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 ______________________________________________ [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.
