Confidential

1. i.
I support the application of AR (2) by Peter because it increases the user knowledge and information. The application of AR (2) is a disadvantage because quite expensive for everyday application and it is not accessible to small businesses.
ii.
When describing the process, it is important to study both the ACF plot and the PACF plot together. It is anticipated that the ACF plots for AR (2) would show a gradual decrease, but the PACF plots are anticipated to show a sudden reduction after the first substantial delays. The ACF should show a sudden decline after a certain number of delays (q), while the PACF should show a geometric or progressive decline in trend for the ACF and PACF plots to indicate the opposite of what we expect to see in an MA process. In other words, the ACF should show the reverse of what we expect to see in an MA process.
iii.
The proposed Elizabeth’s MA (1) is suitable because it’s smooth nature and lower probability of producing false signal. The smoothness allows the users of the MA (1) to clearly identify trends in the market. The lower probability of false signal allows its users to be in a better position to identify opportunities in the market.
iv.
Model one
Xt = 0.77Xt−1 + 0.21Xt−2 +Zt
X – 0.98x = Z
0.02 (0.8661) = Z
= 0.0173
Model two
Yt = Zt −0.19Zt−1, where Yt = Xt −Xt−1
Thus:-
Zt −0.19Zt−1 = Xt −Xt−1
z = 0.81(0.8806)
= 0.7133
v.
Model one
Using the value obtained 0.0173 and the aic as follows:-
=0.0173 * 547.89
= 9.4785
Model two
Using the obtained value 0.7133 and the aic as follows:-
= 0.7133 * 543.46
= 387.65
vi.
We are analyzing this specific variable with the help of a model called ARIMA (1,1,1), which has both an AR and an MA term. One of the possible explanations for the linear trend in the data is a first difference.
2.
i.
Xt = 6 – 0.7 + 0.5 + 0.9
= 6.7
ii.
iii.
Let k be 2 which is equal or greater to 2
3.097 = 0.7p2-1
0.7P1 = 3.097
P1 = 4.4243
iv.
The ARIMA (1, 1) model is better than the AR (1) model based on the inclusion of 0.5Zt−1 in the ARIMA (1, 1) model. The ARIMA (1, 1) model performs better in short period of time as compared to AR (1) that doesn’t perform better in short period.
3.
i.
Xt = 26+0.5Xt−1 −0.8Xt−2 +Zt +0.6Zt−2 −0.4Zt−3
t = 300
Thus:-
X300 = 26+0.5X300−1 −0.8X300−2 +Z300 +0.6Z300−2 −0.4Z300−3
X300 = 26+0.5X299 −0.8X298 +Z300 +0.6Z298 −0.4Z297
ii.
The model in long-term profitability is 20 when l → ∞ because of the factor 1.22.
iii.
The increament from 300 to 301 is subjected to an increase and so forth:-
X301 = 1.2 + 1(0.3) = 1.5
X 302 = 1.2 + 2(0.3) = 1.8
X 303 = 1.2 + 3(0.3) = 2.1
iv.
The two-step-ahead forecast error is 2(0.3) = 0.6 and the variance is 0.3