1 / 52

TM 745 Forecasting for Business & Technology

TM 745 Forecasting for Business & Technology. ARIMA (Box-Jenkins). =. Y. observatio. n. (. realizatio. n. ). at. t. t. =. +. e. Pattern. t. ARIMA Models. =. Y. function. of. past. values. t. +. random. shocks. =. e. +. e. f. (. Y. ,. ). -. -. t. k. t. k.

celina
Download Presentation

TM 745 Forecasting for Business & Technology

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. TM 745Forecasting for Business & Technology ARIMA(Box-Jenkins)

  2. = Y observatio n ( realizatio n ) at t t = + e Pattern t ARIMA Models

  3. = Y function of past values t + random shocks = e + e f ( Y , ) - - t k t k t Assumptions

  4. Notation AR I MA p d q p = order of autoregressive d = order of integration (differencing) q = order of moving average

  5. = q + f + e Y Y - 1 1 t o t t $ = q + f Y Y - 1 1 t o t Processes • ARIMA (1, 0, 0)

  6. = q + f + e Y Y - 1 1 t o t t f £ | | 1 f Þ 1 drift / trend f Processes • ARIMA (1, 0, 0) • Autoregressive • Stationarity

  7. AutoRegressive Process

  8. = m - q e + e Y - 1 1 t t t $ = m - q e Y - 1 1 t t Moving Average Process • ARIMA (0, 0, 1)

  9. Moving Average Process

  10. = + e Y Y - 1 t t t $ = Y Y - 1 t t Integrated Processes • ARIMA (0, 1, 0)

  11. Integrated Process

  12. = + q + e Y Y - 1 t t o t $ = + q Y Y - 1 t t o Deterministic Trend • ARIMA (0, 1, ,0) 1

  13. Deterministic Trend

  14. Model Identification • (Figure 7-1, p. 277) • AutoRegressive ARIMA(1, 0, 0), f=0.8 ACF(1) = f = 0.8 ACF(2) = f2 = 0.64 ACF(3) = f3 = 0.512 ACF(4) = f4 = 0.410

  15. Model Identification • (Figure 7-1, p. 277) • AutoRegressive ARIMA(1, 0, 0), f=0.4 ACF(1) = f = 0.4 ACF(2) = f2 = 0.16 ACF(3) = f3 = 0.064 ACF(4) = f4 = 0.0256

  16. Model Identification • (Figure 7-1, p. 277) • AutoRegressive ARIMA(1, 0, 0), f=-0.8 ACF(1) = f = -0.8 ACF(2) = f2 = 0.64 ACF(3) = f3 = -0.512 ACF(4) = f4 = 0.410

  17. Example; Stock Prices

  18. Example; ACF

  19. Example; PACF

  20. = + e Y Y - 1 t t t $ = Y Y - 1 t t e = - Y Y - 1 t t t Example; Stock Prices • Based on ACF and PACF, appears to be a random walk

  21. Example; Stock Prices

  22. Example; Stock Prices

  23. Non-Stationarity • Two Primary Types • Non-Stationary in Level (Mean) • Non-Stationary Variance

  24. Example; FAD Apparel

  25. Example; FAD Apparel ACF and PACF indicate random walk or more likely, a non- stationary process

  26. = - z Y Y - 1 t t t FAD Apparel (First Differences)

  27. Fad Apparel (1st Diff) ACF and PACF suggest MA model on zt.

  28. Fad Ex; Fitted Model

  29. Fad Ex; ARIMA(0, 1, 1)

  30. FAD Residuals

  31. Non-Stationary Variance

  32. Power Consumption (1st diff)

  33. 2 s µ Y Y t = Y KY - 1 t t Power Consumption • Variance of series is proportional to level • Common in product demand, economy, stock market

  34. = + ln( Y ) ln( Y ) ln( K ) - 1 t t - = ln( Y ) ln( Y ) ln( K ) - = 1 t t Y KY - 1 t t = - ¢ z ln( Y ) ln( Y ) - 1 t t t Transformation

  35. Power Transformed

  36. Power Transformed

  37. Power Transformed

  38. Power Transformed (diff)

  39. AutoCorrelation (Zt) Suggests MA(1) with q = -0.15 ARIMA(0,1,1)

  40. = e - q e ¢ z - 1 1 t t t - = e - q e ln( Y ) ln( Y ) - - 1 1 1 t t t t = + ln( Y ) ln( Y ) 0 . 15 e - - 1 1 t t t Power Model

  41. Fitted Model

  42. Residuals

  43. Fitted Model

  44. = f + f + e Y Y Y - - t 1 t 1 2 t 2 t = f + f + e 2 Y Y Y Y Y Y - - - - - t 1 t 1 t 1 2 t 1 t 2 t t 1 = f + f + e 2 E [ Y Y ] E [ Y ] E [ Y Y ] E [ Y ] - - - - - t 1 t 1 t 1 2 t 1 t 2 t t 1 PACF’s (Partial Derivation) Consider an AR(2) process, Multiply both sides by Yt-1 Take Expectations

  45. = f + f Cov ( Y Y ) Var ( Y ) Cov( Y Y ) - - - - 1 1 1 2 1 2 t t t t t = f + f + e 2 E [ Y Y ] E [ Y ] E [ Y Y ] E [ Y ] - - - - - t 1 t 1 t 1 2 t 1 t 2 t t 1 = f + f Cov ( Y Y ) Var ( Y ) Cov ( Y Y ) - - - 1 1 1 2 1 t t t t t PACF’s (Partial Derivation) by stationarity, independence by stationarity, Cov(Yt-1Yt-2) = Cov(YtYt-1)

  46. f Cov ( Y Y ) cov( Y Y ) = f + - - 1 2 1 t t t t 1 Var ( Y ) Var ( Y ) t t = f + f ACF ( 1 ) ACF ( 1 ) 1 2 = f + f Cov ( Y Y ) Var ( Y ) Cov ( Y Y ) - - - 1 1 1 2 1 t t t t t PACF’s (Partial Derivation) Dividing both sides by Var(Yt-1)

  47. = f + f ACF ( 1 ) ACF ( 1 ) 1 2 PACF’s (Partial Derivation) Total AC = Direct AC + Indirect AC ACF = PACF + Indirect

  48. = f + f + e Y Y Y - - t 1 t 1 2 t 2 t = f f + e 2 Y Y Y Y Y Y + - - - - - t 2 t 1 t 2 t 1 t 2 t t 2 2 PACF’s (Partial Derivation) Consider an AR(2) process, Multiply both sides by Yt-2

  49. = f f + e 2 Y Y Y Y Y Y + - - - - - t 2 t 1 t 2 t 1 t 2 t t 2 2 = f ACF ( 2 ) + ACF f ( 2 ) 1 2 PACF’s (Partial Derivation) Taking expectations, miracle 37b

  50. = f + f ACF ( 1 ) ACF ( 1 ) 1 2 = f ACF ( 2 ) ACF + f ( 2 ) 1 2 PACF’s (Partial Derivation) Two eqs, two unknowns Solve for f1, f2 from ACF(1), ACF(2) Estimate f1 from PACF(1)

More Related