# Confidence Intervals for ARMA+GARCH forecasts

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.

• May 10 at 14:57

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.