similar to: Beta distribution- help needed

Hi, sorry for repeating the question but this is kind of important to me and i don't know whom should i ask. So as noted before when I do a parameter fit to the beta distr i get: fitdist(vectNorm,"beta"); Fitting of the distribution ' beta ' by maximum likelihood Parameters: estimate Std. Error shape1 2.148779 0.1458042 shape2 810.067515 61.8608126 Warning

Dear all, I need to plot an cumulative distribution plot of a variable and then to fit a distribution to that, probably a weibull or lognormal. I have plotted the ecdf as > plot(ecdf(x)) but I haven't managed to fit the distribution. I have as well attached the data. I would appreciate if you could help me on that. Thank you. Kind regards Maria -------------- next part --------------

Hi, Interpretation problem ! so what i did is by using the: >fit1 <- fitdist(vectNorm,"beta") Warning messages: 1: In dbeta(x, shape1, shape2, log) : NaNs produced 2: In dbeta(x, shape1, shape2, log) : NaNs produced 3: In dbeta(x, shape1, shape2, log) : NaNs produced 4: In dbeta(x, shape1, shape2, log) : NaNs produced 5: In dbeta(x, shape1, shape2, log) : NaNs produced 6: In

Hi, I am trying fit certain data into Beta distribution. I get the error saying "Error in fitdist(discrete_random_variable_c, "beta", start = NULL, fix.arg = NULL) : the function mle failed to estimate the parameters, with the error code 100" Below is the sorted data that I am trying to fit. Where am I going wrong. Thanks a lot for any help. Vinod

>From the documentation I have found, it seems that one of the functions from package plyr, or a combination of functions like split and lapply would allow me to have a really short R script to analyze all my data (I have reduced it to a couple hundred thousand records with about half a dozen records. I get the same result from ddply and split/lapply: >

Dear Sirs, The below listed code fits a gamma and a pareto distribution to a data set danishuni. However the distributions are not appropriate to fit both tails of the data set hence a mixed distribution is required which has ben defined as "mixgampar" as shown below. library(fitdistrplus) x<- danishuni$Loss fgam<- fitdist(x,"gamma",lower=0) fpar<-

Hi, I am trying to estiamte parameters for gamma distribution using mle for below data using fitdist & fitdistr functions which are from "fitdistrplus" & "MASS"packages . I am getting errors for both functions. Can someone please let me know how to overcome this issue?? data y1<- c(256656, 76376, 6467673, 46446, 3400, 3100, 5760, 4562, 8000, 512, 4545, 4562,

Hi all, I'm currently working on the fitdistrplus package (that basically fit distributions). There is something I do not understand about the generic function summary. In the current version on CRAN, there is no NAMESPACE saying S3method(summary, fitdist) . However if we use summary on an object send by fitdist function it works fine... According to R-lang, we have " The most

Capture the results of the apply command into an object and then work with that. Here is one way to do it: > res <- apply(C, 2, fitdist, "gamma") > out <- c( res$A$estimate["shape"], res$B$estimate["shape"], res$A$estimate["rate"], res$B$estimate["rate"]) > names(out) <- c("A shape","B shape","A

Hi, I am trying to fit a Johnson SB distribution using fitdist function in fitdistrplus Library. I have defined the Johnson SB distribution from ( http://www.ntrand.com/johnson-sb-distribution/) . But it gives me the follwing errors. Any help would be appreciated #xi = xi #lambda =l #delta =d #gamma = g djohn = function(x,xi,l,d,g) (d/(l*sqrt(2*pi)*((x-xi)/l)*(1-((x-xi)/l))))*exp[-0.5*(g +

Hi, Let say I have data by two columns A and B, and I have fit each column using the gamma distribution by 'fitdist' . I just want the result show only the shape and rate only. Eg: library(fitdistrplus) A <-c(1,2,3,4,5) B<-c(6,7,8,9,10) C <-cbind(A,B) apply(C, 2, fitdist, "gamma") Output show like this: $A Fitting of the distribution ' gamma ' by maximum

The data.frame is constructed by one of the following functions: funweek <- function(df) if (length(df$elapsed_time) > 5) { rv = fitdist(df$elapsed_time,"exp") rv$year = df$sale_year[1] rv$sample = df$sale_week[1] rv$granularity = "week" rv } funmonth <- function(df) if (length(df$elapsed_time) > 5) { rv =

I must have missed something simple, but still, i don't know what. I obtained my basic data as follows: x <- sprintf("SELECT m_id,sale_date,YEAR(sale_date) AS sale_year,WEEK(sale_date) AS sale_week,return_type,0.0001 + DATEDIFF(return_date,sale_date) AS elapsed_time FROM `merchants2`.`risk_input` WHERE DATEDIFF(return_date,sale_date) IS NOT NULL") moreinfo <- dbGetQuery(con,

Hello everyone, I was trying to do some distribution fitting with a numerical field called Tolls. The sample size = 999 rows. Basically I assigned the Toll data to a new variable K by doing: k<-dtest$Toll After that, tried to fit a gamma distribution by doing: fitG<-fitdist(k, "gamma") Then the following messages showed (oh and I checked for empty rows before doing this):

Dear all, I would like to use the following function fitdist(data, distr, method=c("mle", "mme", "qme", "mge"), start=NULL, fix.arg=NULL, ...) for many different distr values like distr=c("norm","lnorm","pois") (just a small example) and take back into a list the parameter name which is what is inside distr plus what the

Hello, I have detailed (with pictures and whatnot) my question on my weblog at http://www.cs-ed.org/blogs/mjadud/archives/2005/05/a_question_abou.html The short version of the question is this: When I ask 'fitdistr' to try and fit my distribution as a "weibull" distribution, it comes up with some rather wacky parameters. If I take the same distribution, and do something

Hi Folks! Just started using the gamlss package and I tried a simple code example (see below). Why the negative sigma? John > y <- rt(100, df=1)> m1<-fitDist(y, type="realline")Warning messages:1: In MLE(ll3, start = list(eta.mu = eta.mu, eta.sigma = eta.sigma, : possible convergence problem: optim gave code=1 false convergence (8)2: In MLE(ll4, start = list(eta.mu =

Hi there, I have fitted a sample (with size 20) to a normal and/or logistic distribution using fitdistr() in MASS or fitdist() in fitdistrplus package. It's easy to get the parameter estimates. Now, I hope to report the confidence interval for those parameter estimates. However, I don't find a function that could give the confidence interval in R. I hope to write a function, however,

Hi all, I am trying to fit a distribution to some data about survival times. I am interested only in a specific interval, e.g., while the data lies in the interval (0,...., 600), I want the best for the interval (0,..., 24). I have tried both fitdistr (MASS package) and fitdist (from the fitdistrplus package), but I could not get them working, e.g. fitdistr(left, "weibull", upper=24)

Hi, I've got an ugly but fairly simple function: mdevstdev <- function(a){ l <- dnorm(a)/(1-pnorm(a)) integrand <- function(z)(abs(z-l)*dnorm(z)) inted <- integrate(integrand, a, Inf) inted[[1]]/((1- pnorm(a))*sqrt((1 + a*l - l^2))) } I wanted to quickly produce a graph of this over the range [-3,3] so I used: plotit <-function(x=seq(-3,3,0.01),...){