Some Illustrations of Econometric Problems

Topic1

Econometrics attempts to measure quantitatively the concepts

developed by Economic theory

and use the measures to prove or disprove the latter.

Problem1: Estimating the demand curve of a product and measuring the price elasticity of demand at a single point

Step1:Collection of data (after sortingvarious problems associated with it. )

One additional problem: How do you Know that this data is suitable for estimating a demand curve?

The data needs to come from a period such that consumer income prices of related goods and consumer tastes and preferenceshad all remained constant.

Solution: Adjust data and/or throw some of it out

This is known as theIdentification Problem.

Step2: Identify that

PED d(lnQ)/d(lnP)

Step3: Change the numbers in the dataset to the natural log form.

That is, change P =2 to lnP = ln2 etc.

Step4: Propose the regression model

Recognize that = dlnQ/dlnP = PED

lnQ = lnP +

Step5:  Impose the restrictions of the Classical Linear Regression Model

Perform a linear regression of lnQ on lnP and estimate

Problem2: Do we suffer from money illusion? Testing the homogeneity property of degree 0 of a demand function

Theory :Demand stays unchanged if all prices as well as consumer income changeby the same proportion.

The rational consumer does not suffer from money illusion.

The demand function is homogeneous of degree 0.

Procedure of Test:

Step 1: Estimate

lnQ = lnP + lnp1 + lnp2 +

….+ n lnpn + lnY

Test the hypothesis :

n

Problem3: Does a production function exhibit Constant, Increasing or Decreasing returns to scale?

A Cobb-Douglas production function:

Q = ALaKb A, a, b >0

CRS if a+b =1, IRS if a+b > 1 DRS otherwise.

Step1: Rewrite the production function as

 lnQ = lnA + alnL + blnK

Step2: Collect data on Q, L and K  

Step3: Transform each number to its natural log form

Step4: Run a linear regression of lnQ on lnL and lnK and estimate the coefficients a and b.

Step5: Test the hypotheses H0 : a+b =1 versus H1: a+b >1 ; and/or H0 : a+b =1 versus H1: a+b < 1 

Diagnostic tests

An airliner suspects that the demand relationship post September 11 is not the same as it was before

How does it verify this?

Chow test

Use the sum of squared errors or squared residuals, or RSS, to evaluate how good an estimated regression line

High RSS means poor fit and vice versa

Idea:

If the old model is no longer applicable then the use of newly acquired data will produce a larger RSS, compared to the original value of the RSS

So reject the null hypothesis that the demand is unchanged if the new RSS is too large compared to the old value

A statistic called an F-statistic and a distribution called an F-distribution

is used to quantify the notion of ‘too large’

Revision of Probability Theory

Topic2

Suppose that an experiment is scheduled to be undertaken

What is the chance of a success?

What is the chance of a failure?

Success and Failure are called Outcomes of the experiment or Events

Events may be made up of elementaryevents

Experiment: Tossing a dice

An elementary event : Number 3 shows up

An event : A number less than 3 shows up

Question: What is the probability of getting a 3?

Answer: 1/6 (assuming all six outcomes are equally likely )

This approach is known asClassical Probability

If we could not assume that all the eventswere equally likely, we might proceed as follows:

If the experiment was done a large number of times, say 100, and number 3 came up 21 times, then(Probability of getting a 3) = 21/100

This is the Relative Frequency approachto assess probability

Assigning probability values according to One’s own beliefs is the Subjective probability method

We shall follow the Relative Frequency approach

That is, use past data to assess probability

Probability Theory  The probability of an event e is

the number   P(e) = f/N where

f is the frequency of the event occurring N is the total frequency and N is ‘large’.

  

The sample space S is the grey area = 1

Event A: Red oval

Event B: Blue Triangle

A’: Everything that is still grey (not A)

Red and White and Blue: A or B (A B)

White Area: A and B (A B)

A and B are not mutually exclusive

The sample space S is the grey area = 1

Event A: Red oval Event B: Blue Triangle

A and B are mutually exclusive

Axiomatic Probability is a branch of probability theory based on the following axioms.

(1) 0 P(e) 1 (2) If e, f are a pair of events that are

mutually exclusive then probability both e and f occur is zero

P(e and f) = 0  

(3) P(e) = 1 – P(e’)

  where e’ is the event “ not e”

  (4) P(S) = 1

  that is, the sample space S contains all events that can possibly occur

  (5) P() = 0 where is the non-event

  That is, an event not contained within S will not occur.

= P(A or B) + P(A and B)

P(B) = Triangle Area P(A) = Oval Area

P(A or B) = r + w + b P(A and B) = w So, P(A) + P(B)

= r + w + b + w

The Addition Rule for 2 events A and B

P(A or B) = P(A) + P(B) - P( A and B)

Probability Distributions

Outcome

Deterministic Random

Non-numeric Numeric

• Throwing a dice and noting the number on the side up has a numeric outcome.

• Let Y be “the result of throwing a dice”

• Y is a random variable because Y can take any of the values 1,2,3,4,5 and 6.

The probability distribution of Y (assuming a fair dice) is given by the Table below:

Y= y P(Y=y) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

Formally, we denote by xi (for i = 1,2,

….n) the possible values taken by the random variable X. If p(xi) is the

probability assigned to xi, then

 p(xi) = 1 (1)

 

The Expected Value of a random variable Y ( E(Y) ) is the value the variable is most likely to take, on averageThe Expected Value therefore is also called the Average or Mean of the Probability Distribution.

• The Standard Deviation of a random

• variable Y ( Ymeasures its dispersion around the expected value.

• The Variance (2Y) is the square of the

standard deviation.

Expectation:

The Expected value of X, written E(X) is the weighted average of the values X can take. Using notations,

 E(X) ≡p(xi)xi (2)

• Calculations : For the probability distribution

• discussed,

Y= y P(Y=y) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

E(Y) = (1/6) *1 +(1/6) * 2 +(1/6) * 3 +(1/6) * 4 +(1/6) * 5 +(1/6) * 6 = 3.5

2Y

= (1/6) *(1-3.5)2 +(1/6) * (2 -3.5)2 +(1/6) * (3 -3.5)2 +(1/6) * (4 -3.5)2 +(1/6) * (5 -3.5)2 +(1/6) * (6 -3.5)2 = 2.917

Y = 2.917= 1.708

The expected value of a constant k is k itself

E(k) = k The idea is that if I am always going to get the same number, say 5, then the expected value is 5

The expected value of a function g(X) is given by

E(g(x)) ≡ p(xi)g(xi) (3)

Example: Suppose that I stand to win the money value of the square of the number that shows up on the dice.

I throw a 2 I win 4, and if I throw 6, I get 36, etc.

E(X2) = p(xi)xi2 = 1/6* 12 + 1/6* 22 +

1/6* 32+ 1/6* 42+ 1/6* 52+ 1/6* 62 = 15.167

E(kX) = kE(X) where k is a constant

Variance of X, ( X ) is the spread

around the mean value of X, x . So  

X ≡ E(X - x)

2 where x is mean of X or E(X). 

E(X-x )2 = E(X2 ) – 2E(Xx )+ E(x

2)

= E(X2 ) – x (X)+ E(x2)

= E(X2 ) – x x + x2

X = E(X2 ) –x

2

(X-x )2 = X2 - 2Xx + x

2

Standard deviation of X, X =

In the dice-throwing example, x = 3.5 and E(X2) = 15.167. So  

X = 15.167 – (3.5)2 = 2.917

and so Y = aX

 

Theory: If Y ≡ aX + b where a and b are constants, then

 Y = aX + b ; Y = a2

X

The risk and the return of 10 shares is ten times that of holding one share of the same company. 

= aE(X) + b

Proof: E(Y) = E(aX +b) = E(aX) + E(b)

= ax + b 

= E(aX + b - ax –b)2

 

Y = E(Y-Y)2

= E(aX - ax )2  

= E(a2X2 ) – ax (X)+ E(a2 x2)

= E(aX)2) – 2E(aXax )+ E(ax)2 =

= a2 E(X2 ) – a2 x x + a2 x2

= a2E(X2 ) – a2x2

= a2 (E(X2 ) – x2)

= a2X

Continuous random variables

Each possible value of the random variable x has zero probability but a positive probability density

The probability density function (pdf), is denoted by f(x)

f(x) assigns a probability density

to each possible value x the random

variable X may take.

X

The f(.) function assigns a vertical distance to each value of x

x

Probability Density f(x)

The integral of the pdf on an interval is the probability that the random variable takes a value within this interval.

P(a X b )

= a b f(x)dx

The total probability must be 1

= - + f(x)dx

P(-X )

= 1

Example: A pdf is given by f(x) where

 f(x) = 3x2 for 0 ≤ x ≤ 1

= 0 otherwise

 

= 0 1 3x2dx

= [x3]0,1 = 13 – 03 = 1

 

0 1 f(x)dx

Proof that the function is indeed a pdf:

E(X) = X xf(x)dx

= 0 1 x*3x2 dx

= 0 1 3x3dx

= [(3/4)x4]0,1 = 0.75

The mode is the value of X that has the maximum density. So it is 1.

The median m solves

0 m 3x2dx = 0.5

m3 = 0.5 so that m = 0.794.

[x3]0,m = 0.5

(X) = 0 1 (X-0.75)23x2dx

= 0 1 3x4dx - 0 1 4.5x3dx

+

0 1 1.6875x2dx

= 0.6 -1.125 + 0.5625 = 0.0375