# Understanding the GARCH(1,1) model: the constant, the ARCH term and the GARCH term

I would like some help with a GARCH(1,1) volatility modeling.

I am working with the assumption the volatility is the weighted sum of three factors: Long run variance + $n-1$ squared return + $n-1$ variance

If that is accurate, my doubt is, what is the difference between the 1st and 3rd parts of the equation? I was reading it as $n-1$ variance being the historical variance of the moving window I am using. However, that seems to me the same as long run variance.

Can anyone clarify that for me?

• Possible duplicate of How to interpret GARCH parameters? – Matifou Nov 18 '17 at 17:27
• I do not think this is exactly a duplicate, because the answers in the other thread do not address the precise question specified here. – Richard Hardy Nov 21 '17 at 19:11

## 1 Answer

A GARCH(1,1) model is \begin{aligned} y_t &= \mu_t + u_t, \\ \mu_t &= \dots \text{(e.g. a constant or an ARMA equation without the term $u_t$)}, \\ u_t &= \sigma_t \varepsilon_t, \\ \sigma_t^2 &= \omega + \alpha_1 u_{t-1}^2 + \beta_1 \sigma_{t-1}^2, \\ \varepsilon_t &\sim i.i.d(0,1). \\ \end{aligned} The three components in the conditional variance equation you refer to are $\omega$, $u_{t-1}^2$, and $\sigma_{t-1}^2$. Your question seems to be, how is $\omega$ different from $\sigma_{t-1}^2$?

First, note that $\omega$ is not the long-run variance; the latter actually is $\sigma_{LR}^2:=\frac{\omega}{1-(\alpha_1+\beta_1)}$. $\omega$ is an offset term, the lowest value the variance can achieve in any time period, and is related to the long-run variance as $\omega=\sigma_{LR}^2(1-(\alpha_1+\beta_1))$.

Second, $\sigma_{t-1}^2$ is not the historical variance of the moving window; it is instantaneous variance at time $t-1$.

• I hope this answers your question. Feel free to ask for further clarification. – Richard Hardy Nov 21 '17 at 19:25
• Hi, thank you very much for helping on this. I do have some follow up doubts. The instantaneous variance you mean is variance between t-1 and t-2? And w is still not very clear to me. Sorry i am having trouble with formatting questions still. – Luiza Nov 21 '17 at 23:19
• @Luiza, no problem, glad to help! Regarding instantaneous variance, it depends on how you imagine the underlying process. If it is a discrete-time process, then instantaneous variance is at a particular time point $t-1$ because nothing happens in between time points; this is what I had in mind. If it is a continuous-time process, then you are correct. Regarding formatting, you may click on "edit" and see the underlying code of any post you find relevant; you can find the code behind the formulas this way. – Richard Hardy Nov 22 '17 at 6:24
• @Luiza, so what do you think of my answer? FYI, satisfactory answers may be accepted by clicking on the tick mark to the left. Unsatisfactory answers do not need to be accepted. That is how Cross Validated works. – Richard Hardy Nov 23 '17 at 8:25
• I am still a little confused regarding w. But your answer has most certainly helped me. Sorry about not accepting it earlier. Thanks again! – Luiza Nov 23 '17 at 8:33