401-assign1-ans-14w

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Economics 401

Assignment 1Must be handed in, in class, Tuesday 14 January 2014

Answer all questions in the space provided. Please write clearly and concisely.

1. What does it mean to say that a utility function, u(·), represents a preference relationon some choice set X ? Prove that if u(·) represents preference relation , this preference

relation must be complete and transitive.

ANSWER

We will work with the “at least as good as” preference relation, , because this is thebasic one used in MWG. u(·), represents on some choice set X if

for all a, b ∈ X, a b if and only if u (a) ≥ u (b) .

Now we want to prove that if u(·) represents preference relation , this preference relationmust be complete and transitive.

Complete: complete means that for any a, b ∈ X either a b or b a or both. If u(·)represents preference relation then u(a) and u(b) exist and they are real numbers, whichare themselves complete. Either u(a) ≥ u(b) in which case a b or u(b) ≥ u(a) in whichcase b a, or possibly u(a) = u(b) in which case a ∼ b.

Transitive: transitive means that for any a,b,c ∈ X where we know a b and b c, weare allowed to deduce that a c. And, again, if u(·) represents preference relation on X the transitivity of will follow from the transitivity of the real numbers. Thus, a b andb c implies u(a) ≥ u(b) ≥ u(c). Since these are real numbers we know that u(a) ≥ u(c)and thus a c.

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2. Suppose X = 2+ and (a1, a2) (b1, b2) when a1 > b1, or a1 = b1 and a2 > b2. Is this

preference relation complete, transitive and continuous? Defend your answers carefully.

ANSWER

This is the lexicographic preference relation with good 1 dominant. It is complete andtransitive but not continuous.

Completeness

Let a and b be any two distinct points in 2+. If a1 > b1 then a b; if b1 > a1 then b a;

if a1 = b1 and a2 > b2 then a b; and, finally, if a1 = b1 and b2 > a2 then b a. Since thesefour cases are the only ones possible, must be complete.

Transitivity

Let a,b,c be any three elements in 2+ where a b and b c. We have to prove

that a c. By the definition of lexicographic preferences and the transitivity of the realnumbers, a1 ≥ c1. If a1 > c1 then a c; if a1 = c1 then a1 = b1 = c1 and then it must bethat a2 > b2 > c2 and thus again a c.

Continuity

Let

{xn}∞

n=1 ⊂ 2

+, {yn}∞

n=1 ⊂ 2

+, xn yn, for all n

limn →∞

xn = x and limn →∞

yn = y.

Then if is continuous x y. Use the following counterexample to show that is not

continuous.

xn = (1/n, 0) , x = (0, 0) , yn = (0, 1) , y = (0, 1)

Here

xn yn, for all n

But

y x.

3. Use mathematical notation to define the weak axiom of revealed preference. Statewhether or not each of the following choice structures satisfies WARP and defend youranswers.

(a) X = {a,b,c},B = {B1, B2, B3} , B1 = {a, b} , B2 = {b, c} , B3 = {c, a} , C (B1) ={a} , C (B2) = {b} , C (B3) = {c};

(b) X = {a,b,c},B = {B1, B2} , B1 = {a, b} , B2 = {a,b,c} , C (B1) = {a} , C (B2) = {c};

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(c) X = {a,b,c,d},B = {B1, B2} , B1 = {a,b,c} , B2 = {a,b,d} , C (B1) = {a, c} , C (B2) ={a, d};

ANSWER

WARP holds in some choice structure (B , C (·)) if:

Assume B1, B2 ∈ B , {a, b} ⊂ B1 ∩B2; a ∈ C (B1) and b ∈ C (B2) ⇒ a ∈ C (B2).It follows from this definition that WARP has no leverage unless our observations give us

at least two budget sets with two elements in common — say a and b — AND a is amongstthe best elements in one budget set and b is amongst the best elements in the other budgetset.

These assumptions are not met in any of the cases above so WARP is trivially satisfied.

4. MWG, 1.B.3

ANSWER

Given that u : X → represents on X we know:

for all x, y ∈ X, x y if and only if u (x) ≥ u (y) .

To complete the proof note that if f (·) is a strictly increasing function defined on thereal numbers then

for all x, y ∈ X, x y if and only if u (x) ≥ u (y) if and only if f (u (x)) ≥ f (u (y))

5. MWG, 1.B.4

ANSWER

Proof: Denote(∗) for all x, y ∈ X, x ∼ y if and only if u (x) = u (y)(∗∗) for all x, y ∈ X, x y if and only if u (x) > u (y)Given that (∗) and (∗∗) are true we need to prove the definition in MWG 1.B.3 holds.

Note that x y implies either x y in which case u (x) > u (y) or x ∼ y in which caseu (x) = u (y); combining the two options then x y implies u (x) ≥ u (y). Conversely,

u (x) ≥ u (y) implies either u (x) > u (y) in which case x y or u (x) = u (y) in which casex ∼ y. So combining the two options then u (x) ≥ u (y) implies x y.

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