similar to: Simulating a variable following an arbitrary distribution

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 =

Full_Name: Ray Kiddy Version: R 2.5.0 GUI 1.19 (4308) OS: Mac OS X 10.4.9 Submission from: (NULL) (71.139.33.60) There is no section in the FAQ equivalent to "Using R". The "R Basics" is too basic. I would say it addresses "meta-R" questions, such as installing. I had a FAQ that I would think would be a basic "Using R" question, namely: "How does

Hola. Tengo un problema con el formato de salida de un objeto "numeric", que los expresa en formato científico o exponencial. > range(total$ImpTotal) [1] 5.66 806907887.10 > Valores <- quantile(total$ImpTotal, c(0, 0.01, 0.02, 0.025, 0.03, 0.05, 0.1, 0.25, 0.5, 0.75, 0.9, 0.95, 0.96, 0.97, 0.975, 0.98, 0.99), na.rm=TRUE) > class(Valores) [1] "numeric" >

Cordial saludo. Estoy trabajando con una densidad conjunta algo elaborada (suma y cociente de dos variables aleatorias positivas usando la cópula Frank) y requiero encontrar la distribución marginal de cada una de las variables. A manera de ejemplo (no es el caso específico, que tiene mucho más código), si partimos de la distribución exponencial bivariada #----- theta<-0.5

Hi R-users, Recently, I use maxNR function to find maximizer. I have error appears as follows Error in maxNRCompute(fn = fn, grad = grad, hess = hess, start = start, : NA in the initial gradient My code is mu=2 s=1 n=300 library(maxLik) set.seed(1004) x<-rcauchy(n,mu,s) loglik<-function(mu) { log(prod(dcauchy(x,mu,s))) } maxNR(loglik,start=median(x))$estimate Does anyone know how

Hi, I have the following R-code: pdf(file="test.pdf") hist(rcauchy(100)) dev.off() How can I set the size of the histogram in my pdf? I tried to do with par(...) but I didn't find the right parameter yet. Thanks, Max -- View this message in context: http://www.nabble.com/set-size-of-a-plot-tp22477083p22477083.html Sent from the R help mailing list archive at Nabble.com.

Hola, ¿qué tal? Tengo una pregunta de esta

Colegas: Necesito hacer una serie de ajustes de un modelo de decaimiento exponencial a unos datos de concentración de compuestos fluorescentes contra el tiempo. Para la mayoría de los experimentos tengo tres réplicas por cada tiempo, y haciendo los gráficos correspondientes parece haber diferencias muy grandes en la dispersión de los datos, siendo generalmente mas grandes al principio del

Hola a todos, Soy nuevo en R y estoy intentando modelizar una serie de datos no estacionarios usand la distribucion Generalizada de Valores Extremos GEV. ¿Podriais indicarme como se modeliza una tendencia polinómica (cuadrática, por ejemplo) en alguno de los 3 parámetros (situación, escala o forma)? He encontrado documentación a cerca de modelización linear o exponencial, pero no acabo de

Dear list, I would like to generate random numbers from a Voigt distribution, hopefully in a way as simple as getting random numbers from a normal distribution with 'rnorm'. Is there any package to do this? Speed is an issue in this application. Or, as the Voigt distribution is a convolution of a Gaussian and a Lorentzian, can I simply combine random numbers from rnorm and rcauchy in some

Hello, I would like to make a nonlinar fit (exactly the exponencial fit) to the data. But my data set is not ideal at all, so any robust method (such as LTS) would be bettre then LS. Could you advice me, please, if there is any R package or R function which provides the nonlinear robust regression? Thank you Eva Gelnarova

HI Every one Could some one provide me definitions of following bivariate distributions gamma, exponencial, Weibull, half-normal , Rayleigh, Erlang,chi-square thanks A.S. Qureshi

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Folks, I wonder if anything could be said about the distribution of a random variate x, where x = N(0,1)/N(0,1) Obviously x is pathological because it could be 0/0. If we exclude this point, so the set is {x/(0/0)}, does x have a well defined distribution? or does it exist a distribution that approximates x. (The case could be generalized of course to N(mu1, sigma1)/N(mu2, sigma2) and one

Full_Name: Madeleine Yeh Version: 1.9.1 OS: AIX 5.2 Submission from: (NULL) (151.121.225.1) After compiling R-1.9.1 on AIX 5.2 using the IBM cc compiler, I ran the checks. One of them failed. Here is the output from running the check solo. root@svweb:/fsapps/test/build/R/1.9.1/R-1.9.1/tests/Examples: ># ../../bin/R --vanilla < stats-Ex.R R : Copyright 2004, The R

On Wed, 2008-05-14 at 13:26 -0400, David Vandevoorde wrote: > On May 14, 2008, at 10:46 AM, Jonathan S. Shapiro wrote: > > On Wed, 2008-05-14 at 23:23 +0900, Neil Booth wrote: > >> Jonathan S. Shapiro wrote:- > >>> Is there a requirement somewhere in the C *Language* Specification > >>> that > >>> ties all of this together in the required

On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include

Has anyone got vim to have syntax highlighting with R function codes? I know there's something similar that works with emacs (ESS or something like that), but I was wondering if anyone knew an equivalent that worked with vim. Thank you, -- []'s mentus at gmx.de Bitte l?cheln! Fotogalerie online mit GMX ohne eigene Homepage!

I noticed some very weird behaviour of the function: options(digits=n), where n is the number of digits you would expect to get in R calculations. Let's take a example: > options(digits=4) > getdata(caso.pool.k3.r3.e2) [1] 6.053 2.641 -3.639 14.259 6.082 Which works fine... now, trying again, with different data: > options(digits=4) > getdata(controle.pool.k3.r3.e2)