similar to: MLE for two random variables

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]]

Hello, maybe that my Question is a "beginner"-Question, but up to now, my research didn't bring any useful result. I'm trying to fit a distribution (e.g. lognormal) to a given set of data (ML-Estimation). I KNOW about my data that there is a truncation for all data below a well known threshold. Is there an R-solution for an ML-estimation for this kind of data-problem? As

Hi , I am trying to get some estimator based on lognormal distribution when we have left,interval, and right censored data. Since, there is now avalible pakage in R can help me in this, I had to write my own code using Newton Raphson method which requires first and second derivative of log likelihood but my problem after runing the code is the estimators were too high. with this email ,I provide

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 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

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

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

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]]

Hi, I am trying to get the maximum likelihood estimator for lognormal distribution with censored data;when we have left, interval and right censord. I built my code in R, by writing the deriving of log likelihood function and using newton raphson method but my estimators were too high " overestimation", where the values exceed the 1000 in some runing of my code. is there any one can

So I am trying to use the mle command (from stats4 package) to estimate a number of parameters using data but it keeps throwing up this error message: Error in solve.default(oout$hessian) : Lapack routine dgesv: system is exactly singular: U[1,1] = 0 This error sometimes indicates that the list of starting values is too far from optimum but this is unlikely since I picked values close to where

Hello, I want to find the parameters mu and sigma that minimize the following function. It's important, that mu and sigma are strictly positive. ----------------- optimiere = function(fmean,smean,d,x,mu,sigma) { merk = c() for (i in 1:length(d)) merk=c(merk,1/(d[i]^2)*(d[i]-1/(fmean*(1-plnorm(x[i],mu,sigma))))^2) return(sum(merk)) } ----------------- To do that I'm using the nlm

In theory, dlnorm(x, ...) == dnorm(log(x), ...)/x, x>0. Unfortunately, when sd=0, dlnorm and plnorm return NaN, while dnorm returns (if(x != mean)0 else Inf) and pnorm returns (if(x<mean)0 else 1). [A numerical optimization, maxLik{maxLik}, reported the NaNs for me.] help('dlnorm') says, "dlnorm is calculated from the definition (in ?Details?). [pqr]lnorm are based on the

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

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",

I am trying to fit a lognormal distribution on interval-censored data. Some of my intervals have a lower bound of zero. Unfortunately, it seems like survreg() cannot deal with lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate

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, I would like to write a function that finds parameters of a log-normal distribution with a 1-alpha CI of (x_lcl, x_ucl): However, I don't know how to optimize for the two unknown parameters. Here is my unsuccessful attempt to find a lognormal distribution with a 90%CI of 1,20: prior <- function(x_lcl, x_ucl, alpha, mean, var) { a <- (plnorm(x_lcl, mean, var) - (alpha/2))^2 b

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

Dear ALL I have a list of data below 0.80010 0.72299 0.69893 0.99597 0.89200 0.69312 0.73613 1.13559 0.85009 0.85804 0.73324 1.04826 0.84002 1.76330 0.71980 0.89416 0.89450 0.98670 0.83571 0.73833 0.66549 0.93641 0.80418 0.95285 0.76876 0.82588 1.09394 1.00195 1.14976 0.80008 1.11947 1.09484 0.81494 0.68696 0.82364 0.84390 0.71402 0.80293 1.02873 all of them are ninty. Nowaday, i try to find a

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