Writing ARCH process as AR process

Question

How can I show that the ARCH(1) process for $y$:

$$y_t=\varepsilon_t=z_t\sigma_t$$ $$\sigma^2_t=\omega+\alpha y^2_{t-1}$$

is actually AR(1) process for $y$:

$$y^2_t=\omega+\alpha y^2_{t-1}+v_t$$

where, $v_t=\sigma^2_t(z^2_t-1)$

I know it is pretty straightforward, but I don't really understand what is this $v_t$ term we are introducing and how can we interpret it in this context?

My calculations:

$y^2_t=z^2_t \sigma^2_t$

$y^2_t=z^2_t(\omega+\alpha y^2_{t-1})$

$z^2_t\sigma^2_t=z^2_t(\omega+\alpha y^2_{t-1})$

$v_t=y^2_t-E(y^2_t|I_{t-1})=y^2_t-\sigma^2_t$

$y^2_t-v_t=\omega+\alpha y^2_{t-1}$

$y^2_t=\omega+\alpha y^2_{t-1}+y^2_t-\sigma^2_t$

$y^2_t=\omega+\alpha y^2_{t-1}+z^2_t\sigma^2_t-\sigma^2_t$

$y^2_t=\omega+\alpha y^2_{t-1}+\sigma^2_t(z^2_t-1)$

Answer 1

In an AR(1) model we have the error $\varepsilon_t$ satisfying $$ \varepsilon_t = y_t - \mu_t, $$ where $\mu_t = \varphi y_{t-1}$ is the expected value of $y_t$ given the information available at time $t-1$.

Similarly, in an ARCH(1) model we have the error $v_t$ satisfying $$ v_t = y_t^2 - \sigma_t^2, $$ where $\sigma_t^2 = \mathbb{E}(y_t^2-\mathbb{E}(y_t)^2) = \mathbb{E}(y_t^2-0) = \mathbb{E}(y_t^2)$ is the expected value of $y_t^2$ given the information available at time $t-1$. So that is the analogy between $\varepsilon_t$ and $v_t$ in AR(1) and ARCH(1) models, respectively.

but I don't really understand what is this $v_t$ term we are introducing and how can we interpret it in this context?

$v_t$ is the deviation of $y_t^2$ from its conditional mean given the information available at time $t-1$ -- similarly as to how $\varepsilon_t$ is the deviation of $y_t$ from its conditional mean.