Showing that a GARCH(1, 1) model is an ARMA(1, 1) process for squared errors

Question

Consider a GARCH(1, 1) model: $$ \sigma_t^2 = \alpha_0 + \alpha_1 \epsilon_{t-1}^2 + \beta \sigma_{t-1}^2 $$

Where $ \sigma_t $ is the conditional variance at time $ t $, $ \epsilon_{t}^2 $ is the error term in time $t$. In "In Introductory Econometrics for Finance" by Brooks (pg. 418/674), section 8.8, it is shown that this can be represented as an ARMA(1, 1) model for the squared errors. That is, the above can be written as:

$$ \epsilon_t^2 = \alpha_0 + ( \alpha_1 + \beta) \epsilon_{t-1}^2 - \beta w_{t-1} + w_t $$

To do so, the author starts with the statement:

$$ w_t = \epsilon_t^2 - \sigma_t^2 $$

What does this statement mean and how is it valid?

Answer 1

The derivation is as follows: You start with the conditional variance equation: \begin{equation} \sigma_t^2=\alpha_0+\alpha_1\epsilon_{t-1}^2+\beta_1\sigma_{t-1}^2 \end{equation} Now, add $w_t=\epsilon_t^2-\sigma_t^2$ on both sides. You obtain: \begin{align} \sigma_t^2+w_t&=\alpha_0+\alpha_1\epsilon_{t-1}^2+\beta_1\sigma_{t-1}^2+w_t \\ \sigma_t^2+\epsilon_t^2-\sigma_t^2&=\alpha_0+\alpha_1\epsilon_{t-1}^2+\beta_1\sigma_{t-1}^2+w_t\\ \epsilon_t^2&=\alpha_0+\alpha_1\epsilon_{t-1}^2+\beta_1\sigma_{t-1}^2+ w_t \\ \end{align} Notice that $w_{t-1}=\epsilon_{t-1}^2-\sigma_{t-1}^2$ and therefore $\sigma_{t-1}^2=\epsilon_{t-1}^2-w_{t-1}$. You obtain: \begin{align} \epsilon_t^2&=\alpha_0+\alpha_1\epsilon_{t-1}^2+\beta_1(\epsilon_{t-1}^2-w_{t-1})+ w_t \\ \epsilon_t^2&=\alpha_0+(\alpha_1+\beta_1)\epsilon_{t-1}^2 -\beta_1w_{t-1} +w_t \end{align} This is an ARMA(1,1) for the squared shocks if $w_t$ is a white noise process. Check that $E(w_t)=0$ and $Cov(w_t,w_{t+h})=0 , h\geq 1$. If $E(\epsilon_t^4)<\infty$ than $V(w_t)<\infty$ and $w_t$ is a weak white noise process. One thing that gets obvious when looking at this equation is that the $\epsilon_t$ are uncorrelated but not independent because the squared shocks follow an ARMA(1,1) process. Furthermore, it is now easy to derive the ACF of $\epsilon_t^2$. You can see that the ACF is always positive and converges at rate $\alpha_1+\beta_1$ to zero.