1. Some Illustrations of Econometric Problems Topic1

2. Econometrics attempts to measure quantitatively 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 Recognize that  = dlnQ/dlnP = PED lnQ = ll nP + 

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

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 =   lnP +   lnp 1 +   lnp 2 + ….+  n lnp n +  lnY Test the hypothesis :        n 

11. Problem3: Does a production function exhibit Constant, Increasing or Decreasing returns to scale? A Cobb-Douglas production function: Q = AL a K b 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. Revision of Probability Theory Topic2

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

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/N where 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 Event B: Blue Triangle A’: Everything that is still grey (not A)

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

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

28. 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

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(  ) = 0 where  is the non-event That is, an event not contained within S will not occur.

30. = 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)

31. Probability Distributions

32. Outcome Deterministic Random Non-numeric 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 x i (for i = 1,2,….n) the possible values taken by the random variable X. If p(x i ) is the probability assigned to x i, then  p(x i ) = 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 Deviation of a random variable Y (  Y  measures its dispersion around the expected value. The Variance (  2 Y ) is the square of the standard deviation.

38. Expectation: The Expected value of X, written E(X) is the weighted average of the values X can take. Using notations, E(X) ≡  p(x i )x i (2)

39. 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  2 Y = (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

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)) ≡  p(x i )g(x i ) (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(X 2 ) =  p(x i )x i 2 = 1/6* 1 2 + 1/6* 2 2 + 1/6* 3 2 + 1/6* 4 2 + 1/6* 5 2 + 1/6* 6 2 = 15.167

43. 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).

44. E(X-  x ) 2 = E(X 2 ) – 2E(X  x )+ E(  x 2 ) = E(X 2 ) –  x  (X)+ E(  x 2 ) = E(X 2 ) –  x  x +  x 2   X = E(X 2 ) –  x 2 (X-  x ) 2 = X 2 - 2X  x +  x 2

45. Standard deviation of X,  X =   In the dice-throwing example,  x = 3.5 and E(X 2 ) = 15.167. So   X = 15.167 – (3.5) 2 = 2.917

46. and so  Y = a  X Theory: If Y ≡ aX + b where a and b are constants, then  Y = a  X + b ;   Y = a 2   X The risk and the return of 10 shares is ten times that of holding one share of the same company.

47. = aE(X) + b Proof: E(Y) = E(aX +b) = E(aX) + E(b)  = a  x + b

48. = E(aX + b - a  x –b) 2   Y = E(Y-  Y ) 2  = E(aX - a  x ) 2

49. = E(a 2 X 2 ) –  a   x  (X)+ E(a 2  x 2 ) = E(aX) 2 ) – 2E(aXa  x )+ E(a  x ) 2 =  = a 2 E(X 2 ) –  a 2  x  x + a 2  x 2 = a 2 E(X 2 ) – a 2  x 2 = a 2 (E(X 2 ) –  x 2 ) = a 2   X

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

51. 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.

52. XX The f(.) function assigns a vertical distance to each value of x x Probability Density f(x)

53. 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

55. The total probability must be 1 = -   +  f(x)dx P(-  X  ) = 1

56. Example: A pdf is given by f(x) where f(x) = 3x 2 for 0 ≤ x ≤ 1 = 0 otherwise = 0  1 3x 2 dx = [x 3 ] 0,1 = 1 3 – 0 3 = 1 0  1 f(x)dx Proof that the function is indeed a pdf:

57. E(X) =  X xf(x)dx = 0  1 x*3x 2 dx = 0  1 3x 3 dx = [(3/4)x 4 ] 0,1 = 0.75

58. The mode is the value of X that has the maximum density. So it is 1. The median m solves 0  m 3x 2 dx = 0.5 m 3 = 0.5 so that m = 0.794. [x 3 ] 0,m = 0.5

59. (   X ) = 0  1 (X-0.75) 2 3x 2 dx = 0  1 3x 4 dx - 0  1 4.5x 3 dx + 0  1 1.6875x 2 dx = 0.6 -1.125 + 0.5625 = 0.0375