1. Economics of the Firm Consumer Demand Analysis

2. Today’s Plan: Motivation Refresher on Probability and Statistics Refresher on Regression analysis Example of Regression Analysis Cross Section Estimation Forecasting Questions

3. A demand curve tells us a lot about the customers that we face. Movie Tickets Sold (in Thousands) Price of a movie ticket Average Income (in Thousands) Price of a DVD (A substitute to the movies) Every $1 increase in the price of a movie ticket lowers sales by 4,000 tickets Every $1,000 increase in average income raises sales by 6,000 tickets Every $1 increase in the price of a DVD raises sales by 2,000 tickets

4. We can use a demand curve to forecast sales and revenues…

5. Demand curves slope downwards – this reflects the negative relationship between price and quantity. Elasticity of Demand measures this effect quantitatively Movie Tickets Price $8.00 480 A 1% rise in price lowers sales by.067%

6. For any fixed price, demand (typically) responds positively to increases in income. Income Elasticity measures this effect quantitatively Movie Tickets Price $8.00 480 A 1% rise in average income raises sales by.6%

7. Cross price elasticity refers to the impact on demand of another price changing Movie Tickets Price $4.00 480 A positive cross price elasticity refers to a substitute while a negative cross price elasticity refers to a compliment A 1% rise in DVD prices raises sales by.05%

8. Application: At the revenue maximizing price, elasticity should be -1

9. We can then re-calculate elasticity A 1% rise in price lowers sales by 1% A 1% rise in average income raises sales by 1.125% A 1% rise in DVD prices raises sales by.09%

10. Suppose that average income rose by 8%. By how much could you raise price without losing any sales?

11. All We need is a demand curve!!

12. What are the odds that a fair coin flip results in a head? What are the odds that the toss of a fair die results in a 5? What are the odds that tomorrow’s temperature is 95 degrees?

13. The answer to all these questions come from a probability distribution Head Tail 1/2 Probability 16 1/6 Probability 2345 A probability distribution is a collection of probabilities describing the odds of any particular event

14. Probability Temperature MeanMean+1SDMean+2SDMean -1SDMean-2SD 2.5% 13.5%34% 13.5% We generally assume a Normal Distribution which can be characterized by a mean (average) and standard deviation (measure of dispersion)

15. Annual Temperature in South Bend has a mean of 59 degrees and a standard deviation of 18 degrees. Probability Temperature 5977954123 95 degrees is 2 standard deviations to the right – there is a 2.5% chance the temperature is 95 or greater (97.5% chance it is cooler than 95) Can’t we do a little better than this?

16. Conditional distributions give us probabilities conditional on some observable information – the temperature in South Bend conditional on the month of July has a mean of 84 with a standard deviation of 7. Probability Temperature 8491987770 95 degrees falls a little more than one standard deviation away (there approximately a 16% chance that the temperature is 95 or greater) 95 Conditioning on month gives us a more accurate probabilities!

17. We know that there should be a “true” probability distribution that governs the outcome of a coin toss (assuming a fair coin) Suppose that we were to flip a coin over and over again and after each flip, we calculate the percentage of heads & tails That is, if we collect “enough” data, we can eventually learn the truth! (Sample Statistic)(True Probability)

18. We can follow the same process for the temperature in South Bend Temperature ~ We could find this distribution by collecting temperature data for south bend Sample Mean (Average) Sample Variance Note: Standard Deviation is the square root of the variance.

19. Mean = 8 Std. Dev. = 2 Mean = $ 12,000 Std. Dev. = $ 2,000 Suppose we know that the value of a car is determined by its age Value = $20,000 - $1,000 (Age) Car Age Value

20. We could also use this to forecast: Value = $20,000 - $1,000 (Age) How much should a six year old car be worth ? Value = $20,000 - $1,000 (6) = $14,000 Note: There is NO uncertainty in this prediction.

21. Value = a + b * (Age) + error We want to choose ‘a’ and ‘b’ to minimize the error! a Slope = b Searching for the truth….a linear regression Error

22. Regression Results VariableCoefficientsStandard Errort Stat Intercept12,35465318.9 Age- 85480-10.60 Value = $12,354 - $854 * (Age) + error We have our estimate of “the truth” Intercept (a) Mean = $12,354 Std. Dev. = $653 Age (b) Mean = -$854 Std. Dev. = $80 T-Stats bigger than 2 in absolute value are considered statistically significant!

23. Regression Statistics R Squared0.36 Standard Error2250 Error Term Mean = 0 Std, Dev = $2,250 Percentage of value variance explained by age We also have some statistics about the error term

24. We can now forecast the value of a 6 year old car Salary = $12,354 - $854 * (Age) + error 6 Mean = $12,354 Std. Dev. = $653 Mean = $854 Std. Dev. = $ 80 Mean = $0 Std. Dev. = $2,250 (Recall, The Average Car age is 8 years)

25. +95% -95% Age Value Note that your forecast error will always be smallest at the sample mean! Also, your forecast gets worse at an increasing rate as you depart from the mean Forecast Interval

26. What are the odds that Pat Buchanan received 3,407 votes from Palm Beach County in 2000? An applied example…

27. The Strategy: Estimate a relationship for Pat Buchanan’s votes using every county EXCEPT Palm Beach Using Palm Beach data, forecast Pat Buchanan’s vote total for Palm Beach Pat Buchanan’s Votes “Are a function of” Observable Demographics

28. The Data: Demographic Data By County CountyBlack (%) Age 65 (%) Hispanic (%) College (%) Income (000s) Buchanan Votes Total Votes Alachua21.89.44.734.626.526284,966 Baker16.87.71.55.727.6738,128 What variables do you think should affect Pat Buchanan’s Vote total? % of County that is college educated # of votes gained/lost for each percentage point increase in college educated population # of Buchanan votes

29. Parameterab Value5.3514.95 Standard Error58.53.84 T-Statistic.093.89 Results R-Square =.19 The distribution for ‘b’ has a mean of 15 and a standard deviation of 4 15 There is a 95% chance that the value for ‘b’ lies between 23 and 7 CountyCollege (%) Predicted Votes Actual Votes Error Alachua34.6522262260 Baker5.7907317 0 Plug in Values for College % to get vote predictions 19% of the variation in Buchanan’s votes across counties is explained by college education Each percentage point increase in college educated (i.e. from 10% to 11%) raises Buchanan’s vote total by 15

30. CountyCollege (%)Buchanan Votes Log of Buchanan Votes Alachua34.62625.57 Baker5.7734.29 Lets try something a little different… % of County that is college educated Percentage increase/decease in votes for each percentage point increase in college educated population Log of Buchanan votes

31. Parameterab Value3.45.09 Standard Error.27.02 T-Statistic12.65.4 Results R-Square =.31 The distribution for ‘b’ has a mean of.09 and a standard deviation of.02.09 There is a 95% chance that the value for ‘b’ lies between.13 and.05 CountyCollege (%) Predicted Votes Actual Votes Error Alachua34.6902262640 Baker5.75573-18 0 Plug in Values for College % to get vote predictions 31% of the variation in Buchanan’s votes across counties is explained by college education Each percentage point increase in college educated (i.e. from 10% to 11%) raises Buchanan’s vote total by.09%

32. CountyCollege (%)Buchanan Votes Log of College (%) Alachua34.62623.54 Baker5.7731.74 How about this… Log of % of County that is college educated Gain/ Loss in votes for each percentage increase in college educated population # of Buchanan votes

33. Parameterab Value-424252 Standard Error13954 T-Statistic-3.054.6 Results R-Square =.25 The distribution for ‘b’ has a mean of 252 and a standard deviation of 54.09 There is a 95% chance that the value for ‘b’ lies between 360 and 144 CountyCollege (%) Predicted Votes Actual Votes Error Alachua34.6469262207 Baker5.71573-58 0 Plug in Values for College % to get vote predictions 25% of the variation in Buchanan’s votes across counties is explained by college education Each percentage increase in college educated (i.e. from 30% to 30.3%) raises Buchanan’s vote total by 252 votes

34. CountyCollege (%) Buchanan Votes Log of College (%)Log of Buchanan Votes Alachua34.62623.545.57 Baker5.7731.744.29 One More… Log of % of County that is college educated Percentage gain/Loss in votes for each percentage increase in college educated population Log of Buchanan votes

35. Parameterab Value.711.61 Standard Error.63.24 T-Statistic1.136.53 Results R-Square =.40 The distribution for ‘b’ has a mean of 1.61 and a standard deviation of.24.09 There is a 95% chance that the value for ‘b’ lies between 2 and 1.13 CountyCollege (%) Predicted Votes Actual Votes Error Alachua34.6624262362 Baker5.73473-39 0 Plug in Values for College % to get vote predictions 40% of the variation in Buchanan’s votes across counties is explained by college education Each percentage increase in college educated (i.e. from 30% to 30.3%) raises Buchanan’s vote total by 1.61%

36. It turns out the regression with the best fit looks like this. CountyBlack (%) Age 65 (%) Hispanic (%) College (%) Income (000s) Buchanan Votes Total Votes Alachua21.89.44.734.626.526284,966 Baker16.87.71.55.727.6738,128 Parameters to be estimated Error term Buchanan Votes Total Votes *100

37. The Results: VariableCoefficientStandard Errort - statistic Intercept2.146.3965.48 Black (%)-.0132.0057-2.88 Age 65 (%)-.0415.0057-5.93 Hispanic (%)-.0349.0050-6.08 College (%)-.0193.0068-1.99 Income (000s)-.0658.00113-4.58 Now, we can make a forecast! R Squared =.73 CountyPredicted Votes Actual Votes Error Alachua520262258 Baker5573-18 CountyBlack (%)Age 65 (%)Hispanic (%)College (%)Income (000s) Buchanan Votes Total Votes Alachua21.89.44.734.626.526284,966 Baker16.87.71.55.727.6738,128

38. CountyBlack (%) Age 65 (%) Hispanic (%) College (%) Income (000s) Buchanan Votes Total Votes Palm Beach21.823.69.822.133.53,407431,621 This would be our prediction for Pat Buchanan’s vote total!

39. Probability LN(%Votes) There is a 95% chance that the log of Buchanan’s vote percentage lies in this range -2.004 – 2*(.2556)-2.004 + 2*(.2556) = -2.5152= -1.4928 We know that the log of Buchanan’s vote percentage is distributed normally with a mean of -2.004 and with a standard deviation of.2556

40. Probability % of Votes There is a 95% chance that Buchanan’s vote percentage lies in this range Next, lets convert the Logs to vote percentages

41. Probability Votes There is a 95% chance that Buchanan’s total vote lies in this range Finally, we can convert to actual votes 3,407 votes turns out to be 7 standard deviations away from our forecast!!!

42. Back to the original problem. We know that the quantity of some good or service demanded should be related to some basic variables Quantity Price Quantity Demanded Price Income Other “Demand Shifters” “ Is a function of”

43. Time Demand Factors t t+1t-1 Cross Sectional estimation holds the time period constant and estimates the variation in demand resulting from variation in the demand factors For example: can we estimate demand for Pepsi in South Bend by looking at selected statistics for South bend

44. CityPriceAverage Income (Thousands) Competitor’s Price Advertising Expenditures (Thousands) Total Sales Granger1.0221.9341.482.3679,809 Mishawaka2.5635.7962.5326.922130,835 Suppose that we have the following data for sales in 200 different Indiana cities Lets begin by estimating a basic demand curve – quantity demanded is a linear function of price. Change in quantity demanded per $ change in price (to be estimated)

45. Regression Results VariableCoefficientStandard Errort Stat Intercept155,04218,1338.55 Price (X)-46,0877214-6.39 That is, we have estimated the following equation Regression Statistics R Squared.17 Standard Error48,074 Every dollar increase in price lowers sales by 46,087 units.

46. Values For South Bend Price of Pepsi$1.37 91,903 We can now use this estimated demand curve along with price in South Bend to estimate demand in South Bend

47. 91,903 $1.37 We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity

48. CityPriceAverage Income (Thousands) Competitor’s Price Advertising Expenditures (Thousands) Total Sales Granger1.0221.9341.482.3679,809 Mishawaka2.5635.7962.5326.922130,835 As we did earlier, we can experiment with different functional forms by using logs Using logs changes the interpretation of the coefficients. Change in quantity demanded per percentage change in price (to be estimated)

49. Regression Results VariableCoefficientStandard Errort Stat Intercept133,13314,8928.93 Price (X)-103,97316,407-6.33 That is, we have estimated the following equation Regression Statistics R Squared.17 Standard Error48,140 Every 1% increase in price lowers sales by 103,973 units.

50. Values For South Bend Price of Pepsi$1.37 Log of Price.31 $1.37 100,402 We can now use this estimated demand curve along with price in South Bend to estimate demand in South Bend

51. We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity $1.37 100,402

52. CityPriceAverage Income (Thousands) Competitor’s Price Advertising Expenditures (Thousands) Total Sales Granger1.0221.9341.482.3679,809 Mishawaka2.5635.7962.5326.922130,835 As we did earlier, we can experiment with different functional forms by using logs Percentage change in quantity demanded per $ change in price (to be estimated) Using logs changes the interpretation of the coefficients.

53. Regression Results VariableCoefficientStandard Errort Stat Intercept13.3438.1 Price (X)-1.22.13-8.98 That is, we have estimated the following equation Regression Statistics R Squared.28 Standard Error.90 Every $1 increase in price lowers sales by 1.22%.

54. Values For South Bend Price of Pepsi$1.37 83,283 We can now use this estimated demand curve along with price in South Bend to estimate demand in South Bend

55. We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity $1.37 83,283

56. CityPriceAverage Income (Thousands) Competitor’s Price Advertising Expenditures (Thousands) Total Sales Granger1.0221.9341.482.3679,809 Mishawaka2.5635.7962.5326.922130,835 As we did earlier, we can experiment with different functional forms by using logs Percentage change in quantity demanded per percentage change in price (to be estimated) Using logs changes the interpretation of the coefficients.

57. Regression Results VariableCoefficientStandard Errort Stat Intercept12.3.2842.9 Price (X)-2.60.31-8.21 That is, we have estimated the following equation Regression Statistics R Squared.25 Standard Error.93 Every 1% increase in price lowers sales by 2.6%.

58. Values For South Bend Price of Pepsi$1.37 Log of Price.31 $1.37 72,402 We can now use this estimated demand curve along with price in South Bend to estimate demand in South Bend

59. We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity $1.37 83,283

60. We can add as many variables as we want in whatever combination. The goal is to look for the best fit. % change in Sales per $ change in price % change in Sales per % change in income % change in Sales per % change in competitor’s price Regression Results VariableCoefficientStandard Errort Stat Intercept5.981.294.63 Price-1.29.12-10.79 Log of Income1.46.344.29 Log of Competitor’s Price2.00.345.80 R Squared:.46

61. Values For South Bend Price of Pepsi$1.37 Log of Income3.81 Log of Competitor’s Price.80 $1.37 87,142 Now we can make a prediction and calculate elasticities

62. Time Demand Factors t t+1t-1 We could use a cross sectional regression to forecast quantity demanded out into the future, but it would take a lot of information! Estimate a demand curve using data at some point in time Use the estimated demand curve and forecasts of data to forecast quantity demanded

63. Time Demand Factors t t+1t-1 Time Series estimation ignores the demand factors constant and estimates the variation in demand over time For example: can we predict demand for Pepsi in South Bend next year by looking at how demand varies across time

64. Time series estimation leaves the demand factors constant and looks at variations in demand over time. Essentially, we want to separate demand changes into various frequencies Trend: Long term movements in demand (i.e. demand for movie tickets grows by an average of 6% per year) Business Cycle: Movements in demand related to the state of the economy (i.e. demand for movie tickets grows by more than 6% during economic expansions and less than 6% during recessions) Seasonal: Movements in demand related to time of year. (i.e. demand for movie tickets is highest in the summer and around Christmas

65. Suppose that you work for a local power company. You have been asked to forecast energy demand for the upcoming year. You have data over the previous 4 years: Time PeriodQuantity (millions of kilowatt hours) 2003:111 2003:215 2003:312 2003:414 2004:112 2004:217 2004:313 2004:416 2005:114 2005:218 2005:315 2005:417 2006:115 2006:220 2006:316 2006:419

66. First, let’s plot the data…what do you see? This data seems to have a linear trend

67. A linear trend takes the following form: Forecasted value at time t (note: time periods are quarters and time zero is 2003:1) Time period: t = 0 is 2003:1 and periods are quarters Estimated value for time zero Estimated quarterly growth (in kilowatt hours)

68. Regression Results VariableCoefficientStandard Errort Stat Intercept11.9.95312.5 Time Trend.394.0994.00 Regression Statistics R Squared.53 Standard Error 1.82 Observations16 Lets forecast electricity usage at the mean time period (t = 8)

69. Here’s a plot of our regression line with our error bands…again, note that the forecast error will be lowest at the mean time period T = 8

70. Sample We can use this linear trend model to predict as far out as we want, but note that the error involved gets worse!

71. Time PeriodActualPredictedError 2003:11112.29-1.29 2003:21512.682.31 2003:31213.08-1.08 2003:41413.47.52 2004:11213.87-1.87 2004:21714.262.73 2004:31314.66-1.65 2004:41615.05.94 2005:11415.44-1.44 2005:21815.842.15 2005:31516.23-1.23 2005:41716.63.37 2006:11517.02-2.02 2006:22017.412.58 2006:31617.81-1.81 2006:41918.20.79 One method of evaluating a forecast is to calculate the root mean squared error Number of Observations Sum of squared forecast errors

72. Lets take another look at the data…it seems that there is a regular pattern… Q2 We are systematically under predicting usage in the second quarter

73. Time PeriodActualPredictedRatioAdjusted 2003:11112.29.8912.29(.87)=10.90 2003:21512.681.1812.68(1.16) = 14.77 2003:31213.08.9113.08(.91) = 11.86 2003:41413.471.0313.47(1.04) = 14.04 2004:11213.87.8713.87(.87) = 12.30 2004:21714.261.1914.26(1.16) = 16.61 2004:31314.66.8814.66(.91) = 13.29 2004:41615.051.0615.05(1.04) = 15.68 2005:11415.44.9115.44(.87) = 13.70 2005:21815.841.1415.84(1.16) = 18.45 2005:31516.23.9216.23(.91) = 14.72 2005:41716.631.0216.63(1.04) = 17.33 2006:11517.02.8817.02(.87) = 15.10 2006:22017.411.1417.41(1.16) = 20.28 2006:31617.81.8917.81(.91) = 16.15 2006:41918.201.0418.20(1.04) = 18.96 Average Ratios Q1 =.87 Q2 = 1.16 Q3 =.91 Q4 = 1.04 We can adjust for this seasonal component…

74. Now, we have a pretty good fit!!

75. Recall our prediction for period 76 ( Year 2022 Q4)

76. We could also account for seasonal variation by using dummy variables Note: we only need three quarter dummies. If the observation is from quarter 4, then

77. Regression Results VariableCoefficientStandard Errort Stat Intercept12.75.22656.38 Time Trend.375.016822.2 D1-2.375.219-10.83 D21.75.2158.1 D3-2.125.213-9.93 Regression Statistics R Squared.99 Standard Error.30 Observations16 Note the much better fit!!

78. Time PeriodActualRatio MethodDummy Variables 2003:11110.9010.75 2003:21514.7715.25 2003:31211.8611.75 2003:41414.0414.25 2004:11212.3012.25 2004:21716.6116.75 2004:31313.2913.25 2004:41615.6815.75 2005:11413.7013.75 2005:21818.4518.25 2005:31514.7214.75 2005:41717.3317.25 2006:11515.1015.25 2006:22020.2819.75 2006:31616.1516.25 2006:41918.9618.75 Ratio Method Dummy Variables

79. Recall our prediction for period 76 ( Year 2022 Q4)

80. Recall, our trend line took the form… This parameter is measuring quarterly change in electricity demand in millions of kilowatt hours. Often times, its more realistic to assume that demand grows by a constant percentage rather that a constant quantity. For example, if we knew that electricity demand grew by g% per quarter, then our forecasting equation would take the form

81. If we wish to estimate this equation, we have a little work to do… Note: this growth rate is in decimal form If we convert our data to natural logs, we get the following linear relationship that can be estimated

82. Regression Results VariableCoefficientStandard Errort Stat Intercept2.49.06339.6 Time Trend.026.0064.06 Regression Statistics R Squared.54 Standard Error.1197 Observations16 Lets forecast electricity usage at the mean time period (t = 8) BE CAREFUL….THESE NUMBERS ARE LOGS !!!

83. The natural log of forecasted demand is 2.698. Therefore, to get the actual demand forecast, use the exponential function Likewise, with the error bands…a 95% confidence interval is +/- 2 SD

84. Again, here is a plot of our forecasts with the error bands T = 8

85. Errors is growth rates compound quickly!!

86. Let’s try one…suppose that we are interested in forecasting gasoline prices. We have the following historical data. (the data is monthly from April 1993 – June 2010) Does a linear (constant cents per gallon growth per year) look reasonable?

87. Let’s suppose we assume a linear trend. Then we are estimating the following linear regression: Price at time t Price at April 1993Number of months from April 1993 monthly growth in cents per gallon Regression Results VariableCoefficientStandard Errort Stat Intercept.67.0512.19 Time Trend.010.000423.19 R Squared=.72

88. We can check for the presence of a seasonal cycle by adding seasonal dummy variables: Cents per gallon impact of quarter I relative to quarter 4 Regression Results VariableCoefficientStandard Errort Stat Intercept.58.078.28 Time Trend.01.000423.7 D1-.03.075-.43 D2.15.0742.06 D3.16.0752.20 R Squared=.74

89. If we wanted to remove the seasonal component, we could by subtracting the seasonal dummy off each gas price Seasonalizing DatePrice Regression coefficient Seasonalized data 1993 – 041.05.15.90 1993 - 071.06.1690 1993 - 101.060 1994 - 01.98-.031.01 1994 - 041.00.15.85 2 nd Quarter 3 rd Quarter 4 th Quarter 1 st Quarter 2 nd Quarter

90. Note: Once the seasonal component has been removed, all that should be left is trend, cycle, and noise. We could check this: Seasonalized Price Series Regression Results VariableCoefficientStandard Errort Stat Intercept.587.0511.06 Time Trend.010.000423.92 Seasonalized Price Series Regression Results VariableCoefficientStandard Errort Stat Intercept.587.078.28 Time Trend.010.000423.7 D10.0750 D20.0740 D30.0750

91. The regression we have in place gives us the trend plus the seasonal component of the data TrendSeasonal If we subtract our predicted price (from the regression) from the actual price, we will have isolated the business cycle and noise Business Cycle Component DateActual Price Predicted Price (From regression) Business Cycle Component 1993 - 041.050.752.297 1993 - 051.071.763.308 1993 - 061.075773.301 1993 - 071.064.797.267 1993 - 081.048.807.240 Predicted

92. We can plot this and compare it with business cycle dates Actual Price Predicted Price

93. Data Breakdown DateActual PriceTrendSeasonalBusiness Cycle 1993 - 041.050.60.15.30 1993 - 051.071.61.15.31 1993 - 061.075.62.15.30 1993 - 071.064.63.16.27 1993 - 081.048.64.16.24 Regression Results VariableCoefficientStandard Errort Stat Intercept.58.078.28 Time Trend.01.000423.7 D1-.03.075-.43 D2.15.0742.06 D3.16.0752.20

94. Perhaps an exponential trend would work better… An exponential trend would indicate constant percentage growth rather than cents per gallon.

95. We already know that there is a seasonal component, so we can start with dummy variables Percentage price impact of quarter I relative to quarter 4 Regression Results VariableCoefficientStandard Errort Stat Intercept-.14.03-4.64 Time Trend.005.000129.9 D1-.02.032-.59 D2.06.0322.07 D3.07.0322.19 R Squared=.81 Monthly growth rate

96. If we wanted to remove the seasonal component, we could by subtracting the seasonal dummy off each gas price, but now, the price is in logs Seasonalizing DatePriceLog of Price Regression coefficient Log of Seasonalized data Seasonalized Price 1993 – 041.05.049.06-.019.98 1993 - 071.06.062.07-.010.99 1993 - 101.06.0620 1.06 1994 - 01.98-.013-.02.0061.00 1994 - 041.00.005.06-.062.94 2 nd Quarter 3 rd Quarter 4 th Quarter 1 st Quarter 2 nd Quarter Example:

97. The regression we have in place gives us the trend plus the seasonal component of the data Trend Seasonal If we subtract our predicted price (from the regression) from the actual price, we will have isolated the business cycle and noise Business Cycle Component DateActual Price Predicted Log Price (From regression) Predicted Price Business Cycle Component 1993 - 041.050-.069.93.12 1993 - 051.071-.063.94.13 1993 - 061.075-.057.94.13 1993 – 071.064-.047.95.11 1993 - 081.048-.041.96.09 Predicted Log of Price

98. As you can see, very similar results Actual Price Predicted Price

99. In either case, we could make a forecast for gasoline prices next year. Lets say, April 2011. Forecasting Data DateTime PeriodQuarter April 20112172 OR

100. QuarterMarket Share 120 222 323 424 518 623 719 817 922 1023 1118 1223 Consider a new forecasting problem. You are asked to forecast a company’s market share for the 13 th quarter. There doesn’t seem to be any discernable trend here…

101. Smoothing techniques are often used when data exhibits no trend or seasonal/cyclical component. They are used to filter out short term noise in the data. QuarterMarket Share MA(3)MA(5) 120 222 323 42421.67 51823 6 21.6721.4 71921.6722 8172021.4 92219.6720.2 102319.3319.8 111820.6720.8 12232119.8 A moving average of length N is equal to the average value over the previous N periods

102. The longer the moving average, the smoother the forecasts are…

103. QuarterMarket Share MA(3)MA(5) 120 222 323 42421.67 51823 6 21.6721.4 71921.6722 8172021.4 92219.6720.2 102319.3319.8 111820.6720.8 12232119.8 Calculating forecasts is straightforward… MA(3) MA(5) So, how do we choose N??

104. QuarterMarket Share MA(3)Squared Error MA(5)Squared Error 120 222 323 42421.675.4289 5182325 62321.671.768921.42.56 71921.677.1289229 81720921.419.36 92219.675.428920.23.24 102319.3313.468919.810.24 111820.677.128920.87.84 122321419.810.24 Total = 78.3534Total = 62.48

105. Exponential smoothing involves a forecast equation that takes the following form Forecast for time t+1 Actual value at time t Forecast for time t Smoothing parameter Note: when w = 1, your forecast is equal to the previous value. When w = 0, your forecast is a constant.

106. QuarterMarket Share W=.3W=.5 12021.0 22220.720.5 32321.121.3 42421.722.2 51822.423.1 62321.120.6 71921.721.8 81720.920.4 92219.718.7 102320.4 111821.221.7 122320.219.9 For exponential smoothing, we need to choose a value for the weighting formula as well as an initial forecast Usually, the initial forecast is chosen to equal the sample average

107. As was mentioned earlier, the smaller w will produce a smoother forecast

108. Calculating forecasts is straightforward… W=.3 W=.5 So, how do we choose W?? QuarterMarket Share W=.3W=.5 12021.0 22220.720.5 32321.121.3 42421.722.2 51822.423.1 62321.120.6 71921.721.8 81720.920.4 92219.718.7 102320.4 111821.221.7 122320.219.9

109. QuarterMarket Share W =.3Squared Error W=.5Squared Error 12021.01 1 22220.71.6920.52.25 32321.13.6121.32.89 42421.75.2922.23.24 51822.419.3623.126.01 62321.13.6120.65.76 71921.77.2921.87.84 81720.915.2120.411.56 92219.75.2918.710.89 102320.46.7620.46.76 111821.210.2421.713.69 122320.27.8419.99.61 Total = 87.19Total = 101.5