similar to: Is my understanding of rlnorm correct?

Hi list, Does anyone know the code behind rlnorm(n, meanlog = 0, sdlog = 1)? I am going to write it in c#. thanks Alireza [[alternative HTML version deleted]]

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

Dear all, I have a code that generates data vectors within R. For example assume: z <- rlnorm(1000, meanlog = 0, sdlog = 1) Every time a vector has been generated I would like to export it into a csv file. So my idea is something as follows: for (i in 1:100) { z <- rlnorm(1000, meanlog = 0, sdlog = 1) write.csv(z, "c:/z_i.csv") Where "z_i.csv" is a filename that is

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 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 >=

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 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 =

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 've a problem to have a lognorm distribution with mean=1 and var (or sigma)=1. rlnorm(1000,0,0) rlnorm(1000,1,1) rlnorm(1000,0,1) .... ? Can you help me?

Dear All,   thanks to Berend, my question posted yesturday was solved succesfully here: http://r.789695.n4.nabble.com/hep-on-arithmetic-covariance-conversion-to-log-covariance-td4646068.html . I posted the question with the assumption of using the results with rlnorm.rplus() from compositions. Unfortunatelly, I am not getting reasonable enough outcome. Am I applying the results wrongfully? The

Dear Carlos, Duncan and everyone You may have already sorted the matter by now, but since I have not seen anything posted since Duncan's reply, here I go. I apologize in advance for the spam, if it turns out I've missed some post. I think the test and the implementation of the truncgof package are just fine. I've done Carlos' experiment (repeatedly generating samples and testing

Dear All, I know that this topic has been already discussed on this list (see e.g. http://markmail.org/message/bq2bdxwblwl4rpgf?q=r+fit+truncated+lognormal&page=1&refer=2ufc4fb2eftfwwml#query:r%20fit%20truncated%20lognormal+page:1+mid:7wxgkdxhixotorr5+state:results for the case of weibull distribution), but I am experiencing some problems. I deal with truncated distributions (that this to

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'

The density of a lognormal should be 0 for negative arguments, but > dlnorm(-1) [1] NaN Warning message: NaNs produced in: dlnorm(x, meanlog, sdlog, log) A simple fix is to change dlnorm's definition to: function (x, meanlog = 0, sdlog = 1, log = FALSE) .Internal(dlnorm(x*(x>0), meanlog, sdlog, log)) It might be faster to put the same sort of adjustment into the internal code, but

Dear all, Is there a way to generate K numbers of integer (K = 10^6). The maximum value of the integer is 200,000 and minimum is 1. And the occurrences of this integer follows a lognormal distribution. - Gundala Viswanath Jakarta - Indonesia

Hello, I've read the other posts with regard to "chisq.test" and "goodness of fit" and am still missing something. 1. I create a simple vector of randomly generated lognormal values with mean=0 and sd=1; >d1 <- rlnorm(100,meanlog=0,sdlog=1); 2. I also create a vector of probabilities that are expected for a lognormal distribution. I suspect this is the culprit. >pr

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,

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, ...) #

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