Lecture: Testing Stationarity: Structural Change Problem

Transcription

1 Lecture: Testing Stationarity: Structural Change Problem Applied Econometrics Jozef Barunik IES, FSV, UK Summer Semester 2009/2010 Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

2 Introduction Outline Outline of the today s talk Notes about Stability and Structural Breaks. Stationarity when structural breaks are present. Examples: U.S. output, Czech unemployment. Other tools for testing stability - Chow test from a time series perspective. Appendix: Dickey-Fuller, Monte Carlo and how is it with the strange critical values story. Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

3 Stability and Structural Change Stability and Structural Change Implicit assumption of almost all econometric models is constant coefficients over time/over the sample. Aim of econometric model is usually to recover the true data generating process (often referred as DGP) - the structure of the data to allow for forecasting... If not fulfilled: form of misspecification error. That means the model is not appropriate representation of the data Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

4 Augmented Dickey-Fuller test fails in case of structural break: bias towards non-rejection of unit root. Suppose AR(1) process with coef. 0.5 and the level shock: y t = 0.5y t 1 + ɛ t + D L, (1) where D L is a dummy variable such that D L = 0 for t = 1,..., 50 and D L = 3 for t = 51,..., 100 Note the L subscript indicates the level change Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

5 Is this process stationary? Figure: AR(1) with φ 1 = 0.5 and level shock Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

6 What about y t = α 0 + α 1 t + ɛ t fit? Figure: AR(1) with φ 1 = 0.5 and level shock Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

7 Estimation will be misspecified by fitting y t = α 0 + α 1 y t 1 + ɛ t and will tend to mimic the trend line. Thus α 1 estimate will be biased toward unity. This bias in α 1 means that the Dickey-Fuller test will be biased toward accepting the null hypothesis of a unit root, even though the series is stationary within each of the subperiods. Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

8 In practice, the structural change may not be as apparent as the break from our artificial example. Problem: Accepting unit-root hypothesis even though the series is (trend-)stationary within each period. An artificial problem that can be solved easily after visual inspection? Might be... But the distinction between trend and difference stationary matters. Why? Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

9 If the time series is Trend Stationary (TS), the tend is the most important feature of the long-term behavior of that particular series. So that knowing that time series is TS even with some structural breaks might have a great impact on forecasts ( predpoved line on the picture). Discussion about the U.S. GDP in Nelson-Plosser (1982), Perron (1989)... Lecture: Testing Stationarity: Structural Change Summer ProblemSemester 2009/ / 21

10 Trend Stationarity and Structural Change Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

11 Simple approach to test stationarity in the presence of a structural break is to split the time series to segments and test ADF on each of them Caveats: Loosing degrees of freedom (twice so high number of parameters). Prior knowledge about the break needed. Perron s test (Perron, 1989) solves these problems. Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

12 Perron s test Perron s test Assume a structural change at time t = τ + 1. Then, H 1 : y t = α 0 + y t 1 + µ 1 D P + ɛ t, A 1 : y t = α 0 + a 2 t + µ 2 D L + ɛ t, where D P is pulse dummy such that D P = 1 if t = τ + 1 and zero otherwise. and D L represents a level dummy such that D L = 1 if t > τ and zero otherwise. Under the null H 1, {y t } is a unit root process with one-time jump in the level in period t = τ + 1 Under the alternative H A, {y t } is trend-stationary with a one jump in the intercept. Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

13 Perron s test Perron s test Perron s test thus allows to distinguish between difference stationarity and trend stationarity. To test the H 0 we need to estimate following equation: y t = α 0 + α 1 y t 1 + α 2 t + µ 1 D L + µ 2 D P + k β i y t i + ɛ t, (2) and compare the t-statistic for the null α 1 = 1 with the critical values by Perron(1989): i=1 Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

14 Perron s test Perron s test y t = α 0 + α 1 y t 1 + α 2 t + µ 1 D L + µ 2 D P + Why not simply t-statistics. k β i y t i + ɛ t, (3) The logic as in ADF test it can be proved that in case of RW the OLS method fails to estimate the α 1 correctly. Because of autocorrelation there is a significant bias to underestimate the value. i=1 Critical values derived using Monte Carlo method. Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

15 Perron s test Perron s test - Extensions y t = α 0 + α 1 y t 1 + α 2 t + µ 1 D L + µ 2 D P + k β i y t i + ɛ t, (4) Furthermore Perron extended his test to more general examples that allow for a change in slope and combination of both. i=1 For trend: trend dummy D T starting from τ + 1: y t = α 0 + α 1 y t 1 + α 2 t + γd T + k β i y t i + ɛ t, (5) Further extensions for more complex transitions from one level to another. i=1 Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

16 U-R test U-R test with structural breaks in JMulti Slightly more advanced version following Lanne, Lutkepohl and Saikkonen (2002): Deterministic part estimated via GLS, substracted from original series and the resulting series tested in a ADF test fashion. Lanne et al. (2002) showed that their test performs better on small samples. Again own critical values, reported in each test window. Next slide shows types of the structural shift included in JMulti. Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

17 Level shifts types in JMulti U-R test Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

18 U-R test Level shifts types in JMulti y t = α 0 + α 1 f t (θ)γ +... Exponential shift: t < τ... 0 t τ... (1 exp( θ(t τ + 1))) Rational shift: t < τ... 0, t = τ... γ 1 t τ... (γ 1 + T τ j=1 θi 1 (θγ 1 + γ 2 )) θ determines speed of the transition, the larger, the faster. Next slide: some examples Note that in Jmulti the break date means start of the transition, the Jmulti functions are not centered as on the next picture. Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

19 Level shifts types in JMulti U-R test Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

20 Conclusion Conclusion Distinction between Trend- and Difference- stationarity important for forecasting: the knowledge about trend is very valuable. Sometimes the picture about the nature of stationarity is distorted by the presence of structural breaks (the UR tests don t work in these cases). Perron s test solves this problem. As in ADF critical values derived from Monte Carlo simulations. This leads us to the more general question of stability, which is usually tested using the Chow Test. Another tests and different perspectives will be presented during next lectures and seminars. One problem connected with structural breaks: forecasting. Usually structural breaks become evident after some delay. Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21

21 Examples during the Seminar Conclusion Thank you for your Attention! Lecture: Testing Stationarity: Structural Change Summer Problem Semester 2009/ / 21