I have been looking at the GARCH chapter in Brooks book and he shows how to transform a GARCH(1,1) into a an ARMA(1,1).Basically starting from:

$$h_t=w+\beta h_{t-1}+\alpha \varepsilon^2_{t-1} $$ he suggests defining $v_t =\varepsilon^2_t - h_t$ $\rightarrow$ $h_t=\varepsilon^2_t-v_t$ and replacing in the above equation:

$$ \varepsilon^2_t-v_t=w+\beta \varepsilon^2_{t-1}-\beta v_{t-1}+\alpha \varepsilon^2_{t-1} $$ and get $$\varepsilon^2_t=w+(\alpha+\beta)\varepsilon^2_{t-1}+\beta v_{t-1}+v_t$$ I am really curious where does the term $v_t$ comes from? is it just the diffence between the actual value and the fitted line $h_t$?

Thank you!


1 Answer 1


$v_t$ is the difference between $\varepsilon_t^2$ and its conditional mean $h_t$. This is a necessary element in an ARMA model.

Recall that an ARMA model contains autoregressive terms, an error and lags of the error. The error term in an ARMA model is the difference between the actual value of the series and its conditional mean. E.g. in ARMA(1,1), \begin{aligned} x_t &= \varphi_1 x_{t-1} + \theta_1 \varepsilon_{t-1} + \varepsilon_t, \\ \mathbb{E}(x_t|I_{t-1}) &= \varphi_1 x_{t-1} + \theta_1 \varepsilon_{t-1}, \\ \varepsilon_t &= x_t - \mathbb{E}(x_t|I_{t-1}), \\ \end{aligned} where $I_{t-1}$ is the information available at time $t-1$.

Therefore, your last equation specifies a valid ARMA model for $\varepsilon_t^2$: it has an autoregressive term $\varepsilon_{t-1}^2$, an error term $v_t$ (which is the difference between the dependent variable $\varepsilon_t^2$ and its conditional mean) and its lag $v_{t-1}$.


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