Economics Vault

Microeconomics I #1: Several theorems in Choice Theory

2011-09-09T20:22:00.000-07:00

Another technical post before I switch back to my unfinished work on Freefall over the weekend. This time is something micro.

Some preliminaries:

Preference is a binary relation ≿ between x,y∈X, the set of all possible choices such that x≿y if and only if "x is at least as good as y".
Preference is complete if ∀x,y, either x≿y or y≿x or both. Preference is transitive if x≿y and y≿z imply x≿z.
Rational choice theory assumes that preferences are both complete and transitive.
Now, suppose we have established a preference ≿, and B is the set of available choices, define the choice rule C(B;≿)={x∈B: x≿y for all y∈B}. Notice C(B;≿) may not be singleton. When B is finite, C(B;≿) is non-empty; however, when B is infinite, C(B;≿) can be empty.
For a given choice rule C(X), the Houthaker's Axiom of Revealed Preference (HARP) states if x,y∈A∩B with x∈C(A) and y∈C(B), then x∈C(B) and y∈C(A).

Some propositions:

If a predetermined preference ≿ is complete and transitive over the choice set X, then the choice rule C(X;≿) satisfies HARP.
Now suppose we only have revealed preferences in the forms of non-empty choice rule C, then there is a complete and transitive preference ≿ with C(A)=C(A;≿) for all subsets A of X if and only if C satisfies HARP.

It is easier if we can represent a preference relation by a utility function.

A utility function, u, is a mapping from the choice set X to the real numbers. Hence values of the utility function cannot be positive or negative infinity.
We say a utility function u represents a preference relation ≿ (or the preference relation ≿ affords the utility function u) if there is an equivalence between x≿y and u(x)≥u(y).

Some more propositions on utility representation of preferences:

If the choice set X is finite, then every complete and transitive preference has a utility representation.

Proof: let n be the size of X, label elements in X with subscripts x_1, x_2,...,x_n, with x_n≿x_(n-1)≿...≿x_2≿x_1. Now let u(x_i)=i. Clearly u is a valid utility representation for the preference ≿. QED

If the choice set X is countable, then every complete and transitive preference has a utility representation.

Proof: Let X={x_1,x_2,...}. Define the set "No Better Than" of some element x, NBT(x)={y∈X: x≿y}. By observation, we know ∀x,y∈X, either NBT(x)⊆NBT(y) or NBT(y)⊆NBT(x). Now define the utility function, u(x)=∑(0.5)^n, where n is the subscript of x_n for all x_n∈NBT(x). Then it is simple to verify (1) u(x) is well-define, in particular is convergent; (2) u is indeed a valid utility representation for the preference ≿. QED

Note, the above proposition holds for any complete and transitive preferences over a countable choice set X, but not any arbitrary infinite set X. Consider for example X=[0,1]×[0,1], for any two points x=(a,b) and y=(c,d) in X, x≿y if either a>c or a=c but b≥d. This lexicographic preference has no proper utility representation even though it is complete and transitive.

Proof: By the definition of the preference, let α<β∈[0,1], then (β,1)≿(β,0)≿(α,1)≿(α,0). Suppose there is a utility representation u affording this preference, then u(β,1)≥u(β,0)≥u(α,1)≥u(α,0). In fact these inequalities hold strictly. Now consider the two disjoint open intervals, A=(u(α,0),u(α,1)) and B=(u(β,0),u(β,1)). Due to the density of the rational numbers, there exist rational numbers p,q such that p∈A and q∈B. But there are uncountable pairs of real numbers (α,β) with α<β∈[0,1], hence uncountable disjoint open intervals of those forms. This implies there are uncountable rational numbers in [0,1], but it contradicts with the fact that rational numbers are countable. Hence, by contradiction, such utility representation does not exist for this lexicographic preference defined on X. QED

Sufficient and necessary condition: ≿ over X has a utility representation if and only if ∃X*⊆X with X* being countable such that ∀x,y∈X with x≻y, ∃x*∈X* with x≿x*≻y.

Econometrics I #2: Borel sigma-algebra

2011-09-06T22:28:00.000-07:00

Borel σ-algebra is a special type of σ-algebra defined over a metric space. Here we only concern the Borel σ-algebra over the real line.

Definition 1: The Borel σ-algebra over the reals, denoted as B, is the σ-algebra generated by all open sets. Elements of Borel σ-algebra are called Borel sets.

Definition 2: A subset E of the reals is an F_σ set if it is a countable union of closed sets; E is a G_δ set if it is a countable intersection of open sets.

Some consequences:

Every G_δ set is a Borel set, by property of a σ-algebra. Since the complement of a G_δ set is an F_σ set, an F_σ set is a Borel set.
Semi-open intervals of the form [a,b) (equivalently (a,b]) are both F_σ and G_δ, since [a,b)=∪[a,b-1/n]=∩(a-1/n,b). Hence [a,b) is a Borel set for any real number a and b.

An alternative definition 3 of Borel σ-algebra (Marcelo Moreira, Columbia): The Borel σ-algebra (over the reals) is the smallest σ-algebra containing sets of the form (-∞,x].

I want to show that the two definitions are in fact equivalent.

Proof:
First we need to observe the following lemma.

Lemma:

Suppose A and B are two σ-algebras over the same space. Further, B is the σ-algebra generated by F, a set of subsets in that space. Then A contains B if and only if A contains its generators (i.e. A contains elements of F).

Main Proposition:

Let A denote the Borel σ-algebra in definition 1, i.e. A=σ(all open sets over reals). Let B denote the Borel σ-algebra in definition 3, i.e. B=σ((-∞,x]:x is real). Therefore, we need to prove A=B as sets and in fact we need to prove the two-way inclusions of the two sets.

First, to show B is a subset of A, by the lemma, we just need to show its generator (-∞,x] is in A. But clearly (-∞,x] is G_δ, hence Borel in definition 1, a.k.a an element of A.

Then, to show A is a subset of B, similarly by the lemma, we need to show an arbitrary open interval (a,b) is in B. Notice the semi-open interval (a,b]=(-∞,b]∩{complement of (-∞,a]} for all real number a and b, hence it is in B. Then the open interval (a,b)=∪(a,b-1/n]. Since every (a,b-1/n] is in B, the countable union is in B, a.k.a (a,b) is in B for any real number a and b. With easy and similar treatment of the special cases (-∞,b), (a,+∞) and (-∞,+∞), we conclude that every open set is in B. Hence by lemma, A is a subset of B.

With the two inclusions, A=B and hence the two definitions are equivalent. QED

Econometrics I #1: Sigma-algebra

2011-09-06T19:05:00.000-07:00

An interesting concept in this first econometrics lecture (besides the routine probability/statistics stuff) was the notion of the σ-algebra.

Definition: Let B denote a set of some subsets of the sample space Ω. Then B is a σ-algebra, if

Φ is an element of B;
If x is in B, then the complement of x in Ω is in B;
A countable (finite/countably infinite) union of elements in B is an element in B.

Some direct results from this definition:

B is a subset of the power set of Ω, hence the power set of Ω is a σ-algebra.
Ω is in B, since Ω is the complement of Φ.
The trivial σ-algebra would be the set {Φ,Ω}.
A countable intersection of elements in B is another element in B, since this intersection is equivalent to the complement of the countable union of the complements of those elements. (De Morgan's Law)

For an arbitrary collection F of subsets of Ω, there exists a unique smallest σ-algebra B that contains F. B is called the σ-algebra generated by F, or σ(F).

Proof:

First the proof of the following lemma.

Lemma: For a fixed Ω, any nonempty intersection of its σ-algebras is a σ-algebra.

Clearly the intersection is a subset of the power set of Ω, containing Φ. (Property 1 checked) If x is an element in the intersection, x is an element of all the σ-algebras, and so is x-complement. Hence x-complement is in the intersection. (Property 2 checked) Similarly, take a countable union of elements in the intersection, since all these elements are in the intersection, they are in every σ-algebras, and so is their union. Hence this union is in the intersection. (Property 3 checked). QED

Main Proposition:

Let A be the set of all σ-algebras of Ω containing F. Clearly A is not empty because the power set of Ω is in A. Then consider the intersection of all elements in A, call it G. G is apparently nonempty. Since every element of A contains F, G contains F. Since every element is a σ-algebra, by the lemma, G is a σ-algebra. And lastly, every element in A contains G, hence G is the smallest element in A. Or put it in another way, G is the smallest σ-algebra containing F, a.k.a G=σ(F). QED

This is the first post concerning some technical stuff, and I find it quite interesting. I will post more on Borel sigma-algebra the next time. Enjoy.

On Freefall #1

2011-09-02T21:08:00.000-07:00

In this book, Joseph E. Stiglitz provided a holistic and in-depth analysis of the 2008 Great Recession. In this post (and many following ones), I would like to recapitulate the main theses of the book with my interpretations.

The causes

Many critics and commentators held Wall Streeters, and in particular their insatiable greed, responsible for this economic epidemic. However, Stiglitz believed that Wall Streeters were no greedier than an ordinary American. He saw far deeper causes than greed - a part of human nature.

Misaligned Incentives. This is a fundamental problem underlying many other problems in the financial sector. The banks and other financial intermediaries were supposed to be the middlemen between lenders and borrowers, easing the process of searching and matching and providing some basic functions such as credit assessment. However, these financial institutions had no incentive to ease this process by reducing transaction costs because these costs were indeed revenues for them. One such example was the failure to provide an efficient Electronic Payment System which was absolutely viable given the current advancement in technology. Similarly, mortgage dealers, who saw extra profits from collecting mortgage restructuring fees, would encourage borrowers to take on mortgages beyond their ability to repay and later offer them mortgage refinancing.
Conflicting Interests. When commercial banks were allowed to merge with investment banks, a severe problem of conflicting interests arose. Typically commercial banks should be more risk averse since they were responsible for their depositors' savings. On the other hand, investment banks tended to be more risk taking as they sought more profits. When these two different types of banks merged, the new entities, with more funds to play with, often undertook many gambling behaviors, putting depositors' money at stake.
Ill-conceived Financial Innovations. Innovations were supposed to bring more good than evil to the society as a whole. In particular, financial innovations should help people manage risk better and make the process easier. However, due to the mismatch between duties and incentives of the financial institutions, many complicated financial innovations such as collateralized debt obligations (CDO) and credit default swaps (CDS) in fact increased the risks enormously (due to the hidden systematic risks that were not accounted for by their ratings) and rendered risk assessment almost impossible (due to their complexity). These innovations helped build record high accounting profits, but the institutions' actual positions were elusive.
Deregulation. Since Clinton's term, special interest groups in the financial sector had gained enough political power and successfully shielded the industry from governmental scrutiny. Furthermore, the regulators of the various financial institutions were themselves from Wall Street; they often shared a common interest with those being monitored. This lack of effective regulation made possible all these crafty innovations and mergers of commercial and investment banks on top of many illicit fraudulent behaviors.
Loose Monetary Policies. Federal Reserve had long kept the interest rate at low level, flooding the economy with liquidity. This discouraged savings and encouraged more risky investments. In particular, financing through mortgage and mortgage restructuring appeared very lucrative when the interest rate was low, and helped sustain a normally unsustainable level of consumption.
False Belief in Free Markets. The fervent belief in free markets was the ideological cause of this crisis. Free market fundamentalists were predominant in both Wall Street and Washington D.C. They supported the deregulation movement which eventually led to an over-sized, inefficient and irresponsible financial sector.

Opening Post

2011-09-01T16:23:00.000-07:00

As suggested by the name of this blog, I intend to archive all my economics readings, thoughts and writings here for today's record and tomorrow's reference, in an efficient and orderly manner.

It is primarily a personal blog, but critical yet constructive comments are welcome!