I have a simple question about the GARCH model.
We know that the $\alpha$ and $\beta$ parameters of the models are fitted for the local volatility of each time $t$ as follows:
$$\sigma_t^2= \alpha_0 + \sum_{i=1}^q \alpha_i \varepsilon_{t-i} + \sum_{i=1}^p \beta_i \sigma_{t-i}^2$$
with $\varepsilon_t=\sigma_t z_t$ and $z_t \sim N(0,1)$
However, the wikipedia article says that the process $y$ is behaving as follows:
$$y_t=a_0 + \sum_{i=1}^q a_i y_{t-i} + \varepsilon_t$$
I just wanted to make sure that here we assume that $y_t$ is the "original" time series values, and hence that $\varepsilon_t$ was the "return" of the time series at time $t$. Is that correct?
In a financial application, would $y_t$ be the price or the return at time $t$?
EDIT:
As the answers indicates that $y_t$ models the returns, I'm a bit surprised because usually you use the maximum log-likelihood:
$$\log L = -\frac{1}{2} \sum_{i=1}^n \left(\log (2 \pi) + \log (\sigma_{i}^2) + \frac{y_i^2}{\sigma_i^2} \right)$$
But this is only true if $y_i \sim N(0,\sigma_i^2)$
Now clearly with the setup presented above $\text{Var}(y_i) = \sigma_i^2$, but if $a \neq 0 ~ \forall i$, then $E[y_i] = a_0 + \sum_{i=1}^q a_i y_{t-i} \neq 0$
Is it because the log-likelihood is computed assuming $a=0 \forall i$?
Best Answer
This is pretty common notation:
$y_t$ is the return at $t$.
$\varepsilon_t$ is residual from modeling the returns as an $AR(q)$ process as shown in the equations.
Taken together, you have an $AR(q)-GARCH(p,q)$ model there (with slight abuse of notation as we have $q$ twice).