## TSS module 9 Time series differencing

 Author Message NEAS Supreme Being         Group: Administrators Posts: 4.3K, Visits: 1.3K TSS module 9 Time series differencing Cryer and Chan show how differencing converts a non-stationary ARIMA process to a stationary time series.  ** Question 1.2: Linear over three time points Suppose that Yt = Mt + et     with      Mt = Mt-1 + åt  The time series M(t) is linear over three consecutive points, so the least squares estimator of M(t) is ⅓ (Yt-1 + Yt + Yt+1)    Which of the following is stationary? A.      Y(t)B.       Y(t)C.       2Y(t)D.        Y(t)E.      Y(t) – Y(t-1) + Y(t-2)  Answer 1.2: C If Mt is linear over three points, the best estimate for Mt is the centered moving average of three Yt values:  If we remove the trend, the time series is stationary.   Simplify the right hand side of this equation:  = –⅓ (Yt+1 –2 Yt + Yt-1) = –⅓ ∇(∇Yt+1) = –⅓ ∇2(Yt+1) See Cryer and Chan, chapter 5, page 91  **Question 1.3: First difference  Suppose Yt = Mt + et and Mt = Mt-1 + åt  What is ∇Yt (the first difference of Yt)?  A.      åtB.      etC.      åt – et – et-1D.     åt + et – et-1E.      åt + et + et-1 Answer 1.3: D (See Cryer and Chan, chapter 5, page 90, Equation 5.1.9) ∇Yt = Yt – Yt-1 = Mt + et – (Mt-1 + et-1) = åt + et – et-1 Final exam problems give Yt = k × Mt + et (where k is a scalar) and the variances of ei and åt. You must work out variances, auto-covariances, and autocorrelations.  **Question 1.4: ARIMA(p,1,q) process An ARIMA(p,1,q) process Yt has first differences ∇ Yt that are an ARMA(p,q) process with parameters öi and èj. The ARIMA(p,1,q) process is written as a non-stationary moving average process ARMA(p+1, q). What is the coefficient of Yt-2 in the non-stationary ARMA(p+1, q) process?   A.      1 – ö2 B.      ö1 – ö2 C.      ö2 – ö1 D.     ö2 – 1E.      1 – ö1 – ö2  Answer 1.4: C  See equation 5.2.2 on page 92. Intuition: The ARIMA(p,q) process is Yt – Yt-1 = ö1 (Yt-1 – Yt-2) + ö2 (Yt-2 – Yt-3) + … + öp (Yt-p – Yt-p-1) + åt – è1 åt-1 – è2 åt=2 – … – èq åt-q  Rewrite this as Yt = (1 + ö1) Yt-1 + (ö2 – ö1) Yt-2 + (ö3 – ö2) Yt-3 + … + (öp – öp-1) Yt-p + + åt – è1 åt-1 – è2 åt=2 – … – èq åt-q   **Question 1.5: ARIMA(p,1,q) process An ARIMA(p,1,q) process Yt has first differences ∇ Yt that are an ARMA(p,q) process with parameters öi and èj. The ARIMA(p,1,q) process is written as a non-stationary moving average process ARMA(p+1, q). What is the coefficient of åt-2 in the non-stationary ARMA(p+1, q) process?   A.      –è2 B.      1 – è2 C.      è1 – è2 D.       è2 – è1 E.      è2 – 1 Answer 1.5: A See equation 5.2.2 on page 92. Intuition: The ARIMA(p,q) process is Yt – Yt-1 = ö1 (Yt-1 – Yt-2) + ö2 (Yt-2 – Yt-3) + … + öp (Yt-p – Yt-p-1) + åt – è1 åt-1 – è2 åt=2 – … – èq åt-q  Rewrite this as Yt = (1 + ö1) Yt-1 + (ö2 – ö1) Yt-2 + (ö3 – ö2) Yt-3 + … + (öp – öp-1) Yt-p + + åt – è1 åt-1 – è2 åt=2 – … – èq åt-q  The coefficients of the error terms åt-k remain unchanged; they are the negatives of the moving average parameters.    Attachments TSS module 9 Time series differencing df.pdf (1.9K views, 72.00 KB) NNact Forum Newbie         Group: Forum Members Posts: 1, Visits: 1 Hi NEAS, In Q 1.3 you note: "Final exam problems give Yt = k × Mt + et (where k is a scalar) and the variances of ei and åt. You must work out variances, auto-covariances, and autocorrelations."Can you possibly post an example and work through the solution?  I'd like to make sure I understand what the exam problems will be like on this section.  Thank you. minnie53053 Junior Member         Group: Forum Members Posts: 11, Visits: 1 NEAS:Question 1.2: Linear over three time pointsSince Yt = Mt + et with Mt = Mt-1 + sigma t, then ,delta Yt =delta Mt + delta et =sigma t +et-e(t-1)So, delta Yt is stationary?
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