COINTEGRATION*

Multivariate Unit Root ProcessesGenerally we cannot reject the null hypothesis, that many time series have unit roots. For example,log consumption and log output are both non-stationary, but log consumption log output is stationary. This situation is called cointegration. The practical problem is that when we have cointegration, asymptotics change completely. Furthermore, we really do not have enough data to definitively tell whether or not we have cointegrated series.*

Multivariate Unit Root ProcessesIn a univariate nonstationary time series Yt is said to be integrated of order d, I(d), if its (d1)th difference is nonstationary but d-th difference is stationary.If Yt is nonstationary but Yt=(1B)Yt is stationary, then Yt is integrated of order 1.Yt~I(1) but Yt~I(0)*

Multivariate Unit Root ProcessesIn many time series, integrated processes are considered together and they form equilibrium relationships.Short-term and long-term interest ratesIncome and consumptionThese leads to the concept of cointegration.The idea behind the cointegration is that although multivariate time series is integrated, certain linear transformations of the time series may be stationary.

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Granger Causality TestsAccording to Granger, causality can be further sub-divided into long-run and short-run causality.This requires the use of error correction models or VECMs, depending on the approach for determining causality.Long-run causality is determined by the error correction term, whereby if it is significant, then it indicates evidence of long run causality from the explanatory variable to the dependent variable.Short-run causality is determined as before, with a test on the joint significance of the lagged explanatory variables, using an F-test or Wald test.

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Granger Causality TestsBefore the ECM can be formed, there first has to be evidence of cointegration, given that cointegration implies a significant error correction term, cointegration can be viewed as an indirect test of long-run causality.It is possible to have evidence of long-run causality, but not short-run causality and vice versa.In multivariate causality tests, the testing of long-run causality between two variables is more problematic, as it is impossible to tell which explanatory variable is causing the causality through the error correction term.

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SPURIOUS REGRESSIONIf we regress a y series with unit root on regressors who also have unit roots the usual t tests on regression coefficients show statistically significant regressions, even if in reality it is not so.The Spurious Regression Problem can appear with I(0) series (see Granger, Hyung and Jeon (1998)). This is telling us that the problem is generated by using WRONG CRITICAL VALUES!!!! In a Spurious Regression the errors would be correlated and the standard t-statistic will be wrongly calculated because the variance of the errors is not consistently estimated. In the I(0) case the solution is:*

SPURIOUS REGRESSIONHow do we detect a Spurious Regression (between I(1) series)?Looking at the correlogram of the residuals and also by testing for a unit root on them.How do we convert a Spurious Regression into a valid regression?By taking differences.Does this solve the SPR problem?It solves the statistical problems but not the economic interpretation of the regression. Think that by taking differences we are loosing information and also that it is not the same information contained in a regression involving growth rates than in a regression involved the levels of the variables.*

SPURIOUS REGRESSIONTypical symptom: High R2, t-values, F-value, but low DW

1. Egyptian infant mortality rate (Y), 1971-1990, annual data, on Gross aggregate income of American farmers (I) and Total Honduran money supply (M)Y ^ = 179.9 - .2952 I - .0439 M, R2 = .918, DW = .4752, F = 95.17 (16.63) (-2.32) (-4.26) Corr = .8858, -.9113, -.9445

2. US Export Index (Y), 1960-1990, annual data, on Australian males life expectancy (X)Y ^ = -2943. + 45.7974 X, R2 = .916, DW = .3599, F = 315.2(-16.70) (17.76) Corr = .9570*

SPURIOUS REGRESSION3. US Defense Expenditure (Y), 1971-1990, annual data, on Population of South African (X)Y ^= -368.99 + .0179 X, R2 = .940, DW = .4069, F = 280.69(-11.34) (16.75) Corr = .9694

4. Total Crime Rates in the US (Y), 1971-1991, annual data, on Life expectancy of South Africa (X)Y ^= -24569 + 628.9 X, R2 = .811, DW = .5061, F = 81.72(-6.03) (9.04) Corr = .9008

5. Population of South Africa (Y), 1971-1990, annual data, on Total R&D expenditure in the US (X)Y ^= 21698.7 + 111.58 X, R2 = .974, DW = .3037, F = 696.96(59.44) (26.40) Corr = .9873

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SPURIOUS REGRESSIONDoes it make sense a regression between two I(1) variables?Yes if the regression errors are I(0).Can this be possible?The same question asked David Hendry to Clive Granger time ago. Clive answered NO WAY!!!!! but he also said that he would think about. In the plane trip back home to San Diego, Clive thought about it and concluded that YES IT IS POSSIBLE. It is possible when both variables share the same source of the I(1)ness (co-I(1)), when both variables move together in the long-run (co-move), ... when both variables are COINTEGRATED!*

COINTEGRATIONAn mx1 vector time series Yt is said to be cointegrated of order (d, b), CI(d,b) where 0

COINTEGRATIONMore generally, if the mx1 vector series Yt contains more than two components, each being I(1), then there may exist k (

EXAMPLEConsider the following system of processes

where the three error terms are uncorrelated white noise processes. Clearly, all those three processes are individually I(1). Let yt=(x1t,x2t,x3t) and =(1,1,2), then yt=a1t. which is a I(0) process. Another cointegrating relationship is between x2t and x3t. So, we can let *= (0, 1,3), then *yt = a2t is also I(0).*

VAR with CointegrationLet Yt be mx1. Suppose we estimate VAR(p)

or Let say we have a unit root. Then, we can write

This is like a multivariate version of the augmented Dickey- Fuller test *

VAR with CointegrationRearranging the equation

where Rank((1)I)

VAR with CointegrationNote that if you run OLS in differences, then the modeled is misspecified and the results will be biased. What can you do?(a)If you know the location of the unit roots and cointegration relations, then you can run the VECM by doing OLS of Yt on lags of Y and Yt1.(b)If you know nothing, then you can either (i) run OLS in levels, or (ii) test (many times) to estimate cointegrating relations, and run VECM. The problem with this approach is that you are testing many times and you are estimating cointegrating relationships. This leads to poor finite sample properties. *

Residual Based Tests of the Null of No CointegrationProcedures designed to distinguish a system without cointegration from a system with at least one cointegrating relationship; they do not estimate the number of cointegrating vectors (the k). Tests are conditional on a pretest for unit roots in each of the variables.When the cointegration vector is known: construct the hypothesized linear combination that is stationary, treat it as data, and apply a Dickey-Fuller unit root test to that linear combination. The null hypothesis is that there is a unit root, or no cointegration.*

Residual Based Tests of the Null of No CointegrationWhen the cointegration vector is not known: Assume that, if there exists a cointegrating relation, the coefficient on Y1t is nonzero, allowing us to express the static regression equation as

You can apply a unit root test to the estimated OLS residual from estimation of the above equation, butInclude a constant in the static regression if the alternative allows for a nonzero mean in ut Include a trend in the static regression if the alternative is stochastic cointegration, i.e., a nonzero trend for AYt.

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Residual Based Tests of the Null of No CointegrationThe first step in testing cointegration is to test the null hypothesis of a unit root in each component series Yit individually using the univariate unit root tests.If the hypothesis is rejected, then the next step is to test cointegration among the components, i.e., to test whether Yt is stationary.*

Residual Based Tests of the Null of No CointegrationIn practice, the cointegration vector is unknown. One way to test the existence of cointegration is the regression method (Engle&Granger, 1986, 1987).If Yt=(Y1t,Y2t,,Ymt) is cointegrated, Yt is stationary where =(1, 2,, m). Then, (1/1) is also a cointegrated vector where 10.*

Residual Based Tests of the Null of No CointegrationConsider the regression model for Y1t

and check whether t is I(1) or I(0).If t~I(1), then Yt is not cointegrated.If t~I(0), then Yt is cointegrated with a normalizing cointegrating vector =(1,1,, m1) .

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Residual Based Tests of the Null of No CointegrationIn testing the error series for nonstationary,Calculate the OLS estimateUse the residual series for the test using the standard ADF or PP . if t~I(0). H0: =1 vs H1:

Residual Based Tests of the Null of No Cointegrationt-statistic:

The critical values are obtained by simulation (Engle&Granger, 1987).Level of significance1%5% p=1 4.07 3.37 p>1 3.73 3.17If T

The Johansen Trace and Maximal Eigenvalue TestsTo test whether the variables are cointegrated or not, one of the well-known tests is the Johansen trace test. The Johansen test is used to test for the existence of cointegration and is based on the estimation of the ECM by the maximum likelihood, under various assumptions about the trend or intercepting parameters, and the number k of cointegrating vectors, and then conducting likelihood ratio tests. *

The Johansen Trace and Maximal Eigenvalue TestsAssuming that the ECM errors are independent Nm[0, ] distribution, and given the cointegrating restrictions on the trend or intercept parameters, the maximum likelihood Lmax(k) is a function of the cointegration rank k. The trace test is based on the log-likelihood ratio ln[Lmax(k)/Lmax(k)], and is conducted sequentially for k = m-1,...,1,0. The name comes from the fact that the test statistics involved are the trace (the sum of the diagonal elements) of a diagonal matrix of generalized eigenvalues. This test examines the null hypothesis that the cointegration rank is less than or equal to k, against the alternative that the cointegration rank is greater than k. If the trace is greater than the critical value for a certain rank, then the null hypothesis that the cointegration rank is equal to k is rejected.*

The Johansen Trace and Maximal Eigenvalue TestsConsider a non-stationary cointegrated VAR(p) model

where at are normally distributed with mean 0 and covariance matrix . In a series of influential papers, Johansen (1988, 1991), and Johansen and Juselius (1990) proposed practical full maximum likelihood estimation and testing approaches based on the error correction representation (ECM).

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The Johansen Trace and Maximal Eigenvalue TestsConsider the ECM

where ,dt is a vector of deterministic variables, such as constant and seasonal dummy variables,are mm, ,A and are mk parameter matrices, the are i.i.d. Nm(0, ) errors, and det( ) has all of its roots outside the unit circle. *

The Johansen Trace and Maximal Eigenvalue TestsThis ECM is based on the Engle-Granger (1987) error correction representation theorem for cointegrated systems, and the asymptotic inference involved is related to the work of Sims, Stock, and Watson (1990). By step-wise concentrating all the parameter matrices in the likelihood function out except for the matrix A, Johansen shows that the maximum likelihood estimator of A can be derived as the solution of a generalized eigenvalue problem. Likelihood ratio tests of hypotheses about the number of cointegrating vectors can then be based on these eigenvalues. Moreover, Johansen (1988) also proposes likelihood ratio tests for linear restrictions on these cointegrating vectors.

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The Johansen Trace and Maximal Eigenvalue TestsThe Johansen test for the existence of cointegration is based on the estimation of the above ECM by the maximum likelihood and is used to test the hypothesis , where k is less than m. This formulation shows that I(1) models form nested sequence modelswhere H(m) is the unrestricted VAR model or I(0) model, and H(0) corresponds to the restriction =0, which is the VAR model for indifferences. Since , it is equivalent to test that A and are of full column rank k, the number of independent cointegrating vectors that forms the matrix A. The test has been named the Johansen trace test because the likelihood ratio test statistic is the trace of a diagonal matrix of generalized eigenvalues from . *

Sequential tests:i. H0: k=0, cannot be rejected stop(at most zero coint) rejected next testii. H0: k

The Johansen Trace and Maximal Eigenvalue Tests(i) Rank k = m: all variables in x are I(0), not an interesting case to start with.(ii) Rank k = 0: there are no linear combinations of x that are I(0), no cointegration exists, and is full of zeros. Model on differenced series(iii) Rank k (m-1): up to (m-1) cointegration relationships xt-k.i.e. k (m-1) rows of form k linearly independent combinations of variables in x, each of which is I(0); alternatively (m-k) nonstationary vectors forming I(1) stochastic trends.

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The Johansen Trace and Maximal Eigenvalue TestsUnder some regularity conditions, we can write the cointegrated process as an Error Correction Model (ECM):

where is the difference operator , the at's are i.i.d. N(0, ).

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The Johansen Trace and Maximal Eigenvalue TestsWe can write this ECM as

where , , ,The likelihood ratio statistic for hypothesis is given by

where denotes the eigenvalues of and are ordered by

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The Johansen Trace and Maximal Eigenvalue TestsWhere

If the test statistics are greater than the critical value for rank k, then the null hypothesis that the cointegration rank is equal to k is rejected.*

The Johansen Trace and Maximal Eigenvalue TestsThe statistic ln has the following limiting distribution which can be expressed in terms of a mk dimensional Brownian motion as

The percentiles of the asymptotic distribution for the trace statistic are tabulated in Johansen (1988, Table 1) using simulation analysis.

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The Johansen Trace and Maximal Eigenvalue TestsAn alternative LR statistic, given by

and called the maximal eigenvalue statistic, examines the null hypothesis of k cointegrating vectors versus the alternative k+1 cointegrating vectors. The asymptotic distribution of this statistic is given by the maximum eigenvalue of the stochastic matrix in

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Analysis of U.S. Economic Variables (From SAS Online Doc)Consider the following four-dimensional system of U.S. economic variables. Quarterly data for the years 1954 to 1987 are used (Ltkepohl 1993, Table E.3.). The following statements plot the series and proceed with the VARMAX procedure. *

SAS Codesymbol1 v=none height=1 c=black; symbol2 v=none height=1 c=black; title 'Analysis of U.S. Economic Variables'; data us_money; date=intnx( 'qtr', '01jan54'd, _n_-1 ); format date yyq. ; input y1 y2 y3 y4 @@; y1=log(y1); y2=log(y2); label y1='log(real money stock M1)' y2='log(GNP in bil. of 1982 dollars)' y3='Discount rate on 91-day T-bills' y4='Yield on 20-year Treasury bonds'; datalines; ... data lines omitted ... ;

legend1 across=1 frame label=none; *

SAS Code (Contd.)proc gplot data=us_money;symbol1 i = join l = 1; symbol2 i = join l = 2; axis2 label = (a=-90 r=90 " "); plot y1 * date = 1 y2 * date = 2 / overlay vaxis=axis2 legend=legend1; run; proc gplot data=us_money; symbol1 i = join l = 1; symbol2 i = join l = 2; axis2 label = (a=-90 r=90 " "); plot y3 * date = 1 y4 * date = 2 / overlay vaxis=axis2 legend=legend1; run; proc varmax data=us_money; id date interval=qtr; model y1-y4 / p=2 lagmax=6 dftest print=(iarr(3)) cointtest=(johansen=(iorder=2)) ecm=(rank=1 normalize=y1); cointeg rank=1 normalize=y1 exogeneity; run;*

SAS OutputThis example performs the Dickey-Fuller test for stationarity, the Johansen cointegrated test integrated order 2, and the exogeneity test. The VECM(2) fits the data. From the outputs shown below, you can see that the series has unit roots and is cointegrated in rank 1 with integrated order 1. The fitted VECM(2) is given as *

SAS Output*

SAS Output*

Dickey-Fuller Unit Root TestsVariableTypeRhoPr < RhoTauPr < Tauy1Zero Mean0.050.69341.140.9343Single Mean-2.970.6572-0.760.8260Trend-5.910.7454-1.340.8725y2Zero Mean0.130.71245.140.9999Single Mean-0.430.9309-0.790.8176Trend-9.210.4787-2.160.5063y3Zero Mean-1.280.4255-0.690.4182Single Mean-8.860.1700-2.270.1842Trend-18.970.0742-2.860.1803y4Zero Mean0.400.78030.450.8100Single Mean-2.790.6790-1.290.6328Trend-12.120.2923-2.330.4170

SAS Output *

Cointegration Rank Test for I(2)r\k-r-s4321Trace of I(1)5% CV of I(1)0384.60903214.37904107.9378237.0252355.963347.211219.6239589.2150827.3260920.654229.38273.6177922.132792.647715.34338.294350.01493.845% CV I(2)47.2100029.3800015.340003.84000

SAS Output *

Long-Run Parameter Beta Estimates When RANK=1Variable1y11.00000y2-0.46458y314.51619y4-9.35520

Adjustment Coefficient Alpha Estimates When RANK=1Variable1y1-0.01396y2-0.02811y3-0.00215y40.00510

Diagnostic Checks *

Schematic Representation of Cross Correlations of ResidualsVariable/Lag0123456y1++......++......+.....--....y2++++........................y3.+++....+.-...++-...........y4.+++..........+.............+is>2*stderror,-isChiSq31653.90

Diagnostic Checks *

Univariate Model ANOVA DiagnosticsVariableR-SquareStandard DeviationF ValuePr>Fy10.67540.0071232.51Fy12.134187.190.02751.620.2053y22.040031.200.54831.230.2697y31.86892253.76

Diagnostic Checks *

Univariate Model AR DiagnosticsVariableAR1AR2AR3AR4F ValuePr>FF ValuePr>FF ValuePr>FF ValuePr>Fy10.680.41262.980.05422.010.11542.480.0473y20.050.81850.120.88420.410.74530.300.8762y30.560.45472.860.06104.830.00323.710.0069y40.010.93400.160.85591.210.31030.950.4358

Diagnostic Checks *Whether each variable is the weak exogeneity of other variables. The variable y1 is not the weak exogeneity of other variables, y2, y3, and y4; the variable y2 is not the weak exogeneity of other variables, y1, y3, and y4. If a variable can be taken as "given" without losing information for the purpose of statistical inference, it call weak exogenous.Weak exogeneityLong-run noncausality

Testing Weak Exogeneity of Each VariablesVariableDFChi-SquarePr>ChiSqy116.550.0105y2112.540.0004y310.090.7695y411.810.1786