VECM Vector Error COrrection model

7/28/2019 VECM Vector Error COrrection model

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VECM


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First we test to see if variables are stationary I(0).If not they are assumed to have a unit root andbe I(1).

If a set of variables are all I(1) they should not beestimated using ordinary regression analysis, butbetween them there may be one or moreequilibrium relationships. We can both estimate

how many and what they are (calledcointegrating vectors) using Johansenstechnique.


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If a set of variables are found to have one ormore cointegrating vectors then a suitableestimation technique is a VECM (Vector Error

Correction Model) which adjusts to both shortrun changes in variables and deviations fromequilibrium.

In what follows we work back to front.Starting with the VECMs, then Johansenstechnique than stationarity.


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We have data on monthly unemployment

rates in Indiana, Illinois, Kentucky, and

Missouri from

January 1978 through December 2003. We

suspect that factor mobility will keep the

unemployment

rates in equilibrium. The following graph plots

the data.


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use http://www.stata-press.com/data/r11/urates,clear

line missouri indiana kentucky illinois t

Note the form of the above line to draw the linegraph; then the variables which will be plotted;finally t the time variable against which they are

all plotted

For further info press the help key, then line


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2

4

6

8

10

12

1980m1 1985m1 1990m1 1995m1 2000m1 2005m1t

missouri indiana

kentucky illinois


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The graph shows that although the series do appear to movetogether, the relationship is not that. There are periods whenIndiana has the highest rate and others when Indiana has thelowest rate.

Although the Kentucky rate moves closely with the other series formost of the sample, there is a period in the mid-1980s when theunemployment rate in Kentucky does not fall at the same rate asthe other series.

We will model the series with two cointegrating equations and nolinear or quadratic time trends in the original series.

For now we use the noetable option to suppress displaying theshort-run estimation table.


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vec missouri indiana kentucky illinois, trend(rconstant)

rank(2) lags(4) noetable

_ce2 2 195.6324 0.0000_ce1 2 133.3885 0.0000Equation Parms chi2 P>chi2Cointegrating equationsDet(Sigma_ml) = 7.83e-07 SBIC = -1.555184Log likelihood = 417.1314 HQIC = -2.005818

AIC = -2.306048Sample: 1978m5 - 2003m12 No. of obs = 308Vector error-correction model

_cons 2.92857 .6743122 4.34 0.000 1.606942 4.250197illinois -1.51962 .2804792 -5.42 0.000 -2.069349 -.9698907kentucky .2059473 .2718678 0.76 0.449 -.3269038 .7387985indiana 1 . . . . .missouri -1.11e-16 . . . . .

_ce2_cons -.3880707 .4974323 -0.78 0.435 -1.36302 .5868787illinois -1.135152 .2069063 -5.49 0.000 -1.540681 -.7296235

kentucky .3493902 .2005537 1.74 0.081 -.0436879 .7424683indiana (omitted)missouri 1 . . . . ._ce1

beta Coef. Std. Err. z P>|z| [95% Conf. Interval]Johansen normalization restrictions imposed


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Except for the coefficients on kentucky in the

two cointegrating equations and the constant

term in the first, all the parameters are

significant at the 5% level.

We can refit the model with the Johansen

normalization and the overidentifying

constraint that the coefficient on kentucky inthe second cointegrating equation is zero.


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constraint 1 [_ce1]missouri = 1

constraint 2 [_ce1]indiana = 0

constraint 3 [_ce2]missouri = 0constraint 4 [_ce2]indiana = 1

constraint 5 [_ce2]kentucky = 0

vec missouri indiana kentucky illinois,trend(rconstant) rank(2) lags(4) noetable

bconstraints(1/5)


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constraint 1 [_ce1]missouri = 1

Constraint number 1, [_ce1] tells us which

equation and missouri=1 sets constraint.


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LR test of identifying restrictions: chi2( 1) = .3139 Prob > chi2 = 0.575_cons 2.937016 .6448924 4.55 0.000 1.67305 4.200982illinois -1.314265 .0907071 -14.49 0.000 -1.492048 -1.136483

kentucky (omitted)indiana 1 . . . . .missouri (omitted)_ce2

_cons -.3891102 .4726968 -0.82 0.410 -1.315579 .5373586illinois -1.037453 .1734165 -5.98 0.000 -1.377343 -.6975626kentucky .2521685 .1649653 1.53 0.126 -.0711576 .5754946indiana (omitted)missouri 1 . . . . ._ce1

beta Coef. Std. Err. z P>|z| [95% Conf. Interval]

The test of the overidentifying restriction does not reject the null hypothesis that the restriction

is valid, and the p-value on the coefficient on kentucky in the first cointegrating equation

indicates that it is not significant. We will leave the variable in the model and attribute the lack

of significance to whatever caused the kentucky series to temporarily rise above the others from

1985 until 1990, though we could instead consider removing kentucky from the model.


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Next, we look at the estimates of the

adjustment parameters. In the output below,

we replay the previous results.

vec missouri indiana kentucky illinois,

trend(rconstant) rank(2) lags(4)

bconstraints(1/5)


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Results for D_Missouri

L3D. -.0323928 .0490934 -0.66 0.509 -.1286141 .0638285L2D. .0086696 .0493593 0.18 0.861 -.0880728 .1054119LD. .050437 .0491142 1.03 0.304 -.0458251 .1466992illinois

L3D. .0212794 .0470264 0.45 0.651 -.0708907 .1134495L2D. .0611493 .0473822 1.29 0.197 -.0317182 .1540168LD. .0169935 .0475225 0.36 0.721 -.0761489 .1101359kentucky

L3D. .024743 .0536092 0.46 0.644 -.0803291 .1298151L2D. -.0071074 .0530023 -0.13 0.893 -.11099 .0967752LD. .000313 .0526959 0.01 0.995 -.102969 .1035951

indianaL3D. .1996762 .0604606 3.30 0.001 .0811755 .3181768L2D. .0463021 .061306 0.76 0.450 -.0738555 .1664596LD. .2391442 .0597768 4.00 0.000 .1219839 .3563045missouriL1. .0405613 .0112417 3.61 0.000 .018528 .0625946_ce2L1. -.0683152 .0185763 -3.68 0.000 -.1047242 -.0319063_ce1

D_missouriCoef. Std. Err. z P>|z| [95% Conf. Interval]


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Interpretation

L1. .0405613 .0112417 3.61 0.000 .018528 .0625946_ce2L1. -.0683152 .0185763 -3.68 0.000 -.1047242 -.0319063

_ce1D_missouri

If the error term in the first cointegration relation is positive unemployment in

Missouri FALLS.

If the error term in the second cointegrating regression is positive then unemployment

in Missouri INCREASES.

The first cointegrating regression is Missouri + 0.425Kentucky 1.037Illinois -0.389 = Error


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Missouri + 0.425Kentucky 1.037Illinois -0.389

= Error

Viewed in this context if the error term ispositive then unemployment in Missouri canbe viewed as being above equilibrium, same

for Kentucky, but for Illinois it is belowequilibrium (because if we increase Illinois theerror term falls)

To get back to equilibrium we needunemployment to fall in Missouri.


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As we can see from the regression this is what

we get.

D_missouri is the change in unemployment in

Missouri i.e. DUMt = UmtUmt-1

The coefficient on _ce1 L1 (_ce1 : the error

term from the first cointegrating regression;

L1 lagged one period) is -0.068 and significant

at the 1% level.

L1. -.0683152 .0185763 -3.68 0.000 -.1047242 -.0319063_ce1D_missouri


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Thus if in period t-1 the error term in _ce1 waspositive, which we can see can be seen asunemployment in Missouri being too high

compared to the equilibrium relationship withthe other two states, then it will fall.

The bigger the (negative) coefficient on _ce1L1 the more rapid is the correction. If it = -1then the entire error is corrected for in thefollowing period.


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Let us look at the second cointegrating

regression

This can be written as:

Error=Indiana -1.342Illinois + 2.93

_cons 2.937016 .6448924 4.55 0.000 1.67305 4.200982illinois -1.314265 .0907071 -14.49 0.000 -1.492048 -1.136483kentucky (omitted)indiana 1 . . . . .missouri (omitted)_ce2


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And its impact in the VECM (Vector

Error Correction Model)

We can see its positive and significant.

Unemployment in Missouri increases if this is

error term is positive. But why? Missouri does

not enter the second cointegrating vector. Sowhy does unemployment in it respond to it?

L1. .0405613 .0112417 3.61 0.000 .018528 .0625946ce2


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Indiana =1.342Illinois + 2.93 + Error

Well its a little convoluted, but if the error termis positive it suggests that unemployment inIllinois is below equilibrium (and may increase as

a consequence). Now from first cointegratingvector:

Missouri = -0.425Kentucky + 1.037Illinois, ifIllinois unemployment is to increase then the

error term in the first cointegrating vector will fall(perhaps going negative).


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Let us look at the second equation for

Indiana

L1. .0325804 .0133713 2.44 0.015 .0063732 .0587877 _ce2L1. -.0342096 .0220955 -1.55 0.122 -.0775159 .0090967_ce1

D_indiana


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Let us look at the second equation for

Indiana

The error term from _ce1 is not significant,

but that from _ce2 is and it is positive._ce2 Is

Error=Indiana -1.342Illinois + 2.93

L1. .0325804 .0133713 2.44 0.015 .0063732 .0587877 _ce2L1. -.0342096 .0220955 -1.55 0.122 -.0775159 .0090967_ce1

D_indiana


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Now this does not make much sense if rgw

error term is positive unemployment in

Indiana needs to fall to restore equilibrium.

Yet the coefficient on it is positive indicatingthe opposite.


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Another View


trend(rconstant) rank(2) lags(4) bconstraints(1/5)

matrix cerr=e(beta)

display cerr[1,1]

display cerr[1,3]

display cerr[1,5]

display cerr[1,9]

drop cerr1 cerr2


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matrix cerr=e(beta) saves the coefficients from

the two cointgretaing regressions in a vector

cerr.

cerr[1,1] is the first, cerr[1,9] is the penultimate

coefficient in the second equation

Thus: display cerr[1,9] gives: -1.3142654, the

coefficient on Illinois in _ce2


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Generate the error terms for the two

equations

generate cerr1= cerr[1,5]+ cerr[1,1]*missouri +

cerr[1,2]*indiana + cerr[1,3]*kentucky +

cerr[1,4]*illinois

generate cerr2= cerr[1,10]+ cerr[1,6]*missouri +

cerr[1,7]*indiana + cerr[1,8]*kentucky +

cerr[1,9]*illinois


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Now this:

regress D.missouri LD.missouri LD.indiana

LD.kentucky LD.illinois L2D.missouri L2D.indiana

L2D.kentucky L2D.illinois L3D.missouriL3D.indiana L3D.kentucky L3D.illinois L.cerr1

L.cerr2

Is almost equivalent to this:vec missouri indiana kentucky illinois,

trend(rconstant) rank(2) lags(4) bconstraints(1/5)


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I say almost because the VEC estimates both

equations jointly and the regressions are

slightly different, but very slightly.

Note to if we have a slightly different short run

structure then the cointegrating vectors

change which is a little unsatisfactory


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For example compare:


trend(rconstant) rank(2) lags(2)bconstraints(1/5)

vec missouri indiana kentucky illinois,trend(rconstant) rank(2) lags(4)bconstraints(1/5)


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Short Run dynamics

Lets look at the rest of the equation, below is

for D.missouri

The one period lag is significant as is the 3

period lag. That is it responds to its own

lagged values.

L3D. .1996762 .0604606 3.30 0.001 .0811755 .3181768L2D. .0463021 .061306 0.76 0.450 -.0738555 .1664596LD. .2391442 .0597768 4.00 0.000 .1219839 .3563045

ssouri


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But not to those of Indianna.

L3D. .024743 .0536092 0.46 0.644 -.0803291 .1298151 L2D. -.0071074 .0530023 -0.13 0.893 -.11099 .0967752LD. .000313 .0526959 0.01 0.995 -.102969 .1035951 indiana


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This has been based on an example in

the STATA manual, but..

There are more variables. Lets try the

regression in full:

vec missouri indiana kentucky illinois arkansas

ten, trend(rconstant) rank(2) lags(3)


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Tennessee appears related to nothing,

so..

_cons 2.832045 .7736451 3.66 0.000 1.315728 4.348361 tenn -.1928471 .2691262 -0.72 0.474 -.7203248 .3346307 arkansas -.522436 .3243762 -1.61 0.107 -1.158202 .1133296 illinois -1.304683 .2924498 -4.46 0.000 -1.877874 -.7314915 kentucky .6949351 .3402238 2.04 0.041 .0281088 1.361762 indiana 1 . . . . .missouri -1.11e-16 . . . . ._ce2

_cons -.2406195 .9257524 -0.26 0.795 -2.055061 1.573822 tenn -.2175613 .3220395 -0.68 0.499 -.8487471 .4136245 arkansas -1.463609 .3881522 -3.77 0.000 -2.224373 -.7028443 illinois -.5534946 .3499487 -1.58 0.114 -1.239382 .1323923 kentucky 1.399639 .4071156 3.44 0.001 .6017075 2.197571 indiana (dropped)

missouri 1 . . . . ._ce1


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vec missouri indiana kentucky illinois arkansas

ten, trend(rconstant) rank(3) lags(3)

but Tennessee still remains unrelated to

anything


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We can see from the map that Tennessee is on

the South east fringe of this group and it

would be interesting to bring in North

Carolina, Alabama and Georgia.


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Johansens methoodology

vecrank implements three types of methods fordetermining r, the number of cointegratingequations in a VECM. The first is Johansenstrace statistic method. The second is hismaximum eigenvalue statistic method. Thethird method chooses r to minimize aninformation criterion.

All three methods are based on Johansensmaximum likelihood (ML) estimator of theparameters of a cointegrating VECM.


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webuse balance2

We have quarterly data on the natural logs of

aggregate consumption, investment, and GDPinthe United States from the first quarter of 1959through the fourth quarter of 1982. As discussedin King et al. (1991), the balanced-growth

hypothesis in economics implies that we wouldexpect to find two cointegrating equations amongthese three variables.


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describe

c double %10.0g ln(consumption)i double %10.0g ln(investment)y double %10.0g ln(gdp)consump float %9.0ginv float %9.0gt int %tqgdp float %9.0gvariable name type format label variable labelstorage display value


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In this example, because the trace statistic at r = 0 of 46.1492

exceeds its critical value of 29.68, we reject the null hypothesis of

no cointegrating equations.

Similarly, because the trace statistic at r = 1 of 17.581 exceeds its

critical value of 15.41, we reject the null hypothesis that there is

one or fewer cointegrating equation.

In contrast, because the trace statistic at r = 2 of 3.3465 is less than

its critical value of 3.76, we cannot reject the null hypothesis thatthere are two or fewer cointegrating equations.

3 48 1254.1787 0.036112 47 1252.5055 0.14480 3.3465* 3.761 44 1245.3882 0.26943 17.5810 15.410 39 1231.1041 . 46.1492 29.68

rank parms LL eigenvalue statistic valuemaximum trace critical5%

Sample: 1960q2 - 1982q4 Lags = 5Trend: constant Number of obs = 91Johansen tests for cointegration

. vecrank y i c, lags(5)


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Because Johansens method for estimating r isto accept as the actual r the first r for whichthe null hypothesis is not rejected, we accept r

= 2 as our estimate of the number ofcointegrating equations between these threevariables.

The * by the trace statistic at r = 2 indicates

that this is the value of r selected byJohansens multiple-trace test procedure


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vecrank y i c, lags(5) level99

In the previous example, we used the default 5% criticalvalues. We can estimate r with 1% critical valuesinstead by specifying the level99 option.

The output indicates that switching from the 5% tothe 1% level changes the resulting estimate from r =2 to r = 1.

3 48 1254.1787 0.036112 47 1252.5055 0.14480 3.3465 6.651 44 1245.3882 0.26943 17.5810* 20.040 39 1231.1041 . 46.1492 35.65rank parms LL eigenvalue statistic value

maximum trace critical1%Sample: 1960q2 - 1982q4 Lags = 5Trend: constant Number of obs = 91Johansen tests for cointegration


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The maximum eigenvalue

statistic

A second test. This assumes a given r under the nullhypothesis and test this against the alternativethat there are r+1 cointegrating equations.Johansen (1995, chap. 6, 11, and 12) derives anLR test of the null of r cointegrating relationsagainst the alternative of r+1 cointegratingrelations.

This method is used less often than the tracestatistic method, but often both test statistics arereported.


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vecrank y I c, lags(5) max levela

The levela option obtains both the 5% and 1%

critical values.

3 48 1254.1787 0.036112 47 1252.5055 0.14480 3.3465 3.76 6.651 44 1245.3882 0.26943 14.2346 14.07 18.630 39 1231.1041 28.5682 20.97 25.52

rank parms LL eigenvalue statistic value valuemaximum max 5% critical 1% critical 3 48 1254.1787 0.03611

2 47 1252.5055 0.14480 3.3465*5 3.76 6.651 44 1245.3882 0.26943 17.5810*1 15.41 20.040 39 1231.1041 46.1492 29.68 35.65rank parms LL eigenvalue statistic value value

maximum trace 5% critical 1% criticalSample: 1960q2 - 1982q4 Lags = 5Trend: constant Number of obs = 91

Johansen tests for cointegration


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The test statistics are often referred to as lambda trace

and lambda max respectively

We print out both tests in this table the eigenvalue ones are in thesecond half of the table.

The test is for r versus r+1 cointegrating vectors.

In this example, because the trace statistic at r = 0 of 46.1492exceeds its critical value of 29.68, we reject the null hypothesis of

no cointegrating equations. Similarly, because the trace statistic at r = 1 of 17.581 exceeds its

critical value of 15.41, we reject the null hypothesis that there isone or fewer cointegrating equation. In contrast, because the tracestatistic at r = 2 of 3.3465 is less than its critical value of 3.76, wecannot reject the null hypothesis that there are two or fewer

cointegrating equations. The net result is we conclude there are 2 cointegrating vectors.


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Stationarity

Intuitively a variable is stationary (I(0) integrated to order nought) ifits characteristics do not change over time, e.g. variance, covarianceand mean is unchanging.

Another way of looking at it is that


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Stationarity

Among the earliest tests proposed is the oneby Dickey and Fuller (1979), though mostresearchers now use an improved variant

called the augmented DickeyFuller testinstead of the original version.

Other common unit-root tests implemented inStata include the DFGLS test of Elliot

Rothenberg, and Stock (1996) and thePhillipsPerron (1988) test.


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webuse air2

dfuller air

The test statistics is less negative than any of the critical

values and hence we cannot reject the nullhypothesis that the variable exhibits a unit root and

is thus not stationary

MacKinnon approximate p-value for Z(t) = 0.4065Z(t) -1.748 -3.496 -2.887 -2.577

Statistic Value Value ValueTest 1% Critical 5% Critical 10% Critical

Interpolated Dickey-Fuller

Dickey-Fuller test for unit root Number of obs = 143


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dfuller air, lags(3) trend

This is a similar regression, but includes 3

lagged values and a trend term. It is now

stationary. What has made the difference?




Augmented Dickey-Fuller test for unit root Number of obs = 140


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The inclusion of the trend term




Dickey-Fuller test for unit root Number of obs = 143. dfuller air, trend




Augmented Dickey-Fuller test for unit root Number of obs = 140. dfuller air, lags(3)


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dfuller air, lags(3) trend regres

_cons 44.49164 7.78335 5.72 0.000 29.09753 59.88575 _trend 1.407534 .2098378 6.71 0.000 .9925118 1.822557 L3D. .14511 .0879922 1.65 0.101 -.0289232 .3191433 L2D. .095912 .0876692 1.09 0.276 -.0774825 .2693065

LD. .5572871 .0799894 6.97 0.000 .399082 .7154923L1. -.5217089 .0752195 -6.94 0.000 -.67048 -.3729379air

D.air Coef. Std. Err. t P>|t| [95% Conf. Interval]




Augmented Dickey-Fuller test for unit root Number of obs = 140

This is the test statistic, the coefficient on air(t-1) =L1.air

Lagged values

of D.air


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The regression basically regresses the change

in the variable (D.air) on lagged changes an

the lagged value of air plus a constant and

time trend.

The inclusion of lagged D.Air values makes this

the augmented Dickey-Fuller test, i.e. it is

what differentiates ir from the Dickey Fullertest.


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pperron air

Phillips and Perrons test statistics can be viewed as Dickey

Fuller statistics that have been made robust to serial

correlation by using the NeweyWest (1987)

heteroskedasticity- and autocorrelation-consistent covariance

matrix estimator.

MacKinnon approximate p-value for Z(t) = 0.3588Z(t) -1.844 -3.496 -2.887 -2.577Z(rho) -6.564 -19.943 -13.786 -11.057

Statistic Value Value Value

Test 1% Critical 5% Critical 10% CriticalInterpolated Dickey-Fuller

Newey-West lags = 4Phillips-Perron test for unit root Number of obs = 143


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Z(rho) is the main statistic we are interested in

as it is similar to the ADF test statistic.


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DFGLS Test

webuse lutkepohl2dfgls dln_inv

dfgls tests for a unit root in a time series. It performs the modifiedDickeyFuller t test (known as the DF-GLS test) proposed by Elliott,

Rothenberg, and Stock (1996). Essentially, the test is an augmentedDickeyFuller test, similar to the test performed by Statas dfullercommand, except that the time series is transformed via a generalizedleast squares (GLS) regression before performing the test.

Elliott, Rothenberg, and Stock and later studies have shown that this testhas significantly greater power than the previous versions of theaugmented DickeyFuller test.

Documents

VECM Vector Error COrrection model