Why in an ARMA-GARCH model for a stationary series $r$ (without $c$ for simplicity) is $r_{forecast} = ARMA + \sqrt{GARCH} \cdot inn$ and not $r_{forecast} = ARMA \pm \sqrt{GARCH} \cdot inn$? The latter would seem to be even more intuitive. Does the GARCH process'root square not produce the double result $\pm$?


That is because the innovation $inn$ is a standard normally distributed random variable and therefore it can actually take positive or negative values. Without the innovation it would be a deterministic model (i.e. no risk). So, de facto it is +/- but mathematically you have to write $+\sqrt{GARCH} \cdot inn$ because $inn$ itself contributes the sign and can be either positive or negative.

  • $\begingroup$ The error distribution may also be other than normal, but in any case its support includes both positive and negative values, so you are right. $\endgroup$ – Richard Hardy Sep 1 '18 at 19:20
  • $\begingroup$ @Richard Hardy normally distributed was probably in the review that I accepted because it put some order in my formulas (written through mobile version). I have to stress that you may assume that innovations are normally distributed but that assumption is almost always wrong, unless you try to forecast very long-term and low-frequency returns. If you deal with high frequency returns (daily included, and often multiday), then you have a distribution which is absolutey divergent from a normal, usually a fat-tail distribution which is often skewed. I will try to re-edit at home. $\endgroup$ – Fr1 Sep 1 '18 at 23:28
  • $\begingroup$ If one wants to use a parametric innov distribution, then a t distribution would be less wrong than a normal but still wrong in many cases due to skewness,that is not explicitly modeled by the standard version of symmetric t. That is why Kernels are fitted and used. $\endgroup$ – Fr1 Sep 1 '18 at 23:30
  • $\begingroup$ Indeed it was in the review, but I would get rid of it (or at least state that this is just one option and not necessarily a good one). I agree with all the rest, too. $\endgroup$ – Richard Hardy Sep 2 '18 at 7:38

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