Confidence Intervals for ARMA+GARCH forecasts

Question

I have fitted an ARMA(1,1)+GARCH(1,1) model to my logreturns series. When it comes to my standarized error's distribution however, I have opted for a Skewed Generalized Error Distribution, because of the much better fit.

My model is therefore:

$$r_t \cdot (1 - \phi_1 \cdot B) = \epsilon_t \cdot (1 + \theta_1 \cdot B)$$

where ${B}$ is the lag-operator and:

$$\epsilon_t = \sigma_t \cdot e_t,$$

$$\sigma_t = \sqrt{\omega + \alpha_1 \cdot \epsilon^2_{t-1} + \beta_1 \cdot \sigma^2_{t-1}}$$

Finally:

$$e_t \stackrel{iid}{\sim} SKED (mean=0, variance =1, skew, shape) $$

At the moment of forecasting (I use R and the rugarch package), I have a point-estimate and a sigma estimate.

Instead of having "one-sigma estimates" I would like to have actual confidence intervals. To calculate this I would of coarse have to know the distribution of ${\epsilon_t}$.

My actual question is: how are these distributed? I suppose that they are also SGED but with other parameters. Could someone confirm this? Or maybe show some convenient function to extract this intervals.

Thanks in advance.

Answer 1

How are these distributed? $\epsilon_{t+1}\sim\text{SGED}(\mu_{t+1},\sigma_{t+1},\text{skew},\text{shape})$.

For a $(1-\alpha)$ level $1$-step-ahead forecast interval that is consistent with the model

• obtain the $\alpha/2$ and $1-\alpha/2$ quantiles of the distribution of the standardized innovation $e$ (regardless of the time index, since $e_t$s are i.i.d.),
• multiply them by $\sigma_{t+1}$ and
• add $\mu_{t+1}$ (due to the ARMA part of the model) to each.

The resulting two points will be the end points of the interval. This may not be the shortest $(1-\alpha)$ level forecast interval if the distribution of $e$ is asymmetric, but it will have the correct coverage.

See also "How to make $h$-step interval forecasts from an ARMA-GARCH model?".

Answer 2

If you just want a confidence interval for the sigma squared produced by the garch notice that the most popular approach is to assume that the distribution of the estimator is unknown and, as such, use non-parametric methods like bootstrapping. For more info read this answer which is also consistent with other sources like this . On bootstrapping meaning I also suggest this and this