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I'm dealing with a GARCH-M model that I've estimated using R and EViews. Here are its mean and variance equations.

Mean equation:

$$ y_t=\mu + \rho \sigma^2_t + \varepsilon_t $$

Variance equation:

$$ \sigma^2_t = \omega + \alpha \varepsilon_{t-1}^2 + \beta \sigma^2_{t-1} + T$$

where T is a dummy variable containing 0 and 1 to indicate structural change.

Here is my EViews result:

eviews result http://s25.postimg.org/m8gc8y6ql/eviews_result.jpg

And my R code is as follows:

#get data
re=read.table("return.csv",sep=",",header=TRUE)
......
xts<-as.xts(re[,-1],order.by=re[,1])
......
T<- as.matrix(xts[,2])

#GARCH specification
garchspec<- ugarchspec(variance.model = list(model = "sGARCH", garchOrder = c(1, 1), 
                       submodel = NULL, external.regressors = T, 
                       variance.targeting = FALSE), 
                       mean.model = list(armaOrder = c(0, 0), include.mean = TRUE,
                       archm = TRUE, archpow = 1, arfima = FALSE, 
                       external.regressors = NULL, archex = FALSE), 
                       distribution.model = "ged",
                       start.pars = list(), fixed.pars = list())

#fitting
fit<-ugarchfit(spec=garchspec, data=xts[,1], out.sample = 0,solver="solnp",
               solver.control = list(trace=0), fit.control = 
               list(stationarity = 1, fixed.se = 0, scale = 0, rec.init = 0.7))
show(fit)

It gives these results:

R result http://s25.postimg.org/ai2erkdy5/R_result.jpg

As you can see, the dummy variable (denoted by vxreg1) is totally insignificant using rugarch in R contrary to a 2.58% p-value in the EViews result. Other estimates have some differences with their counterparts, but they are all minor.

I checked the vignette of rugarch package for many times and cannot find any mistakes in the syntax, and R didn't show any error as well. I wonder what the problem is. I really appreciate it if you can solve my problem.

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  • $\begingroup$ the 7th parameter is the GED parameter which is shown in the EViews result $\endgroup$ – Junnan May 12 '14 at 16:39
  • $\begingroup$ OK, but did you notice that one is using z-test and the other one is using the t-test? Maybe its better to ask this question in SO. $\endgroup$ – Stat May 12 '14 at 16:44
  • $\begingroup$ I noticed that difference, but I don't think it will have much influence on the results. $\endgroup$ – Junnan May 12 '14 at 16:53
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The coefficients are of comparable magnitudes in both models. While I do not know the actual implementations in R and Eviews, I am pretty sure that both implementations numerically maximize the (log-)likelihood function. The difference between the results may lay in different convergence criteria for numerical optimization algortihms used in the softwares. This implies that one set of the results may be more precise than the other, while the other implementation may run faster.

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