[R-SIG-Finance] rugarch VaR calculation "manually"

[Neuman Co]

thanks a lot for your help, but
"Use the 'sigma' and 'fitted' methods"

But these are the fitted values for the volatility and the final
fitted values. But to calculate the VaR I need the 1 step ahead
forecast of the cond. sigmas and the 1 step ahead forecast of the
cond. mean. The cond.mean is not equivalent to the fitted values? And
the fitted values are not one step ahead forecasts?

2013/5/7 alexios ghalanos <alexios at 4dscape.com>:
> Hello,
>
>
> On 07/05/2013 12:15, Neuman Co wrote:
>>
>> I am using the rugarch package in R and I have some questions:
>>
>> I want to use the rugarch package to calculate the VaR.
>>
>> I used the following code to to fit a certain model:
>>
>> spec2<-ugarchspec(variance.model = list(model = "sGARCH", garchOrder =
>> c(1, 1)),
>> mean.model = list(armaOrder = c(5, 5), include.mean = FALSE),
>> distribution.model =
>> "norm",fixed.pars=list(ar1=0,ar2=0,ar3=0,ma1=0,ma2=0,ma3=0))
>>
>> model2<-ugarchfit(spec=spec2,data=mydata)
>>
>> Now I can look at the 2.5 % VaR with the following command:
>> plot(model)
>>
>> and choosing the second plot.
>>
>> Now my first question is: How can I get the 1.0% VaR VALUES, so not
>> the plot, but
>> the values/numbers?
>
> Use the 'quantile' method i.e. 'quantile(model2, 0.01)'...also applies to
> uGARCHforecast, uGARCHsim etc It IS documented.
>
>>
>> In case of the normal distribution, one can easily do the calculation of
>> the VaR with using the forecasted conditional volatility and the
>> forecasted
>> conditional mean:
>>
>> I use the ugarchforecast command and with that I can get the cond.
>> volatility
>> and cond. mean (my mean equation is an modified ARMA(5,5), see above in
>> the spec
>> command):
>>
>> forecast = ugarchforecast(spec, data, n.roll = , n.ahead = , out.sample=)
>>
>> # conditional mean
>> cmu = as.numeric(as.data.frame(forecast, which = "series",
>> rollframe="all", aligned = FALSE))
>> # conditional sigma
>> csigma = as.numeric(as.data.frame(forecast, which = "sigma",
>> rollframe="all", aligned = FALSE))
>>
> NO. 'as.data.frame' has long been deprecated. Use the 'sigma' and 'fitted'
> methods (and make sure you upgrade to latest version of rugarch).
>
>
>> I can calculate the VaR by using the property, that the normal
>> distribution
>> is part of the location-scale distribution families
>>
>> # use location+scaling transformation property of normal distribution:
>> VaR = qnorm(0.01)*csigma + cmu
>>
>> My second question belongs to the n.roll and out.sample command. I had
>> a look at the
>> description
>> http://www.inside-r.org/packages/cran/rugarch/docs/ugarchforecast
>> but I did not understand the n.roll and out.sample command. I want to
>> calculate the
>> daily VaR, so I need one step ahead predicitons and I do not want to
>> reestimate
>> the model every time step. So what does it mean "to roll the forecast
>> 1 step " and
>> what is out.sample?
>
> out.sample, which is used in the estimation stage retains 'out.sample' data
> (i.e. they are not used in the estimation) so that a rolling forecast can
> then be performed using this data. 'Rolling' means using data at time T-p
> (p=lag) to create a conditional 1-ahead forecast at time T. For the
> ugarchforecast method, this means using only estimates of the parameters
> from length(data)-out.sample. There is no re-estimation taking place (this
> is only done in the ugarchroll method).
> For n.ahead>1, this becomes an unconditional forecast. Equivalently, you can
> append new data to your old dataset and use the ugarchfilter method.
>
>>
>> My third question(s) is (are): How to calculate the VaR in case of a
>> standardized
>> hyperbolic distribution? Can I still calculate it like in the normal case
>> or does it not work anymore (I am not sure, if the sdhyp belongs to the
>> location-scale family).
>>
>> Even if it does work (so if the sdhyp belongs to the family of
>> location-scale distributions) how do I calculate it in case of a
>> distribution, which does not belong to the location-scale family?
>> (I mean, if I cannot calculate it via VaR = uncmean + sigma* z_alpha how
>> do I have to calculate it). Which distribution implemented in the rugarch
>> package does not belong to the location-scale family?
>>
> ALL distributions in the rugarch package are represented in a location- and
> scale- invariant parameterization since this is a key property required in
> working with the standardized residuals in the density function (i.e. the
> subtraction of the mean and scaling by the volatility).
> The standardized Generalized Hyperbolic distribution does indeed have this
> property, and details are available in the vignette. See also paper by
> Blaesild (http://biomet.oxfordjournals.org/content/68/1/251.short) for the
> linear transformation (aX+b) property.
>
> If working with the standard (NOT standardized) version of the GH
> distribution (\lambda, \alpha, \beta, \delta, \mu) you need to apply the
> location/scaling transformation as the example below shows which is
> equivalent to just using the location/scaling transformation in the
> standardized version:
> #################################################
> library(rugarch)
> # zeta = shape
> zeta = 0.4
> # rho = skew
> rho = 0.9
> # lambda = GIG mixing distribution shape parameter
> lambda=-1
> # POSITIVE scaling factor (sigma is in any always positive)
> # mean
> scaling = 0.02
> # sigma
> location = 0.001
>
> # standardized transformation based on (0,1,\rho,\zeta)
> # parameterization:
> x1 = scaling*qdist("ghyp", seq(0.001, 0.5, length.out=100), shape = zeta,
> skew = rho, lambda = lambda)+location
> # Equivalent to standard transformation:
> # First obtain the standard parameters (which have a mean of zero
> # and sigma of 1).
> parms = rugarch:::.paramGH(zeta, rho, lambda)
> x2 = rugarch:::.qgh( seq(0.001, 0.5, length.out=100), alpha =
> parms[1]/abs(scaling), beta = parms[2]/abs(scaling),
> delta=abs(scaling)*parms[3], mu = (scaling)*parms[4]+location, lambda =
> lambda)
> all.equal(x1, x2)
> # Notice the approximation error in the calculation of the quantile for #
> which there is no closed form solution (and rugarch uses a tolerance
> # value of .Machine$double.eps^0.25)
> # Load the GeneralizedHyperbolic package of Scott to adjust the
> # tolerance:
> library(GeneralizedHyperbolic)
>
> x1 = location+scaling*qghyp(seq(0.001, 0.5, length.out=100), mu = parms[4],
> delta = parms[3], alpha =  parms[1], beta = parms[2], lambda = lambda,
> lower.tail = TRUE, method = c("spline", "integrate")[2], nInterpol = 501,
> subdivisions = 500, uniTol = 2e-12)
> # equivalent to:
> x2 = qghyp(seq(0.001, 0.5, length.out=100), mu =
> (scaling)*parms[4]+location, delta = abs(scaling)*parms[3], alpha =
> parms[1]/abs(scaling), beta = parms[2]/abs(scaling), lambda = lambda,
> lower.tail = TRUE, method = c("spline", "integrate")[2], nInterpol = 501,
> subdivisions = 500, uniTol = 2e-12)
>
> all.equal(x1, x2)
> #################################################
>
>
>
>> Thanks a lot for your help!
>>
>
> Regards,
>
> Alexios
>>
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