2
$\begingroup$

I have got two time series on which I have run the Johansen test for cointegration in R (package: urca). From the results I can see that no series is mean reverting and there is the cointegration at r=0 at 90% confidence level (is that correct interpretation?)

Using the results I want to build a model

$$y_{1t}=β_2 \ \mathbf{y_{2t}} + u_t, \quad \quad where \ \ \ u_t∼I(0) \quad \quad (12.4)$$ $$\mathbf{y_{2t}}=y_{2t-1}+vt, \quad \quad where \ \ \ v_t∼I(0) \quad \quad (12.5)$$ as presented here http://faculty.washington.edu/ezivot/econ584/notes/cointegration.pdf p.435

I assume that the $u_t$ can be seen as the modeled difference between series (spread). How could I calculate the half life of the spread. Should it be log(2)/Eigenvalue as suggested in the comment to this question? https://quant.stackexchange.com/questions/2076/how-to-interpret-the-eigenmatrix-from-a-johansen-cointegration-test

Which number from the below example should be taken to calculate half life?
What number should be taken for $\beta_2$?

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

Test type: maximal eigenvalue statistic (lambda max) , with linear trend 

Eigenvalues (lambda):
[1] 0.00308977956 0.00008189485

Values of teststatistic and critical values of test:

          test 10pct  5pct  1pct
r <= 1 |  0.37  6.50  8.18 11.65
r = 0  | 13.82 12.91 14.90 19.19

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

           A.l2     B.l2
A.l2  1.0000000 1.00000000
B.l2 -0.8064909 0.00237543

Weights W:
(This is the loading matrix)

             A.l2        B.l2
A.d 0.00005520918 -0.0003176665
B.d 0.00728082027 -0.0002809040
$\endgroup$

1 Answer 1

0
$\begingroup$

Michal, to calculate the half life of the deviation from the long run relationship, you need to estimate the VECM form found in pg 439 equations 12.12 and 12.13 of the link you provided. For the first variable, the half life can be calculated as $$log(0.5)/(1-\alpha_1)$$. Similarly, the half life of the second variable can be calculated as $$log(0.5)/(1-\alpha_2)$$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.