# Johansen test for cointegration

I am testing for cointegration using the Johansen test. I have seen questions like how to interpret the test results, but when I am interpreting mine I have some doubts. In my results r = 3 since 4.10 < 10.49, so I cannot form a stationary series. It is the same for r = 2 and r = 1. But for r = 0, 86.12 > 59.14, so there is a stationary combination.

But r = 0 implies that there are zero cointegrating vectors. Does that mean that my data are not cointegrated and therefore I cannot build a VECM?

> cointegration <- ca.jo(Canada, type="trace",ecdet="trend",spec="transitory")
> summary(cointegration)

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

Test type: trace statistic , with linear trend in cointegration

Eigenvalues (lambda):
[1]  4.483918e-01  2.323995e-01  1.313250e-01  4.877895e-02 -1.859499e-17

Values of teststatistic and critical values of test:

test 10pct  5pct  1pct
r <= 3 |  4.10 10.49 12.25 16.26
r <= 2 | 15.65 22.76 25.32 30.45
r <= 1 | 37.33 39.06 42.44 48.45
r = 0  | 86.12 59.14 62.99 70.05

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

e.l1    prod.l1       rw.l1        U.l1    trend.l1
e.l1      1.0000000  1.0000000  1.00000000  1.00000000  1.00000000
prod.l1   0.3685667 -0.1582521  2.01545971  0.06122231 -0.09644538
rw.l1    -0.1369713 -0.5035147 -0.08233586 -0.15589592 -0.47523051
U.l1      3.2569951  2.4162383  2.98414327  1.57795960  1.54780259
trend.l1 -0.1539863  0.1477376 -0.53596432 -0.20898570  0.16907450

Weights W:

e.l1     prod.l1       rw.l1        U.l1      trend.l1
e.d     0.01520061  0.10989739  0.04306410 -0.01664954 -6.999563e-13
prod.d  0.06282619  0.17899905 -0.05415524 -0.10283813 -5.525444e-12
rw.d   -0.22958927  0.17308184 -0.03869293  0.06509098 -6.034107e-12
U.d    -0.05230297 -0.08731406 -0.01833898 -0.03719022  1.367902e-12


In Johansen cointegration test, the null hypothesis for the eigenvalue test is that there are $r+1$ cointegration relations.
The test is therefore sequential: you test first for $r=0$, then $r=1$, etc.
The test concludes on the value of $r$ when the test fails to reject $H_0$ for the first time. In your case, the test fails to reject the null hypothesis for the first time when $r=1$.
• Regarding your first paragraph, isn't it the alternative rather than the null hypothesis? Isn't the null r<=0, r<=1, etc., as listed in the first column of the table in the middle of the output? – Richard Hardy Jan 26 '16 at 19:06