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I'm a newbie in GARCH models. I tried to realize ARMA(p, q)-GARCH(u, v) model via fGarch.

So, 2 main questions.

1) Can I use BIC/AIC for selection best model for all (p, q)-(u, v) models? So, is it correct to compare BICs from, for example, ARMA(2, 1)-GARCH(1, 2) and ARMA(1, 3)-GARCH(2, 1)? If not, what should I do in this case (best way)? [Added] And is it correct to compare BICs/AICs for different ARMA-GARCH model, based on different distributions (sged vs t-studentm for example)?

2) I want to use distributions which are not normal. They may be skewed and so on. I know formula for $r_{t}$, that $r_{t} = z_{t}\sigma_{t}$, where $z_{t}$ iid, zero mean and unit variation. For example, I have same model, fitted values for $r_{t}$ and $\sigma_{t}$ by the same ARMA-GARCH model I choose as best. And I choose skewed normal distribution when I fitted model. So, how should I calculate quantiles? Is it correct to get quantiles of $z_{t}$ as a skewed normal with 0 mean and unit variation and calculate conf.levels as $r_{t}^{fitted} +- q_{1,2}*\sigma_{t}^{fitted}$, where $q_{1,2}$ are quantiles, which I get by skewed normal distribution (sure, zero mean and unit variance)?

Thank you.

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2 Answers 2

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1) The answer is yes: you can use the AIC/BIC to select the best model.

2) You can have confidence intervals by: $r_t^{fitted} \pm 2* \sigma_t^{fitted}$ so that you have a confidence interval of $\pm 2$ times the conditional standard deviation.

You can see a plot of this by: plot(your_garchFit_object) and typing 3 to select the plot of the series with the confidence bounds.

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  • $\begingroup$ so, what should I do if I need 0.95 or 0.99 quantiles? And if distribution skewed? I can't take only "2" for confidence intervals... I don't know why it was chosen in standard fGarch plot function for fitted model. $\endgroup$
    – Dmitriy
    Commented Mar 22, 2016 at 11:44
  • $\begingroup$ To have the quantiles of any distribution (e.g. the skew-normal) you have to use the specific inverse Cumulative Distribution Function for the distribution you need, calculated on the probability you need (i.e. 0.95 or 0.99) $\endgroup$
    – simmy
    Commented Mar 22, 2016 at 11:55
  • $\begingroup$ So, my assertions about $z_{t}$ (which has a distribution I choose when I calibrate the model) and calculation $q_{1,2}$ as the same quantiles for $z_{t}$ are correct, aren't they? $\endgroup$
    – Dmitriy
    Commented Mar 22, 2016 at 12:00
  • $\begingroup$ I am very sorry: I have misunderstood your question. Your assertions are right. Again, sorry. $\endgroup$
    – simmy
    Commented Mar 22, 2016 at 12:16
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  1. Information Criteria estimate the quality of a model based on the likelihood / the numbers of parameters (or degree of freedom) and the number of observations. It is a measure of goodness-of-fit and so it may suffers of overfitting problems. You can use it to compare any models (even with different errors distributions) however you may risk to select the model which fit the best to your past data (to a specific scenario). An additional way is is to select the "best" model based on its predictive power by using out-of-sample forecast.

  2. The confidence interval is given by : conditional mean +/- $\sigma_{t} \times q $ where $\sigma$ is the conditional variance at time $t$ and $q$ is the quantile obtained via the inverse CDF based on the parameters you obtained during the estimation and the confidence interval. So if you use a Skew Normal Dist, the Inverse CDF must also be based on the shape parameter ($\alpha$) you obtained. At the end the $z_{t}$ have a zero mean and unit variance distribution but you need to use the specific estimated shape parameter to obtain the quantile. When we use a density different that the normal one, it is always re-parameterize in such a way that the innovation process has zero mean and unit variance. (For instance if you use the student t distribution the quantile is obtained via the $t(0,1,v)$ student density with $v$ the degree of freedom)

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  • $\begingroup$ Damn, I have too few reputation, can't vote for answers.(( $\endgroup$
    – Dmitriy
    Commented Mar 23, 2016 at 10:58

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