Question 5. Consider the following time series: Y = -3t + e + 0.25e-1, How we can make the series stationary? A) By log transformation. B) By power transformation. C) By first difference of the series. D) By second difference of the series. 6. How is the following process written using backshift notation? Y - Y-1 = e - 0.64-1 A) (1 - B)Y, = (1 -0.6B) B) BY = (1 -0.6B)e C) B(Y, -Y-1) = 0.6 Bei D) None of the above. 7. What is the name of the following process? Y, -Y-1 = e - 0.6-1 A) IMA(2,2) B) IMA(1,1) C) ARI(1,1) D) None of the above.

987ERV The Asker · Advanced Mathematics

Transcribed Image Text: 5. Consider the following time series: Y = -3t + e + 0.25e-1, How we can make the series stationary? A) By log transformation. B) By power transformation. C) By first difference of the series. D) By second difference of the series. 6. How is the following process written using backshift notation? Y - Y-1 = e - 0.64-1 A) (1 - B)Y, = (1 -0.6B) B) BY = (1 -0.6B)e C) B(Y, -Y-1) = 0.6 Bei D) None of the above. 7. What is the name of the following process? Y, -Y-1 = e - 0.6-1 A) IMA(2,2) B) IMA(1,1) C) ARI(1,1) D) None of the above.

More

Community Answer

Honor Code

Solved 1 Answer

S.5) A stationary time series does not depend on present time.Since given model hes e_{+}and e_{t-1} factors only, the series can be made stationary usingCorrect option \rightarrow C) By First difference of the series6) Given process \rightarrowy_{t}-y_{t-1}=e_{t}-0,6 e_{t-1}using backshift notetion, B y_{t}=y_{t-1}\begin{array}{l}y_{t}-B Y_{t}=e_{t}-0,6 B e_{t} \\y_{t}(1-B)=(1-0,6 B) e_{t}\end{array}Correct optiòn \rightarrow(A)7) Proces given is y_{t}-y_{t-1}=e_{t}-0,6 e_{t-1}The IMA (1,1) is a rendom walk \operatorname{model} with an additional - lagged error term Qet-1 correct oftion \rightarrow B)   NOTE** - If you have any doubts or need extra explanation, please comment. Thank you. ...