Hello, > > In Reply To > compare two matrices > Apr 10, 2012; 9:26am — by Kehl Dániel Kehl Dániel > Dear Members, > > I have two estimated transition matrices and I want to compare them. > In fact I want to check the hypothesis if they come from the same process. > I tried to look for some test but all I found was independence test of > contingency tables. >
There is a discussion on markov chain transition matrices in http://www-sre.wu-wien.ac.at/ersa/ersaconfs/ersa02/cd-rom/papers/024.pdf > > The following code shows that the usual chi-squared test statistic does > not follow chisq distribution. > Like discussed in the paper above, you must multiply the ratio (M1-M2)^2/M2 by the frequencies, and get an asymptotic chi-square distribution. More or less like this: set.seed(1) for (i in 1:MCRepl){ n <- 10000 t1 <- table(sample(1:9, size = n, replace = TRUE)) M1 <- matrix(t1, 3, 3) M1b <- M1/rowSums(M1) t2 <- table(sample(1:9, size = n, replace = TRUE)) M2 <- matrix(t2, 3, 3) M2b <- M2/rowSums(M2) khi12[i] <- sum(M1*(M1b-M2b)^2/M2b) } x <- seq(floor(min(khi12)), ceiling(max(khi12)), length=MCRepl) y <- dchisq(x, df=prod(dim(M1b) - 1)) ylim <- range(c(y, density(khi12)$y)) dev.new() hist(khi12, prob=TRUE, ylim=ylim) lines(density(khi12)) lines(x, y, col="blue", lty="dotted") The loop could be a bit simplified, I've kept M2 and M2b for clarity sake. Hope this helps, Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/compare-two-matrices-tp4544922p4546339.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.
