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I have been working on a time series where after the first difference, I observe heteroskedasticity. To handle the situation, I found that ARCH/GARCH models are used typically.

When I read about the procedure, they say that the time series is first fitted with a conditional mean model like AR or ARMA and ARCH/GARCH model is applied to the residuals of the fitted AR/ARMA model.

My questions are:

  1. Why do we have to fit AR/ARMA?

  2. Why do we have to apply ARCH/GARCH to the residuals? Is that done to model the volatility in the residuals or the volatility in the actual data (differenced data)?

  3. If it is used to model the volatility of the residuals, how is that going to help in modeling the volatility in the actual data?

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    $\begingroup$ ARMA is a mean model, whereas GARCH is a variance model. If the underlying time series is known to be 0 mean, then we can apply GARCH directly. Otherwise the mean needs to be estimated. If not, the dynamics of mean will leak into variance estimation and we will not be able to distinguish them. For this reason they recommend to fit ARMA first (and subtract this, which gives you the residuals). However this is not the best approach. The best approach is to fit an ARMA-GARCH model in one shot. It is statistically more efficient. $\endgroup$ – Cagdas Ozgenc Jul 7 '17 at 19:37
  • $\begingroup$ Thank you for your response Cagdas Ozgenc. Now it really makes sense. $\endgroup$ – Dhineshkumar Jul 8 '17 at 4:57
  • $\begingroup$ @Dhineshkumar, what do you think about my answer? Is it clear? (I see you have not accepted it.) $\endgroup$ – Richard Hardy Aug 29 '17 at 12:00
  • $\begingroup$ @RichardHardy sorry sir. I have my internal exams going on this month, so I did not work out those equation on my own. I will just work out within a day or two and let you know. Thank you for your response. $\endgroup$ – Dhineshkumar Aug 30 '17 at 0:56
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On the one hand, GARCH is a model for the conditional distribution of the time series $y_t$:

\begin{aligned} y_t &\sim d(\mu_t,\sigma_t^2), \\ \mu_t &= \dots \text{(e.g. some ARMA equation)} \\ \sigma_t^2 &= \omega + \alpha_1 ( y_{t-1} - \mu_{t-1} )^2 + \dotsc + \alpha_s ( y_{t-s} - \mu_{t-s} )^2 + \beta_1 \sigma_{t-1}^2 + \dotsc + \beta_r \sigma_{t-r}^2, \\ \varepsilon_t &:= \frac{y_t-\mu_t}{\sigma_t} \sim i.i.d(0,1). \\ \end{aligned}

GARCH specifically characterizes the conditional variance equation, but this inevitably depends on some equation for the conditional mean; otherwise the conditional variance would be undefined. This hopefully answers your questions 1. and 3.

On the other hand, GARCH happens to characterize the conditional variance of the residuals from the conditional mean model, $u_t$, as well. The exact same model as above can be represented as follows:

\begin{aligned} y_t &= \mu_t + u_t, \\ \mu_t &= \dots \text{(e.g. some ARMA equation)} \\ u_t &= \sigma_t \varepsilon_t, \\ \sigma_t^2 &= \omega + \alpha_1 u_{t-1}^2 + \dotsc + \alpha_s u_{t-s}^2 + \beta_1 \sigma_{t-1}^2 + \dotsc + \beta_r \sigma_{t-r}^2, \\ \varepsilon_t &\sim i.i.d(0,1), \\ \end{aligned}

and $\sigma_t^2$ is the conditional variance of the residual $u_t$ since $\text{Var}(u_t|I_{t-1})=\sigma_t^2$ where $I_{t-1}$ is information available at time $t-1$. This hopefully answers your question 2.

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  • $\begingroup$ I understood the reasoning for 1st question. For other two questions, please check whether have I understood properly from your answer. GARCH is used to model the variance of the residuals, which in turn helps in modeling the volatility of the actual time series because the conditional mean of the time series is already known by applying ARMA/ARIMA. So, just by adding the mean model from ARMA/ARIMA to the variance model from GARCH would produce the actual time series. Is that right? $\endgroup$ – Dhineshkumar Sep 10 '17 at 14:27
  • $\begingroup$ @Dhineshkumar, Yes, ARMA+GARCH gives you the conditional distribution of the series. The ARMA part gives the point forecasts, the GARCH part gives the density forecasts around these point forecasts. $\endgroup$ – Richard Hardy Sep 10 '17 at 14:55
  • $\begingroup$ Can you please tell me what does density forecast around point forecast mean? $\endgroup$ – Dhineshkumar Sep 10 '17 at 14:59
  • $\begingroup$ @Dhineshkumar, Density or distribution, it is the same. By having it, you can have forecast intervals with arbitrary coverage. Density contains all the probabilistic information you could wish for. $\endgroup$ – Richard Hardy Sep 10 '17 at 16:50

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