similar to: square root of inverse

hi , lik <- function(nO, nA, nB, nAB){ loglik <- function(par) { p=par[1] q=par[2] r <- 1 - p - q if (c(p,q,r) > rep(0,3) && c(p,q,r) < rep(1,3) ) { -(2 * nO * log (r) + nA * log (p^2 + 2 * p * r) + nB * log (q^2 + 2 * q * r) + nAB * (log(2) +log(p) +log(q))) } else NA } loglik }

x <- iris$Sepal.Length[1:50]/iris$Sepal.Width[1:50] y <- iris$Petal.Length[1:50]/iris$Petal.Width[1:50] I want to check whether (x,y) follows a bivariate normal distribution or not, using density plot or scatter plot. Is it possible to plot a bivariate density in R. I cant find any. Arindam Fadikar M.Stat Indian Statistical Institute. New Delhi, India [[alternative HTML version

Dear all, Given data of a continuous frequency distribution with class intervals .. how to enter such data so that we can use the normal mean, median and mode functions to calculate them and also can draw histogram .. for example how to enter this data in R Height range[image: ↓]<http://en.wikipedia.org/wiki/Frequency_distribution#> Number of students[image: ↓]

It works! But Once I have the square root of this matrix, how do I convert it to a real (not imaginary) matrix which has the same property? Is that possible? Best, Simon >----Messaggio originale---- >Da: p.dalgaard at biostat.ku.dk >Data: 21-nov-2009 18.56 >A: "Charles C. Berry"<cberry at tajo.ucsd.edu> >Cc: "simona.racioppi at

Dear Peter, thank you very much for your answer. My problem is that I need to calculate the following quantity: solve(chol(A)%*%Y) Y is a 3*3 diagonal matrix and A is a 3*3 matrix. Unfortunately one eigenvalue of A is negative. I can anyway take the square root of A but when I multiply it by Y, the imaginary part of the square root of A is dropped, and I do not get the right answer. I tried

hi all.. i just noticed while patching ulog(patch-o-matic) with kernel-2.4.17--->patch not applied ! would be gratefull if someone can help on this ( Gurus ''am not a programmer pls.):-( arindam

Hi, I am new to virtualization and I am having the obvious hassle of choosing between Xen and KVM. Can Xen experts help me with this? Regards, Arindam _______________________________________________ Xen-users mailing list Xen-users@lists.xen.org http://lists.xen.org/xen-users

Hi, I have been trying to debug Xen using kdb. However, I am unable to successfully boot dom0 after compiling and installing kdb. Software/hardware config: Base distribution - Debian Lenny (64-bit) on a Dell desktop (Optiplex 780) with a serial cable to another machine. Xen 4.0.1 (without kdb) works fine. Here are the steps I followed to set up Xen-unstable with kdb. 1. Cloned code from

Hi, I installed xen using yum in fedora 16. When I boot grub shows me a lot xen botting options. Are the all same or have different funtionality? also in xl info I have xen_changeset : unavailable, what does it mean? I also can not copy host CPU configuration in guest, it gives Error copying host CPU: No host CPU reported in capabilities Traceback (most recent call last): File

Martyn, No, that didn't work. One other thing in the mix (which I don't think is the issue) is that I call one of the C-entry points of expm. So the DESCRIPTION file imports expm, the NAMESPACE file imports expm, and the init.c file is #include "R.h" #include "R_ext/Rdynload.h" /* Interface to expm package. */ typedef enum {Ward_2, Ward_1, Ward_buggy_octave}

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I am trying to import Indian National Sample Survey Data. It is ASCII flat and an example is below 001460306025001117101011117111201*01*000000211 270104070204093 002460306025001117101011117111201*02*0000008011116104910519572 022 2600 1150 30 55 170 4005 003460306025001117101011117111201*03* 0000111039204112 222 Previous post don't seem to provide much help. I have an document

This message is about a curious difference in timing between two ways of computing the same function. One uses expm, so is expected to be a bit slower, but "a bit" turned out to be a factor of >1000. The code is below. We would be grateful if anyone can point out any egregious bad practice in our code, or enlighten us on why one approach is so much slower than the other. The problem

Martin, I am aware of --vanilla; I use it myself for some testing. In this case R_LIBS_USER was set externally (part of my login) and does not involve any of the R scripts. That means it is inherited by any subprocess. For example: tmt1495% R --vanilla --no-environ R version 3.3.1 (2016-06-21) -- "Bug in Your Hair" Copyright (C) 2016 The R Foundation for Statistical Computing

Dear Friends, I do require centos 4.2 x86_64 cds here in delhi,India. If anyone of you can arrange that for me i will be vary grateful. The requirement is urgent. Thanks for your time and effort. -- Regards Abhishek Jain -------------------------------------------------------------------- mail2web - Check your email from the web at http://mail2web.com/ .

Hi, My name is Laura. I'm a PhD student at Queen's University Belfast and have just started learning R. I was wondering if somebody could help me to see where I am going wrong in my code for estimating the parameters [mu1, mu2, lambda1] of a 2-phase Coxian Distribution. cox2.lik<-function(theta, y){ mu1<-theta[1] mu2<-theta[2] lambda1<-theta[3]

I have a local library which depends on the expm library. The expm library is loaded into my personal space and I have the environment variable R_LIBS_USER set appropriately. The command "library(expm)" works just fine from the command line, and in fact the package works if I do the source() and dyn.load() commands by hand. The following sequence works: tmt% R CMD build

Hi Could anybody give me a bit of advice on some code I'm having trouble with? I've been trying to calculate the loglikelihood of a function iterated over a data of time values and I seem to be experiencing difficulty when I use the function expm(). Here's an example of what I am trying to do y<-c(5,10) #vector of 2 survival times p<-Matrix(c(1,0),1,2) #1x2 matrix

hi all At last i got sucess !.. but am confused y it didnt work earlier..the difference today was that i reinstalled RH7.2 & complied kernel 2.4.16(not 17).. rest was same..... & the bandwidth too is under control!.. is it normal for to get more than said bandwidth--i mean i restricted a network with 8Kbit(with same script as below) but still was able to get a download(ftp) of

Can the expm function be used to calculate the exponential of a matrix where the matrix is multiplied by a vector in a data frame? For example for (i in 1:length(y)){ expmN<-expm(Q*y[i]) Q<-Matrix(c(1, 2, 3, 4), 2, 2) return(expmN) } Sorry I am new to R and I have been finding it difficult to get information on calculating Matrix exponentials in R. Many thanks in