lec1-40001-2014(3) (1)

Consumer Theory

ECON40001/90063

March 3-5, 2014

ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 1 / 24

Goals

Existence of utility function

Existence and uniqueness of Marshallian (=empirically observable)demand functions.

Constrained Optimization (Lagrange’s method).


Consumption Set (Choice Set)

n goods are indexed i = 1, .., n;

Quantity of good i is denoted xi , xi ∈ R+;

A typical vector of goods the individual can consume, a consumptionbundle, is denoted

x = (x1, ..., xi , .., xn) ∈ Rn+(≡ R+ × ...× R+︸︷︷︸

=n times

);

X ≡ Rn+ is the consumption set. Note that we assume X ≡ Rn

+

because we think of a consumption set as all bundles we are able tothink about (e.g. me being F1 driver), rather than something that isachievable.

R+ is the set of all non-negative real numbers;

In x, bold implies that it is a vector.


Primitives: Preferences (Economics)

Faced with the choice between x1 and x2, an individual is able to expressher preferences, e.g. x1 � x2 (x1 is weakly preferred to x2; weakly means

that an individual may be indifferent between x1 and x2.)We feel that there are certain regularities in how people make their choices.We will try to capture these regularities in our mathematical theory.

Source: wikipedia


Primitives: Preferences (Mathematics)

Consumption set X

An abstract binary relation � on set X

Indifference: x1 � x2 and x2 � x1. Denoted as x1 ∼ x2.Strict preference: x1 � x2, but not x2 � x1. Denoted as x1 � x2.

Note: Speaking Economics, you should think of x1 � x2 as a YES answer to thequestion “Do you weakly prefer x1 to x2?” and you should think of x1 ∼ x2 as aYES answer to the question “Do you weakly prefer x1 to x2?” AND a YESanswer to the question “Do you weakly prefer x2 to x1?”.You should not think of x1 ∼ x2 as a YES answer to the question “Are youindifferent between x1 and x2. If you do, it implies that I introduced two wiggles,� and ∼. If so, I need to have axioms both on � and on ∼. Axioms are theobjects that fill these wiggles with meaning; before I impose axioms, these wigglesare completely meaningless objects.

I do not want many axioms. I would rather describe (via axioms) a single wiggle,

and then tell you what ∼ and � means using a single well-defined wiggle.


Primitives: Axioms on �.In English, there are some reqularities in how we answer to the question“Do you weakly prefer x1 to x2?” Before we impose any axioms on �, inMathematics there are no regularities; in fact, before we impose axioms wecan write f (x1, x2) and, mathematically, it would make no difference.

Axiom 1, Completeness: For all x1, x2, either x1 � x2 or x2 � x1 orboth.

Axiom 2, Transitivity: For any three bundles x1, x2, and x3, if x1 � x2

and x2 � x3, then x1 � x3.

For an abstract binary relation �, if completeness and transitivity issatisfied, we will call the binary relation preferences.

Note: what we done here is that we came a step closer matching our

mathematical notation, �, to what me mean in English when we say

“preferences”.


Axioms 1 and 2: discussion

Completeness: what do you like better, a live chicken or a boiled egg?

Failure of transitivity: (redwine, cheese) � (whitewine, cheese);(beer , cheese) � (redwine, cheese), (whitewine, cheese) � (beer , cheese).

Note that (redwine,meat) � (whitewine,meat);

(beer , crackers) � (redwine, crackers), (whitewine, cheese) � (beer , cheese) does

not violate transitivity.

However, there is a major problem with an individual who violatestransitivity in a consistent way (not as a mistake)


Violation of transitivity leads to a money pump

x1 � x2, x2 � x3, x3 � x1

Start with x3; a penny for x2; a penny for x1; a penny for x3.The same bundle, but three pennies less. Repeat infinitely.


A note on Sneetches

Preferences of individual P (initially a plain-belly sneetch): same belly asindividual 2. Preferences of individual S (initially a star-belly sneetch):belly that is different from individual 1.

Formal model:X = {0, 1} × {0, 1} (The first number refers to whether a plain-belly-type of

sneetch has a star (1) or not (0) and the second number refers to whether a

star-belly sneetch has a star (1) or not (0). The consumption bundle (1, 1) means

that both sneetches have stars.)

(1, 1) �P (0, 1) and (0, 0) �P (1, 0) These are preferences of a plain-belly

sneetch. This is indicated by superscript P. Note that these preferences arenot completely specified; other combinations of 0 and 1 are possible.(1, 0) �S (1, 1) and (0, 1) �S (1, 1)

There is no violation of transitivity. Note that we do not compare plain-belly

and star-belly sneetches. They are two different individuals and we check the

violation of transitivity separately for each of them. Further details on the story:

Dr. Seuss, “Sneetches”ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 9 / 24

Axioms 3 and 4, I

Axioms 3 and 4 are more technical axioms. If they are violated, someresults can still be recovered. Hence, we do not discuss them.

Axiom 3, Continuity: there are no sudden changes in preferences, such that1 orange is better than 0.99999 apple, but 1 apple is better than 1 orange.

Mathematically: consider a sequence of pieces of apples, 0.9; 0.99; 0.999;0.9999. Suppose that each element of this sequence is weakly worse thanan orange. This sequence converges to a whole apple. Continuity wouldthen imply that an apple is weakly worse than an orange.


Axioms 3 and 4, II

Mathematically: consider a sequence of pieces of apples, 0.9; 0.99; 0.999; 0.9999.

Suppose that each element of this sequence is weakly worse than an orange. This

sequence converges to a whole apple. Continuity would then imply that an apple

is weakly worse than an orange.

Formally: Consider a sequence yn ∈ X such that x � yn and yn → y.Then x � y. Consider a sequence zn ∈ X such that zn � x and zn → z.Then z � y.

Alternative definition (as in JR): for any x ∈ X , the sets� (x) = {y ∈ X |x � y} (lower contour set) and � (x) = {z ∈ X |z � x}(upper contour set) are closed.

If it is not obvious to you that these two definitions are equivalent, please have alook at the definition of a closed set and convince yourself that this is true. Wewill use open/closed sets later, so it would be useful to be familiar with theconcepts.

Notation: {a ∈ A|B} means all a from A such that B is true.ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 11 / 24

Axioms 3 and 4, III

Axiom 4, strict monotonicity: For all x0, x1 ∈ X , if x0 ≥ x1, then x0 � x1,while if x0 >> x1, then x0 � x1.

Note that Axiom 4 allows an individual not to care about a particulargood; she can be indifferent between (3, 1) and (3, 5), but she cannotprefer (3, 1) to (3, 5).

A note on ≥, >>x0 and x1 are vectors. How do we compare (3, 1) and (1, 3)?

We say that x0 ≥ x1 if some components of x0 are larger than components of x1

and none are smaller. We say that x0 >> x1 if all components of x0 are larger

than components of x1. Neither applies to (3, 1) and (1, 3); I have defined a

partial order in Rn.


Utility function as representation of preferences

It is not very convenient to work with an abstract binary relation (e.g.maximisation). Can we simplify the representation of preferences?

Utility function u : X → R represent preference relation �, if for eachx0, x1 ∈ Rn

+, u(x0) ≥ u(x1)⇔ x0 � x1

Note. Suppose u(x) = x1 + x2 represents preference relation �. Let uslook at v(x) = x21 + 2x1x2 + x2. Utility function v must represent thesame preference relation because

x21 + 2x1x2 + x2 = (x1 + x2)2

For any two bundles x0, x1, whenever u(x0) > u(x1), we know that

v(x0) =[u(x0)

]2>[u(x1)

]2= v(x1)

For any increasing function f : R→ R, u(x) and f (u(x)) represent thesame preference relation (more details: Theorem 1.2 in JR).

“How to make a person twice as happy” does not make sense. “How much ofgood 2 is needed to compensate for the loss of one unit of good 1” makes sense.

f : A→ B: f takes an element from A and maps it to an element in B, f (a) = b.ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 13 / 24

Existence of utility function, proof, step 1

Theorem (Theorem 1.1, JR)

If preference relation � is continuous and strictly monotonic, there exists acontinuous function u : X → R, which represents �.

Proof outline:

1 Pick a vector e, e.g. e = (1, 1).2 For any bundle x, find a number, such that a person is indifferent

between x and a corresponding fraction of e. For example, how (2, 3)and (1, 1) compare? There may be a number, e.g. 2.5, such that aperson is indifferent between (2, 3) and (2.5, 2.5).Why do we need this? A utility function needs to map into R, so weare looking for a number.

3 Once we have found a number for each x, we call it a utility functionu. We need to check that u indeed represent preferences �; that is,x1 � x2 if and only if u(x1) ≥ u(x2), where u(x1) and u(x2) are thenumbers we have constructed.

This is an example of a constructive proof.ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 14 / 24

With Axioms 1-4, continuous utility function exists



Idea of the existence proof: pick a bundle e (for example (1, 1, . . . , 1)) and

measure any other bundle x in relation to this bundle (how much of bundle e is

needed to make an individual indifferent between it and x?)

Why is it not obvious? Consider lexicographic preferences: for any twobundles x0, x1 ∈ R2

+, x0 � x1 if

x01 > x11 orif x01 = x11 , then x02 > x12

How would you measure bundle (1, 3) in terms of the bundle (1, 1)? Weknow that (1, 3) � (1, 1), but if we improve (1, 1) a bit, e.g.(1.001, 1.001), suddenly (1.001, 1.001) � (1, 3) (example hints at the

importance of continuity).ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 15 / 24

Existence of utility function, proof, step 1



Proof:For each x, define u(x) ∈ R (note, this is just a number) as a number suchthat u(x)e ∼ x.

Is u(x) well defined? That is, does it exist for any x and is it unique (ifnot, we would need to specify a selection rule to pick a particular number)


Number u(x) exists, I

Consider two sets, A = {t|te � x} and B = {t|x � te}. Note that bothare the sets in R. We would intuitively think that one such set is [τ,∞)and another is [0, τ ] and τ would be exactly u(x). We need to show itusing axioms.Observation 1: if we can show that there is t such that t ∈ A ∩ B, then twould define u(x). We show that such t exists in three steps:


Number u(x) exists, II

1 Note that strict monotonicity implies that if te � x, and τ > t, it must bethat τe � x. Also, if x � te and τ < t, it must be that x � τe.

2 From step 1, it follows that set A must be of the form (t,∞) or [t,∞),where t is an arbitrary number. From continuity, the set A must be closed,so it is [t,∞) (I am sloppy here, because the sets A and � (x) are not thesame and we assumed that � (x) is closed, not A). Set B is [0, t].

3 Suppose t < t; that is, there is t such that it is not in A or in B. Fromcompleteness, it must be the case that either (i) te � x or (ii) x � te. Yet,if (i) holds, t must belong to A and if (ii) holds, it must belong to B. Thus,we conclude that t ≥ t t ≥ t

4 As t ≥ t t ≥ t, t (as well as t) belongs to both A and B. The number tdefines u(x).

In summary: for every x we have found a way to associate a number. Note, that

we still have to show that these numbers represent preferences �. Before we turn

to that, we need to show that if we pick t, we would not get a different number.ECON40001/90063 L1-2, Consumer Theory March 3-5, 2014 18 / 24

Number u(x) is unique

Suppose that there are two numbers, t and t such that te ∼ x ∼ te. Bytransitivity, te ∼ te. By strict monotonicity, t = t (Note how it would not

be true if, at some point, an individual would not have wanted any more goods)


u(x) represent preferences �

Consider two bundles, x1 and x2. We have devised a way to construct anumber for each of these bundles, u(x1) and u(x2). Can we be sure thatx1 � x2 if and only if (denoted by ⇔) u(x1) ≥> u(x2).

By construction: u(x1)e ∼ x1 � x2 ∼ u(x2)e. By transitivity,u(x1)e � u(x2)e. By strict monotonicity, u(x1) ≥ u(x2). Hence, ifx1 � x2, u(x1) ≥ u(x2). To show the reverse direction, we repeat the stepsin reverse order.

We skip continuity argument.


Lexicographic preferences, Axioms 1, 2 and 4

We conjectured that lexicographic preferences � on R2 cannot berepresented by a continuous utility function. This conjecture is correct, butrequires a proof. We will check that lexicographic preferences do notsatisfy the conditions of Theorem 1.1 and find the place where the prooffails. Note that this does not prove that lexicographic preferences cannotbe represented, because Theorem 1.1 is not if and only if statement.

1 � is a preference relation.1 Completeness: for any two bundles, we have specified a procedure how

to compare them2 Transitivity: Suppose there are three bundles, x1, x2, x3 such that

x1 � x2 and x2 � x3. We need to show that x1 � x3.If x1 � x2, it means that either x11 ≥ x21 and, if x11 = x21 , x12 ≥ x22 .If x2 � x3, it means that either x21 ≥ x31 and, if x21 = x31 , x22 ≥ x32 .Combining these two observations, we see that it must be the case thatx11 ≥ x31 and, if x11 = x31 , then x12 ≥ x32 . That means that x1 � x3.

2 Strict monotonicity is also satisfied (trivially).


Lexicographic preferences, Axiom 3

Lexicographic preferences are not continuous.Consider set � (1, 3) = {(x1, x2) ∈ R2|x1 ≤ 1 or x1 = 1, x2 ≤ 3}A compliment to this set is C = {(x1, x2) ∈ R2|x1 > 1 or x1 = 1, x2 > 3}.This is not an open set; that is, we cannot find a small ball around anypoint that is contained within the set.

Consider a point (1.01, 2). Take a ball of radius 0.001 around thispoint. Note that the “worst” point for an agent within this ball is(1.009, 1.999). Note that this point belongs to C. Hence, all otherpoints (which are better) also belong to C.

Consider a point (1, 4). Take a ball of radius ε around this point.Note that the “worst” point for an agent within this ball is(1− ε, 4− ε). Note that, for any ε, this point does not belong to C.Hence, the set C is not open and the set � (1, 3) is not closed.


Lexicographic preferences, failure of the proof

In step 2 on page 16 we use continuity to show that set A is [t,∞). Forpoint (1, 4), this set is, actually, (1,∞) because (1, 4) � e, but for anyγ > 1, γe � (1, 4).

Note that if we do not require that u maps into R, but, instead, require itto map into R2, then the function is already given, which is these numbersthemselves. That is, x = (x1, x2) ∈ X maps into u(x) = (x1, x2) ∈ R2.


Working with proofs

Check where assumptions are used. In this proof, I have put an emphasisevery time I have used a particular assumption. We can ask a question:what would happen if we do not make this assumption? Why the proofwould fail?

Consider strict monotonicity in step 1. What do I need it for? “Havingmore of everything is better”. Note that this uses only the second part ofstrict monotonicity. We do not need, in this particular place, theassumption that giving more of a single good cannot make you worse.

Working out the proof with a particular example is very useful (takingsome weired case, such as lexicographic preferences, that is expected tofail somewhere, is even more useful).


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