I am confused. I don't understand the difference a ARMA and a GARCH process.. to me there are the same no ?
Here is the (G)ARCH(p, q) process
$$\sigma_t^2 =
\underbrace{
\underbrace{
\alpha_0
+ \sum_{i=1}^q \alpha_ir_{t-i}^2}
_{ARCH}
+ \sum_{i=1}^p\beta_i\sigma_{t-i}^2}
_{GARCH}$$
And here is the ARMA($p, q$):
$$ X_t = c + \varepsilon_t + \sum_{i=1}^p \varphi_i X_{t-i} + \sum_{i=1}^q \theta_i \varepsilon_{t-i}.\,$$
Is the ARMA simply an extension of the GARCH, GARCH being used only for returns and with the assumption $r = \sigma\varepsilon$ where $\varepsilon$ follows a strong white process?
Best Answer
You are conflating the features of a process with its representation. Consider the (return) process $(Y_t)_{t=0}^\infty$.
$$ \begin{align} \mathbb{E}(Y_t \mid \mathcal{I}_t) &= \alpha_0 + \sum_{j=1}^p \alpha_j Y_{t-j}+ \sum_{k=1}^q \beta_k\epsilon_{t-k}\\ \end{align} $$ Here, $\mathcal{I}_t$ is the information set at time $t$, which is the $\sigma$-algebra generated by the lagged values of the outcome process $(Y_t)$.
Note in particular the first equivalence $ \mathbb{V}(Y_t \mid \mathcal{I}_t)= \mathbb{V}(\epsilon_t \mid \mathcal{I}_t)$.
Aside: Based on this representation, you can write $$ \epsilon_t \equiv \sigma_t Z_t $$ where $Z_t$ is a strong white noise process, but this follows from the way the process is defined.