NBS8257 Formative Questions - Solutions N.B. the following discusses the solutions, but should not be taken as ‘model answers’, of a format to be imitated. Q1. We have the regression model and results ∆jpsuzukit = c+ γjpsuzukit−1 + δt+ ut ∆jpsuzukit = 3636.04− 0.6799044jpsuzukit−1 + 157.0193t which is the Dickey-Fuller test equation for the hypotheses H0 : γ ≥ 0 and H1 : γ < 0 (and jpsuzuki stationary around a deterministic trend, given the inclusion of t in the test model). Given the t-statistic associated with γˆ of −11.368 we can reject the null hypothesis (see critical values given in question). Given the conclusion above, the series could be made covariance-stationary by linear detrending. One possible issue with the test regression would be serial correlation of the residuals and incorrect standard errors. This could be addressed by augmenting the test model with lags of the differences. Q2. Given (as an example, as the question refers to ARIMA models, but other examples would be acceptable): yt = β0 + β1yt−1 + · · ·+ βpyt−p + et then the Breusch-Godfrey auxiliary regression would be eˆt = η1yt−1 + · · ·+ ηpyt−p + ξ1eˆt−1 + . . . ξmeˆt−m 1 with null hypothesis ξ1 = · · · = ξm = 0. The test is used in the context of ARIMA models for two reasons. Firstly, in the presence of serial correlation we might choose to use robust standard- errors for the calculation of t-tests etc. Secondly, serial correlation in an AR model will lead to inconsistency of OLS. Q3. ˆEXP t = 0.71 + 0.84GNPt ∆eˆt = 0.2∆eˆt−1 − 0.5eˆt−1 With a t-stat of −7 (note that the negative standard error cannot be correct) we can be confident that we would reject the null of a unit root in the residuals and conclude that the residuals (errors) are stationary and that therefore EXP and GNP are co-integrated. The lagged difference serves the same purpose here as in the general ADF test (as in Q1 above). The ECM would then be ∆EXPt = c+ γ(EXPt−1 − 0.71− 0.84GNPt−1) (1) ∆EXPt = c+ γ(EXPt−1 − 0.71− 0.84GNPt−1) + β1∆EXPt−1 + β2∆GNPt + . . . (2) 1 is the minimum ECM but typically we would opt for something akin to 2, with an attempt to allow for short-run influences also. The value of γ determines whether EXP moves in the direction of equilibrium; given that we found evidence of co-integration, we would expect this to be corroborated by a negative value of γ. Q4. yt = 2.3 + 0.7yt−1 + 0.2yt−2 + ut. Then stationarity can be determined by solving the characteristic equa- tion. 2 (1− 0.7L− 0.2L2)yt = ut −0.2z2 − 0.7z + 1 = 0 z = 0.7±√(−0.7)2 − 4 · (−0.2) · 1 −0.4 = 0.7±√1.29 −0.4 = −4.6, 1.1 Both roots of the characteristic equation are greater than 1 in absolute value, hence the process is covariance-stationary. 3



























































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