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When i run the same johansen test across Python and R I get very different critical values.

If I normalize the two columns in evec in the Python result I get fairly similar eigenvectors as the ones in R (first and third eigenvector at least). So far so good.

But the test statistics and critical values in the Python results seem strange. As far as I understand, for the first test (r=0) I can't reject the null hypothesis there is no co-integration. But for the second test (r=1) there is strong evidence (>5%) for rejecting the null hypothesis there is no co-integration. So at the same time the rank of r is both zero and greater than 1? (In other words, there are zero and >1 stationary combinations of the input data?)

Or should I simply ignore the second test if the first test isn't statistically significant?

In R the results look more straight forward. Both the first and second test are statistically insignificant.

So to summarize:

  • Are the critical values in Python incorrect?
  • Or am I simply misinterpreting the Python results?

coint_johansen in statsmodels.tsa.vector_ar.vecm in Python

import statsmodels.tsa.vector_ar.vecm as stvv
model = stvv.coint_johansen(data, 0, 2)

print("\nnormalized eigenvector 0\n", model.evec[:,0] / model.evec[:,0][0])
print("\nnormalized eigenvector 1\n", model.evec[:,1] / model.evec[:,1][0])
print("\ntest statistics\n", model.lr1[0], model.lr1[1])
print("\ncritical values\n", model.cvt[0], model.cvt[1])
print("\neig\n", model.eig)
print("\nevec\n", model.evec)
print("\nlr1\n", model.lr1)
print("\nlr2\n", model.lr2)
print("\ncvt\n", model.cvt)
print("\ncvm\n", model.cvm)
print("\nind\n", model.ind)

normalized eigenvector 0
 [ 1.         -0.18956975]

normalized eigenvector 1
 [ 1.         -1.56557504]

test statistics
 12.170080461590794 4.8892460854155075

critical values
 [13.4294 15.4943 19.9349] [2.7055 3.8415 6.6349]

eig
 [0.0095468  0.00642099]

evec
 [[ 0.04683733 -0.00754275]
 [-0.00887894  0.01180875]]

lr1
 [12.17008046  4.88924609]

lr2
 [7.28083438 4.88924609]

cvt
 [[13.4294 15.4943 19.9349]
 [ 2.7055  3.8415  6.6349]]

cvm
 [[12.2971 14.2639 18.52  ]
 [ 2.7055  3.8415  6.6349]]

ind
 [0 1]

ca.jo in urca library in R

> model = ca.jo(data.frame(data$a, data$b), type="trace", ecdet="const",K=2,spec="longrun")
> summary(model)

###################### 
# Johansen-Procedure # 
###################### 

Test type: trace statistic , without linear trend and constant in cointegration 

Eigenvalues (lambda):
[1] 1.184071e-02 7.329430e-03 5.204170e-18

Values of teststatistic and critical values of test:

          test 10pct  5pct  1pct
r <= 1 |  5.59  7.52  9.24 12.97
r = 0  | 14.64 17.85 19.96 24.60

Eigenvectors, normalised to first column:
(These are the cointegration relations)

              data.a.l2     data.b.l2    constant
data.a.l2     1.0000000      1.000000    1.000000
data.b.l2    -0.2446287      1.633518   -1.478432
constant  -1441.2146598 -15496.470449 7340.786442

Weights W:
(This is the loading matrix)

           data.a.l2    data.b.l2      constant
data.a.d -0.02386740 -0.001330821  2.330213e-16
data.b.d  0.01599476 -0.005125451 -6.495089e-16
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  • $\begingroup$ Would be interesting to hear why... $\endgroup$
    – confused
    Jul 1, 2020 at 7:43
  • $\begingroup$ Maybe: github.com/statsmodels/statsmodels/issues/5355 m see end of discussion for similar differences between urca and matlab, GRETL. My main takeaway was that urca is basing critical values on cases or assumptions that make the test more consevative. $\endgroup$
    – Josef
    Jan 2, 2021 at 15:27

1 Answer 1

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The difference may come from the VECM specification. You have used spec = "longrun" in your R function whereas, if I am not mistaken, the Python function uses what would be spec = "transitory" in ca.jo.

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