[R] garch results is different other soft

Spencer Graves spencer.graves at pdf.com
Thu Aug 17 05:35:02 CEST 2006

      I do not believe there are any garch functions in the core R.  
RSiteSearch("garch", "tseries") just returned 21 hits, identifying garch 
capabilities in three different packages (tseries, fSeries and 
fOptions), plus mentions of garch in the Ecdat package. 

      I would be interested in a sensible reply to your question, and I 
believe many others would as well.  However, without more specifics, I 
don't feel I can afford the time to investigate this further.  You could 
help this by providing commented, minimal, self-contained, reproducible 
code, mentioning which 'garch' function you used in which package and 
which data readily available to all R users (as suggested in the posting 
guide "www.R-project.org/posting-guide.html").  If you do this, please 
tell us which capability of which other software you used for 
comparison, and compare the answers obtained from the other software and 
the R function(s) you used. 

      You might even be able to answer the question yourself by reading 
the code:  R is open source, and you can view the R code for any 
function just by typing the function name at a command prompt.  If part 
of the work is done in a compiled language, you will be able to see 
that, and the GNU license guarantees that you can get the source.  
Moreover, the 'debug' function in R makes it fairly easy to walk through 
a function line by line, looking at what it does at each step of the way. 

      I know this doesn't answer your question, but I hope it helps. 

      Best Wishes,
      Spencer Graves

Yong Xiao wrote:
> Hi
> I compared garch results in R with those give by other software and found
> that their coefficients are different from each other. So I wondered that a
> convention the garch funcion in R takes.
> By testing the output, I noticed it seems that garch function in R by
> default takes such a convention:
> y(t) = c + sigma(t) where c=0 and sigma(t) = a(0) + a(1)*epsilon^2 +
> b(1)*sigma(t-1)^2.
> I also checked the standard deviation series, i.e., the
> output$fitted.values, and noticed that the first element (the starting
> variance) is NA. I feel puzzled because in other software, the
> starting variance is estimated together with a(0), a(1) and b(1) by ML
> method.
> Could any clear this puzzle for me? Many thanks!
> 	[[alternative HTML version deleted]]
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