Steps of estimating VECM and interpretation of the results

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Steps of estimating VECM and interpretation of the results

Postby alex7134126 on Fri Feb 24, 2012 12:35 am

Hi, I'm currently studying the relation between construction tender price index and some economic variables (real GDP, building approvals, price index of private housing). I have searched relevant topics on this forum and resources on the internet and tried to estimate a VAR model on the data. But I am now facing the problem of interpretation of the coefficients estimated so I would like to ask for help here. First I would like to see if my works are correctly done:

The data are collected quarterly and covers the periods from year 1997 to 2010. I found that all variables are integrated at order 1so I decided to run VAR on them.

At first I ran VAR on the first order differenced data and see the lag-length criteria:
===================================================
Lag LogL LR FPE AIC SC HQ

0 -1305.396 NA 6.56e+17 52.37584 52.52880 52.43409
1 -1262.789 76.69339 2.27e+17 51.31154 52.07635 51.60279
2 -1227.069 58.57934 1.04e+17 50.52278 51.89944* 51.04702
3 -1204.974 32.70173 8.46e+16 50.27895 52.26745 51.03618
4 -1171.179 44.60934* 4.42e+16* 49.56715 52.16750 50.55738*
5 -1161.688 11.00912 6.40e+16 49.82753 53.03973 51.05075
6 -1147.021 14.66676 8.02e+16 49.88086 53.70490 51.33707
7 -1117.144 25.09694 6.01e+16 49.32576 53.76166 51.01498
8 -1093.219 16.26918 6.54e+16 49.00875* 54.05649 50.93096
====================================================
According to the third column, I decided to use lag length of four.

Then I ran Johansen Cointegration test on the variables at level with lag interval (1 4) and Eviews reports:
Unrestricted Cointegration Rank Test (Trace)
====================================================
Hypothesized Trace 0.05
No. of CE(s) Eigenvalue Statistic Critical Value Prob.**

None * 0.373539 60.00961 47.85613 0.0024
At most 1 * 0.300718 34.75552 29.79707 0.0124
At most 2 0.246171 15.43969 15.49471 0.0510
At most 3 0.003325 0.179835 3.841466 0.6715

Trace test indicates 2 cointegrating eqn(s) at the 0.05 level
====================================================
Since then I applied VECM on the variables at level, with lag interval (1 4) and 2 cointegrating equations, and the result is:

Cointegrating Eq: CointEq1 CointEq2

DLS(-1) 1.000000 0.000000

GDP(-1) 0.000000 1.000000

CONSENT(-1) 0.840727 345.3397
(0.16585) (32.6895)
[ 5.06924] [ 10.5642]

PIDOMESTIC(-1) -7.522969 -1238.059
(1.31409) (259.013)
[-5.72484] [-4.77991]

C -624.6280 -352319.6
(131.294) (25878.6)
[-4.75747] [-13.6143]

Error Correction: D(DLS) D(GDP) D(CONSENT) D(PIDOMESTIC)

CointEq1 -0.087126 -1.310316 -0.752511 -0.016387
(0.03805) (10.6109) (0.21696) (0.00627)
[-2.28981] [-0.12349] [-3.46850] [-2.61509]

CointEq2 0.000316 -0.123061 -0.001585 4.82E-05
(0.00020) (0.05454) (0.00112) (3.2E-05)
[ 1.61563] [-2.25614] [-1.42120] [ 1.49597]

D(DLS(-1)) 0.239486 35.02476 1.454408 -0.029682
(0.15763) (43.9577) (0.89878) (0.02596)
[ 1.51932] [ 0.79678] [ 1.61820] [-1.14340]

D(DLS(-2)) -0.113142 22.30902 0.152244 0.001768
(0.16567) (46.1998) (0.94462) (0.02728)
[-0.68295] [ 0.48288] [ 0.16117] [ 0.06482]

D(DLS(-3)) 0.156774 -27.36019 0.370964 -0.017493
(0.15664) (43.6818) (0.89314) (0.02580)
[ 1.00087] [-0.62635] [ 0.41535] [-0.67813]

D(DLS(-4)) 0.128427 42.62697 -0.143566 0.046587
(0.14463) (40.3344) (0.82470) (0.02382)
[ 0.88794] [ 1.05684] [-0.17408] [ 1.95586]

D(GDP(-1)) 0.000130 0.118840 0.003525 4.27E-05
(0.00037) (0.10380) (0.00212) (6.1E-05)
[ 0.34815] [ 1.14491] [ 1.66082] [ 0.69676]

D(GDP(-2)) 0.000193 0.007038 0.002379 6.52E-05
(0.00037) (0.10243) (0.00209) (6.0E-05)
[ 0.52447] [ 0.06871] [ 1.13572] [ 1.07823]

D(GDP(-3)) 0.000468 0.015970 0.001642 2.83E-05
(0.00033) (0.09310) (0.00190) (5.5E-05)
[ 1.40194] [ 0.17154] [ 0.86282] [ 0.51531]

D(GDP(-4)) 2.16E-05 0.890051 0.000352 3.44E-06
(0.00034) (0.09461) (0.00193) (5.6E-05)
[ 0.06355] [ 9.40752] [ 0.18201] [ 0.06162]

D(CONSENT(-1)) -0.057579 44.41051 -0.128162 -0.003001
(0.06261) (17.4592) (0.35698) (0.01031)
[-0.91970] [ 2.54367] [-0.35902] [-0.29102]

D(CONSENT(-2)) -0.066064 33.57194 0.020893 0.001181
(0.05468) (15.2474) (0.31176) (0.00900)
[-1.20830] [ 2.20181] [ 0.06702] [ 0.13117]

D(CONSENT(-3)) -0.017503 21.20179 0.152667 -0.003753
(0.04248) (11.8471) (0.24223) (0.00700)
[-0.41201] [ 1.78963] [ 0.63025] [-0.53638]

D(CONSENT(-4)) 0.012376 14.66215 0.210645 0.004571
(0.02810) (7.83662) (0.16023) (0.00463)
[ 0.44041] [ 1.87098] [ 1.31463] [ 0.98770]

D(PIDOMESTIC(-1)) 1.209503 412.5680 0.925190 0.619721
(0.88909) (247.940) (5.06951) (0.14642)
[ 1.36039] [ 1.66398] [ 0.18250] [ 4.23248]

D(PIDOMESTIC(-2)) 0.437685 -303.5912 -11.05572 -0.067530
(1.07281) (299.176) (6.11712) (0.17668)
[ 0.40798] [-1.01476] [-1.80734] [-0.38222]

D(PIDOMESTIC(-3)) -1.060077 179.0560 -6.118109 -0.228047
(0.95429) (266.124) (5.44131) (0.15716)
[-1.11085] [ 0.67283] [-1.12438] [-1.45106]

D(PIDOMESTIC(-4)) -0.937314 -543.9680 -4.354184 -0.209890
(0.90725) (253.005) (5.17307) (0.14941)
[-1.03314] [-2.15003] [-0.84170] [-1.40478]

R-squared 0.633543 0.904778 0.673666 0.738878
Adj. R-squared 0.460494 0.859812 0.519564 0.615571
Sum sq. resids 26443.66 2.06E+09 859737.7 717.1915
S.E. equation 27.10251 7558.094 154.5367 4.463405
F-statistic 3.661056 20.12135 4.371559 5.992167
Log likelihood -243.8550 -547.9153 -337.8584 -146.4544
Akaike AIC 9.698332 20.95983 13.17994 6.090903
Schwarz SC 10.36133 21.62282 13.84294 6.753897
Mean dependent 3.648148 3346.574 -5.844444 0.018519
S.D. dependent 36.89872 20186.29 222.9534 7.198770
============================================

At this moment I would like to ask whether the steps of estimating this VECM is correct?
If not, could someone provide some guide on how should I do for the estimation?
alex7134126
 
Posts: 4
Joined: Fri Feb 24, 2012 12:16 am

Re: Steps of estimating VECM and interpretation of the resul

Postby alex7134126 on Fri Feb 24, 2012 12:44 am

Further if the steps are correct, my next concern is that how should I interpret the results.

As I know, the coefficients of cointeq are the speed of adjustments which adjust the particular variable by deviation from equilibrium of previous period (is it correct?). Someone said in theory, the coefficients of cointeq should lie between -1 and 0, but in the results above, some of them are not satisfying this condition. Also, some of their t-values are close to zero indicating insignificance. So how should I interpret these insignificant coefficients? Or if my purpose is to forecast the variables for a few steps later, should I drop those insignificant coefficients?

On the other hand, should I drop the insignificant coefficients in the error correction part?

Sorry I am very new to this modeling technique and hope someone could provide some ideas.
Thank you very much!

Alex
alex7134126
 
Posts: 4
Joined: Fri Feb 24, 2012 12:16 am

Re: Steps of estimating VECM and interpretation of the resul

Postby Mbila on Fri Aug 03, 2012 5:34 am

I wish someone could assist us on this because I have the same question, and I wanted also to know if we run VECM in levels or differences?

Thanks a lot.
Mbila
 
Posts: 3
Joined: Fri May 18, 2012 7:38 am

Re: Steps of estimating VECM and interpretation of the resul

Postby obicna89 on Thu Jan 17, 2013 1:38 pm

if you look at VAR(p) model, the VEC representation of this model is p-1 (the p refers to the lags). so if you estimate first a VAR(1) model, this means that: Z(t)=a+b*Z(t-1)+e(t), where Z(t) is vector of the variables you are working with, a is the vector of constants, etc.
the VEC representation of VAR(1) is deltaZ(t)=c+d*Z(t-1)+u(t), and here you have deltaZ(t) - the differenced values in a vector, so you have to put a zero in the lag criteria for VEC model, because it refers to number of lagged values of deltaZ(t), and you don't have any of them on the right hand side of the VEC representation of VAR(1)

edit: and you don't difference the data for the VAR. you use them without differencing first. do everything up to the johansen test - and if the rank of the d matrix is full - all of the variables are stationary and you do just the VAR procedure, if the rank is not full - the variables are not stationary but a cointegration exists and you do the VEC procedure. if the rank is null - variables are not stationary but there is no cointegration, then you have to difference the data and do the VAR procedure over differenced data
obicna89
 
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