Some Illustrations of Econometric Problems

1. Topic1 Some Illustrations of Econometric Problems

2. Econometrics attempts to measurequantitatively the concepts developed by Economic theory and use the measures to prove or disprove the latter.

3. Problem1:Estimating the demand curve of a product and measuring the price elasticity of demand at a single point Step1:Collection of data (after sorting various problems associated with it. )

4. 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.

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

6. 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.

7. Step4: Propose the regression model lnQ = a + b lnP + e Recognize that b = dlnQ/dlnP = PED

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

9. 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.

10. 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

11. 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.

12. 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

13. 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

14. 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?

15. 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

16. 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

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

18. Topic2 Revision of Probability Theory

19. Suppose that an experiment is scheduled to be undertaken What is the chance of a success? What is the chance of a failure?

20. 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

21. 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:

22. 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 Frequencyapproach to assess probability Assigning probability values according to One’s own beliefs is the Subjective probabilitymethod

23. We shall follow the Relative Frequency approach That is, use past data to assess probability

24. Probability Theory The probability of an event e is the number P(e) = f/Nwhere f is the frequency of the event occurring N is the total frequency and N is ‘large’.

25. 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

26. White Area: A and B (A B) A and B are not mutually exclusive Red and White and Blue: A or B (A B)

27. The sample space S is the grey area = 1 Event A: Red oval Event B: Blue Triangle A and B aremutually exclusive

28. Axiomatic Probability is a branch ofprobability theory based on the following axioms. (1) 0  P(e)  1 (2) If e, fare a pair of events that are mutually exclusive then probability both e and f occur is zero P(e and f) = 0

29. (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.

30. 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)

31. Probability • Distributions

32. Outcome Random Deterministic Numeric Non-numeric

33. 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.

34. 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

35. 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)

36. 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.

37. The Standard Deviationof a random • variable Y ( sY) measures its dispersion around the expected value. • The Variance (s2Y) is the square of the standard deviation.

38. Expectation: The Expected value of X, written E(X) is the weightedaverage of the values X can take. Using notations, E(X) ≡ Sp(xi)xi (2)

39. 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 • 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

40. 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

41. 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.

42. 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

43. 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). 

44. (X-mx )2 = X2 - 2Xmx + mx2 E(X-mx )2 = E(X2 ) – 2E(Xmx )+ E(mx2) = E(X2 ) – 2mxE(X)+ E(mx2) = E(X2 ) – 2mxmx + mx2 s2X = E(X2 ) –mx2

45. 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

46. Theory: If Y ≡ aX + b where a and b are constants, then mY = amX + b ; s2Y = a2s2X and so sY = a*sX The risk and the return of 10 shares is ten times that of holding one share of the same company.

47. Proof: E(Y) = E(aX +b) = E(aX) + E(b) = aE(X) + b = amx + b

48. s2Y = E(Y-mY)2 = E(aX + b - amx –b)2 = E(aX - amx )2

49. = E(aX)2) – 2E(aXamx )+ E(amx)2 = = E(a2X2 ) – 2a2mxE(X)+ E(a2mx2) = a2 E(X2 ) – 2a2mxmx + a2mx2 = a2E(X2 ) – a2mx2 = a2 (E(X2 ) – mx2) = a2 s2X

50. Continuous random variables Each possible value of the random variable x has zero probability but a positive probability density