1
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

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).

$\endgroup$
6
  • $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.