rugarch VaR calculation "manually"

Applying 'sigma' (conditional volatility) and 'fitted' (conditional mean)
METHODS on a uGARCHforecast object will return the forecast values of those
quantities with appropriately labelled dates (the T+0 time). This too is
documented in the help files, and there was also a blog post about this
('Whats new in rugarch (ver 1.01-5)').

For VaR, as I already mentioned, you can use the 'quantile' method on
any of the returned S4 class objects.

-Alexios


On 07/05/2013 12:51, Neuman Co wrote:

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