# GARCH model prediction

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 $$y_{T+1}$$.

• 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. – Richard Hardy Sep 6 at 11:40

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 $$\sigma_T^2$$: $$\mathbb{E}[y_{T+h}^2\mid\mathcal{F}_T] = \omega\sum_{k=0}^{h-1}(\alpha+\beta)^k + (\alpha+\beta)^{k-1}(\alpha y_{T}^2 +\beta\sigma_T^2)$$
• 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. – Tommaso Ferrari Sep 6 at 10:58
• 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 $T+1$. – Richard Hardy Sep 6 at 11:39
• 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. – Richard Hardy Sep 6 at 12:38