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Econometrics 2 — Fall 2005

Non-Stationary Time Series

and Unit Root Tests

Heino Bohn Nielsen

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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 eff ects.

(2) A process with a stochastic trend or a unit root:Shocks have permanent eff ects.

• Why are unit roots important?

(1) Interesting to know if shocks have permanent or transitory eff 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 regressionand cointegration.

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Outline of the Lecture(1) Diff 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).

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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 = θtY 0 + µ + µ1t + t + θt−1 + θ

2t−2 + θ3t−3 + ...

Note that the mean,

E [Y t] = θtY 0 + µ + µ1t → µ + µ1t for T → ∞,

contains a linear trend, while the variance is constant:

V [Y t] = V [t + θt−1 + θ2t−2 + ...] = σ

2 + θ2σ2 + θ4σ2 + ... = σ2

1− θ2.

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• The original process, Y t, is not stationary.

The deviation from the mean,

yt = Y t −E [Y t] = Y t − µ− µ1t

is a stationary process. The process Y t is called trend-stationary.

• The stochastic part of the process is stationary and shocks have transitory eff ects

We say that the process is mean reverting.

Also, we say that the process has an attractor, namely the mean, µ + µ1t.

• We can analyze deviations from the mean, yt.

From the Frisch-Waugh theorem this is the same as a regression including a trend.

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Shock to a Trend-Stationarity Process

0 10 20 30 40 50 60 70 80 90 100

0.0

2.5

5.0

7.5

10.0

12.5

15.0 y t =0.8⋅ yt −1+ε t

Y t = y

t + 0.1⋅ t

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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 diff erence stationary process.

• If ∆Y t is stationary while Y t is not, Y t is called integrated of first order, I(1).

A process is integrated of order d, I(d), if it contains d unit roots.

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• The solution for Y t is given by

Y t = Y 0 +tX

i=1

∆Y i = Y 0 +tX

i=1

(δ + i) = Y 0 + δt +tX

i=1

i,

with moments

E [Y t] = Y 0 + δt and V [Y t] = t · σ2

Remarks:

(1) The eff 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 eff 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.

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Shock to a Unit Root Process

0 10 20 30 40 50 60 70 80 90 100

-8

-6

-4

-2

0

2 y t = yt −1+ε t

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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 difficult 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?

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The Dickey-Fuller (DF) Test• Idea: Set up an autoregressive model for yt and test if θ(1) = 0.

• Consider the AR(1) regression model

yt = θyt−1 + t.

The unit root null hypothesis against the stationary alternative corresponds to

H 0 : θ = 1 against H A : θ


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Augmented Dickey-Fuller (ADF) test• The DF test is extended to an AR(p) model. Consider an AR(3):

yt = θ1yt−1 + θ2yt−2 + θ3yt−3 + t.

A unit root in θ(L) = 1

−

θ1L−

θ2L

2−

θ3L

3 corresponds to θ(1) = 0

.

• The test is most easily performed by rewriting the model:

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 = −θ3.13 of 25

• The hypothesis θ(1) = 0 again corresponds to

H 0 : π = 0 against H A : π


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Deterministic Terms in the DF Test• The deterministic specification is important:

— We want an adequate model for the data.

— The deterministic specification changes the asymptotic distribution.

• If the variable has a non-zero level, consider a regression model of the form

∆Y t = πY t−1 + c1∆yt−1 + c2∆yt−2 + δ + t.

The ADF test is the t−test, bτ c = bπ/se( bπ).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 + c1∆yt−1 + c2∆yt−2 + δ + γt + t.

The ADF test is the t−test, bτ t = bπ/se( bπ).The critical value at a 5% level is −3.41.

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The DF Distributions

-5 -4 -3 -2 -1 0 1 2 3 4

0.1

0.2

0.3

0.4

0.5

N(0,1)

DF, τ

DF with a constant , τc

DF with a linear trend, τt

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Empirical Example: Danish Bond Rate

1970 1975 1980 1985 1990 1995 2000 2005

-0.025

0.000

0.025

0.050

0.075

0.100

0.125

0.150

0.175

0.200

Bond rate, r t

First difference, ∆r t

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• An AR(4) model gives

∆rt = −0.0093(−0.672)

rt−1 + 0.4033(4.49)

∆rt−1 −0.0192(−0.199)

∆rt−2 −0.0741(−0.817)

∆rt−3 + 0.0007(0.406)

.

Removing insignificant terms produce a model

∆rt = −0.0122(−0.911)

rt−1 + 0.3916(4.70)

∆rt−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 first diff erence.

Deleting insignificant terms we find a preferred model

∆2rt = −0.6193

(−7.49)∆rt−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 rt is non-stationary while ∆rt is stationary.

That is rt ∼I(1)

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A Note of Caution on Deterministic Terms• The way to think about the inclusion of deterministic terms is via the factor repre-

sentation:

yt = θyt−1 + t

Y t = yt + µIt follows that

(Y t − µ) = θ(Y t−1 − µ) + t

Y t = θY t−1 + (1 − θ)µ + t

Y t = θY t−1 + δ + t

which implies a common factor restriction.

• If θ = 1, then implicitly the constant should also be zero, i.e.

δ = (1− θ)µ = 0.

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• The common factor is not imposed by the normal t−test. Consider

Y t = θY t−1 + δ + t.

The hypotheses

H 0 : θ = 1 against H A : θ


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• A simple alternative is to consider the combined hypothesis

H ∗0 : π = δ = 0.

• The hypothesis H ∗0

can be tested by running the two regressions

H A : ∆Y t = πY t−1 + δ + t

H ∗0 : ∆Y t = t,

and perform a likelihood ratio test

τ LR = T · log

µRSS0

RSSA

¶= −2 (loglik0 − loglikA) ,

where RSS0 and RSSA denote the residual sum of squares.

The 5% critical value is 9.13.

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Same Point with a Trend• The same point could be made with a trend term

∆Y t = πY t−1 + δ + γt + t.

Here, the common factor restriction implies that if π = 0 then γ = 0.

• Since we do not impose the restriction under the null, the trend will accumulate.

A quadratic trend is allowed under H 0, but only a linear trend under H A.

• A solution is to impose the combined hypothesis

H ∗0 : π = γ = 0.

This is done by running the two regressions

H A : ∆Y t = πY t−1 + δ + γt + t

H ∗0 : ∆Y t = δ + t,

and perform a likelihood ratio test. The 5% critical value for this test is 12.39.

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Special Events• Unit root tests assess whether shocks have transitory or permanent eff ects.

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 eff ect.

Even if the series is stationary, such that normal shocks have transitory eff ects, the

presence of a break will make it look like the shocks have permanent eff ects.

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.

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A Reversed Test: KPSS• Sometimes it is convenient to have stationarity as the null hypothesis.

KPSS (Kwiatkowski, Phillips, Schmidt, and Shin) Test.

• Assume there is no trend. The point of departure is a DGP of the form

Y t = ξ t + et,

where et is stationary and ξ t is a random walk, i.e.

ξ t = ξ t−1 + vt, vt ∼ IID(0, σ2v).

If the variance is zero, σ2v = 0, then ξ t = ξ 0 for all t and Y t is stationary.

Use a simple regression:

Y t = bµ + bet, (∗)to find the estimated stochastic component. Under the null, bet is stationary.

• That observation can be used to design a test:

H 0 : σ2v = 0 against H A : σ

2v > 0.

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• The test statistic is given by

KPSS = 1

T 2 ·

PT t=1 S

2t bσ2

∞

,

where S t =

Pts=1

bes is a partial sum;

bσ2∞

is a HAC estimator of the variance of

bet.

(This is an LM test for constant parameters against a RW parameter).

• The regression in (∗) can be augmented with a linear trend. Critical values:

Deterministic terms Critical valuesin regression (∗) 0.10 0.05 0.01Constant 0.347 0.463 0.739Constant and trend 0.119 0.146 0.216

• Can be used for confirmatory analysis:

KPSS:Rejection of I(0) Non-rejection of I(0)

DF: Rejection of I(1) ? I(0)Non-rejection of I(1) I(1) ?

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