similar to: returns from dnorm and dmvnorm

Dear R community, I was trying to estimate density at point zero of a multivariate distribution (9 dimensions) and for this I was using a multinormal approximation and the function dmvnorm , gtools package. To have a sense of the error I tried to look the mismatch between a unidimensional version of my distribution and estimate density at point zero with function density, dmvnorm and dnorm. At

I am experiencing strange results using dmvnorm. I define a scaled distance matrix from the coordinates bellow and then calculate a covariance matrix using a spherical correlation function. Then with certain combinations of range and sill parameters dmvnorm is returning values greater than 1. Surely the results of dmvnorm should be in the interval 0:1 (or do I just nead a holiday?). In addition

Hello, everyone, I have a question about integration of product of two densities. Here is the sample code; however the mean of first density is a function of another random variable, which is to be integrated. ## f=function(x) {dmvnorm(c(0.6, 0.8), mean=c(0.75, 0.75/x))*dnorm(x, mean=0.6, sd=0.15)} integrate(f, lower=-Inf, upper=Inf) ## error message Error in dmvnorm(c(0.6, 0.8), mean = c(0.75,

Dear R users, i try to use function dmvnorm(x, mean, sigma, log=FALSE) from R package mvtnorm to calculate the probability of x under the multivariate normal distribution with mean equal to mean and covariance matrix sigma. I become the following Error in solve.default(cov, ...) : system is computationally singular: reciprocal condition number = 1.81093e-19 What could be the reason of it?

Summary: + I am writing an R extension that needs to call dmvnorm more than 10,000 times during a model fitting computation. + My extension uses openmp for parallel execution. + As of R 3.0, it is no longer permitted for threads to call the R interpreter because there is a stack overflow check that always trips because the thread's stack is different from what R is expecting. +

Howdee, One of my student spotted something I can't explain: a probability >1 vs a normal probability density function. > dnorm(x=1, mean=1, sd=0.4) [1] 0.9973557 > dnorm(x=1, mean=1, sd=0.39) [1] 1.022929 > dnorm(x=1, mean=1, sd=0.3) [1] 1.329808 > dnorm(x=1, mean=1, sd=0.1) [1] 3.989423 > dnorm(x=1, mean=1, sd=0.01) [1] 39.89423 > dnorm(x=1, mean=1, sd=0.001) [1]

I am using curve3d in the emdbook package to graph a gaussian copula density function generated via the copula package. Unfortunately, it appears that emdbook masks dmvnorm from the package mvtnorm in a way that prohibits copula from generating the gaussian copula. (Sounds very confusing!) For example, > library(copula) > f<-function(x,y) dcopula(normalCopula(0),c(x,y)) >

Greets, I'm trying to use nlminb() to estimate the parameters of a bivariate normal sample and during one of the iterations it passes a parameter vector to the likelihood function resulting in an invalid covariance matrix that causes dmvnorm() to throw an error. Thus, it seems I need to somehow communicate to nlminb() that the final three parameters in my parameter vector are used to

Package mvtnorm provides dmvnorm, pmvnorm that can be used to compute Pr(X=x,Y=y) and Pr(X<x,Y<y) for a bivariate normal. Are there functions that would compute Pr(X<x,Y=y)? I'm currently using "integrate" with dmvnorm but it is too slow.

Does anybody know of a function that implements the derivative (gradient) of the multivariate normal density with respect to the *parameters*? It?s easy enough to implement myself, but I?d like to avoid reinventing the wheel (with some bugs) if possible. Here?s a simple example of the result I?d like, using numerical differentiation: library(mvtnorm) library(numDeriv) f=function(pars, xx, yy)

I have the following function, a MSE calc based on some Multivariate normals: MV.MSE<-function(n,EP,X,S){ (dmvnorm(X,mean=rep(0,2),I+S+EP)-dmvnorm(X,mean=rep(0,2),I+S))^2 + 1/n*(dmvnorm(X,mean=rep(0,2),1+S+EP/2)*det(4*pi*EP)^-.5- (dmvnorm(X,mean=rep(0,2),I+S+EP ))^2)} I can get the MV.MSE for given values of the function e.g

Dear All, I am trying to convert some S-plus code that I have to run MCMC into R-code. The program works in S-plus, but runs slowly. I have managed to source the program into R. R recognizes that the program is there; for example, it will display the code when I type the function name at the prompt. However, the program will not run. When I try to run the program, I get the following error

I have the code for the bivariate Gaussian copula. It is written with for-loops, it works, but I wonder if there is a way to vectorize the function. I don't see how outer() can be used in this case, but maybe one can use mapply() or Vectorize() in some way? Could anyone help me, please? ## Density of Gauss Copula rho <- 0.5 #corr R <- rbind(c(1,rho),c(rho,1)) #vcov matrix id <-

sorry for cross-posting Dear all, I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about "how different" these distributions are (as t-statistics for univariate distributions). In order to visualize what I mean hear a little

Hello: The example "integrate(dnorm,0,20000)" says it "fails on many systems". I just got 0 from it, when I should have gotten either an error or something close to 0.5. I got this with R 2.12.0 under both Windows Vista_x64 and Linux (Fedora 13); see the results from Windows below. I thought you might want to know. Thanks for all your work in creating

Hi, All: Is there a way to identify whether any users are using a particular package in a shared network R installation? I ask, because we have such a multiple-user installation and when I tried to install a package using Rgui that was in use by Rterm on a single-user installation, 'install.packages' deleted the existing package but failed to install the new version;

Hello all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context:

I'm trying to provide different parameters to the integrate function for various probability functions. I'm using dnorm as the simplest example here. For instance integrate(dnorm, -1.96, 1.96) produces the correct answer for a normal distribution with mean 0 and standard deviation 1. I've tried two ways to use mean=2.0 and standard deviation 1, but with no luck. The examples follow.

Dear R-users Idea: Plot a dnorm line using specific mean/sd to complete a histogram (skewed). xs:range of y-values, ys: dnorm function Problem: I expected to multiply the ys function with the sample size (n=250-300). I was wondering about a factor between 12'000 and 30'000 to match the size of the dnorm line with the specific histogram. Thanks Sibylle hist(Biotree[Ld,]$Height2008,

I am attempting to wrap the histogram function in my own custom function, so that I can quickly generate some standard plots. A part of what I want to do is to draw a normal curve over the histogram: > x <- rnorm(1000) > hist(x, freq=F) > curve(dnorm(x), lty=3, add=T) (for normal use, x would be a vector of empirical values, but the rnorm() function works for testing) That