On Fri, Jun 03, 2011 at 01:54:33PM -0700, Ned Dochtermann wrote: > Petr, > This is the code I used for your suggestion: > > k<-6;kk<-(k*(k-1))/2 > x<-matrix(0,5000,kk) > for(i in 1:5000){ > A.1<-matrix(0,k,k) > rs<-runif(kk,min=-1,max=1) > A.1[lower.tri(A.1)]<-rs > A.1[upper.tri(A.1)]<-t(A.1)[upper.tri(A.1)] > cors.i<-diag(k) > t<-.001-min(Re(eigen(A.1)$values)) > new.cor<-cov2cor(A.1+(t*cors.i)) > x[i,]<-new.cor[lower.tri(new.cor)]} > hist(c(x)); max(c(x)); median(c(x)) > > This, unfortunately, does not maintain the desired distribution of > correlations.
Hello. On the contrary to what i thought originally, there are solutions also for the case of the correlation matrix. The first solution creates a singular correlation matrix (of rank 3), but the nondiagonal entries have exactly the uniform distribution on [-1, 1], since the scalar product of two independent uniformly distributed unit vectors in R^3 has the uniform distribution on [-1, 1]. x <- matrix(rnorm(18), nrow=6, ncol=3) x <- x/sqrt(rowSums(x^2)) a <- x %*% t(x) The next solution produces a correlation matrix of full rank, whose non-diagonal entries have distribution very close to the uniform on [-1, 1]. KS test finds a difference only with sample size more than 50'000. w <- c(0.01459422, 0.01830718, 0.04066405, 0.50148488, 0.60330865, 0.61832829) x <- matrix(rnorm(36), nrow=6, ncol=6) %*% diag(w) x <- x/sqrt(rowSums(x^2)) a <- x %*% t(x) Hope this helps. Petr Savicky. ______________________________________________ R-help@r-project.org 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.