Econometrics II Non-Stationary Time Series and Unit Root Testing

university of copenhagen departmentofeconomics

Morten Nyboe Tabor university of copenhagen departmentofeconomics Course Outline: Non-Stationary Time Series and Unit Root Testing

1 Stationarity and Deviation from Stationarity Trend-Stationarity Level Shifts and Structural Breaks Variance Changes Unit Roots

2 An Autoregressive Unit Root Process Stationary AR(1) Autoregression with a Unit Root Deterministic Terms Stationary vs. unit root AR(1)

3 Dickey-Fuller Unit Root Testing Test in an AR(1) Test in an AR(p) Test with a Constant Term Test with a Trend Term

4 Empirical Examples

5 Further Issues

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 2/35 1. Stationarity and Deviation from Stationarity university of copenhagen departmentofeconomics Stationarity

• The main assumption on the time series data so far has been stationarity. Recall the deﬁnition:

Weak stationarity A time series is called weakly stationary if

E(yt ) = µ 2 V (yt ) = E((yt − µ) ) = γ0 Cov(yt , yt−k ) = E((yt − µ)(yt−k − µ)) = γk for k = 1, 2, ...

• This can be violated in diﬀerent ways. Examples of non-stationarity:

1 Deterministic trends (trend stationarity).

2 Level shifts.

3 Variance changes.

4 Unit roots (stochastic trends).

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 4/35 university of copenhagen departmentofeconomics Four Non-StationaryFour Non-Stationary Time Series Time Series

(A) Stationary and trend-stationary process (B) Process with a level shift 10

yt 5 5

zt 0 0

0 50 100 150 200 0 50 100 150 200

5 10

0 5

0 -5

0 50 100 150 200 0 50 100 150 200

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Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 5/35 university of copenhagen departmentofeconomics (A) Trend-Stationarity

• Observation: Many macro-economic variables are trending. How should we model the relationship between trending variables?

• Assume that zt is stationary and that yt is zt plus a deterministic linear trend, e.g.,

zt = θzt−1 + t , | θ | < 1, and yt = zt + µ0 + µ1t.

• Remarks:

1 yt has a trending mean, E(yt ) = µ0 + µ1t, and is non-stationary.

2 The de-trended variable, zt = yt − µ0 − µ1t, is stationary. yt is trend-stationary.

3 We may analyze the OLS-detrended variable, bzt = yt − µb0 − µb1t. Standard asymptotics apply to regressions with bzt .

4 This is equivalent to extending the regression with a deterministic trend, e.g., yt = β0 + β1 · xt + β3 · t + t .

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 6/35 university of copenhagen departmentofeconomics (B) Level Shifts and Structural Breaks

• Another type of non-stationarity is due to changes in parameters, e.g., a level shift: µ1 for t = 1, 2, ..., T0 E(yt ) = µ2 for t = T0 + 1, T0 + 2, ..., T .

• If each sub-sample is stationary, then there are two modelling approaches:

1 Include a dummy variable

n 0 for t = 1, 2, ..., T0 Dt = 1 for t = T0 + 1, T0 + 2, ..., T in the regression model,

yt = β0 + β1 · xt + β3 · Dt + t .

If yt − β3 · Dt is stationary, standard asymptotics apply.

2 Analyze the two sub-samples separately. This is particularly relevant if we think that more parameters have changed.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 7/35 university of copenhagen departmentofeconomics (C) Variance Changes

• A third type of non-stationary is related to changes in the variance. An example is yt = 0.5 · yt−1 + t , where N(0, 1) for t = 1, 2, ..., T0 t ∼ N(0, 5) for t = T0 + 1, T0 + 2, ..., T The interpretation is that the time series covers diﬀerent 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.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 8/35 Socrative Question 1

We simulate M = 10, 000 replications of the AR(1) model,

yt = θyt−1 + t , where t ∼ iidN(0, 1) and y0 = 0.

We simulate for both θ = 0.5 and θ = 1. For an increasing sample√ size, T = {10, 20, ..., 500}, we consider the distribution of θb and T (θb− θ).

Q: What are the properties of θb when θ = 1?

(A) It is consistent and asymptotically normal.

(B) It is consistent, but not asymptotically normal.

(D) It is inconsistent and not asymptotically normal.

(E) Don’t know.

Please go to www.socrative.com, click Student login, and enter room id Econometrics2. university of copenhagen departmentofeconomics (D) Unit Roots

• If there is a unit root in(D) an autoregressive Unit Roots model, no standard asymptotics apply! If there is a unit root in an autoregressive model, no standard asymptotics apply! • Consider the DGP Consider the DGP yt = θyt−1 + t , t ∼ N(0, 1),  =  1 +  (0 1) − ∼ forfort=1= 12,2, ...,500500,,and and0 =0y0 .= Consider 0. Consider the distribution the distribution of . of θb. • Note:Note: the theshape shape, location, locationand andvariance varianceof the of distributions. the distributions. • b

(C) Distribution of ^ for =0.5 (D) Distribution of ^ for =1 ^ 10.0 Distribution of  Distribution of ^ 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

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Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 9/35

2. An Autoregressive Unit Root Process university of copenhagen departmentofeconomics Properties of a Stationary AR(1)

• Consider the AR(1) model

yt = θyt−1 + t , t = 1, 2, ..., T .

The characteristic polynomial is θ(z) = 1 − θz, with characteristic root −1 −1 z1 = θ and inverse root φ1 = z1 = θ. The stationarity condition is that |φ1| = |θ| < 1. • Recall the solution

2 t−1 t yt = t + θt−1 + θ t−2 + ... + θ 1 + θ y0,

i where θ → 0 for i → ∞. Shocks have only transitory eﬀects; yt has an attractor (mean reversion). • Note the properties

t E(yt ) = θ y0 → 0 σ2 V (y ) = σ2 + θ2σ2 + θ4σ2 + ... + θt−1σ → t 1 − θ2 s ρs = Corr(yt , yt−s ) = θ

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 11/35 university of copenhagen departmentofeconomics Simulated Example Simulated Example

(A) Shock to a stationary process,  = 0.8 (B) Shock to a unit root process,  = 1

10 5

0 0

0 20 40 60 80 100 0 20 40 60 80 100

0.5 0.5

0 5 10 15 20 25 0 5 10 15 20 25

10 of 28

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 12/35 Socrative Question 2

Consider the following models for yt :

yt = 0.8yt−1 + 0.2yt−2 + t , (1)

∆yt = t , (2)

yt = 0.2yt−1 − 1.2yt−2 + t , (3)

yt = 0.2yt−1 + 1.2yt−2 − 0.4yt−3 + t , (4)

Q: In which of the models (1)-(4) is yt a unit-root process?

(A) None of them are unit-root processes.

(B) (1) and (2).

(D) They are all unit-root processes.

(E) Don’t know. university of copenhagen departmentofeconomics Autoregression with a Unit Root

• Consider the AR(1) with θ = 1, i.e.,

yt = yt−1 + t .

The characteristic polynomial is θ(z) = 1 − z. There is a unit root, θ(1) = 0. • The solution is given by:

yt = y0 + ∆y1 + ∆y2 + ... + ∆yt−1 + t

= y0 + 1 + 2 + ... + t t X = y0 + i . i=1 • Note the remarkable diﬀerences between θ = 1 and a stationary process, |θ| < 1 :

1 The eﬀect of the initial value, y0, stays in the process. And E(yt | y0) = y0.

2 Shocks, , have permanent eﬀects. t P Accumulate to a random walk component, i , called a stochastic trend.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 13/35 Socrative Question 3

Consider the AR(1) process with a unit root,

yt = yt−1 + t ,

2 where t ∼ iid(0, σ ).

Q: What is the variance of yt conditional on the initial value?

σ2 (A) V (yt |y0) = 1−θ2 2 (B) V (yt |y0) = σ 2 (C) V (yt |y0) = tσ 2 (D) V (yt |y0) = y0 + tσ

(E) Don’t know. university of copenhagen departmentofeconomics

3 The variance increases,

t X 2 V (yt |y0) = V i |y0 = t · σ . i=1 The process is clearly non-stationary.

4 The covariance, Cov(yt , yt−s |y0), is given by

2 E((yt −y0)(yt−s −y0)|y0) = E((1+2+...+t )(1+2+...+t−s )|y0) = (t−s)σ .

The autocorrelation is

2 r Cov(yt , yt−s |y0) (t − s)σ t − s Corr(yt , yt−s |y0) = = = , p p 2 2 V (yt |y0) · V (yt−s |y0) tσ · (t − s)σ t

which dies out very slowly with s.

5 The first difference, ∆yt = t , is stationary. yt is called integrated of first order, I(1).

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 14/35 Socrative Question 4

Consider the unit root AR(1) process with a constant term and a linear trend,

yt = δ + γ · t + yt−1 + t , t = 1, 2, ..., T .

Q. What is the eﬀect of the deterministic terms δ and γ · t on yt ?

(A) δ determines the level of yt , while γ · t is a linear trend in yt .

(B) Both δ and γ · t accumulate into a linear trend in yt .

(C) The constant term accumulates into the linear trend, δt, in yt , 2 while γ · t accumulates into the quadratic trend, γt , in yt .

(D) The constant term accumulates into the linear trend, δt, in yt , 4 while γ · t accumulates into the quadratic trend, γt , in yt .

(E) Don’t know.

(Do you find this deterministic specification relevant for economic time series? And what is the effect of a dummy variable or a shift-dummy?) university of copenhagen departmentofeconomics Deterministic Terms

• To model actual time series we include deterministic terms, e.g.,

yt = δ + θyt−1 + t .

With a unit root, the terms accumulate! • If |θ| < 1, the solution is

t t X i 2 t−1 yt = θ y0 + θ t−i + (1 + θ + θ + ... + θ )δ, i=0

where the mean converges to (1 + θ + θ2 + ...)δ = δ/(1 − θ). • If θ = 1, the solution is

t t X X yt = y0 + (δ + i ) = y0 + 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.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 15/35 university of copenhagen departmentofeconomics Stationary vs. unit root AR(1)

Properties of the processes Stationary AR(1) Unit root AR(1)

Process yt = δ + θyt−1 + t , |θ| < 1, yt = δ + yt−1 + t , 2 2 t ∼ IID(0, σ ) t ∼ IID(0, σ ) t Pt−1 i Pt−1 i Pt (1) MA Repr. yt = θ y0 + θ δ + θ t−i yt = y0 + δt + i i=0 i=0 i=1

(2) Role of y0 Dies out Stays in the process (3) Role of δ Effect on the level Accummulates to a linear trend (4) Shocks Transitory effects Permanent effects (5) Variance Converges to a constant Increases with t (6) Autocorrelation Coverges exponentially to zero Converges slowly to zero Does not depend on t Depends on t (7) OLS estimator Consistent (Super)Consistent for θ Asymptotically normal Non-normal asymp. distribution

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 16/35

Socrative Question 5

Consider the AR(1) process with a unit root,

yt = δ + yt−1 + t , t = 1, 2, ..., T ,

2 and assume that t ∼ NIID(0, σ ).

Q. What is the distribution of the forecast of yT +2 conditional on the information set, IT ?

2 (A) yT +2|T ∼ N(0, σ ). 2 (B) yT +2|T ∼ N(y0 + (T + 2)δ, (T + 2)σ ). 2 (C) yT +2|T ∼ N(yT + δ, σ ). 2 (D) yT +2|T ∼ N(yT + 2δ, 2σ ). (E) Don’t know. 3. Dickey-Fuller Unit Root Testing university of copenhagen departmentofeconomics 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!

1 Careful statistical model What kinds of deterministic components are relevant: constant or trend?

2 Hypothesis We compare two models, H0 and HA. What are the properties of the model under the null and under the alternative. Are both models relevant?

3 Test statistic What is a relevant test statistic for H0 | HA?

4 Asymptotic distribution The asymptotic distribution for a unit root test is non-standard.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 18/35 university of copenhagen departmentofeconomics Dickey-Fuller Test in an AR(1)

• Consider an AR(1) model

yt = θyt−1 + t .

The unit root hypothesis is θ(1) = 1 − θ = 0. The one-sided test against stationarity: H0 : θ = 1 against HA : −1 < θ < 1. • An equivalent formulation is

∆yt = πyt−1 + t ,

where π = θ − 1 = −θ(1). The hypothesis θ(1) = 0 translates into

H0 : π = 0 against HA : −2 < π < 0.

• The Dickey-Fuller (DF) test statistic is simply the t-ratio, i.e.,

θb− 1 πb τb = = se(θb) se(πb) The asymptotic distribution is Dickey-Fuller, DF, and not N(0, 1).

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 19/35 university of copenhagen departmentofeconomics

Quantiles Distribution 1% 2.5% 5% 10% Quantile N(0, 1) −2.33 −1.96 −1.64 −1.28 Distribution 1% 25% 5% 10% (0DF 1) −2332.56 −21.9623 −1.94164 −1.62128 DF DFc −−2563.43 −−32.2312 −2−.86194 −2.57−162 − − − − DFDFl −3433.96 −3.1266 −3.41286 −3.13257 − − − − DF 396 366 341 313 − − − − (A) Dickey-Fuller distributions 0.6 DF DFl c DF 0.4 N(0,1)

0.2

0.0 -4 -2 0 2 4 16 of 28

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 20/35 Socrative Question 6

Consider the estimated model,

∆yt = −0.0003 −0.202 ·yt−1 + t (1) [−0.219] [−2.42] b with t-values in square brackets.

Q. What is your conclusion to the Dickey-Fuller test?

(A) We reject the null of a unit root process.

(B) We cannot reject the null of a unit root process.

(D) We cannot reject the null of a stationary process.

(E) Don’t know. Socrative Question 7

The estimated AR(3) model,

rt = 0.0005 + 1.391rt−1 − 0.446rt−2 + 0.045rt−3 + bt , (1) can be re-written as:

∆rt = δ + πrt−1 + c1∆rt−1 + c2∆rt−2 + bt . (2)

Q. What are the estimates of π and c2 in (2)?

(A) The estimates cannot be derived from (1). (B) πb = −0.008 and bc2 = −0.045. (C) πb = −0.9 and bc2 = −0.015. (D) πb = 0.388 and bc2 = 0.045. (E) Don’t know.

Please go to www.socrative.com, click Student login, and enter room id Econometrics2. university of copenhagen departmentofeconomics Dickey-Fuller Test in an AR(p)

• For the AR(p) process we derive the Augmented Dickey-Fuller (ADF) test.

• Consider the case of p = 3 lags:

yt = θ1yt−1 + θ2yt−2 + θ3yt−3 + t .

2 3 A unit root in θ(z) = 1 − θ1z − θ2z − θ3z corresponds to θ(1) = 0. To avoid testing a restriction on 1 − θ1 − θ2 − θ3, the model is rewritten as

yt − yt−1 = (θ1 − 1)yt−1 + θ2yt−2 + θ3yt−3 + t

yt − yt−1 = (θ1 − 1)yt−1 + (θ2 + θ3)yt−2 + θ3(yt−3 − yt−2) + t

yt − yt−1 = (θ1 + θ2 + θ3 − 1)yt−1 + (θ2 + θ3)(yt−2 − yt−1)

+θ3(yt−3 − yt−2) + t

∆yt = πyt−1 + c1∆yt−1 + c2∆yt−2 + t ,

where

π = θ1 + θ2 + θ3 − 1 = −θ(1), c1 = − (θ2 + θ3) , c2 = −θ3.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 21/35 Socrative Question 8 Consider the estimates of the model,

∆rt = δ + πrt−1 + c1∆rt−1 + c2∆rt−2 + c3∆rt−3 + t (std.err) [p-val.] (1) (2) (3) (4) δ 0.000 0.001 0.001 −0.000 (0.001) (0.001) (0.001) (0.001) c1 0.395 0.399 0.383 . (0.081) (0.081) (0.075) c2 −0.015 −0.045 .. (0.087) (0.082) c3 −0.079 ... (0.082) π −0.007 −0.008 −0.009 −0.003 (0.011) (0.010) (0.010) (0.011) Log-lik. 568.366 567.879 567.723 555.554 No autocorr. 1-5 [0.24] [0.65] [0.49] [0.00]

Q. What type of process is rt ?

(A) A conclusion cannot be made based on the presented results.

(B) A stationary process.

(D) An explosive process.

(E) Don’t know. university of copenhagen departmentofeconomics Dickey-Fuller Test in an AR(p)

• The hypothesis for θ(1) = 0 is unchanged:

H0 : π = 0 against HA : −2 < π < 0.

The t−test statistic τπ=0 again follows the DF-distribution.

• Remarks:

1 It is only the test for π = 0 that follows the DF distribution. Tests on c1 and c2 are N(0, 1).

2 We use the normal tools to determine the appropriate lag-length: general-to-speciﬁc testing or information criteria. Lag-length cannot be decided by Box-Jenkins identiﬁcation (PACF).

3 Verbeek suggests to calculate the DF test for all values of p. ... but why should we look at inferior or misspeciﬁed models? Find the best model and test in that.

2 4 The tests are based on the assumption that t ∼ IID(0, σ ). This is checked by applying the usual misspeciﬁcation tests.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 22/35 Socrative Question 9 Consider the AR(3) model rewritten as,

2 ∆yt = δ + πyt−1 + ∆yt−1 + ∆yt−2 + εt , εt ∼ NIID(0, σ ). for t = 1, 2, ..., T and with initial values y−2, y−1, and y0 given.

Q: Which processes for yt does the Augmented Dickey-Fuller test,

H0 : π = 0 against HA : −2 < π < 0. compare under the null and the alternative?

(A) A unit root process without a drift under H0 against a mean-zero stationary procees under HA.

(B) A unit root process without a drift under H0 against a stationary procees with non-zero mean under HA.

(D) A unit root process with a drift under H0 against a trend-stationary procees under HA.

(E) Don’t know. university of copenhagen departmentofeconomics Dickey-Fuller Test with a Constant Term

• We need deterministic variables to model actual time series, E(yt ) 6= 0. The DF regression with a constant term (and p = 3 lags again) is

∆yt = δ + πyt−1 + c1∆yt−1 + c2∆yt−2 + t . (∗)

The hypothesis is unchanged H0 : π = 0, and as a test statistic we can use

πb τbc = . se(πb) • Remarks:

1 The constant term in the regression changes the asymptotic distribution. The relevant distribution, DFc , is shifted to the left of DF.

2 Under the null hypothesis, π = 0, the constant gives a trend in yt . We have n µ + stationary process for |θ| < 1 yt = y0 + random walk + δt for θ = 1 That is not a natural comparison. We implicitly assume that δ = 0 if θ = 1.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 23/35 university of copenhagen departmentofeconomics Dickey-Fuller Test with a Constant Term

∗ • A more satisfactory hypothesis is H0 : π = δ = 0, i.e., compare (∗) with

∆yt = c1∆yt−1 + c2∆yt−2 + t . (∗∗)

• The joint hypothesis can be tested by a LR test,

LR(π = δ = 0) = −2 · (log L0 − log LA) ,

where log L0 and log LA denote the log-likelihood values from (∗) and (∗∗). 2 The LR statistic follows a non-standard distribution, DFc , under the null.

Quantiles Distribution 1% 2.5% 5% 10% χ2(2) 9.21 7.38 5.99 4.61 2 DFc 12.73 10.73 9.13 7.50 2 DFl 16.39 14.13 12.39 10.56

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 24/35 university of copenhagen departmentofeconomics Dickey-Fuller Test with a Trend Term • For trending variables, the relevant alternative is often trend-stationarity. We use ∆yt = δ + γt + πyt−1 + c1∆yt−1 + c2∆yt−2 + t . (#)

The hypothesis is still H0 : π = 0, and the DF t−test is πb τbl = . se(πb) The presence of a trend shifts the asymptotic distribution, DFl , further to the left. • For the AR(1) with θ = 1,

yt = δ + γt + yt−1 + t t γ γ X = δ + t + t2 + + y . 2 2 i 0 i=1 • To avoid the accumulation of the trend under π = 0, we may consider the ∗ joint hypothesis, H0 : π = γ = 0, i.e., to compare (#) with

∆yt = δ + c1∆yt−1 + c2∆yt−2 + t . (##)

The LR test is LR(π = γ = 0) = −2 · (log L0 − log LA), which follows a 2 DFl .

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 25/35 4. Empirical Examples university of copenhagen departmentofeconomics Empirical Example: Danish Bond Rate

0.200

0.175 Bond rate r 0.150 First difference ∆r 0.125

0.100

0.075

0.050

0.025

0.000

-0.025 1970 1975 1980 1985 1990 1995 2000 2005 2010

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 27/35 ∗ • To test the joint hypothesis, H0 : π = δ = 0, we estimate the model under the null, ∆rt = 0.3787∆rt−1, (5.09)

with log L0 = 567.14. The LR test is given by

LR(π = δ = 0) = −2 · (log L0 − log LA) = −2 · (567.14 − 567.72) = 1.16,

2 where the 5% critical value is 9.13 in DFc . The LR test does not reject 2 the null of a unit root (1.16 9.13 = DFc at 5%)

university of copenhagen departmentofeconomics Empirical Example: Danish Bond Rate • An AR(4) model for 1972 : 1 − 2010 : 4 gives (t−values):

∆rt = −0.0069rt−1 + 0.3945∆rt−1 − 0.0152∆rt−2 − 0.0795∆rt−3 + 0.0003. (−0.66) (4.86) (−0.17) (−0.97) (0.29)

Removing insigniﬁcant terms produces a model

∆rt = −0.0091rt−1 + 0.3832∆rt−1 + 0.0006, (−0.88) (5.08) (0.52)

with log LA = 567.72. The DF t−test is τbc = −0.88. We do not reject the hypothesis of a unit root (DFc = −2.86 at a 5% level).

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 28/35 university of copenhagen departmentofeconomics Empirical Example: Danish Bond Rate • An AR(4) model for 1972 : 1 − 2010 : 4 gives (t−values):

∆rt = −0.0069rt−1 + 0.3945∆rt−1 − 0.0152∆rt−2 − 0.0795∆rt−3 + 0.0003. (−0.66) (4.86) (−0.17) (−0.97) (0.29)

Removing insigniﬁcant terms produces a model

∆rt = −0.0091rt−1 + 0.3832∆rt−1 + 0.0006, (−0.88) (5.08) (0.52)

with log LA = 567.72. The DF t−test is τbc = −0.88. We do not reject the hypothesis of a unit root (DFc = −2.86 at a 5% level). ∗ • To test the joint hypothesis, H0 : π = δ = 0, we estimate the model under the null, ∆rt = 0.3787∆rt−1, (5.09)

with log L0 = 567.14. The LR test is given by

LR(π = δ = 0) = −2 · (log L0 − log LA) = −2 · (567.14 − 567.72) = 1.16,

2 where the 5% critical value is 9.13 in DFc . The LR test does not reject 2 the null of a unit root (1.16 9.13 = DFc at 5%)

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 28/35 university of copenhagen departmentofeconomics EmpiricalEmpirical Examples: Examples: Trend-Stationarity Trend-Stationarity

(A) Log of Danish productivity (B) Log of Danish private consumption 1.00 6.4

0.75 6.2

0.50

6.0 0.25

1970 1980 1990 2000 1970 1980 1990 2000 (C) Productivity, deviation from trend (D) Consumption, deviation from trend

0.05 0.05

0.00 0.00

-0.05 -0.05 1970 1980 1990 2000 1970 1980 1990 2000

24 of 28

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 29/35 ∗ • To test the joint hypothesis, H0 : π = γ = 0, we run the regression under the null ∆LPRODt = 0.0057 + t , (3.48)

with log L0 = 348.63. The LR test is

LR(π = γ = 0) = −2 · (log L0 − log LA) = −2 · (348.63 − 366.09) = 34.92.

2 This is larger than the critical value of 12.39 in DFl at 5%, so we reject the unit root hypothesis.

university of copenhagen departmentofeconomics Empirical Example: Danish Productivity

• To test whether log-productivity is trend-stationary we use an AR(1) regression

∆LPRODt = 0.091 + 0.0024t − 0.439LPRODt−1 + t , (6.58) (6.15) (−6.22)

with log LA = 366.09. The DF t−test is given by τbl = −6.22 −3.96 (1% critical value). Here we reject the hypothesis of a unit root and conclude that productivity is trend-stationary.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 30/35 university of copenhagen departmentofeconomics Empirical Example: Danish Productivity

• To test whether log-productivity is trend-stationary we use an AR(1) regression

∆LPRODt = 0.091 + 0.0024t − 0.439LPRODt−1 + t , (6.58) (6.15) (−6.22)

with log LA = 366.09. The DF t−test is given by τbl = −6.22 −3.96 (1% critical value). Here we reject the hypothesis of a unit root and conclude that productivity is trend-stationary.

∗ • To test the joint hypothesis, H0 : π = γ = 0, we run the regression under the null ∆LPRODt = 0.0057 + t , (3.48)

with log L0 = 348.63. The LR test is

LR(π = γ = 0) = −2 · (log L0 − log LA) = −2 · (348.63 − 366.09) = 34.92.

2 This is larger than the critical value of 12.39 in DFl at 5%, so we reject the unit root hypothesis.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 30/35 ∗ • To test the joint hypothesis, H0 : π = γ = 0, we use the regression under the null, ∆LCONSt = 0.0046 − 0.274∆LCONSt−1 + t , (2.97) (−3.29)

with log L0 = 355.87. The LR test for a unit root is given by

LR(π = γ = 0) = −2 · (log L0 − log LA) = −2 · (355.87 − 359.23) = 6.72,

2 which is smaller than 12.39 (critical value DFl at 5% level), so we cannot reject the hypothesis of a unit root in consumption.

university of copenhagen departmentofeconomics Empirical Example: Danish Consumption

• To test if log-consumption is trend-stationary we use the regression

∆LCONSt = 0.764+ 0.0004t − 0.129LCONSt−1 − 0.209∆LCONSt−1 +t , (2.57) (2.58) (−2.56) (−2.43)

with log LA = 359.23. The Dickey-Fuller t−test is given by τbl = −2.56, which is not signiﬁcant in the DFl distribution (5% critical value is −3.41). We conclude that private consumption seems to have a unit-root.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 31/35 university of copenhagen departmentofeconomics Empirical Example: Danish Consumption

• To test if log-consumption is trend-stationary we use the regression

∆LCONSt = 0.764+ 0.0004t − 0.129LCONSt−1 − 0.209∆LCONSt−1 +t , (2.57) (2.58) (−2.56) (−2.43)

∗ • To test the joint hypothesis, H0 : π = γ = 0, we use the regression under the null, ∆LCONSt = 0.0046 − 0.274∆LCONSt−1 + t , (2.97) (−3.29)

with log L0 = 355.87. The LR test for a unit root is given by

LR(π = γ = 0) = −2 · (log L0 − log LA) = −2 · (355.87 − 359.23) = 6.72,

2 which is smaller than 12.39 (critical value DFl at 5% level), so we cannot reject the hypothesis of a unit root in consumption.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 31/35 5. Further Issues university of copenhagen departmentofeconomics The ProblemThe of Low Problem Power of Low Power It• isIt di isﬃcult diﬃcult to distinguish to distinguish unit unitroots roots from from large large stationary stationary roots. roots. • AlwaysAlways be careful be careful in conclusions. in conclusions. • Consider time series generated from the two models Consider time series generated from the two models • ∆yt = −0.2 · yt−1 + 0.05 · t + t

∆∆xt == 0.0252 +t1. +005  +  − · − · ∆ =025 +   Hard to tell apart in practice. We need many observations to be sure. Hard to tell apart in practice. We need many observations to be sure. Graphs of Yt :

(A) Trend-stationary and unit root process (B) Trend-stationary and unit root process 30 Yt = 0.2Yt1  0.05t  t Yt = 0.2Yt1  0.05t  t Yt = 0.25  t Yt = 0.25  t 100 20

50 10

0 0 0 20 40 60 80 100 0 100 200 300 400 500

27 of 28

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 33/35 university of copenhagen departmentofeconomics Special Events

• Unit root tests assess whether shocks have transitory or permanent eﬀects. The conclusions are sensitive to a few large shocks.

• Consider a one-time change in the mean of the series, a so-called break. This is one large shock with a permanent effect. Even if the series is stationary, such that normal shocks have transitory effects, the presence of a break will make it look like the shocks have permanent effects. That may bias the conclusion towards a unit root.

• Consider a few large outliers, i.e., a single strange observations. The series may look more mean reverting than it actually is. That may bias the results towards stationarity.

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 34/35 university of copenhagen departmentofeconomics

Summary of the Tests Testing for a unit root in an autoregressive model:

Model Test Statistic DF LR 2 ∆yt = πyt−1 + t DF DF 2 ∆yt = πyt−1 + c1∆yt−1 + c2∆yt−2 + t DF DF 2 ∆yt = δ + πyt−1 + c1∆yt−1 + c2∆yt−2 + t DFc DFc 2 ∆yt = δ + γ t + πyt−1 + c1∆yt−1 + c2∆yt−2 + t DFl DFl

The null hypothesis is always: • For the (augmented) Dicky-Fuller (DF) test:

H0 : π = 0 vs. HA = −2 < π < 0.

• For the likelihood-ratio (LR) test:

∗ ∗ ∗ H0 : π = 0, or H0 : π = δ = 0, or H0 : π = γ = 0.

Tests on the remaining coeﬃcients c1 and c2 are N(0, 1).

Econometrics II — Non-Stationary Time Series and Unit Root Testing— Slide 35/35