Statistics for Business and Economics Module 2: Regression and time series analysis Spring 2010 Lecture 8: Time Series Analysis and Forecasting 2 Priyantha Wijayatunga, Department of Statistics, Umeå University These materials are altered ones from copyrighted lecture slides (© 2009 W.H. Freeman and Company) from the homepage of the book: The Practice of Business Statistics Using Data for Decisions :Second Edition by Moore, McCabe, Duckworth and Alwan.

Time series models Reference to the book: chapter 13.2  Looking for autocorrelation in residuals  Autoregressive models  Moving average models  Exponential smoothing models

Autocorrelation  The residuals from a regression model that uses time as an explanatory variable should be examined for signs of autocorrelation.  If there is an autocorrelation among residual then they are not independent, therefore our regression models for time series data are not good!

Autocorrelation Time plot of residuals from the exponential trend model for DVD player sales  Starting from the left, negative residuals tend to be followed by another negative residual.  In the middle, positive residuals tend to be followed by another positive residual.  On the right, negative residuals tend to be followed by another negative residual.  This pattern indicates positive autocorrelation.

Lagged residual plot Lagged residual plot of residuals from the exponential trend model for DVD player sales. Plot indicates positive autocorrelation.

Positive First Order Autocorrelation Residuals Time Positive first order autocorrelation occurs when consecutive residuals tend to be similar. 0 + y t

Durbin–Watson Test Durbin–Watson test is used to test first order autocorrelation Test statistic If d is small (d < 2 but close to 0) then it may implies positive autocorrelation (positive residuals are followed by positive residuals and/or negative residuals are followed by negative residuals) If d is big (d > 2 but close to 4) then it implies negative autocorreation (residuals basically alternate in their sign) There is a table for Durbin–Watson test but book does not include it!

 Autocorrelation among the errors of the regression model provides opportunity to produce accurate forecasts.  In a stationary time series (no trend and no seasonality) correlation between consecutive residuals leads to the following first order autoregressive model called AR(1): y t =  0 +  1 y t-1 +  t Autoregressive time series models take advantage of the linear relationship between successive values of a time series to predict future values. Example: Price of a certain stock today may be heavily dependent that of yesterday’s but if we know the yesterday’s price, it may be the case that prices of the stock two and more days back are irrelevant! Autoregressive models

Moving average models  The autoregressive model looks back only one time period and uses that value in the forecast for the current time period.  Moving average models use the average of the last several values of the time series to forecast the next value.

Small example: Moving average–span 4 TimeSales($m)MA(4) 1999Q Q Q Q Q12220,5 2000Q22821, Q31721, Q Q12422,5 2001Q Q31823, Q42524, Q124,33333 Note that when we use span same equal to the lenght of the seasonal cycle we get the linear treand model! Our prediction for the 2002Q1 is not so good

Exponential smoothing model  Criticisms can be made against moving average models.  First, our forecast for the next time period ignores all but the last k observations. Second, the data values are all weighted equally.  Exponential smoothing models address these criticisms.

Exponential smoothing model How the expeontial smoothing method take into account all the past observations for predicting the future

Small ‘w’ provides a lot of smoothing Big ‘w’ provides a little smoothing The exponential smoothing model can be used to produce forecasts when the time series…  exhibits gradual(not a sharp) trend  no cyclical effects  no seasonal effects

Example timepricesmooothp 2000Q150 = Q25150=0.6*50+(1-0,6)* Q34950,6=0.6*51+(1-0,6)* Q45349,64=0.6*49+(1-0,6)*50,6 2001Q15251, Q25451, Q35553, Q45354, Q15453, Q25653, Q35555, Q45255, Q153,21928=0.6*52+(1-0,6)*55,0482