1. Demand Estimation and Forecasting Finance 30210: Managerial Economics

2. 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?

3. 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

4. The distribution for temperature in south bend is a bit more complicated because there are so many possible outcomes, but the concept is the same Probability Temperature We generally assume a Normal Distribution which can be characterized by a mean (average) and standard deviation (measure of dispersion) Mean Standard Deviation

5. Probability Temperature Without some math, we can’t find the probability of a specific outcome, but we can easily divide up the distribution MeanMean+1SDMean+2SDMean -1SDMean-2SD 2.5% 13.5%34% 13.5%

6. 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?

7. 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!

8. 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)

9. 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.

10. Mean = 1 Variance = 4 Std. Dev. = 2 Probability distributions are scalable 3 X = Mean = 3 Variance = 36 (3*3*4) Std. Dev. = 6 Some useful properties of probability distributions

11. Mean = 1 Variance = 1 Std. Dev. = 1 Probability distributions are additive + Mean = 2 Variance = 9 Std. Dev. = 3 COV = 2 = Mean = 3 Variance = 14 (1 + 9 + 2*2) Std. Dev. = 3.7

12. Mean = 8 Variance = 4 Std. Dev. = 2 Mean = $ 12,000 Variance = 4,000,000 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

13. 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.

14. Searching for the truth…. You believe that there is a relationship between age and value, but you don’t know what it is…. 1.Collect data on values and age 2.Estimate the relationship between them Note that while the true distribution of age is N(8,4), our collected sample will not be N(8,4). This sampling error will create errors in our estimates!!

15. Value = a + b * (Age) + error We want to choose ‘a’ and ‘b’ to minimize the error! a Slope = b

16. 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!

17. Regression Statistics R Squared0.36 Standard Error2250 Error Term Mean = 0 Std. Dev = $2,250 Percentage of value variance explained by age

18. We can now forecast the value of a 6 year old car Value = $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)

19. +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

20. What are the odds that Pat Buchanan received 3,407 votes from Palm Beach County in 2000?

21. 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

22. 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

23. 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

24. 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

25. 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%

26. 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

27. 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

28. 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

29. 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%

30. 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

31. 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

32. 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!

33. 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

34. 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

35. 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!!!

36. 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”

37. 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

38. 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)

39. 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.

40. Values For South Bend Price of Pepsi$1.37 91,903

41. 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 Adding logs changes the interpretation of the coefficients Change in quantity demanded per percentage change in price (to be estimated)

42. 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.

43. Values For South Bend Price of Pepsi$1.37 Log of Price.31 $1.37 100,402

44. 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) Adding logs changes the interpretation of the coefficients

45. 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%.

46. 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

47. 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) Adding logs changes the interpretation of the coefficients

48. 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%.

49. Values For South Bend Price of Pepsi$1.37 Log of Price.31 $1.37 72,402

50. 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

51. 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

52. 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

53. Time Demand Factors t t+1t-1 Time Series estimation ignores the demand factors 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

54. Time series estimation ignores the demand factors 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

55. 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

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

57. 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 millions of kilowatt hours)

58. 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)

59. 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

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

61. Lets take another look at the data…it seems that there is a regular pattern… Q2 There appears to be a seasonal cycle…

62. 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 One seasonal adjustment process is to adjust each quarter by the average of actual to predicted For each observation: Calculate the ratio of actual to predicted Average the ratios by quarter Use the average ration to adjust each predicted value

63. Time PeriodActualAdjustedError 2003:11110.90 -0.1 2003:21514.77 -0.23 2003:31211.86 -0.14 2003:41414.04 0.04 2004:11212.30 0.3 2004:21716.61 -0.39 2004:31313.29 0.29 2004:41615.68 -0.32 2005:11413.70 -0.3 2005:21818.45 0.45 2005:31514.72 -0.28 2005:41717.33 0.33 2006:11515.10 0.1 2006:22020.28 0.28 2006:31616.15 0.15 2006:41918.96 -0.04 With the seasonal adjustment, we don’t have any statistics to judge goodness of fit. One method of evaluating a forecast is to calculate the root mean squared error Number of Observations Sum of squared forecast errors

64. Looks pretty good…

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

66. 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

67. 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!!

68. 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

69. A plot confirms the similarity of the methods

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

71. 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

72. 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

73. 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 !!!

74. 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

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

76. Errors is growth rates compound quickly!!

77. 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?

78. 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 dollars per gallon Regression Results VariableCoefficientStandard Errort Stat Intercept.67.0512.19 Time Trend.010.000423.19 R Squared=.72

79. We can check for the presence of a seasonal cycle by adding seasonal dummy variables: dollars 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

80. 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

81. 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

82. 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

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

84. Data Breakdown DateActual PriceTrendSeasonalBusiness Cycle 1993 - 041.050.58.15.320 1993 - 051.071.59.15.331 1993 - 061.075.60.15.325 1993 - 071.064.61.16.294 1993 - 081.048.62.16.268 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

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

86. 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

87. 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:

88. 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

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

90. In either case, we could make a forecast for gasoline prices next year. Lets say, April 2011. Forecasting Data DateTime PeriodQuarter April 20112172 OR By the way, the actual price in April 2011 was $3.80

91. 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…

92. 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

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

94. 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??

95. 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

96. 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.

97. 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

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

99. 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

100. 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