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/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? 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!
rugarch VaR calculation "manually"
12 messages · Alexios Ghalanos, Neuman Co, Brian G. Peterson
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
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 <alexios at 4dscape.com>:
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
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 at model$modeldata$data, fit at 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 at model$modeldata$data, fit at model$modeldata$index), : Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl f?r Funktion 'plot': Fehler in xts(fit at model$modeldata$data, fit at 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 f?r '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 <neumancohu at gmail.com>:
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 <alexios at 4dscape.com>:
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$double.eps^0.25) # Load the GeneralizedHyperbolic package of Scott to adjust the # tolerance: library(GeneralizedHyperbolic) x1 = location+scaling*qghyp(seq(0.001, 0.5, length.out=100), mu = parms[4], delta = parms[3], alpha = parms[1], beta = parms[2], lambda = lambda, lower.tail = TRUE, method = c("spline", "integrate")[2], nInterpol = 501, subdivisions = 500, uniTol = 2e-12) # equivalent to: x2 = qghyp(seq(0.001, 0.5, length.out=100), mu = (scaling)*parms[4]+location, delta = abs(scaling)*parms[3], alpha = parms[1]/abs(scaling), beta = parms[2]/abs(scaling), lambda = lambda, lower.tail = TRUE, method = c("spline", "integrate")[2], nInterpol = 501, subdivisions = 500, uniTol = 2e-12) all.equal(x1, x2) #################################################
Thanks a lot for your help!
Regards, Alexios
_______________________________________________ R-SIG-Finance at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance -- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.
I mean, I only try to get the one-step ahead in-sample predictions. So I do NOT want to leave out any observations in the estimation stage? I want to use all observations, estimate the GARCH and use the estimated parameters and data to get one step ahead in sample predictions so I can use these one-step-ahead-in-sample predictions of my cond. volatility and cond. mean to calculate the VaR. 2013/5/8 Neuman Co <neumancohu at gmail.com>:
Also http://www.unstarched.net/2013/02/27/whats-new-in-rugarch-ver-1-01-5/ does not work if I try it. If I do c(is(sigma((fit))), is(fitted(fit)), is(residuals(fit))) I do not get xts xts xts but numeric vector and so on? If I try plot(xts(fit at model$modeldata$data, fit at model$modeldata$index), auto.grid = FALSE, minor.ticks = FALSE, main = 'S&P500 Conditional Mean') lines(fitted(fit), col = 2) grid() I get the error messages Fehler in plot(xts(fit at model$modeldata$data, fit at model$modeldata$index), : Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl f?r Funktion 'plot': Fehler in xts(fit at model$modeldata$data, fit at model$modeldata$index) : order.by requires an appropriate time-based object and sigma(f) is also not working? Fehler in UseMethod("sigma") : nicht anwendbare Methode f?r '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 <neumancohu at gmail.com>:
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 <alexios at 4dscape.com>:
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
Please TRY to send one question, wait for a reply and then ask another rather than sending 3 emails in a row. 1. The only reason that you would not get an xts returned object in the example below is that you are using an old version of rugarch. Make sure you download the latest version (which requires R>=3.0.0) and update your packages. It is also good practice to send the output of sessionInfo() in such cases. 2. I'm not sure what you mean by "one step ahead in sample predictions". - If you want the 1-ahead forecast use ugarchforecast(alvnomodelgarch, n.ahead = 1) - If you want the in-sample estimated values use 'fitted' and 'sigma' (in which case the terminology you used in not 'standard'). 3. There are enough examples in 'rugarch.tests' folder in the source distribution, online blog examples, previous mailing list posts, other online resources, for you to be able to do what you need. Finally, I am more likely to respond to future requests for help if you provide minimally reproducible examples without sending zip files of your data (this is rarely needed), showed that you made an effort to read the documentation, and have signed your emails with your real name. Regards, Alexios
On 08/05/2013 11:28, Neuman Co wrote:
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 <neumancohu at gmail.com>:
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 at model$modeldata$data, fit at 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 at model$modeldata$data, fit at model$modeldata$index), : Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl f?r Funktion 'plot': Fehler in xts(fit at model$modeldata$data, fit at 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 f?r '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 <neumancohu at gmail.com>:
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 <alexios at 4dscape.com>:
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
well but this is exactly the point, that ugarchforecast(alvnomodelgarch, n.ahead = 1) just gives me one forecast, but I want to have this from my beginning of the time series up to the end. So I do not want to exclude these observations in my fitting. I want to use all observations for my fitting and then do one step ahead in sample predictions. So e.g. I have a time series of 1000 observations. I to the fitting with the 1000 observations and then I want to predict the one step ahead in-sample forecasts of the cond. mean and the cond. volatility? I do not understand the thing with the fitted and the sigma. Correct me if I am wrong, but these are the fitted values, these are not 1-step-ahead forecasts? I try to give a more simpler example to make the understanding easier: consider a simple linear regression: The fitted values of y at time point t use the realization of x at time point t. The one step ahead forecast uses the realization of timepoint minus 1. So I do NOT want to have the fitted values of the cond. volatility and the cond. mean (in my case the cond. mean is zero, because I have specified no model for it), but the one step ahead forecasts of it. So R should do the n.ahead=1 not only for the last value, but from the beginning up to the end? I mean I only get the forecast of the next day, the 2013-03-04, but I want to have it from the beginning up to the end? 2013/5/8 alexios ghalanos <alexios at 4dscape.com>:
Please TRY to send one question, wait for a reply and then ask another rather than sending 3 emails in a row. 1. The only reason that you would not get an xts returned object in the example below is that you are using an old version of rugarch. Make sure you download the latest version (which requires R>=3.0.0) and update your packages. It is also good practice to send the output of sessionInfo() in such cases. 2. I'm not sure what you mean by "one step ahead in sample predictions". - If you want the 1-ahead forecast use ugarchforecast(alvnomodelgarch, n.ahead = 1) - If you want the in-sample estimated values use 'fitted' and 'sigma' (in which case the terminology you used in not 'standard'). 3. There are enough examples in 'rugarch.tests' folder in the source distribution, online blog examples, previous mailing list posts, other online resources, for you to be able to do what you need. Finally, I am more likely to respond to future requests for help if you provide minimally reproducible examples without sending zip files of your data (this is rarely needed), showed that you made an effort to read the documentation, and have signed your emails with your real name. Regards, Alexios On 08/05/2013 11:28, Neuman Co wrote:
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 <neumancohu at gmail.com>:
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 at model$modeldata$data, fit at model$modeldata$index), auto.grid = FALSE, minor.ticks = FALSE, main = 'S&P500 Conditional Mean') lines(fitted(fit), col = 2) grid() I get the error messages Fehler in plot(xts(fit at model$modeldata$data, fit at model$modeldata$index), : Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl f?r Funktion 'plot': Fehler in xts(fit at model$modeldata$data, fit at model$modeldata$index) : order.by requires an appropriate time-based object and sigma(f) is also not working? Fehler in UseMethod("sigma") : nicht anwendbare Methode f?r '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 <neumancohu at gmail.com>:
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 <alexios at 4dscape.com>:
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
On 05/08/2013 06:06 AM, Neuman Co wrote:
I want to use all observations for my fitting and then do one step ahead in sample predictions.
There is no such thing as a one step ahead in-sample *prediction*. Please stop and think a moment about how absurd that is before you waste more of everyone's time. At each time t, I may make a prediction for time t+1 ugarchroll will give you a set of rolling one or n step ahead *predictions*, as Alexios has also already told you. If you want the in-sample *estimate* from the fitted model values (it is not a prediction), then you want the 'fitted' and 'sigma' values, as Alexios has told you at least twice now. ...and I still don't see you signing your name. Brian
Brian G. Peterson http://braverock.com/brian/ Ph: 773-459-4973 IM: bgpbraverock
That may be the case for your linear regression example, but certainly NOT the case in the ARMA-GARCH models. Everything is based on information upto time T-1. There is no "contemporaneous" data, as a reading of the vignette (where the models are clearly explained and formulated) shows. Therefore, and as pointed out before, USE the fitted, sigma, and quantile methods. -Alexios
On 08/05/2013 12:06, Neuman Co wrote:
well but this is exactly the point, that ugarchforecast(alvnomodelgarch, n.ahead = 1) just gives me one forecast, but I want to have this from my beginning of the time series up to the end. So I do not want to exclude these observations in my fitting. I want to use all observations for my fitting and then do one step ahead in sample predictions. So e.g. I have a time series of 1000 observations. I to the fitting with the 1000 observations and then I want to predict the one step ahead in-sample forecasts of the cond. mean and the cond. volatility? I do not understand the thing with the fitted and the sigma. Correct me if I am wrong, but these are the fitted values, these are not 1-step-ahead forecasts? I try to give a more simpler example to make the understanding easier: consider a simple linear regression: The fitted values of y at time point t use the realization of x at time point t. The one step ahead forecast uses the realization of timepoint minus 1. So I do NOT want to have the fitted values of the cond. volatility and the cond. mean (in my case the cond. mean is zero, because I have specified no model for it), but the one step ahead forecasts of it. So R should do the n.ahead=1 not only for the last value, but from the beginning up to the end? I mean I only get the forecast of the next day, the 2013-03-04, but I want to have it from the beginning up to the end? 2013/5/8 alexios ghalanos <alexios at 4dscape.com>:
Please TRY to send one question, wait for a reply and then ask another rather than sending 3 emails in a row. 1. The only reason that you would not get an xts returned object in the example below is that you are using an old version of rugarch. Make sure you download the latest version (which requires R>=3.0.0) and update your packages. It is also good practice to send the output of sessionInfo() in such cases. 2. I'm not sure what you mean by "one step ahead in sample predictions". - If you want the 1-ahead forecast use ugarchforecast(alvnomodelgarch, n.ahead = 1) - If you want the in-sample estimated values use 'fitted' and 'sigma' (in which case the terminology you used in not 'standard'). 3. There are enough examples in 'rugarch.tests' folder in the source distribution, online blog examples, previous mailing list posts, other online resources, for you to be able to do what you need. Finally, I am more likely to respond to future requests for help if you provide minimally reproducible examples without sending zip files of your data (this is rarely needed), showed that you made an effort to read the documentation, and have signed your emails with your real name. Regards, Alexios On 08/05/2013 11:28, Neuman Co wrote:
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 <neumancohu at gmail.com>:
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 at model$modeldata$data, fit at 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 at model$modeldata$data, fit at model$modeldata$index), : Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl f?r Funktion 'plot': Fehler in xts(fit at model$modeldata$data, fit at 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 f?r '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 <neumancohu at gmail.com>:
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 <alexios at 4dscape.com>:
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
_______________________________________________ R-SIG-Finance at r-project.org 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.
Thanks a lot for your help, ok, I added the signature in my mail and
the r sig list accout.
But to be honest neither I am not really understanding you nor are
your hints helping me. From someone other, I was told to use the
following code to get what I want:
(I add my model here specification again, so you do not have to search
in my posts before)
-----------------------------------------------------------------------------------
library(rugarch)
alvdatemod<-alvdate[-1]
alvgarchdata<-data.frame(alvlloss,row.names=alvdatemod)
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=alvgarchdata)
#plot(alvnomodelgarch)
-------------------------------------------------------------------------------------
and here comes the relevant part:
spec = getspec(alvnomodelgarch);
setfixed(spec) <- as.list(coef(alvnomodelgarch));
forecast = ugarchforecast(spec, n.ahead = 1, n.roll = 2579, data =
alvgarchdata[1:2580, ,drop=FALSE], out.sample = 2579);
sigma(forecast);
fitted(forecast)
I get the (german) error message after the sigma command:
Fehler in UseMethod("sigma") :
nicht anwendbare Methode f?r 'sigma' auf Objekt der Klasse
"c('uGARCHforecast', 'GARCHforecast', 'rGARCH')" angewendet
And the forecast specification is also not what I want, if you just
look at forecast, you will see, that again I just got one value, this
is NOT what I want!
Also, I am not understanding why I have to use this n.roll number and
this out.sample number. Also, why can't I just use my dataset, but
instead use [1:2580, ,drop=FALSE] ? But these are just some smaller
unimportant questions.
Later on, I want to calculate the VaR and compare it to the quantile
method, therefore I would continue with the following idea, first,
calculate VaR:
VaR=fitted(forecast)+sigma(forecast)*qnorm(0.99)
and then compare it to the quantile method:
quantile(forecast,probs=0.99)
but it seemed, that I also did a syntax error here.
2013/5/8 alexios ghalanos <alexios at 4dscape.com>:
That may be the case for your linear regression example, but certainly NOT the case in the ARMA-GARCH models. Everything is based on information upto time T-1. There is no "contemporaneous" data, as a reading of the vignette (where the models are clearly explained and formulated) shows. Therefore, and as pointed out before, USE the fitted, sigma, and quantile methods. -Alexios On 08/05/2013 12:06, Neuman Co wrote:
well but this is exactly the point, that ugarchforecast(alvnomodelgarch, n.ahead = 1) just gives me one forecast, but I want to have this from my beginning of the time series up to the end. So I do not want to exclude these observations in my fitting. I want to use all observations for my fitting and then do one step ahead in sample predictions. So e.g. I have a time series of 1000 observations. I to the fitting with the 1000 observations and then I want to predict the one step ahead in-sample forecasts of the cond. mean and the cond. volatility? I do not understand the thing with the fitted and the sigma. Correct me if I am wrong, but these are the fitted values, these are not 1-step-ahead forecasts? I try to give a more simpler example to make the understanding easier: consider a simple linear regression: The fitted values of y at time point t use the realization of x at time point t. The one step ahead forecast uses the realization of timepoint minus 1. So I do NOT want to have the fitted values of the cond. volatility and the cond. mean (in my case the cond. mean is zero, because I have specified no model for it), but the one step ahead forecasts of it. So R should do the n.ahead=1 not only for the last value, but from the beginning up to the end? I mean I only get the forecast of the next day, the 2013-03-04, but I want to have it from the beginning up to the end? 2013/5/8 alexios ghalanos <alexios at 4dscape.com>:
Please TRY to send one question, wait for a reply and then ask another rather than sending 3 emails in a row. 1. The only reason that you would not get an xts returned object in the example below is that you are using an old version of rugarch. Make sure you download the latest version (which requires R>=3.0.0) and update your packages. It is also good practice to send the output of sessionInfo() in such cases. 2. I'm not sure what you mean by "one step ahead in sample predictions". - If you want the 1-ahead forecast use ugarchforecast(alvnomodelgarch, n.ahead = 1) - If you want the in-sample estimated values use 'fitted' and 'sigma' (in which case the terminology you used in not 'standard'). 3. There are enough examples in 'rugarch.tests' folder in the source distribution, online blog examples, previous mailing list posts, other online resources, for you to be able to do what you need. Finally, I am more likely to respond to future requests for help if you provide minimally reproducible examples without sending zip files of your data (this is rarely needed), showed that you made an effort to read the documentation, and have signed your emails with your real name. Regards, Alexios On 08/05/2013 11:28, Neuman Co wrote:
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 <neumancohu at gmail.com>:
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 at model$modeldata$data, fit at model$modeldata$index), auto.grid = FALSE, minor.ticks = FALSE, main = 'S&P500 Conditional Mean') lines(fitted(fit), col = 2) grid() I get the error messages Fehler in plot(xts(fit at model$modeldata$data, fit at model$modeldata$index), : Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl f?r Funktion 'plot': Fehler in xts(fit at model$modeldata$data, fit at model$modeldata$index) : order.by requires an appropriate time-based object and sigma(f) is also not working? Fehler in UseMethod("sigma") : nicht anwendbare Methode f?r '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 <neumancohu at gmail.com>:
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 <alexios at 4dscape.com>:
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
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-- Neumann, Conrad