Solved – VECM – interpreting output from cajorls()

cointegrationrvector-autoregressionvector-error-correction-model

I am a bit puzzled on how to interpret the test results cajorls() from the urca-package. This function returns the OLS regressions of a restricted VECM, i.e. it returns a list object with elements of class ‘lm’ containing the restricted VECM and a matrix object with the normalised cointegrating relationships. The output is as follows:

$rlm

Call:
lm(formula = substitute(form1), data = data.mat)

Coefficients:
                rstar.ger2.d  rstar.nl2.d  rstar.fr2.d  rstar.sp2.d  rstar.it2.d
ect1            -0.0017846     0.0153640    0.0040054   -0.0069455   -0.0001016 
ect2             0.0107730    -0.1374979    0.1362569    0.0798872    0.0291866 
rstar.ger2.dl1   0.0177601    -0.0071459    0.1840403    0.1453076    0.0388310 
rstar.nl2.dl1    0.3616836     0.0006218   -0.2059346   -0.0004955    0.0281643 
rstar.fr2.dl1    0.1159788     0.0083223   -0.0885601    0.0916455    0.0959205 
rstar.sp2.dl1    0.0494191    -0.0308582   -0.0448139   -0.1467505    0.0539171 
rstar.it2.dl1    0.1004643     0.0284516    0.2993759    0.2938969   -0.0073476 
rstar.ger2.dl2  -0.0048584    -0.0060393    0.1232667   -0.0615904    0.0292541 
rstar.nl2.dl2    0.3012379    -0.1223988   -0.1139923   -0.2024982   -0.1418449 
rstar.fr2.dl2    0.0033811     0.0109410   -0.2276895   -0.0343024   -0.1120294 
rstar.sp2.dl2   -0.0187626     0.0241342   -0.0209384    0.2175508    0.0923392 
rstar.it2.dl2    0.0096038    -0.1689995   -0.0548127    0.3646600    0.0143355 


$beta
                   ect1          ect2
rstar.ger2.l3  1.000000  6.938894e-18
rstar.nl2.l3   0.000000  1.000000e+00
rstar.fr2.l3  -7.743012 -9.253440e-01
rstar.sp2.l3   9.681516  6.076362e-01
rstar.it2.l3  -6.690274 -8.328822e-01
constant      -3.566494 -1.377382e+00

Where the Johansen maximum eigenvalue test indicated that there is one cointegrating relationship at the 1% level, and two cointegrating relationships at the 5% level.

How can I translate these results to a coherent story for my thesis? What does this output tell me?

Your help is invaluable!

PS. VECM specification:

VECMcoeff = VECM(combined, lag = 2, r = 2, include = "const", 
            beta = NULL, estim = "ML", LRinclude = "const", exogen = NUL

Best Answer

I am afraid that it is too late for your thesis. You can get more detailed information (t-statistics, p-values, etc.) if you use the function summary. To be more clear, If you put summary(vecm$rlm), you will get a lot more information. Here vecm is the name you gave to your estimated cajorls function. I hope this is what you are looking for.

Related Solutions

Solved – Understanding vec2var conversion in R

The vec2var() function will provide an object of class vec2var. You cannot convert it further to the class varest for standard VAR, but there are still many methods that will work on it, check it with:

methods(class = "vec2var")

Will return:

[1] fevd fitted irf logLik Phi plot predict print Psi residuals

So you can run irf(), fevd() and so on, to "perform VAR analyses on this output" as you want. For exampe, if you want to run the IRF, do:

plot(irf(vec2var(x)))

The only thing you cannot get easily are standard errors for the coefficients of the (constrained) VAR representation.

Solved – Interpreting VECM result

The Johansen procedure which you performed is used to decide how many cointegrating relationships you have. To perform it, first you need to choose appropriate number of lags, which can be done by using function VARselect from package vars. You should also choose whether constant, or trend terms are estimated besides cointegration relationship or not. This decision is usually made based on data or/and model.

After that, the main result of Johansen procedure is the table of test statistics:

          test  10pct 5pct  1pct    
r <= 3 |  1.48  7.52  9.24 12.97    
r <= 2 |  6.89 13.75 15.67 20.20    
r <= 1 | 10.19 19.77 22.00 26.81    
r = 0  | 19.07 25.56 28.14 33.24

It should be read from the bottom to the top. The last row has the hypothesis that there are no cointegrating relationships, and in your case this hypothesis is not rejected. This means that there are no cointegrating relationships in your data and it is not possible to use VEC model with this data, and you should use VAR on the first differences, since all of your time series has unit root.

If you are really sure that there is a long term relationship in your data, then check that you are using the correct number of lags and appropriate dummy variables (constant, trend, seasonal dummies, etc) and then rerun the Johansen procedure again.

As I said the table should be read from bottom to the top. You start at the last row and if the hypothesis is rejected (the test statistic is larger than critical values), then you move one row up, until the test statistic is not rejected. The number of row counting from the bottom, for which the hypothesis is accepted is the number of cointegrating relationships minus one, i.e. the last row means zero, the second last one, etc.

If you only want to use the model for forecasting, then you can convert the resulting VEC model to VAR using function vec2var, where you supply as arguments the output of ca.jo and the number of cointegration relationships. You can then forecast from the resulting model with the function predict, using the argument n.ahead to indicate how many steps you want to forecast.

Best Answer

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