*
* Example 7.3 from pp 295-296
* Value at Risk calculation
*
open data d-ibmln98.dat
data(format=free,org=columns) 1 9190 ibmlog
set ibmlog = .01*ibmlog
*
* GARCH with normal residuals
*
garch(p=1,q=1,regressors,hseries=h,resids=res) / ibmlog
# constant ibmlog{2}
*
compute r9191=%beta(1)+%beta(2)*ibmlog(9189)
compute h9191=%beta(3)+%beta(4)*res(9190)**2+%beta(5)*h(9190)
*
* %invnormal(.05) and %invnormal(.01) will give the left tail quantiles for the Normal
*
disp "5% VaR on 10000000" -10000000*(r9191+%invnormal(.05)*sqrt(h9191))
disp "1% VaR on 10000000" -10000000*(r9191+%invnormal(.01)*sqrt(h9191))
*
* GARCH with t(5)
*
garch(p=1,q=1,regressors,distrib=t,shape=5,hseries=h,resids=res) / ibmlog
# constant ibmlog{2}
compute r9191=%beta(1)+%beta(2)*ibmlog(9189)
compute h9191=%beta(3)+%beta(4)*res(9190)**2+%beta(5)*h(9190)
*
* %invtcdf gives the left tail quantiles for the t. Note that sqrt(h9191) is our
* estimate for the variance of the distribution at 9191. Since the t with n
* degrees of freedom has a variance of n/(n-2), we have to correct for this to
* get a distribution with the correct variance
*
disp "5% VaR on 10000000" -10000000*(r9191+%invtcdf(.05,5)*sqrt(h9191)/sqrt(5.0/3.0))
disp "1% VaR on 10000000" -10000000*(r9191+%invtcdf(.01,5)*sqrt(h9191)/sqrt(5.0/3.0))