Some Illustrations of

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Topic1
Some Illustrations of Econometric Problems
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Econometrics attempts to measure quantitatively
the concepts developed by Economic theory
and use the measures to prove or disprove the
latter.
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Problem1 Estimating the demand curve of a
product and measuring the price elasticity of
demand at a single point
Step1Collection of data (after sorting various
problems associated with it. )
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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 preferences
• had all remained constant.

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Solution Adjust data and/or throw some of it
out
This is known as the
Identification Problem.
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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.

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Step4 Propose the regression model
lnQ a b lnP e
Recognize that b dlnQ/dlnP PED

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Step5  Impose the restrictions of the Classical
Linear Regression Model
Perform a linear regression of lnQ on lnP and
estimate b

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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 change by the same
proportion.
The rational consumer does not suffer from money
illusion.

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The demand function is homogeneous of degree 0.
Procedure of Test
Step 1 Estimate
lnQ a b0lnP b1lnp1 b2lnp2 . bn lnpn
glnY
Test the hypothesis  
b0 b1 b2 .... bn g 0

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Problem3 Does a production function exhibit
Constant, Increasing or Decreasing returns to
scale?
A Cobb-Douglas production function
Q ALaKb A, a, b gt0
CRS if ab 1, IRS if ab gt 1 DRS otherwise.

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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

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Step4 Run a linear regression of lnQ on lnL and
lnK and estimate the coefficients a and b.
Step5 Test the hypotheses   H0 ab 1 versus
H1 ab gt1 and/or  H0 ab 1 versus H1 ab
lt 1  

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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?

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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

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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

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A statistic called an F-statistic and a
distribution called an F-distribution is used to
quantify the notion of too large

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Topic2
Revision of Probability Theory
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Suppose that an experiment is scheduled to be
undertaken
What is the chance of a success?
What is the chance of a failure?
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Success and Failure are called Outcomes of the
experiment or Events
Events may be made up of elementary events
Experiment Tossing a dice
An elementary event Number 3
shows up
An event A number less than 3 shows up
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Question What is the probability of getting a 3?
Answer 1/6 (assuming all six outcomes are
equally likely )
This approach is known as Classical Probability
If we could not assume that all the events were
equally likely, we might proceed as follows
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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 approach to assess
probability
Assigning probability values according to Ones
own beliefs is the Subjective probability method
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We shall follow the Relative Frequency approach
That is, use past data to assess probability
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• 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.
•  
•  

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The sample space S is the grey area 1
Event A Red oval
A Everything that is still grey (not A)
Event B Blue Triangle
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White Area A and B (A ?B)
A and B are not mutually exclusive
Red and White and Blue A or B (A ?B)
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The sample space S is the grey area 1
Event A Red oval
Event B Blue Triangle
A and B are mutually exclusive
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• 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
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• (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(f) 0 where f is the non-event
•   That is, an event not contained within S will
  not occur.

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P(B) Triangle Area
P(A) Oval Area
P(A and B) w
P(A or B) r w b
So,
P(A) P(B)
r w b w
P(A or B) P(A and B)
The Addition Rule for 2 events A and B P(A or B)
P(A) P(B) - P( A and B)
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Probability Distributions
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Outcome
Random
Deterministic
Numeric
Non-numeric
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• 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.

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The probability distribution of Y (assuming a
fair dice) is given by the Table below
Y y P(Yy) 1
1/6 2
1/6 3 1/6 4
1/6 5
1/6 6 1/6

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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
  Sp(xi) 1 (1)  
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The Expected Value of a random variable Y ( E(Y)
) is the value the variable is most likely to
take, on average
The Expected Value therefore is also called the
Average or Mean of the Probability Distribution.
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• The Standard Deviation of a random
• variable Y ( sY ) measures its dispersion around
  the expected value.
• The Variance (s2Y) is the square of the standard
  deviation.

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Expectation The Expected value of X, written
E(X) is the weighted average of the values X
can take. Using notations,  E(X) Sp(xi)xi
(2)
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• Calculations For the probability distribution
• discussed,

sY ?2.917 1.708
s2Y (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
E(Y) (1/6) 1 (1/6) 2 (1/6) 3 (1/6)
4 (1/6) 5 (1/6) 6 3.5
Y y P(Yy) 1 1/6 2 1/6 3
1/6 4 1/6 5 1/6 6
1/6
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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
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The expected value of a function g(X) is given
by  E(g(x)) Sp(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.
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I throw a 2 I win 4, and if I throw 6, I get 36,
etc.
E(X2) Sp(xi)xi2 1/6 12 1/6 22 1/6 32
1/6 42 1/6 52 1/6 62 15.167
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E(kX) kE(X) where k is a constant
Variance of X, ( s2X ) is the spread around the
mean value of X, mx . So   s2X E(X - mx)2
where mx is mean of X or E(X). 
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(X-mx )2 X2 - 2Xmx mx 2
E(X-mx )2 E(X2 ) 2E(Xmx ) E(mx2)
E(X2 ) 2mx E(X) E(mx2)
E(X2 ) 2mx mx mx2
s2X E(X2 ) mx2
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In the dice-throwing example, mx 3.5 and E(X2)
15.167. So   s2X 15.167 (3.5)2 2.917
Standard deviation of X, sX ?(2.917) 1.707
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Theory If Y aX b where a and b are
constants, then  mY amX b s2Y a2 s2X
and so sY asX  
The risk and the return of 10 shares is ten
times that of holding one share of the same
company.  
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Proof E(Y) E(aX b) E(aX) E(b)
aE(X) b
amx b 

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s2Y E(Y-mY)2
E(aX b - amx b)2  

E(aX - amx )2  
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E(aX)2) 2E(aXamx ) E(amx)2
E(a2X2 ) 2a2mx E(X) E(a2 mx2)
a2 E(X2 ) 2a2 mx mx a2 mx2
a2E(X2 ) a2mx2 a2 (E(X2 ) mx2) a2 s2X

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Continuous random variables

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

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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.

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Probability Density f(x)
mX
x
The f(.) function assigns a vertical distance to
each value of x
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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
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The total probability must be 1
P(-? ?X ? ? )
- ? ? ? f(x)dx
1
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Example A pdf is given by f(x) where  f(x)
3x2 for 0 x 1 0
otherwise
Proof that the function is indeed a pdf
  0 ? 1 3x2dx
  0 ? 1 f(x)dx
x30,1 13 03 1
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E(X) ?X xf(x)dx
0 ? 1 x3x2 dx
0 ? 1 3x3dx
(3/4)x40,1 0.75
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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
x30,m 0.5
m3 0.5 so that m 0.794.
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(s2X) 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