similar to: Minor bug in dlnorm (PR#1781)

Hello, I tried to fit a lognormal distribution by using optim. But sadly the output seems to be incorrect. Who can tell me where the "bug" is? test = rlnorm(100,5,3) logL = function(parm, x,...) -sum(log(dlnorm(x,parm,...))) start = list(meanlog=5, sdlog=3) optim(start,logL,x=test)$par Carsten. [[alternative HTML version deleted]]

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal

I am trying to fit a lognormal distribution to a set of data and test its goodness of fit with regard to predicted values. I managed to get so far: > y <- c(2,6,2,3,6,7,6,10,11,6,12,9,15,11,15,8,9,12,6,5) > library(MASS) > fitdistr(y,"lognormal",start=list(meanlog=0.1,sdlog=0.1)) meanlog sdlog 1.94810515 0.57091032 (0.12765945) (0.09034437) But I would

Hello, I am trying to compute the density of a variable k that is either (1) Normally distributed; (2) Log-Normally distributed; or (3) follows proportional error distribution. I tried to search R-help and the answer for normal distribution was easy to find (please see 1c). I am not sure if my formula for dlnorm is correct (please see 2c below)? I really don't know what function to use for the

Hi R lovers! I'd like to know how to use the parameter 'start' in the function fitdistr() obviously I have to provide the initial value of the parameter to optimize except in the case of a certain set of given distribution Indeed according to the help file for fitdistr " For the following named distributions, reasonable starting values will be computed if `start'

Hi Again, I think I've solved my problem, but please tell me if you think I'm wrong, or you can see a better way! A plot of the integrand showed a very sharp peak, so I was running into the integrand "feature" mentioned in the note. I resolved it by limiting the range of integration as shown here: -------------------------------------------------- function (x, meanlog = 0,

rlnorm takes two 'shaping' parameters: meanlog and sdlog. meanlog would appear from the documentation to be the log of the mean. eg if the desired mean is 1 then meanlog=0. So to generate random values that fit a lognormal distribution I would do this: rlnorm(N , meanlog = log(mean) , sdlog = log(sd)) But when I check the mean I don't get it when sdlog>0. Interestingly I

Hi R lovers Here is a numerical vector test > test [1] 206 53 124 112 92 77 118 75 48 176 90 74 107 126 99 84 114 147 99 114 99 84 99 99 99 99 99 104 1 159 100 53 [33] 132 82 85 106 136 99 110 82 99 99 89 107 99 68 130 99 99 110 99 95 153 93 136 51 103 95 99 72 99 50 110 37 [65] 102 104 92 90 94 99 76 81 109 91 98 96 104 104 93 99 125 89

Hi again, Adding further information to my own query, this function gets to the core of the problem, which I think lies in the behaviour of 'integrate'. ------------------------------------- function (x, meanlog = 0, sdlog = 1, ...) { require(stats) integrand <- function(t, x, meanlog, sdlog) dpois(x,t)*dlnorm(t, meanlog, sdlog) mapply(function(x, meanlog, sdlog, ...) #

Full_Name: Anthony Gichangi Version: 1.90 OS: Windows XP Pro Submission from: (NULL) (130.225.131.206) The function rlnorm generates negative values for lognormal distribution. x- rlnorm(1000, meanlog = 0.6931472, sdlog = 1) Regards Anthony

Hola Ruben, Sí precisamente es lo que comentas, en matemáticas no se suele llamar bucketización (este término se emplea más en informática) sino datos agrupados. Pero la idea es la que tu mismo dices. Respecto a las gráficas que has puesto, me han aclarado mucho sobre el tema, gracias. Si realizo lo mismo, por ejemplo con nbucket=1000 sigo obteniendo un p-valor de 1. Es decir, que casi le

Hi! I have following data which is left truncated say at 10. I am trying to estimate the parameters of the Truncated Lognormal distribution to this data as given below. (I have referred to R code appearing in an earlier post - http://finzi.psych.upenn.edu/Rhelp10/2008-October/176136.html) library(MASS) x <- c(600.62,153.05,70.26,530.42,3440.29,97.45,174.51,168.47, 116.63,36.51, 219.77,

DeaR all, I do not have a clue with is the following NOT working like I expect to do... (and I cannot find any answer at CRAN)... # This one is for my sample > x _ seq(3,10) # This two for parameters > a _ seq(2,4) > b _ seq(2,5) # This one for the likelihood of a sample >f _ function(a,b) { + prod(dlnorm(x,meanlog=a,sdlog=b)) + } > outer(a,b,f) [,1] [,2]

Hi all, this might be a stupid question, but still. Everytime i find some new function it's prettty easy to understand how to use the syntax and to perform a text. Even the general idea of what the function does is pretty easy to understand, but i can not find an explanation (detailed explanation) of the R output for each function. For example, a function fitdistr() in MASS package i

Hello, I have a question regarding the use of an offset term with survreg(), in the Survival library. In particular, I am trying to figure out on what scale the offset term should be. Here's a simple example with no censoring and no coefficients: --------- y = rlnorm(1000, meanlog = 10, sdlog = 2) delta = rep(1, 1000) int = rep(1, 1000) survreg(Surv(y,delta)~offset(10*int), dist =

Dear R-helpers, I was testing the truncgof CRAN package, found something that looked like a bug, and did my job: contacted the maintainer. But he did not reply, so I am resending my query here. I installed package truncgof and run the example for function ad.test. I got the following output: set.seed(123) treshold <- 10 xc <- rlnorm(100, 2, 2) # complete sample xt <- xc[xc >=

Hi, just for the record, although I don't think it's relevant (!) ------------------------------------- > sessionInfo() R version 2.6.0 (2007-10-03) i386-pc-mingw32 locale: LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.1252;LC_MONETARY=English_United Kingdom.1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252 attached base packages: [1] stats4 splines

Hello R-Experts, I want to calculate values like 15^200 or 17^300 in R. In normal case it can calculate the small values of b (a^b). I have fixed width = 10000 and digits = 22 but still answers are Inf. How to deal the cases like these? Thanks in advance. Regards, rehena [[alternative HTML version deleted]]

Tom Cohen <tom.cohen78@yahoo.se> skrev: Thanks Prof Brian for your suggestion. I should know that for right-skewed data, one should generate the samples from a lognormal. My problem is that x and y are two instruments that were thought to be measured the same thing but somehow show a wide confidence interval of the difference between the two intruments.This may be true that these

I am trying to plot a lognormal density curve on top of an existing histogram. Can anybody suggest a simple way to do this? Even if someone could just explain how to plot a regular normal density curve on top of an existing histogram, it would be a big help. Also, is there some way to search through the R-help archives other than simple browsing? Thank you so much. Your help and time is greatly