Post on 23-Oct-2016
description
Advanced Microeconomics – Assignment 1
Problem 1: Axioms on preferences
!! ≿ !!, !! ≻ !!
By definition, !! ≻ !! means !! ≿ !! but not !! ≿ !!
So, !! ≿ !! ≿ !!
By transitivity, this means !! ≿ !!
Now, suppose instead we have !! ≿ !!. Since !! ≿ !!, by transitivity !! ≿ !!
But there is a contradiction here because we know that !! ≿ !!
So, we have shown that by transitivity !! ≿ !! is true, and by contradiction !! ≿ !! cannot hold.
Therefore, we conclude !! ≻ !!
Problem 2: Utility Functions
1. The binary relation ≿ on the consumption set X is called a preference relation if it satisfies Axioms 1 and 2. Axiom 1 Completeness: For all !! and !! in X, either !! ≿ !! or !! ≿ !!.
a. Since bundle !! is better than !! whenever !! − !! ! ≥ !! −!! !, it is possible to describe all the relationships between bundles. This satisfies axiom 1.
Axiom 2 Transitivity: For any three elements !!, !! and !! in X, if !! ≿ !! and !! ≿ !!, then !! ≿ !!.
a. Suppose there was a third bundle !!, given !! is better than !! whenever !! − !! ! ≥ !! − !! !, by transitivity this means that !! ≿ !! if !! − !! ! ≥ !! − !! !. Therefore transitivity is satisfied.
Thus, ≿ represents preference relations.
2. Axiom 3 Continuity: For all ! ∈ ℝ!!, the ‘at least as good as’ set, ≿ (!),
and the ‘no better than’ set, ≾ (!), are closed in ℝ!!.
Axiom 4 Strict Monotonicity: For all !!, !! ∈ ℝ!
!, if !! ≥ !!, then !! ≿ !!, while if !! ≫ !!, then !! ≻ !!.
a. For Axiom 3, ≿ !! = {!! ∈ !|!! ≿ !!} when !! − !! ! ≤ !! − !! !
≾ !! = {!! ∈ !|!! ≾ !!} when !! − !! ! ≥ !! − !! ! This reads as; the upper contour set (!!) that is weakly preferred to set !! is closed when !! − !! ! ≤ !! − !! ! is satisfied, and the lower contour set (!!) that is no better than to set !! is closed when !! − !! ! ≥ !! − !! ! is satisfied. Based on this definition, the continuity axiom is met.
b. For Axiom 4, given ! ∈ [0,1] and that bundle !! is weakly preferred to !! whenever !! − !! ! ≥ !! − !! !, if !! contains 0.7 units and !! contains 0.6 units, !! ≿ !!, despite !! > !!. So in this case, monotonicity is not satisfied since more is not necessarily better. For axiom 4 to be satisfied, the set must be restricted to the region ! ∈ [0,0.5], so that the more of one bundle (e.g. !! > !!) necessitates in the preference of !! ≿ !!.
3. Theorem 1.1: If the binary relation ≿ is complete, transitive, continuous, and strictly monotonic, there exists a continuous real-‐valued function, !: ℝ!
! → ℝ, which represents ≿.
a. From part 2.2, it has been established that over the region ! ∈ [0,1], axiom 4 is violated. However, theorem 1.1 only states sufficient conditions for the existence of a !, which does not necessarily mean that the theorem does not still allow for a ! to exist.
4. In proving theorem 1.1, sets A and B are defined as: ! ≡ ! ≥ 0 !" ≿ ! ! ≡ {! ≥ 0|!" ≾ !} where ! ≡ (1), and !∗ ∈ ! ∩ ! will define ! ! !~! when ! ! = !∗ such that ! ∈ [0,1], and !! ≿ !! whenever !! − !! ! ≥ !! − !! !. The proof fails in determining the uniqueness of ! ! . Suppose there is only one number such that !!!~!~!!!, by transitivity this means !!!~!!!. But it is no longer true that holds when strict monotonicity is violated because !! = !!, !! < !! and !! > !! all can represent !!!~!!!.
5. From part 2.4, a utility function can be proven to exist. In the region of ! ∈ [0,1], since strict monotonicity is violated, set A is true when ! > ! is in the interval of [0,0.5] and when ! < ! is in the interval of [0.5,1]. Similarly for set B, it is true when ! < ! in the interval of [0,0.5] and when ! > ! is in the interval of [0.5,1], with all sets closed given the assumption of continuity. Suppose there is a ! that doesn’t exist in the set of A and B. From completeness, it must be the case that either !" ≿ ! or !" ≾ !. But this must mean that ! ∈ ! ∪ ! for either case to hold. Thus, by proof of contradiction, there is at least one !∗ that represents ! ! .