TS Module 19 Seasonal models basics
(The attached PDF file has better formatting.)
Time series practice problems seasonal models
*Question 19.1: Seasonal moving average process
A time series is generated by Yt = et – et-12.
What is ñ12?
A. –1.0
B. –0.5
C. 0
D. +0.5
E. +1.0
Answer 19.1: B
The seasonal correlation is –Θ / (1+ Θ2) = –1 / (1 + 12) = –0.5
(See Cryer and Chan, page 229: equation 10.1.3)
*Question 19.2: Seasonal moving average process
A seasonal moving average process has a characteristic polynomial of (1 – èx)(1 – Θx12), with è = 0.4, Θ = 0.5, and ó2e = 4
What is ã0?
A. 0.20
B. 0.80
C. 1.60
D. 3.20
E. 5.80
Answer 19.2: E
(P230: equation 10.2.2: ã0 = (1 + è2)(1+ Θ2) ó2e )
(1.16 × 1.25 × 4 = 5.80
*Question 19.3: Seasonal moving average process
A seasonal moving average process has a characteristic polynomial of (1 – èx)(1 – Θx12), with è = 0.4, Θ = 0.5, and ó2e = 4
What is ñ12?
A. –0.50
B. –0.40
C. 0
D. +0.40
E. +0.50
Answer 19.3: B
(P231: equation 10.2.5: ñ12 = –Θ / (1+ Θ2) = –0.5 / 1.25 = –0.4
*Question 19.4: Seasonal ARMA process
A seasonal ARMA process Yt = Ö Yt-12 + et – è et-1 has Ö = –0.707, è = 1.414, and ó2e = 4
What is ã0?
A. 2
B. 4
C. 8
D. 16
E. 24
Answer 19.4: E
See Cryer and Chan, page 232, equation 10.2.11:
ã0 = ó2e × (1 + è2) / (1 – Ö2) = 4 × (1 + 2) / (1 – ½) = 24
The variance is the product of the seasonal autoregressive process and the non-seasonal moving average process.
*Question 19.5: Seasonal ARMA process
A seasonal ARMA process Yt = Ö Yt-12 + et – è et-1 has Ö = –0.707, è = 1.414, and ó2e = 4
What is ñ11 – ñ13 ?
A. –1.00
B. –0.50
C. 0
D. +0.50
E. +1.00
Answer 19.5: C
(See Cryer and Chan, page 232, equation 10.2.11: ñ12k-1 = ñ12k+1 = – Ök è / (1 + è2)
For a stationary process, ñ1 = ñ–1. We form the autocorrelations for the 12 months seasonal lags, and then add or subtract one month for the non-seasonal lag.
*Question 19.6: Non-stationary seasonal ARIMA process
Suppose Yt = St + et, with ó2e = 2 and St = St-s + åt , with ó2ε = 1
What is the variance of ∇s Yt?
A. 1
B. 2
C. 3
D. 4
E. 5
Answer 19.6: E
∇s Yt = Yt – Yt-s
➾ ∇s Yt = St – St-s + et – et-s = åt + et – et-s
➾ variance ∇s Yt = 1 + 2 + 2 = 5
*Question 19.7: Seasonal AR(1)12 process
A store’s toy sales in constant dollars follow a seasonal AR(1)12 process: Yt = Ö Yt-12 + et.
Sales are $10,000 in January 20X1 and $100,000 in December 20X1.
Projected sales are $110,000 in December 20X2.
What are projected sales for January 20X3?
A. $10,000
B. $11,000
C. $11,100
D. $12,000
E. $12,100
Answer 19.7: E
The December projection is one period ahead.
The January projection is two periods ahead.
(See Cryer and Chan, page 241, last line; Ö = 1.1)
Note that we take logarithms and first differences to make this process stationary.