search for: syy

...(Y)==F) (Y<-as.matrix(Y)) if(is.matrix(X)==F) (X<-as.matrix(X)) if(is.matrix(saM)==F) (saM<-as.matrix(saM)) RX<-rent.matrix(X,1)$Rentabilidade.tipo RY<-rent.matrix(Y,1)$Rentabilidade.tipo r2m<-matrix(0,nrow=ncol(Y),ncol=1) RSS<-matrix(0,ncol=ncol(Y),nrow=1) SYY<-matrix(0,ncol=ncol(Y),nrow=1) for (i in 1:ncol(RY)) { RSS[,i]<-(t(RY[,i])%*%RY[,i])-(saM[i,]%*%(t(RX)%*%RX)%*%t(saM)[,i]) SYY[,i]<-sum((RY[,i]-mean(RY[,i]))^2) r2m[i,]<-1-(RSS[,i]/SYY[,i])*((nrow(RY))/(nrow(RY)-ncol(saM)-1)) } dimnames(r2m)<-list(colnam...

...stein-uhlenbeck-model/ this is the mean-reverting process http://r.789695.n4.nabble.com/file/n4637271/process.txt process.txt and this is the script that i am using....... ouFit.ML=function(spread) { n=length(spread) delta=n/n Sx=sum(spread[1:n-1]) Sy=sum(spread[2:n]) Sxx=(Sx)^2 Syy=(Sy)^2 Sxy=Sx*Sy mu = (Sy*Sxx - Sx*Sxy) / ( n*(Sxx - Sxy) - (Sx^2 - Sx*Sy) ) lambda = -log( (Sxy - mu*Sx - mu*Sy + n*mu^2) / (Sxx -2*mu*Sx + n*mu^2) )/delta a = exp(-lambda*delta) sigmah2 = (Syy - 2*a*Sxy + a^2*Sxx - 2*mu*(1-a)*(Sy - a*Sx) + n*mu^2*(1-a)^2)/n; sigma = sqrt((sigmah2)*2*l...

...(Y<-as.matrix(Y)) if(is.matrix(X)==F) (X<-as.matrix(X)) if(is.matrix(saM)==F) (saM<-as.matrix(saM)) RX<-rent.matrix(X,1)$Rentabilidade.tipo RY<-rent.matrix(Y,1)$Rentabilidade.tipo r2m<-matrix(0,nrow=ncol(Y),ncol=1) RSS<-matrix(0,ncol=ncol(Y),nrow=1) SYY<-matrix(0,ncol=ncol(Y),nrow=1) for (i in 1:ncol(RY)) { RSS[,i]<-(t(RY[,i])%*%RY[,i])-(saM[i,]%*%(t(RX)%*%RX)%*%t(saM)[,i]) SYY[,i]<-sum((RY[,i]-mean(RY[,i]))^2) r2m[i,]<-1-(RSS[,i]/SYY[,i])*((nrow(RY))/(nrow(RY)-ncol(saM)-1)) } dimnames(r2m)<...

...so it may be necessary/advisable to variance stabilize. Is their a best way to go about doing this in R? I am tempted to do something like the following: Use the logistic transformation on each proportion. Next do weighted least squares. Finally, adjust the beta coef by multiplying by sqrt(Sxx/Syy). However, this doesn't take into account the variance in the independant variable. Maybe its better too work with counts of successes instead of proportions? Is their a better way and how do I do it best with R? Thanks in advance for any help. ___________________________________________...

...ptember 29, 2017 10:26 To: devel at lists.fedoraproject.org "Selinux List at Fedora Project" <selinux at lists.fedoraproject.org> -------------------------------------------------------------------------- On 09/29/2017 03:57 PM, Alexander Bokovoy wrote: > On pe, 29 syys 2017, Lukas Vrabec wrote: >> I'm planning change the default value of httpd_graceful_shutdown >> boolean in Fedora Rawhide because of improving SELinux configuration. >> Rawhide builds with this change will be available in ~5 days. >> >> Together with Dan Walsh, w...

Hi, I'm having problems while working with large data sets with R 1.5.1 in windows 2000. Given a integer matrix size of 30 columns and 15000 rows my function should return a boolean matrix size of about 5000 rows and 15000 columns. First of all I tried to run this function on computer with 256 MB of RAM. I increased memory limit of R with memory.limit() up to 512 MB. I was inspecting

Hello list. I'm trying to rebuild the 2.6.9.67.0.20.EL kernel, but it fails even without modifications. How did I try it? Created a (non-root) build environment (not a mock ) Installed the kernel.scr.rpm and did a rpmbuild -ba --target=`uname -m` kernel-2.6.spec 2> prep-err.log | tee prep-out.log The build failed at the end: Processing files: kernel-xenU-devel-2.6.9-67.0.20.EL Checking

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