Again, a big thanks for your answer. On this webpage: http://financetrainingcourse.com/education/2010/03/master-class-calculating-value-at-risk-var-final-steps/
I found, that I have to rescale by dividing the weights calculated in Step B2 by 1-?n The "?" is the lambda, since the webpage cannot display it, I also found it on another webpage, therefore, I changed my code to the following: dummy2<-lambda^(a-1)/(1-lambda^100)*t(datamatrix[(i-a),]-t(meanvalues))%*%(datamatrix[(i-a),]-t(meanvalues)) Do you think this is correct? One further question: You also told me, that I do not have to initialize my dummy2, what does this mean? I wrote dummy2<-0 because I have to create this variable before using it for the loop? 2013/6/2 Berend Hasselman <[email protected]>: > > On 02-06-2013, at 19:03, Neuman Co <[email protected]> wrote: > >> Thanks a lot for your answer, one more question: >> I now use 100 values, so not infinity values. That means I cut some >> values off, so the weights will not sum up to one. With which factor >> do I have to multiply the (1-lambda)*summe2 to rescale it? So that I >> do not always underestimate the variance anymore? >> > > I don't know but maybe something like this > > 1/sum(lambda^((1:100)-1))/(1-lambda) > > which in your case is 1.000027 > > Berend > >> 2013/6/2 Berend Hasselman <[email protected]>: >>> >>> On 02-06-2013, at 15:17, Neuman Co <[email protected]> wrote: >>> >>>> Hi, >>>> since I want to calculate the VaR of a portfolio consiting of 4 assets >>>> (returns saved into "eonreturn","henkelreturn" and so on) I have to >>>> estimate the covariance matrix. I do not want to take the rectangular >>>> version with equal weights, but the exponentially weighted moving >>>> average in a multivariate version. I want to estimate a covariance >>>> matrix at every time point t. Then I want to comput the VaR at this >>>> time point t. Afterwards, I will look at the exceedances and do a >>>> backtest. >>>> >>>> I tried to implement it as follows (data attached): >>>> >>>> lambda<-0.9 >>>> >>>> summe2<-0 >>>> dummy2<-0 >>>> covestiexpo<-list(NA) >>>> meanvalues<-NA >>>> for(i in 101:length(eonreturn)){ >>>> meanvalues<-matrix(c(mean(eonreturn[(i-100):(i-1)]),mean(henkelreturn[(i-100):(i-1)]),mean(siemensreturn[(i-100):(i-1)]),mean(adidasreturn[(i-100):(i-1)])),4) >>>> for(a in 1:100){ >>>> dummy2<-lambda^(a-1)*t(datamatrix[(i-a),]-t(meanvalues))%*%(datamatrix[(i-a),]-t(meanvalues)) >>>> summe2<-summe2+dummy2 >>>> } >>>> covestiexpo[[i]]<-(1-lambda)*summe2 >>>> } >>>> >>>> >>>> So the covestieexpo[[101]] would be the covariance estimate for the >>>> 101th day, taking into account the last 100 observations. Now, the >>>> problem is, that there seems to be something wrong, since the >>>> covariance estimates are cleraly wrong, they seem to be too big. At >>>> the beginning, compared to the normal covariance estimate the >>>> difference is as follows: >>>> >>>> covestiexpo[[101]] >>>> [,1] [,2] [,3] [,4] >>>> [1,] 0.004559042 0.002346775 0.004379735 0.003068916 >>>> [2,] 0.002346775 0.001978469 0.002536891 0.001909276 >>>> [3,] 0.004379735 0.002536891 0.005531590 0.003259803 >>>> [4,] 0.003068916 0.001909276 0.003259803 0.003140198 >>>> >>>> >>>> >>>> compared to cov(datamatrix[1:100,]) >>>> [,1] [,2] [,3] [,4] >>>> [1,] 0.0018118239 0.0007432779 0.0015301070 0.001119120 >>>> [2,] 0.0007432779 0.0008355960 0.0009281029 0.000754449 >>>> [3,] 0.0015301070 0.0009281029 0.0021073171 0.001269626 >>>> [4,] 0.0011191199 0.0007544490 0.0012696257 0.001325716 >>>> >>>> So already here, it is obvious, that something is not correct, if I >>>> look at a period far ahead: >>>> >>>> covestiexpo[[1200]] >>>> >>>> [,1] [,2] [,3] [,4] >>>> [1,] 0.5312575 0.1939061 0.3419379 0.2475233 >>>> [2,] 0.1939061 0.3204951 0.2303478 0.2022423 >>>> [3,] 0.3419379 0.2303478 0.5288435 0.2943051 >>>> [4,] 0.2475233 0.2022423 0.2943051 0.4599648 >>>> >>>> >>>> you can see, that the values are way too large, so where is my mistake? >>> >>> Without actual data this is an unverifiable statement. >>> But you probably have to move the statement >>> >>> summe2 <- 0 >>> >>> to inside the i-forloop just before the a-forloop. >>> >>> summe2 <- 0 >>> for(a in 1:100){ >>> … >>> >>> so that summe2 is initialized to 0 every time you use it as an accumulator >>> in the a-forloop. >>> Furthermore there is no need to initialize dummy2. It gets overwritten >>> continuously. >>> >>> Berend >>> >>> >> >> >> >> -- >> Neumann, Conrad > -- Neumann, Conrad ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
