3
$\begingroup$

I have done fitted a DCC-GARCH model using the dccfit function from the "rmgarch" package in R. The output is below:

*---------------------------------*
*          DCC GARCH Fit          *
*---------------------------------*

Distribution         :  mvnorm
Model                :  DCC(1,1)
No. Parameters       :  62
[VAR GARCH DCC UncQ] : [0+32+2+28]
No. Series           :  8
No. Obs.             :  240
Log-Likelihood       :  4896.6
Av.Log-Likelihood    :  20.4 

Optimal Parameters
-----------------------------------
                  Estimate   Std. Error  t value  Pr(>|t|)
[FTSE100].mu      0.005599    0.003457 1.6195e+00 0.105339
[FTSE100].omega   0.000100    0.000160 6.2312e-01 0.533205
[FTSE100].alpha1  0.176637    0.124341 1.4206e+00 0.155436
[FTSE100].beta1   0.807578    0.072324 1.1166e+01 0.000000
[MSUSAML].mu      0.007760    0.003077 2.5219e+00 0.011673
[MSUSAML].omega   0.000056    0.000053 1.0484e+00 0.294455
[MSUSAML].alpha1  0.092896    0.040348 2.3023e+00 0.021316
[MSUSAML].beta1   0.886704    0.028933 3.0647e+01 0.000000
[MSEXUK.].mu      0.009228    0.003421 2.6976e+00 0.006984
[MSEXUK.].omega   0.000114    0.000189 6.0293e-01 0.546552
[MSEXUK.].alpha1  0.070957    0.046983 1.5103e+00 0.130978
[MSEXUK.].beta1   0.889084    0.091959 9.6682e+00 0.000000
[DAXINDX].mu      0.010099    0.004489 2.2496e+00 0.024474
[DAXINDX].omega   0.001005    0.000794 1.2650e+00 0.205864
[DAXINDX].alpha1  0.191733    0.113491 1.6894e+00 0.091142
[DAXINDX].beta1   0.600585    0.225184 2.6671e+00 0.007651
[BMUK10Y].mu      0.001496    0.001295 1.1548e+00 0.248181
[BMUK10Y].omega   0.000000    0.000027 0.0000e+00 1.000000
[BMUK10Y].alpha1  0.025774    0.174068 1.4807e-01 0.882287
[BMUK10Y].beta1   0.969964    0.178467 5.4350e+00 0.000000
[BMUS10Y].mu      0.001069    0.001481 7.2147e-01 0.470623
[BMUS10Y].omega   0.000021    0.000014 1.4980e+00 0.134123
[BMUS10Y].alpha1  0.025983    0.024924 1.0425e+00 0.297181
[BMUS10Y].beta1   0.928892    0.037850 2.4542e+01 0.000000
[BMBD10Y].mu      0.000893    0.001088 8.2098e-01 0.411657
[BMBD10Y].omega   0.000000    0.000000 1.2974e-01 0.896774
[BMBD10Y].alpha1  0.000000    0.000089 7.8000e-05 0.999938
[BMBD10Y].beta1   0.999000    0.000075 1.3363e+04 0.000000
[LHUSTRY].mu      0.000170    0.000950 1.7931e-01 0.857694
[LHUSTRY].omega   0.000007    0.000000 2.2820e+01 0.000000
[LHUSTRY].alpha1  0.024463    0.001250 1.9571e+01 0.000000
[LHUSTRY].beta1   0.941022    0.005656 1.6638e+02 0.000000
[Joint]dcca1      0.017443    0.005703 3.0584e+00 0.002225
[Joint]dccb1      0.942324    0.012105 7.7843e+01 0.000000

Information Criteria
---------------------

Akaike       -40.288
Bayes        -39.389
Shibata      -40.388
Hannan-Quinn -39.926

Can someone tell me what is the meaning of Pr(>|t|)? Is it the p value for the parameter? If it is, then I have lots of insignificant parameters which indicates a very bad model I have there. I have tried run examples from the rmgarch.tests folder as well but the Pr(>|t|) values for the example are also big (greater than 0.05). What can I do here?

$\endgroup$
3
  • 1
    $\begingroup$ Those are p-values. Maybe the variables just aren't related to the response. Be aware that you also seem to have 62 parameters, but only 240 data, so you may just not have enough information. $\endgroup$ Commented Jan 25, 2016 at 16:44
  • $\begingroup$ So, does it means it is better for me to have a longer period of observation? @gung $\endgroup$
    – nsaa
    Commented Jan 25, 2016 at 17:29
  • $\begingroup$ I don't know. Someone who knows this material better will have to answer. $\endgroup$ Commented Jan 25, 2016 at 17:38

1 Answer 1

4
$\begingroup$

Yes, the column Pr(>|t|) are the $p$-values.

You should mostly care about the joint significance of (1) alpha1 and beta1 for each of the series and (2) the joint significance of dcca1 and dccb1.

  • (1) will tell you whether the GARCH(1,1) "makes sense" for the given series. If alpha1 and beta1 are jointly insignificant, you may be better off using constant conditional variance rather than GARCH(1,1).
  • (2) will tell you whether DCC "makes sense" for the system of series. If dcca1 and dccb1 are jointly insignificant, you may be better off using a constant conditional correlation model rather than DCC(1,1).

You may not care that much about the significance of mu; it is the intercept of the conditional mean model, and there are reasons (not specific to GARCH modelling) for keeping the intercept in even though it is not significant.

Meanwhile, you want to keep omega in the model regardless of its significance unless alpha1+beta1=1, otherwise the absence of omega generates funny patterns in conditional variance -- see this answer for details.

$\endgroup$
8
  • $\begingroup$ From my results,most of the 'alpha1' has value greater than 0.1 which is not significant. So, does it means that I am better to try other model such as Constant Conditional Correlation? $\endgroup$
    – nsaa
    Commented Jan 25, 2016 at 19:59
  • $\begingroup$ The reason for this exercise is because I would like to forecast the mean return and variance covariance matrix so that I can use them for portfolio optimization. $\endgroup$
    – nsaa
    Commented Jan 25, 2016 at 20:09
  • $\begingroup$ Well, I indicated in my answer that you should look at the joint significance of alpha1 and beta1 rather than alpha1 alone. Besides, alpha1 is a parameter of the (univariate) GARCH model rather than the DCC part of the DCC-GARCH model. Meanwhile, it is the joint significance of dcca1 and dccb1 that indicate whether DCC makes more sense than CCC. $\endgroup$ Commented Jan 25, 2016 at 20:32
  • $\begingroup$ I will try to run F test to see the joint significance of these parameters. Thanks! $\endgroup$
    – nsaa
    Commented Jan 26, 2016 at 11:47
  • $\begingroup$ Is there any reference or articles that I can refer to saying that we need to look at the joint significant of alpha1 and beta1 and we may not care much about the significance of mu and omega? Thanks $\endgroup$
    – nsaa
    Commented Jan 26, 2016 at 14:04

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.