Econometrics 2 Non-Stationary Time Series and Unit Root Tests Heino Bohn Nielsen 1 of 28
Outline (1) Deviations from stationarity: • Trends. • Level shifts. • Variance changes. • Unit roots. (2) More on autoregressive unit root processes. (3) Dickey-Fuller unit root test. • AR(1). • AR(p). • Deterministic terms. (4) Further issues. 2 of 28
Stationarity • The main assumption on the time series data so far has been stationarity. Recall the de fi nition: A time series is called weakly stationary if E [ y t ] = µ V [ y t ] = E [( y t − µ ) 2 ] = γ 0 Cov [ y t , y t − k ] = E [( y t − µ ) ( y t − k − µ )] = γ k for k = 1 , 2 , ... • This can be violated in di ff erent ways. Examples of non-stationarity: (A) Deterministic trends (trend stationarity). (B) Level shifts. (C) Variance changes. (D) Unit roots (stochastic trends). 3 of 28
Four Non-Stationary Time Series (A) Stationary and trend-stationary process (B) Process with a level shift 10 x t 5 5 ~ x t 0 0 0 50 100 150 200 0 50 100 150 200 (C) Process with a change in the variance (D) Unit root process 10 5 5 0 0 -5 0 50 100 150 200 0 50 100 150 200 4 of 28
(A) Trend-Stationarity • Observation: Many macro-economic variables are trending. We need a model for a trending variable, y t . • Assume that x t is stationary and that y t is x t plus a deterministic linear trend, e.g. x t = θx t − 1 + � t , | θ | < 1 , y t = x t + µ 0 + µ 1 t. • Remarks: (1) y t has a trending mean, E [ y t ] = µ 0 + µ 1 t , and is non-stationary. (2) The de-trended variable, b x t = y t − b µ 0 − b µ 1 t , is stationary. We say that y t is trend-stationary . (3) The stochastic part is stationary and standard asymptotics apply to b x t . (4) Solution: Extend the regression with a deterministic trend, e.g. y t = β 0 + β 1 · z t + β 3 · t + � t . 5 of 28
(B) Level Shifts and Structural Breaks • Another type of non-stationarity in due to changes in parameters, e.g. a level shift: ½ µ 1 for t = 1 , 2 , ..., T 0 E [ y t ] = µ 2 for t = T 0 + 1 , T 0 + 2 , ..., T. • If each sub-sample is stationary, then there are two modelling approaches: (1) Include a dummy variable ½ 0 for t = 1 , 2 , ..., T 0 D t = 1 for t = T 0 + 1 , T 0 + 2 , ..., T in the regression model, y t = β 0 + β 1 · z t + β 3 · D t + � t . If y t − β 3 · D t is stationary, standard asymptotics apply. (2) Analyze the two sub-samples separately. This is particularly relevant if we think that more parameters have changed. 6 of 28
(C) Changing Variances • A third type of non-stationary is related to changes in the variance. An example is y t = 0 . 5 · y t − 1 + � t , where ½ N (0 , 1) for t = 1 , 2 , ..., T 0 � t ∼ N (0 , 5) for t = T 0 + 1 , T 0 + 2 , ..., T The interpretation is that the time series covers di ff erent regimes. • A natural solution is to model the regimes separately. • Alternatively we can try to model the variance. We return to so-called ARCH models for changing variance later. 7 of 28
(D) Unit Roots • If there is a unit root in an autoregressive model, no standard asymptotics apply! Consider the DGP y t = θy t − 1 + � t , � t ∼ N (0 , 1) , for t = 1 , 2 , ..., 500 , and y 0 = 0 . Consider the distribution of b θ . • Note: the shape, location and variance of the distributions. (C) Distribution of ^ (D) Distribution of ^ θ for θ =0.5 θ for θ =1 Distribution of ^ θ Distribution of ^ θ 10.0 N(s=0.0389) N(s=0.00588) 100 7.5 5.0 50 2.5 0.0 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.97 0.98 0.99 1.00 1.01 8 of 28
Properties of a Stationary AR(1) • Consider the AR(1) model y t = θy t − 1 + � t , t = 1 , 2 , ..., T. The characteristic polynomial is θ ( z ) = 1 − θz , with characteristic root z 1 = θ − 1 and inverse root φ 1 = z − 1 = θ . The stationarity condition is that | θ | < 1 . 1 • Recall the solution y t = � t + θ� t − 1 + θ 2 � t − 2 + ... + θ t − 1 � 1 + θ t y 0 , where θ s → 0 . Shocks have only transitory e ff ects; y t has an attractor. • Note the properties E [ y t ] = θ t y 0 → 0 σ 2 V [ y t ] = σ 2 + θ 2 σ 2 + θ 4 σ 2 + ... + θ t − 1 σ → 1 − θ 2 ρ s = Corr ( y t , y t − s ) = θ s 9 of 28
Simulated Example θ = 0.8 θ = 1 (A) Shock to a stationary process, (B) Shock to a unit root process, 10 10 5 0 0 0 20 40 60 80 100 0 20 40 60 80 100 θ =0.8 θ =1 (C) ACF for stationary process, (D) ACF for unit root process, 1.0 1.0 0.5 0.5 0 5 10 15 20 25 0 5 10 15 20 25 10 of 28
Autoregression with a Unit Root • Consider the AR(1) with θ = 1 , i.e. y t = y t − 1 + � t . The characteristic polynomial is θ ( z ) = 1 − z . There is a unit root, θ (1) = 0 . • The solution is given by X t y t = y 0 + ∆ y 1 + ∆ y 2 + ... + ∆ y t = y 0 + � 1 + � 2 + ... + � t = y 0 + � i . i =1 • Note the remarkable di ff erences between θ = 1 and a stationary process, | θ | < 1 : (1) The e ff ect of the initial value, y 0 , stays in the process. And E [ y t | y 0 ] = y 0 . (2) Shocks, � t , have permanent e ff ects. Accumulated to a random walk component, P � i , called a stochastic trend. 11 of 28
(3) The variance increases, hX t i = t · σ 2 . V [ y t ] = V i =1 � i The process is clearly non-stationary. (4) The covariance, Cov ( y t , y t − s ) , is given by E [( y t − y 0 )( y t − s − y 0 )] = E [( � 1 + � 2 + ... + � t )( � 1 + � 2 + ... + � t − s )] = ( t − s ) σ 2 . The autocorrelation is r ( t − s ) σ 2 Cov ( y t , y t − s ) t − s p p Corr ( y t , y t − s ) = = tσ 2 · ( t − s ) σ 2 = , t V [ y t ] · V [ y t − s ] which dies out very slowly with s . (5) The fi rst di ff erence, ∆ y t = � t , is stationary. y t is called integrated of fi rst order, I (1) . 12 of 28
Deterministic Terms • To model actual time series we include deterministic terms, e.g. y t = δ + θy t − 1 + � t . With a unit root, the terms accumulate! • If | θ | < 1 , the solution is X t θ i � t − i + (1 + θ + θ 2 + ... ) δ, y t = θ t y 0 + i =0 where the mean converges to (1 + θ + θ 2 + ... ) δ → δ/ (1 − θ ) . • If θ = 1 , the solution is X t X t y t = y 0 + ( δ + � i ) = y 0 + δt + � i . i =1 i =1 The constant term produces a deterministic linear trend: Random walk with drift. Note the parallel between a deterministic and a stochastic trend. 13 of 28
Unit Root Testing • Estimate an autoregressive model and test whether θ (1) = 0 , i.e. whether z = 1 is a root in the autoregressive polynomial. • This is a straightforward hypothesis test! We compare two relevant models: H 0 and H A . The complication is that the asymptotic distributions are not standard. • Issues: (1) Remember to specify the statistical model carefully! (2) What kinds of deterministic components are relevant: constant or trend? (3) What are the properties of the model under the null and under the alternative. Are both H 0 and H A relevant? (4) What is the relevant distribution of the test statistic? 14 of 28
Dickey-Fuller Test in an AR(1) • Consider an AR(1) model y t = θy t − 1 + � t . The unit root hypothesis is θ (1) = 1 − θ = 0 . The one-sided test against stationarity: H 0 : θ = 1 against H A : − 1 < θ < 1 . • An equivalent formulation is ∆ y t = πy t − 1 + � t , where π = θ − 1 = − θ (1) . The hypothesis θ (1) = 0 translates into H 0 : π = 0 H A : − 2 < π < 0 . against • The Dickey-Fuller (DF) test statistic is simply the t − ratio, i.e. b b θ − 1 π b τ = = π ) . se ( b se ( b θ ) The asymptotic distribution is Dickey-Fuller, DF, and not N (0 , 1) . 15 of 28
Quantile Distribution 1% 2 . 5% 5% 10% N (0 , 1) − 2 . 33 − 1 . 96 − 1 . 64 − 1 . 28 − 2 . 56 − 2 . 23 − 1 . 94 − 1 . 62 DF − 3 . 43 − 3 . 12 − 2 . 86 − 2 . 57 DF c − 3 . 96 − 3 . 66 − 3 . 41 − 3 . 13 DF l (A) Dickey-Fuller distributions 0.6 DF c DF l DF N(0,1) 0.4 0.2 0.0 -4 -2 0 2 4 16 of 28
Dickey-Fuller Test in an AR(p) • The DF distribution is derived under the assumption that � t is IID. For the AR(p) process we derive the Augmented Dickey-Fuller (ADF) test. • Consider the case of p = 3 lags: y t = θ 1 y t − 1 + θ 2 y t − 2 + θ 3 y t − 3 + � t . A unit root in θ ( z ) = 1 − θ 1 z − θ 2 z 2 − θ 3 z 3 corresponds to θ (1) = 0 . To avoid testing a restriction on 1 − θ 1 − θ 2 − θ 3 , the model is rewritten as 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 . 17 of 28
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