CODEWITHSUNDEEP
rtime-seriesfgarch
Claudio Moneo
Claudio Moneo

Reputation: 569

R Output of fGarch

I am modelling a time series as a GARCH(1,1)-process:

Garch

And the z_t are t-distributed.

In R, I do this in the fGarch-package via

model <- garchFit(formula = ~garch(1,1), cond.dist = "std", data=r)

Is this correct?

Now, I would like to understand the output of this to check my formula.

Obviously, model@fit$coefs gives me the coefficients and model@fitted gives me the fitted r_t.

But how do I get the fitted sigma_t and z_t?

Upvotes: 1

Views: 759

Answers (2)

Rui Barradas
Rui Barradas

Reputation: 76402

I believe that the best way is to define extractor functions when generics are not available and methods when generics already exist.

The first two functions extract the values of interest from the fitted objects.

get_sigma_t <- function(x, ...){
  [email protected]
}
get_z_t <- function(x, ...){
  x@fit$series$z
}

Here a logLik method for objects of class "fGARCH" is defined.

logLik.fGARCH <- function(x, ...){
  x@fit$value
}

Now use the functions, including the method. The data comes from the first example in help("garchFit").

N <- 200
r <- as.vector(garchSim(garchSpec(rseed = 1985), n = N)[,1])
model <- garchFit(~ garch(1, 1), data = r, trace = FALSE)

get_sigma_t(model) # output not shown
get_z_t(model)     # output not shown

logLik(model)
#LogLikelihood 
#    -861.9494

Note also that methods coef and fitted exist, there is no need for model@fitted or model@fit$coefs, like is written in the question.

fitted(model)  # much simpler
coef(model)
#          mu        omega       alpha1        beta1 
#3.541769e-05 1.081941e-06 8.885493e-02 8.120038e-01

Upvotes: 1

akrun
akrun

Reputation: 886928

It is a list structure. Can find the structure with

str(model)

From the structure, it is easier to extract with $ or @

model@fit$series$z
[email protected]

Upvotes: 1

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