Profession Assistance with Econometrics Assignment on Forecasting
Here we are focused on providing help with forecasting, which is very common in most econometrics assignment. Given the formula, the forecaster is wrong. The reason is: since the data is monthly data, the monthly percentage change in IP should just be the simple percentage change, i.e., we won't multiply it by 1200. Multiplying it by 1200 has turned it to an annualized percentage change. The forecast will be
The test statistics is 1.4, which is less than the critical value of 2. For the econometrics assignment, we conclude that the coefficient is not statistically significant. The critical value for the QLR test at 5 restrictions and 5% level of significance is 3.66, which is less than the test statistics value of 3.94. We reject the null hypothesis at 5% level. There is a break in the data. The formula for BIC is
The BIC and AIC result is shown below
| Lag | SSR | k | e^k/n | e^2k/n | BIC | AIC |
| 0 | 19533 | 1 | 1.002981 | 1.00597 | 58.3072 | 58.481 |
| 1 | 18643 | 2 | 1.00597 | 1.011976 | 55.81637 | 56.1496 |
| 2 | 17377 | 3 | 1.008969 | 1.018018 | 52.18109 | 52.64908 |
| 3 | 16285 | 4 | 1.011976 | 1.024095 | 49.0477 | 49.63509 |
| 4 | 15842 | 5 | 1.014992 | 1.030209 | 47.85568 | 48.57314 |
| 5 | 15824 | 6 | 1.018018 | 1.03636 | 47.94378 | 48.80761 |
| 6 | 15824 | 7 | 1.021052 | 1.042547 | 48.08668 | 49.09899 |
The number of lags that should be added is four since it has the lowest BIC value. The result is the same if AIC is used.
Help with Econometrics Homework on the Test Statistic
We test the significance of ln(IPt-1)
The test statistics were conducted by our econometrics homework helper, and this is -1.93, which is less than the critical value of 2. We conclude that there is a unit root in the series. Yes, the specification is correct. Even though ln(IP) is not stationary, but Y is stationary because its formula contains a differenced ln(IP). The critical value of F at 5% level and (4, 9) degree of freedom are 3.633. Since the test statistic value of 4.16 is larger than the critical value in the econometrics homework, therefore, have to go with the alternative hypothesis. Therefore, interest rates help in the prediction of IP growth. The critical value of F at 5% level and (4, 9) degree of freedom are 3.633. Since the test statistic value of 1.52 is lower than the critical value. We, therefore, shall not reject the null hypothesis statement that IP growth does not granger the cause of interest rates. Therefore, we conclude that IP growth does not help to predict interest rates.
Online econometrics tutors fitting an AR(2) model
The inflation rate is measured in percentage per year.
The online econometrics tutor identified that indeed, inflation has a stochastic trend as we can see from the plot that, wecan see that the series fluctuated and does not revert to a constant mean. Bi
The plot above is the time series plot of dinfl. We see from the plot that successive values follows each order closely, which means there is significant autocorrelation and is thus consistent with the significant serial correlation computed in part (i) Results using AR(2) Model
The regression result above shows that the coefficient of ∆Inflt-1is significantly different from 0 because the p-value on the coefficient is less than 1% (p<0.001). Since the coefficient of ∆Inflt-1is significant, understanding the rate of change in inflation this quarter helps in predicting the inflation change next quarter. Since the AR(2) term is significant, it means the AR(2) term is not redundant. We will use AIC and BIC to select the best model between AR(1) and AR(2).
| lag | SSR | k | e^k/n | e^2k/n | BIC | AIC |
| 1 | 3771.817 | 2 | 1.01005 | 1.020201 | 19.04862 | 19.24006 |
| 2 | 3508.778 | 3 | 1.015113 | 1.030455 | 17.80903 | 18.07818 |
The BIC and AIC result is shown below
| lag | ll | D | N | BIC | AIC |
| 0 | -584.108 | 1 | 200 | 1170.517 | 1170.216 |
| 1 | -577.521 | 2 | 200 | 1159.644 | 1159.042 |
| 2 | -570.366 | 3 | 200 | 1147.635 | 1146.732 |
| 3 | -570.358 | 4 | 200 | 1149.919 | 1148.715 |
| 4 | -569.144 | 5 | 200 | 1149.792 | 1148.287 |
| 5 | -567.897 | 6 | 200 | 1149.6 | 1147.794 |
| 6 | -567.102 | 7 | 200 | 1150.311 | 1148.203 |
| 7 | -567.013 | 8 | 200 | 1152.434 | 1150.025 |
| 8 | -566.929 | 9 | 200 | 1154.568 | 1151.859 |
Stochastic Trend
The p-value for the test is 0.0819>0.05. This means we have to accept the null hypothesis for a unit root. We conclude there is a stochastic trend in inflation. From the time series plot of inflation in ai, we noticed there is no deterministic trend in the inflation series. This is deduced from the fact that the plot does not maintain either an upward or downward trend throughout the period. This suggests there is no need to include trends in the ADF test. Moreover, the result when the trend is included displayed below shows that the p-value on the trend term is 0.066>0.05, which means the trend term is not significant. Thus, we conclude that the ADF test based on Equation (14.31) is preferred to the test on Equation (14.32), which tested for the stochastic trend in Infl.
The lag selection criteria all picked lag 2 as the optimal lag. Therefore, we should have used not to have used more or fewer lags. Even though the test suggests that there is a unit root in inflation, however, inflation itself is a differenced variable from the price index for personal consumption expenditures (PCEP), which is an I(1) variable which means inflation ought to be stationary and the AR(2) model. Moreover, checking the p-value of the test, we would see that we could have rejected the null hypothesis of unit root if we had set a higher level of significance (10%). es
The p-value of the test is 0.5461>0.05, which connotes the non-rejection of the null hypothesis that there is no structural break. With the absence of a structural break, we can conclude that the AR(2) model stable
