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?


1 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.

  • $\begingroup$ Thanks - so $ w_t = \epsilon^2_t - \sigma_t^2 $ here in your derivation is just notation? $\endgroup$
    – WorldGov
    Mar 5, 2021 at 14:34
  • 1
    $\begingroup$ Yes, this differs from book to book. $\endgroup$
    – Lars
    Mar 6, 2021 at 8:25
  • $\begingroup$ @Lars: I always wondered how that was the case so thanks. $\endgroup$
    – mlofton
    Apr 17 at 15:44
  • $\begingroup$ @Lars . I went over your answer carefully and it was clear. Thank you. But I had one question. You say that the $\epsilon_t$ are uncorrelated. I tried to prove this but was not successful. Then I thought about it and that doesn't make sense to me because $\epsilon_t$ is a function of $u_t^2$ ( and $\sigma^2_t$ which is not observed ) which is a function of $\epsilon_{t-1}$. So it seems to me that the $\epsilon_t$ have to be correlated atleast at lag one. Thanks for any clarification of that statement. $\endgroup$
    – mlofton
    Apr 18 at 18:55
  • $\begingroup$ To proof that $\epsilon_t$ is uncorrelated, note that ($h \geq 1$): $Cov(\epsilon_t,\epsilon_{t-h})=E(\epsilon_t\epsilon_{t-h})=E(E(\epsilon_t\epsilon_{t-h} \vert \mathcal F_{t-1}))=E(\epsilon_{t-h}E(\epsilon_t \vert \mathcal F_{t-1}))=0$ as $E(\epsilon_{t} \vert \mathcal F_{t-1})=E(\sigma_{t}u_{t} \vert \mathcal F_{t-1})=\sigma_{t}E(u_{t} \vert \mathcal F_{t-1})=0$. $\endgroup$
    – Lars
    Apr 19 at 10:42

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