# GARCH specification - why are $\sigma_t^2$ and $\epsilon_t^2$ not the same?

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?

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).
• +1 However, You seem to be using both $\epsilon$ and $\varepsilon$ to represent the same thing. Nov 4, 2018 at 23:04
• @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. Nov 5, 2018 at 10:33