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Suppose that the stationary series $r_t$ is well fitted by an $ARMA(p,q)+c$ and $GARCH(r,s)$ model, where $GARCH(r,s) = \sigma_t ^2$

If in the testing sample I have to graphically compare the estimated $GARCH (r,s)$ with the actual conditional variance series in order to better visualize the goodness of fit, is it more useful (and maybe correct) directly comparing the $GARCH(r,s)$ series with the $r^2$ series (as actual conditional variance approximation), or the $\sigma_t = \sqrt{\sigma_t^2} = \sqrt{GARCH(r,s)}$ series with the absolute values of $r_t$ ? Or is it equal?

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If the series has a significant ARMA structure on the conditional mean, if you want to evaluate the only GARCH specification assuming ARMA is fine, then you have to display GARCH estimates against the $(r - ARMA_{forecast})^2$ because via GARCH you are trying to estimate the conditional variance of return around a conditional mean represented by ARMA. So you can’t plot the square of $r$ series against the only GARCH structure because you are missing the conditional mean represented by ARMA. Remember that daily returns $r$ are generally representable as $r= \mu_r+ \sigma_r \cdot innov$ where $innov$ is assumed to be a standard normally distributed $N\left(0;1\right)$. ARMA is used for the conditional mean structure and GARCH for the conditional standard deviation. So your forecast is $r_{forecast}=ARMA+\sqrt{Garch}\cdot innov$, not $r_{forecast}=\sqrt{Garch} \cdot innov$ as you assume if you plot $r^2$ against GARCH.

However, in my answer here I am assuming that you are taking it for sure that your ARMA specification is correct: If you wish to test all your GARCH and ARMA specs at the same time (i.e. your full model), then you plot your $r_{forecasted}$ against $r$. Intuitively, we could say that ARMA will try to predict the sign and GARCH will try to predict the magnitude of the return.

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  • $\begingroup$ Thank you. And is it eventually correct to plot the square root of estimated GARCH against the absolute values of r_t in order to only evaluate the goodness of volatility fit? Are there differences? $\endgroup$ – LeoAn Aug 30 '18 at 14:29
  • $\begingroup$ I would say yes, it’s correct.. you are plotting the conditional forecasted st dev against the actual realized st dev.. so yes. You can either reason on variance or st dev as you wish. $\endgroup$ – Fr1 Aug 30 '18 at 14:40
  • $\begingroup$ (r- ARMA-forecast), as you wrote, is the error term series. Is your answer perhaps justified by the fact that in the ARMA-GARCH structure Var(r_t) = Var(error_t)? Thank you $\endgroup$ – LeoAn Aug 30 '18 at 14:48
  • $\begingroup$ So if the conditional distribution for returns is expressed by r=arma+root(Garch)*innov where innov is iid and has 0 mean and 1 variance, then the conditional variance is Var(r)=Var(arma+root(Garch)*innov) where Arma is a conditionally deterministic component (i.e. you know Arma when you are trying to predict) Garch does not contain innovations so var(Garch*innov) is just (Garch^2)*Var(innov) and Var(innov) is 1 by assumotion $\endgroup$ – Fr1 Aug 30 '18 at 14:53
  • $\begingroup$ Therefore is the definitive answer yes? $\endgroup$ – LeoAn Aug 30 '18 at 15:19

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