Building ARIMA Models
In an ARIMA model, we do not have a priori for forecasting model before model identification takes place. ARIMA helps us to choose a “right model” to best fit the time series. Put it in a flow chart:
Demonstration to find "right ARIMA model (p, d, q)" to fit the time series through trial and error:
Firstly, download the excel file called "exchange_rate" from the "Sample Data" of Econ3600 homepage.
Second, open EVIEWS program in this way: click "File", "New", "Workfile" commands, then in the "Workfile Range", choose "Monthly" and type "1990.01" for the "Start observation" and "2000.07" for "End observation" in the dialogue box. Then, we will get a workfile. Next, import the data from the excel file to generate the following result: (Remember to change "B8" for upper left data cell.)
Double click the variable "yen" to check its data whether it is consistent with the Excel file and choose "View", "Line" to get a general idea about the time series is stationary or no. Also, choose "View", "Correlogram" to get the tentatively identify patterns and model components (i.e. the degree of p, d, q of ARIMA) The resulting graphs are:
From the above graphs, you can see that the time series is likely to have random walk pattern, which random walk up and down in the line graph. Also, in correlogram, the ACFs are suffered from linear decline and there is only one significant spike for PACFs. The graph of correlogram suggests that ARIMA(1, 0, 0) may be an appropriate model. Then, we take the first-difference of "Yen" to see whether the time series becomes stationary before further finding AR(p) and MA(q). (Remember that I(d) is used to get stationary series if necessary.)
To see whether first difference can get level-stationary time series or not, you need to generate it by choosing "GENR", type "dyen=d(yen)". Then, you will get "dyen" item in the "Workfile", and use it to draw a line graph and also get a correlogram graph. the results are:
Now, the first-difference series "DYEN" becomes stationary as showing in line graph and is white noise as shown no significant patterns in the graph of correlgram. And the unit root test also confirms the first-difference becomes stationary. The strong evidents support that the ARIMA(0,1,0) is suitable for the time series. Then, we can construct the ARIMA model as following steps:
Step 1. Choosing "Quick", "Estimate Equation", then specify the mode and type " yen c ar(1)",
click "OK", the result is:
Step 2. choosing " View", "Residual tests", "Correlogram-Q- Statistic" the result is:
(Since there is no significant spikes of ACFs and PACFs, it means that the resduals of this selected ARIMA model are white noise, so that there is no other significant patterns left in the time series, then we can stop at here and don't need to further consider another AR(p) and MA(q))
The criterions to judge for the best model are as follows:
Relatively small of BIC (Schwarz criterion which is measured by nLog(SEE)+kLog(n))
Relatively small of SEE
Relatively high adjust R2
Q- statistics and correlogram show that there is no significant pattern left in the ACFs and PACFs of the residuals, it means the residuals of the selected model are white noise.
You may try another ARIMAs and compare the statistical results as in the following table:
|ARIMA model||BIC||Adjusted R2||SEE|
|(1, 0 , 0)||5.708||0.93476||4.075|
|(1, 0, 1)||5.734||0.93503||4.067|
|(2, 0, 0)||5.725||0.93425||4.047|
|(0, 0, 1)||7.384||0.65598||9.422|
|(0, 0, 2)||6.888||0.79754||7.220|
|(1, 1, 0)||5.724||0.0019||4.108|
|(0, 1, 0)||5.708||0.9347||4.075|
As you can see that ARIMA(1,0,0) is a relatively best model,
Remark: The ARIMA (1, 0, 0) is same as ARIMA (0, 1, 0). The result of ARIMA(0,1,0) is:
In our several trial and error procedures, the ARIMA(1,0,0) or ARIMA(0,1,0) is selected as the best model.
Now, we can express this selected best model as
Students are encouraged to try to find the best ARIMA model for the series of "pound".
Now, let's try another complicated time series. Click Here to continue.