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I can see the R package rugarch allows the estimation of GARCH models with exogenous variables in the specification of the variance model:

\begin{aligned} \epsilon_t &= \sqrt{h_t}\eta_t, \\ h_t &= \alpha_0 + \alpha_1 \epsilon_{t-1}^2 + \alpha_2 h_{t-1} + \alpha_3x_{t-1}, \\ \eta_t &\sim N(0,1) \ \textrm{iid}\ \forall \ t. \\ \end{aligned}

I'm wondering what is the log-likelihood function associated with this model. Is it simply the same as with a normal GARCH:

$$\ln f\left(\epsilon_{t}|h_{t}^{2}\right)=-\frac{1}{2}\left(\ln2\pi+\ln h_{t}^{2}+\frac{\left(\epsilon_t\right)^{2}}{h_{t}^{2}}\right),$$

?

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    $\begingroup$ In the model specification, $\eta_t$ are also i.i.d. It is an important building block in the construction of the likelihood. $\endgroup$ Feb 2, 2023 at 8:51

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We know that $$\eta_{t}\sim N(0,1)$$

So it follows that $$\epsilon_t=\sqrt{h_{t}}\eta_{t}\quad\Longleftrightarrow\quad\epsilon_t\,|\,\epsilon_{t-1}\sim N(0,h_{t})$$

Let $\theta=(\alpha_{0},\alpha_{1},\ldots)$ be your parameter set. We can construct the joint distribution for $\epsilon$ as $$\begin{align} L(\theta)=f(\epsilon_{T},\ldots,\epsilon_{0}\,|\,\theta)&=f(\epsilon_{0};\theta)\cdot \prod_{t=1}^{T}f(\epsilon_{t}\,|\,\epsilon_{t-1};\theta)\\ &=f(\epsilon_{0};\theta)\cdot \prod_{t=1}^{T}\frac{1}{\sqrt{2\pi\cdot h_{t}(\theta)}}\exp\bigg(-\frac{\epsilon_{t}^2}{2\cdot h_{t}(\theta)}\bigg)\\ \end{align}$$

$$\begin{align} \Rightarrow\mathcal{L}(\theta)&=\log L(\theta)\\ &=\log f(\epsilon_{0};\theta)-\frac{T}{2}\log(2\pi)-\frac{1}{2}\sum_{t=1}^{T}\bigg[\log h_{t}(\theta)+\frac{\epsilon_{t}^{2}}{h_{t}(\theta)}\bigg] \end{align}$$

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  • $\begingroup$ Should there be $\Pi_{t=1}^T$ instead of $\sum_{t=1}^T$ in $L(\theta)$? It might also make sense to use $h_t(\theta)$ in place of $h_t$ as otherwise it may look that $\theta$ only enters in the first element of $\mathcal{L}(\theta)$. $\endgroup$ Feb 2, 2023 at 7:29
  • $\begingroup$ And then if I actually wanted to maximize $\mathcal{L}(\theta)$, should I have to express $h_t$ recursively so that $\mathcal{L}(\theta)$ becomes a function of $\epsilon$s and $\theta$ but not $h$s? (I am thinking about a practical implementation.) $\endgroup$ Feb 2, 2023 at 7:30
  • $\begingroup$ Hey, thanks. What is $\log f(\epsilon_0,\theta)$ in the example above? Thanks $\endgroup$
    – user603
    Feb 2, 2023 at 8:24
  • $\begingroup$ Any thoughts about my second comment? $\endgroup$ Feb 6, 2023 at 13:27
  • $\begingroup$ @RichardHardy I haven't given it much thought or tried to implement it myself, but that seems reasonable to me? $\endgroup$ Feb 7, 2023 at 6:19

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