X θ , X 1 ] such that 1 ¯ F (X 1 ) = F (X θ ) + F ′ (X θ )(X 1 − X θ ) + F ′′ (X 1 )(X 1 − X θ )2 2 1 ¯ F (X 0 ) = F (X θ ) + F ′ (X θ )(X 0 − X θ ) + F ′′ (X 0 )(X 0 − X θ )2 2 1 ¯ ¯ ⇒ F θ = F (X θ ) + θ(1 − θ)(X 1 − X 0 )2 [F ′′ (X 0 ) + F ′′ (X 1 )]. 2 Theorem 2: If F (X) is concave and F ′ (X 0 ) = 0, then X 0 is a global maximum. If F (X) is strictly concave and F ′ (X 0 ) = 0, then X 0 is a unique global maximum. Proof: By theorem 1, X 0 must be a local maximum. If it is not a global maximum, then there exists X 1 such that F (X 1 ) > F (X 0 ), which implies that F (X θ ) > F (X 0 ) for θ close to 0. Therefore, X 0 is not a local maximum, a contradiction. Remark 1 (boundary/corner solution): The boundary or corner condition F ′ (X) ≤ 0,

59 XF ′ (X) = 0 (or F ′ (X) ≥ 0, (X − a)F ′ (X) = 0) becomes suﬃcient for global maximum. Remark 2 (minimization problem): For the minimization problem, we replace concavity with convexity and F ′′ (X) < 0 with F ′′ (X) > 0. If F (X) is convex and F ′ (X ∗ ) = 0, then X ∗ is a global minimum. If F (X) is strictly convex and F ′ (X ∗ ) = 0, then X ∗ is a unique global minimum. Remark 3: The sum of two concave functions is concave. The product of two concave function is not necesarily concave. X a is strictly concave if a < 1, strictly convex if a > 1. eX is strictly convex and ln X is strctly concave with X > 0. F
T

F

X a, a > 1

T

F

X a, 0 < a < 1

T

X a, a < 0

EX

EX

EX

F

T

F

eaX

T

F

T

−aX 2 + bX + c

EX

ln X

EX

EX

Remark 4: A concave function does not have to be diﬀerentiable, but it must be continuous on the interior points. 7.16 Indeterminate forms and L’Hˆpital’s rule o

g(x) , g(a) = h(a) = 0 and g(x) and h(x) be continuous at x = a. f (a) is h(x) 0 not deﬁned because it is . However, limx→a f (x) can be calculated. 0 Let f (x) = g(x) g(x) − g(a) g(a + ∆x) − g(a) g ′ (a) lim f (x) = lim = lim = lim = ′ . x→a x→a h(x) x→a h(x) − h(a) ∆x→0 h(a + ∆x) − h(a) h (a) The same procedure also works for the case with g(a) = h(a) = ∞. g(x) ln[(ax + 1)/2] Example 1: f (x) = = , h(x) x

60 g(0) = h(0) = 0, h′ (x) = 1, g ′(x) = ⇒ h′ (0) = 1 and g ′ (0) = (ln a)ax . ax + 1

ln a ln a , ⇒ limx→0 f (x) = . 2 2 x g(x) Example 2: f (x) = = x , h′ (x) = ex , g ′ (x) = 1. h(x) e 1 ⇒ h′ (∞) = ∞ and g ′ (∞) = 1, ⇒ limx→∞ f (x) = = 0. ∞ g(x) ln x ′ 1 Example 3: f (x) = = , h (x) = 1, g ′ (x) = . h(x) x x ⇒ limx→0+ h′ (x) = 1 and limx→0+ g ′(x) = ∞, ⇒ limx→0+ f (x) = 7.17 Newton’s method

1 = 0. ∞

We can approximate a root x∗ of a nonlinear equation f (x) = 0 using an algorithm called Newton’s method.
T

f ′ (xn ) =

f (xn ) xn − xn+1 f (xn ) . f ′ (xn )

f (xn )

Recursive formula: xn+1 = xn −
Ex

xn

xn+1 x∗

If f (x) is not too strange, we will have limn→∞ xn = x∗ . Usually, two or three steps would be good enough. x3 − 3 x 1 Example: f (x) = x3 − 3, f ′ (x) = 3x2 , xn+1 = xn − = xn − + 2 . 2 3x 3 x 1 5 5 5 9 331 Starting x0 = 1, x1 = 1 − + 1 = ≈ 1.666. x2 = − + = ≈ 1.47. The 3 3 3 9 25 225 √ true value is x∗ = 3 3 ≈ 1.44225.

61

8

Optimization–Multivariate Case x1 ,...,xn

Suppose that the objective function has n variables: max F (x1 , . . . , xn ) = F (X). A local maximum X ∗ = (x∗ , . . . , x∗ ): ∃ǫ > 0 such that F (X ∗ ) ≥ F (X) for all X 1 n satisfying xi ∈ (x∗ − ǫ, x∗ + ǫ) ∀i. i i ∂F (X c ) c A critical point X = (xc , . . . , xc ): = 0 ∀i. 1 n ∂xi A global maximum X ∗ : F (X ∗ ) ≥ F (X) ∀X. A unique global maximum X ∗ : F (X ∗ ) > F (X) ∀X = X ∗ . The procedure is the same as that of the single variable case. (1) Use FOC to ﬁnd critical points; (2) use SOC to check whether a critical point is a local maximum, or show that F (X) is concave so that a critical point must be a global maximum. If we regard variables other than xi as ﬁxed, then it is a single variable maximization problem and, therefore, we have the necessary conditions FOC: ∂F = 0 and ∂xi SOC: ∂2F < 0, i = 1, . . . , n. ∂x2 i

However, since there are n×n second order derivatives (the Hessian matrix H(F )) and the SOC above does not consider the cross-derivatives, we have a reason to suspect that the SOC is not suﬃcient. In the next section we will provide a counter-example and give a true SOC. 8.1 SOC

SOC of variable-wise maximization is wrong Example: max F (x1 , x2 ) = −x2 + 4x1 x2 − x2 . 1 2 x1 ,x2

FOC: F1 = −2x1 + 4x2 = 0, F2 = 4x1 − 2x2 = 0, ⇒ x1 = x2 = 0. SOC? F11 = F22 = −2 < 0. x2
T

(0,0) is a saddle point. -4
£ £ £

£ £

£

£

£

F

T

4 1

When xj = 0, xi = 0 is a maximum.

$$$ $$$ $ $ £ $$ E x1 $$ $$ £ $$ -1 $$ 1 £ $$ -4 £ 4 £ £ £ £ £ £ £

-1 £

E xi

62 F (0, 0) = 0 < F (k, k) = 2k 2 for all k = 0. ⇒ (0, 0) is not a local maximum! The true SOC should take into consideration the possibility that when x1 and x2 increase simultaneously F may increase even if individual changes cannot increase F . SOC of sequential maximization: We can solve the maximization problem sequentially. That is, for a given x2 , we can ﬁnd a x1 such that F (x1 , x2 ) is maximized. The FOC and SOC are F1 = 0 and F11 < 0. The solution depends on x2 , x1 = h(x2 ), i.e., we regard x1 as endogenous variable and x2 as an exogenous variable in the ﬁrst stage. Using implicit function rule, dx1 /dx2 = h′ (x2 ) = −F12 /F11 . In the second stage we maximize M(x2 ) ≡ F (h(x2 ), x2 ). The FOC is M ′ (x2 ) = F1 h′ + F2 = F2 = 0 (since F1 = 0). The SOC is
2 M ′′ (x2 ) = F1 h′′ + F11 (h′ )2 + 2F12 h′ + F22 = (−F12 + F11 F22 )/F11 < 0. 2 Therefore, the true SOC is: F11 < 0 and F11 F22 − F12 > 0. For the n-variable case, the sequential argument is more complicated and we use Taylor’s expansion.

SOC using Taylor’s expansion: To ﬁnd the true SOC, we can also use Taylor’s theorem to expand F (X) around a critical point X ∗ : F (X) = F (X ∗ ) + F1 (X ∗ )(x1 − x∗ ) + F2 (X ∗ )(x2 − x∗ ) 1 2 1 + (x1 − x∗ , x2 − x∗ ) 1 2 2 F11 (X ∗ ) F12 (X ∗ ) F21 (X ∗ ) F22 (X ∗ ) x1 − x∗ 1 x2 − x∗ 2 + higher order terms.

Since X ∗ is a critical point, F1 (X ∗ ) = F2 (X ∗ ) = 0 and we have the approximation for X close to X ∗ : 1 F (X) − F (X ∗ ) ≈ (x1 − x∗ , x2 − x∗ ) 1 2 2 True SOC: v ′ H ∗ v < 0 for all v = 0. F11 (X ∗ ) F12 (X ∗ ) F21 (X ∗ ) F22 (X ∗ ) x1 − x∗ 1 x2 − x∗ 2 1 ≡ v ′ H ∗ v. 2

In the next section, we will derive a systematic method to test the true SOC. 8.2 Quadratic forms and their signs

Since vi vj = vj vi , we can assume that aij = aji so that A is symmetrical. 1 −1 v1 2 2 Example 1: (v1 − v2 )2 = v1 − 2v1 v2 + v2 = (v1 , v2 ) . −1 1 v2

A quadratic form in n variables is a second degree homogenous function.    a11 · · · a1n v1 n n  . .  .  f (v1 , . . . , vn ) = aij vi vj = (v1 , . . . , vn ) . . . . .  .  ≡ v ′ Av. . . . i=1 j=1

an1 · · · ann

vn

63 0 1 1 0 v1 . v2 

Example 2: 2v1 v2 = (v1 , v2 )

v ′ Av v ′ Av v ′ Av v ′ Av (1) (2) (3) (4)

  1 0 3 x1 Example 3: x2 + 2x2 + 6x1 x3 = (x1 , x2 , x3 ) 0 2 0  x2 . 1 2 3 0 0 x3 is is is is called called called called

negative semideﬁnite if v ′ Av ≤ 0 for all v ∈ Rn . negative deﬁnite if v ′ Av < 0 for all v = 0. positive semideﬁnite if v ′ Av ≥ 0 for all v ∈ Rn . positive deﬁnite if v ′ Av > 0 for all v = 0.

2 2 −v1 − 2v2 < 0, if v1 = 0 or v2 = 0 ⇒ negative. −(v1 − v2 )2 ≤ 0, = 0 if v1 = v2 ⇒ negative semideﬁnite. (v1 + v2 )2 ≥ 0, = 0 if v1 = −v2 ⇒ positive semideﬁnite. 2 2 v1 + v2 > 0, if v1 = 0 or v2 = 0 ⇒ positive. > 2 2 > (5) v1 − v2 0 if |v1 | |v | ⇒ neither positive nor negative deﬁnite. < < 2 > = (6) v1 v2 0 if sign(v1 ) sign(v2 ) ⇒ neither positive nor negative deﬁnite. < = Notice that if v ′ Av is negative (positive) deﬁnite then it must be negative (positive) semideﬁnite. Testing the sign of a quadratic form: a11 a12 v1 2 2 n = 2: v ′ Av = (v1 , v2 ) = a11 v1 + 2a12 v1 v2 + a22 v2 = a21 a22 v2

a11

a2 2 a2 a12 a12 2 2 v2 v1 + 2 v1 v2 + 12 v2 + − 12 + a22 v2 = a11 v1 + a11 a2 a11 a11 11

2

a11 a12 a21 a22 2 + v2 . a11

Negative deﬁnite ⇔ a11 < 0 and Positive deﬁnite ⇔ a11 > 0 and

a11 a21 a11 a21

a12 = a11 a22 − a2 > 0. 12 a22 a12 = a11 a22 − a2 > 0. 12 a22

For negative or positive semideﬁnite, replace strict inequalities with semi-inequalities. The proofs for them are more diﬃcult and discussed in Lecture 13.
2 2 Example 1: F (v1 , v2 ) = −2v1 + 8v1 v2 − 2v2 = (v1 , v2 )

−2 4 = −12 < 0 ⇒ neither positive nor negative. The matrix is 4 −2 the Hessian of F of the counter-example in section 1. Therefore, the counter-example violates the SOC. −2 < 0 but

−2 4 4 −2

v1 v2

, a11 =

64  a11 · · · a1n  . .  General n: v ′ Av = (v1 , . . . , vn ) . . . . .  . . an1 · · · ann  a11 a12 a21 a22 + a11  v1 . .  = a11 (v1 + · · ·)2 .  vn

a11 a12 a13 a21 a22 a23 a31 a32 a33 (v3 +· · ·)2 +· · ·+ (v2 +· · ·)2 + a11 a12 a21 a22  

a11 . . .

a11 · · · a1n . .. . . ··· . an1 · · · ann

a(n−1)1

··· a1(n−1) . .. . . . · · · a(n−1)(n−1)

2 vn .

a11 · · · a1k  . .. .  k-th order principle minor of A: A(k) ≡  . . . , k = 1, . . . , n. . . ak1 · · · akk (2) |A | |A(3) | |A(n) | Negative deﬁnite ⇔ A(1) = a11 < 0, (1) < 0, (2) < 0, . . ., (n−1) < 0. |A | |A | |A | (2) (3) (n) |A | |A | |A | Positive deﬁnite ⇔ A(1) = a11 > 0, (1) > 0, (2) > 0, . . ., (n−1) > 0. |A | |A | |A | a11 a12 Negative deﬁnite ⇔ |A(1) | = a11 < 0, |A(2) | = > 0, |A(3) | = a21 a22 a11 · · · a1n a11 a12 a13 . .. . n (n) n . a21 a22 a23 < 0, · · ·, (−1) |A | = (−1) > 0. . . . . a31 a32 a33 an1 · · · ann a11 a12 Positive deﬁnite ⇔ |A(1) | = a11 > 0, |A(2) | = > 0, |A(3) | = a21 a22 a11 · · · a1n a11 a12 a13 . .. . (n) . a21 a22 a23 > 0, · · ·, |A | = > 0. . . . . a31 a32 a33 an1 · · · ann The conditions for semideﬁnite are more complicated and will be discussed in Lecture 13 using the concept of eigenvalues of a square matrix.

8.3

SOC again

From the last two sections, the SOC for a local maximum can be summarized as: F11 (X ∗ ) F12 (X ∗ ) SOC: v ′ HF (X ∗ )v is negative deﬁnite ⇒ F11 (X ∗ ) < 0, = F21 (X ∗ ) F22 (X ∗ ) F11 (X ∗ ) F12 (X ∗ ) F13 (X ∗ ) 2 F11 F22 − F12 > 0, F21 (X ∗ ) F22 (X ∗ ) F23 (X ∗ ) < 0, . . . F31 (X ∗ ) F32 (X ∗ ) F33 (X ∗ )
2 Example: max F (x1 , x2 ) = 3x1 + 3x1 x2 − 3x1 − x3 . (A cubic function in 2 vari2 ables.)

65 FOC: F1 = 3 + 3x2 − 6x1 = 0 and F2 = 3x1 − 3x2 = 0. 2   1 x1 1   Two critical points: =  41  and . x2 1 − 2 F11 (X) F12 (X) −6 3 Hessian matrix: H(x1 , x2 ) = = ; |H 1 (X)| = F21 (X) F22 (X) 3 −6x2 −6 3 F11 (X) = −6 < 0, |H 2(X)| = = 36x2 − 9. 3 −6x2  1 1 1  4  |H 2 ( , − )| = −27 < 0 ⇒  1  is not a local max. 4 2 − 2 1 |H 2 (1, 1)| = 27 > 0 ⇒ is a local max. F (1, 1) = 2. It is not a global maximum 1 because F →∞ when x2 → − ∞.
2 Remark 1: F11 < 0 and F11 F22 − F12 > 0 together implies F22 < 0. Remark 2: If |H k (X ∗ )| = 0 for some 1 ≤ k ≤ n then X ∗ is a degenerate critical point. Although we can still check whether v ′ Hv is negative semideﬁnite, it is insuﬃcient for a local maximum. We have to check the third or higher order derivatives to determine whether X ∗ is a local maximum. It is much more diﬃcult than the single variable case with F ′′ = 0.

8.4

Joint products

2 2 Example: C(q1 , q2 ) = 2q1 + 3q2 − q1 q2 FOC: p1 − 4q1 + q2 = 0, p2 + q1 − 6q2 = 0 ⇒ supply functions q1 = (6p1 + p2 )/23 and q2 = (p1 + 4p2 )/23. −C11 −C12 −4 1 SOC: −C11 = −4 < 0, = 23 > 0. = −C21 −C22 1 −6

A competitive producer produces two joint products. (eg., gasoline and its by products or cars and tructs, etc.) Cost function: C(q1 , q2 ). Proﬁt function: Π(q1 , q2 ; p1 , p2 ) = p1 q1 + p2 q2 − C(q1 , q2 ). FOC: Π1 = p1 − C1 = 0, Π2 = p2 − C2 = 0; or pi = MCi . −C11 −C12 SOC: Π11 = −C11 < 0, ≡ ∆ > 0. −C21 −C22

Comparative statics: Total diﬀerential of FOC:  C11 C12 C21 C22 dq1 dq2 = dp1 dp2

∂q1  ⇒  ∂p1 ∂q2 ∂p1

 ∂q1 ∂p2  = 1 ∂q2  ∆ ∂p2

C22 −C12 −C21 C11

.

66 SOC ⇒ C11 > 0, C22 > 0, ∆ > 0 ⇒ ∂q2 ∂q1 > 0, > 0. ∂p1 ∂p2

∂q1 ∂q2 and are positive if the joint products are beneﬁcially to each other in the ∂p2 ∂p1 production so that C12 < 0. 8.5 Monopoly price discrimination

Example: f1 = a − bq1 , f2 = α − βq2 , and C(Q) = 0.5Q2 = 0.5(q1 + q2 )2 . ′ ′ ′′ ′′ f1 = −b, f2 = −β, f1 = f2 = 0, C ′ = Q = q1 + q2 , and C ′′ = 1. 1 + 2b 1 q1 a FOC: a − 2bq1 = q1 + q2 = α − 2βq2 ⇒ = 1 1 + 2β q2 α 1 q1 a(1 + 2β) − α ⇒ = . q2 (1 + 2b)(1 + 2β) − 1 α(1 + 2b) − a SOC: −2b − 1 < 0 and ∆ = (1 + 2b)(1 + 2β) − 1 > 0. p
T

A monopoly sells its product in two separable markets. Cost function: C(Q) = C(q1 + q2 ) Inverse market demands: p1 = f1 (q1 ) and p2 = f2 (q2 ) Proﬁt function: Π(q1 , q2 ) = p1 q1 + p2 q2 − C(q1 + q2 ) = q1 f1 (q1 ) + q2 f2 (q2 ) − C(q1 + q2 ) ′ ′ FOC: Π1 = f1 (q1 ) + q1 f1 (q1 ) −C ′ (q1 + q2 ) = 0, Π2 = f2 (q2 ) + q2 f2 (q2 ) −C ′ (q1 + q2 ) = 0; or MR1 = MR2 = MC. ′ ′′ 2f1 + q1 f1 − C ′′ −C ′′ ′ ′′ SOC: Π11 = 2f1 + q1 f1 − C ′′ < 0, ≡ ∆ > 0. ′′ ′ ′′ −C 2f2 + q2 f2 − C ′′

MC∗

d d d d d d MR1+2 d d MR d 2 d MR1
∗ q1 ∗ q2

MC

MC = MR1+2 ⇒ Q∗ , MC∗
∗ MC∗ = MR1 ⇒ q1 ∗ MC∗ = MR2 ⇒ q2

Q

∗

Eq

8.6

SR supply vs LR supply - Le Chˆtelier principle a

A competitive producer employs a variable input x1 and a ﬁxed input x2 . Assume that the input prices are both equal to 1, w1 = w2 = 1. Production function: q = f (x1 , x2 ), assume MPi = fi > 0, fii < 0, fij > 0,

67
2 f11 f22 > f12 . Proﬁt function: Π(x1 , x2 ; p) = pf (x1 , x2 ) − x1 − x2 .

Short-run problem (only x1 can be adjusted, x2 is ﬁxed): SR FOC: Π1 = pf1 − 1 = 0, or w1 = VMP1 . SOC: Π11 = pf11 < 0, 2 dx1 −f1 dq s −f1 −1 Comparative statics: f1 dp+pf11 dx1 = 0 ⇒ = > 0, = = 3 > 0. dp pf11 dp pf11 p f11 Long-run problem (both x1 and x2 can be adjusted): LR FOC: Π1 = pf1 − 1 = 0, Π2 = pf2 − 1 = 0; or wi = VMPi . pf11 pf12 SOC: Π11 = pf11 < 0, ≡ ∆ > 0. pf21 pf22 f1 dp pf11 pf12 dx1 0 Comparative statics: + = f2 dp pf21 pf22 dx2 0 −1 dq L −(f11 + f22 − 2f12 ) dx1 /dp f1 f22 − f2 f12 ⇒ = , = 2 2 dx2 /dp f2 f11 − f1 f21 p(f11 f22 − f12 ) dp p3 (f11 f22 − f12 ) dq s dq L > . Le Chˆtelier principle: a dp dp Example: f (x1 , x2 ) = 3x1 x2 (homogenous of degree 2/3) −2/3 1/3 1/3 −2/3 LR FOC: px1 x2 = 1 = px1 x2 ⇒ x1 = x2 = p3 q L = 3p2 . −2/3 SR FOC (assuming x2 = 1): px1 ¯ = 1 ⇒ x1 = p3/2 q s = 3p1/2 . L s η (LR supply elasticity) = 2 > η (SR supply elasticity) = 1/2. L p π ¨ π (p) s 0.5 T T ¨ S (p) = 3p ¨
¨ ¨¨ ¨¨ ¨¨
1/3 1/3

π s (p; x2 ) ¯ h(p)

S L (p) = 3p2
Eq Ep

p ¯

Envelop theorem, Hotelling’s lemma, and Le Chˆtelier principle a From the SR problem we ﬁrst derive the SR variable input demand function x1 = x1 (p, x2 ). ¯ Then the SR supply function is obtained by substituting into the production function q s = f (x1 (p, x2 ), x2 ) ≡ S s (p; x2 ). ¯ ¯ ¯ The SR proﬁt function is π s (p, x2 ) = pq s − x1 (p, x2 ) − x2 . ¯ ¯ ¯ ∂π s Hotelling’s lemma: By envelop theorem, = S s (p, x2 ). ¯ ∂p From the LR problem we ﬁrst derive the input demand functions x1 = x1 (p) and x2 = x2 (p). Then the LR supply function is obtained by substituting into the production function

68 q L = f (x1 (p), x2 (p)) ≡ S L (p). The LR proﬁt function is π L (p) = pq L − x1 (p) − x2 (p). ∂π L (Also Hotelling’s lemma) By envelop theorem, = S L (p). ∂p Notice that π L (p) = π s (p; x2 (p)). Let x2 = x2 (¯), deﬁne h(p) ≡ π L (p) − π s (p, x2 ). h(p) ≥ 0 because in the LR, ¯ p ¯ the producer can adjust x2 to achieve a higher proﬁt level. Also, h(¯) = 0 because π L (¯) = π s (¯; x2 (¯)) = π s (¯; x2 ). p p p p p ¯ ∂2 πL ∂2πs ′′ Therefore, h(p) has a minimum at p = p and the SOC is h (¯) > 0 = ¯ p − 2 > 0, ∂p2 ∂p which implies dq L dq s Le Chˆtelier principle: a − > 0. dp dp 8.7 Concavity and Convexity

Similar to the single variable case, we deﬁne the concepts of concavity and convexity for 2-variable functions F (X) = F (x1 , x2 ) by deﬁning G+ , G− ⊂ R2 s follows. F F G+ ≡ {(x1 , x2 , y)| y ≥ F (x1 , x2 ), (x1 , x2 ) ∈ R2 }, F G− ≡ {(x1 , x2 , y)| y ≤ F (x1 , x2 ), (x1 , x2 ) ∈ R2 }. F If G+ (G− ) is a convex set, then we say F (X) is a convex function (a concave funcF F tion). If F (X) is deﬁned only for nonnegative values x1 , x2 ≥ 0, the deﬁnition is similar. (The extension to n-variable functions is similar.) Equivalently, given X 0 = F 1 ≡ F (X 1 ), we deﬁne x0 1 x0 2 , X1 = x1 1 x1 2 ≡ , 0 ≤ θ ≤ 1, F 0 ≡ F (X 0 ), xθ 1 xθ 2 , F (X θ ) = F ((1 − θ)X 0 +

X θ ≡ (1 − θ)X 0 + θX 1 =

θX 1 ) F θ ≡ (1 − θ)F (X 0 ) + θF (X 1 ) = (1 − θ)F 0 + θF 1 . x2
T

(1 − θ)x0 + θx1 1 1 (1 − θ)x0 + θx1 2 2

x0 2 xθ 2 x1 2

d d Xθ d d d d X1

X0

X θ is located on the straight line connecting X 0 and X 1 , when θ shifts from 0 to 1,
E x1

x0 1

xθ 1

x1 1

X θ shifts from X 0 to X 1 .

On 3-dimensional (x1 –x2 –F ) space, (X θ , F θ ) is located on the straight line connecting (X 0 , F 0 ) and (X 1 , F 1), when θ shifts from 0 to 1, (X θ , F θ ) shifts from (X 0 , F 0 ) to (X 1 , F 1 ). On the other hand, (X θ , F (X θ )) shifts along the surface representing the graph of F (X).

69

F (X) F (X) F (X) F (X)

is is is is

strictly concave ⇒ if for all X 0 , X 1 and θ ∈ (0, 1), F (X θ ) > F θ . concave ⇒ if for all X 0 , X 1 and θ ∈ [0, 1], F (X θ ) ≥ F θ . strictly convex ⇒ if for all X 0 , X 1 and θ ∈ (0, 1), F (X θ ) < F θ . convex ⇒ if for all X 0 , X 1 and θ ∈ [0, 1], F (X θ ) ≤ F θ .
1/3 1/3

Example: 9x1 x2 . Assume that F is twice diﬀerentiable. Theorem 1: F (X) is concave, ⇔ v ′ HF v is negative semideﬁnite for all X. v ′ HF v is negative deﬁnite for all X ⇒ F (X) is strictly concave. ¯ ¯ Proof: By Taylor’s theorem, there exist X 0 ∈ [X 0 , X θ ] and X 1 ∈ [X θ , X 1 ] such that 1 ¯ F (X 1) = F (X θ ) + ∇F (X θ )(X 1 − X θ ) + (X 1 − X θ )′ HF (X 1 )(X 1 − X θ ) 2 1 ¯ F (X 0) = F (X θ ) + ∇F (X θ )(X 0 − X θ ) + (X 0 − X θ )′ HF (X 0 )(X 0 − X θ ) 2 θ(1 − θ) 1 ¯ ¯ ⇒ F θ = F (X θ ) + (X − X 0 )′ [HF (X 0 ) + HF (X 1 )](X 1 − X 0 ). 2 Theorem 2: If F (X) is concave and ∇F (X 0 ) = 0, then X 0 is a global maximum. If F (X) is strictly concave and ∇F (X 0 ) = 0, then X 0 is a unique global maximum. Proof: By theorem 1, X 0 must be a local maximum. If it is not a global maximum, then there exists X 1 such that F (X 1 ) > F (X 0 ), which implies that F (X θ ) > F (X 0 ) for θ close to 0. Therefore, X 0 is not a local maximum, a contradiction. Remark 1 (boundary or corner solution): The boundary or corner condition Fi (X) ≤ 0, xi Fi (X) = 0 (or Fi (X) ≥ 0, (xi − ai )Fi (X) = 0) becomes suﬃcient for global maximum. Remark 2 (minimization problem): For the minimization problem, we replace concavity with convexity and negative deﬁnite with positive deﬁnite. If F (X) is convex and ∇F (X ∗ ) = 0, then X ∗ is a global minimum. If F (X) is strictly convex and ∇F (X ∗ ) = 0, then X ∗ is a unique global minimum. 8.8 Learning and utility maximization

Consumer B’s utility function is U = u(x, k) + m − h(k), x, m, k ≥ 0,

where x is the quantity of commodity X consumed, k is B’s knowledge regard2 ing the consumption of X, m is money, u(x, k) is the utility obtained, ∂ u < 0, ∂x2 ∂2u > 0 (marginal utility of X increases with k), and h(k) is the disutility of acquir∂x∂k ing/maintaining k, h′ > 0, h′′ > 0. Assume that B has 100 dollar to spend and the price of X is Px = 1 so that m = 100 − x and U = u(x, k) + 100 − x − h(k). Assume

70 also that k is ﬁxed in the short run. The short run utility maximization problem is ∂u ∂2u max u(x, k) + 100 − x − h(k), ⇒ FOC: − 1 = 0, SOC: < 0. x ∂x ∂x2 The short run comparative static dx is derived from FOC as dk ∂ 2 u/∂x2 dx =− 2 > 0, dk ∂ u/∂x∂k that is, consumer B will consume more of X if B’s knowledge of X increases. In the long run consumer B will change k to attain higher utility level. The long run utility maximization problem is max u(x, k) + 100 − x − h(k), x,k ⇒ FOC:

∂u − 1 = 0, ∂x

∂u − h′ (k) = 0. ∂k

The SOC is satisﬁed if we assume that u(x, k) is concave and h(k) is convex (h′′ (k) > 0), because F (x, k) ≡ u(x, k) + 100 − x − h(k), x, k ≥ 0 is strictly concave then. Consider now the speciﬁc case when u(x, k) = 3x2/3 k 1/3 and h(k) = 0.5k 2 .

1. Calculate consumer B’s short run consumption of X, xs = x(k). (In this part, you may ignore the nonnegative constraint m = 100 − x ≥ 0 and the possibility of a corner solution.) x = 1/(8k). 2. Calculate the long run consumption of X, xL . x = 32 3. Derive the short run value function V (k) ≡ u(x(k), k) + 100 − x(k) − h(k). 4. Solve the maximization problem maxk V (k). k = 4. 5. Explain how the SOC is satisﬁed and why the solution is the unique global maximum. (demand functions) Consider now the general case when Px = p so that U = 3x2/3 k 1/3 + 100 − px − 0.5k 2 . 1. Calculate consumer B’s short run demand function of X, xs = x(p; k). (Warning: There is a nonnegative constraint m = 100 − px ≥ 0 and you have to consider both interior and corner cases.) √ √ xs (p) = 8kp−3 (100/p) if p ≤ 2k/5 (p < 2k/5). 2. Calculate the long run demand function of X, xL (p). and the optimal level of K, kL (p). (Both interior and corner cases should be considered too.) xL (p) = 32p−5 (100/p) if p > (8/25)1/4 (p < (8/25)1/4), kL (p) = [xL (p)]2/5 .

71 8.9 Homogeneous and homothetic functions

A homogeneous function of degree k is f (x1 , . . . , xn ) such that f (hx1 , . . . , hxn ) = hk f (x1 , . . . , xn ) ∀h > 0. If f is homogeneous of degree k1 and g homogeneous of degree k2 , then f g is homogef neous of degree k1 + k2 , is homogeneous of degree k1 − k2 , and f m is homogeneous g of degree mk1 . x2 x2 If Q = F (x1 , x2 ) is homogeneous of degree 1, then Q = x1 F (1, ) ≡ x1 f . If x1 x1 x2 x2 m = H(x1 , x2 ) is homogeneous of degree 0, then m = H(1, ) ≡ h . x1 x1 Euler theorem: If f (x1 , . . . , xn ) is homogeneous of degree k, then x1 f1 + . . . + xn fn = kf If f is homogeneous of degree k, then fi is homogenous of degree k − 1. Examples: 1. Cobb-Douglas function Q = Axα xβ is homogeneous of degree α + β. 1 2 k 2. CES function Q = {axρ + bxρ } ρ is homogenous of degree k. 1 2 3. A quadratic form x′ Ax is homogeneous of degree 2. 4. Leontief function Q = min {ax1 , bx2 } is homogeneous of degree 1. Homothetic functions: If f is homogeneous and g = H(f ), H is a monotonic increasing function, then g is called a homothetic function. Example: Q = α ln x1 + β ln x2 = ln xα xβ 1 2 is homothetic. x2 . x1

The MRS of a homothetic function depends only on the ratio 8.10 Problems

1. Use Newton’s method to ﬁnd the root of the nonlinear equation X 3 +2X +2 = 0 accurate to 2 digits. 1 − e−aX e2X − eX 1 − 2−X 2. Find (a) lim , (b) lim+ , (c) lim . X→0 X→0 X→0 X X X 3. Given the total cost function C(Q) = eaQ+b , use L’hˆpital’s rule to ﬁnd the o + AVC at Q = 0 . 4. Let z = x1 x2 + x2 + 3x2 + x2 x3 + x2 . 1 2 3 (a) Use matrix multiplication to represent z. (b) Determine whether z is positive deﬁnite or negative deﬁnite. (c) Find the extreme value of z. Check whether it is a maximum or a minimum.

72 Q3 + K, where K 3K

5. The cost function of a competitive producer is C(Q; K) = is, say, the plant size (a ﬁxed factor in the short run).

(a) At which output level, the SAC curve attains a minimum? (b) Suppose that the equilibrium price is p = 100. The proﬁt is Π = 100Q − Q3 − K. For a given K, ﬁnd the supply quantity Q(K) such that SR 3K proﬁt is maximized. (c) Calculate the SR maximizing proﬁt π(K). (d) Find the LR optimal K = K ∗ to maximize π(K). (e) Calculate the LR supply quantity Q∗ = Q(K ∗ ). (f) Now solve the 2-variable maximization problem Q3 max Π(Q, K) = pQ − C(Q) = 100Q − − K. Q,K≥0 3K and show that Π(Q, K) is concave so that the solution is the unique global maximum. 6. A competitive ﬁrm produces two joint products. The total cost function is 2 2 C(q1 , q2 ) = 2q1 + 3q2 − 4q1 q2 . (a) Use the ﬁrst order conditions for proﬁt maximization to derive the supply functions. (b) Check that the second order conditions are satisﬁed. 7. Check whether the function f (x, y) = ex+y is concave, convex, or neither. 8. (a) Check whether f (x, y) = 2 ln x+3 ln y −x−2y is concave, convex, or neither. (Assume that x > 0 and y > 0.) (b) Find the critical point of f . (c) Is the critical point a local maximum, a global maximum, or neither? 9. Suppose that a monopoly can produce any level of output at a constant marginal cost of $c per unit. Assume that the monopoly sells its goods in two diﬀerent markets which are separated by some distance. The demand curve in the ﬁrst market is given by Q1 = exp[−aP1 ] and the curve in the second market is given by Q2 = exp[−bP2 ]. If the monopolist wants to maximize its total proﬁts, what level of output should be produced in each market and what price will prevail in each market? Check that your answer is the unique global maximum. (Hints: 1. P1 = −(1/a) ln Q1 and P2 = −(1/b) ln Q2 . 2. Π = P1 Q1 + P2 Q2 − (Q1 + Q2 )c = −(1/a)Q1 ln Q1 − (1/b)Q2 ln Q2 − (Q1 + Q2 )c is strictly concave.) 10. The production function of a competitive ﬁrm is given by q = f (x1 , x2 ) =
1 1 3 3 3x1 x2 , where x1 is a variable input and x2 is a ﬁxed input. Assume that the prices of the output and the ﬁxed input are p = w2 = 1. In the short run, the amount of the ﬁxed input is given by x2 = x2 . The proﬁt function of the ¯ competitive ﬁrm is given by π = f (x1 , x2 ) − w1 x1 − x2 . ¯ ¯

73 (a) State the FOC for the SR proﬁt maximization and calculate the SR input S S 1 ∂x demand function xS = x1 (w1 ) and the SR input demand elasticity wS ∂w11 . 1 x
1

(b) Now consider the LR situation when x2 can be adjusted. State the FOC for LR proﬁt maximization and calculate the LR input demand funtion L w1 ∂x xL = xL (w1 ) and the LR input demand elasticity xL ∂w11 . 1 1
1

(c) Verify the Le Chˆtelier principle: a

L w1 ∂x1 xL ∂w1 1

>

S w1 ∂x1 xS ∂w1 1

.

(d) Show that the LR proﬁt is a strictly concave function of (x1 , x2 ) for x1 , x2 > 0 and therefore the solution must be the unique global maximum. 11. Let U(x, y) = xa y + xy 2 , x, y, a > 0. (a) For what value(s) of a U(x, y) is homogeneous?. (b) For what value(s) of a U(x, y) is homothetic?

74

9

Optimization and Equality Constraints and Nonlinear Programming

In some maximization problems, the agents can choose only values of (x1 , . . . , xn ) that satisfy certain equalities. For example, a consumer has a budget constraint p1 x1 + · · · + pn xn = m. x1 ,...,xn

max U(x) = U(x1 , . . . , xn ) subject to p1 x1 + · · · + pn xn = m.

The procedure is the same as before. (1) Use FOC to ﬁnd critical points; (2) use SOC to check whether a critical point is a local maximum, or show that U(X) is quasi-concave so that a critical point must be a global maximum. Deﬁne B = {(x1 , . . . , xn ) such that p1 x1 + · · · + pn xn = m}. A local maximum X ∗ = (x∗ , . . . , x∗ ): ∃ǫ > 0 such that U(X ∗ ) ≥ U(X) for all X ∈ B 1 n satisfying xi ∈ (x∗ − ǫ, x∗ + ǫ) ∀i. i i A critical point: A X ∗ ∈ B satisfying the FOC for a local maximum. A global maximum X ∗ : F (X ∗ ) ≥ F (X) ∀X ∈ B. A unique global maximum X ∗ : F (X ∗ ) > F (X) ∀X ∈ B, X = X ∗ . To deﬁne the concept of a critical point, we have to know what is the FOC ﬁrst. 9.1 FOC and SOC for a constraint maximization

Consider the 2-variable utility maximization problem: max U(x1 , x2 ) subject to p1 x1 + p2 x2 = m. x1 ,x2

m − p1 x1 dx2 p1 = h(x1 ), = − = h′ (x1 ), and it p2 dx1 p2 becomes a single variable maximization problem: Using the budget constraint, x2 = max U x1 x1 ,

m − p1 x1 p2

, FOC:

dU p1 = U1 + U2 − dx1 p2 p1 p2
2

= 0, SOC:

d2 U < 0. dx2 1

d2 U p1 = U11 − 2 U12 + 2 dx1 p2 By FOC, U1 U2 = ≡ λ (MU of $1). p1 p2

U22 =

−1 p2 2

0 −p1 −p2 −p1 U11 U12 . −p2 U21 U22

FOC: U1 = p1 λ, U2 = p2 λ, SOC: Alternatively, we can deﬁne Lagrangian:

0 U1 U2 U1 U11 U12 U2 U21 U22

> 0.

L ≡ U(x1 , x2 ) + λ(m − p1 x1 − p2 x2 )

75 ∂L ∂L ∂L = U1 − λp1 = 0, L2 = = U2 − λp2 = 0, Lλ = = ∂x1 ∂x2 ∂λ m − p1 x1 − p2 x2 = 0. 0 −p1 −p2 SOC: −p1 L11 L12 > 0. −p2 L21 L22 FOC: L1 = max F (x1 , x2 ) subject to g(x1 , x2 ) = 0. x1 ,x2

general 2-variable with 1-constraint case:

Using the constraint, x2 = h(x1 ), variable maximization problem: max F (x1 , h(x1 )) FOC: x1 dx2 g1 = − = h′ (x1 ), and it becomes a single dx1 g2

dF d2 F = F1 + F2 h′ (x1 ) = 0, SOC: < 0. dx1 dx2 1

d2 F d 2 = (F1 + F2 h′ ) = F11 + 2h′ F12 + (h′ ) F22 + F2 h′′ 2 dx1 dx1 h′′ = ⇒ d2 F = dx2 1 0 d dx1 − g1 g2 = −1 g11 + 2g12 h′ + g22 (h′ )2 . g2 F2 F2 g12 h′ + F22 − g22 (h′ )2 g2 g2 0 g1 g2 g1 g2 F2 F2 F11 − g11 F12 − g12 g2 g2 F2 F2 F21 − g21 F22 − g22 g2 g2

F11 −

F2 g11 g2

+ 2 F12 −

−h′ 1 F2 F2 −h′ F11 − g11 F12 − g12 −1 g2 g2 =− = 2 g2 F2 F2 1 F21 − g21 F22 − g22 g2 g2 F2 F1 By FOC, = ≡ λ (Lagrange multiplier). g1 g2 Alternatively, we can deﬁne Lagrangian:

L ≡ F (x1 , x2 ) − λg(x1 , x2 ) FOC: L1 = 0. SOC: ∂L ∂L ∂L = F1 − λg1 = 0, L2 = = F2 − λg2 = 0, Lλ = = −g(x1 , x2 ) = ∂x1 ∂x2 ∂λ > 0.

n-variable 1-constraint case: x1 ,...,xn

0 g1 g2 g1 L11 L12 g2 L21 L22

max F (x1 , . . . , xn ) subject to g(x1 , . . . , xn ) = 0. L ≡ F (x1 , . . . , xn ) − λg(x1 , . . . , xn )

76 FOC: Li = SOC: ∂L ∂L = Fi − λgi = 0, i = 1, . . . , n, Lλ = = −g(x1 , . . . , xn ) = 0. ∂xi ∂λ 0 −g1 −g2 −g3 0 −g1 −g2 0 −g1 −g1 L11 L12 L13 < 0, −g1 L11 L12 > 0 < 0, etc. −g1 L11 −g2 L21 L22 L23 −g2 L21 L22 −g3 L31 L32 L33

9.2

Examples

2 Example 1: max F (x1 , x2 ) = −x1 − x2 subject to x1 + x2 = 1. 2 L = −x2 − x2 + λ(1 − x1 − x2 ). 1 2 FOC: L1 = −2x1 − λ = 0, 2 = 2 − λ = 0 and Lλ = 1 − 2x1 − 2x2 = 0. L −2x 1 x1  2  Critical point: =  1 , λ = −1. x2 2 0 1 1 SOC: 1 −2 0 = 4 > 0. 1 0 −2 1/2 ⇒ is a local maximum. 1/2

x2

T d d dr d d d

x2

T

E x1

d d d dr d d d d d d d d d d d d d d d d d d r d d d d d d d d

E x1

Example 2: max F (x1 , x2 ) = x1 + x2 subject to x2 + x2 = 1. 1 2 L = x1 + x2 + λ(1 − x2 − x2 ). 1 2 FOC: L1 = 1 − 2λx1 = 0, L2 = 1 − 2λx2 = 0 and Lλ = 1 − x2 − x2 = 0. 1 2 √ √ Two critical points: x1 = x2 = λ = 1/ 2 and x1 = x2 = λ = −1/ 2. 0 −2x1 −2x2 0 SOC: −2x1 −2λ = 8λ(x2 + x2 ). 1 2 −2x2 0 −2λ √ √ ⇒ x1 = x2 = λ = 1/ 2 is a local maximum and x1 = x2 = λ = −1/ 2 is a local minimum.

77 9.3 Cost minimization and cost function

1−a min C(x1 , x2 ; w1, w2 ) = w1 x1 + w2 x2 subject to xa x2 = q, 0 < a < 1. 1 L = w1 x1 + w2 x2 + λ(q − xa x1−a ). 1 2 FOC: L1 = w1 − aλxa−1 x1−a = 0, L2 = w2 − (1 − a)λxa x−a = 0, Lλ = q − xa x1−a = 0. 1 2 1 2 1 2 1−a a ax2 aw2 (1 − a)w1 w1 ⇒ = ⇒ x1 = q , x2 = q . w2 (1 − a)x1 (1 − a)w1 aw2 SOC:

0
1−a −axa−1 x2 1 −(1 − a)xa x−a 1 2

1−a a −axa−1 x2 −(1 − a)x1 x−a 1 2 a−2 1−a a(1 − a)λx1 x2 −a(1 − a)λxa−1 x−a 1 2 a−1 −a −a(1 − a)λx1 x2 a(1 − a)λxa x−a−1 1 2 1−a a

a(1 − a)q 3 λ =− < 0. (x1 x2 )2

(1 − a)w1 , x2 = q is a local minimum. aw2 1−a a a 1−a a 1−a The total cost is C(w1 , w2 , q) = q + w1 w2 . 1−a a 9.4 Utility maximization and demand function

aw2 ⇒ x1 = q (1 − a)w1

max U(x1 , x2 ) = a ln x1 + b ln x2 subject to p1 x1 + p2 x2 = m. L = a ln x1 + b ln x2 + λ(m − p1 x1 − p2 x2 ). a b − λp1 = 0, L2 = − λp2 = 0 and Lλ = m − p1 x1 − p2 x2 = 0. FOC: L1 = x1 x2 a x2 p1 am bm ⇒ = ⇒ x1 = , x2 = b x1 p2 (a + b)p1 (a + b)p2 0 −p1 −p2 −a ap2 bp2 −p1 0 SOC: = 22 + 21 > 0. x2 1 x1 x2 −b −p2 0 x2 2 am bm ⇒ x1 = , x2 = is a local maximum. (a + b)p1 (a + b)p2 9.5 Quasi-concavity and quasi-convexity

As discussed in the intermediate microeconomics course, if the MRS is strictly decreasing along an indiﬀerence curve (indiﬀerence curve is convex toward the origin), then the utility maximization has a unique solution. A utility function U(x1 , x2 ) U1 is quasi-concave if MRS (= ) is decreasing along every indiﬀerence curve. In U2 case MRS is strictly decreasing, the utility function is strictly quasi-concave. If U(x1 , x2 ) is (strictly) quasi-concave, then a critical point must be a (unique) global maximum. Two ways to determine whether U(x1 , x2 ) is quasi-concave: (1) the set {(x1 , x2 ) ∈ ¯ ¯ R2 U(x1 , x2 ) ≥ U } is convex for all U . (Every indiﬀerence curve is convex toward the

78 origin.) d (2) dx1 curve.)

−

U1 U2

> 0. (MRS is strictly decreasing along every indiﬀerence
¯ U (x1 ,x2 )=U

+ In sections 6.14 and 7.7 we used F (X) to deﬁne two sets, GF and G− , and we say F − that F (X) is concave (convex) if GF (G+ ) is a convex set. Now we say that F (X) is F + quasi-concave (quasi-convex) if every y-cross-section of G− (GF ) is a convex set. A F y-cross-section is formally deﬁned as

G− (¯) ≡ {(x1 , x2 )| F (x1 , x2 ) ≥ y } ⊂ R2 , ¯ F y

G+ (¯) ≡ {(x1 , x2 )| F (x1 , x2 ) ≤ y } ⊂ R2 . ¯ F y

Clearly, G− is the union of all G− (¯): G− = ∪y G− (¯). ¯ F y F F y F Formal deﬁnition: F (x1 , . . . , xn ) is quasi-concave if ∀X 0 , X 1 ∈ A and 0 ≤ θ ≤ 1, F (X θ ) ≥ min{F (X 0), F (X 1 )}. F (x1 , . . . , xn ) is strictly quasi-concave if ∀X 0 = X 1 ∈ A and 0 < θ < 1, F (X θ ) > min{F (X 0), F (X 1 )}. x2 x2
T T

E x1

d d d d d d

E x1

If F is (strictly) concave, then F must be (strictly) quasi-concave. Proof: If F (X θ ) < min{F (X 0), F (X 1 )}, then F (X θ ) < (1 − θ)F (X 0 ) + θF (X 1 ). Geometrically, if G− is convex, then G− (¯) must be convex for every y . ¯ F F y If F is quasi-concave, it is not necessarily that F is concave. Counterexample: F (x1 x2 ) = x1 x2 , x1 , x2 > 0 is strictly quasi-concave but not concave. In this case, every G− (¯) is convex but still G− is not convex. F y F 0 F1 , |B 2 | ≡ F1 F11 0 F1 F2 F1 F11 F12 , |B 3 | ≡ F2 F21 F22 0 F1 F2 F3 F1 F11 F12 F13 , F2 F21 F22 F23 F3 F31 F32 F33

Bordered Hessian: |B 1 | ≡ etc.

Theorem 1: Suppose that F is twice diﬀerentiable. If F is quasi-concave, then |B 2 | ≥ 0, |B 3 | ≤ 0, etc. Conversely, if |B 1 | < 0, |B 2 | > 0, |B 3 | < 0, etc., then F is strictly quasi-concave. |B 2 | dMRS Proof (n = 2): = 3 and therefore F is quasi-concave if and only if |B 2 | ≥ 0. dx1 F2

79 Consider the following maximization problem with a linear constraint. x1 ,...,xn

max F (x1 , . . . , xn ) subject to a1x1 + · · · + an xn = b.

Theorem 2: If F is (strictly) quasi-concave and X 0 satisﬁes FOC, then X 0 is a (unique) global maximum. Proof: By theorem 1, X 0 must be a local maximum. Suppose there exists X 1 satisfying the linear constraint with U(X 1 ) > U(X 0 ). Then U(X θ ) > U(X 0 ), a contradiction. Theorem 3: A monotonic increasing transformation of a quasi-concave function is a quasi-concave function. A quasi-concave function is a monotonic increasing transformation of a concave function. Proof: A monotonic increasing transformation does not change the sets {(x1 , x2 ) ∈ ¯ R2 U(x1 , x2 ) ≥ U}. To show the opposite, suppose that f (x1 , x2 ) is quasi-concave. Deﬁne a monotonic transformation as g(x1 , x2 ) = H(f (x1 , x2 )) where H −1(g) = f (x, x). 9.6 Elasticity of Substitution

Consider a production function Q = F (x1 , x2 ). w1 F1 x2 ¯ Cost minimization ⇒ = ≡ θ. Let ≡ r. On an isoquant F (x1 , x2 ) = Q, w2 F2 x1 θ dr . there is a relationship r = φ(θ). The elasticity of substitution is deﬁned as σ ≡ r dθ w1 If the input price ratio θ = increases by 1%, a competitive producer will increase w2 x2 its input ratio r = by σ%. x1 k 1 . Example: For a CES function Q = {axρ + bxρ } ρ , σ = 1 2 1−ρ 9.7 Problems a11 a12 a21 a22  0 c1 c2 , B =  c1 a11 a12 , and C = c2 a21 a22  −c2 c1 .

1. Let A =

(b) Calculate the product C ′ AC. Does |B| have any relationship with C ′ AC? 2. Consider the utility function U(x1 , x2 ) = xα xβ . 1 2 (a) Calculate the marginal utilities U1 = (b) Calculate the hessian matrix H = ∂U ∂U and U2 = . ∂x1 ∂x2 U11 U12 . U21 U22

(a) Calculate the determinant |B|.

80  0 U1 U2 (c) Calculate the determinant of the matrix B =  U1 U11 U12 . U2 U21 U22 (d) Let C = −U2 U1 . Calculate the product C’HC. 

3. Use the Lagrangian multiplier method to ﬁnd the critical points of f (x, y) = x + y subject to x2 + y 2 = 2. Then use the bordered Hessian to determine which point is a maximum and which is a minimum. (There are two critical points.) 4. Determine whether f (x, y) = xy is concave, convex, quasiconcave, or quasiconvex. (x > 0 and y > 0.) 5. Let U(x, y), U, x, y > 0, be a homogenous of degree 1 and concave function with Uxy = 0. (a) Show that V (x, y) = [U(x, y)]a is strictly concave if 0 < a < 1 and V (x, y) is strictly quasi-concave for all a > 0. Hint: Uxx Uyy = [Uxy ]2 . (b) Show that F (x, y) = xβ + y β , x, y > 0 and −∞ < β < 1 is homogenous of degree 1 and concave if a = 1. (c) Determine the range of a so that F (x, y) is strictly concave. (d) Determine the range of a so that F (x, y) is strictly quasi-concave. a/β 81 9.8 Nonlinear Programming

The general nonlinear programming problem is:   g 1 (x1 , . . . , xn ) ≤ b1    . . . max F (x1 , . . . , xn ) subject to x1 ,...,xn  g m (x1 , . . . , xn ) ≤ bm    x1 , . . . , xn ≥ 0.

In equality constraint problems, the number of constraints should be less than the number of policy variables, m < n. For nonlinear programming problems, there is no such a restriction, m can be greater than or equal to n. In vector notation, max F (x) x subject to

g(x) ≤ b,

x ≥ 0.

9.9

Kuhn-Tucker condition

Deﬁne the Lagrangian function as m L(x, y) = F (x) + y(b − g(x)) = F (x1 , . . . , xn ) +

j=1

yj (bj − g j (x1 , . . . , xn )).

Kuhn-Tucker condition: The FOC is given by the Kuhn-Tucker conditions: ∂L ∂F = − ∂xi ∂xi ∂g j yj ≤ 0, ∂xi j=1 yj m xi

∂L = 0 xi ≥ 0, ∂xi

i = 1, . . . , n

∂L = bj − g j (x) ≥ 0, ∂yj

∂L = 0 yj ≥ 0, j = 1, . . . , m ∂yj

Kuhn Tucker theorem: x∗ solves the nonlinear programming problem if (x∗ , y ∗) solves the saddle point problem: L(x, y ∗ ) ≤ L(x∗ , y ∗) ≤ L(x∗ , y) for all x ≥ 0, y ≥ 0,

Conversely, suppose that f (x) is a concave function and g j (x) are convex functions (concave programming) and the constraints satisfy the constraint qualiﬁcation condition that there is some point in the opportunity set which satisﬁes all the inequality constraints as strict inequalities, i.e., there exists a vector x0 ≥ 0 such that g j (x0 ) < bj , j = 1, . . . , m, then x∗ solves the nonlinear programming problem only if there is a y ∗ for which (x∗ , y ∗ ) solves the saddle point problem. If constraint qualiﬁcation is not satisﬁed, it is possible that a solution does not satisfy the K-T condition. If it is satisﬁed, then the K-T condition will be necessary. For the case of concave programming, it is also suﬃcient.

82 In economics applications, however, it is not convenient to use K-T condition to ﬁnd the solution. In stead, we ﬁrst solve the equality constraint version of the problem and then use K-T condition to check or modify the solution when some constraints are violated. The K-T condition for minimization problems: the inequalities reversed. 9.10 Examples

Example 1. (Joint product proﬁt maximization) The cost function of a competitive producer producing 2 joint products is c(x1 , x2 ) = x2 + x1 x2 + x2 . The proﬁt function 1 2 is given by π(p1 , p2 ) = p1 x1 + p2 x2 − (x2 + x1 x2 + x2 ). 1 2 x1 ≥0,x2 ≥0

max

f (x1 , x2 ) = p1 x1 + p2 x2 − (x2 + x1 x2 − x2 ) 1 2

K-T condition: f1 = p1 − 2x1 − x2 ≤ 0, f2 = p2 − x1 − 2x2 ≤ 0, xi fi = 0, i = 1, 2. Case 1. p1 /2 < p2 < 2p1 . x1 = (2p1 − p2 )/3, x2 = (2p2 − p1 )/3. Case 2. 2p1 < p2 . x1 = 0, x2 = p2 /2. Case 3. 2p2 < p1 . x1 = p1 /2, x2 = 0.

Example 2. The production function of a producer is given by q = (x1 +1)(x2 +1)−1. For q = 8, calculate the cost function c(w1 , w2 ). x1 ≥0,x2 ≥0

min

w1 x1 + w2 x2

subject to

− [(x1 + 1)(x2 + 1) − 1] ≥ −8

Lagrangian function: L = w1 x1 + w1 x2 + λ[(x1 + 1)(x2 + 1) − 9]. K-T conditions: L1 = w1 − λ(x2 − 1) ≥ 0, L2 = w2 − λ(x2 − 1) ≥ 0, xi Li = 0, i = 1, 2, and Lλ = (x1 + 1)(x2 + 1) − 9 ≥ 0, λLλ = 0. Case 1. w1 /9 < w2 < 9w1 . √ x1 = 9w2 /w1 − 1, x2 = 9w1 /w2 − 1 and c(w1 , w2 ) = 6 w1 w2 − w1 − w2 . Case 2. 9w1 < w2 . x1 = 8, x2 = 0, and c(w1 , w2 ) = 8w1 . Case 3. 9w2 < w1 . x1 = 0, x2 = 8, c(w1 , w2 ) = 8w2 .

83 Example 3. The utility function of a consumer is U(x1 , x2 ) = x1 (x2 +1). The market price is p1 = p2 = 1 and the consumer has $11. Therefore the budget constraint is x1 + x2 ≤ 11. Suppose that both products are under rationing. Besides the money price, the consumer has to pay ρi rationing points for each unit of product i consumed. Assume that ρ1 = 1 and ρ2 = 2 and the consumer has q rationing points. The rationing point constraint is x1 + 2x2 ≤ q. The utility maximization problem is given by max U(x1 , x2 ) = x1 (x2 + 1) x1 ,x2

subject to

x1 + x2 ≤ 11,

x1 + 2x2 ≤ q,

x1 , x2 ≥ 0.

Lagrangian function: L = x1 (x2 + 1) + λ1 (11 − x1 − x2 ) + λ2 (q − x1 − 2x2 ). K-T conditions: L1 = x2 + 1 − λ1 − λ2 ≤ 0, L2 = x1 − λ1 − 2λ2 ≤ 0, xi Li = 0, i = 1, 2, and Lλ1 = 11 − x1 − x2 ≥ 0, Lλ2 = q − x1 − 2x2 ≥ 0 λi Lλi = 0.

Case 1: q < 2. x1 = q, x2 = 0, λ1 = 0, and λ2 = 1. Case 2: 2 ≤ q ≤ 14. x1 = (q + 2)/2, x2 = (q − 2)/4, λ1 = 0, and λ2 = (q + 2)/4. Case 3: 14 < q ≤ 16. x1 = 22 − q, x2 = q − 11, λ1 = 3(q − 14), and λ2 = 2(16 − q). Case 4: 16 < q. x1 = 6, x2 = 5, λ1 = 6, and λ2 = 0.

9.11

Problems

√ 1. Given the individual utility function U(X, Y ) = 2 X + Y , a) show that U is quasi-concave for X ≥ 0 and Y ≥ 0, b) state the Kuhn-Tucker conditions of the following problem: √ max 2 X + Y
X≥0, Y ≥0

s. t. PX X + PY Y ≤ I,
Y c) derive the demand functions X(PX , PY , I) and Y (PX , PY , I) for the case I ≥ PX , check that the K-T conditions are satisﬁed, 2 PY d) and do the same for I < PX . e) Given that I = 1 and PY = 1, derive the ordinary demand function X = D(PX ). f) Are your answers in (c) and (d) global maximum? Unique global maximum? Why

P2

84 or why not? 2. A farm has a total amount of agricultural land of one acre. It can produce two crops, √ (C) and lettuce (L), according to the production functions C = NC and corn L = 2 NL respectively, where NC (NL ) is land used in corn (lettuce) production. The prices of corn and lettuce are p and q respectively. Thus, if the farm uses NC of land in corn production and NL in lettuce production, (NC ≥ 0, NL ≥ 0, and √ NC + NL ≤ 1) its total revenue is pNC + 2q NL . a) Suppose the farm is interested in maximizing its revenue. State the revenue maximization problem and the Kuhn-Tucker conditions. b) Given that q > p > 0, how much of each output will the farm produce? Check that the K-T conditions are satisﬁed. c) Given that p ≥ q > 0, do the same as (b). 3. Suppose that a ﬁrm has two activities producing two goods “meat” (M) and “egg” (E) from the same input “chicken” (C) according to the production functions 1/2 M = CM and E = CE , where CM (respectively CE ) ≥ 0 is the q. Suppose in the ¯ short run, the ﬁrm has C units of chicken that it must take as given and suppose that the ﬁrm faces prices p > 0, q > 0 of meat and egg respectively. 0.5 a) Show that the proﬁt function π = pCM + qCE is quasi-concave in (CM , CE ). b) Write down the short run proﬁt maximization problem. c) State the Kuhn-Tucker conditions. d) Derive the short run supply functions. (There are two cases.) e) Is your solution a global maximum? Explain. 4. State the Kuhn-Tucker conditions of the following nonlinear programming problem max s. t. U(X, Y ) = 3 ln X + ln Y 2X + Y ≤ 24 X + 2Y ≤ 24 X ≥ 0, Y ≥ 0.

Show that X = 9, Y = 6, λ1 = 1/6, and λ2 = 0 satisfy the Kuhn-Tucker conditions. What is the economic interpretations of λ1 = 1/6 and λ2 = 0 if the ﬁrst constraint is interpreted as the income constraint and the second constraint as the rationing point constraint of a utility maximization problem? 9.12 Linear Programming – A Graphic Approach

Example 1 (A Production Problem). A manufacturer produces tables x1 and desks x2 . Each table requires 2.5 hours for assembling (A), 3 hours for buﬃng (B), and 1 hour for crating (C). Each desks requires 1 hour for assembling (A), 3 hours for buﬃng (B), and 2 hours for crating (C). The ﬁrm can use no more than 20 hours for assembling, 30 hours for buﬃng, and 16 hours for crating each week. Its proﬁt margin is $3 per table and $4 per desk.

85

max Π subject to 2.5x1 + x2 3x1 + 3x2 x1 + 2x2 x1 , x2

= ≤ ≤ ≤ ≥

3x1 + 4x2 20 30 16 0.

(4) (5) (6) (7) (8)

extreme point: The intersection of two constraints. extreme point theorem: If an optimal feasible value of the objective function exists, it will be found at one of the extreme points. In the example, There are 10 extreme points, but only 5 are feasible: (0 ,0), (8, 0), (6 2 , 3 1 ), (4, 6), and (0, 8), called basic feasible solutions. At (4, 6), Π = 36 is the 3 3 optimal. x2 v T v v v v

y2

T

v d Zd v rZ v rd Z r Z dr v rq Z v dZ (4,6) rr dv rr Z v dZ rr vdZZ rr v d Z rr r v d Z

v v v v v

E x1

e e de e d e d ed HH e d e HH d e Hd H Hd e H (9,3) e d q HH e d HH e d H H e d H

Ey

Example 2 (The Diet Problem). A farmer wants to see that her herd gets the minimum daily requirement of three basic nutrients A, B, and C. Daily requirements are 14 for A, 12 for B, and 18 for C. Product y1 has 2 units of A and 1 unit each of B and C; product y2 has 1 unit each of A and B and 3 units of C. The cost of y1 is $2, and the cost of y2 is $4. min c subject to 2y1 + y2 y1 + y2 y1 + 3y2 y1 , y2 = ≥ ≥ ≥ ≥ 2y1 + 4y2 14 12 18 0. (9) (10) (11) (12) (13)

86 Slack and surplus variables: To ﬁnd basic solutions, equations are needed. This is done by incorporating a separate slack or surplus variable si into each inequality. In example 1, the system becomes 2.5x1 + x2 + s1 = 20 In matrix for,   2.5 1 1 0 0    3 3 0 1 0   1 2 0 0 1   x1 x2 s1 s2 s3     20   =  30  .   16 3x1 + 3x2 + s2 = 30 x1 + 2x2 + s3 = 16.

In example 2, the inequalities are ”≥” and the surplus variables are substracted: 2y1 + y2 − s1 = 14 In matrix for,   2 1 −1 0 0   1 1 0 −1 0    1 3 0 0 −1    y1 y2 s1 s2 s3     14   =  12  .   18 y1 + y2 − s2 = 12 y1 + 3y2 − s3 = 18.

For a system of m equations and n variables, where n > m, a solution in which at least n − m variables equal to zero is an extreme point. Thus by setting n − mand solving the m equations for the remaining m variables, an extreme point can be found. There are n!/m!(n − m)! such solutions. 9.13 Linear programming – The simplex algorithm

The algorithm moves from one basic feasible solution to another, always improving upon the previous solutions, until the optimal solution is reached. In each step, those variables set equal to zero are called not in the basis and those not set equal to zero are called in the basis. Let use example one to illustrate the procedure. 1. The initial Simplex Tableau x1 x2 2.5 1 3 3 1 2 -3 -4 s1 1 0 0 0 s2 0 1 0 0 s3 0 0 1 0 Constant 20 30 16 0

The ﬁrst basic feasible solution can be read from the tableau as x1 = 0, x2 = 0, s1 = 20, s2 = 30, and s3 = 16. The value of Π is zero. 2. The Pivot Element and a change of Basis

87 (a) The negative indicator with the largest absolute value determines the variable to enter the basis. Here it is x2 . The x2 column is called the pivot column (j-th column). (b) The variable to be eliminated is determined by the smallest displacement ratio. Displacement ratios are found by dividing the elements of the constant column by the elements of the pivot column. Here the smallest is 16/2=8 and row 3 is the pivot row (i-th row). The pivot element is 2. 3. (Pivoting) and we are going to move to the new basic solution with s3 = 0 and x2 > 0. First, divides every element of the pivoting row by the pivoting element (2 in this example) to make the pivoting element equal to 1. Then subtracts akj times the pivoting row from k-th row to make the j-th column a unit vector. (This procedure is called the Gaussian elimination method, usually used in solving simultaneous equations.) After pivoting, the Tableau becomes x1 x2 s1 s2 s3 Constant 2 0 1 0 -.5 12 1.5 0 0 1 -1.5 6 .5 1 0 0 .5 8 -1 0 0 0 2 32 The basic solution is x2 = 8, s1 = 12, s2 = 6, and x1 = s3 = 0. The value of Π is 32. 4. (Optimization) Repeat steps 2-3 until a maximum is reached. In the example, x1 column is the new pivoting column. The second row is the pivoting row and 1.5 is the pivoting element. After pivoting, the tableau becomes x1 x2 s1 s2 s3 Constant 0 0 1 − 4 1.5 4 3 2 1 0 0 -1 4 3 1 0 1 0 −3 1 6 2 0 0 0 1 36 3 The basic solution is x1 = 4, x2 = 6, s1 = 4, and s2 = s3 = 0. The value of Π is 36. Since there is no more negative indicators, the process stops and the basic solution is the optimal. s1 = 4 > 0 and s2 = s3 = 0 means that the ﬁrst constraint is not binding but the second and third are binding. The indicators for s2 and s3 , t2 = 2 and t3 = 1 are called the shadow values, representing the 3 marginal contributions of increasing one hour for buﬃng or crating. Because y1 = y2 = 0 is not feasible, the simplex algorithm for minimization problem is more complex. Usually, we solve its dual problem.

88 9.14 Linear programming – The dual problem

To every linear programming problem there corresponds a dual problem. If the original problem, called the primal problem, is max F = cx x subject to

Ax ≤ b,

x≥0

then the dual problem is min G = yb y subject to

yA ≥ c,

y≥0

where 

 a11 . . . a1n  . . , .. .  A= . . . . am1 . . . amn

Existence theorem: A necessary and suﬃcient condition for the existence of a solution is that the opportunity sets of both the primal and dual problems are nonempty. Proof: Suppose x, y are feasible. Then Ax ≤ b, yA ≥ c. It follows that F (x) = cx ≤ yAx and G(y) = yb ≥ yAx. Therefore, F (x) ≤ G(y). Duality theorem: A necessary and suﬃcient condition for a feasible vector to represent a solution is that there exists a feasible vector for the dual problem for which the values of the objective functions of both problems are equal. Complementary slackness theorem: A necessary and suﬃcient condition for feasible vectors x∗ , y ∗ to solve the dual problems is that they satisfy the complementary slackness condition: (c − y ∗ A)x∗ = 0 Proof: Use Kuhn-Tucker theorem. Dual of the Diet Problem max c∗ subject to 2x1 + x2 + x1 x1 + x2 + 3x3 x1 , x2 , x3 = ≤ ≤ ≥ 14x1 + 12x2 + 18x3 2 4 0. (14) (15) (16) (17) y ∗(b − Ax∗ ) = 0.

 x1  .  x =  . , . xn



 b1  .  b =  . , . bm



c = (c1 , . . . , cn ),

y = (y1 , . . . , ym ).

x1 , x2 , x3 , is interpreted as the imputed value of nutrient A, B, C, respectively.

89

10
10.1

General Equilibrium and Game Theory
Utility maximization and demand function

A consumer wants to maximize his utility function subject to his budget constraint: max U(x1 , . . . , xn ) subj. to p1 x1 + · · · + pn xn = I.

Endogenous variables: x1 , . . . , xn Exogenous variables: p1 , . . . , pn , I (the consumer is a price taker) Solution is the demand functions xk = Dk (p1 , . . . , pn , I), k = 1, . . . , n Example: max U(x1 , x2 ) = a ln x1 + b ln x2 subject to p1 x1 + p2 x2 = m. L = a ln x1 + b ln x2 + λ(m − p1 x1 − p2 x2 ). a b FOC: L1 = − λp1 = 0, L2 = − λp2 = 0 and Lλ = m − p1 x1 − p2 x2 = 0. x1 x2 p1 am bm a x2 = ⇒ x1 = , x2 = ⇒ b x1 p2 (a + b)p1 (a + b)p2 0 −p1 −p2 −a ap2 bp2 0 −p1 SOC: = 22 + 21 > 0. x2 1 x1 x2 −b −p2 0 x2 2 am bm ⇒ x1 = , x2 = is a local maximum. (a + b)p1 (a + b)p2 10.2 Proﬁt maximization and supply function

A producer’s production technology can be represented by a production function q = f (x1 , . . . , xn ). Given the prices, the producer maximizes his proﬁts: max Π(x1 , . . . , xn ; p, p1 , . . . , pn ) = pf (x1 , . . . , xn ) − p1 x1 − · · · − pn xn Exogenous variables: p, p1 , . . . , pn (the producer is a price taker) Solution is the supply function q = S(p, p1 , . . . , pn ) and the input demand functions, xk = Xk (p, p1 , . . . , pn ) k = 1, . . . , n √ √ √ √ Example: q = f (x1 , x2 ) = 2 x1 + 2 x2 and Π(x1 , x2 ; p, p1 , p2 ) = p(2 x1 + 2 x2 ) − p1 x1 − p2 x2 , √ √ max p(2 x1 + 2 x2 ) − p1 x1 − p2 x2 x1 .x2

FOC:

∂Π p ∂Π p = √ − p1 = 0 and = √ − p2 = 0. ∂x1 x1 ∂x2 x2 ⇒ x1 = (p/p1 )2 , x2 = (p/p2 )2 (input demand functions) and q = 2(p/p1 ) + 2(p/p2 ) = 2p( p11 + p12 ) (the supply function)

90 Π = p2 ( p11 + SOC:
1 ) p2

is negative deﬁnite. 10.3

∂2Π  ∂x2 1   ∂2Π ∂x1 ∂x2 

  −p ∂2Π ∂x1 ∂x2   2x3/2 = 1 ∂2Π   0 ∂x2 1

0 2x2
3/2

  −p 



Transformation function and proﬁt maximization

In more general cases, the technology of a producer is represented by a transformation j j j j j function: F j (y1 , . . . , yn ) = 0, where (y1 , . . . , yn ) is called a production plan, if yk > 0 j (yk ) then k is an output (input) of j. Example: a producer produces two outputs, y1 and y2 , using one input y3 . Its technology is given by the transformation function (y1 )2 + (y2 )2 + y3 = 0. Its proﬁt is Π = p1 y1 + p2 y2 + p3 y3 . The maximization problem is y1 ,y2 ,y3

max p1 y1 + p2 y2 + p3 y3

subject to (y1 )2 + (y2 )2 + y3 = 0.

To solve the maximization problem, we can eliminate y3 : x = −y3 = (y1 )2 + (y2 )2 > 0 and max p1 y1 + p2 y2 − p3 [(y1 )2 + (y2 )2 ]. y1 ,y2

The solution is: y1 = p1 /(2p3 ), y2 = p2 /(2p3 ) (the supply functions of y1 and y2 ), and x = −y3 = [p1 /(2p3 )]2 + [p2 /(2p3 )]2 (the input demand function for y3 ). 10.4 The concept of an abstract economy and a competitive equilibrium

Commodity space: Assume that there are n commodities. The commodity space is n R+ = {(x1 , . . . , xn ); xk ≥ 0} Economy: There are I consumers, J producers, with initial endowments of commodities ω = (ω1 , . . . , ωn ). i Consumer i has a utility function U i (x1 , . . . , xi ), i = 1, . . . , I. n j j Producer j has a production transformation function F j (y1 , . . . , yn ) = 0, A price system: (p1 , . . . , pn ). A private ownership economy: Endowments and ﬁrms (producers) are owned by consumers. i i Consumer i’s endowment is ω i = (ω1 , . . . , ωn ), I ω i = ω. i=1 I Consumer i’s share of ﬁrm j is θij ≥ 0, i=1 θij = 1. j j An allocation: xi = (xi , . . . , xi ), i = 1, . . . , I, and y j = (y1 , . . . , yn ), j = 1, . . . , J. 1 n

91

A competitive equilibrium: A combination of a price system p = (¯1 , . . . , pn ) and an allocation ({¯i }i=1,...,I , {¯j }j=1,...,J ) ¯ p ¯ x y such that 1. i xi = ω + j y j (feasibility condition). ¯ ¯ j j 2. y maximizes Π , j = 1, . . . , J and xi maximizes U i , subject to i’s budget con¯ ¯ i i 1 i straint p1 x1 + . . . + pn xn = p1 ω1 + . . . + pn ωn + θi1 Π1 + . . . + θiJ ΠJ . Existence Theorem: Suppose that the utility functions are all quasi-concave and the production transformation functions satisfy some theoretic conditions, then a competitive equilibrium exists. Welfare Theorems: A competitive equilibrium is eﬃcient and an eﬃcient allocation can be achieved as a competitive equilibrium through certain income transfers. Constant returns to scale economies and non-substitution theorem: Suppose there is only one nonproduced input, this input is indispensable to production, there is no joint production, and the production functions exhibits constant returns to scale. Then the competitive equilibrium price system is determined by the production side only.

92 10.5 Multi-person Decision Problem and Game Theory

In this chapter, we consider the situation when there are n > 1 persons with diﬀerent objective (utility) functions; that is, diﬀerent persons have diﬀerent preferences over possible outcomes. There are two cases: 1. Game theory: The outcome depends on the behavior of all the persons involved. Each person has some control over the outcome; that is, each person controls certain strategic variables. Each one’s utility depends on the decisions of all persons. We want to study how persons make decisions. 2. Public Choice: Persons have to make decision collectively, eg., by voting. We consider only game theory here. Game theory: the study of conﬂict and cooperation between persons with diﬀerent objective functions. Example (a 3-person game): The accuracy of shooting of A, B, C is 1/3, 2/3, 1, respectively. Each person wants to kill the other two to become the only survivor. They shoot in turn starting A. Question: What is the best strategy for A? 10.6 Ingredients and classiﬁcations of games

A game is a collection of rules known to all players which determine what players may do and the outcomes and payoﬀs resulting from their choices. The ingredients of a game: 1. Players: Persons having some inﬂuences upon possible income (decision makers). 2. Moves: decision points in the game at which players must make choices between alternatives (personal moves) and randomization points (called nature’s moves). 3. A play: A complete record of the choices made at moves by the players and realizations of randomization. 4. Outcomes and payoﬀs: a play results in an outcome, which in turn determines the rewords to players. Classiﬁcations of games: 1. according to number of players: 2-person games – conﬂict and cooperation possibilities. n-person games – coalition formation possibilities in addition. inﬁnite-players’ games – corresponding to perfect competition in economics. 2. according to number of strategies: ﬁnite – strategy (matrix) games, each person has a ﬁnite number of strategies,

93 payoﬀ functions can be represented by matrices. inﬁnite – strategy (continuous or discontinuous payoﬀ functions) games like duopoly games. 3. according to sum of payoﬀs: 0-sum games – conﬂict is unavoidable. non-zero sum games – possibilities for cooperation. 4. according to preplay negotiation possibility: non-cooperative games – each person makes unilateral decisions. cooperative games – players form coalitions and decide the redistribution of aggregate payoﬀs. 10.7 The extensive form and normal form of a game

Extensive form: The rules of a game can be represented by a game tree. The ingredients of a game tree are: 1. Players 2. Nodes: they are players’ decision points (personal moves) and randomization points (nature’s moves). 3. Information sets of player i: each player’s decision points are partitioned into information sets. An information set consists of decision points that player i can not distinguish when making decisions. 4. Arcs (choices): Every point in an information set should have the same number of choices. 5. Randomization probabilities (of arcs following each randomization points). 6. Outcomes (end points) 7. Payoﬀs: The gains to players assigned to each outcome. A pure strategy of player i: An instruction that assigns a choice for each information set of player i. Total number of pure strategies of player i: the product of the numbers of choices of all information sets of player i. Once we identify the pure strategy set of each player, we can represent the game in normal form (also called strategic form). 1. Strategy sets for each player: S1 = {s1 , . . . , sm }, S2 = {σ1 , . . . , σn }. 2. Payoﬀ matrices: π1 (si , σj ) = aij , π2 (si , σj ) = bij . A = [aij ], B = [bij ]. Normal form: d II d d

I s1 . . .

σ1 (a11 , b11 ) . . . (am1 , bm1 )

... ... .. .

σn (a1n , b1n ) . . .

sm

. . . (amn , bmn )

94 10.8 Examples

Example 1: A perfect information game 1 L R 2 2 d r l L d R d d

1 9

9 6

3 7

8 2

d

II d Example 2: Prisoners’ dilemma game 1 R L ¨ 2 © L dd L dd R R 4 4 0 5 5 0 1 1

S1 = { L, R }, S2 = { Ll, Lr, Rl, Rr }.

I d Ll Lr Rl Rr L (1,9) (9,6) (1,9) (9,6) R (3,7)* (8,2) (3,7) (8,2)

d

II d d

S1 = { L, R }, S2 = { L, R }. Example 3: Hijack game 1 R L 2 L d R d −1 2 2 −10 −2 −10 S1 = { L, R }, S2 = { L, R }.
1/2 ¨rr1/2 ¨ rr 1 1¨¨ e R ¡d r L l e ¡ ¨d 2 ¡e ¡e © d L¡ e R L¡ e R 4

I L R

L R (4,4) (0,5) (5,0) (1,1)*

d

II d d

I L R

L R (-1,2) (-1,2)* (2,-2)* (-10,-10)

Example 4: A simpliﬁed stock price manipulation game 0 d II d I d

2

7 5

5 7

4 5

4 2

3 7

S1 = { Ll, Lr, Rl, Rr }, S2 = { L, R }.

Ll Lr Rl Rr

L R (4, 3.5) (4, 2) (3.5, 4.5) (3.5, 4.5) (5.5, 5)* (4.5, 4.5) (5,6) (4,7)

Remark: Each extensive form game corresponds a normal form game. However, diﬀerent extensive form games may have the same normal form.

95 10.9 Strategy pair and pure strategy Nash equilibrium

1. A Strategy Pair: (si , σj ). Given a strategy pair, there corresponds a payoﬀ pair (aij , bij ). 2. A Nash equilibrium: A strategy pair (si∗ , σj∗) such that ai∗j∗ ≥ aij∗ and bi∗j∗ ≥ bi∗j for all (i, j). Therefore, there is no incentives for each player to deviate from the equilibrium strategy. ai∗j∗ and bi∗j∗ are called the equilibrium payoﬀ. The equilibrium payoﬀs of the examples are marked each with a star in the normal form. Remark 1: It is possible that a game does no have a pure strategy Nash equilibrium. Also, a game can have more than one Nash equilibria. Remark 2: Notice that the concept of a Nash equilibrium is deﬁned for a normal form game. For a game in extensive form (a game tree), we have to ﬁnd the normal form before we can ﬁnd the Nash equilibria. 10.10 Subgames and subgame perfect Nash equilibria

1. Subgame: A subgame in a game tree is a part of the tree consisting of all the nodes and arcs following a node that form a game by itself. 2. Within an extensive form game, we can identify some subgames. 3. Also, each pure strategy of a player induces a pure strategy for every subgame. 4. Subgame perfect Nash equilibrium: A Nash equilibrium is called subgame perfect if it induces a Nash equilibrium strategy pair for every subgame. 5. Backward induction: To ﬁnd a subgame perfect equilibrium, usually we work backward. We ﬁnd Nash equilibria for lowest level (smallest) subgames and replace the subgames by its Nash equilibrium payoﬀs. In this way, the size of the game is reduced step by step until we end up with the equilibrium payoﬀs. All the equilibria, except the equilibrium strategy pair (L,R) in the hijack game, are subgame perfect. Remark: The concept of a subgame perfect Nash equilibrium is deﬁned only for an extensive form game. 10.10.1 Perfect information game and Zemelo’s Theorem

An extensive form game is called perfect information if every information set consists only one node. Every perfect information game has a pure strategy subgame perfect Nash Equilibrium.

96 10.10.2 Perfect recall game and Kuhn’s Theorem

A local strategy at an information set u ∈ Ui : A probability distribution over the choice set at Uij . A behavior strategy: A function which assigns a local strategy for each u ∈ Ui . The set of behavior strategies is a subset of the set of mixed strategies. Kuhn’s Theorem: In every extensive game with perfect recall, a strategically equivalent behavior strategy can be found for every mixed strategy. However, in a non-perfect recall game, a mixed strategy may do better than behavior strategies because in a behavior strategy the local strategies are independent whereas they can be correlated in a mixed strategy.
¨rr A 2-person 0-sum non-perfect recall game. 1/2 ¨ rr 1/2 ¨ ¨ 1 1 1 1 rr u¨¨ 2 11 NE is (µ∗ , µ∗ ) = ( ac ⊕ bd, A ⊕ B). 1 2 2 2 2 2 a d b A dd B µ∗ is not a behavioral strategy. d 1 ¨ d d u12 © d d 1 −1 d d c d c d d d −1 1 ¡e ¡e ¡e 0 ¡ e −2 0¡ e ¡ e2 e ¡ e ¡ e ¡ −2 0 2 0

0

10.10.3

Reduction of a game

Redundant strategy: A pure strategy is redundant if it is strategically identical to another strategy. Reduced normal form: The normal form without redundant strategies. Equivalent normal form: Two normal forms are equivalent if they have the same reduced normal form. Equivalent extensive form: Two extensive forms are equivalent if their normal forms are equivalent.

10.11

Continuous games and the duopoly game

In many applications, S1 and S2 are inﬁnite subsets of Rm and Rn Player 1 controls m variables and player 2 controls n variables (however, each player has inﬁnite many strtategies). The normal form of a game is represented by two functions Π1 = Π1 (x; y) and Π2 = Π2 (x; y), where x ∈ S1 ⊂ Rm and y ∈ S2 ⊂ Rn . To simplify the presentation, assume that m = n = 1. A strategic pair is (x, y) ∈ S1 × S2 . A Nash equilibrium is a pair (x∗ , y ∗) such that Π1 (x∗ , y ∗ ) ≥ Π1 (x, y ∗ ) and Π2 (x∗ , y ∗) ≥ Π2 (x∗ , y) for all x ∈ S1 y ∈ S2 .

97 Consider the case when Πi are continuously diﬀerentiable and Π1 is strictly concave in x and Π2 strictly concave in y (so that we do not have to worry about the SOC’s). Reaction functions and Nash equilibrium: To player 1, x is his endogenous variable and y is his exogenous variable. For each y chosen by player 2, player 1 will choose a x ∈ S1 to maximize his objective function Π1 . This relationship deﬁnes a behavioral equation x = R1 (y) which can be obtained by solving the FOC for player 1, Π1 (x; y) = 0. Similarly, player 2 regards y as enx dogenous and x exogenous and wants to maximize Π2 for a given x chosen by player 1. Player 2’s reaction function (behavioral equation) y = R2 (x) is obtained by solving 2 Πy (x; y) = 0. A Nash equilibrium is an intersection of the two reaction functions. The FOC for a Nash equilibrium is given by Π1 (x∗ ; y ∗ ) = 0 and Π2 (x∗ ; y ∗) = 0. x y Duopoly game: There are two sellers (ﬁrm 1 and ﬁrm 2) of a product. The (inverse) market demand function is P = a − Q. The marginal production costs are c1 and c2 , respectively. Assume that each ﬁrm regards the other ﬁrm’s output as given (not aﬀected by his output quantity). The situation deﬁnes a 2-person game as follows: Each ﬁrm i controls his own output quantity qi . (q1 , q2 ) together determine the market price P = a − (q1 + q2 ) which in turn determines the proﬁt of each ﬁrm: Π1 (q1 , q2 ) = (P −c1 )q1 = (a−c1 −q1 −q2 )q1 and Π2 (q1 , q2 ) = (P −c2 )q2 = (a−c2 −q1 −q2 )q2 The FOC are ∂Π1 /∂q1 = a − c1 − q2 − 2q1 = 0 and ∂Π2 /∂q2 = a − c2 − q1 − 2q2 = 0. The reaction functions are q1 = 0.5(a − c1 − q2 ) and q2 = 0.5(a − c2 − q1 ). ∗ ∗ The Cournot Nash equilibrium is (q1 , q2 ) = ((a − 2c1 + c2 )/3, (a − 2c2 + c1 )/3) with P ∗ = (a + c1 + c2 )/3. (We have to assume that a − 2c1 + c2 , a − 2c2 + c1 ≥ 0.) 10.11.1 A simple bargaining model

Two players, John and Paul, have $ 1 to divide between them. They agree to spend at most two days negotiating over the division. The ﬁrst day, John will make an oﬀer, Paul either accepts or comes back with a counteroﬀer the second day. If they cannot reach an agreement in two days, both players get zero. John (Paul) discounts payoﬀs in the future at a rate of α (β) per day. A subgame perfect equilibrium of this bargaining game can be derived using backward induction. 1. On the second day, John would accept any non-negative counteroﬀer made by Paul. Therefore, Paul would make proposal of getting the whole $ 1 himself and John would get $ 0. 2. On the ﬁrst day, John should make an oﬀer such that Paul gets an amount equivalent to getting $ 1 the second day, otherwise Paul will reject the oﬀer. Therefore, John will propose of 1 − β for himself and β for Paul and Paul will accept the oﬀer.

98

An example of a subgame non-perfect Nash equilibrium is that John proposes of getting 1-0.5β for himself and 0.5β for Paul and refuses to accept any counteroﬀer made by Paul. In this equilibrium, Paul is threatened by John’s incredible threat and accepts only one half of what he should have had in a perfect equilibrium. 10.12 2-person 0-sum game

1. B = −A so that aij + bij = 0. 2. Maxmin strategy: If player 1 plays si , then the minimum he will have is minj aij , called the security level of strategy si . A possible guideline for player 1 is to choose a strategy such that the security level is maximized: Player 1 chooses si∗ so that minj ai∗j ≥ minj aij for all i. Similarly, since bij = −aij , Player 2 chooses σj ∗ so that maxi aij∗ ≤ maxi aij for all j. 3. Saddle point: If ai∗j∗ = maxi minj aij = minj maxi aij , then (si∗ , σj∗ ) is called a saddle point. If a saddle point exists, then it is a Nash equilibrium. A1 = 2 1 4 −1 0 6 A2 = 1 0 0 1

In example A1 , maxi minj aij = minj maxi aij = 1 (s1 , σ2 ) is a saddle point and hence a Nash equilibrium. In A2 , maxi minj aij = 0 = minj maxi aij = 1 and no saddle point exists. If there is no saddle points, then there is no pure strategy equilibrium. 4. Mixed strategy for player i: A probability distribution over Si . p = (p1 , . . . , pm ), q = (q1 , . . . , qn )′ . (p, q) is a mixed strategy pair. Given (p, q), the expected payoﬀ of player 1 is pAq. A mixed strategy Nash equilibrium (p∗ , q ∗ ) is such that p∗ Aq ∗ ≥ pAq ∗ and p∗ Aq ∗ ≤ p∗ Aq for all p and all q. 5. Security level of a mixed strategy: Given player 1’s strategy p, there is a pure strategy of player 2 so that the expected payoﬀ to player 1 is minimized, just as in the case of a pure strategy of player 1. t(p) ≡ min{ j pi ai1 , . . . , i i

pi ain }.

The problem of ﬁnding the maxmin mixed strategy (to ﬁnd p∗ to maximize t(p)) can be stated as max t subj. to p i

pi ai1 ≥ t, . . . ,

i

pi ain ≥ t,

pi = 1. i 6. Linear programming problem: The above problem can be transformed into a linear programming problem as follows: (a) Add a positive constant to each element of A to insure that t(p) > 0 for all p. (b) Deﬁne yi ≡ pi /t(p) and

99 replace the problem of max t(p) with the problem of min 1/t(p) = constraints become i yi ai1 ≥ 1, . . . , i yi ain ≥ 1. y1 ,...,ym ≥0

i

yi . The

min

y1 + . . . + ym subj. to i yi ai1 ≥ 1, . . . ,

i

yi ain ≥ 1

7. Duality: It turns out that player 2’s minmax problem can be transformed similarly and becomes the dual of player 1’s linear programming problem. The existence of a mixed strategy Nash equilibrium is then proved by using the duality theorem in linear programming. 1 0 . 0 1 To ﬁnd player 2’s equilibrium mixed strategy, we solve the linear programming problem: x1 ,x2 ≥0

Example (tossing coin game): A =

max x1 + x2

subj. to x1 ≤ 1 x2 ≤ 1.

The solution is x1 = x2 = 1 and therefore the equilibrium strategy for player 2 is ∗ ∗ q1 = q2 = 0.5. x2 y2
T T d d d dr d d d d d d d dr d d d d

1

1
E x1

1

1

E y1

Player 1’s equilibrium mixed strategy is obtained by solving the dual to the linear programming problem: y1 ,y2 ≥0

min y1 + y2

subj. to y1 ≥ 1 y2 ≥ 1.

The solution is p∗ = p∗ = 0.5. 1 2

10.13

Mixed strategy equilibria for non-zero sum games

The idea of a mixed strategy equilibrium is also applicable to a non-zero sum game. Similar to the simplex algorism for the 0-sum games, there is a Lemke algorism. Example (Game of Chicken)

100 1 N S ¨ 2 © S dd S dd N N 0 0 −3 3 3 −3 −9 −9

d II d I d

Swerve Don’t

Swerve (0,0) (3,-3)*

Don’t (-3,3)* (-9,-9)

S1 = { S, N }, S2 = { S, N }. There are two pure strategy NE: (S, N) and (N, S). There is also a mixed strategy NE. Suppose player 2 plays a mixed strategy (q, 1 − q). If player 1 plays S, his expected payoﬀ is Π1 (S) = 0q + (−3)(1 − q). If he plays N, his expected payoﬀ is Π1 (N) = 3q + (−9)(1 − q). For a mixed strategy NE, Π1 (S) = Π1 (N), therefore, q = 2 . 3 ∗ ∗ The mixed strategy is symmetrical: (p∗ , p∗ ) = (q1 , q2 ) = ( 2 , 1 ). 1 2 3 3 Example (Battle of sex Game) 1 O B ¨ 2 © B dd B dd O O 5 4 0 0 0 0 4 5

d II d I d

Ball game Opera Ball game (5,4)* (0,0) Opera (0,0) (4,5)*

S1 = { B, O }, S2 = { B, O }. There are two pure strategy NE: (B, B) and (O, O). There is also a mixed strategy NE. Suppose player 2 plays a mixed strategy (q, 1 − q). If player 1 plays B, his expected payoﬀ is Π1 (B) = 5q + (0)(1 − q). If he plays O, his expected payoﬀ is Π1 (O) = 0q +(4)(1−q). For a mixed strategy NE, Π1 (B) = Π1 (O), 4 therefore, q = 9 . 4 ∗ ∗ The mixed strategy is: (p∗ , p∗ ) = ( 5 , 9 ) and (q1 , q2 ) = ( 4 , 5 ). 1 2 9 9 9 10.14 Cooperative Game and Characteristic form

2-person 0-sum games are strictly competitive. If player 1 gains $ 1, player 2 will loss $ 1 and therefore no cooperation is possible. For other games, usually some cooperation is possible. The concept of a Nash equilibrium is deﬁned for the situation when no explicit cooperation is allowed. In general, a Nash equilibrium is not eﬃcient (not Pareto optimal). When binding agreements on strategies chosen can be contracted before the play of the game and transfers of payoﬀs among players after a play of the game is possible, players will negotiate to coordinate their strategies and redistribute the payoﬀs to achieve better results. In such a situation, the determination of strategies is not the key issue. The problem becomes the formation of coalitions and the distribution of payoﬀs.

101

Characteristic form of a game: The player set: N = {1, 2, . . . , n}. A coalition is a subset of N: S ⊂ N. A characteristic function v speciﬁes the maximum total payoﬀ of each coalition. Consider the case of a 3-person game. There are 8 subsets of N = {1, 2, 3}, namely, φ, (1), (2), (3), (12), (13), (23), (123). Therefore, a characteristic form game is determined by 8 values v(φ), v(1), v(2), v(3), v(12), v(13), v(2 Super-additivity: If A ∩ B = φ, then v(A ∪ B) ≥ v(A) + v(B). An imputation is a payoﬀ distribution (x1 , x2 , x3 ). Individual rationality: xi ≥ v(i). Group rationality: i∈S xi ≥ v(S). Core C: the set of imputations that satisfy individual rationality and group rationality for all S. Marginal contribution of player i in a coalition S ∪ i: v(S ∪ i) − v(S) Shapley value of player i is an weighted average of all marginal contributions |S|!(n − |S| − 1)! [v(S ∪ i) − v(S)]. n!

πi =
S⊂N

Example: v(φ) = v(1) = v(2) = v(3) = 0, v(12) = v(13) = v(23) = 0.5, v(123) = 1. C = {(x1 , x2 , x3 ), xi ≥ 0, xi + xj ≥ 0.5, x1 + x2 + x3 = 1}. Both (0.3, 0.3, 0.4) and (0.2, 0.4, 0.4) are in C. The Shapley values are (π1 , π2 , π3 ) = ( 1 , 1 , 1 ). 3 3 3 Remark 1: The core of a game can be empty. However, the Shapley values are uniquely determined. Remark 2: Another related concept is the von-Neumann Morgenstern solution. See CH 6 of Intriligator’s Mathematical Optimization and Economic Theory for the motivations of these concepts.

10.15

The Nash bargaining solution for a nontransferable 2-person cooperative game

In a nontransferable cooperative game, after-play redistributions of payoﬀs are impossible and therefore the concepts of core and Shapley values are not suitable. For the case of 2-person games, the concept of Nash bargaining solutions are useful. Let F ⊂ R2 be the feasible set of payoﬀs if the two players can reach an agreement and Ti the payoﬀ of player i if the negotiation breaks down. Ti is called the threat point of player i. The Nash bargaining solution (x∗ , x∗ ) is deﬁned to be the solution 1 2 to the following problem:

102 x2

T
(x1 ,x2 )∈F

max (x1 − T1 )(x2 − T2 )

x∗ 2 T2 T1 x∗ 1
E x1

See CH 6 of Intriligator’s book for the motivations of the solution concept. 10.16 Problems two-person 0-sum game: σ1 σ2 σ3 4 3 -2 3 4 10 7 6 8

1. Consider the following I \ II s1 s2 s3

(a) Find the max min strategy of player I smax min and the min max strategy of player II σmin max . (b) Is the strategy pair (smax min , σmin max ) a Nash equilibrium of the game? (c) What are the equilibrium payoﬀs? 2. Find the maxmin strategy (smax min ) and the minmax strategy (σmin max ) of the following two-person 0-sum game: I \ II σ1 σ2 s1 -3 6 s2 8 -2 s3 6 3 Is the strategy pair (smax min , σmin max ) a Nash equilibrium? If not, use simplex method to ﬁnd the mixed strategy Nash equilibrium. 3. Find the (mixed strategy) Nash Equilibrium of the following two-person game: I \ II H T H (-2, 2) (2, -1) T (2, -2) (-1,2) 4. Suppose that two ﬁrms producing a homogenous product face a linear demand curve P = a−bQ = a−b(q1 +q2 ) and that both have the same constant marginal costs c. For a given quantity pair (q1 , q2 ), the proﬁts are Πi = qi (P − c) = qi (a − bq1 − bq2 − c), i = 1, 2. Find the Cournot Nash equilibrium output of each ﬁrm. 5. Suppose that in a two-person cooperative game without side payments, if the two players reach an agreement, they can get (Π1 , Π2 ) such that Π2 + Π2 = 47 1

103 and if no agreement is reached, player 1 will get T1 = 3 and player 2 will get T2 = 2. (a) Find the Nash solution of the game. (b) Do the same for the case when side payments are possible. Also answer how the side payments should be done? 6. A singer (player 1), a pianist (player 2), and a drummer (player 3) are oﬀered $ 1,000 to play together by a night club owner. The owner would alternatively pay $ 800 the singer-piano duo, $ 650 the piano drums duo, and $ 300 the piano alone. The night club is not interested in any other combination. Howeover, the singer-drums duo makes $ 500 and the singer alone gets $ 200 a night in a restaurant. The drums alone can make no proﬁt. (a) Write down the characteristic form of the cooperative game with side payments. (b) Find the Shapley values of the game. (c) Characterize the core.

Mathematical Economics

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