similar to: Truncated Lognormal Distribution

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

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 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, 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

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 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

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

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

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

I run the following: library(actuar) x <- seq(0, 22, 0.5) fl <- discretize(plnorm(x, 2.1), from = 0, to = 22, step = 0.5, method ="lower") Fs <- aggregateDist("recursive", model.freq = "poisson",model.sev = fl, lambda = 10, x.scale = 0.5) Warning message: In panjer(fx = model.sev, dist = dist, p0 = p0, x.scale = x.scale, : maximum number of recursions

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

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

Dear R-users I've got the next problem: I've got this *function*: fitcond=function(x,densfun,pcorte,start,...){ myfn <- function(parm,x,pcorte,...) -sum(log(dens(parm,x,pcorte,...))) Call <- match.call(expand.dots = TRUE) if (missing(start)) start <- NULL dots <- names(list(...)) dots <- dots[!is.element(dots, c("upper",

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

Hi All, Did anyone else have a problem like this? I am sorry if its a small issue, I seem to not understand what to do to get rid of this error. > Sigma [1] 0.1939025 > MuRest [1] 8.512772 > TauZero [1] 0.1 > curve(qlnorm(x,-TauZero+MuRest, Sigma,lower.tail=F), xlim=c(4000,9000), ylim=c(0,.99),xlab="", ylab="") Warning message: In qlnorm(p, meanlog, sdlog,

Hi, I am trying to learn lognormal mixture models with EM. I was wondering how does one compute the log likelihood. The current implementation I have is as follows, which perform really bad in learning the mixture models. __BEGIN__ # compute probably density of lognormal. dens <- function(lambda, theta, k){ temp<-NULL meanl=theta[1:k] sdl=theta[(k+1):(2*k)]

Hi, I'm playing around with gamlss and don't entirely understand the sigma result from an attempted lognormal fit. In the example below, I've created lognormal data with mu=10 and sigma=2. When I try a gamlss fit, I get an estimated mu=9.947 and sigma=0.69 The mu estimate seems in the ballpark, but sigma is very low. I get similar results on repeated trials and with Normal and

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 am not certain even how to search the archives for this particular question, so if there is an obvious answer, please smack me with a large halibut and send me to the URLs. I have been experimenting with fitting curves by using both maximum likelihood and maximum spacing estimation techniques. Originally, I have been writing distribution-specific functions in 'R' which work