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Econometrics 2 — Fall 2004

Non-Stationary Time Series and Unit Root Tests

Heino Bohn Nielsen

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Outline of the Lecture

(1) Introduction. (2) Trend stationarity. (3) Unit root processes. (4) Unit root tests. (5) Dickey-Fuller test. (6) Deterministic terms. (7) An alternative test (KPSS).

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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 effects.

(2) A process with a stochastic trend or a unit root:

Shocks have permanent effects.

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Trend Stationarity

Consider the stationary AR(1) model with a deterministic linear trend term

Yt = θYt−1 + δ + γt + t, t = 1, 2, ..., T,

(∗) where |θ| < 1, and Y0 is an initial value.

The solution for Yt (MA-representation) has the form

Yt = θtY0 + µ + µ1t + t + θt−1 + θ2t−2 + θ3t−3 + ...

The first moments are given by

E[Yt] = θtY0 + µ + µ1t → µ + µ1t

for

T → ∞, V [Yt] = σ2 1 − θ2.

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SLIDE 3

The original process, Yt, is not stationary!
The deviation from the mean,

yt = Yt − E[Yt] = Yt − µ − µ1t

is a stationary process. The process

Yt = yt + µ + µ1t

is denoted trend-stationary.

Shocks to the process are transitory.

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

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Shock to a Trend Stationarity Process

10 20 30 40 50 60 70 80 90 100 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5

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SLIDE 4

Unit Root Processes

Consider the AR(1) model with a unit root, θ = 1 :

Yt = Yt−1 + δ + t, t = 1, 2, ..., T,

(∗∗)

r

∆Yt = δ + t,

where Y0 is the initial value.

Note that θ(L) = (1 − L) is not invertible and Yt is non-stationary.
The process ∆Yt is stationary. We denote Yt a difference stationary process.
If ∆Yt is stationary while Yt is not, Yt is called integrated of first order, I(1).

A process is integrated of order d, I(d) if ∆dYt is stationary while Yt is not.

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The solution for Yt is given by

Yt = Y0 +

t

X

i=1

∆Yi = Y0 +

t

X

i=1

(δ + i) = Y0 + δt +

t

X

i=1

i.

The moments are given by

E[Yt] = Y0 + δt V [Yt] = V hXt

i=1 i

i = t · σ2

Remarks:

(1) The effect of the initial value, Y0, stays in the process. (2) The innovations, t, are accumulated to a random walk, P i.

This is the stochastic trend. Shocks have permanent effects.

(3) The constant δ is accumulated to a linear trend in Yt. The process in (∗∗) is

denoted a random walk with drift.

(4) The process has no attractor. (5) The variance of Yt grows with t.

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Shock to a Unit Root Process

10 20 30 40 50 60 70 80 90 100 10 20 30 40 50

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Why are Unit Roots Important?

Interesting to know if shocks have permanent or transitory effects.

Example: If workers ability deteriorates in unemployment, shocks to unemployment could be permanent. This is the hypothesis of hysteresis.

It is important for forecasting to know if the process has an attractor.
The standard distributions for inference in regression models change!

Consider the regression model

yt = x0

tβ + t.

— If yt and xt are stationary, b

β is (asymptotically) normally distributed.

— If yt and xt have unit roots, b

β is not necessarily normal!

Later we look at spurious regression and cointegration.

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Unit Root Tests

A good way to think about unit root tests:

We compare two relevant models: H0 and HA.

(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: H0 is a unit root, HA is stationarity. (2) KPSS test: H0 is stationarity, HA 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

Set up an autoregressive model for yt and test if it has a unit root.

Note that MA models are approximated by AR models.

Consider the AR(1) regression model

yt = θyt−1 + t.

The unit root null hypothesis against the stationary alternative corresponds to

H0 : θ = 1

against

HA : θ < 1.

If explosive models are relevant, the alternative could be two-sided:

HB : θ 6= 1,

but rarely seen in practice.

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Alternatively, the model can be formulated as

∆yt = (θ − 1)yt−1 + t = πyt−1 + t,

where π = θ − 1. The hypothesis translates into

H0 : π = 0

against

HA : π < 0.

The Dickey-Fuller (DF) test is simply the t− test of H0 :

b τ = b θ − 1

se(b

θ) = b π

se(b

π).

The asymptotic distribution of b

τ is not normal!

The distribution depends on the deterministic components. In the simple case, the 5% critical value is −1.95 and not −1.65.

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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.

It is convenient to rewrite the model 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, c1 = − (θ2 + θ3) ,

and

c2 = −θ3.

A unit root in θ(L) = 1 − θ1L − θ2L2 − θ3L3 corresponds to θ(1) = 0, i.e.

H0 : π = 0

against

HA : π < 0.

The t−test for H0 is denoted the augmented Dickey-Fuller (ADF) test.

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To determine the number of lags, k, we can use the normal procedures.

— General-to-specific testing: Start with kmax and delete insignificant lags. — Estimate possible models and use information criteria.

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 models.

An alternative to the ADF test is to use HAC standard errors for the DF test.

Phillips-Perron non-parametric correction for autocorrelation. Likely to be inferior in small samples.

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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

∆Yt = πYt−1 + c1∆yt−1 + c2∆yt−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

∆Yt = πYt−1 + c1∆yt−1 + c2∆yt−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.

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The DF Distributions

5
4
3
2
1

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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A Note of Caution on Deterministic Terms

The way to think about the inclusion of deterministic terms is via the factor

representation:

Yt = yt + µ yt = θyt−1 + t

It follows that

(Yt − µ) = θ(Yt−1 − µ) + t Yt = θYt−1 + (1 − θ)µ + t Yt = θYt−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

Yt = θYt−1 + δ + t.

The hypotheses

H0 : θ = 1

against

HA : θ < 1,

imply

HA : Yt = µ + stationary process H0 : Yt = Y0 + δt + stochastic trend.

We compare a model with a linear trend against a model with a non-zero level!

Potentially difficult to interpret.

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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

HA : ∆Yt = πYt−1 + δ + t H∗ : ∆Yt = t,

and perform a likelihood ratio test

τ LR = T · log µRSS0 RSSA ¶ ,

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

∆Yt = πYt−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 H0, but only a linear trend under HA.

A solution is to impose the combined hypothesis

H∗

0 : π = γ = 0.

This is done by running the two regressions

HA : ∆Yt = πYt−1 + δ + γt + t H∗ : ∆Yt = 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 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.

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

Sometimes it is convenient to have stationarity as the null hypothesis.
Decompose the time series into

Yt = xt + zt,

where zt is stationary and xt is a random walk, i.e.

xt = xt−1 + vt, vt ∼ iid(0, σ2

v).

The null hypothesis that Yt is stationary implies that σ2

v = 0.

Under H0, it holds that xt is a constant and the stochastic part can be isolated

as

Yt = b µ + et.

(∗)

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The test statistic is given by

KPSS = T −2 PT

t=1 S2 t

b σ2

∞

,

where St = Pt

t=1 et is a partial sum, and b

σ2

∞ is a HAC estimator of the variance

f et.
The 5% critical value is 0.463 in the case with a constant term.
The regression in (∗) can be augmented with a linear trend.

In this case the critical value is 0.146.

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