Non-Stationary Time Series and Unit Root Tests Heino Bohn Nielsen - PDF document

Econometrics 2 — Fall 2005 Non-Stationary Time Series and Unit Root Tests Heino Bohn Nielsen 1 of 25 Introduction • Many economic time series are trending. • Important to distinguish between two important cases: (1) A stationary process with a deterministic trend: Shocks have transitory e ff ects. (2) A process with a stochastic trend or a unit root: Shocks have permanent e ff ects. • Why are unit roots important? (1) Interesting to know if shocks have permanent or transitory e ff ects. (2) It is important for forecasting to know if the process has an attractor. (3) Stationarity was required to get a LLN and a CLT to hold. For unit root processes, many asymptotic distributions change! Later we look at regressions involving unit root processes: spurious regression and cointegration. 2 of 25

Outline of the Lecture (1) Di ff erence between trend stationarity and unit root processes. (2) Unit root testing. (3) Dickey-Fuller test. (4) Caution on deterministic terms. (5) An alternative test (KPSS). 3 of 25 Trend Stationarity • Consider a stationary AR(1) model with a deterministic linear trend term: Y t = θY t − 1 + δ + γt + � t , t = 1 , 2 , ..., T, ( ∗ ) where | θ | < 1 , and Y 0 is an initial value. • The solution for Y t (MA-representation) has the form Y t = θ t Y 0 + µ + µ 1 t + � t + θ� t − 1 + θ 2 � t − 2 + θ 3 � t − 3 + ... Note that the mean, E [ Y t ] = θ t Y 0 + µ + µ 1 t → µ + µ 1 t for T → ∞ , contains a linear trend, while the variance is constant: σ 2 V [ Y t ] = V [ � t + θ� t − 1 + θ 2 � t − 2 + ... ] = σ 2 + θ 2 σ 2 + θ 4 σ 2 + ... = 1 − θ 2 . 4 of 25

• The original process, Y t , is not stationary. The deviation from the mean, y t = Y t − E [ Y t ] = Y t − µ − µ 1 t is a stationary process. The process Y t is called trend-stationary. • The stochastic part of the process is stationary and shocks have transitory e ff ects We say that the process is mean reverting. Also, we say that the process has an attractor, namely the mean, µ + µ 1 t . • We can analyze deviations from the mean, y t . From the Frisch-Waugh theorem this is the same as a regression including a trend. 5 of 25 Shock to a Trend-Stationarity Process y t =0.8 ⋅ y t − 1 +ε t 15.0 Y t = y t + 0.1 ⋅ t 12.5 10.0 7.5 5.0 2.5 0.0 0 10 20 30 40 50 60 70 80 90 100 6 of 25

Unit Root Processes • Consider the AR(1) model with a unit root, θ = 1 : Y t = Y t − 1 + δ + � t , t = 1 , 2 , ..., T, ( ∗∗ ) or ∆ Y t = δ + � t , where Y 0 is the initial value. • Note that z = 1 is a root in the autoregressive polynomial, θ ( L ) = (1 − L ) . θ ( L ) is not invertible and Y t is non-stationary. • The process ∆ Y t is stationary. We denote Y t a di ff erence stationary process. • If ∆ Y t is stationary while Y t is not, Y t is called integrated of fi rst order, I(1). A process is integrated of order d , I( d ), if it contains d unit roots. 7 of 25 • The solution for Y t is given by X t X t X t Y t = Y 0 + ∆ Y i = Y 0 + ( δ + � i ) = Y 0 + δt + � i , i =1 i =1 i =1 with moments V [ Y t ] = t · σ 2 E [ Y t ] = Y 0 + δt and Remarks: (1) The e ff ect of the initial value, Y 0 , stays in the process. (2) The innovations, � t , are accumulated to a random walk, P � i . This is denoted a stochastic trend. Note that shocks have permanent e ff ects. (3) The constant δ is accumulated to a linear trend in Y t . The process in ( ∗∗ ) is denoted a random walk with drift. (4) The variance of Y t grows with t . (5) The process has no attractor. 8 of 25

Shock to a Unit Root Process y t = y t − 1 +ε t 2 0 -2 -4 -6 -8 0 10 20 30 40 50 60 70 80 90 100 9 of 25 Unit Root Tests • A good way to think about unit root tests: We compare two relevant models : H 0 and H A . (1) What are the properties of the two models? (2) Do they adequately describe the data? (3) Which one is the null hypothesis? • Consider two alternative test: (1) Dickey-Fuller test: H 0 is a unit root, H A is stationarity. (2) KPSS test: H 0 is stationarity, H A is a unit root. • Often di ffi cult to distinguish in practice (Unit root tests have low power). Many economic time series are persistent, but is the root 0 . 95 or 1 . 0 ? 10 of 25

The Dickey-Fuller (DF) Test • Idea: Set up an autoregressive model for y t and test if θ (1) = 0 . • Consider the AR(1) regression model y t = θy t − 1 + � t . The unit root null hypothesis against the stationary alternative corresponds to H 0 : θ = 1 against H A : θ < 1 . • Alternatively, the model can be formulated as ∆ y t = ( θ − 1) y t − 1 + � t = πy t − 1 + � t , where π = θ − 1 = θ (1) . The unit root hypothesis translates into H 0 : π = 0 against H A : π < 0 . 11 of 25 • The Dickey-Fuller (DF) test is simply the t − test for H 0 : b b θ − 1 π τ = b = π ) . se ( b se ( b θ ) The asymptotic distribution of b τ is not normal! The distribution depends on the deterministic components. In the simple case, the 5% critical value (one-sided) is − 1 . 95 and not − 1 . 65 . Remarks: (1) The distribution only holds if the errors � t are IID (check that!) If autocorrelation, allow more lags. (2) In most cases, MA components are approximated by AR lags. The distribution for the test of θ (1) = 0 also holds in an ARMA model. 12 of 25

Augmented Dickey-Fuller (ADF) test • The DF test is extended to an AR(p) model. Consider an AR(3): y t = θ 1 y t − 1 + θ 2 y t − 2 + θ 3 y t − 3 + � t . A unit root in θ ( L ) = 1 − θ 1 L − θ 2 L 2 − θ 3 L 3 corresponds to θ (1) = 0 . • The test is most easily performed by rewriting the model: y t − y t − 1 = ( θ 1 − 1) y t − 1 + θ 2 y t − 2 + θ 3 y t − 3 + � t y t − y t − 1 = ( θ 1 − 1) y t − 1 + ( θ 2 + θ 3 ) y t − 2 + θ 3 ( y t − 3 − y t − 2 ) + � t y t − y t − 1 = ( θ 1 + θ 2 + θ 3 − 1) y t − 1 + ( θ 2 + θ 3 )( y t − 2 − y t − 1 ) + θ 3 ( y t − 3 − y t − 2 ) + � t ∆ y t = πy t − 1 + c 1 ∆ y t − 1 + c 2 ∆ y t − 2 + � t , where π = θ 1 + θ 2 + θ 3 − 1 = − θ (1) c 1 = − ( θ 2 + θ 3 ) c 2 = − θ 3 . 13 of 25 • The hypothesis θ (1) = 0 again corresponds to H 0 : π = 0 against H A : π < 0 . The t − test for H 0 is denoted the augmented Dickey-Fuller (ADF) test. • To determine the number of lags, k , we can use the normal procedures. — General-to-speci fi c testing: Start with k max and delete insigni fi cant lags. — Estimate possible models and use information criteria. — Make sure there is no autocorrelation. • Verbeek suggests to calculate the DF test for all values of k . This is a robustness check, but be careful! Why would we look at tests based on inferior/misspeci fi ed models? • An alternative to the ADF test is to correct the DF test for autocorrelation. Phillips-Perron non-parametric correction based on HAC standard errors. Quite complicated and likely to be inferior in small samples. 14 of 25

Deterministic Terms in the DF Test • The deterministic speci fi cation is important: — We want an adequate model for the data. — The deterministic speci fi cation changes the asymptotic distribution. • If the variable has a non-zero level, consider a regression model of the form ∆ Y t = πY t − 1 + c 1 ∆ y t − 1 + c 2 ∆ y t − 2 + δ + � t . The ADF test is the t − test, b τ c = b π/ se ( b π ) . The critical value at a 5% level is − 2 . 86 . • If the variable has a deterministic trend, consider a regression model of the form ∆ Y t = πY t − 1 + c 1 ∆ y t − 1 + c 2 ∆ y t − 2 + δ + γt + � t . The ADF test is the t − test, b τ t = b π/ se ( b π ) . The critical value at a 5% level is − 3 . 41 . 15 of 25 The DF Distributions DF with a linear trend, τ t 0.5 DF with a constant, τ c DF, τ 0.4 N(0,1) 0.3 0.2 0.1 -5 -4 -3 -2 -1 0 1 2 3 4 16 of 25

Empirical Example: Danish Bond Rate 0.200 0.175 0.150 Bond rate, r t 0.125 0.100 0.075 0.050 First difference, ∆ r t 0.025 0.000 -0.025 1970 1975 1980 1985 1990 1995 2000 2005 17 of 25 • An AR(4) model gives ∆ r t = − 0 . 0093 ( − 0 . 672) r t − 1 + 0 . 4033 (4 . 49) ∆ r t − 1 − 0 . 0192 ( − 0 . 199) ∆ r t − 2 − 0 . 0741 ( − 0 . 817) ∆ r t − 3 + 0 . 0007 (0 . 406) . Removing insigni fi cant terms produce a model ∆ r t = − 0 . 0122 ( − 0 . 911) r t − 1 + 0 . 3916 (4 . 70) ∆ r t − 1 + 0 . 0011 (0 . 647) . The 5% critical value ( T = 100 ) is − 2 . 89 , so we do not reject the null of a unit root. • We can also test for a unit root in the fi rst di ff erence. Deleting insigni fi cant terms we fi nd a preferred model ∆ 2 r t = − 0 . 6193 ( − 7 . 49) ∆ r t − 1 − 0 . 00033 . ( − 0 . 534) Here we safely reject the null hypothesis of a unit root ( − 7 . 49 ¿ − 2 . 89) . • Based on the test we concluce that r t is non-stationary while ∆ r t is stationary. That is r t ∼ I (1) 18 of 25