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I'm studying the Markov representation of a GARCH(p,q) process, i.e.

$$\boldsymbol{v_t} = \boldsymbol{u_t} + M_t\boldsymbol{v_{t-1}}$$

where \begin{equation} \boldsymbol{u_t} = \begin{pmatrix} \alpha_0\eta^2_t \\ 0 \\ \vdots \\ \alpha_0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \quad \text{and} \quad \boldsymbol{v_t} = \begin{pmatrix} \epsilon^2_t \\ \epsilon^2_{t-1} \\ \vdots \\ \epsilon^2_{t-(q-1)} \\ \sigma^2_t \\ \vdots \\ \sigma^2_{t-(p-1)} \end{pmatrix}. \end{equation}

and \begin{equation} M_t = \begin{pmatrix} \alpha_1\eta^2_t & \alpha_2\eta^2_t & \dots & & \alpha_q\eta^2_t & \beta_1\eta^2_t & \beta_2\eta^2_t & \dots & & \beta_p\eta^2_t \\ 1 & 0 & \dots & & 0 & 0 & 0 & \dots & & 0 \\ 0 & 1 & \dots & & 0 & 0 & 0 & \dots & & 0 \\ \vdots & \vdots & \ddots & & \vdots & \vdots & \vdots & \ddots & & \vdots \\ 0 & 0 & \dots & 1 & 0 & 0 & 0 & \dots & 0 & 0 \\ \alpha_1 & \alpha_2 & \dots & & \alpha_q & \beta_1 & \beta_2 & \dots & & \beta_p \\ 0 & 0 & \dots & & 0 & 1 & 0 & \dots & & 0 \\ 0 & 0 & \dots & & 0 & 0 & 1 & \dots & & 0 \\ \vdots & \vdots & \ddots & & \vdots & \vdots & \vdots & \ddots & & \vdots \\ 0 & 0 & \dots & 0 & 0 & 0 & 0 & \dots & 1 & 0 \end{pmatrix} \end{equation} and I do not really understand the objective of its representation. When I start multiplicating the vector $v_{t-1}$ and the matrix $M_t$, then I will get another vector obviously. This vector also contains the "classic" GARCH(p,q) equation but the first $q-1$ values equate $$\epsilon_{t-i}, i\in\{1, ..., q-1\}$$ with either themselves or, in the first row, $\epsilon_t$ with the coefficients and the squared random innovation $\eta^2_t$. And I just do not understand why I need that and not just the GARCH(p,q) equation. Why do I do that?

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  • $\begingroup$ Why did you include the var tag? $\endgroup$ – Richard Hardy Apr 4 '16 at 9:28
  • $\begingroup$ Because the first equation defines a first-order vector autoregressive model. $\endgroup$ – Taufi Apr 4 '16 at 10:31
  • $\begingroup$ There is not error term in that equation, so it does not look quite like a VAR model. $\endgroup$ – Richard Hardy Mar 15 '17 at 17:02
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Sorry for answering my own question but I now know what caused me trouble there. The first row is included to account for the equation $$ \epsilon_t = \sigma_t\eta_t $$ which is the other part of the GARCH(p,q) model specification. This is easily seen when factoring out $\eta^2_t$ and then taking the root. The first $q-2$ rows are just there to keep the vector $v_t$ well-defined when multiplying with $M_t$.

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