Multistep ahead forecasts in GARCH equations

Question

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?

Answer 1

$k$-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.

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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 $\mathbb{E}_t\left[u^2_{t+1}\right] \mathbb{E}_t\left[I_{\{u_{t+1}<0\}}\right] = \frac{1}{2}\mathbb{E}_t\left[h_{t+1}\right]$. $^{[1]}$

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

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

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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:

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

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[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.