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When regressing a variable on a constant of 1, the coefficient of this constant is the mean. However, when I specified that the residuals follow a GARCH(1,1) model, the coefficient of the constant does not anymore represent the mean in my model. In my opinion this makes no sense since the residuals are defined to have a mean of zero. The mu in the example below should be the same as the mean of R_d.

Small Example in R:

library(rugarch)
library(quantmod)
getSymbols('C', from = '2000-01-01')
C = adjustOHLC(C, use.Adjusted = TRUE)
R_d = ROC(Cl(C), na.pad = FALSE)
mean(R_d)
-0.000436420257283668

spec = ugarchspec(mean.model = list(armaOrder = c(0, 0)), variance.model = list(model = 'sGARCH', garchOrder = c(1, 1)), distribution = 'norm')
fit = ugarchfit(data = R_d, spec = spec)
coef(fit)

mu: 0.000430648533256351
omega: 2.11481883824743e-06
alpha1: 0.0871525584932368
beta1: 0.911847414938857

For egarch or gjr, I get also different results for mu..

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  • $\begingroup$ It really depends on how they implement it. I was looking at the documentation for rugarch and I see two things that stand out. 1) In mean.model, try replacing armaOrder with include.mean and see if there result is different. 2) there is a method called uncmean, not sure if that gives anything different. $\endgroup$ – John Jun 7 '16 at 17:23
  • $\begingroup$ Thanks for your replay. I added the include.mean at the GARCH model, the results is the same since the default value is TRUE. The method uncmean just returns the value of mu. $\endgroup$ – Filippo Scopel Jun 7 '16 at 17:56
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When calculating the simple arithmetic mean, each observation has an equal weight:

$$ \hat \mu^{simple} = \frac{1}{T}\sum_{t=1}^T x_t.$$

If the observations are $i.i.d.$, $\hat \mu^{simple}$ is an efficient estimator of the population mean.

When estimating the mean of a GARCH process, $\hat \mu^{simple}$ is no longer efficient. It makes sense to downweight the noisy observations relative to the low-noise observations so as to gain efficiency. So you fit a GARCH model and use inverse fitted standard deviations as weights:

$$ \hat \mu^{GARCH} = \frac{1}{\sum_{t=1}^T \hat \sigma_t}\sum_{t=1}^T \frac{x_t}{\hat \sigma_t}.$$

Generally the case of estimated weights gives a different number than the case of equal weights:

$$\hat \mu^{simple} \not\equiv \hat \mu^{GARCH}.$$

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  • $\begingroup$ I am glad I could help! $\endgroup$ – Richard Hardy Jun 10 '16 at 14:58

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