Question

After preliminary tests were performed on Ghana’s annual inflation rate data spanning

from 1948 to 2015; eight competing models were fitted. The competing models were:

ARIMA(1,1,0), ARIMA(2,1,0), ARIMA(1,1,2), ARIMA(2,1,1), and

Model 1: ARIMA(1,1,0)

Coefficients: ar1 -0.3609 s.e. 0.1134

ARIMA(0,1,1), ARIMA(0,1,2), ARIMA(1,1,1), ARIMA(2,1,2). The output from R are given below:

sigma^2 estimated as 6.761e+09: log likelihood=-852.89 AIC=1709.77 AICc=1709.96 BIC=1714.18

Model 2: ARIMA(2,1,0)

Coefficients:

ar1 ar2

-0.4023 -0.1163 s.e. 0.1211 0.1231

sigma^2 estimated as 6.772e+09: log likelihood=-852.44 AIC=1710.89 AICc=1711.27 BIC=1717.5

Model 3: ARIMA(0,1,1)

Coefficients: ma1 -0.3589 s.e. 0.1032

sigma^2 estimated as 6.749e+09: log likelihood=-852.83 AIC=1709.65 AICc=1709.84 BIC=1714.06

Model 4: ARIMA(0,1,2)

Coefficients:

ma1 ma2

-0.3978 0.0996 s.e. 0.1213 0.1280

sigma^2 estimated as 6.789e+09: log likelihood=-852.52 AIC=1711.05 AICc=1711.43 BIC=1717.66

S. Twumasi-Ankrah

Model 5: ARIMA(1,1,1)

Coefficients:

ar1 ma1

-0.1905 -0.2007 s.e. 0.2755 0.2677

sigma^2 estimated as 6.807e+09: log likelihood=-852.61 AIC=1711.22 AICc=1711.6 BIC=1717.83

Model 6: ARIMA(1,1,2)

Coefficients:

ar1 ma1 ma2

0.1143 -0.5095 0.1375 s.e. 0.8895 0.8752 0.3012

sigma^2 estimated as 6.893e+09: log likelihood=-852.51 AIC=1713.03 AICc=1713.67 BIC=1721.85

Model 7: ARIMA(2,1,1)

Coefficients:

ar1 ar2 ma1

-1.1155 -0.4033 0.7384 s.e. 0.2254 0.1146 0.2265

sigma^2 estimated as 6.632e+09: log likelihood=-851.29 AIC=1710.57 AICc=1711.22 BIC=1719.39

Model 8: ARIMA(2,1,2)

Coefficients:

ar1 ar2 ma1 ma2

-1.2477 -0.5734 0.9035 0.2039 s.e. 0.5603 0.8223 0.7474 1.0497

sigma^2 estimated as 6.676e+09: log likelihood=-851 AIC=1712.01 AICc=1712.99 BIC=1723.03

i. Based on the output above, which model would you choose and why?

ii. Write out your chosen model (hint: use backshift notation)

Answer 1

Now Generally while choosing any model we have couple of things which should be taken care of-

One is that parsimony must be maintained which means the model must have less mumber of parameters as possible to avoid complications in further forecasting and secondly the residual diagnosis which includes Independence of errors, Normality assumptions of residuals, mean must be zero, variance must be constant that is homoskedascity.

Here nothing is given regarding Residual analysis so I will Judge a model only on basis of parsimony and AICc or BIC scores.

Now the model ARIMA(1, 1,0) and ARIMA(0, 1,1) has lowest AICc and BIC. (Neglect minor difference). So we can choose any one of the both models and parsimony is also maintained here.