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BEKK stability check
Posted: Wed Jan 25, 2017 8:30 am
by fatemaalaali
Hi,
I am applying VAR(1)-GARCH(1,1) model using the following code:
Code: Select all
system(model=var1)
variables RUSRet GOLDRET WTIRet
lags 1
det constant
end(system)
garch(p=1,q=1,model=var1,mv=BEKK,asymmetric,pmethod=simplex,piters=10,rseries=rs,mvhseries=hhs,$
stdresids=zu,derives=dd)
@mvqstat(lags=15)
# zu
@mvarchtest(lags=15)
# zu
@regcrits
I would like to check for the stationarity of the model using the roots of|I-AL-BL|=0. How can I get these roots?
Many thanks for your help.
BEKK stability check
Posted: Wed Jan 25, 2017 9:36 am
by TomDoan
I'm assuming you're talking about the roots of the GARCH recursion, and not the VAR. After the GARCH instruction:
@MVGARCHtoVECH(mv=BEKK)
eigen(cvalues=cv) %%vech_a+%%vech_b
disp "Eigenvalues from BEKK-t" *.### cv
(The I-A-B roots are 1- those).
BEKK stability check
Posted: Wed Jan 25, 2017 2:11 pm
by fatemaalaali
Many thanks for your prompt reply.
I have used the posted code and got the following:
Eigenvalues from BEKK-t (0.965,0.000) (0.948,0.000) (0.941,0.000) (0.921,0.000) (0.905,0.000) (0.889,0.000)
Do these results mean that the model is not stationary?
BEKK stability check
Posted: Wed Jan 25, 2017 2:24 pm
by TomDoan
Stationary is the wrong word. (Very strange-looking GARCH models can be stationary---see D. Nelson(1990)). And those are all stable roots (less than 1 in absolute value), so there is no obvious problem.
BEKK stability check
Posted: Wed Jan 25, 2017 4:42 pm
by fatemaalaali
Thank you very much Tom.
I got confused because I was reading the following attached paper. On page 279, in the notes of the table. They got roots that are greater than 1.
Many thanks for your clarification. That was helpful.
BEKK stability check
Posted: Wed Jan 25, 2017 4:56 pm
by TomDoan
Those are equivalent statements. What you did computes the eigenvalues of A+B. Those are the reciprocals of the roots of |I-(A+B)z|=0 so if the first are less than one, the latter are greater than one.
BEKK stability check
Posted: Thu Jan 26, 2017 4:10 am
by fatemaalaali
Got it now.
Thank you very much Tom.
Re: BEKK stability check
Posted: Sat Sep 23, 2017 9:38 am
by mengqi
Hi Tom,
In the above example: (0.965,0.000) (0.948,0.000) (0.941,0.000) (0.921,0.000) (0.905,0.000) (0.889,0.000).
these eigenvalues are calculated for each conditional variance covariance equations? So each of these single equations have stable root?
I have the following results: Eigenvalues from BEKK-t (1.076,-0.000) (1.020,0.000) (0.966,0.000)
is it possible to calculate the dominant root for the whole model? or there is no point of doing so because my results are not good.
Thank you.
Re: BEKK stability check
Posted: Sat Sep 23, 2017 12:23 pm
by TomDoan
mengqi wrote:Hi Tom,
In the above example: (0.965,0.000) (0.948,0.000) (0.941,0.000) (0.921,0.000) (0.905,0.000) (0.889,0.000).
these eigenvalues are calculated for each conditional variance covariance equations? So each of these single equations have stable root?
No. Those are determined by the whole system. The entire system is stable.
mengqi wrote:
I have the following results: Eigenvalues from BEKK-t (1.076,-0.000) (1.020,0.000) (0.966,0.000)
is it possible to calculate the dominant root for the whole model? or there is no point of doing so because my results are not good.
The dominant root is 1.076. If you forecast the covariance matrix from pretty much any set of initial conditions, eventually the variances will go up at an exponential rate of 1.076. If it were 1.02, I wouldn't be all that concerned (that's probably within sampling error), but 1.076 is pretty big.