similar to: folded normal distribution in R

Full_Name: Long Qu Version: 2.3.1 OS: Windows XP Submission from: (NULL) (64.113.93.235) The QQ-plot of two versions of simulating noncentral F-distributed random numbers has quite different scales: > qqplot(rf(1000,2,15,3),qf(runif(1000),2,15,3)) The rf() function reads: > rf function (n, df1, df2, ncp = 0) { if (ncp == 0) .Internal(rf(n, df1, df2)) else rchisq(n, df1,

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 <<insert bug report here>> Reproduced on Debian and Windows ... On 2.4.x if you execute set.seed(12345) t.1 <- rt(n = 1000, df = 20) set.seed(12345) t.2 <- rt(n = 1000, df = 20, ncp = 0) all.equal(t.1, t.2) ## Not close to true This appears to be due to the fact that in 2.4.x rt is now rt function (n, df, ncp = 0) { if

Hi, I would like to know if there is an algorithm in R for testing if a data set as a normal destribution. Thank you for your time, Pedro Marques

Hi there: I'd thought these two versions of noncentral t-distribution are essentially the same: > qqplot(rt(1000,df=20,ncp=3),qt(runif(1000),df=20,ncp=3)) But, the scales of the x-axis and the y-axis are quite different according to the QQ-plot. Did I make any mistakes somewhere? Thanks, Long ---------------------------------

help(qt) states that: "ncp non-centrality parameter delta; currently except for rt(), only for abs(ncp) <= 37.62" so I would expect that calling qt with non-centrality parameter exceeding 37.62 should fail, instead e.g. calling > mapply(function(x) qt(p = 0.9, df = 55, ncp = x),35:45) gives: [1] 40.21448 41.35293 42.49164 43.68862 44.82945 45.97048 47.11170 48.25310 [9]

Hi everyone, I am new to R, but it's really great and helped me a lot! But now I have 2 questions. It would be great, if someone can help me: 1. I want to integrate a normal distribution, given a median and sd. The integrate function works great BUT the first argument has to be a function so I do integrate(dnorm,0,1) and it works with standard m. and sd. But I have the m and sd given.

Dear all Just for fun, I have just downloaded the paper mentioned below and checked it with R-1.6.1. Everything is ok with exception of Table 2b, where I get always 1 instead of 0.5: > pbinom(1e15,2e15,0.5) [1] 1 Which value should be correct? Best regards Christian Stratowa ============================================== Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead

Not real data. It was gererated randomly. The original codes are the following: par(mfrow=c(2,1)) n <- 500 ######################### #DATA GENERATING PROCESS# ######################### x1 <- rnorm(n,0,1) x2 <- rchisq(n,df=3,ncp=0)-3 sigma <- 1 u1 <- rnorm(n,0,sigma) ylatent1 <-x1+x2+u1 y1 <- (ylatent1 >=0) # create the binary indicator ####################### #THE

Hi! This is my first time posting. I've read the general rules and guidelines, but please bear with me if I make some fatal error in posting. Anyway, I have a continuous response and 29 predictors made up of continuous variables and nominal and ordinal categorical variables. I'd like to do lasso on these, but I get an error. The way I am using "lars" doesn't allow for the

Hi all, I'm trying to superimpose a rchisq density line over a histogram with something like: hist(alnlength) lines(density(rchisq(length(alnlength), 4)),col="red") But the rchisq line won't appear anywhere, Anyone knows what I am missing here? Thanks in advance, Albert.

Hi Folks, I note that, while the "chisq" functions dchisq(x, df, ncp=0, log = FALSE) pchisq(q, df, ncp=0, lower.tail = TRUE, log.p = FALSE) qchisq(p, df, ncp=0, lower.tail = TRUE, log.p = FALSE) rchisq(n, df, ncp=0) all have a slot for the non-centrality parameter "ncp", of the functions for the t and F distributions: dt(x, df, log = FALSE)

Hi! This is going to be a real newbie question, but I can't figure it out. I'm trying to plot densities of various functions of chi-square. A simple chi-square plot I can do with dchisq(). But e.g. chi.sq/degrees of freedom I only know how to do using density(rchisq()/df). For example: plot(1, type="n", xlab="", ylab="", xlim=c(0,2), ylim=c(0,7)) for (i

Dear R-users, i have a simple problem maybe, but i don't see the solution. i want to find the entry-wise closest element of an vector compared with another. ind1<-c(1,4,10) ind2<-c(3,5,11) for (i in length(ind2):1) { print(which.min(abs(ind1-ind2[i]))) } for ind2[3] it should be ind1[3] 10, for ind2[2] it should be ind1[2] 4 and for ind2[1] it should be ind1[1] 1. but with the

Dear R useRs, with the following piece of code i try to find the first value which can be calculated without warnings `test` <- function(a) { repeat { ## hide warnings suppressWarnings(log(a)) if (exists("last.warning", envir = .GlobalEnv)) { a <- a + 0.1 ## clear existing warnings rm("last.warning", envir = .GlobalEnv) }

Hi, I have written out the log-likelihood function to fit some data I have (called ONES20) to the non-central chi-squared distribution. >library(stats4) >ll<-function(lambda,k){x<-ONES20; 25573*0.5*lambda-25573*log(2)-sum(-x/2)-log((x/lambda)^(0.25*k-0.5))-log(besselI(sqrt(lambda*x),0.5*k-1,expon.scaled=FALSE))} > est<-mle(minuslog=ll,start=list(lambda=0.05,k=0.006))

Hi, I am running the following code based on the cpm vignette's code. I believe the code is syntactically correct but it just seems to hang R. I can get this to run if I set the sims to 100 but with 2000 it just hangs. Any ideas why? Thanks, Chris library(cpm) cpmTypes <- c("Kolmogorov-Smirnov","Mann-Whitney","Cramer-von-Mises") changeMagnitudes <- c(1, 2,

Dear R users, i try to integrate lgamma from 0 to Inf. But here i get the message "roundoff error is detected in the extrapolation table", if i use 1.0e120 instead of Inf the computation works, but this is against the suggestion of integrates help information to use Inf explicitly. Using stirlings approximation doesnt bring the solution too. ## Stirlings approximation lgammaApprox

Hi, I need some help on integrating a function that is a vector. I have a function - vector which each element is different. And, naturally, function integrate() does not work I checked the article of U. Ligges and J. Fox (2008) about code optimization "How Can I Avoid This Loop or Make It Faster?" on http://promberger.info/files/rnews-vectorvsloops2008.pdf. Their advice did not help

Is there any R function that computes the convolution of the double exponential distribution? If not, is there a good way to integrate ((q+x)^n)*exp(-2x) over x from 0 to Inf for any value of q and for any positive integer n? I need to perform the integration within a function with q and n as arguments. The function integrate() is giving me this message: "evaluation of function gave a

Hello, I'm taking samples from certain distributions and drawing a density distribution over the histogram of the samples It works fine for the chi-square and for the normal, but not for the cauchy. Any idea what I'm doing wrong? Thanks x <- rchisq(10000, df = 4) hist(x, freq = FALSE, breaks=100) curve(dchisq(x, df = 4), col = 2, add = TRUE) x <- rnorm(10000) hist(x, freq =