Questions regarding cointegration tests (likelihood ratio test of identifying restrictions) - Time Series

[Daniel Earp]

Hello,

I've been having some doubts regarding this subject, my professor has given me and my class some exercises to do regarding VAR, VEC, cointegration etc, he gives each student the a particular set of data and posed a few questions, one of them he tells us to find whether or not the 3 series he gave us were cointegrated, for that I did:

. vecrank Y ER M, lags (3) max ic levela

                       Johansen tests for cointegration                        
Trend: constant                                         Number of obs =      71
Sample:  1998q2 - 2015q4                                         Lags =       3
-------------------------------------------------------------------------------
maximum                                      trace     5% critical  1% critical
  rank    parms       LL       eigenvalue  statistic      value        value
    0      21     -779.64073                 17.8110*1*5  29.68        35.65
    1      26     -774.80921     0.12724      8.1479      15.41        20.04
    2      29     -770.94587     0.10311      0.4212       3.76         6.65
    3      30     -770.73525     0.00592
-------------------------------------------------------------------------------
maximum                                       max      5% critical  1% critical
  rank    parms       LL       eigenvalue  statistic      value        value
    0      21     -779.64073                  9.6630      20.97        25.52
    1      26     -774.80921     0.12724      7.7267      14.07        18.63
    2      29     -770.94587     0.10311      0.4212       3.76         6.65
    3      30     -770.73525     0.00592
-------------------------------------------------------------------------------
maximum
  rank    parms       LL       eigenvalue     SBIC       HQIC       AIC
    0      21     -779.64073                 23.2225*   22.8194*  22.55326
    1      26     -774.80921     0.12724    23.38659   22.88751   22.55801
    2      29     -770.94587     0.10311    23.45788   22.90121   22.53369
    3      30     -770.73525     0.00592    23.51198   22.93612   22.55592
-------------------------------------------------------------------

From that test I concluded that there were no cointegration, the following questions assume that the student found cointegration between the series, so my professor said that those who didn't find any cointegration should do the following questions "as if" they had found one cointegration vector. The question was to "test whether or not the cointegration vector of Y, ER and M could be [1, -1, 1], so I did the following:

constraint define 1 [_ce1]ER = 1
constraint define 2 [_ce1]M = -1

vec Y ER M, lags(3) bconstraints(1/2)

Iteration 1:     log likelihood = -775.84756
Iteration 2:     log likelihood = -775.77803
Iteration 3:     log likelihood = -775.77713
Iteration 4:     log likelihood = -775.77711
Iteration 5:     log likelihood = -775.77711
Iteration 6:     log likelihood = -775.77711
Iteration 7:     log likelihood = -775.77711
Iteration 8:     log likelihood = -775.77711
Iteration 9:     log likelihood = -775.77711

Vector error-correction model

Sample:  1998q2 - 2015q4                        Number of obs     =         71
                                                AIC               =    22.5571
Log likelihood = -775.7771                      HQIC              =   22.87393
Det(Sigma_ml)  =  621098.6                      SBIC              =   23.35382

Equation           Parms      RMSE     R-sq      chi2     P>chi2
----------------------------------------------------------------
D_Y                   8      2.6647   0.6010   94.90711   0.0000
D_ER                  8     .152769   0.0943   6.558798   0.5849
D_M                   8     3434.37   0.3102   28.33345   0.0004
----------------------------------------------------------------

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_Y          |
        _ce1 |
         L1. |   .0000916   .0000789     1.16   0.246    -.0000632    .0002463
             |
           Y |
         LD. |  -.2792682   .1295411    -2.16   0.031    -.5331641   -.0253723
        L2D. |  -.7996009    .125757    -6.36   0.000     -1.04608   -.5531217
             |
          ER |
         LD. |  -.1294314   2.231254    -0.06   0.954     -4.50261    4.243747
        L2D. |  -3.547765   2.274089    -1.56   0.119    -8.004897    .9093669
             |
           M |
         LD. |   .0003968   .0001434     2.77   0.006     .0001159    .0006778
        L2D. |    .000038    .000131     0.29   0.771    -.0002186    .0002947
             |
       _cons |    1.31649   .3962059     3.32   0.001     .5399403    2.093039
-------------+----------------------------------------------------------------
D_ER         |
        _ce1 |
         L1. |  -.0000102   4.53e-06    -2.26   0.024    -.0000191   -1.37e-06
             |
           Y |
         LD. |   .0075074   .0074267     1.01   0.312    -.0070487    .0220634
        L2D. |  -.0018011   .0072097    -0.25   0.803    -.0159319    .0123297
             |
          ER |
         LD. |  -.0652166   .1279195    -0.51   0.610    -.3159343    .1855011
        L2D. |  -.0677055   .1303753    -0.52   0.604    -.3232363    .1878253
             |
           M |
         LD. |  -6.31e-06   8.22e-06    -0.77   0.442    -.0000224    9.80e-06
        L2D. |   4.44e-07   7.51e-06     0.06   0.953    -.0000143    .0000152
             |
       _cons |   .0259104   .0227148     1.14   0.254    -.0186097    .0704306
-------------+----------------------------------------------------------------
D_M          |
        _ce1 |
         L1. |   .1889534   .1017487     1.86   0.063    -.0104705    .3883772
             |
           Y |
         LD. |  -64.16371   166.9578    -0.38   0.701    -391.3949    263.0675
        L2D. |  -476.8146   162.0807    -2.94   0.003    -794.4869   -159.1423
             |
          ER |
         LD. |  -3259.871    2875.73    -1.13   0.257    -8896.199    2376.456
        L2D. |  -1869.447   2930.937    -0.64   0.524    -7613.978    3875.083
             |
           M |
         LD. |   .3444399   .1847622     1.86   0.062    -.0176874    .7065671
        L2D. |   .0707883   .1687823     0.42   0.675    -.2600189    .4015955
             |
       _cons |  -.0006366    510.646    -0.00   1.000    -1000.848    1000.847
------------------------------------------------------------------------------

Cointegrating equations

Equation           Parms    chi2     P>chi2
-------------------------------------------
_ce1                  1   127.8627   0.0000
-------------------------------------------

Identification:  beta is overidentified

 ( 1)  [_ce1]ER = 1
 ( 2)  [_ce1]M = -1
------------------------------------------------------------------------------
        beta |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_ce1         |
           Y |   695.5911   61.51514    11.31   0.000     575.0236    816.1586
          ER |          1          .        .       .            .           .
           M |         -1          .        .       .            .           .
       _cons |  -62438.07          .        .       .            .           .
------------------------------------------------------------------------------
LR test of identifying restrictions:  chi2(  1) =   1.936  Prob > chi2 = 0.164

I understand that the foot note tells me that I cannot reject the null hypothesis that the coefficients are equal to 1, -1 and 1, since the p-value is 0.164, so that would mean that the cointegration vector is [1, -1, 1], but like I said above, the Johansen tests told me that there was no cointegration whatsoever, that makes no sense to me! Shouldn't I be able to reject the null? Could anyone tell me what i'm doing wrong? Am I completely off here?

Thank you!

[Daniel Earp]

Anybody?

[Eric de Souza]

There's something wrong with the trace statistic:the number cannot be 17.8110*1*5.
You are correct in saying that if there is no cointegration the linear combination of Y, ER and M is an I(1) variable and any test on their coefficients would require non-standard tables.
But supposing that there is cointegration, you are not testing what your professor asked for: you need an extra constraint.
constraint define 1 [_ce1]ER = -1 (and not 1) constraint define 2 [_ce1]M = 1 (and not -1) constraint define 3 [_ce1]Y = 1 As you can see from the end of the output you reproduce, the coefficients do not correspond to what you wish to impose

[Daniel Earp]

There's something wrong with the trace statistic:the number cannot be 17.8110*1*5.
You are correct in saying that if there is no cointegration the linear combination of Y, ER and M is an I(1) variable and any test on their coefficients would require non-standard tables.
But supposing that there is cointegration, you are not testing what your professor asked for: you need an extra constraint.
constraint define 1 [_ce1]ER = -1 (and not 1) constraint define 2 [_ce1]M = 1 (and not -1) constraint define 3 [_ce1]Y = 1 As you can see from the end of the output you reproduce, the coefficients do not correspond to what you wish to impose

Hello, Eric,

First of all, thank you for the answer!

The trace statistic is actually just 17.8110, for some reason the "*1*5" appear in blue letters right next to the statistic, in a link form, when I click on it it takes me to the help tab with a message saying

" The * indicates the rank selected by a sequence of trace statistics

As discussed in [TS] vecrank and Johansen (1995), performing a sequence of trace tests at a given significance level produces an estimate of the
number of cointegrating equations. The * indicates that this estimator has selected the number of cointegrating equations corresponding to this row
of the table. "

As for your suggestion, if I try to put constraints on the 3 variables it won't work, this is what happens:

 constraint define 5 [_ce1]ER=-1
constraint define 6 [_ce1]M = 1
constraint define 7 [_ce1]Y=1
vec Y ER M, lags(3) bconstraints(5 6 7)
there are at least as many constraints as parameters

My sub question is regarding what you said about non-standard tables, please tell me if I understood correctly: Since the johansen tests concluded that there is no cointegration whatsoever between the series, doing a likelihood ratio test to see if the "cointegration vector" = [1, -1, 1] makes no sense because the series wouldn't follow a chi-squared distribution?

[Eric de Souza]

Then try

constraint define [_ce1]Y = - [_ce1]ER
constraint define [_ce1]Y = [_ce1]M

As to non-standard, if the combination b1*Y(t-1) + b2*ER(t-1) + b3*M(t-1) is non-stationary, that is, not cointegrated, then any test of restrictions involving b1, b2 and b2 is an non-standard test, that is, requires special statistical tables.
Added on edit: don't forget to number the constraints

[Daniel Earp]

Then try

constraint define [_ce1]Y = - [_ce1]ER
constraint define [_ce1]Y = [_ce1]M

As to non-standard, if the combination b1*Y(t-1) + b2*ER(t-1) + b3*M(t-1) is non-stationary, that is, not cointegrated, then any test of restrictions involving b1, b2 and b2 is an non-standard test, that is, requires special statistical tables.
Added on edit: don't forget to number the constraints

I'm afraid that solution didn't work either... Do you know how I could do that non standard test?
When I do constraint for 2 of the variables am I not already doing it for the 3, since one will be normalized to 1?

[Eric de Souza]

Saying it doesn't work doesnet get me very far. I have no idea what you typed and what you got. It works for me.

Identification:  beta is underidentified

 ( 1)  [_ce1]m + [_ce1]p = 0
 ( 2)  [_ce1]m + [_ce1]y = 0
------------------------------------------------------------------------------
        beta |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_ce1         |
           m |   .7783569   .2067281     3.77   0.000     .3731772    1.183537
           p |  -.7783569   .2067281    -3.77   0.000    -1.183537   -.3731772
           y |  -.7783569   .2067281    -3.77   0.000    -1.183537   -.3731772
       _cons |   .0442434          .        .       .            .           .
------------------------------------------------------------------------------

Adjustment parameters

Equation           Parms    chi2     P>chi2
-------------------------------------------
D_m                   1   10.81401   0.0010
D_p                   1   .0848638   0.7708
D_y                   1   2.002648   0.1570
-------------------------------------------
------------------------------------------------------------------------------
       alpha |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_m          |
        _ce1 |
         L1. |  -.0426555   .0129712    -3.29   0.001    -.0680786   -.0172323
-------------+----------------------------------------------------------------
D_p          |
        _ce1 |
         L1. |  -.0016759   .0057527    -0.29   0.771     -.012951    .0095993
-------------+----------------------------------------------------------------
D_y          |
        _ce1 |
         L1. |    .012988   .0091778     1.42   0.157    -.0050002    .0309762
------------------------------------------------------------------------------
LR test of identifying restrictions: chi2(2) = 10.51       Prob > chi2 = 0.005

As you can see the coefficients of m p and y are the same in absolute value .
The commands I issued were:

use ukm.dta
var m p y , lags(1/2)
vecrank m p y, lags(2)
constraint 1 [_ce1]m = - [_ce1]p
constraint 2 [_ce1]m = - [_ce1]y
vec m p y, rank(1) alpha bconstraints(1 2) dforce

On edit: will be back tomorrow

[Daniel Earp]

I'm sorry I wasn't more clear! Now I followed what you wrote and I think it worked.. I used the exact same commands as you posted above and got this:

Iteration 1:     log likelihood = -827.05206
Iteration 2:     log likelihood = -827.05206

Vector error-correction model

Sample:  1998q1 - 2015q4                        Number of obs     =         72
                                                AIC               =   23.39034
Log likelihood = -827.0521                      HQIC              =   23.57916
Det(Sigma_ml)  =   1905085                      SBIC              =   23.86464

Equation           Parms      RMSE     R-sq      chi2     P>chi2
----------------------------------------------------------------
D_Y                   5     4.04925   0.0621   4.433771   0.4888
D_ER                  5     .154722   0.0120   .8136976   0.9762
D_M                   5     3763.32   0.1241   9.490264   0.0910
----------------------------------------------------------------

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_Y          |
        _ce1 |
         L1. |  -.4206815   .4354299    -0.97   0.334    -1.274108    .4327455
             |
           Y |
         LD. |    .002349   .1773968     0.01   0.989    -.3453424    .3500404
             |
          ER |
         LD. |  -2.560336   3.254236    -0.79   0.431    -8.938521    3.817849
             |
           M |
         LD. |   .0000245   .0001855     0.13   0.895     -.000339     .000388
             |
       _cons |   .6496446     .50305     1.29   0.197    -.3363152    1.635604
-------------+----------------------------------------------------------------
D_ER         |
        _ce1 |
         L1. |    .012808   .0166378     0.77   0.441    -.0198016    .0454176
             |
           Y |
         LD. |   .0019735   .0067784     0.29   0.771    -.0113118    .0152589
             |
          ER |
         LD. |  -.0005144   .1243449    -0.00   0.997    -.2442259    .2431971
             |
           M |
         LD. |  -2.26e-07   7.09e-06    -0.03   0.975    -.0000141    .0000137
             |
       _cons |  -.0014228   .0192216    -0.07   0.941    -.0390965    .0362509
-------------+----------------------------------------------------------------
D_M          |
        _ce1 |
         L1. |   -446.722   404.6829    -1.10   0.270    -1239.886     346.442
             |
           Y |
         LD. |   157.4399   164.8703     0.95   0.340       -165.7    480.5797
             |
          ER |
         LD. |  -5439.613   3024.445    -1.80   0.072    -11367.42    488.1893
             |
           M |
         LD. |   .0680095   .1723847     0.39   0.693    -.2698583    .4058773
             |
       _cons |  -.0006118   467.5281    -0.00   1.000    -916.3389    916.3377
------------------------------------------------------------------------------

Cointegrating equations

Equation           Parms    chi2     P>chi2
-------------------------------------------
_ce1                  1   1.587516   0.2077
-------------------------------------------

Identification:  beta is underidentified

 ( 1)  [_ce1]Y + [_ce1]ER = 0
 ( 2)  [_ce1]Y + [_ce1]M = 0
------------------------------------------------------------------------------
        beta |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_ce1         |
           Y |  -.0000612   .0000485    -1.26   0.208    -.0001563     .000034
          ER |   .0000612   .0000485     1.26   0.208     -.000034    .0001563
           M |   .0000612   .0000485     1.26   0.208     -.000034    .0001563
       _cons |  -2.138665          .        .       .            .           .
------------------------------------------------------------------------------

Adjustment parameters

Equation           Parms    chi2     P>chi2
-------------------------------------------
D_Y                   1   .9334053   0.3340
D_ER                  1    .592612   0.4414
D_M                   1   1.218554   0.2696
-------------------------------------------
------------------------------------------------------------------------------
       alpha |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_Y          |
        _ce1 |
         L1. |  -.4206815   .4354299    -0.97   0.334    -1.274108    .4327455
-------------+----------------------------------------------------------------
D_ER         |
        _ce1 |
         L1. |    .012808   .0166378     0.77   0.441    -.0198016    .0454176
-------------+----------------------------------------------------------------
D_M          |
        _ce1 |
         L1. |   -446.722   404.6829    -1.10   0.270    -1239.886     346.442
------------------------------------------------------------------------------
LR test of identifying restrictions:  chi2(  2) =   8.058  Prob > chi2 = 0.018

The chi statistic (8.058) and pvalue (0.018) tell me that I can reject the null hypothesis that the coefficients are [1,-1,1], is that correct?

Thank you very much!

[Eric de Souza]

Your results confirm what you found with the test for cointegration rank, namlely, that there is no cointegration: The alpha coefficients (those attached to _ce1 in the vec output) have high p-values. In other words, the cointegration relationship can be dropped from all three equations..

The answer to your question at the end is: yes, it is correct.

[Eric de Souza]

I just noticed that you have not imposed the constraints on the cointegration relation correctly.

  ( 1)  [_ce1]Y + [_ce1]ER = 0
   ( 2)  [_ce1]Y + [_ce1]M = 0

is not correct. It should be

  ( 1)  [_ce1]Y + [_ce1]ER = 0
   ( 2)  [_ce1]Y - [_ce1]M = 0

Furthermore , all your coefficients have high p-values.

[Daniel Earp]

Good afternoon! I did what you said, exchanged the signal in the equation, didn't change much, however!

 constraint 1 [_ce1]Y = - [_ce1]ER

.
. constraint 2 [_ce1]Y = [_ce1]M

.
. vec Y ER M, rank(1) alpha bconstraints(1 2) dforce

Iteration 1:     log likelihood = -827.05379
Iteration 2:     log likelihood = -827.05379

Vector error-correction model

Sample:  1998q1 - 2015q4                        Number of obs     =         72
                                                AIC               =   23.39038
Log likelihood = -827.0538                      HQIC              =   23.57921
Det(Sigma_ml)  =   1905177                      SBIC              =   23.86469

Equation           Parms      RMSE     R-sq      chi2     P>chi2
----------------------------------------------------------------
D_Y                   5     4.04927   0.0621   4.433133   0.4889
D_ER                  5     .154725   0.0120      .8114   0.9763
D_M                   5     3763.38   0.1240   9.487961   0.0911
----------------------------------------------------------------

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_Y          |
        _ce1 |
         L1. |  -.4196229   .4344754    -0.97   0.334    -1.271179    .4319332
             |
           Y |
         LD. |   .0023917    .177396     0.01   0.989    -.3452981    .3500816
             |
          ER |
         LD. |  -2.560912   3.254211    -0.79   0.431    -8.939048    3.817225
             |
           M |
         LD. |   .0000244   .0001855     0.13   0.895    -.0003391     .000388
             |
       _cons |   .6495872   .5030747     1.29   0.197    -.3364211    1.635596
-------------+----------------------------------------------------------------
D_ER         |
        _ce1 |
         L1. |   .0127554   .0166016     0.77   0.442    -.0197831    .0452939
             |
           Y |
         LD. |   .0019719   .0067784     0.29   0.771    -.0113135    .0152574
             |
          ER |
         LD. |  -.0004853   .1243455    -0.00   0.997     -.244198    .2432274
             |
           M |
         LD. |  -2.24e-07   7.09e-06    -0.03   0.975    -.0000141    .0000137
             |
       _cons |  -.0014265   .0192228    -0.07   0.941    -.0391026    .0362495
-------------+----------------------------------------------------------------
D_M          |
        _ce1 |
         L1. |  -445.3717   403.8001    -1.10   0.270    -1236.805    346.0619
             |
           Y |
         LD. |   157.4886   164.8713     0.96   0.339    -165.6532    480.6305
             |
          ER |
         LD. |  -5440.353   3024.454    -1.80   0.072    -11368.17    487.4677
             |
           M |
         LD. |   .0679486   .1723831     0.39   0.693     -.269916    .4058133
             |
       _cons |  -.0006121   467.5561    -0.00   1.000    -916.3937    916.3925
------------------------------------------------------------------------------

Cointegrating equations

Equation           Parms    chi2     P>chi2
-------------------------------------------
_ce1                  1   1.584216   0.2082
-------------------------------------------

Identification:  beta is underidentified

 ( 1)  [_ce1]Y + [_ce1]ER = 0
 ( 2)  [_ce1]Y - [_ce1]M = 0
------------------------------------------------------------------------------
        beta |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_ce1         |
           Y |   .0000611   .0000486     1.26   0.208    -.0000341    .0001563
          ER |  -.0000611   .0000486    -1.26   0.208    -.0001563    .0000341
           M |   .0000611   .0000486     1.26   0.208    -.0000341    .0001563
       _cons |  -2.155279          .        .       .            .           .
------------------------------------------------------------------------------

Adjustment parameters

Equation           Parms    chi2     P>chi2
-------------------------------------------
D_Y                   1   .9327989   0.3341
D_ER                  1   .5903219   0.4423
D_M                   1   1.216501   0.2700
-------------------------------------------
------------------------------------------------------------------------------
       alpha |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_Y          |
        _ce1 |
         L1. |  -.4196229   .4344754    -0.97   0.334    -1.271179    .4319332
-------------+----------------------------------------------------------------
D_ER         |
        _ce1 |
         L1. |   .0127554   .0166016     0.77   0.442    -.0197831    .0452939
-------------+----------------------------------------------------------------
D_M          |
        _ce1 |
         L1. |  -445.3717   403.8001    -1.10   0.270    -1236.805    346.0619
------------------------------------------------------------------------------
LR test of identifying restrictions:  chi2(  2) =   8.062  Prob > chi2 = 0.018

.

I also tried doing:

constraint 3 [_ce1]Y = - [_ce1]ER + [_ce1]M

. vec Y ER M, rank(1) alpha bconstraints(3) dforce

Which is what I wanted to "impose" ( I want to test the vector [1,-1,1]), then I got this:

Identification:  beta is underidentified

 ( 1)  [_ce1]Y + [_ce1]ER - [_ce1]M = 0
------------------------------------------------------------------------------
        beta |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_ce1         |
           Y |    .716104   .2760126     2.59   0.009     .1751293    1.257079
          ER |  -.7171676   .2763834    -2.59   0.009    -1.258869   -.1754661
           M |  -.0010636   .0003821    -2.78   0.005    -.0018126   -.0003146
       _cons |  -64.47123          .        .       .            .           .
------------------------------------------------------------------------------

Adjustment parameters

Equation           Parms    chi2     P>chi2
-------------------------------------------
D_Y                   1   .6413788   0.4232
D_ER                  1   5.846295   0.0156
D_M                   1   2.917504   0.0876
-------------------------------------------
------------------------------------------------------------------------------
       alpha |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
D_Y          |
        _ce1 |
         L1. |   .0819512   .1023288     0.80   0.423    -.1186096     .282512
-------------+----------------------------------------------------------------
D_ER         |
        _ce1 |
         L1. |  -.0090871   .0037583    -2.42   0.016    -.0164532   -.0017211
-------------+----------------------------------------------------------------
D_M          |
        _ce1 |
         L1. |   160.1117   93.73838     1.71   0.088    -23.61212    343.8356
------------------------------------------------------------------------------
LR test of identifying restrictions:  chi2(  1) =   1.681  Prob > chi2 = 0.195

In this scenario I have coefficients with small p values so the null cannot be rejected.. but the absolute values aren't the same now.. is this approach wrong?

My professor always used the LR test at the foot of the code to check whether the constraints were valid, but I see you are more concerned with the beta coefficients in the equations, are they the same thing?

[Eric de Souza]

( 1) [_ce1]Y + [_ce1]ER - [_ce1]M = 0 doesn't make any sense. And definitely does not correspond to the restrictions that the coefficient of ER should be minus the coefficient of Y, and that the ceofficient of M should be equal to the coefficient of Y This means that: [_ce1]Y = -[ce1]ER [_ce1]Y = [_ce1]M The test of the restriction is indeed the LR test at the bottom. All I was saying was that none of the coefficients in the VECM appeared to be significantly different from zero

[Daniel Earp]

( 1) [_ce1]Y + [_ce1]ER - [_ce1]M = 0 doesn't make any sense. And definitely does not correspond to the restrictions that the coefficient of ER should be minus the coefficient of Y, and that the ceofficient of M should be equal to the coefficient of Y This means that: [_ce1]Y = -[ce1]ER [_ce1]Y = [_ce1]M The test of the restriction is indeed the LR test at the bottom. All I was saying was that none of the coefficients in the VECM appeared to be significantly different from zero

Yes, you're quite right, I don't know what I was thinking, please ignore that comment.. I'm embarassed by it! Moving on, indeed the option you taught me seems to be the right fit for what I was looking for, so, to sum it up, the alpha (adjusment) coefficients and the [_ce1] coefficeints all have high p values, is that equivalent to analyzing the low p value at the foot test? The 0.018 p-value would mean that we could reject the null hypothesis that the cointegration vector is [1,-1,1], right?

Again, thank you and sorry for any confusion!