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If my one step ahead forecasts from GARCH(1,1)-X are: \begin{equation} \hat{h}_{t+1} = \hat{\alpha}_0 + \hat{\alpha}_1 \hat{u}^2_t + \hat{\beta}_1 \hat{h}_t + \hat{\psi} X_t \end{equation} Where $\hat{\alpha}_0,\, \hat{\alpha}_1,\,\hat{u}^2_t,\, \hat{\beta},\, \hat{h}_t$ and $ \hat{\psi}$ denote the GARCH(1,1)-X estimates of $\alpha_0, \alpha_1,u^2_t, \beta_1, h_t, \psi$ respectively.

Also, the one step ahead forecasts from a GJR-GARCH(1,1) forecasts are: \begin{equation} \hat{h}_{t+1} = \hat{\alpha}_0 + \hat{\alpha}_1 \hat{u}^2_t + \hat{\beta}_1 \hat{h}_t + \hat{\gamma} \hat{u}^2_{t}I_{u_{t}<0} +\hat{\psi} X_t \end{equation} Where $\hat{\alpha}_0,\, \hat{\alpha}_1,\,\hat{u}^2_t,\, \hat{\beta},\, \hat{h}_t,\, \hat{\gamma}$ and $ \hat{\psi}$ denote the GJR-GARCH(1,1)-X estimates of $\alpha_0, \alpha_1,u^2_t, \beta_1, h_t, \gamma$ and $ \psi$ respectively.

How can I write the h step ahead (h>1) for both equations the GARCH(1,1)-X and the GJR-GARCH(1,1)-X?

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    $\begingroup$ AFAIK, multistep ahead forecasts of GARCH-X type models depend on the assumed auxiliary model on the exogenous regressor. Do you have a model assumption on $X_t$? Best thing to do, is to derive the multistep ahead forecasts without the exogenous regressor in both equations, and then see (depending on your assumed model) how the addition of the regressor will affect the recursive forecast equation. $\endgroup$
    – Pleb
    Oct 20, 2022 at 18:49
  • $\begingroup$ I derived the multi-step ahead forecasts (h \textgreater{1}) for the conditional volatility of the GARCH(1,1) model can be written as: \begin{equation} \hat{h}_{t+h} = \sum_{i=0}^{h-1} \left( \hat{\alpha}_1 + \hat{\beta}_1 \right)^{i}\hat{\alpha}_0 + \left(\hat{\alpha_1} + \hat{\beta}_1 \right)^{h-1} + \left(\hat{\alpha}_1 \hat{u}^2_t + \hat{\beta}_1 \hat{h}_t \right) \end{equation} But I am unsure if this is correct, can you please let me know what you think? For multi-step ahead forecasts (h \textgreater{1}) for the GJR-GARCH(1,1), do you have a suggestion on how it should be? $\endgroup$
    – Moataz
    Oct 20, 2022 at 22:01
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    $\begingroup$ that is almost correct. You need to multiply the last two terms in the sum, instead of adding them, ie. $\left(\alpha_1+\beta_1\right)^{h-1} \cdot (\alpha_1 u_t^2 + \beta_1 h_t)$. Then it's correct. For the GJR-GARCH model, if we assume that the distribution of $u_t$ is symmetric about zero, then you can recover a similar recursive formula: $$ h_{t+h} = \sum_{i=0}^{h-1} \alpha_0 \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^{i} + \left( \alpha_1 + \frac{\gamma}{2} + \beta_1\right)^{h-1} \cdot \left(\alpha_1 u_t^2 + \gamma I_{\{u_t < 0\}} u_t^2 + \beta_1 h_t \right).$$ $\endgroup$
    – Pleb
    Oct 20, 2022 at 22:45
  • $\begingroup$ Thank you for your help!! $\endgroup$
    – Moataz
    Oct 20, 2022 at 23:25
  • $\begingroup$ Hi Pleb, if you do not mind can you share with me the derivations for GJR-GARCH h step ahead conditional volatility? $\endgroup$
    – Moataz
    Oct 21, 2022 at 7:37

1 Answer 1

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$k$-step ahead forecasts of the GJR-GARCH model:

Let us briefly define the demeaned return-process following your notation: \begin{align*} r_{t+1} &= u_{t+1}\\ u_{t+1} &= \sqrt{h_{t+1}} z_{t+1}, \end{align*} where $z_{t+1} \overset{iid}{\sim} D(0,1)$ is a standardized distribution and $h_{t+1}$ follows the GJR-GARCH model:

$$ h_{t+1} = \alpha_0 + \alpha_1u_t^2 + \beta_1 h_t + \gamma u_t^2 I_{\{u_t<0\}}. $$

The 1-step ahead forecasts for GARCH models are known at time $t$ per construction, and as such, we will focus our attention to the 2- and 3-step ahead forecasts for the GJR-GARCH model.


We attain the 2-step ahead forecast following the same argumentation as provided in [1]:

\begin{align} \mathbb{E}_t\left[h_{t+2}\right] &= \alpha_0 + \alpha_1 \mathbb{E}_t\left[u^2_{t+1}\right] + \gamma \mathbb{E}_t\left[u^2_{t+1} I_{\{u_{t+1}<0\}}\right] + \beta_1 \mathbb{E}_t\left[h_{t+1}\right]\\ &=\alpha_0 + \alpha_1 \mathbb{E}_t\left[h_{t+1}\right] + \gamma \mathbb{E}_t\left[h_{t+1}\right] \mathbb{E}_t\left[I_{\{u_{t+1}<0\}}\right] + \beta_1 \mathbb{E}_t\left[h_{t+1}\right]\\ &\overset{\star}{=} \alpha_0 + \alpha_1 \mathbb{E}_t\left[h_{t+1}\right] + \frac{\gamma}{2} \mathbb{E}_t\left[h_{t+1}\right] + \beta_1 \mathbb{E}_t\left[h_{t+1}\right]\\ &= \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)\mathbb{E}_t\left[h_{t+1}\right]\\ &= \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right) \left(\alpha_0 + \alpha_1u_t^2 + \beta_1 h_t + \gamma u_t^2 I_{\{u_t<0\}}\right)\\ &= \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right) \alpha_0 +\left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right) \left(\alpha_1u_t^2 + \beta_1 h_t + \gamma u_t^2 I_{\{u_t<0\}}\right) \end{align}

where we in $(\star)$ have assumed that the distribution of $u_t$ is symmetric about 0, such that $\mathbb{E}_t\left[u^2_{t+1}\right] \mathbb{E}_t\left[I_{\{u_{t+1}<0\}}\right] = \frac{1}{2}\mathbb{E}_t\left[h_{t+1}\right]$. $^{[1]}$


You can derive the 3-step ahead forecast in a similar fashion and get the following:

$\displaystyle{ \begin{align*} \mathbb{E}_t\left[h_{t+3}\right] &= \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right) \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^2 \mathbb{E}_t\left[h_{t+1}\right]\\ &=\alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right) \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^2\left(\alpha_0 + \alpha_1u_t^2 + \beta_1 h_t + \gamma u_t^2 I_{\{u_t<0\}}\right)\\ &=\alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right) \alpha_0 + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^2 \alpha_0 \\ &+ \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^2 \left( \alpha_1u_t^2 + \beta_1 h_t + \gamma u_t^2 I_{\{u_t<0\}}\right) \end{align*} }$


From these two examples we can observe the recursive characteristics for the multistep ahead forecast equation. Thus the $k$-step ahead forecast for $k \geq 2$ is given by:

$$ \mathbb{E}_t\left[h_{t+k}\right] = \sum_{i=0}^{k-1} \alpha_0 \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^i + \left(\alpha_1 + \frac{\gamma}{2} + \beta_1\right)^{k-1} \left(\alpha_1u_t^2 + \beta_1 h_t + \gamma u_t^2 I_{\{u_t<0\}}\right) $$


[1]: The equality and corresponding derivations are found on pp. 28 - 29 in Zivot, E. (2009). Practical issues in the analysis of univariate GARCH models. if you want to cite a source.

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