Exhibit 5
Unit Root Test for Nonstationarity and the Test for Heteroskedasticity
Unit root test statistic | −18.7402 |
Unit root test critical value at the 5% level of significance | −2.89 |
Heteroskedasticity test statistic | 2.016733 |
Heteroskedasticity test critical value at the 5% level of significance | 1.96 |
Q. Based on the results reported in Exhibit 5, the AR(1) model is best described as having:
- a unit root.
- heteroskedasticity in the error term variance.
- reliable standard errors.
B is correct. Because the unit root test statistic (–18.7402) is smaller than the critical value (–2.89), the AR(1) model does not exhibit a unit root. The test for heteroskedasticity, however, suggests that the error term variances are heteroskedastic. The heteroskedasticity test statistic (2.016733) is greater than the critical value (1.96). A more sophisticated approach, such as generalized least squares, is needed.
A is incorrect because the significantly negative test statistic strongly suggests the absence of a unit root.
C is incorrect because when a model exhibits ARCH, the standard errors for the regression parameters will not be correct.
my question is-
for testing hetroskedasticity in multiple regression we used a chi2 distribution but for testing ARCH in AR models we use t distribution but for concluding abt hetroskedasticity we have to accept h0 or reject it i mean what is the process of testing?
Yes, the testing process is different in Multiple Regression and in Time Series.
If this is your doubt, I repeat, the process is different in MR and Time Series. If this is not your doubt, then I am unclear as to what is your doubt.
In the above problem, there is a Dickey Fuller Test for Unit Root –
H0 : G1 greater than equal to 0
H1 : G1 less than 0
Since the Unit Root test stat, i.e. -18.7402 is less than the critical value of -2.89, H0 rejected, so there is no problem of Unit Root.
The hateroskedasticity test concerns running a regression of et^2 over et-1^2 and the slope coefficient is A1, which we wish to be 0, so that there is no CH.
H0 : a1 equal to 0
H1 : a1 not equal to 0
Here, the hateroskedasticity statistic i.e. 2.016733 is greater than the critical value of 1.96. Hence H0 rejected. And this is an ARCH series. So the answer is option B.