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I was analyzing a GARCH(1,1) process. In particular, let's say that I have a process ${y_t}$, with $t \in {1,2,...,T}$. I have created a GARCH process that can be written as:

$\sigma_t^2 = \omega + \alpha y_{t-1}^2 + \beta \sigma_{t-1}^2$,

with $t \in{1,2,...,T}$. After that, I maximize the Log Likelihood of the model, obtaining the three parameters, namely $\hat{\alpha}, \hat{\beta}, \hat{\omega}$.

Now I would like to use these estimated parameters for forecasting volatility. In particular, I can get

$\sigma_{T+1}^2 = \hat{\omega} + \hat{\alpha} y_{T}^2 + \hat{\beta} \sigma_{T}^2$

that is, I can get the estimated volatility at $(T+1)$. It is not clear in my mind how can I get, for example, $\sigma_{T+2}^2$. According to the formulation above, I shall use $y_{T+1}^2$, but my series stops at $T$. How can I get the forecasted value $y_{T+1}$?

I have found in literature that for a GARCH(1,1) and $k>2$

$\mathbb{E}_t[\sigma_{t+k}^2] = \sum_{i=0}^{k-2} (\hat{\alpha}+\hat{\beta})^i\hat{\omega} + (\hat{\alpha}+\hat{\beta})^{k-1}\sigma_{T+1}^2$

Thus I can use the forecasted value for $\sigma_{t+k}^2$ and, inverting the formulation of GARCH(1,1), get the forecasted value of $y_{T+1}$.

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  • $\begingroup$ I understand that you want a point, interval or density forecast of $y_{T+1}$. You already have the forecasted variance. If you add a distributional assumption for standardized innovations and a model for the conditional mean, you will be able to obtain a density forecast for $y_{T+1}$ from which you can derive any interval or point forecast you like. $\endgroup$ – Richard Hardy Sep 6 at 11:40
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Assuming you have a model of the type $y_t = \sigma_t\varepsilon_t$, with $\sigma_t^2 = \omega + \alpha y_{t-1}^2 + \beta\sigma_{t-1}^2$, that would make your $y_t$ have the martingale-difference property with respect to its history before time $T$. In other words, optimal predictions of $y_{T+h}$, for $h\geq1$, are zero. This is because $$\mathbb{E}[y_t\mid\mathcal{F}_{T-1}] = \mathbb{E}[\sigma_t\varepsilon_t\mid\mathcal{F}_{T-1}] = \sigma_t\mathbb{E}[\varepsilon_t\mid\mathcal{F}_{T-1}] = \sigma_t\mathbb{E}[\varepsilon_t] = 0$$

Now, based on your history $t\in\{1, 2,\ldots,T\}$, an approximate forecast of $y_{T+1}$ also functions as an estimate of the squared volatility at time $T+1$, given by $$ \hat{\sigma}_{T+1}^2 = \hat{\mathbb{E}}[y_{T+1}^2\mid\mathcal{F}_{T}] = \omega + \alpha y_{T}^2 + \beta\hat{\sigma}_{T}^2 $$ If you look at this as a recursive scheme for one-step ahead volatility forecasting, then you can also look at $h>1$ steps ahead with the information available at $T$. Both $y_{T+h}$ and $\sigma_{T+h}$ are random variables, and their predictions coincide: $$ \mathbb{E}[y_{T+h}^2\mid\mathcal{F}_T] = \mathbb{E}[\sigma_{T+h}^2\mid\mathcal{F}_T] = \omega + \alpha\mathbb{E}[y_{T+h-1}^2\mid\mathcal{F}_T] + \beta\mathbb{E}[\sigma_{T+h-1}^2\mid\mathcal{F}_T]$$ $$= \omega + (\alpha+\beta)\mathbb{E}[y_{T+h-1}^2\mid\mathcal{F}_T]$$ Working backwards, this leads to a general formula which only needs $y_T^2$ and $\sigma_T^2$: $$ \mathbb{E}[y_{T+h}^2\mid\mathcal{F}_T] = \omega\sum_{k=0}^{h-1}(\alpha+\beta)^k + (\alpha+\beta)^{k-1}(\alpha y_{T}^2 +\beta\sigma_T^2) $$


In my answer I have quoted heavily -- almost verbatim, in fact -- from the book Quantitative Risk Management, by McNeil, Frey, and Embrechts. Chapter 4.4.1 in particular offers a lot of good explanations that might help you better understand some of the mechanics.

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  • $\begingroup$ Dear Emil, thanks for your precise answer! After my analysis, I arrive at the conclusion that I don't need the estimation of $y_{T+h}^2$, but only the ones of $\sigma$. In particular,I really appreciate your answer and help. $\endgroup$ – Tommaso Ferrari Sep 6 at 10:58
  • $\begingroup$ Both $y_{T+h}$ and $\sigma_{T+h}$ are random variables. Hmm, in classical statistics parameters (such as $\sigma_{T+h}$) are not random variables. Moreover, predictions of $y_{T+h}$ and $\sigma_{T+h}$ do not coincide, though those of $y_{T+h}^2$ and $\sigma_{T+h}$ may. The same applies also to an approximate forecast of $y_{T+1}$ also functions as an estimate of the squared volatility at time $T+1$. $\endgroup$ – Richard Hardy Sep 6 at 11:39
  • $\begingroup$ On the other hand, I see why $\sigma_{T+h}$ can be considered a random variable given only the information set $\mathcal{F}_{T+h-2}$ or its subset. $\endgroup$ – Richard Hardy Sep 6 at 12:38
  • $\begingroup$ Please, have a look at my new post regarding the way in which I can get the Garch model volatilities assuming t-student distribution! $\endgroup$ – Tommaso Ferrari Sep 11 at 19:31

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