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I try to model currency rates volatility using GARCH models through the RUGARCH package in R.

Starting from the observed currency rate series, I compute the log-return through:

data <- diff(log(series)) # Log-return

Then (after some statistical analysis) I decide to use a GARCH(1,1) model with a skew-student distribution, hence I use

spec_final <- ugarchspec(mean.model=list(armaOrder=c(0,0),include.mean=FALSE),variance.model=list(model="sGARCH",garchOrder=c(1,1)),distribution.model="sstd")
fit_final <- ugarchfit(spec_final,data=data)

I then try to simulate future outcomes of this series with an horizon of 260 days with the code

horizon <- 260
exp(diffinv(ugarchsim(fit_final,n.sim=horizon)@simulation$seriesSim))[horizon+1]

If I perform this a great number of times (200,000) I can compute quantiles. More especially I see that the quantile at 0.5% is equal to 0.605 and the quantile at 99.5% is equal to 1.623. The distribution has a mean very close to 1 but is not symmetric.

I would like to understand why there is a lack of symmetry in the simulated distribution, even if the GARCH model is known to be symmetric. It does not happen only for one currency but for all those I tried to model. This is really an issue to me as I do not have any particular reason to explain why the model predicts larger upward shocks than downward shocks.

Thanks.

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If log returns have a symmetric distribution, prices will have a positively skewed distribution, since exponentiating induces positive skew.

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  • $\begingroup$ Thanks. But I do not understand why in most of the articles I have seen, people use this log-return transformation on their time series without any evidence of skewness. It's like an automatic trick. Anyway, in my case, this skewness is not really what I want. What alternative do I have to this transformation ? Thanks. $\endgroup$ – Ludo Jan 13 '15 at 7:33
  • $\begingroup$ @Ludo Check out this: symmys.com/node/85 $\endgroup$ – John Jan 13 '15 at 14:36
  • $\begingroup$ @Ludo Modelling the proportional changes x(t+1)/x(t)-1 as normal instead of the log changes log(x(t+1)/x(t)) as normal allows x(t+1) to be negative, which is often (for example with stock prices) not allowed. $\endgroup$ – Fortranner Jan 13 '15 at 17:32

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