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)
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. åt
B. et
C. åt – et – et-1
D. åt + et – et-1
E. 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 – 1
E. 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.