lecnotes econ analiyses

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ECON 381 SC Foundations Of Economic Analysis

2003

John HillasUniversity of Auckland


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Contents

Chapter 1. Logic, Sets, Functions, and Spaces 11. Introduction 12. Logic 13. Some Set Theory 14. Functions 2

5. Spaces 26. Metric Spaces and Continuous Functions 27. Open sets, Compact Sets, and the Weierstrauss Theorem 28. Linear Spaces and Convex Sets 2

Chapter 2. Linear Algebra 31. The Space R n 32. Linear Functions from R n to R m 53. Matrices and Matrix Algebra 64. Matrices as Representations of Linear Functions 85. Linear Functions from R n to R n and Square Matrices 106. Inverse Functions and Inverse Matrices 117. Changes of Basis 118. The Trace and the Determinant 149. Calculating and Using Determinants 1510. Eigenvalues and Eigenvectors 20

Chapter 3. Choice, Preference, and Utility 231. Formal Model of Motivated Choice 232. The Properties of Choice Structures 243. Preference Relations and Preference Based Choice 244. The Relationship Between the Choice Based Approach and the Preference

Based Approach 255. Representing Preferences by Utility Functions 266. Consumer Behaviour: Optimisation Subject to a Budget Constraint 27

Chapter 4. Consumer Behaviour: Optimisation Subject to the Budget Constraint 291. Constrained Maximisation 292. The Theorem of the Maximum 333. The Envelope Theorem 354. Applications to Microeconomic Theory 38

Chapter 5. Decision Making Under Uncertainty 431. Revision of Probability Theory 432. Preferences Over Lotteries 463. Expected Utility Functions 47

Chapter 6. Multiple Agent Models I: Introduction to Noncooperative Game Theory 511. Normal Form Games 51

2. Extensive Form Games 55i


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ii CONTENTS

3. Existence of Equilibrium 59

Chapter 7. Multiple Agent Models II: Introduction to General Equilibrium Theory 611. The basic model of a competitive economy 61


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CHAPTER 1

Logic, Sets, Functions, and Spaces

1. Introduction

This chapter is very incomplete. It includes at the moment only some material that weshall use in the next chapter, and not even all of that. Never mind, perhaps there will bemore soon.

2. Logic

Exercises.

3. Some Set Theory

Set theory was developed in the second half of the 19th century and is at the veryfoundation of modern mathematics. We shall not be concerned here with the developmentof this theory. Rather we shall only give the basic language of set theory and outline someof the very basic operations on sets. Exercise 1 of this section gives a description of oneof the most famous paradoxes of what is called naive set theory. This paradox is known asRussells Paradox.

We start by dening a set to be a collection of objects or elements . We shall, in thissection, denote sets by capital letters and elements of sets by lower case letters. If theelement a is in the set A we write a A. If every element of the set B is also in the set A wecall B a subset of the set A and write B A. We shall also say the A contains B . If A and Bhave exactly the same elements then we say they are equal or identical. Alternatively wecould say A = B if and only if A B and B A. If B A and B = A then we say that B isa proper subset of A or that A strictly contains B .

In order to avoid the paradoxes such as the one referred to in the rst paragraph weshall always assume that in whatever situation we are discussing there is some given set U called the universal set which contains all of the sets with which we shall deal. Exercise 1asks you to show why such an assumption will avoid Russells Paradox.

We customarily enclose our specication of a set by braces. In order to specify a set wemay simply list the elements. For example to specify the set D which contains the numbers

1,2, and 3 we may write D = {1, 2, 3}. Alternatively we may dene the set by specifyinga property that identies the elements. For example we may specify the same set D by D = { x | x is an integer and 0 < x < 4}. Notice that this second method is more powerful.We could not, for example, list all the integers. (Since there are an innite number of themwe would die before we nished.)

For any two sets A and B we dene the union of A and B to be that set which containsexactly all of the elements of A and all the elements of B. We denote the union of A and B by A B. Similarly we dene the intersection of A and B to be that set which containsexactly those elements which are in both A and B. We denote the intersection of A and Bby A B. Thus we have

A B = { x | x A or x B}

A B = { x | x A and x B}.1


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2 1. LOGIC, SETS, FUNCTIONS, AND SPACES

Just as the number zero is extremely useful so the concept of a set that has no elementsis extremely useful also. This set we call the empty set or the null set and denote by / 0. To

see one use of the empty set notice that having such a concept allows the intersection of two sets be well dened whether or not the sets have any elements in common.

We also introduce the concept of a Cartesian product. If we have two sets, say A and B,the Cartesian product, A B, is the set of all ordered pairs, (a , b) such that a is an elementof A and b is an element of B. Symbolically we write

A B = {(a , b) | a A and b B}.

Exercises.

4. Functions

Exercises.

5. Spaces

Exercises.

6. Metric Spaces and Continuous Functions

Exercises.

7. Open sets, Compact Sets, and the Weierstrauss Theorem

Exercises.

8. Linear Spaces and Convex Sets

Exercises.


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CHAPTER 2

Linear Algebra

1. The Space R n

In the previous chapter we introduced the concept of a linear space or a vector space. 1

We shall now examine in some detail one example of such a space. This is the spaceof all ordered n-tuples ( x1, x2, . . . , xn) where each xi is a real number. We call this spacen-dimensional real space and denote it R n .

Remember from the previous chapter that to dene a vector space we not only needto dene the points in that space but also to dene how we add such points and how wemultiple such points by scalars. In the case of R n we do this element by element in then-tuple or vector. That is,

( x1, x2, . . . , xn) + ( y1, y2, . . . , yn) = ( x1 + y1, x2 + y2, . . . , xn + yn)

and ( x1, x2, . . . , xn) = ( x1, x2, . . . , xn).

Let us consider the case that n = 2, that is, the case of R 2 . In this case we can visualisethe space as in the following diagram. The vector ( x1, x2) is represented by the point thatis x1 units along from the point (0, 0) in the horizontal direction and x2 units up from (0, 0)

in the vertical direction.

E

T

x1

x2

q (1, 2)

1

2

Figure 1

1Of,course, we havent actually done any previous chapter from this book and, if youve checked theprevious chapter on the web site, you will know that the previous chapter is rather incomplete. Never mind.Perhaps it will be there for next years students. Sigh. And even more sighing, since the previous was actuallywritten for last years notes.

3


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4 2. LINEAR ALGEBRA

Let us for the moment continue our discussion in R 2 . Notice that we are implicitlywriting a vector ( x1, x2) as a sum x1 v

1 + x2 v2 where v1 is the unit vector in the rst

direction and v2 is the unit vector in the second direction. Suppose that instead we consid-ered the vectors u1 = ( 2, 1) = 2 v1 + 1 v2 and u2 = ( 1, 2) = 1 v1 + 2 v2. We couldhave written any vector ( x1, x2) instead as z1 u

1 + z2 u2 where z1 = ( 2 x1 x2)/ 3 and

z2 = ( 2 x2 x1)/ 3. That is, for any vector in R2 we can uniquely write that vector in terms

of u1 and u2 . Is there anything that is special about u1 and u2 that allows us to make thisclaim? There must be since we can easily nd other vectors for which this would not havebeen true. (For example, (1, 2) and (2, 4).)

The property of the pair of vectors u1 and u2 is that they are independent. That is, wecannot write either as a multiple of the other. More generally in n dimensions we would saythat we cannot write any of the vectors as a linear combination of the others, or equivalentlyas the following denition.

DEFINITION 1. The vectors x1, . . . , xk all in R n are linearly independent if it is notpossible to nd scalars 1, . . . , k not all zero such that

1 x1 + + k x

k = 0.

Notice that we do not as a matter of denition require that k = n or even that k n. Westate as a result that if k > n then the collection x1, . . . , xk cannot be linearly independent.(In a real maths course we would, of course, have proved this.)

COMMENT 1. If you examine the denition above you will notice that there is nowherethat we actually need to assume that our vectors are in R n . We can in fact apply the samedenition of linear independence to any vector space. This allows us to dene the conceptof the dimension of an arbitrary vector space as the maximal number of linearly indepen-

dent vectors in that space. In the case of Rn

we obtain that the dimension is in fact n.EXERCISE 1. Suppose that x1, . . . , xk all in R n are linearly independent and that the

vector y in R n is equal to 1 x1 + + k x

k . Show that this is the only way that y can beexpressed as a linear combination of the xis. (That is show that if y = 1 x

1 + + k xk

then 1 = 1, . . . , k = k .)

The set of all vectors that can be written as a linear combination of the vectors x1, . . . , xk is called the span of those vectors. If x1, . . . , xk are linearly independent and if the span of x1, . . . , xk is all of R n then the collection { x1, . . . , xk } is called a basis for R n . (Of course,in this case we must have k = n.) Any vector in R n can be uniquely represented as a linearcombination of the vectors x1, . . . , xk . We shall later see that it can sometimes be useful tochoose a particular basis in which to represent the vectors with which we deal.

It may be that we have a collection of vectors { x1, . . . , xk } whose span is not all of R n . In this case we call the span of { x1, . . . , xk } a linear subspace of R n . Alternatively wesay that X R n is a linear subspace of R n if X is closed under vector addition and scalarmultiplication. That is, if for all x, yX the vector x + y is also in X and for all xX and R the vector x is in X . If the span of x1, . . . , xk is X and if x1, . . . , xk are linearlyindependent then we say that these vectors are a basis for the linear subspace X . In thiscase the dimension of the linear subspace X is k . In general the dimension of the span of x1, . . . , xk is equal to the maximum number of linearly independent vectors in x1, . . . , xk .

Finally, we comment that R n is a metric space with metric d : R 2n R + dened by

d (( x1, . . . , xn), ( y1, . . . , yn)) = ( x1 y1)2 + + ( xn yn)2.There are many other metrics we could dene on this space but this is the standard one.


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2. LINEAR FUNCTIONS FROM R n TO R m 5

2. Linear Functions from R n to R m

In the previous section we introduced the space R n . Here we shall discuss functionsfrom one such space to another (possibly of different dimension). The concept of continu-ity that we introduced for metric spaces is immediately applicable here. We shall be mainlyconcerned here with an even narrower class of functions, namely, the linear functions.

DEFINITION 2. A function f : R n R m is said to be a linear function if it satisesthe following two properties:

(1) f ( x+ y) = f ( x) + f ( y) for all x, yR n , and(2) f ( x) = f ( x) for all xR n and R .

COMMENT 2. When considering functions of a single real variable, that is, functionsfrom R to R functions of the form f ( x) = ax + b where a and b are xed constants aresometimes called linear functions. It is easy to see that if b = 0 then such functions donot satisfy the conditions given above. We shall call such functions afne functions. More

generally we shall call a function g : R n R m an afne function if it is the sum of a linearfunction f : R n R m and a constant bR m. That is, if for any xR n g( x) = f ( x) + b.

Let us now suppose that we have two linear functions f : R n R m and g : R n R m.It is straightforward to show that the function ( f + g) : R n R m dened by ( f + g)( x) = f ( x) + g( x) is also a linear function. Similarly if we have a linear function f : R n R mand a constant R the function ( f ) : R n R m dened by ( f )( x) = f ( x) is alinear function. If f : R n R m and g : R m R k are linear functions then the composite function g f : R n R k dened by g f ( x) = g( f ( x)) is again a linear function. Finally,if f : R n R n is not only linear, but also one-to-one and onto so that it has an inverse f 1 : R n R n then the inverse function is also a linear function.

EXERCISE 2. Prove the facts stated in the previous paragraph.

Recall in the previous section we dened the notion of a linear subspace. A linearfunction f : R n R m denes two important subspaces, the image of f , denoted Im ( f ) R m, and the kernel of f , denoted Ker ( f ) R n . The image of f is the set of all vectors inR m such that f maps some vector in R n to that vector, that is,

Im( f ) = { yR m | xR n such that y = f ( x) }.The kernel of f is the set of all vectors in R n that are mapped by the function f to the zerovector in R m, that is,

Ker( f ) = { xR n | f ( x) = 0 }.The kernel of f is sometimes called the null space of f .

It is intuitively clear that the dimension of Im ( f ) is no more than n. (It is of course nomore than m since it is contained in R m.) Of course, in general it may be less than n, forexample if m < n or if f mapped all points in R n to the zero vector in R m. (You shouldsatisfy yourself that this function is indeed a linear function.) However if the dimensionof Im( f ) is indeed less than n it means that the function has mapped the n-dimensionalspace R n into a linear space of lower dimension and that in the process some dimensionshave been lost. The linearity of f means that a linear subspace of dimension equal to thenumber of dimensions that have been lost must have been collapsed to the zero vector (andthat translates of this linear subspace have been collapsed to single points). Thus we cansay that

dim(Im( f )) + dim(Ker( f )) = n.In the following section we shall introduce the notion of a matrix and dene various

operations on matrices. If you are like me when I rst came across matrices, these deni-tions may seem somewhat arbitrary and mysterious. However, we shall see that matricesmay be viewed as representations of linear functions and that when viewed in this way the

operations we dene on matrices are completely natural.


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6 2. LINEAR ALGEBRA

3. Matrices and Matrix Algebra

A matrix is dened as a rectangular array of numbers. If the matrix contains m rowsand n columns it is called an m n matrix (read m by n matrix). The element in the ithrow and the jth column is called the i j th element. We typically enclose a matrix in squarebrackets [ ] and write it as

a 11 . . . a 1n...

. . ....

am1 . . . a mn.

In the case that m = n we call the matrix a square matrix. If m = 1 the matrix contains asingle row and we call it a row vector . If n = 1 the matrix contains a single column and wecall it a column vector . For most purposes we do not distinguish between a 1 1 matrix[a ] and the scalar a .

Just as we dened the operation of vector addition and the multiplication of a vector by

a scalar we dene similar operations for matrices. In order to be able to add two matriceswe require that the matrices be of the same dimension. That is, if matrix A is of dimensionm n we shall be able to add the matrix B to it if and only if B is also of dimension m n.If this condition is met then we add matrices simply by adding the corresponding elementsof each matrix to obtain the new m n matrix A+ B. That is,

a 11 . . . a 1n...

. . ....

a m1 . . . amn

+b11 . . . b1n

.... . .

...bm1 . . . bmn

=a 11 + b11 . . . a 1n + b1n

.... . .

...am1 + bm1 . . . a mn + bmn

.

We can see that this denition of matrix addition satises many of the same propertiesof the addition of scalars. If A, B, and C are all m n matrices then

(1) A+ B = B + A,

(2) ( A+ B) + C = A+ ( B+ C ),(3) there is a zero matrix 0 such that for any m n matrix A we have A+ 0 = 0 + A =

A, and(4) there is a matrix A such that A+ ( A) = ( A) + A = 0.

Of course, the zero matrix referred to in 3 is simply the m n matrix consisting of allzeros (this is called a null matrix ) and the matrix A referred to in 4 is the matrix obtainedfrom A by replacing each element of A by its negative, that is,

a11 . . . a 1n

.... . .

...a m1 . . . a mn

= a11 . . . a 1n

.... . .

... a m1 . . . amn

.

Now, given a scalar in R and an m n matrix A we dene the product of and A which we write A to be the matrix in which each element is replaced by times thatelement, that is,

a 11 . . . a 1n

.... . .

...am1 . . . a mn

= a 11 . . . a1n

.... . .

... a m1 . . . a mn

.

So far the denitions of matrix operations have all seemed the most natural ones. Wenow come to dening matrix multiplication. Perhaps here the denition seems somewhatless natural. However in the next section we shall see that the denition we shall give is infact very natural when we view matrices as representations of linear functions.

We dene matrix multiplication of A times B written as AB where A is an m n matrixand B is a p q matrix only when n = p. In this case the product AB is dened to be anm q matrix in which the element in the ith row and jth column is nk = 1 a ik bk j . That is,


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3. MATRICES AND MATRIX ALGEBRA 7

to nd the term to go in the ith row and the jth column of the product matrix AB we takethe ith row of the matrix A which will be a row vector with n elements and the jth column

of the matrix B which will be a column vector with n elements. We then multiply eachelement of the rst vector by the corresponding element of the second and add all theseproducts. Thus

a11 . . . a 1n...

. . ....

a m1 . . . a mn

b11 . . . b1q...

. . ....

bn1 . . . bnq

=

nk = 1 a 1k bk 1 . . . nk = 1 a1k bkq

.... . .

...nk = 1 a mk bk 1 . . .

nk = 1 a mk bkq

.

For example

a b cd e f

p qr st v

= ap + br + ct aq + bs + cvd p + er + f t dq + es + f v .

We dene the identity matrix of order n to be the n n matrix that has 1s on its maindiagonal and zeros elsewhere that is, whose i j th element is 1 if i = j and zero if i = j. Wedenote this matrix by I n or, if the order is clear from the context, simply I . That is,

I =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

.

It is easy to see that if A is an m n matrix then AI n = A and I m A = A. In fact, we couldequally well dene the identity matrix to be that matrix that satises these properties forall such matrices A in which case it would be easy to show that there was a unique matrixsatisfying this property, namely, the matrix we dened above.

Consider an m n matrix A. The columns of A are m-dimensional vectors, that is,elements of R m and the rows of A are elements of R n . Thus we can ask if the n columnsare linearly independent and similarly if the m rows are linearly independent. In fact weask: What is the maximum number of linearly independent columns of A? It turns out thatthis is the same as the maximum number of linearly independent rows of A. We call the

number the rank of the matrix A.


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8 2. LINEAR ALGEBRA

4. Matrices as Representations of Linear Functions

Let us suppose that we have a particular linear function f : R n R m. We have sug-gested in the previous section that such a function can necessarily be represented as multi-plication by some matrix. We shall now show that this is true. Moreover we shall do so byexplicitly constructing the appropriate matrix.

Let us write the n-dimensional vector x as a column vector

x =

x1 x2

... xn

.

Now, notice that we can write the vector x as a sum ni= 1 xiei, where ei is the ith unit vector,

that is, the vector with 1 in the ith place and zeros elsewhere. That is,

x1 x2

... xn

= x1

10...0

+ x2

01...0

+ + xn

00...1

.

Now from the linearity of the function f we can write

f ( x) = f (n

i= 1

xiei)

=n

i= 1

f ( xiei)

=n

i= 1

xi f (ei).

But, what is f (ei)? Remember that ei is a unit vector in R n and that f maps vectors in R nto vectors in R m. Thus f (ei) is the image in R m of the vector ei. Let us write f (ei) as

a 1ia 2i...

a mi

.

Thus

f ( x) =n

i= 1

xi f (ei)

= x1

a 11a 21

...a m1

+ x2

a 12a 22

...a m2

+ + xn

a 1na 2n

...a mn

=

ni= 1 a 1i xini= 1 a 2i xi

...ni= 1 a mi xi

and this is exactly what we would have obtained had we multiplied the matrices


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4. MATRICES AS LINEAR FUNCTIONS 9

a 11 a12 . . . a 1na 21 a22 . . . a 2n

... ... . . . ...a m1 am2 . . . a mn

x1 x2

... xn

.

Thus we have not only shown that a linear function is necessarily represented by multi-plication by a matrix we have also shown how to nd the appropriate matrix. It is preciselythe matrix whose n columns are the images under the function of the n unit vectors in R n .

EXERCISE 3. Find the matrices that represent the following linear functions from R 2

to R 2 .

(1) a clockwise rotation of / 2 (90 ),(2) a reection in the x1 axis,(3) a reection in the line x2 = x1 (that is, the 45 line),(4) a counter clockwise rotation of / 4 (45 ), and(5) a reection in the line x2 = x1 followed by a counter clockwise rotation of / 4.

Recall that in Section 2 we dened, for any f , g : R n R m and R , the functions( f + g) and ( f ). In Section 3 we dened the sum of two m n matrices A and B, andthe product of a scalar with the matrix A. Let us instead dene the sum of A and B asfollows.

Let f : R n R m be the linear function represented by the matrix A and g : R n R mbe the linear function represented by the matrix B. Now dene the matrix ( A+ B) to be thematrix that represents the linear function ( f + g). Similarly let the matrix A be the matrixthat represents the linear function ( f ).

EXERCISE 4. Prove that the matrices ( A + B) and A dened in the previous para-graph coincide with the matrices dened in Section 3.

We can also see that the denition we gave of matrix multiplication is precisely theright denition if we mean multiplication of matrices to mean the composition of the linearfunctions that the matrices represent. To be more precise let f : R n R m and g : R m R k be linear functions and let A and B be the m n and k m matrices that represent them.Let (g f ) : R n R k be the composite function dened in Section 2. Now let us denethe product BA to be that matrix that represents the linear function (g f ).


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10 2. LINEAR ALGEBRA

Now since the matrix A represents the function f and B represents g we have

(g f )( x) = g( f ( x))

= g

a11 a 12 . . . a 1na21 a 22 . . . a 2n

......

. . ....

a m1 am2 . . . a mn

x1 x2

... xn

= g


...ni= 1 a mi xi

=

b11 b12 . . . b1mb21 b22 . . . b2m

... ... . . . ...bk 1 bk 2 . . . bkm


...ni= 1 a mi xi

=

m j= 1 b1 j ni= 1 a ji xi

m j= 1 b2 j ni= 1 a ji xi

...m j= 1 bk j

ni= 1 a ji xi

=

ni= 1 m j= 1 b1 ja ji xi

ni= 1 m j= 1 b2 ja ji xi

...n

i= 1m

j= 1b

k ja

ji x

i

=

m j= 1 b1 ja j1 m j= 1 b1 ja j2 . . .

m j= 1 b1 ja jn

m j= 1 b2 ja j1 m j= 1 b2 ja j2 . . .

m j= 1 b2 ja jn

......

. . ....

m j= 1 bk ja j1 m j= 1 bk ja j2 . . .

m j= 1 bk ja jn

x1 x2

... xn

.

And this last is the product of the matrix we dened in Section 3 to be BA with the columnvector x. As we have claimed the denition of matrix multiplication we gave in Section 3was not arbitrary but rather was forced on us by our decision to regard the multiplicationof two matrices as corresponding to the composition of the linear functions the matricesrepresented.

Recall that the columns of the matrix A that represented the linear function f : R n R m

were precisely the images of the unit vectors inR n

under f . The linearity of f meansthat the image of any point in R n is in the span of the images of these unit vectors andsimilarly that any point in the span of the images is the image of some point in R n . ThusIm( f ) is equal to the span of the columns of A. Now, the dimension of the span of thecolumns of A is equal to the maximum number of linearly independent columns in A, thatis, to the rank of A.

5. Linear Functions from R n to R n and Square Matrices

In the remainder of this chapter we look more closely at an important subclass of linearfunctions and the matrices that represent them, viz the functions that map R n to itself. Fromwhat we have already said we see immediately that the matrix representing such a linearfunction will have the same number of rows as it has columns. We call such a matrix a

square matrix.


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7. CHANGES OF BASIS 11

If the linear function f : R n R n is one-to-one and onto then the function f hasan inverse f 1 . In Exercise 2 you showed that this function too was linear. A matrix

that represents a linear function that is one-to-one and onto is called a nonsingular matrix .Alternatively we can say that an n n matrix is nonsingular if the rank of the matrix is n. Tosee these two statements are equivalent note rst that if f is one-to-one then Ker ( f ) = {0}.(This is the trivial direction of Exercise 5.) But this means that dim (Ker( f )) = 0 and sodim(Im( f )) = n. And, as we argued at the end of the previous section this is the same asthe rank of matrix that represents f .

EXERCISE 5. Show that the linear function f : R n R m is one-to-one if and only if Ker( f ) = {0}.

EXERCISE 6. Show that the linear function f : R n R n is one-to-one if and only if it is onto.

6. Inverse Functions and Inverse Matrices

In the previous section we discussed briey the idea of the inverse of a linear function f : R n R n . This allows us a very easy denition of the inverse of a square matrix A. Theinverse of A is the matrix that represents the linear function that is the inverse function of the linear function that A represents. We write the inverse of the matrix A as A 1 . Thus amatrix will have an inverse if and only if the linear function that the matrix represents hasan inverse, that is, if and only if the linear function is one-to-one and onto. We saw in theprevious section that this will occur if and only if the kernel of the function is {0} whichin turn occurs if and only if the image of f is of full dimension, that is, is all of R n . This isthe same as the matrix being of full rank, that is, of rank n.

As with the ideas we have discussed earlier we can express the idea of a matrix inversepurely in terms of matrices without reference to the linear function that they represent.Given an n n matrix A we dene the inverse of A to be a matrix B such that BA = I nwhere I n is the n n identity matrix discussed in Section 3. Such a matrix B will exist if and only if the matrix A is nonsingular. Moreover, if such a matrix B exists then it is alsotrue that AB = I n , that is, ( A 1) 1 = A.

In Section 9 we shall see one method for calculating inverses of general n n matrices.Here we shall simply describe how to calculate the inverse of a 2 2 matrix. Suppose thatwe have the matrix

A = a bc d .

The inverse of this matrix is

1ad bc

d b c a .

EXERCISE 7. Show that the matrix A is of full rank if and only if ad bc = 0.

EXERCISE 8. Check that the matrix given is, in fact, the inverse of A.

7. Changes of Basis

We have until now implicitly assumed that there is no ambiguity when we speak of the vector ( x1, x2, . . . , xn). Sometimes there may indeed be an obvious meaning to sucha vector. However when we dene a linear space all that are really specied are whatstraight lines are and where zero is. In particular, we do not necessarily have dened inan unambiguous way where the axes are or what a unit length along each axis is. Inother words we may not have a set of basis vectors specied.

Even when we do have, or have decided on, a set of basis vectors we may wish to re-

dene our description of the linear space with which we are dealing so as to use a different


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set of basis vectors. Let us suppose that we have an n-dimensional space, even R n say, witha given set of basis vectors v1, v2, . . . , vn and that we wish instead to describe the space in

terms of the linearly independent vectors b1, b2, . . . , bn wherebi = b1iv

1 + b2iv2 + + bni v

n .

Now, if we had the description of a point in terms of the new coordinate vectors, e.g.,as

z1b1 + z2b

2 + + znbn

then we can easily convert this to a description in terms of the original basis vectors. Wewould simply substitute the formula for bi in terms of the e js into the previous formulagiving

n

i= 1

b1i zi v1 +

n

i= 1

b2i zi v2 + +

n

i= 1

bni zi vn

or, in our previous notationni= 1 b1i zini= 1 b2i zi

...(ni= 1 bni zi)

.

But this is simply the product

b11 b12 . . . b1nb21 b22 . . . b2n

......

. . ....

bn1 bn2 . . . bnn

z1 z2...

zn

.

That is, if we are given an n-tuple of real numbers that describe a vector in terms of thenew basis vectors b1, b2, . . . , bn and we wish to nd the n-tuple that describes the vectorin terms of the original basis vectors we simply multiply the ntuple we are given, writtenas a column vector by the matrix whose columns are the new basis vectors b1, b2, . . . , bn .We shall call this matrix B. We see among other things that changing the basis is a linearoperation.

Now, if we were given the information in terms of theoriginal basis vectors and wantedto write it in terms of the new basis vectors what should we do? Since we dont have theoriginal basis vectors written in terms of the new basis vectors this is not immediatelyobvious. However we do know that if we were to do it and then were to carry out theoperation described in the previous paragraph we would be back with what we started.Further we know that the operation is a linear operation that maps n-tuples to n-tuples

and so is represented by multiplication by an n n matrix. That is we multiply the n-tuple written as a column vector by the matrix that when multiplied by B gives the identitymatrix, that is, the matrix B 1 . If we are given a vector of the form

x1v1 + x2v

2 + + xnvn

and we wish to express it in terms of the vectors b1, b2, . . . , bn we calculate

b11 b12 . . . b1nb21 b22 . . . b2n

......

. . ....

bn1 bn2 . . . bnn

1 x1 x2

... xn

.

Suppose now that we consider a linear function f : R n R n and that we have origi-nally described R n in terms of the basis vectors v1, v2, . . . , vn where vi is the vector with 1


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of having the function take a particular value is invariant for all particular values we maychoose.

Two particularly important invariants are the trace of a square matrix and the determi-nant of a square matrix. We examine these in more detail in the following section.

8. The Trace and the Determinant

In this section we dene two important real valued functions on the space of n nmatrices, the trace and the determinant. Both of these concepts have geometric interpreta-tions. However, while the trace is easy to calculate (much easier than the determinant) itsgeometric interpretation is rather hard to see. Thus we shall not go into it. On the otherhand the determinant while being somewhat harder to calculate has a very clear geometricinterpretation. In Section 9 we shall examine in some detail how to calculate determinants.In this section we shall be content to discuss one denition and the geometric intuition of the determinant.

Given an n n matrix A the trace of A, written tr( A) is the sum of the elements on themain diagonal, that is,

tr

a 11 a 12 . . . a 1na 21 a 22 . . . a 2n

......

. . ....

a n1 a n2 . . . a nn

=n

i= 1

a ii .

EXERCISE 11. For the matrices given in Exercise 10 conrm that tr ( A) = tr( B 1 AB).

It is easy to see that the trace is a linear function on the space of all n n matrices, thatis, that for all A and B n n matrices and for all R

(1) tr( A+ B) = tr( A) + tr( B),

and

(2) tr( A) = tr( A).

We can also see that if A and B are both n n matrices then tr ( AB) = tr( BA). In fact, if A is an m n matrix and B is an n m matrix this is still true. This will often be extremelyuseful in calculating the trace of a product.

EXERCISE 12. From the denition of matrix multiplication show that if A is an m nmatrix and B is an n m matrix that tr ( AB) = tr( BA). [Hint: Look at the denition of matrix multiplication in Section 2. Then write the determinant of the product matrix usingsummation notation. Finally change the order of summation.]

The determinant, unlike the trace is not a linear function of the matrix. It does howeverhave some linear structure. If we x all columns of the matrix except one and look atthe determinant as a function of only this column then the determinant is linear in thissingle column. Moreover this is true whatever the column we choose. Let us write thedeterminant of the n n matrix A as det( A). Let us also write the matrix A as [a 1, a 2, . . . , a n]where a i is the ith column of the matrix A. Thus our claim is that for all n n matrices A,for all i = 1, 2, . . . n, for all n vectors b, and for all R

det([a1, . . . , a i 1, a i + b, a i+ 1, . . . , an]) = det([a1, . . . , a i 1, a i, a i+ 1, . . . , a n])

+ det([a 1, . . . , a i 1, b, a i+ 1, . . . , a n])(3)

and

(4) det([a 1, . . . , a i 1, a i, a i+ 1, . . . , a n]) = det([a1, . . . , a i 1, a i, a i+ 1, . . . , a n]).

We express this by saying that the determinant is a multilinear function .


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9. CALCULATING AND USING DETERMINANTS 15

Also the determinant is such that any n n matrix that is not of full rank, that is, of rank n, has a zero determinant. In fact, given that the determinant is a multilinear function

if we simply say that any matrix in which one column is the same as one of its neighbourshas a zero determinant this implies the stronger statement that we made. We already seeone use of calculating determinants. A matrix is nonsingular if and only if its determinantis nonzero.

The two properties of being multilinear and zero whenever two neighbouring columnsare the same already almost uniquely identify the determinant. Notice however that if thedeterminant satises these two properties then so does any constant times the determinant.To uniquely dene the determinant we tie down this constant by assuming that det ( I ) = 1.

Though we havent proved that it is so, these three properties uniquely dene the deter-minant. That is, there is one and only one function with these three properties. We call thisfunction the determinant. In Section 9 we shall discuss a number of other useful propertiesof the determinant. Remember that this additional properties are not really additional factsabout the determinant. They can all be derived from the three properties we have givenhere.

Let us now look to the geometric interpretation of the determinant. Let us rst think about what linear transformations can do to the space R n . Since we have already said thata linear transformation that is not onto is represented by a matrix with a zero determinantlet us think about linear transformations that are onto, that is, that do not map R n into alinear space of lower dimension. Such transformations can rotate the space around zero.They can stretch the space in different directions. And they can ip the space over.In the latter case all objects will become mirror images of themselves. We call lineartransformations that make such a mirror image orientation reversing and those that dontorientation preserving . A matrix that represents an orientation preserving linear functionhas a positive determinant while a matrix that represents an orientation reversing linearfunction has a negative determinant. Thus we have a geometric interpretation of the sign

of the determinant.The absolute size of the determinant represents how much bigger or smaller the linear

function makes objects. More precisely it gives the volume of the image of the unithypercube under the transformation. The word volume is in quotes because it is the volumewith which we are familiar only when n = 3. If n = 2 then it is area, while if n > 3 then itis the full dimensional analog in R n of volume in R 3 .

EXERCISE 13. Consider the matrix

3 11 2 .

In a diagram show the image under the linear function that this matrix represents of the

unit square, that is, the square whose corners are the points (0,0), (1,0), (0,1), and (1,1).Calculate the area of that image. Do the same for the matrix

4 1 1 1 .

In the light of Exercise 10, comment on the answers you calculated.

9. Calculating and Using Determinants

We have already used the concepts of the inverse of a matrix and the determinantof a matrix. The purpose of this section is to cover some of the cookbook" aspects of

calculating inverses and determinants.


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Suppose that we have an n n matrix

A =a

11 . . .a

1n.... . .

...a n1 . . . a nn

then we shall use | A| ora 11 . . . a 1n...

. . ....

a n1 . . . a nn

as an alternative notation for det ( A). Always remember that

a 11 . . . a 1n...

.. .

.

..a n1 . . . a nn

is not a matrix but rather a real number. For the case n = 2 we dene

det( A) = a 11 a 12a 21 a 22

as a 11 a22 a 21a 12 . It is possible to also give a convenient formula for the determinant of a 3 3 matrix. However, rather than doing this, we shall immediately consider the case of an n n matrix.

By the minor of an element of the matrix A we mean the determinant (remember areal number) of the matrix obtained from the matrix A by deleting the row and columncontaining the element in question. We denote the minor of the element a i j by the symbol| M i j | . Thus, for example,

| M 11 | =a 22 . . . a2n...

. . ....

a n2 . . . ann

.

EXERCISE 14. Write out the minors of a general 3 3 matrix.

We now dene the cofactor of an element to be either plus or minus the minor of theelement, being plus if the sum of indices of the element is even and minus if it is odd. Wedenote the cofactor of the element a i j by the symbol |C i j | . Thus |C i j | = | M i j | if i + j is evenand |C i j | = | M i j | if i + j is odd. Or,

|C i j | = ( 1)i+ j| M i j |.

We now dene the determinant of an n n matrix A,

det( A) = | A| =a 11 . . . a 1n...

. . ....

a n1 . . . a nn

to be n j= 1 a 1 j|C 1 j|. This is the sum of n terms, each one of which is the product of anelement of the rst row of the matrix and the cofactor of that element.

EXERCISE 15. Dene the determinant of the 1 1 matrix [a ] to be a . (What else couldwe dene it to be?) Show that the denition given above corresponds with the denition

we gave earlier for 2 2 matrices.


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EXERCISE 16. Calculate the determinants of the following 3 3 matrices.

(a)1 2 33 6 94 5 7

(b)1 5 21 4 30 1 2

(c)1 1 05 4 12 3 2

(d)1 0 00 1 00 0 1

(e)2 5 21 5 30 1 3

EXERCISE 17. Show that the determinant of the identity matrix, det ( I n) is 1 for allvalues of n. [Hint: Show that it is true for I 2. Then show that if it is true for I n 1 then it istrue for I n .]

One might ask what was special about the rst row that we took elements of that rowmultiplied them by their cofactors and added them up. Why not the second row, or the rst

column? It will follow from a number of properties of determinants we list below that infact we could have used any row or column and we would have arrived at the same answer.

EXERCISE 18. Expand the matrix given in Exercise 16(b) in terms of the 2nd and3rd rows and in terms of each column and check that the resulting answer agrees with theanswer you obtained originally.

We now have a way of calculating the determinant of any matrix. To nd the deter-minant of an n n matrix we have to calculate n determinants of size (n 1) (n 1).This is clearly a fairly computationally costly procedure. However there are often ways toeconomise on the computation.

EXERCISE 19. Evaluate the determinants of the following matrices

(a)

1 8 0 72 3 4 61 6 0 10 5 0 8

(b)

4 7 0 45 6 1 80 0 9 01 3 1 4

[Hint: Think carefully about which column or row to use in the expansion.]

We shall now list a number of properties of determinants. These properties implythat, as we stated above, it does not matter which row or column we use to expand thedeterminant. Further these properties will give us a series of transformations we mayperform on a matrix without altering its determinant. This will allow us to calculate adeterminant by rst transforming the matrix to one whose determinant is easier to calculateand then calculating the determinant of the easier matrix.

PROPERTY 1. The determinant of a matrix equals the determinant of its transpose.

| A| = | A |

PROPERTY 2. Interchanging two rows (or two columns) of a matrix changes its signbut not its absolute value. For example,

c d a b = cb ad = (ad cb) =

a bc d .

PROPERTY 3. Multiplying one row (or column) of a matrix by a constant willchange the value of the determinant -fold. For example,

a 11 . . . a1n...

. . ....

a n1 . . . ann

= a 11 . . . a 1n...

. . ....

a n1 . . . a nn

.


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EXERCISE 20. Check Property 3 for the cases n = 2 and n = 3.

COROLLARY 1. | A| = n | A| (where A is an n n matrix).

COROLLARY 2. | A| = | A| if n is even. | A| = | A| if n is odd.

PROPERTY 4. Adding a multiple of any row (column) to any other row (column) doesnot alter the value of the determinant.

EXERCISE 21. Check that1 5 21 4 30 1 2

=1 5 + 3 2 21 4 + 3 3 30 1 + 3 2 2

=1 + ( 2) 1 5 + ( 2) 4 2 + ( 2) 31 4 30 1 2

.

PROPERTY 5. If one row (or column) is a constant times another row (or column) thenthe determinant the matrix is zero.

EXERCISE 22. Show that Property 5 follows from Properties 3 and 4.

We can strengthen Property 5 to obtain the following.

PROPERTY 5 . The determinant of a matrix is zero if and only if the matrix is not of full rank.

EXERCISE 23. Explain why Property 5 is a strengthening of Property 5, that is, why5 implies 5.

These properties allow us to calculate determinants more easily. Given an n n matrix A the basic strategy one follows is to use the above properties, particularly Property 4 tond a matrix with the same determinant as A in which one row (or column) has only onenon-zero element. Then, rather than calculating n determinants of size (n 1) (n 1)one only needs to calculate one. One then does the same thing for the (n 1) (n 1)determinant that needs to be calculated, and so on.

There are a number of reasons we are interested in determinants. One is that they giveus one method of calculating the inverse of a nonsingular matrix. (Recall that there is noinverse of a singular matrix.) They also give us a method, known as Cramers Rule, forsolving systems of linear equations. Before proceeding with this it is useful to state onefurther property of determinants.

PROPERTY 6. If one expands a matrix in terms of one row (or column) and the cofac-

tors of a different row (or column) then the answer is always zero. That isn j= 1

a i j |C k j | = 0

whenever i = k . Alson

i= 1

a i j |C ik | = 0

whenever j = k .

EXERCISE 24. Verify Property 6 for the matrix

4 1 25 2 1

1 0 3

.


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Let us dene the matrix of cofactors C to be the matrix [|C i j |] whose i j th element isthe cofactor of the i j th element of A. Now we dene the adjoint matrix of A to be the

transpose of the matrix of cofactors of A. That isadj( A) = C .

It is straightforward to see (using Property 6) that Aadj( A) = | A| I n = adj( A) A. That is, A 1 = 1| A| adj( A). Notice that this is well dened if and only if | A| = 0. We now have amethod of nding the inverse of any nonsingular square matrix.

EXERCISE 25. Use this method to nd the inverses of the following matrices

(a)3 1 21 0 34 0 2

(b)4 2 17 3 32 0 1

(c)1 5 21 4 30 1 2

.

Knowing how to invert matrices we thus know how to solve a system of n linearequations in n unknowns. For we can express the n equations in matrix notation as Ax = bwhere A is an n n matrix of coefcients, x is an n 1 vector of unknowns, and b is ann 1 vector of constants. Thus we can solve the system of equations as x = A 1 Ax = A 1b.

Sometimes, particularly if we are not interested in all of the xs it is convenient to useanother method of solving the equations. This method is known as Cramers Rule. Let ussuppose that we wish to solve the above system of equations, that is, Ax = b. Let us denethe matrix Ai to be the matrix obtained from A by replacing the ith column of A by thevector b. Then the solution is given by

xi =| Ai || A|

.

EXERCISE 26. Derive Cramers Rule. [Hint: We know that the solution to the systemof equations is solved by x = ( 1/ | A|)adj( A)b. This gives a formula for xi . Show that thisformula is the same as that given by xi = | Ai |/ | A| .]

EXERCISE 27. Solve the following system of equations (i) by matrix inversion and(ii) by Cramers Rule

(a)2 x1 x2 = 2

3 x2 + 2 x3 = 165 x1 + 3 x3 = 21

(b) x1 + x2 + x3 = 1

x1 x2 + x3 = 1 x1 + x2 + x3 = 1

.

EXERCISE 28. Recall that we claimed that the determinant was an invariant. Conrmthis by calculating (directly) det ( A) and det( B 1 AB) where

B =1 0 11 1 22 1 1

and A =1 0 00 2 00 0 3

.

EXERCISE 29. An nth order determinant of the form

a 11 0 0 . . . 0a 21 a 22 0 . . . 0a 31 a 32 a33 . . . 0

......

.... . .

...a n1 a n2 an3 . . . ann

is called triangular . Evaluate this determinant. [Hint: Expand the determinant in terms of its rst row. Expand the resulting (n 1) (n 1) determinant in terms of its rst row,

and so on.]


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10. Eigenvalues and Eigenvectors

Suppose that we have a linear function f : R n R n . When we look at how f deformsR n one natural question to look at is: Where does f send some linear subspace? In particu-lar we might ask if there are any linear subspaces that f maps to themselves. We call suchlinear subspaces invariant linear subspaces . Of course the space R n itself and the zero di-mensional space {0} are invariant linear subspaces. The real question is whether there areany others. Clearly, for some linear transformations there are no other invariant subspaces.For example, a clockwise rotation of / 4 in R 2 has no invariant subspaces other than R 2itself and {0}.

A particularly important class of invariant linear subspaces are the one dimensionalones. A one dimensional linear subspace is specied by one nonzero vector, say x. Thenthe subspace is { x | R }. Let us call this subspace L( x). If L( x) is an invariant linearsubspace of f and if x L( x) then there is some value such that f ( x) = x. Moreover thevalue of for which this is true will be the same whatever value of x we choose in L( x).

Now if we x the set of basis vectors and thus the matrix A that represents f we havethat if x is in a one dimensional invariant linear subspace of f then there is some Rsuch that

Ax = x.Again we can dene this notion without reference to linear functions. Given a matrix A if we can nd a pair x, with x = 0 that satisfy the above equation we call x an eigenvector of the matrix A and the associated eigenvalue . (Sometimes these are called characteristicvectors and values.)

EXERCISE 30. Show that the eigenvalues of a matrix are an invariant, that is, thatthey depend only on the linear function the matrix represents and not on the choice of basisvectors. Show also that the eigenvectors of a matrix are not an invariant. Explain why thedependence of the eigenvectors on the particular basis is exactly what we would expect andargue that is some sense they are indeed invariant.

Now we can rewrite the equation Ax = x as

( A I n) x = 0.

If x, solve this equation and x = 0 then we have a nonzero linear combination of thecolumns of A I n equal to zero. This means that the columns of A I n are not linearlyindependent and so det ( A I n) = 0, that is,

det

a11 a 12 . . . a1na 21 a 22 . . . a2n

......

. . ....

a n1 a n2 . . . a nn

= 0.

Now, the left hand side of this last equation is a polynomial of degree n in , that is, apolynomial in in which n is the highest power of that appears with nonzero coefcient.It is called the characteristic polynomial and the equation is called the characteristic equa-tion. Now this equation may, or may not, have a solution in real numbers. In general, bythe fundamental theorem of algebra the equation has n solutions, perhaps not all distinct, inthe complex numbers. If the matrix A happens to be symmetric (that is, if a i j = a ji for all iand j) then all of its eigenvalues are real. If the eigenvalues are all distinct (that is, differentfrom each other) then we are in a particularly well behaved situation. As a prelude we statethe following result.

THEOREM 1. Given an n n matrix A suppose that we have m eigenvectors of A x1, x2, . . . , xm with corresponding eigenvalues 1, 2, . . . , m . If i = j whenever i = j then

x1, x2, . . . , xm are linearly independent.


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10. EIGENVALUES AND EIGENVECTORS 21

An implication of this theorem is that an n n matrix cannot have more than n eigen-vectors with distinct eigenvalues. Further this theorem allows us to see that if an n n

matrix has n distinct eigenvalues then it is possible to nd a basis for R n in which the lin-ear function that the matrix represents is represented by a diagonal matrix. Equivalentlywe can nd a matrix B such that B 1 AB is a diagonal matrix.

To see this let b1, b2, . . . , bn be n linearly independent eigenvectors with associatedeigenvalues 1, 2, . . . , n . Let B be the matrix whose columns are the vectors b

1, b2, . . . , bn .Since these vectors are linearly independent the matrix B has an inverse. Now

B 1 AB = B 1[ Ab1 Ab2 . . . Abn]

= B 1[ 1b1 2b

2 . . . nbn]

= [ 1 B 1b1 2 B

1b2 . . . n B 1bn]

=

1 0 . . . 00 2 . . . 0... ... . . . ...0 0 . . . n

.


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CHAPTER 3

Choice, Preference, and Utility

We shall start this course with an introduction to the formal modelling of motivatedchoice. There is an abundance of experimental evidence to suggest that, at least in somecircumstances, the decision making of actual people is inuenced by a myriad of appar-ently irrelevant considerations. It is nevertheless traditional in economicsand I think forvery good reasonto start, and often to go no further than, what is usually called rational

choice, which I am calling here motivated choice .The treatment here follows largely the approach and uses the notation of ? The style

shall be rather formal. For better or worse, modern economic theory is formal, sometimes,perhaps, more formal than it needs to be. Whatever style you will choose for your ownwork, you will need to be able to read and work with formal models to access the literaturethat exists.

Before getting into the formal modelling and assumptions of the next section I shalldiscuss here an assumption that is buried in the setup. In the next section we shall denea choice rule as a function that for any choice situation species which of the availableelements might be chosen. A choice situation will be dened by specifying which elementsare available to be selected, that is, by specifying the available set of choices. Notice thatthere is already an assumption buried here. By specifying only the set, and not saying

how it is presented to the decision maker we are already making assumptions. And thoseassumptions are not necessarily realistic. There is a good deal of evidence to suggest thatthe manner a choice is presented to a decision maker might have a good deal of inuenceon the actual decision she comes to.

1. Formal Model of Motivated Choice

We start by specifying the set of all alternatives the decision-maker might be asked tochoose among. For the moment we shall just specify it as an abstract set X . Each elementof the set corresponds to the completely specied consequence of a choice. These mightbe very complicated consequences. One example of such a consequence is a consumptionplan, specifying how much of each good an individual will consume. Another is the pro-duction plan of a rm. In another setting the consequences might be lotteries, giving someoutcome in one realisation and another outcome in a second. In other parts of the coursewe shall look in more detail at some examples of sets of alternatives in which a bit morestructure is assumed. Here however we shall simply let X be an arbitrary set. We shall,however, assume for the moment that X is nite.

While X is the set of all alternatives the decision-maker will typically be offered achoice from some smaller subset, say B X . We shall call such a set B a budget set. Thefamily of all such budget sets that the decision-maker might confront we shall denote B .Notice that this might not be the set of all subsets. For example in choosing which subjectsto enrol in this year not all combinations were available to you. There were considerationsof prerequisites and pairs of subjects which could not both be taken. (If X is nite we couldactually let B be the family of all nonempty subsets of X . But we shall include the extragenerality for later use.)

For a given decision maker a choice structure is a pair (B ,C ()) where B is as de-scribed above and C () is a choice rule that associates to every budget set B in B a chosen

23


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24 3. CHOICE, PREFERENCE, AND UTILITY

nonempty subset of B. Formally we say that C is a correspondence from B to X and write

C : B X

such that for all B in B we have C ( B) B. Informally the choice rule C () tells us for eachbudget set B which elements of B are (or could be) chosen by the decision-maker. Noticethat we make the requirements that C ( B) be nonempty and that it be a subset of B part of the denition of what a choice function is. In the next section we discuss an importantproperty that we do not make part of the denition. This reects the view that it simplywould not make much sense to consider choice functions that did not choose anything orthat chose elements that were not available. On the other hand it is quite coherent to think of a decision-maker choosing in ways that do not satisfy the property given in the nextsection.

2. The Properties of Choice Structures

There are many ways that we might think of a decision-maker choosing. In orderthat you should realise that there is something substantive about the properties that weshall discuss we shall rst give some examples that will not satisfy the property we shallformulate below.

A decision-maker might choose the element that is most like the average of all theelements with which he is confronted. (This, of course, assumes that we have some wayof averaging the elements.) Or he might choose the one most dissimilar from the rest.(Similarly, this assumes we have some notion of similarity.)

Both of these methods of choosing might seem a little strange to you. This is probablybecause you have in mind already some, perhaps informal, idea that the decision-makerschoice is based on some underlying preference of the decision-maker. We shall, in thenext section explicitly introduce preferences. Lets see rst however if we can do anythingwithout explicitly talking about preferences (at least in our formal model).

Suppose that we were observing the decision-maker and we saw her, when confrontedwith a particular budget set choose a particular alternative x and reject another that was alsoavailable y. If we knew already that her choice was based on some underlying preferenceover the alternatives then we might say that she had revealed a preference for x over y. Thuswe would be a bit surprised if we were every to see her choosing y when x was available toher.

We formalise this in the following denition.

DEFINITION 3. The choice structure (B ,C ())satises the weak axiom of revealed preference if whenever there is some budget set B in B containing both x and y with x inC ( B) it is the case for any B in B that contains both x and y if y is in C ( B ) then x is alsoin C ( B ).

3. Preference Relations and Preference Based Choice

Another approach to choice starts by taking not the decision-makers choice behaviouras a primitive but by positing from the start that the decision-maker has a preference re-lation on the elements of X . A preference relation is a list of all those pairs ( x, y) forwhich it is true that x is at least as good for the decision-maker as y. Thus if for a particularpair ( x , y ) it was the case that x was as good as y then ( x , y ) would be in the set . Weactually denote this by the somewhat more suggestive notation x y , which we read x isat least as good as y .

We could do everything we want to about preferences in terms of the relation , andfrom some points of view it might be good to do things that way. However it is very

convenient, and sometimes instructive to introduce two other binary relations on X :


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4. CHOICE AND PREFERENCES 25

(1) the strict preference relation dened by

x y if and only if ( x y and not y x

which we read x is preferred to y.(2) the indifference relation dened by

x y if and only if ( x y and y x

which we read x is indifferent to y.It is also possible to start with the strict preference relation as the primitive and to

dene the (weak) preference relation and the indifference relation in terms of that.There is a bit more to what we would intuitively think of as preferences that we have

so far (which, after all, is almost nothing). We call a preference relation rational if it isboth complete and transitive .

DEFINITION 4. The preference relation is rational if

(1) (Completeness ) for all x, y in X either x y or y x (or both)(2) (Transitivity ) for all x, y, z in X if x y and y z then x z.

The properties that dene rational preference relations also have easily stated implica-tions for the derived strict preference and indifference relations.

PROPOSITION 1. If is rational then(1) is both irreexive (it is not the case that x x for any x in X) and transitive

(for all x , y, z in X if x y and y z then x z).(2) is reexive (x x for all x in X), transitive (for all x , y, z in X if x y and y z

then x z), and symmetric (for all x , y in X if x y then y x).(3) for all x , y, z in X if x y and y z then x z.(4) for all x , y, z in X if x y and y z then x z.

4. The Relationship Between the Choice Based Approach and the Preference BasedApproach

We return to the choice based approach. Given a choice structure (B ,C ())we candene preference relation based on it, the so-called revealed preference relation, .

DEFINITION 5. Given a choice structure (B ,C ()) the revealed preference relation is dened by x y if and only if there is some B in B such that x, y are in B and x is inC ( B).

We read x y as x is revealed at least as good as y o r x is revealed weakly preferredto y. Note that may not be either transitive or complete. We also say that x is revealed(strictly) preferred to y if there is some B in B such that x, y are in B and x is in C ( B) and y is not. We can then restate Denition 3 as follows.

DEFINITION 1 . The choice structure (B ,C ())satises the weak axiom of revealedpreference if whenever x is revealed at least as good as y then y is not revealed preferred to x.

There are two questions we might ask regarding the relationship between the twoapproaches.

(1) If the decision-maker has a rational preference ordering do the preference max-imising choices necessarily generate a choice structure that satises the weak axiom?

(2) If the decision-makers choices facing a family of budget sets B are representedby a choice structure (B ,C ()) that satises the weak axiom is there necessarilya rational preference relation that is consistent with these choices?


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The answer to the rst question is unambiguously yes. To answer the second ques-tion in the afrmative requires some assumptions on B .

To analyse the problem formally we need a bit more notation.Suppose that the decision-maker has a rational preference relation on X . For any

nonempty subset of alternatives B X we let C ( B, ) be her preference maximisingchoices, i.e.,

C ( B, ) = { x B | x y for every y B}.

PROPOSITION 2. Suppose that is a rational preference relation and that B is suchthat C ( B, ) is nonempty for all B in B . Then the choice structure (B ,C (, )) satisesthe weak axiom of revealed preference.

PROOF. Suppose that for some B in B we have x, y in B and x in C ( B, ). By thedenition of C ( B, ) this means that x y. Now consider some other B in B with x, y in B and y in C ( B , ). Thus y z for all z in B . But since x is in C ( B, ) and y is in B we

have x y. Also since is rational it is transitive and so x z for all z in B . And so x isin C ( B , ) as well, as we require.

The relation between C (, ) and the preferences is quite clear and is not problem-atic. The relation between a choice structure (B ,C ())and the revealed preference relation is a little less clear, even if is rational. Can you think of an example in which

is rational but doesnt seem to quite reect the choice structure? [Hint: Try to have

contain all ordered pairs, and yet not have the choice structure be that of complete indif-ference.] To say something systematic it is convenient to say precisely what we mean by apreference relation representing or rationalising a choice structure.

DEFINITION 6. Given a choice structure (B ,C ()) , we say that the rational preferencerelation rationalises C () relative to B if C ( B) = C ( B, ) for all B in B , that is,if

(and B ) generates the choice structure (B ,C ()) .We are now in a position to answer the second question raised earlier.

PROPOSITION 3. If (B ,C ()) is a choice structure satisfying the weak axiom and if B includes all nonempty subsets of X containing three or fewer elements then there is arational preference relation that rationalisesC () relative to B . that is, C ( B) = C ( B, ) for all B in B . Furthermore this rational preference relation is the only preference relationthat does so.

The proof of this proposition is not too difcult. You can read it in ? if you are up toit. (Its Proposition 1.D.2 there.) We shall not prove it here.

5. Representing Preferences by Utility Functions

In the previous sections we have discussed choice and preference, two extremely im-portant concepts in economics. However there is another concept that, if not more im-portant, is more widely used and better known, that of utility. How is the idea of utilityrelated to the two concepts that we have discussed already. Just as we saw in the previoussection that preferences could rationalise or represent a choice structure so too a utilityfunction can represent a preference relation. In most modern choice theory this is all autility function does.

A utility function u( x) assigns a real number to each element of the set of alternatives X . We interpret the utility function to mean that the decision-maker prefers one alternativeto another if and only if the rst is assigned a higher utility level.

DEFINITION 7. A function u : X R is a utility function representing the preferencerelation if, for all x and y in X , x y if and only if u( x) u( y).


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6. CONSUMER OPTIMISATION 27

We can now ask about the relationship between the preference relation being rationaland it being represented by a utility function. Again, in one direction we get a very stark

answer.PROPOSITION 4. If a preference relation can be represented by a utility function

then it is rational.

Again, we shall not prove this here. Its not very difcult. You can nd the proof in ?and one of the homework exercises guides you through the proof.

There are a number of circumstances in which the converse is true. In particular, if theset of alternatives X is nite then any rational preference relation can be represented by autility function. In cases in which X is not nite one needs additional conditions in orderto guarantee that the preferences cab be so represented. We shall return to this question alittle when we discuss the consumers decision problem in a little more detail in the nextsection.

6. Consumer Behaviour: Optimisation Subject to a Budget Constraint

We now look at a special case of our theory so far, that of a consumer choosing whichbundle of commodities to consume. For much of our discussion we shall focus on thecase in which there are only two commodities. When we look at the more general case weshall denote the number of commodities by L and index the commodities by = 1, . . . , L.Thus the vector ( x1, x2, . . . , x , . . . , x L) represents the situation in which the consumer con-sumes (or perhaps, depending on our interpretation, plans to consume) the quantity x1 of commodity 1, x2 of commodity 2, and so on.

In discussing the consumers problem we shall normally assume that the consumermust consume nonnegative amounts of each commodity. If we make no other assumptionsthen the set of alternatives X would be

X = R L+ = {( x1, x2, . . . , x L) | x 0 for all }.

If we made other restrictions on what the consumption bundles could be we might obtaina strict subset of R L+ . A number of such examples are given in the homework.

Having discussed the set of alternatives we turn now to look at budget sets. Thoughin many examples actual budget sets may well be more complicated we shall restrict ourattention here to what are known as Walrasian or competitive budget sets.

DEFINITION 8. Given prices of commodities p = ( p1, p2, . . . , p L) in R L++ (the strictly

positive L-vectors) and wealth w 0 the Walrasian, or competitive budget set B ( p, w) ={ x X | p x w} is the set of all feasible consumption bundles for a consumer with wealthw facing prices p.

For the consumers problem the family of budget sets B is precisely the Walrasianbudget sets. That is,

B = { B( p, w) | pR L++ and w 0}.

We shall dene the Walrasian demand correspondence x( p, w) to be the rule that asso-ciates to any pricewealth pair ( p, w) those consumption bundles chosen by the consumergiven the budget set B( p, w). Now, since we are thinking of only the budget set as relevantfor the consumer and not anything else about the pricewealth pair this means that anychanges to the price-wealth pair that do not change the budget set should not change thedemanded bundles. This is reected in the assumption that the Walrasian demand corre-spondence is homogeneous of degree zero .

DEFINITION 9. The Walrasian demand correspondence is homogeneous of degree zero if x( p, w) = x( p, w) for any p R L++ , w 0 and > 0.


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We think of the commodity space X as including all possible uses of the consumerswealth. Thus if the consumer were not to spend all of her wealth she would be forgoing the

possibility of increasing her consumption with no alternate use of her wealth. We couldallow such situations, but it would limit what we could say about the consumers behaviour.Recognising that it is a substantive assumption, though rather a mild one, we assume thatthe consumer spends all of her wealth. We call this assumption Walras law.

DEFINITION 10. The Walrasian demand correspondence satises Walras law if p x = w for any p R L++ , w 0 and x x( p, w).

We now assume that the Walrasian demand correspondence is single valued, that is,that it is a function. There are a number of essentially adding up restrictions implead by thehomogeneity of degree zero and Walras law. Consider the requirement of homogeneity x( p, w) x( p, w) = 0 for all > 0. Suppose we differentiate this with respect to andevaluate the result at = 1. We obtain the following result.

PROPOSITION 5. If the Walrasian demand function x ( p, w) is homogeneous of degree zero then for all p and w

L

k = 1

x ( p, w) pk

pk + x ( p, w)

ww = 0 for = 1, . . . , L.

If you know matrix notation you can say this a bit more simply.

D p x( p, w) p + Dw x( p, w)w = 0.

Now consider Walras law p x( p, w) = w for all p and w. Suppose we differentiatethis with respect to price. We obtain the following result.

PROPOSITION 6. If the Walrasian demand function x ( p, w) satises Walras law then for all p and w

L

= 1

p x ( p, w) pk

+ xk ( p, w) = 0 for k = 1, . . . , L.

Or in matrix notation p D p x( p, w) p + x( p, w)T = 0T.

Consider again Walras law p x( p, w) = w for all p and w and differentiate this timewith respect to w. We obtain the following result.

PROPOSITION 7. If the Walrasian demand function x ( p, w) satises Walras law then for all p and w

L

= 1

p x ( p, w)

w= 1.

Or in matrix notation p Dw x( p, w) p = 1.

DEFINITION 11. The Walrasian demand function x( p, w) satises the weak axiom of revealed preference if the following property holds for any two price wealth situations( p, w) and ( p , w :

If p x( p , w ) w and x( p , w ) = x( p, w) then p x( p, w) > w

Let us describe this in words. Suppose that we have had a change in prices and wealth(say from ( p, w) to ( p , w )). The weak axiom then says that if the newly chosen bundle isdifferent from the old bundle and the new bundle had been affordable in the old situationthen the previously chosen bundle must no longer be affordable.


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CHAPTER 4

Consumer Behaviour: Optimisation Subject to the BudgetConstraint

1. Constrained Maximisation

1.1. Lagrange Multipliers. Consider the problem of a consumer who seeks to dis-tribute his income across the purchase of the two goods that he consumes, subject to the

constraint that he spends no more than his total income. Let us denote the amount of therst good that he buys x1 and the amount of the second good x2, the prices of the twogoods p1 and p2 , and the consumers income y. The utility that the consumer obtains fromconsuming x1 units of good 1 and x2 of good two is denoted u( x1, x2). Thus the consumersproblem is to maximise u( x1, x2) subject to the constraint that p1 x1 + p2 x2 y. (We shallsoon write p1 x1 + p2 x2 = y, i.e., we shall assume that the consumer must spend all of hisincome.) Before discussing the solution of this problem lets write it in a more mathemat-ical way.

(5)max x1 , x2

u( x1, x2)

subject to p1 x1 + p2 x2 = y

We read this Choose x1

and x2

to maximise u( x1, x

2) subject to the constraint that p

1 x

1+

p2 x2 = y.Let us assume, as usual, that the indifference curves (i.e., the sets of points ( x1, x2) for

which u( x1, x2) is a constant) are convex to the origin. Let us also assume that the indif-ference curves are nice and smooth. Then the point ( x1, x2) that solves the maximisationproblem ( 5) is the point at which the indifference curve is tangent to the budget line asgiven in Figure 1.

One thing we can say about the solution is that at the point ( x1, x2) it must be true thatthe marginal utility with respect to good 1 divided by the price of good 1 must equal themarginal utility with respect to good 2 divided by the price of good 2. For if this werenot true then the consumer could, by decreasing the consumption of the good for whichthis ratio was lower and increasing the consumption of the other good, increase his utility.Marginal utilities are, of course, just the partial derivatives of the utility function. Thus wehave

(6) u x1

( x1, x2)

p1=

u x2

( x1, x2)

p2.

The argument we have just made seems very economic. It is easy to give an alternateargument that does not explicitly refer to the economic intuition. Let xu2 be the functionthat denes the indifference curve through the point ( x1, x2), i.e.,

u( x1, xu2( x1)) u u( x

1, x2).

Now, totally differentiating this identity gives

u

x1 ( x

1, xu

2( x

1)) +

u

x2 ( x

1, xu

2( x

1))

dxu2dx1 (

x1) =

0.

29


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30 4. CONSUMER BEHAVIOUR: OPTIMISATION SUBJECT TO THE BUDGET CONSTRAINT

E

T

d d

d d

d d

d d

d d

d d

d d d

q q q q q q q q q q q q q q q q

q q q q q q q q q q q q q q q

x2

x2

x1 x1

u( x1, x2) = u

p1 x1 + p2 x2 = y

Figure 1

That is,

dxu2dx

1

( x1) = u x1

( x1, xu2( x1))

u

x2 ( x

1, xu

2( x

1))

.

Now xu2( x1) = x2 . Thus the slope of the indifference curve at the point ( x1, x2)

dxu2dx1

( x1) = u x1

( x1, x2) u x2

( x1, x2).

Also, the slope of the budget line is p1 p2 . Combining these two results again gives result(6).

Since we also have another equation that ( x1, x2) must satisfy, viz

(7) p1 x1 + p2 x

2 = y

we have two equations in two unknowns and we can (if we know what the utility function

is and what p1, p2 , and y are) go happily away and solve the problem. (This isnt quite truebut we shall not go into that at this point.) What we shall develop is a systemic and usefulway to obtain the conditions ( 6) and (7). Let us rst denote the common value of the ratiosin (6) by . That is,

u x1

( x1, x2)

p1= =

u x2

( x1, x2)

p2and we can rewrite this and ( 7) as

(8)

u x1

( x1, x2) p1 = 0

u x2

( x1, x2) p2 = 0

y p1 x1 p2 x2 = 0.


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1. CONSTRAINED MAXIMISATION 31

Now we have three equations in x1, x2 , and the new articial or auxiliary variable . Againwe can, perhaps, solve these equations for x1, x2 , and . Consider the following function

(9) L ( x1, x2, ) = u( x1, x2) + ( y p1 x1 p2 x2)

This function is known as the Lagrangian. Now, if we calculate L x1 , L x2

, and, L , andset the results equal to zero we obtain exactly the equations given in ( 8). We now describethis technique in a somewhat more general way.

Suppose that we have the following maximisation problem

(10)max

x1 ,..., xn f ( x1, . . . , xn)

subject to g( x1, . . . , xn) = c

and we let

(11) L ( x1, . . . , xn , ) = f ( x1, . . . , xn) + (c g( x1, . . . , xn))

then if ( x1, . . . , xn) solves (10) there is a value of , say such that

L xi

( x1, . . . , xn , ) = 0 i = 1, . . . , n(12)

L

( x1, . . . , xn , ) = 0.(13)

Notice that the conditions ( 12) are precisely the rst order conditions for choosing x1, . . . , xn to maximise L , once has been chosen. This provides an intuition into thismethod of solving the constrained maximisation problem. In the constrained problem wehave told the decision maker that he must satisfy g( x1, . . . , xn) = c and that he should chooseamong all points that satisfy this constraint the point at which f ( x1, . . . , xn) is greatest. Wearrive at the same answer if we tell the decision maker to choose any point he wishes butthat for each unit by which he violates the constraint g( x1, . . . , xn) = c we shall take away units from his payoff. Of course we must be careful to choose to be the correct value. If we choose too small the decision maker may choose to violate his constrainte.g., if wemade the penalty for spending more than the consumers income very small the consumerwould choose to consume more goods than he could afford and to pay the penalty in utilityterms. On the other hand if we choose too large the decision maker may violate hisconstraint in the other direction, e.g., the consumer would choose not to spend any of hisincome and just receive units of utility for each unit of his income.

It is possible to give a more general statement of this technique, allowing for multiple

constraints. (Of course, we should always have fewer constraints than we have variables.)Suppose we have more than one constraint. Consider the problem

max x1 ,..., xn

f ( x1, . . . , xn)

subject to g1( x1, . . . , xn) = c1...

...

gm( x1, . . . , xn) = cm.

Again we construct the Lagrangian

(14)L ( x1, . . . , xn , 1, . . . , m) = f ( x1, . . . , xn)

+ 1(c1 g1( x1, . . . , xn)) + + m(cm gm( x1, . . . , xn))


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32 4. CONSUMER BEHAVIOUR: OPTIMISATION SUBJECT TO THE BUDGET CONSTRAINT

and again if ( x1, . . . , xn) solves (14) there are values of , say 1 , . . . , m such that

(15)

L xi ( x

1, . . . , x

n ,

1 , . . . ,

m) = 0 i = 1, . . . , n L j

( x1, . . . , xn , 1 , . . . , m) = 0 j = 1, . . . , m.

1.2. Caveats and Extensions. Notice that we have been referring to the set of condi-tions which a solution to the maximisation problem must satisfy. (We call such conditionsnecessary conditions .) So far we have not even claimed that there necessarily is a solutionto the maximisation problem. There are many examples of maximisation problems whichhave no solution. One example of an unconstrained problem with no solution is

(16) max x

2 x

maximise over the choice of x the function 2 x. Clearly the greater we make x the greater is2 x, and so, since there is no upper bound on x there is no maximum. Thus we might wantto restrict maximisation problems to those in which we choose x from some bounded set.Again, this is not enough. Consider the problem

(17) max0 x 1

1/ x .

The smaller we make x the greater is 1 / x and yet at zero 1 / x is not even dened. We coulddene the function to take on some value at zero, say 7. But then the function would notbe continuous. Or we could leave zero out of the feasible set for x, say 0 < x 1. Thenthe set of feasible x is not closed. Since there would obviously still be no solution to themaximisation problem in these cases we shall want to restrict maximisation problems tothose in which we choose x to maximise some continuous function from some closed (andbecause of the previous example) bounded set. (We call a set of numbers, or more generally

a set of vectors, that is both closed and bounded a compact set.) Is there anything else thatcould go wrong? No! The following result says that if the function to be maximised iscontinuous and the set over which we are choosing is both closed and bounded, i.e., iscompact, then there is a solution to the maximisation problem.

THEOREM 2 (The Weierstrauss Theorem). Let S be a compact set. Let f be a continu-ous function that takes each point in S to a real number. (We usually write: let f : S R becontinuous.) Then there is some x in S at which the function is maximised. More precisely,there is some x in S such that f ( x) f ( x) for any x in S.

Notice that in dening such compact sets we typically use inequalities, such as x 0.However in Section 1 we did not consider such constraints, but rather considered onlyequality constraints. However, even in the example of utility maximisation at the beginning

of Section 1.1, there were implicitly constraints on x1 and x2 of the form x1 0, x2 0.

A truly satisfactory treatment would make such constraints explicit. It is possible to ex-plicitly treat the maximisation problem with inequality constraints, at the price of a littleadditional complexity. We shall return to this question later in the book.

Also, notice that had we wished to solve a minimisation problem we could have trans-formed the problem into a maximisation problem by simply multiplying the objective func-tion by 1. That is, if we wish to minimise f ( x) we could do so by maximising f ( x).As an exercise write out the conditions analogous to the conditions ( 8) for the case that wewanted to minimise u( x). Notice that if x1 , x2 , and satisfy the original equations then x1 , x2 , and satisfy the new equations. Thus we cannot tell whether there is a maximumat ( x1 , x2) or a minimum. This corresponds to the fact that in the case of a function of asingle variable over an unconstrained domain at a maximum we require the rst derivative


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2. THE THEOREM OF THE MAXIMUM 33

to be zero, but that to know for sure that we have a maximum we must look at the secondderivative. We shall not develop the analogous conditions for the constrained problem with

many variables here. However, again, we shall return to it later in the book.

2. The Theorem of the Maximum

Often in economics we are not so much interested in what the solution to a particularmaximisation problem is but rather wish to know how the solution to a parameterisedproblem depends on the parameters. Thus in our rst example of utility maximisationwe might be interested not so much in what the solution to the maximisation problem iswhen p1 = 2, p2 = 7, and y = 25, but rather in how the solution depends on p1, p2, and y. (That is, we might be interested in the demand function.) Sometimes we shall also beinterested in how the maximised function depends on the parametersin the example howthe maximised utility depends on p1 , p2 , and y.

This raises a number of questions. In order for us to speak meaningfully of a de-

mand function it should be the case that the maximisation problem has a unique solution.Further, we would like to know if the demand function is continuousor even if it isdifferentiable. Consider again the problem ( 14), but this time let us explicitly add someparameters.

(18)

max x1 ,..., xn

f ( x1, . . . , xn , a 1, . . . , ak )

subject to g1( x1, . . . , xn , a 1, . . . , a k ) = c1...

...

gm( x1, . . . , xn , a 1, . . . , a k ) = cmIn order to be able to say whether or not the problem has a unique solution it is useful

to know something about the shape or curvature of the functions f and g. We say a functionis concave if for any two points in the domain of the function the value of function at aweighted average of the two points is greater than the weighted average of the value of thefunction at the two points. We say the function is convex if the value of the function at theaverage is less than the average of the values. The following denition makes this a littlemore explicit. (In both denitions x = ( x1, . . . , xn) is a vector.)

DEFINITION 12. A function f is concave if for any x and x with x = x and for any t such that 0 < t < 1 we have f (tx +( 1 t ) x ) t f ( x)+( 1 t ) f ( x ). The function is strictlyconcave if f (tx + ( 1 t ) x ) > t f ( x) + ( 1 t ) f ( x ).

A function f is convex if for any x and x with x = x and for any t such that 0 0there is > 0 such that if (a 1, . . . , a k ) is within of (a 1, . . . , ak ) then G(a 1, . . . , a k ) is within of G(a1, . . . , ak ).

It is, unfortunately, not the case that the continuity of the functions g j necessarily im-

plies the continuity of the feasible set. (Exercise 2 asks you to construct a counterexample.)REMARK 1. It is possible to dene two weaker notions of continuity, which we call

upper hemicontinuity and lower hemicontinuity. A correspondence is in fact continuous inthe way we have dened it if it is both upper hemicontinuous and lower hemicontinuous.

We are now in a position to state the Theorem of the Maximum. We assume that f isa continuous function, that G is a continuous correspondence, and that for any (a1, . . . , ak )the set G(a 1, . . . , a k ) is compact. The Weierstrauss Theorem thus guarantees that there is asolution to the maximisation problem ( 21) for any (a1, . . . , a k ).

THEOREM 4 (Theorem of the Maximum). Suppose that f ( x1, . . . , xn , a1, . . . , ak ) iscontinuous (in ( x1, . . . , xn , a 1, . . . , a k )), that G (a 1, . . . , a k ) is a continuous correspondence,and that for any (a 1, . . . , a k ) the set G (a 1, . . . , a k ) is compact. Then

(1) v(a 1, . . . , a k ) is continuous, and


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3. THE ENVELOPE THEOREM 35

(2) if ( x1(a 1, . . . , a k ), . . . , xn(a1, . . . , ak )) are (single valued) functions then they are

also continuous.

Later in the course we s

lecnotes econ analiyses

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