# VECM ## VECM

• • First we test to see if variables are stationary I(0). If not they are assumed to have a unit root and be I(1). If a set of variables are all I(1) they should not be estimated using ordinary regression analysis, but between them there may be one or more equilibrium relationships. We can both estimate how many and what they are (called cointegrating vectors) using Johansen’s technique.

• • If a set of variables are found to have one or more cointegrating vectors then a suitable estimation technique is a VECM (Vector Error Correction Model) which adjusts to both short run changes in variables and deviations from equilibrium.

In what follows we work back to front. Starting with the VECMs, then Johansen’s technique than stationarity.

• • • 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.

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 line graph; 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

1980m1 1985m1 1990m1 t missouri kentucky 1995m1 indiana illinois 2000m1 2005m1

• The graph shows that although the series do appear to move together, the relationship is not that. There are periods when Indiana has the highest rate and others when Indiana has the lowest rate. • Although the Kentucky rate moves closely with the other series for most of the sample, there is a period in the mid-1980s when the unemployment rate in Kentucky does not fall at the same rate as the other series.

• We will model the series with two cointegrating equations and no linear or quadratic time trends in the original series. • For now we use the noetable option to suppress displaying the short-run estimation table.

vec missouri indiana kentucky illinois, trend(rconstant) rank(2) lags(4) noetable Vector error-correction model Sample: 1978m5 - 2003m12 No. of obs = 308 AIC = -2.306048

Log likelihood = 417.1314 HQIC = -2.005818

Det(Sigma_ml) = 7.83e-07 SBIC = -1.555184

Cointegrating equations Equation Parms chi2 P>chi2 _ce1 2 133.3885 0.0000

_ce2 2 195.6324 0.0000

Johansen normalization restrictions imposed beta Coef. Std. Err. z P>|z| [95% Conf. Interval] _ce1 missouri 1 . . . . .

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

illinois -1.135152 .2069063 -5.49 0.000 -1.540681 -.7296235

_cons -.3880707 .4974323 -0.78 0.435 -1.36302 .5868787

_ce2 missouri -1.11e-16 . . . . .

indiana 1 . . . . .

kentucky .2059473 .2718678 0.76 0.449 -.3269038 .7387985

illinois -1.51962 .2804792 -5.42 0.000 -2.069349 -.9698907

_cons 2.92857 .6743122 4.34 0.000 1.606942 4.250197

• • 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 in the second cointegrating equation is zero.

constraint 1 [_ce1]missouri = 1 constraint 2 [_ce1]indiana = 0 constraint 3 [_ce2]missouri = 0 constraint 4 [_ce2]indiana = 1 constraint 5 [_ce2]kentucky = 0 vec missouri indiana kentucky illinois, trend(rconstant) rank(2) lags(4) noetable bconstraints(1/5)

### constraint 1 [_ce1]missouri = 1

• Constraint number 1, [_ce1] tells us which equation and missouri=1 sets constraint.

beta Coef. Std. Err. z P>|z| [95% Conf. Interval] _ce1 missouri 1 . . . . .

indiana (omitted) kentucky .2521685 .1649653 1.53 0.126 -.0711576 .5754946

illinois -1.037453 .1734165 -5.98 0.000 -1.377343 -.6975626

_cons -.3891102 .4726968 -0.82 0.410 -1.315579 .5373586

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

kentucky (omitted) illinois -1.314265 .0907071 -14.49 0.000 -1.492048 -1.136483

_cons 2.937016 .6448924 4.55 0.000 1.67305 4.200982

LR test of identifying restrictions: chi2( 1) = .3139 Prob > chi2 = 0.575

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.

• 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)

## Results for D_Missouri

Coef. Std. Err. z P>|z| [95% Conf. Interval] D_missouri _ce1 L1. -.0683152 .0185763 -3.68 0.000 -.1047242 -.0319063

_ce2 L1. .0405613 .0112417 3.61 0.000 .018528 .0625946

missouri LD. .2391442 .0597768 4.00 0.000 .1219839 .3563045

L2D. .0463021 .061306 0.76 0.450 -.0738555 .1664596

L3D. .1996762 .0604606 3.30 0.001 .0811755 .3181768

indiana LD. .000313 .0526959 0.01 0.995 -.102969 .1035951

L2D. -.0071074 .0530023 -0.13 0.893 -.11099 .0967752

L3D. .024743 .0536092 0.46 0.644 -.0803291 .1298151

kentucky LD. .0169935 .0475225 0.36 0.721 -.0761489 .1101359

L2D. .0611493 .0473822 1.29 0.197 -.0317182 .1540168

L3D. .0212794 .0470264 0.45 0.651 -.0708907 .1134495

illinois LD. .050437 .0491142 1.03 0.304 -.0458251 .1466992

L2D. .0086696 .0493593 0.18 0.861 -.0880728 .1054119

L3D. -.0323928 .0490934 -0.66 0.509 -.1286141 .0638285

## Interpretation

D_missouri _ce1 L1. -.0683152 .0185763 -3.68 0.000 -.1047242 -.0319063

_ce2 L1. .0405613 .0112417 3.61 0.000 .018528 .0625946

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

Missouri + 0.425Kentucky – 1.037Illinois -0.389 = Error • Viewed in this context if the error term is positive then unemployment in Missouri can be viewed as being above equilibrium, same for Kentucky, but for Illinois it is below equilibrium (because if we increase Illinois the error term falls) • To get back to equilibrium we need unemployment to fall in Missouri.

D_missouri _ce1 L1. -.0683152 .0185763 -3.68 0.000 -.1047242 -.0319063

• • • As we can see from the regression this is what we get.

D_missouri is the change in unemployment in Missouri i.e. DU Mt = U mt – U mt-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.

• • Thus if in period t-1 the error term in _ce1 was positive, which we can see can be seen as unemployment in Missouri being too high compared to the equilibrium relationship with the other two states, then it will fall. The bigger the (negative) coefficient on _ce1 L1 the more rapid is the correction. If it = -1 then the entire error is corrected for in the following period.

### Let us look at the second cointegrating regression

• • This can be written as: Error=Indiana -1.342Illinois + 2.93

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

kentucky (omitted) illinois -1.314265 .0907071 -14.49 0.000 -1.492048 -1.136483

_cons 2.937016 .6448924 4.55 0.000 1.67305 4.200982

### 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. So why does unemployment in it respond to it?

_ce2 L1. .0405613 .0112417 3.61 0.000 .018528 .0625946

## + 2.93 + Error

• • Well it’s a little convoluted, but if the error term is positive it suggests that unemployment in Illinois is vector: below equilibrium (and may increase as a consequence). Now from first cointegrating Missouri = -0.425Kentucky + (perhaps going negative). 1.037Illinois

, if Illinois unemployment is to increase then the error term in the first cointegrating vector will fall

### Let us look at the second equation for Indiana

D_indiana _ce1 L1. -.0342096 .0220955 -1.55 0.122 -.0775159 .0090967

_ce2 L1. .0325804 .0133713 2.44 0.015 .0063732 .0587877

### 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

D_indiana _ce1 L1. -.0342096 .0220955 -1.55 0.122 -.0775159 .0090967

_ce2 L1. .0325804 .0133713 2.44 0.015 .0063732 .0587877

• 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 indicating the opposite.

## 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

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

### 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

Now this: regress D.missouri LD.missouri LD.indiana

LD.kentucky LD.illinois L2D.missouri L2D.indiana L2D.kentucky L2D.illinois L3D.missouri L3D.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)

• • 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

• 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)

## 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.

missouri LD. .2391442 .0597768 4.00 0.000 .1219839 .3563045

L2D. .0463021 .061306 0.76 0.450 -.0738555 .1664596

L3D. .1996762 .0604606 3.30 0.001 .0811755 .3181768

• But not to those of Indianna.

indiana LD. .000313 .0526959 0.01 0.995 -.102969 .1035951

L2D. -.0071074 .0530023 -0.13 0.893 -.11099 .0967752

L3D. .024743 .0536092 0.46 0.644 -.0803291 .1298151

### 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)

### Tennessee appears related to nothing, so..

_ce1 missouri 1 . . . . .

indiana (dropped) kentucky 1.399639 .4071156 3.44 0.001 .6017075 2.197571

illinois -.5534946 .3499487 -1.58 0.114 -1.239382 .1323923

arkansas -1.463609 .3881522 -3.77 0.000 -2.224373 -.7028443

tenn -.2175613 .3220395 -0.68 0.499 -.8487471 .4136245

_cons -.2406195 .9257524 -0.26 0.795 -2.055061 1.573822

_ce2 missouri -1.11e-16 . . . . .

indiana 1 . . . . .

kentucky .6949351 .3402238 2.04 0.041 .0281088 1.361762

illinois -1.304683 .2924498 -4.46 0.000 -1.877874 -.7314915

arkansas -.522436 .3243762 -1.61 0.107 -1.158202 .1133296

tenn -.1928471 .2691262 -0.72 0.474 -.7203248 .3346307

_cons 2.832045 .7736451 3.66 0.000 1.315728 4.348361

vec missouri indiana kentucky illinois arkansas ten, trend(rconstant) rank(3) lags(3) • but Tennessee still remains unrelated to anything

• 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.

## Johansen’s methoodology

• • vecrank implements three types of methods for determining r, the number of cointegrating equations in a VECM. The first is Johansen’s “trace” statistic method. The second is his “maximum eigenvalue” statistic method. The third method chooses r to minimize an information criterion.

All three methods are based on Johansen’s maximum likelihood (ML) estimator of the parameters of a cointegrating VECM.

• webuse balance2 • We have quarterly data on the natural logs of aggregate consumption, investment, and GDP inthe United States from the first quarter of 1959 through the fourth quarter of 1982. As discussed in King et al. (1991), the balanced-growth hypothesis in economics implies that we would expect to find two cointegrating equations among these three variables.

describe storage display value variable name type format label variable label gdp float %9.0g t int %tq inv float %9.0g consump float %9.0g y double %10.0g ln(gdp) i double %10.0g ln(investment) c double %10.0g ln(consumption)

• • • 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 that there are two or fewer cointegrating equations.

. vecrank y i c, lags(5) Johansen tests for cointegration Trend: constant Number of obs = 91 Sample: 1960q2 - 1982q4 Lags = 5 5% maximum trace critical rank parms LL eigenvalue statistic value 0 39 1231.1041 . 46.1492 29.68

1 44 1245.3882 0.26943 17.5810 15.41

2 47 1252.5055 0.14480 3.3465* 3.76

3 48 1254.1787 0.03611

• • Because Johansen’s method for estimating r is to accept as the actual r the first r for which the null hypothesis is not rejected, we accept r = 2 as our estimate of the number of cointegrating equations between these three variables. The “*” by the trace statistic at r = 2 indicates that this is the value of r selected by Johansen’s multiple-trace test procedure

## vecrank y i c, lags(5) level99

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

Johansen tests for cointegration Trend: constant Number of obs = 91 Sample: 1960q2 - 1982q4 Lags = 5 1% maximum trace critical rank parms LL eigenvalue statistic value 0 39 1231.1041 . 46.1492 35.65

1 44 1245.3882 0.26943 17.5810* 20.04

2 47 1252.5055 0.14480 3.3465 6.65

3 48 1254.1787 0.03611

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

### The maximum eigenvalue statistic

• A second test. This assumes a given r under the null hypothesis and test this against the alternative that there are r+1 cointegrating equations. Johansen (1995, chap. 6, 11, and 12) derives an LR test of the null of r cointegrating relations against the alternative of r+1 cointegrating relations. This method is used less often than the trace statistic method, but often both test statistics are reported.

### vecrank y I c, lags(5) max levela

• The levela option obtains both the 5% and 1% critical values.

Johansen tests for cointegration Trend: constant Number of obs = 91 Sample: 1960q2 - 1982q4 Lags = 5 maximum trace 5% critical 1% critical rank parms LL eigenvalue statistic value value 0 39 1231.1041 46.1492 29.68 35.65

1 44 1245.3882 0.26943 17.5810*1 15.41 20.04

2 47 1252.5055 0.14480 3.3465*5 3.76 6.65

3 48 1254.1787 0.03611

maximum max 5% critical 1% critical rank parms LL eigenvalue statistic value value 0 39 1231.1041 28.5682 20.97 25.52

1 44 1245.3882 0.26943 14.2346 14.07 18.63

2 47 1252.5055 0.14480 3.3465 3.76 6.65

3 48 1254.1787 0.03611

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 the second 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 46.1492 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 that there are two or fewer cointegrating equations.

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

## Stationarity

• • • • • • • Intuitively a variable is stationary (I(0) – integrated to order nought) if its characteristics do not change over time, e.g. variance, covariance and mean is unchanging.

Another way of looking at it is that ρ<1 in the following equation for a variable y: Y t = ρY t-1 We do not estimate the above, but subtract Y t-1 from both sides: Y t - Y t-1 = ρY t-1 Y t-1 =(ρ-1)Y t-1 Now in this regression we test that (ρ-1) is significantly negative, which implies ρ<1. If we reject this we say y is I(1) and it has a unit root. In this case doing time series OLS, etc on a variable with variables that are I(1) results in bias. The Johansen method is a suitable alternative.

The above is the Dickey Fuller test (DF), add lagged values of Y t to get rid of serial correlation and we have the augmented - Y t-1 Dickey Fuller test.

## Stationarity

• • Among the earliest tests proposed is the one by Dickey and Fuller (1979), though most researchers now use an improved variant called the augmented Dickey–Fuller test instead of the original version. Other common unit-root tests implemented in Stata include the DF–GLS test of Elliot Rothenberg, and Stock (1996) and the Phillips–Perron (1988) test.

webuse air2 dfuller air The test statistics is less negative than any of the critical values and hence we cannot reject the null hypothesis that the variable exhibits a unit root and is thus not stationary Dickey-Fuller test for unit root Number of obs = 143 Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) -1.748 -3.496 -2.887 -2.577

MacKinnon approximate p-value for Z(t) = 0.4065

## 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 Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) -6.936 -4.027 -3.445 -3.145

MacKinnon approximate p-value for Z(t) = 0.0000

## The inclusion of the trend term

. dfuller air, trend Dickey-Fuller test for unit root Number of obs = 143 Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) -4.639 -4.026 -3.444 -3.144

MacKinnon approximate p-value for Z(t) = 0.0009

. dfuller air, lags(3) Augmented Dickey-Fuller test for unit root Number of obs = 140 Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) -1.536 -3.497 -2.887 -2.577

MacKinnon approximate p-value for Z(t) = 0.5158

## dfuller air, lags(3) trend regres

Augmented Dickey-Fuller test for unit root Number of obs = 140 Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) -6.936 -4.027 -3.445 -3.145

MacKinnon approximate p-value for Z(t) = 0.0000

D.air Coef. Std. Err. t P>|t| [95% Conf. Interval] air L1. -.5217089 .0752195 -6.94 0.000 -.67048 -.3729379

LD. .5572871 .0799894 6.97 0.000 .399082 .7154923

L2D. .095912 .0876692 1.09 0.276 -.0774825 .2693065

L3D. .14511 .0879922 1.65 0.101 -.0289232 .3191433

_trend 1.407534 .2098378 6.71 0.000 .9925118 1.822557

_cons 44.49164 7.78335 5.72 0.000 29.09753 59.88575

Lagged values of D.air

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

• • 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 Fuller test.

## pperron air

• Phillips and Perron’s test statistics can be viewed as Dickey– Fuller statistics that have been made robust to serial correlation by using the Newey–West (1987) heteroskedasticity- and autocorrelation-consistent covariance matrix estimator.

Phillips-Perron test for unit root Number of obs = 143 Newey-West lags = 4 Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(rho) -6.564 -19.943 -13.786 -11.057

Z(t) -1.844 -3.496 -2.887 -2.577

MacKinnon approximate p-value for Z(t) = 0.3588

• Z(rho) is the main statistic we are interested in as it is similar to the ADF test statistic.

## DFGLS Test

webuse lutkepohl2 dfgls dln_inv • dfgls tests for a unit root in a time series. It performs the modified Dickey–Fuller t test (known as the DF-GLS test) proposed by Elliott, Rothenberg, and Stock (1996). Essentially, the test is an augmented Dickey–Fuller test, similar to the test performed by Stata’s dfuller command, except that the time series is transformed via a generalized least squares (GLS) regression before performing the test.

• Elliott, Rothenberg, and Stock and later studies have shown that this test has significantly greater power than the previous versions of the augmented Dickey–Fuller test.