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## rugarch VaR calculation "manually"

 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? 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)) 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/ugarchforecastbut 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? 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? Thanks a lot for your help! _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.
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 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/abs(scaling), beta = parms/abs(scaling), delta=abs(scaling)*parms, mu = (scaling)*parms+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 .Machinedouble.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, delta = parms, alpha = parms, beta = parms, lambda = lambda, lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, subdivisions = 500, uniTol = 2e-12) # equivalent to: x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = (scaling)*parms+location, delta = abs(scaling)*parms, alpha = parms/abs(scaling), beta = parms/abs(scaling), lambda = lambda, lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, subdivisions = 500, uniTol = 2e-12) all.equal(x1, x2) ################################################# > Thanks a lot for your help! > Regards, Alexios > _______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-sig-finance> -- Subscriber-posting only. If you want to post, subscribe first. > -- Also note that this is not the r-help list where general R questions should go. > _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go. Reply | Threaded Open this post in threaded view | ## Re: rugarch VaR calculation "manually"  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 <[hidden email]>: > 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/abs(scaling), beta = parms/abs(scaling), > delta=abs(scaling)*parms, mu = (scaling)*parms+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 .Machinedouble.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, > delta = parms, alpha =  parms, beta = parms, lambda = lambda, > lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, > subdivisions = 500, uniTol = 2e-12) > # equivalent to: > x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = > (scaling)*parms+location, delta = abs(scaling)*parms, alpha = > parms/abs(scaling), beta = parms/abs(scaling), lambda = lambda, > lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, > subdivisions = 500, uniTol = 2e-12) > > all.equal(x1, x2) > ################################################# > > > >> Thanks a lot for your help! >> > > Regards, > > Alexios >> >> _______________________________________________ >> [hidden email] mailing list >> https://stat.ethz.ch/mailman/listinfo/r-sig-finance>> -- Subscriber-posting only. If you want to post, subscribe first. >> -- Also note that this is not the r-help list where general R questions >> should go. >> > _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.
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 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 <[hidden email]>: >> 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/abs(scaling), beta = parms/abs(scaling), >> delta=abs(scaling)*parms, mu = (scaling)*parms+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 .Machinedouble.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, >> delta = parms, alpha = parms, beta = parms, lambda = lambda, >> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >> subdivisions = 500, uniTol = 2e-12) >> # equivalent to: >> x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = >> (scaling)*parms+location, delta = abs(scaling)*parms, alpha = >> parms/abs(scaling), beta = parms/abs(scaling), lambda = lambda, >> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >> subdivisions = 500, uniTol = 2e-12) >> >> all.equal(x1, x2) >> ################################################# >> >> >> >>> Thanks a lot for your help! >>> >> >> Regards, >> >> Alexios >>> >>> _______________________________________________ >>> [hidden email] mailing list >>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance>>> -- Subscriber-posting only. If you want to post, subscribe first. >>> -- Also note that this is not the r-help list where general R questions >>> should go. >>> >> > _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go. Reply | Threaded Open this post in threaded view | ## Re: rugarch VaR calculation "manually"  The data I am working with can be found here: http://uploadeasy.net/upload/2fvhy.rarI fitted the following model: alvnomodel<-ugarchspec(variance.model = list(model = "sGARCH", garchOrder = c(1, 1)), mean.model = list(armaOrder = c(0, 0), include.mean = FALSE), distribution.model = "norm") alvnomodelgarch<-ugarchfit(spec=alvnomodel,data=alvlloss) Now I want to get the one-day ahead forecasts of the cond. volatility and the cond. mean. I try to applay ugrachforecast: ugarchforecast(alvnomodelgarch, n.ahead = 1,out.sample=50,n.roll=10) But I get the error message: ugarchforecast-->error: n.roll must not be greater than out.sample! What is wrong? Also I don't know what values I should take for out.sample and n.roll? I just want to get 1 day ahead forecasts. What values do I need to insert? 2013/5/7 alexios ghalanos <[hidden email]>: > 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 <[hidden email]>: >>> >>> 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/abs(scaling), beta = parms/abs(scaling), >>> delta=abs(scaling)*parms, mu = (scaling)*parms+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 .Machinedouble.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, >>> delta = parms, alpha =  parms, beta = parms, lambda = lambda, >>> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >>> subdivisions = 500, uniTol = 2e-12) >>> # equivalent to: >>> x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = >>> (scaling)*parms+location, delta = abs(scaling)*parms, alpha = >>> parms/abs(scaling), beta = parms/abs(scaling), lambda = lambda, >>> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >>> subdivisions = 500, uniTol = 2e-12) >>> >>> all.equal(x1, x2) >>> ################################################# >>> >>> >>> >>>> Thanks a lot for your help! >>>> >>> >>> Regards, >>> >>> Alexios >>>> >>>> >>>> _______________________________________________ >>>> [hidden email] mailing list >>>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance>>>> -- Subscriber-posting only. If you want to post, subscribe first. >>>> -- Also note that this is not the r-help list where general R questions >>>> should go. >>>> >>> >> > _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.
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 Also http://www.unstarched.net/2013/02/27/whats-new-in-rugarch-ver-1-01-5/does not work if I try it. If I do c(is(sigma((fit))), is(fitted(fit)), is(residuals(fit))) I do not get xts xts xts but numeric vector and so on? If I try plot(xts(fit@model$modeldata$data, fit@model$modeldata$index), auto.grid = FALSE,     minor.ticks = FALSE, main = 'S&P500 Conditional Mean') lines(fitted(fit), col = 2) grid() I get the error messages Fehler in plot(xts(fit@model$modeldata$data, fit@model$modeldata$index),  :   Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl für Funktion 'plot': Fehler in xts(fit@model$modeldata$data, fit@model$modeldata$index) :   order.by requires an appropriate time-based object and sigma(f) is also not working? Fehler in UseMethod("sigma") :   nicht anwendbare Methode für 'sigma' auf Objekt der Klasse "c('uGARCHforecast', 'GARCHforecast', 'rGARCH')" angewendet > I just tried to understand the examples, but they do not work? 2013/5/8 Neuman Co <[hidden email]>: > The data I am working with can be found here: > http://uploadeasy.net/upload/2fvhy.rar> > I fitted the following model: > alvnomodel<-ugarchspec(variance.model = list(model = "sGARCH", > garchOrder = c(1, 1)), > mean.model = list(armaOrder = c(0, 0), include.mean = FALSE), > distribution.model = "norm") > > alvnomodelgarch<-ugarchfit(spec=alvnomodel,data=alvlloss) > > Now I want to get the one-day ahead forecasts of the cond. volatility > and the cond. mean. I try to applay ugrachforecast: > > ugarchforecast(alvnomodelgarch, n.ahead = 1,out.sample=50,n.roll=10) > > But I get the error message: > ugarchforecast-->error: n.roll must not be greater than out.sample! > > What is wrong? > > Also I don't know what values I should take for out.sample and n.roll? > I just want to get 1 day ahead forecasts. What values do I need to > insert? > > 2013/5/7 alexios ghalanos <[hidden email]>: >> 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 <[hidden email]>: >>>> >>>> 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/abs(scaling), beta = parms/abs(scaling), >>>> delta=abs(scaling)*parms, mu = (scaling)*parms+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, >>>> delta = parms, alpha = parms, beta = parms, lambda = lambda, >>>> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >>>> subdivisions = 500, uniTol = 2e-12) >>>> # equivalent to: >>>> x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = >>>> (scaling)*parms+location, delta = abs(scaling)*parms, alpha = >>>> parms/abs(scaling), beta = parms/abs(scaling), lambda = lambda, >>>> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >>>> subdivisions = 500, uniTol = 2e-12) >>>> >>>> all.equal(x1, x2) >>>> ################################################# >>>> >>>> >>>> >>>>> Thanks a lot for your help! >>>>> >>>> >>>> Regards, >>>> >>>> Alexios >>>>> >>>>> >>>>> _______________________________________________ >>>>> [hidden email] mailing list >>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance>>>>> -- Subscriber-posting only. If you want to post, subscribe first. >>>>> -- Also note that this is not the r-help list where general R questions >>>>> should go. >>>>> >>>> >>> >> _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go. Reply | Threaded Open this post in threaded view | ## Re: rugarch VaR calculation "manually"  I mean, I only try to get the one-step ahead in-sample predictions. So I do NOT want to leave out any observations in the estimation stage? I want to use all observations, estimate the GARCH and use the estimated parameters and data to get one step ahead in sample predictions so I can use these one-step-ahead-in-sample predictions of my cond. volatility and cond. mean to calculate the VaR. 2013/5/8 Neuman Co <[hidden email]>: > Also http://www.unstarched.net/2013/02/27/whats-new-in-rugarch-ver-1-01-5/> does not work if I try it. > If I do > c(is(sigma((fit))), is(fitted(fit)), is(residuals(fit))) > I do not get xts xts xts but numeric vector and so on? > > If I try > plot(xts(fit@model$modeldata$data, fit@model$modeldata$index), > auto.grid = FALSE, > minor.ticks = FALSE, main = 'S&P500 Conditional Mean') > lines(fitted(fit), col = 2) > grid() > > I get the error messages > Fehler in plot(xts(fit@model$modeldata$data, fit@model$modeldata$index), : > Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl > für Funktion 'plot': Fehler in xts(fit@model$modeldata$data, > fit@model$modeldataindex) : > order.by requires an appropriate time-based object > > and sigma(f) is also not working? > Fehler in UseMethod("sigma") : > nicht anwendbare Methode für 'sigma' auf Objekt der Klasse > "c('uGARCHforecast', 'GARCHforecast', 'rGARCH')" angewendet >> > > > I just tried to understand the examples, but they do not work? > > 2013/5/8 Neuman Co <[hidden email]>: >> The data I am working with can be found here: >> http://uploadeasy.net/upload/2fvhy.rar>> >> I fitted the following model: >> alvnomodel<-ugarchspec(variance.model = list(model = "sGARCH", >> garchOrder = c(1, 1)), >> mean.model = list(armaOrder = c(0, 0), include.mean = FALSE), >> distribution.model = "norm") >> >> alvnomodelgarch<-ugarchfit(spec=alvnomodel,data=alvlloss) >> >> Now I want to get the one-day ahead forecasts of the cond. volatility >> and the cond. mean. I try to applay ugrachforecast: >> >> ugarchforecast(alvnomodelgarch, n.ahead = 1,out.sample=50,n.roll=10) >> >> But I get the error message: >> ugarchforecast-->error: n.roll must not be greater than out.sample! >> >> What is wrong? >> >> Also I don't know what values I should take for out.sample and n.roll? >> I just want to get 1 day ahead forecasts. What values do I need to >> insert? >> >> 2013/5/7 alexios ghalanos <[hidden email]>: >>> 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 <[hidden email]>: >>>>> >>>>> 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/abs(scaling), beta = parms/abs(scaling), >>>>> delta=abs(scaling)*parms, mu = (scaling)*parms+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 .Machinedouble.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, >>>>> delta = parms, alpha =  parms, beta = parms, lambda = lambda, >>>>> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >>>>> subdivisions = 500, uniTol = 2e-12) >>>>> # equivalent to: >>>>> x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = >>>>> (scaling)*parms+location, delta = abs(scaling)*parms, alpha = >>>>> parms/abs(scaling), beta = parms/abs(scaling), lambda = lambda, >>>>> lower.tail = TRUE, method = c("spline", "integrate"), nInterpol = 501, >>>>> subdivisions = 500, uniTol = 2e-12) >>>>> >>>>> all.equal(x1, x2) >>>>> ################################################# >>>>> >>>>> >>>>> >>>>>> Thanks a lot for your help! >>>>>> >>>>> >>>>> Regards, >>>>> >>>>> Alexios >>>>>> >>>>>> >>>>>> _______________________________________________ >>>>>> [hidden email] mailing list >>>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance>>>>>> -- Subscriber-posting only. If you want to post, subscribe first. >>>>>> -- Also note that this is not the r-help list where general R questions >>>>>> should go. >>>>>> >>>>> >>>> >>> _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.
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