Often times people specify the GARCH model as follows:

$$ \sigma _{t}^{2}=\omega +\alpha _{1}\epsilon _{t-1}^{2}+\cdots +\alpha _{q}\epsilon _{t-q}^{2}+\beta _{1}\sigma _{t-1}^{2}+\cdots +\beta _{p}\sigma _{t-p}^{2}=\omega +\sum _{i=1}^{q}\alpha _{i}\epsilon _{t-i}^{2}+\sum _{i=1}^{p}\beta _{i}\sigma _{t-i}^{2}.$$

I must admit that I am a little confused as to what the difference is between $\sigma_t^2$ and $\epsilon_t^2$. As far as I (though I did) know the squared error terms equal the variance, because of the fact that its mean equal zero.

In other words; what I am wondering is what is the actual "numbers" that is put into $\sigma_t^2$'s place in the equation when estimating?


1 Answer 1


In a GARCH model for a time series $x_t$ we have \begin{aligned} x_t&\sim i.i.D(\mu_t,\sigma_t^2), \\ \mu_t&=... \text{ (conditional mean of } x_t \text{ given past information)} \\ \sigma_t^2&=\omega +\sum _{i=1}^{q}\alpha _{i}\epsilon _{t-i}^{2}+\sum _{i=1}^{p}\beta _{i}\sigma _{t-i}^{2} \text{ (conditional variance of } x_t \text{ given past information)} \end{aligned} where $D$ is some distribution parameterized by the conditional mean $\mu_t$ and conditional variance $\sigma_t^2$, and $\epsilon_t:=x_t-\mu_t$ is an additive error term, a random variable itself. Meanwhile, $\sigma_t^2$ is the conditional variance of $x_t$ and simultaneously of $\epsilon_t$, hence, a parameter (an unknown constant). You are right that $\mathbb{E}(\epsilon_t^2)=\sigma_t^2$, but that does not make for $\epsilon_t^2=\sigma_t^2$.

When we are estimating the model, we treat $\sigma_t^2$ as an unknown parameter and estimate it along with the other parameters such as $\omega$, $\alpha$s and $\beta$s. We are not putting in any numbers in for $\sigma_t^2$ in estimation because conditional variances are unobservable and are never a variable in our dataset (just like the other model parameters).

  • $\begingroup$ +1 However, You seem to be using both $\epsilon$ and $\varepsilon$ to represent the same thing. $\endgroup$
    – Glen_b
    Nov 4, 2018 at 23:04
  • $\begingroup$ @Glen_b, that is a typo, thank you for spotting it. I will correct that. $\endgroup$ Nov 5, 2018 at 7:13
  • $\begingroup$ +1 of course. Interesting, thanks. Regarding the estimation, how is it possible to simultaneously estimate the beta's and the sigma's. $\endgroup$ Nov 5, 2018 at 9:49
  • $\begingroup$ @pApaAPPApapapa, that could well be a separate question. I am not that good in estimation specifics, but I think that $\sigma$s come out as by-products. Each of them can be expressed as in the last equation of my answer, so it suffices to specificy an initial $\sigma_0^2$ and the rest follow from the coefficients $\omega$, $\alpha$s and $\beta$s. Hence, estimation actually targets the latter coefficients. Some form of Kalman filter or EM algorithm can be used, but there are probably other alternatives, too. $\endgroup$ Nov 5, 2018 at 10:33
  • $\begingroup$ I guess that it in that sense is similar to the MA-term variable in an ARMA model. Thanks again. $\endgroup$ Nov 5, 2018 at 11:20

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