Lecture notes on Johansen cointegration
- 1. 1 UNIVERSITY OF PRETORIA FACULTY OF ECONOMIC AND MANAGEMENT SCIENCES DEPARTMENT OF ECONOMICS ECONOMETRICS (EKT 813) PART1, 2005 15th March 2005 MOSES SICHEI LECTURE 8: JOHANSEN COINTEGRATION III Lecture Objectives To see how the ECM looks like for particular cointegrating vectors Testing parameter restrictions on the long-run cointegrating relationship Testing parameter restrictions on the adjustment coefficients (Weak exogeneity test) Get an idea of key steps in applying Johansen procedure Key vocabulary Normalisation Testing economic theory restrictions on in Johansen distance Testing economic theory restrictions on in Johansen Testing of weak exogeneity
- 2. 2 Important articles Johansen,S.(1988), “Statistical Analysis of Cointegrating Vectors” Journal of Economic Dynamics and Control, Vol.12 (June-Sept), 231-254. Stock,J. and Watson, M.(1988), “Testing for Common Trends”, Journal of the American Statistical Association, Vol.83 (Dec.), 1097-1107. Johansen,S. and Juselius, K. (1992), “Testing Structural Hypothesis in a Multivariate Cointegration Analysis of the PPP and UIP for UK”, Journal of Econometrics vol.53, 211-244. 1. EXAMPLES OF WHAT THE ECM VAR SYSTEM LOOKS LIKE FOR PARTICULAR VALUES OF COINTEGRATING VECTORS (r) Let’s use an example given in Enders 2004 (question 4) and file on interest rates. The data used contains interest rates paid on U.S. Tbills, 3-year and 10-years. The data run from 1954:7 to 2002:12. The data is downloaded from http://www.cba.ua.edu/~wenders/. Here we have n = 3 with X comprising the three I(1) time series tb , 3r and 10r . Let k, the lag length of the VAR be 2, and suppose that there are two cointegrating relationships among the elements of X ( 2r ). We include a vector of intercepts in the VAR. Ignoring the vector of dummy variables, the VAR system can be written in levels as t2-t21-t1t +++= XAXAX (1) or in its ECM representation as:
- 3. 3 t 1 t-1 t-1 tX = + X + X + (2) Given the existence of cointegration, we can use the Granger Representation theorem to shown that its ECM is; t 1 t-1 / t-1 tX = + X + X + (3) Writing this in full we have as Equation (4): t t t t t t t t t t t t r r tb r r tb r r tb 3 2 1 1 1 1 322212 312111 3231 2221 1211 1 1 1 333231 232221 131211 3 2 1 10 3 10 3 10 3 (4) Ignoring the first two components (intercept and short-run coefficients) and the disturbance term on the right hand side of the equation (and so looking only at the long run relationships), we focus on the following part of this system:
- 4. 4 1 1 1 322212 312111 3231 2221 1211 10 3 10 3 t t t t t t r r tb r r tb Adjustment coefficients Cointegrating vectors 1tXtX (5) If we leave out the part and focus on the cointegrating vector only, (i.e. terms / X t-1 ). This comprises two cointegrating vectors (r=2) as follows 1 1 1 322212 312111 10 3 t t t r r tb (6) It can be seen, by expanding the expression in Equation 6, that the cointegrating vectors are given by; 103 103 r+r+tb r+r+tb 1-t321-t221-t12 1-t311-t21t11 (7) The elements of (sometimes known as the loading matrix) determine into which equation the cointegrating vectors enter and with what magnitudes.
- 5. 5 If all elements of are non-zero, then all cointegrating vectors (in this case two) enter into each equation. This can be seen by writing equation (5) in scalar algebra form, giving; 1321221123213112111131 1321221122213112111121 1321221121213112111111 10310310 1031033 103103 ttttttt ttttttt ttttttt rrtbrrtbr rrtbrrtbr rrtbrrtbtb (8) In general, if there are r cointegrating relationships among n variables, the r cointegrating relationships will enter into each of the n equations. In our example where r = 2, the 2 separate long run relationships enter into each of the 3 equations. Even if we are only really interested in estimating one of these equations - perhaps the first one - then both cointegrating relationships should enter that equation separately. There are in effect two ECM terms in the equation. This shows more clearly the pitfalls of EG which chooses an estimator that constrains r to be zero or one when actually r is greater than one. The problems this leads us into include the following: If we find r = 1 (as opposed to zero) when in fact r = 2, our dynamic adjustment equation (the second stage of the EG two-step approach, for example) will contain only one ECM term when it should contain 2. The regression model is misspecified by assuming that 012 There are in fact cross-equation restrictions in the n equations of the VAR, because the same two ECM terms (with the same coefficients) into each equation. A systems estimator is needed to estimate this system efficiently (taking account of these restrictions). Even if r = 1, it will still be the case that the single cointegrating relationship could
- 6. 6 enter more than one of the n equations in the VAR system. Ignoring this (or these) cross-equations restrictions will lead to lose efficiency in the estimation procedure. What we regard as the single “cointegrating vector” will be a linear combination of two independent cointegrating vectors. Moreover, we will not be able to identify either of the two underlying vectors from the underlying mixed vector. This can be seen more clearly in the following way. First, rearrange the first of the equations in (8) into the following form: tbt = (1111 + 1212 )tbt-1 + (1121 + 12 22)r3t-1 + (11 31 + 12 22)r10t-1 (9) This can be expressed as 131211 103 tttt rrtbtb (10) Where: 221231113 221221112 121211111 When the E-G first step is used, the ‘long-run’ parameter estimates we obtain are those of the parameters. But these three parameters are linear combinations of the six structural and parameters, which are not identified (they cannot be recovered from the estimates). 2.TESTING PARAMETER RESTRICTIONS ON THE LONG-RUN COINTEGRATING RELATIONSHIP(S) 2.1 NORMALISATION
- 7. 7 As was demonstrated in lecture 7, the Johansen technique allows one to determine how many independent cointegrating relationships exist among the set of variables being considered. However, the estimated parameter values in the r cointegrating relations are not unique. Any linear combination of these r stationary relations will itself be stationary, and so qualify as a cointegrating relationship. The particular estimated parameter values obtained would depend on the normalisation procedure chosen: different normalisations will produce different sets of estimated long run relations. 2.2 REASONS FOR TESTING Oxley et al.(1995:196) (Surveys in econometrics) point out that the usual response by applied researchers (especially macroeconometric modelers faced with a trade off between quantity and quality) who seek to model a single long-run behavioural relationship between the variables included in the x when faced with the result that r>1 is to choose to employ only that cointegrating vector which makes “economic sense”. In other words, they choose the vector where the estimated long- run elasticities correspond closely (in magnitude and sign) to those predicted by economic theory. Such an ad hoc approach is wrong because both the EG and Johansen procedure are explicitly multivariate in the sense that they both postulate an ECM for all the variables involved in the model.
- 8. 8 Neither the EG nor Johansen procedure approaches the problem of partitioning variables into endogenous and (weakly) exogenous, which is central to estimating a single behavioural equation. This means that arbitrary selection of one of the significant cointegrating vectors in order to move from Johansen framework to the estimation of a single structural equation implicitly makes the assumption that the conditional model which is being isolated is valid! The best approach is to test for the restricted forms of the cointegrating vectors. There are two kinds of testing we may wish to do; we may wish to test hypotheses about elements of the parameter matrix; or we may wish to test hypotheses about elements of the parameter matrix. Tests about elements of are concerned with issues about which of the equations in the system the cointegrating vectors enter. Tests about are concerned with restrictions on the parameters within the long run relationships themselves. Tests about are of particular importance as our ultimate objective is to extract estimates of the structural equations, which underlie the reduced form. Recall that the parameter estimates which we obtain after having specified how many cointegrating relationships exist (that is, from the Johansen procedure after choosing r but prior to doing any tests on are the unrestricted reduced form parameter estimates. These are not what we are interested in!
- 9. 9 What we are interest in can only be obtained by trying to deduce likely restricted structural equations (from relevant economic theory e.g. PPP, demand for money, uncovered interest parity conditions, consumption function etc.) and then testing whether the implied restrictions are acceptable. 2.3 RESTRICTIONS ON THE ELEMENTS OF THE CASE WHERE r = 1 If r = 1, the parameter estimates of the single cointegrating relationship can be read directly from the estimated vector. In effect, there is no difference between the reduced form and structural model in this case. However, there is still a normalisation issue to consider. If is a valid long run parameter vector, in the sense that X is a stationary, cointegrating linear combination of the variables in X , then any multiple of will also be a valid cointegrating vector. Looking at the application of Johansen at the last part of the notes; if = (2.49, -3.99, 1.66) , so that ttt rrtbX 1066.1399.349.2 is a cointegrating combination, then if we multiply the cointegrating vector by -0.5, (-1.25, 1.99, -0.83) is also a cointegrating vector with ttt rrtb 1083.0399.125.1 a stationary linear combination. We can choose to “normalise” the cointegrating vector in any way we like. It is conventional to choose a normalisation in which the variable we regard as the dependent variable in the relationship is given a coefficient of -1. Having obtained estimates of the parameters of the single cointegrating relationship, we could then test restrictions
- 10. 10 implied by economic theory on these parameters. For example, we could test the null hypothesis H0: against Ha: There are two restrictions implied by this null hypothesis. A likelihood ratio test can be used to test these restrictions. The point is that once and are determined, the test statistic entails comparing the number of cointegrating vectors under the null with the alternative hypotheses. In order to test restrictions on form the statistic; r i iiT 1 * ˆ1lnˆ1ln (11) This has a 2 distribution with degrees of freedom equal to the number of restrictions placed on . The restriction embedded in the null hypothesis is binding if the calculated value of the test statistic exceeds that in 2 table. 2.4 THE CASE WHERE 1r When 1r , matters are more complicated. It will not usually be sensible to take the unrestricted estimates of the vectors in directly as economically-meaningful long run parameter estimates. There are two reasons why not; (i)There is a normalisation issue, similar in principle to the one explained in the r = 1 case. (ii)What we will (usually) be trying to obtain are estimates of the parameters of the structural equations of the system. It is necessary, therefore, to impose (and test) restrictions on the elements of in an attempt to try and obtain the structural
- 11. 11 relationships between the variables. This is likely to be a difficult exercise in practice. One must be guided here by economic theory - there are no "econometric rules" that can be followed in a mechanical way. An important part of this exercise is likely to be testing exclusion restrictions (i.e. that the parameters associated with particular variables have zero coefficients) suggested by economic theory. For example, in the example we looked at earlier, 103 103 r+r+tb r+r+tb 1-t321-t221-t12 1-t311-t21t11 (12) Assume that we have already concluded that there are 2 cointegrating relationships among the three variables. Economic theory might tell us that one of these is of the form; 0103 r+r+tb 1-t311-t21t11 (13) Whereas the other is of the form 0103 r+r 1-t321-t22 In the second of these, the coefficient on ttb has implicitly been set to zero, and so we might wish to test such an exclusion restriction. Important point: Standard row and column operations on do not entail restrictions on the cointegrating vectors. Variable exclusion within an equation: With multiple cointegrating vectors, you cannot test whether any particular value of 0ij since this assumption does not restrict the cointegrating space.
- 12. 12 Where is an rn matrix, a testable exclusion restriction entails the exclusion or r or more variables from a cointegrating vector. Hence excluding r variables from a cointegrating vector entails only one restriction. If the sample value of the 2 statistic with one degree of freedom exceeds a critical value, reject the null hypothesis that this set of variables contains a cointegrating relationship. Variable exclusion across equations Recall 103 103 r+r+tb r+r+tb 1-t321-t221-t12 1-t311-t21t11 (14) The cross-equation restriction 3231 entails only one restriction on the cointegrating space. This is because it is on the same column. 2.5 WEAK EXOGENEITY AND TESTING RESTRICTIONS ON ELEMENTS OF The elements of the matrix (sometimes known as the loading matrix) determine into which equation the cointegrating vectors enter and with what magnitudes. Put another way, the elements of the matrix determine the existence and magnitudes of the "feedback" effects of the various cointegrating relationships in the individual dynamic equations governing the evolution of ttb , tr3 and tr10 over time. The elements of the matrix relate to the issue of weak exogeneity. In a cointegrated system, if a variable does not respond to the discrepancy from the long-run equilibrium, it is weakly exogenous. In economic theory terms, this implies that there are rigidities (such as regulations, market imperfections etc.), which limit the adjustment process.
- 13. 13 Let the parameters of interest be , the parameters of the r cointegrating vectors. 1321221123213112111131 1321221122213112111121 1321221121213112111111 10310310 1031033 103103 ttttttt ttttttt ttttttt rrtbrrtbr rrtbrrtbr rrtbrrtbtb (15) Suppose, for example, that the third row of consists of zeros; that is, 31 = 32 = 0. Then, it is clear that neither of the two cointegrating relationships will enter into the equation determining tr10 . In this case, the variable tr10 is weakly exogenous for the system as a whole, and it would be valid to model a reduced system of two equations - one determining ttb and the other tr3 - conditional upon tr10 . Put another way, if tr10 is weakly exogenous in this way, no information is lost concerning the parameters of the cointegrating relationships by not explicitly modelling the process determining tr10 jointly with that determining the two "endogenous" variables, ttb and tr3 . In this case, it is valid to condition on 10r . Instead of estimating (8), we estimating the following partial Version of it:
- 14. 14 t t t t t t t t t t t r r tb r r tb r r tb 2 1 1 1 1 322212 312111 2221 1211 1 1 1 333231 232221 131211 02 01 2 1 10 3 10 310 3 (16) Notice that in this partial system, the weakly exogenous variable remains in the long run relations although its short run behaviour is not modelled (there is no equation for tr10 in the VECM system). 2.6 ADVANTAGES OF CONDITIONING ON WEAKLY EXOGENOUS VARIABLES 1. Reduction of number of short run variables in the VECM (those that might only be relevant to the process determining x which is not now modelled). 2. The marginal process for x may have awkward stochastic properties, which will not need to be modelled in the conditional model. If individual elements of are zero, this implies the absence of particular cointegrating relationship(s) in particular equations in the ECM system. This may also have implications for weak exogeneity of variables with respect to parameters of interest. For example, ij=0 implies that the jth cointegrating vector does not enter the ith equation in the VAR (ie the equation system for X it in (2)). 3. JOHANSEN IN PRACTICE
- 15. 15 Let’s use question 4e in Enders 2004:375 as an example. The data INT_RATES.XLS is use and contains interest rates paid on U.S.3-month (Tbill), 3-year (R3) and 10-year U.S.government securities (R10) for the period 1954:7 to 2002:12. This example outlines how the test of Johansen cointegration can be carried out using EVIEW 5 software. We will do a practical on Thursday (2005-03-17). STEP 1 Pretest all variables to assess their order of integration. Plot the variables to see if a linear time trend is likely to be present in the DGP. 0 4 8 12 16 20 55 60 65 70 75 80 85 90 95 00 TBILL 0 4 8 12 16 20 55 60 65 70 75 80 85 90 95 00 R3 2 4 6 8 10 12 14 16 55 60 65 70 75 80 85 90 95 00 R10
- 16. 16 STEP 2: Lags for VAR The lag length can be determined by some of the many information criteria procedures. It is important to avoid too many lags, since the number of parameters grows very fast with the lag length and the information criteria strike a compromise between lag length and number of parameters by minimising a linear combination of the residual sum of squares and the number of parameters If long lag length is required to make white noise residuals, reconsider the choice of variables and look for another important explanatory variable to include in the information set. A summary test statistic that measures the magnitude of the residual autocorrelation is given by Portmanteau test. There are other tests for serial correlation, normality of errors etc. We shall study them under the next topic on VAR. Eviews uses Wald lag-exclusion tests to determine the appropriate lag. In our case we use 12 lags. STEP 3: Deterministic Trend Specification of the VAR The variables may have nonzero means and deterministic and/or stochastic trends. Similarly, the cointegrating equations may have intercepts and deterministic trends. Since the asymptotic distributions of the LR test statistic for cointegration does not have the usual 2 distribution and depends on the assumptions made with respect to deterministic trends, we need to make assumptions regarding the trends underlying our data. To understand the deterministic components, let’s write our ECM as; ttt txx 221 1 1 ~
- 17. 17 Where txx tt ,1,~ 11 , is an orthogonal complement of i.e. 0 1. The model does not allow for constant term which means that all stationary linear combinations will have mean zero.i.e. 0221 . 2. There are no trends whatsoever, but a constant term is allowed in the cointegrating relations. 021 . 3. The model allows for linear trend in each variable but not in the cointegrating relations. 02 . 4. The model allows for linear trends in each variable and in the cointegrating relations 5. The level data have quadratic trends and the cointegrating equations have linear trends. STEP 4: Estimation and determination of rank Estimate the model and determine the rank of matrix in. Date: 03/14/05 Time: 11:22 Sample (adjusted): 1955M02 2002M06 Included observations: 569 after adjustments Trend assumption: No deterministic trend (restricted constant) Series: TBILL R3 R10 Lags interval (in first differences): 1 to 12 Unrestricted Cointegration Rank Test (Trace) Hypothesized Trace 0.05 No. of CE(s) Eigenvalue Statistic Critical Value Prob.** None * 0.055360 50.78106 35.19275 0.0005 At most 1 0.025075 18.37547 20.26184 0.0890 At most 2 0.006876 3.925717 9.164546 0.4233 Trace test indicates 1 cointegrating eqn(s) at the 0.05 level * denotes rejection of the hypothesis at the 0.05 level **MacKinnon-Haug-Michelis (1999) p-values Unrestricted Cointegration Rank Test (Maximum Eigenvalue) Hypothesized Max-Eigen 0.05 No. of CE(s) Eigenvalue Statistic Critical Value Prob.** None * 0.055360 32.40559 22.29962 0.0014 At most 1 0.025075 14.44975 15.89210 0.0831
- 18. 18 At most 2 0.006876 3.925717 9.164546 0.4233 Max-eigenvalue test indicates 1 cointegrating eqn(s) at the 0.05 level * denotes rejection of the hypothesis at the 0.05 level **MacKinnon-Haug-Michelis (1999) p-values The test is done in specific order from the largest eigenvalue to the smallest. We use “Pantula Principle” where we test for significance until you no longer reject the null. The first null is that there is no stationary relations in the data (r=0). We use p-values to make a decision. In our case both tests show that r=1. STEP 5: Test restrictions implied by Economic theory Unrestricted Cointegrating Coefficients (normalized by b'*S11*b=I): TBILL R3 R10 C 2.489580 -3.990997 1.656070 0.942992 0.647520 -3.516469 2.897717 -0.684816 -0.309827 0.630744 -0.730085 2.710185 Unrestricted Adjustment Coefficients (alpha): D(TBILL) -0.015193 0.019238 0.029119 D(R3) 0.032793 0.029437 0.017129 D(R10) 0.040614 0.010333 0.012578 We could treat the evidence that r=1 as validating the EG 2 step procedure and stop there. There is one cointegrating vector ( 131121111 103 ttt rrtb ), which feed into the 3 different equations as follows; 13112111131 13112111121 13112111111 10310 1033 103 tttt tttt tttt rrtbr rrtbr rrtbtb (18)
- 19. 19 Notice that if the loading matrices 03121 , we end up with the conventional EG 2 step model. However, even if there is one cointegrating vector, the EG 2 step procedure does not test this restriction. Suppose we want to normalize 131121111 103 ttt rrtb on the basis of Tbill rate, we make the coefficient of 1ttb be –1 by dividing through 11 as follows; 0103 1 11 31 1 11 21 1 11 11 ttt rrtb which generates 1 11 31 1 11 21 1 103 ttt rrtb (19) We can test the hypothesis 111 in Eviews by typing the commands; B(1,1)=-1 which generates the following results; Restrictions: B(1,1)=-1 Tests of cointegration restrictions: Hypothesized Restricted LR Degrees of No. of CE(s) Log-likehood Statistic Freedom Probability 1 396.4883 NA NA NA 2 403.7132 NA NA NA NA indicates restriction not binding. 1 Cointegrating Equation(s): Convergence achieved after 1 iterations. Restricted cointegrating coefficients (standard error in parentheses) TBILL R3 R10 C -1.000000 1.603080 -0.665200 -0.378776 (0.00000) (0.17503) (0.17763) (0.21407) Adjustment coefficients (standard error in parentheses) D(TBILL) 0.037825 (0.04068)
- 20. 20 D(R3) -0.081641 (0.03337) D(R10) -0.101112 (0.02541) We can put this an ECM as; Crrtbr Crrtbr Crrtbtb tttt tttt tttt 378.010665.03603.1101.010 378.010665.03603.1082.03 378.010665.03603.1037.0 111 111 111 (20) Notice that you can normalise on any other variable depending on the economic theory we would like to test. STEP 6: Exogeneity tests We can conduct exogeneity tests. We can visualise it by looking at; 13112111131 13112111121 13112111111 10310 1033 103 tttt tttt tttt rrtbr rrtbr rrtbtb (21) Suppose we want to test the hypothesis that tbill is exogenous. This is equivalent to claiming that 011 . We can impose this restriction in Eviews as follows A(1,1)=0. The results is; Restrictions: A(1,1)=0 Tests of cointegration restrictions: Hypothesized Restricted LR Degrees of No. of CE(s) Log-likehood Statistic Freedom Probability 1 396.1047 0.767256 1 0.381067 2 403.7132 NA NA NA NA indicates restriction not binding.
- 21. 21 1 Cointegrating Equation(s): Convergence achieved after 9 iterations. Restricted cointegrating coefficients (not all coefficients are identified) TBILL R3 R10 C 2.507323 -4.150214 1.777850 1.044228 Adjustment coefficients (standard error in parentheses) D(TBILL) 0.000000 (0.00000) D(R3) 0.042795 (0.00894) D(R10) 0.046106 (0.00836) This hypothesis cannot be rejected in this case implying that tbill is exogenous and plays no role in the adjustment towards the long run. However, it is still in the long-run equation as shown below; 13112111131 13112111121 131121111 10310 1033 01030 tttt tttt tttt rrtbr rrtbr rrtbtb (22) In other words there are only two short-run equations. What can go wrong in Johansen methodology We need normally distributed white noise residuals. The test is asymptotic and can be sensitive on how we formulate the VAR model in limited samples. The test assumes that there are no structural breaks in the data. Dummies may be important if structural breaks are suspected. If we put a stationary variable in the model, the number of cointegrating vectors increase. The judgement of the modeller is therefore crucial OTHER TOPICS ON JOHANSEN TEST Testing for I(2) relations conditional on the assumption that the model is correctly specified I(1) system. (Quite a tricky issue a beginner)
- 22. 22 Use Johansen method to test if individual variables are I(0) or I(1). Unlike the ADF, the null in this case is I(0) like the KPSS test. This is test 5H in Johansen and Juselius (1992:226).