# GARCH modelling and forecasting

I have a few questions regarding GARCH modelling and forecasting and it would be great if someone could help me.
I am modelling oil spot prices log-return using various GARCH models: GARCH, APARCH, EGARCH... and I am trying to forecast the prices.
I found using ACF and PACF plots that the best model for the series is ARMA(0,1) and then the best model for the error term follows GARCH(1,1) or APARCH(1,1) etc.

Here are my questions:

1)

garch1<-garchFit(~arma(0,1)+garch(1,1),data=brentlog,trace=FALSE,include.mean=TRUE)


I have a doubt whether I am forecasting the volatility of the prices or the actual values of return?

2) Since I am not looking at options, there is no point forecasting the volatility right? Because it won't tell me whether prices will go up or down.

3) Since I have an ARMA(0,1) for my model, my forecasts will always be constant and if I don't include a mean in the model then the forecasts are the same using EGARCH, GARCH, APARCH or any model: it is 0. So is there a point of using those different models in this case?

• anyone has an idea? – user3384794 Jun 2 '15 at 8:47
• Unfortunately, I was not able to answer the question earlier, but I hope the answer is still relevant. I gave a similar answer some time ago, so you might have found it if you searched the GARCH thread. – Richard Hardy Jun 16 '15 at 15:47

I have a doubt whether I am forecasting the volatility of the prices or the actual values of return?

The reference manual for the "fGarch" package tells on p. 30 that method predict will give forecasts for both the conditional mean and the conditional variance. There will be different columns "meanForecast", "meanError", and "standardDeviation" in the function's output. I suppose the first one will contain the forecasts for the conditional mean, which you seem to be interested in.

Since I am not looking at options, there is no point forecasting the volatility right? Because it won't tell me whether prices will go up or down.

You may or may not be interested in forecasting the conditional variance. However, as long as the conditional variance process can be well approximated by some GARCH model, you should account for that. Ignoring the GARCH patterns and (silently) assuming a constant conditional variance will yield inferior forecasts for the conditional mean, because the misspecification of the conditional variance equation will negatively affect the estimation of the conditional mean model.
Thus if (1) you want to have a good forecast for the conditional mean and (2) the conditional variance follows a GARCH process, you should keep the GARCH model.

Since I have an ARMA(0,1) for my model, my forecasts will always be constant and if I don't include a mean in the model then the forecasts are <...> 0.

Yes, they will be constant, but no, the $h$-step-ahead forecast (for $h \geqslant 1$) for the conditional mean is not zero. It rather is

$$\hat{x}_{t+h|t}=\hat{\theta}_1 \hat{\varepsilon}_t,$$

where $\hat{\theta}_1$ is the estimated MA(1) coefficient and $\hat{\varepsilon}_t$ is the estimated innovation at time $t$.
I have assumed away the potential presence of the mean component $\hat{\mu}$ for simplicity.

So is there a point of using those different models in this case?

Without a GARCH model your $h$-step-ahead forecast (for $h \geqslant 1$) will be

$$\hat{x}_{t+h|t}=\hat{\theta}_1 \hat{\varepsilon}_t$$

but with a GARCH model your $h$-step-ahead forecast will be a constant

$$\tilde{x}_{t+h|t}=\tilde{\theta}_1 \tilde{\varepsilon}_t.$$

Note that in general $\hat{\theta}_1 \neq \tilde{\theta}_1$ and $\hat{\varepsilon}_t \neq \tilde{\varepsilon}_t$. This is because the estimates of $\theta$ and $\varepsilon$ from the conditional mean model will not be the same under different specifications of the conditional variance model. Therefore, you will have different forecasts $\hat{x}_{t+h|t} \neq \tilde{x}_{t+h|t}$, and a correct specification of the conditional variance model matters.