1. The file coursework2.xls contains the 100 values of a simulated series entitled Y2. Use this series to perform the following tasks.
i) Plot the sequence against time. Does the series appear to be stationary?
Correlogram of Y2
From the line graph it is clear that the data is stationary. The sequence of the series is plotted against time in order to identify the data pattern in term of stationary or non-stationary. Hence, if the data is found to be stationary it means that the statistics such as standard deviation, mean, relevant to the particular data do not have the tendency to change over time (Brooks, Wang; 2008). One can clearly assume the straight line through which the data has passed over the period of time. Hence, it can be concluded here that the data series appears to be stationary. The pattern of line shown in the graph below shows that the relevant series of data has the stationary pattern as it is clear from the signal consistency in the line graph below.
ii) Report the SACF, SPACF and Ljung-Box portmanteau statistics up to k=24. Plot the SACF and SPACF.
Sample: 1 100
Included observations: 100
Autocorrelation
Partial Correlation
OBS
AC
PAC
Q-Stat
Prob
******|. |
******|. |
1
-0.834
-0.834
71.729
0.000
.|**** |
**|. |
2
0.597
-0.328
108.77
0.000
***|. |
*|. |
3
-0.440
-0.194
129.12
0.000
.|** |
.|. |
4
0.350
-0.015
142.12
0.000
**|. |
*|. |
5
-0.319
-0.140
153.03
0.000
.|** |
.|* |
6
0.332
0.089
164.96
0.000
**|. |
.|. |
7
-0.337
0.000
177.43
0.000
.|** |
.|. |
8
0.317
0.014
188.55
0.000
**|. |
.|. |
9
-0.276
0.017
197.09
0.000
.|* |
*|. |
10
0.179
-0.199
200.72
0.000
*|. |
.|. |
11
-0.084
-0.046
201.53
0.000
.|. |
.|. |
12
0.038
-0.021
201.69
0.000
.|. |
*|. |
13
-0.053
-0.113
202.02
0.000
.|. |
*|. |
14
0.073
-0.082
202.65
0.000
*|. |
*|. |
15
-0.090
-0.066
203.62
0.000
.|* |
.|. |
16
0.101
0.017
204.86
0.000
*|. |
.|* |
17
-0.068
0.099
205.43
0.000
.|. |
.|. |
18
-0.002
-0.061
205.43
0.000
.|. |
.|. |
19
0.049
-0.061
205.73
0.000
.|. |
.|. |
20
-0.055
0.002
206.11
0.000
.|. |
.|. |
21
0.027
-0.057
206.21
0.000
.|. |
*|. |
22
-0.002
-0.106
206.21
0.000
.|. |
.|. |
23
0.006
-0.005
206.21
0.000
.|. |
.|. |
24
-0.034
-0.045
206.37
0.000
SACF and SPACF of the series are obtained and are mentioned in the table above. The correlogram for the series shows that there are 5-6 spikes in ACF, whereas in PACF 1-2 spikes are visible for analysis. This suggests that we might have up to MA (3) and AR (1) specifications. Concerning both ACF and PACF are decaying towards 0 exponentially, this series is of high probability to follow an ARMA (p,q) process. So, the possible models are the ARMA (1, 3), ARMA (1, 2) or ARMA (1, 1) models. We then estimate the three possible models. The command for estimating the ARMA (1, 3) models in EViews is:
Ls Y2 c ar(1) ma(1) ma(2) ma(3)
Similarly, for the ARMA (1, 2), it is:
Ls Y2 c ar(1) ma(1) ma(2)
And for the ARMA (1, 1) it is:
Ls Y2 c ar(1) ma(1)
iii) Identify an ARMA model for this series, explaining carefully all the steps in your argument. [You can consider and examine several tentative models.]
Now, as discussed earlier that the values of autocorrelation function and partial autocorrelation functions shows that the model need to be tested against above three estimates. The details are mentioned below: