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Econometrics 2 — Spring 2005

Non-Stationary Time Series, Cointegration and Spurious Regression

Heino Bohn Nielsen

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Outline

(1) Combinations of non-stationary variables; cointegration defined. (2) Economic equilibrium and error correction. (3) Cointegration regression. (4) Engle-Granger residual based test for no-cointegration. (5) Estimation and testing in the error correction model. (6) Empirical example: Danish interest rates. (7) Spurious regression: Example and simulation.

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SLIDE 2

Combinations of Non-Stationary Variables

Let Wt = ( Yt Xt )0 be two I(1) variables, i.e. they contain stochastic trends:

Yt = Y0 + Xt

i=1 i + stationary process

Xt = X0 + Xt

i=1 ηi + stationary process.

In general, a linear combination will also have a random walk component:

Zt = β0Wt = ¡ 1 −β1 ¢ µ Yt Xt ¶ = Yt − β1 · Xt = Y0 − β1 · X0 + Xt

i=1 i − β1 ·

Xt

i=1 ηi + stationary process.

Important exception: There exist a β = ( 1 −β1 )0, so that Zt is stationary.

In this case we say that Yt and Xt co-integrate with cointegration vector β.

Cointegration occurs if the stochastic trends in Yt and Xt are the same:

Xt

i=1 i = β1 ·

Xt

i=1 ηi.

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Cointegration and Economic Equilibrium

Consider a regression model for two I(1) variables, Yt and Xt, given by

Yt = β0 + β1Xt + t.

(∗) The term, t, is interpretable as the deviation from the relation in (∗).

If Yt and Xt cointegrate, then the deviation

t = Yt − β0 − β1Xt

is a stationary process with attractor zero. Shocks to Yt and Xt have permanent effects; but Yt and Xt co-vary so that t returns to zero. We think of (∗) as defining an equilibrium between Yt and Xt.

If Yt and Xt do not cointegrate, then the deviation t is I(1).

There is no natural interpretation of (∗) as an equilibrium relation.

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SLIDE 3

Empirical Example: Consumption and Income

Time series for log consumption, Ct, and log income, Yt, are likely to be I(1).

Define a vector Wt = ( Ct Yt )0.

Consumption and income are cointegrated with cointegration vector β = ( 1 −1 )0

if the (log-) consumption-income ratio,

Zt = β0Wt = ( 1 −1 ) µ Ct Yt ¶ = Ct − Yt,

is a stationary process. The consumption-income ratio is an equilibrium relation.

1970 1980 1990 2000 6.00 6.25 Consumption and income, logs

Income Consumption

1970 1980 1990 2000

0.15
0.10
0.05

0.00 Income ratio, log(Ct)−log(Yt)

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How is the Equilibrium Sustained?

There must be forces pulling Yt or Xt towards the attractor.
Famous representation theorem: Yt and Xt cointegrate if and only if there exist

an error correction model for either Yt, Xt or both.

As an example, let Zt = Yt−β1Xt be a stationary relation between I(1) variables.

Then there exists a stationary AR model for Zt. Assume for simplicity an AR(2):

Zt = θ1Zt−1 + θ2Zt−2 + vt, θ(1) = 1 − θ1 − θ2 > 0.

This is equivalent to

(Yt − β1Xt) = θ1 (Yt−1 − β1Xt−1) + θ2 (Yt−2 − β1Xt−2) + vt Yt = β1Xt + θ1Yt−1 − θ1β1Xt−1 + θ2Yt−2 − θ2β1Xt−2 + vt,

r

∆Yt = β1∆Xt + θ2β1∆Xt−1 − θ2∆Yt−1 − (1 − θ1 − θ2) {Yt−1 − β1Xt−1} + vt.

In this case we have a common-factor restriction. That is not necessarily true.

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OLS Regression with Cointegrated Series

In the cointegration case there exists a β1 so that the error term, t, in

Yt = α + β1Xt + t.

(∗∗) is stationary.

OLS applied to (∗∗) gives consistent results, so that b

β1 → β1 for T → ∞.

Note that consistency is obtained even if potential dynamic terms are neglected.

This is because the stochastic trends in Yt and Xt dominate. We can even get consistent estimates in the reverse regression

Xt = δ + γ1Yt + ηt.

Unfortunately, it turns out that b

β1 is not asymptotically normal in general.

The normal inferential procedures do not apply to b

β1!

We can use (∗∗) for estimation—not for testing.

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Super-Consistency

For stationary series, the variance of b

β1 declines at a rate of T −1.

For cointegrated I(1) series, the variance of b

β1 declines at a faster rate of T −2.

Intuition: If b

β1 = β1 then t is stationary. If b β1 6= β1 then the error is I(1) and

will have a large variance. The ’information’ on the parameter grows very fast.

0.5 1.0 1.5 10 20 30 Stationary, T=50 0.5 1.0 1.5 10 20 30 Stationary, T=100 0.5 1.0 1.5 10 20 30 Stationary, T=500 0.5 1.0 1.5 10 20 30 Non-Stationary, T=50 0.5 1.0 1.5 10 20 30 Non-Stationary, T=100 0.5 1.0 1.5 10 20 30 Non-Stationary, T=500

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SLIDE 5

Test for No-Cointegration, Known β1

Suppose that we know that Yt and Xt are I(1).

Assume that β1 is known.

The series cointegrate, with cointegration vector β = ( 1 −β1 ), if

Zt = Yt − β1Xt

is stationary.

This can be tested using an ADF unit root test, e.g. the test for H0 : π = 0 in

∆Zt = δ +

p−1

X

i=1

∆Zt−i + πZt−1 + ηt.

The usual DF critical values apply to tπ=0.

Note, that the null, H0 : π = 0, is a unit root, i.e. no cointegration.

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Test for No-Cointegration, Estimated β1

Engle-Granger (1987) two-step procedure.
If β1 is unknown, it can be (super-consistently) estimated in

Yt = β0 + β1Xt + t.

(∗ ∗ ∗)

b β = ( 1 −b β1 ) is a cointegration vector if b t is a stationary process.

This can be tested using a DF unit root test, e.g. the test for H0 : π = 0 in

∆b t =

p−1

X

i=1

∆b t−i + πb t−1 + ηt.

The critical value for tπ=0 depends on the deterministic regressors in (∗ ∗ ∗).

The fact that b

β1 is estimated also change the critical values.

OLS in (∗ ∗ ∗) minimizes the variance of b

t. Look ’as stationary as possible’.

Critical value depends on the number of regressors.

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SLIDE 6

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1

1 2 3 4 5 0.1 0.2 0.3 0.4 0.5 0.6 N(0,1) DF(constant) 1 2 3 4 5 6 7 Estimated parameters in cointegrating regression (with constant in the regression (***))

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Critical values for the Dickey-Fuller test for no-cointegration are given by:

Case 1: A constant term in (∗ ∗ ∗). Number of variables Significance level

incl. Yt

1% 5% 10% 2 −3.90 −3.34 −3.04 3 −4.29 −3.74 −3.45 4 −4.64 −4.10 −3.81 5 −4.96 −4.42 −4.13

Case 2: A constant and a trend in (∗ ∗ ∗). Number of variables Significance level

incl. Yt

1% 5% 10% 2 −4.32 −3.78 −3.50 3 −4.66 −4.12 −3.84 4 −4.97 −4.43 −4.15 5 −5.25 −4.72 −4.43

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SLIDE 7

Dynamic Modelling

Given the estimated cointegrating vector we can define the error correction term

ecmt = b

t = Yt − b β0 − b β1Xt,

which is, per definition, a stationary stochastic variable.

Since b

β1 converges to β1 very fast we can treat it as a fixed number and

formulate an error correction model conditional on ecmt, i.e.

∆Yt = δ + φ0∆Xt + φ1∆Xt−1 + θ∆Yt−1 − γ · ecmt−1 + vt.

Given cointegration, all terms are stationary, and normal inference applies to δ,

φ0, φ1, θ, and γ.

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Estimation of β In the ECM

The estimator of β1 from a static regression is super-consistent, but often biased

due to ignored dynamics.

An alternative estimator is based on an unrestricted error correction model, e.g.

∆Yt = δ + θ∆Yt−1 + φ0∆Xt + φ1∆Xt−1 + γ1Yt−1 + γ2Xt−1 + vt,

(#) which is equivalent to the unrestricted ADL regression

Yt = δ + α1Yt−1 + α2Yt−2 + κ1Xt + κ2Xt−1 + κ3Xt−2 + vt.

An estimate of β1 can be found as the parameter in the long-run solutions:

b β1 = −b γ2 b γ1 = (b κ1 + b κ2 + b κ3) / (1 − b α1 − b α2) .

The main advantage is that the analysis is undertaken in a well-specified model.

The approach is optimal if only Yt error corrects.

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SLIDE 8

Testing for No-Cointegration in the ECM

Due to representation theorem, the null hypothesis of no-cointegration corre-

sponds to the null of no-error-correction.

Several tests have been designed in this spirit.
The most convenient is the so-called PcGive test for no-cointegration.

It is based on the unrestricted ECM:

∆Yt = δ + θ∆Yt−1 + φ0∆Xt + φ1∆Xt−1 + γ1Yt−1 + γ2Xt−1 + t,

(#) and tests the hypothesis

H0 : γ1 = 0

against the cointegration alternative, γ1 < 0.

The distribution of the t−ratio,

tγ1=0 = b γ1 SE(b γ1),

depends on the deterministic terms and the number of regressors in (#).

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Critical values for the PcGive test for no-cointegration are given by:

Case 1: A constant term in (#). Number of variables Significance level

incl. Yt

1% 5% 10% 2 −3.79 −3.21 −2.91 3 −4.09 −3.51 −3.19 4 −4.36 −3.76 −3.44 5 −4.59 −3.99 −3.66

Case 2: A constant and a trend in (#). Number of variables Significance level

incl. Yt

1% 5% 10% 2 −4.25 −3.69 −3.39 3 −4.50 −3.93 −3.62 4 −4.72 −4.14 −3.83 5 −4.93 −4.34 −4.03

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Outline of a Cointegration Analysis

(1) First test individual variables, e.g. Yt and Xt, for unit roots. (2a) Run the static cointegrating regression

Yt = β0 + β1Xt + t.

Note that the t−ratios cannot be used for standard inference. (2b) Test for no-cointegration by testing for a unit root in the residuals, b

t.

(2c) If cointegration is not rejected estimate a dynamic model like

∆Yt = δ + φ0∆Xt + φ0∆Xt−1 + θ∆Yt−1 − γecmt−1 + vt.

(3) Estimate an ADL model

∆Yt = δ + φ0∆Xt + φ0∆Xt−1 + θ∆Yt−1 + γ1Yt−1 + γ2Xt−1 + vt.

Test for no-cointegration with tγ1=0. If cointegration is found, the cointegrating relation is the long-run solution.

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Empirical Example: Danish Interest Rates

Consider two Danish interest rates:

rt : Money market interest rate bt :

Bond Yield for the period t = 1972 : 1 − 2003 : 2.

Test for unit roots in rt and bt (5% critical value is −2.89):

c ∆rt = 0.00638118

(1.35)

− 0.126209

(−1.39)

· ∆rt−1 − 0.234330

(−2.70)

· ∆rt−4 − 0.0826987

(−1.80)

· rt−1 c ∆bt = 0.00116558

(0.658)

+ 0.395115

(4.67)

· ∆bt−1 − 0.0128941

(−0.909)

· bt−1

We cannot reject unit roots and test if st = rt − bt is I(1) (5% crit. value is

−2.89): c ∆st = −0.00848594

(−3.71)

+ 0.207606

(2.56)

· ∆st−3 − 0.379449

(−5.35)

· st−1.

It is easily rejected that bt and rt are not cointegrating.

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SLIDE 10

1970 1980 1990 2000 0.1 0.2 Bond Yield and money market interest rate

Bond yield Money market interest rate

1970 1980 1990 2000

0.05

0.00 0.05 Interest rate spread 1980 1990 2000

0.05

0.00 0.05 Residuals from rt=α+ βbt + εt

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Instead of assuming β1 = 1 we could estimate the coefficient

EQ(1) Modelling IMM by OLS (using PR0312.in7) The estimation sample is: 1974 (3) to 2003 (2) Coefficient Std.Error t-value t-prob Part.R^2 Constant

0.00468506

0.005545

0.845

0.400 0.0062 IBZ 0.845524 0.04495 18.8 0.000 0.7563 sigma 0.0224339 RSS 0.0573738644 R^2 0.756314 F(1,114) = 353.8 [0.000]** log-likelihood 276.885 DW 0.82

no. of observations

116

no. of parameters

2 mean(IMM) 0.0919727 var(IMM) 0.00202967

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SLIDE 11

We could test for a unit root in the residuals (5% crit. value is −3.34):

∆b t = 0.230210

(2.95)

· ∆b t−3 − 0.499443

(−6.77)

·b t−1.

Again we reject no-cointegration.

Finally we could estimate the error correction models based on the spread:

c ∆rt = −0.00774026

(−3.23)

+ 1.17725

(4.55)

· ∆bt − 0.406456

(5.22)

· (rt−1 − bt−1) c ∆bt = −0.00181602

(−2.11)

+ 0.438970

(4.16)

· ∆bt−1 − 0.0673997

(−2.01)

· ∆rt − 0.0638286

(2.22)

· (rt−1 − bt−1)

Note that the short-rate, rt, error corrects, while the bond-yield, bt, does not.

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Spurious Regression

Recall that cointegration is a special case where all stochastic trends cancel.

From an empirical point of view this an exception.

What happens if the variables do not cointegrate?
Assume that Yt and Xt are two totally unrelated I(1) variables.

Then we would like the static regression

Yt = β0 + β1Xt + t,

($) to reveal that β1 = 0 and R2 = 0.

This turns out not to be the case!

The standard regression output will indicate a relation between Yt and Xt. This is called a spurious regression or nonsense regression result.

With non-stationary data we always have to think in terms of cointegration.

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SLIDE 12

A Spurious Regression Example

Consider two presumably unrelated variables:

CONS

Danish private consumption in 1995 prices.

BIRD

Number of breeding cormorants (skarv) in Denmark. and consider a static regression model

log(CONSt) = β0 + β1 · log(BIRDt) + t.

We would expect to get b

β1 ≈ 0 and R2 ≈ 0.

Applying OLS for the T = 20 observations gives the result:

log(CONSt) = 12.145

(80.90) + 0.095 (6.30) · log(BIRDt) + t,

with R2 = 0.688.

It looks like a reasonable model—but it is complete nonsense.

The problem is that t ∼ I(1) and standard results for OLS do not hold.

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1985 1990 1995 2000 12.9 13.0 13.1 13.2 Consumption and breeding birds, logs

Consumption Number of breeding birds

1985 1990 1995 2000

1

1 Regression residuals (cons on birds) 1970 1980 1990 2000 6.00 6.25 Consumption and income

Consumption Income

1970 1980 1990 2000

0.05

0.00 0.05 0.10 Regression residual (cons on income)

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SLIDE 13

Simulation: Stationary Case

Consider first two independent i.i.d. variables:

Yt = 1t Xt = 2t

where

µ 1t 2t ¶ ∼ N µµ ¶ , µ 1 0 0 1 ¶¶ ,

for T = 50, 100, 500.

Here, we get standard results for the regression model

Yt = β0 + β1Xt + t.

0.50
0.25

0.00 0.25 0.50 2.5 5.0 7.5 IID Case, estimates Note the convergence to β1=0 as T diverges. 50 100 500

4
2

2 4 0.2 0.4 IID Case, t-ratios Looks like a N(0,1) for all T. Standard testing. N(0,1) 50 100 500

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Simulation: I(1) Spurious Regression

Now consider two independent random walks

Yt = Yt−1 + 1t Xt = Xt−1 + 2t

where

µ 1t 2t ¶ ∼ N µµ ¶ , µ 1 0 0 1 ¶¶ ,

for T = 50, 100, 500.

Under the null hypothesis, β1 = 0, the residual is I(1).

The condition for consistency is not fulfilled.

3
2
1

1 2 3 0.25 0.50 0.75 I(1) case, estimates Looks unbiased, but NO convergence.

50 100 500

75
50
25

25 50 75 0.025 0.050 0.075 I(1) case, t-ratios The distribution gets increasingly dispersed. Note the scale as compared to a N(0,1)

50 100 500

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