# CS计算机代考程序代写 F70TS2: Time Series Multiple Choice Revision Questions

F70TS2: Time Series Multiple Choice Revision Questions
1. A series of length 200 has 145 turning points. Consider a test of the hypotheses: H0:theseriesisnoise(“random”) v H1:theseriesismoreoscillatorythannoise. With respect to this test, which of the following statements is true?
A: The P-value of the observation “145 TPs” is 1 (100%) and H0 can stand
B: The P-value of the observation “145 TPs” is 0.180 (18.0%) and H0 can stand
C: The P-value of the observation “145 TPs” is 0.047 (4.7%) and H0 is rejected at 5%
D: The P-value of the observation “145 TPs” is 0.018 (1.8%) and H0 is rejected at 2.5%
2. The fitted value at the middle of the range for the 5-point least-squares fit of cubic accuracy is:
A: 1[−1,2,3] B: 1 [3,4,9] C: 1 [−3, 12, 17] D: 1 [−1, 10, 21] 5 21 35 39
3. A stationary time series process {Yt} has variance γ0 = 3 and autocorrelation function given by:
ρk =1ik(0.6)k/2{1+(−1)k}=(0.6)k/2cos(kπ/2), k=0,1,2,… 2
The autocovariance between variables with time lag 4 is:
A: 1.08 B: 1.8 C: 10.8 D: 18
4. Consider the process {Yt} defined by Yt = Zt − 0.9Zt−1 + 0.2Zt−2 where {Zt} is a noise process. By using operators, the process can be rewritten in the form Yt = π1Yt−1 + π2Yt−2 + π3Yt−3 +···+Zt.
The values of π1 and π2 are, respectively:
A: –0.9, –0.61 B: –0.9, 0.61 C: 0.9, –0.61 D: 0.9, 0.61
1

5. The process {Yt} is produced as the result of applying the linear filter 1[−1,2,3] to a noise 5∗∗
process {Zt}. Suppose we represent the (normalised) spectrum of {Yt} , f (ω), by f (ω) = ∞
1 + 􏱜ak cos kω. k=1
Which of the following statements is true?
A: a3 =8/19andak =0fork≥5 B: a3 =−8/19andak =0fork≥5 C: a3 =6/19andak =0fork≥4 D: a3 =−6/19andak =0fork≥4
6. Let {Zt} and {Vt} be independent noise processes, and let {Yt} be defined by Yt = Zt + Vt. {Yt} is a:
A: AR(1) process B: MA(1) process C: MA(2) process D: noise process
7. The area under the unnormalised spectrum f(ω) (from 0 to π) for the process {Yt} defined by Yt = Zt + 0.8Zt−1 + 0.1Zt−2, where {Zt} is a noise process with σZ2 = 4 is:
A: 1 B: 6.6π C: π D: 6.6
8. Consider the process {Yt} defined by Yt − 0.6Yt−1 − αYt−2 = Zt. For what range of values of α is the process stationary?
A: −1<α<0.4 B: −1<α<0.6 C: α>−1 D: α<1 9. The process {Yt} defined by Yt = 0.6Yt−1 + 0.4Yt−2 + Zt − Zt−1 is over-parameterised and can be expressed in simpler terms. It is in fact: A: a stationary AR(1) process B: a stationary AR(2) process C: an invertible MA(1) process D: an invertible MA(2) process 10. How do you get from the autocovariance generating function for a time series process {Yt} to the normalised spectrum for the process? A: divide by σY2 B: replace z by e−iω C: replace z by cos ω D: replace z by e−iω and divide by σY2 2 11. Atimeseriesprocess{Yt}oftheARMAclasshasautocovariancegeneratingfunctionCY(z)= 22.5 − 15(z + z−1) + 5(z2 + z−2). The value of ρ2, the autocorrelation coefficient of lag 2, is: A: 0 B: 1/9 C: 2/9 D: 1/3 12. A process in general linear process form is given by 􏰂∞􏰃 Yt = 1+(α+β)􏱜αk−1Bk Zt, |α|<1,|β|<1 k=1 This process is: A: ARIMA(1,0,1) B: ARIMA(1,1,1) C: ARIMA(1,0,0) D: ARIMA(1,1,0) 13. Which of the following processes are invertible? Process 1: Yt = Zt + 0.5Zt−1 + 0.8Zt−2 Process 2: Yt = 0.6Yt−1 + Zt + 0.9Zt−2 Process 3: Yt = Zt + 0.5Zt−1 − 0.5Zt−2 A: 1,2 only B: 1 only C: all three D: 2 only 14. Aprocess{Wt}isdefinedbyWt =Zt+βZt−1 where0<β<1and{Zt}isanoiseprocess withσZ2 =1.Theprocess{Yt}isdefinedbyYt=Wt−γWt−1where0<γ<1. The value of ρ3, the autocorrelation coefficient of lag 3 of the {Yt} process is: A: βγ B: β2γ C: βγ2 D: 0 15. Consider the process {Yt} defined by (1 − B)2Yt = Zt + βZt−1. The coefficient ak in the representation of {Yt} as a general linear process, namely ∞ Yt =􏱜akZt−k k=0 is given by: A: 1+kβ B: 1+kβ2 C: 1+k(1+β) D: 1+k(1+β2) 3 16. An AR(2) model is fitted to the time series y1, . . . , y50. It has forecast function Yt(k) = 0.1Yt(k − 1) + 0.8Yt(k − 2) , k ≥ 1 Based on this model, the forecasts y50(1) = 49.7 and y50(2) = 50.4 are made. Suppose we later observe y51 = 50.1. What is the updated forecast y51(1)? A: 49.96 B: 50.1 C: 50.4 D: 50.44 17. The model is fitted to a time series y1, . . . , y100, where {Zt} is a noise process with variance estimated to (1 − 0.6B − 0.2B2)(1 − B)Yt = (1 − 0.8B)Zt be σ􏲢Z = 1.65. The forecast y100(2) = 57.9 is obtained. What are the 90% prediction limits 2 for this forecast? A: (55.07, 60.73) B: (55.19, 60.61) C: (55.79, 60.01) D: (55.38, 60.42) 4