Why GARCH Innovations Are Standardized to Zero Mean and Unit Variance

# Error term/Innovation process in ARCH/GARCH processes?


# Error term/Innovation process in ARCH/GARCH processes?












I am wondering about the distribution of the error term/innovation process in a ARCH/GARCH process and its implementation, I am not sure about some points. The basic assumption is

$r_t=\sigma_t*\epsilon_t$

where the $\sigma_t$ is the volatility, modeled by ARCH/GARCH and the $\epsilon_t$ are mostly assumed to be N(0,1).

Now my questions are:

- More sophisticated models drop this assumption. So I can say, e.g. $\epsilon_t$ follows a generalized hyperbolic distribution. So the mean does not need to be zero and the variance does not need to be equal to 1. This is correct, right?

- If I use the rugarch package: It supports different distributional assumptions. But I am not getting the following: So they also drop the assumption of mean zero and variance one? Or are they using something like a "standardized" version?

- Suppose I want to fit a GARCH(1,1) assuming, that the $\epsilon_t$ follow a generalized hyperbolic distribution, but the mean does not have to be zero and the variance does not need to be one. Is rugarch doing a jointly parameter estimation? So in my final output, do I get the parameters of the GARCH process and the parameters of my generalized hyperbolic distribution?

My last question is, how can I implement this?

I guess I have to use the following command:

```
ugarchspec(variance.model = list(model = "sGARCH", garchOrder = c(1, 1), 
submodel = NULL, external.regressors = NULL, variance.targeting = FALSE), 
mean.model = list(armaOrder = c(1, 1), include.mean = TRUE, archm = FALSE, 
archpow = 1, arfima = FALSE, external.regressors = NULL, archex = FALSE), 
distribution.model = "norm", start.pars = list(), fixed.pars = list(), ...)
```

the distribution.model has to be set to `ghyp`. Is this assuming a mean of zero and a variance of one?

I think no, right?

How can I use the hyperbolic distribution for distribution.model?

## Answer by Richard Hardy (score 1)

https://quant.stackexchange.com/a/24569

The standardized error in a GARCH model has unit variance (which is needed for identification) and a zero mean. Whatever the conditional distribution, it is scaled and shifted so as to fit those requirements.

The answers to your questions are:

- No.

- No, they don't drop the assumption of mean zero and variance one; and Yes, they are using something like a "standardized" version.

- Yes, "rugarch" is doing a joint parameter estimation; hence, in the final output you get the parameters of the GARCH process and the parameters of the generalized hyperbolic distribution.

- "How can I implement this?" You can implement this using the function `ugarchspec` just as you wrote.

- "Is this assuming a mean of zero and a variance of one?" Yes, the standardized model error are assumed to have these properties.

- "How can I use the hyperbolic distribution for `distribution.model`?" I don't know, but see below.

For more details, consult the "vignette" for the "rugarch" package in R, especially section 2.3 "Conditional distributions".