VECM Vector Error COrrection model
Transcript of VECM Vector Error COrrection model
-
7/28/2019 VECM Vector Error COrrection model
1/56
VECM
-
7/28/2019 VECM Vector Error COrrection model
2/56
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.
-
7/28/2019 VECM Vector Error COrrection model
3/56
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.
-
7/28/2019 VECM Vector Error COrrection model
4/56
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.
-
7/28/2019 VECM Vector Error COrrection model
5/56
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
-
7/28/2019 VECM Vector Error COrrection model
6/56
2
4
6
8
10
12
1980m1 1985m1 1990m1 1995m1 2000m1 2005m1t
missouri indiana
kentucky illinois
-
7/28/2019 VECM Vector Error COrrection model
7/56
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.
-
7/28/2019 VECM Vector Error COrrection model
8/56
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
-
7/28/2019 VECM Vector Error COrrection model
9/56
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.
-
7/28/2019 VECM Vector Error COrrection model
10/56
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)
-
7/28/2019 VECM Vector Error COrrection model
11/56
constraint 1 [_ce1]missouri = 1
Constraint number 1, [_ce1] tells us which
equation and missouri=1 sets constraint.
-
7/28/2019 VECM Vector Error COrrection model
12/56
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.
-
7/28/2019 VECM Vector Error COrrection model
13/56
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)
-
7/28/2019 VECM Vector Error COrrection model
14/56
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]
-
7/28/2019 VECM Vector Error COrrection model
15/56
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
-
7/28/2019 VECM Vector Error COrrection model
16/56
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.
-
7/28/2019 VECM Vector Error COrrection model
17/56
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
-
7/28/2019 VECM Vector Error COrrection model
18/56
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.
-
7/28/2019 VECM Vector Error COrrection model
19/56
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
-
7/28/2019 VECM Vector Error COrrection model
20/56
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
-
7/28/2019 VECM Vector Error COrrection model
21/56
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).
-
7/28/2019 VECM Vector Error COrrection model
22/56
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
-
7/28/2019 VECM Vector Error COrrection model
23/56
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
-
7/28/2019 VECM Vector Error COrrection model
24/56
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.
-
7/28/2019 VECM Vector Error COrrection model
25/56
Another View
vec missouri indiana kentucky illinois,
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
-
7/28/2019 VECM Vector Error COrrection model
26/56
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
-
7/28/2019 VECM Vector Error COrrection model
27/56
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
-
7/28/2019 VECM Vector Error COrrection model
28/56
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)
-
7/28/2019 VECM Vector Error COrrection model
29/56
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
-
7/28/2019 VECM Vector Error COrrection model
30/56
For example compare:
vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(2)bconstraints(1/5)
vec missouri indiana kentucky illinois,trend(rconstant) rank(2) lags(4)bconstraints(1/5)
-
7/28/2019 VECM Vector Error COrrection model
31/56
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
-
7/28/2019 VECM Vector Error COrrection model
32/56
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
-
7/28/2019 VECM Vector Error COrrection model
33/56
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)
-
7/28/2019 VECM Vector Error COrrection model
34/56
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
-
7/28/2019 VECM Vector Error COrrection model
35/56
vec missouri indiana kentucky illinois arkansas
ten, trend(rconstant) rank(3) lags(3)
but Tennessee still remains unrelated to
anything
-
7/28/2019 VECM Vector Error COrrection model
36/56
-
7/28/2019 VECM Vector Error COrrection model
37/56
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.
-
7/28/2019 VECM Vector Error COrrection model
38/56
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.
-
7/28/2019 VECM Vector Error COrrection model
39/56
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.
-
7/28/2019 VECM Vector Error COrrection model
40/56
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
-
7/28/2019 VECM Vector Error COrrection model
41/56
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)
-
7/28/2019 VECM Vector Error COrrection model
42/56
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
-
7/28/2019 VECM Vector Error COrrection model
43/56
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
-
7/28/2019 VECM Vector Error COrrection model
44/56
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.
-
7/28/2019 VECM Vector Error COrrection model
45/56
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
-
7/28/2019 VECM Vector Error COrrection model
46/56
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.
-
7/28/2019 VECM Vector Error COrrection model
47/56
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
-
7/28/2019 VECM Vector Error COrrection model
48/56
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.
-
7/28/2019 VECM Vector Error COrrection model
49/56
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
-
7/28/2019 VECM Vector Error COrrection model
50/56
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?
MacKinnon approximate p-value for Z(t) = 0.0000Z(t) -6.936 -4.027 -3.445 -3.145
Statistic Value Value ValueTest 1% Critical 5% Critical 10% Critical
Interpolated Dickey-Fuller
Augmented Dickey-Fuller test for unit root Number of obs = 140
-
7/28/2019 VECM Vector Error COrrection model
51/56
The inclusion of the trend term
MacKinnon approximate p-value for Z(t) = 0.0009Z(t) -4.639 -4.026 -3.444 -3.144
Statistic Value Value ValueTest 1% Critical 5% Critical 10% Critical
Interpolated Dickey-Fuller
Dickey-Fuller test for unit root Number of obs = 143. dfuller air, trend
MacKinnon approximate p-value for Z(t) = 0.5158Z(t) -1.536 -3.497 -2.887 -2.577
Statistic Value Value ValueTest 1% Critical 5% Critical 10% Critical
Interpolated Dickey-Fuller
Augmented Dickey-Fuller test for unit root Number of obs = 140. dfuller air, lags(3)
-
7/28/2019 VECM Vector Error COrrection model
52/56
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]
MacKinnon approximate p-value for Z(t) = 0.0000Z(t) -6.936 -4.027 -3.445 -3.145
Statistic Value Value ValueTest 1% Critical 5% Critical 10% Critical
Interpolated Dickey-Fuller
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
-
7/28/2019 VECM Vector Error COrrection model
53/56
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.
-
7/28/2019 VECM Vector Error COrrection model
54/56
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
-
7/28/2019 VECM Vector Error COrrection model
55/56
Z(rho) is the main statistic we are interested in
as it is similar to the ADF test statistic.
-
7/28/2019 VECM Vector Error COrrection model
56/56
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.