similar to: Optim

Hi everyone, I am using R 2.4.1 on a MacOS X ("Tiger") operating system. In the last few day I was trying to estimate the parameters of a mixture of two normal distributions using Maximum Likelihood. The code is from Modern Applied Statistics with S (4th edition), chapter 16 ("Optimization"), the dataset is available under MASS in R. Unfortunately, when I tried out the

Dear R users, I have been trying to obtain the MLE of the following model state 0: y_t = 2 + 0.5 * y_{t-1} + e_t state 1: y_t = 0.5 + 0.9 * y_{t-1} + e_t where e_t ~ iidN(0,1) transition probability between states is 0.2 I've generated some fake data and tried to estimate the parameters using the constrOptim() function but I can't get sensible answers using it. I've tried using

Dear R-list I have been trying to make survreg fit a normal regression model with left truncated data, but unfortunately I am not able to figure out how to do it. The following survreg-call seems to work just fine when the observations are right censored: library(survival) n<-100000 #censored observations x<-rnorm(n) y<-rnorm(n,mean=x) d<-data.frame(x,y) d$ym<-pmin(y,0.5)

I keep getting this error when I try to use the sem package. I and another person who has successfully used the sem package for similar analysis (fMRI effective connectivity) cannot figure out what is wrong with my code. I would appreciate any suggestions. The error message: Error in data.frame(object$coeff, se, z, 2 * (1 - pnorm(abs(z))), par.code) : arguments imply differing

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

On 04-Sep-10 19:27:54, Yi wrote: > Enh, I see. > It totally makes sense. > Thank you for your perfect explanation. > Enjoy the long weekend~ > Yi You're welcome! Earlier I tried an experiment with rejection sampling, which seems to work well for the case where you want mean of the sampled values to exactly be the mean of the range being sampled from. The number of tries, even

Good afternoon, gentlemen! After several days studying and researching on categorical data (various forums with answers from the owner of the library - all incipient) how to interpret the output the function MCMCglmm, come to enlist the help of you, if someone has already worked with MCMCglmm function in the case of variables ordinal dependent. I've read and reread all the pdf's of the

Hello, I'm trying to fit a pretty simple confirmatory factor analysis using the sem package. There's a CFA example in the examples, which is helpful, but the output for my (failing) model is hard to understand. I'd be interested in any other ways to do a CFA in R, if this proves troublesome. The CFA is replicating a 5 uncorrelated-factor structure (for those interested, it is a

Hello, I am trying to do a fit a loglikelihood function with Multivariate distribution via copulas with fitMdvc. The problem is that it doesn't recognize that my beta is a vector of km parameter and when I try to run it it say that the length of my initial values is not the same as the parameter. Can somebody guide me where my mistake is. Thanks, Elisa. #################################

Hi, I'm trying to evaluate a character vector within pnorm. I have a vector with values and names x = c(2,3) names(x) = c("mean", "sd") so that i tried the following temp = paste(names(x), x, sep = "=") #gives #> temp #[1] "mean=2" "sd=3" #Problem is that both values 2 and 3 are taken as values for the mean argument in pnorm pnorm(0,

Hello all, I was wondering if there is an R function to do the following: [*] log(pnorm(x)-pnorm(y)), where x>y. I don't want all the area under the natural log of the normal pdf less than x, I only want the area between y and x. I am aware of the ability to specify log.p=TRUE, which gives me the log of the probability that X<=x. This does not help me, because the following code:

Hello r-group I have a question to the ks.test. I would expect different values for less and greater between data1 and data2. Does anybody could explain my point of misunderstanding the function? data1<-c(8,12,43,70) data2<- c(70,43,12,8) ks.test(data1,"pnorm") ks.test(data1,"pnorm",alternative ="less") #expected < 0.001

Hi all. I may be wrong, (and often am), but in trying to determine how to calculate the erf function, the documentation for 'pnorm' states: ## if you want the so-called 'error function' erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1 ## and the so-called 'complementary error function' erfc <- function(x) 2 * pnorm(x * sqrt(2), lower=FALSE) Should, instead, it read:

I stumbled across this and I am wondering if this is unexpected behavior or if I am missing something. > pnorm(-1.0e+307, log.p=TRUE) [1] -Inf > pnorm(-1.0e+308, log.p=TRUE) [1] NaN Warning message: In pnorm(q, mean, sd, lower.tail, log.p) : NaNs produced > pnorm(-1.0e+309, log.p=TRUE) [1] -Inf I don't know C and am not that skilled with R, so it would be hard for me to look into

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

Hi, I am facing a problem which i think i need to explain it to you with some background. I need to use the Project R pnorm function in Visual Basic 6.0. I have already installed R and this is how i perform and get back the result: > pnorm(2, 15) [1] 6.117164e-39 which is what i need. I have already installed R, i generated the Rmath.DLL file out so i can import it into my VB6 and use it.

Trenkler, Dietrich said: > > I found the following strange behavior using qnorm() and pnorm(): > > > x<-8.21;x-qnorm(pnorm(x)) > [1] 0.0004638484 > > x<-8.28;x-qnorm(pnorm(x)) > [1] 0.07046385 > > x<-8.29;x-qnorm(pnorm(x)) > [1] 0.08046385 > > x<-8.30;x-qnorm(pnorm(x)) > [1] -Inf > qnorm(1-.Machine$double.eps) [1] 8.12589

Dear R-people, I have to compute C - -(pnorm(B)*dnorm(B)*B + dnorm(B)^2)/pnorm(B)^2 This expression seems to be converging to -1 if B approaches to -Inf (although I am unable to prove it). R has no problems until B equals around -28 or less, where both numerator and denominator go to 0 and you get NaN. A simple workaround I did was C <- ifelse(B > -25, -(pnorm(B)*dnorm(B)*B

Hi, I would have thought that these two constructions would produce the same result but they do not. Resp <- rbinom(10, 1, 0.5) Stim <- rep(0:1, 5) mm <- model.matrix(~ Stim) Xb <- mm %*% c(0, 1) ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) pnorm(as.vector(Xb), lower.tail = Resp, log.p = TRUE) > ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) [1] -0.6931472 -1.8410216