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I'm an analyst in financial and insurance fields and whenever I try to fit volatility models I obtain awful results: residuals are often non-stationary (in the unit root sense) and heteroskedastic (so the model doesn't explain volatility).

Do ARCH/GARCH models work with other kind of data, maybe?

Edited on 17/04/2015 15:07 to clarify some points.

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    $\begingroup$ Do you mean a general field (e.g. finance, meteorology,...) where these models work well or a specific data set? In the first case, despite these models may capture some overall features common to some data, it would be hard to expect these models to be enough to fit any sample date set from a given field. In the second case, many academic papers about these models show an application to real data. Reality is not always as clear and beautiful as presented in some of those illustrations, but there you will probably find several data sets and compelling examples. $\endgroup$ – javlacalle Apr 16 '15 at 19:38
  • $\begingroup$ I meant a general field. I understand that exist specific datasets on which ARCH and GARCH fit well (Engle won a Nobel, right?), but I was discussing the general case. $\endgroup$ – Stefano R. Apr 17 '15 at 13:02
  • $\begingroup$ Well, I didn't think this was actually too broad until you said "I was discussing the general case"... I don't see how evidence could be presented that it applies for "the general case" for a whole field without at least a book-length treatment. How could such a case be made in the few paragraphs of a reasonable answer in this format? $\endgroup$ – Glen_b -Reinstate Monica Apr 18 '15 at 0:20
  • $\begingroup$ I don't need that. I just wished that somebody would have told me for example: "I'm a researcher in Biochemistry, we use regularly use GARCH in the analysis of rats' liver cells, and its application is very useful" or something like that. $\endgroup$ – Stefano R. Apr 20 '15 at 7:47
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My experiences with programming/implementing and testing ARCH/GARCH procedures have led me to the conclusion that they must be useful somewhere and someplace but I haven't seen it. Gaussian violations such as unusual values/level shifts/seasonal pulses and local time trends should be used initially to deal with changes in volatility/error variance as they have less serious side effects. After any of these adjustments care might be taken to validate that model parameters are constant over time . Furthermore error variance may not be constant but simpler/less intrusive remedies like Box-Cox and detecting deterministic break points in error variance ala Tsay are much more useful and less destructive. Finally if none of these procedures work then my last gasp would be to throw ARCH/GARCH at the data and then add a ton of holy water. I firmly agree with your findings and conclude that these are methods looking for data or just dissertation topics flying in the wind.

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Some background information first:

Given a dependent variable $y_t$, independent variables $X_t$ and a conditional mean model

$$y_t=\beta X_t+\epsilon_t$$

you can use a GARCH model to model the conditional variance of $\epsilon_t$.

Say you have fit a GARCH model and obtained fitted conditional standard deviations $\hat \sigma_t$. If you scale the residuals $\hat \epsilon_t$ by the inverse of the fitted conditional standard deviations $\hat \sigma_t$, you obtain scaled residuals $\hat u_t:=\frac{\hat \epsilon_t}{\hat \sigma_t}$. You would like these to be "nice". At least they should have no ARCH patterns remaining in them. This can be tested by the Li-Mak test, for example.

1: regarding nonstationary residuals
GARCH model does not produce any residuals -- there is no GARCH-model-residual in the GARCH formula (only lagged errors $\epsilon_t$ from the conditional mean model that are used as regressors in the GARCH model).
But what exactly do you mean by nonstationarity: unit root?; heteroskedasticity?; level shift?

When you mention nonstationary residuals, do you have in mind $\hat u_t$ or $\hat \epsilon_t$, or still something else?

Edit: the type of nonstationarity is unit root. I suspect this is due to a poor model for the conditional mean rather than a failure of GARCH. Since the effect of GARCH on $\hat u_t$ is the scaling of $\hat \epsilon_t$ by $\frac{1}{\hat \sigma_t}$, that only changes the scale of $\hat \epsilon_t$ but cannot introduce a unit root. That is, the unit root must have already been a feature of $\hat \epsilon_t$, and that is a problem of the conditional mean model, not the conditional variance model.

2: regarding heteroskedasticity
More could be said when you clarify what residuals you have in mind.

Edit: residuals in mind are $\hat u_t$. If $\hat u_t$ are conditionally heteroskedastic but the pattern is not of ARCH nature, then you could append the standard GARCH model by explanatory variables to explain the remaining heteroskedasticity.

3: regarding non-normality
$\epsilon_t$ can be non-normal, this is no problem. $u_t$ should match the distribution you assume when fitting a GARCH model (you need to assume a distribution to be able to obtain the likelihood function that will be maximized when fitting the GARCH model). If you assume a normal distribution for $u_t$ but can reject normality for $\hat u_t$ then it's a problem. But you do not need to assume normality. A $t$ distribution with 3 or 4 degrees of freedom has been argued to be more relevant than a normal distribution for financial returns, for example.

4: regarding residuals are often non-stationary, heteroskedastic and not normal, so the model doesn't explain volatility
Eidt (more precise formulation): I am not sure I follow the logical connection here. Since GARCH aims to explain a specific type of conditional heteroskedasticity (not any and all types of CH but autoregressive CH), you should assess it on that basis. If $\hat \epsilon_t$ are autoregressively conditionally heteroskedastic (this can be tested by the ARCH-LM test) but $\hat u_t$ are conditionally homoskedastic (as tested by the Li-Mak test), the GARCH model has done its job.

My experience with GARCH models (admittedly limited) is that they do their job but of course are not a panacea.

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  • $\begingroup$ 1 Unit root 2 With residuals I mean $ \hat {u}$ 3 I should try to look further at that. 4 My quastion was fairly more general (maybe too much), but it was edited by Mr. Kolassa. Now that you make me look at that it's probably more correct if it was restrained just to heteroskedasticity. $\endgroup$ – Stefano R. Apr 17 '15 at 12:51

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