similar to: fitting distributions using fitdistr (MASS)

Good-day Sir, I am R.Language users but am try to? estimate parameter of beta distribution particular dataset but give this error, which is not clear to me: (Initial value in "vmmin" is not finite) beta.fit <- fitdistr(data,densfun=dbeta,shape1=value , shape2=value) kindly assist. expecting your reply:

I have the following problem: I have a vector x of data (0<x<=1 ) with a U-shaped histogram and try to fit a beta distribution using fitdistr. In fact, hist(rbeta(100,0.1,0.1)) looks a lot like my data. The equivalent to the example in the manual sometimes work: > a <- rbeta(100,0.1,0.1) > fitdistr(x=a, "beta", start=list(shape1=0.1,shape2=0.1))1) > shape1

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

When using fitdistr() with the exponential, log-normal and beta distributions, you get the relevent rate, mean, standard deviation, shape1 and shape2 but you get a number bellow those that are in () and I was wandering what exactly those numbers represent and how they relate to the data. Many thanks Mark Miller

This may be a begining question. If so, please bear with me. If I have some data that based on the historgram and other plots it "looks" like a beta distribution. Is there a function or functions within R to help me determine the model parameters for such a distirbution? Similarily for other "common" distirbutions, Poisson(lambda), Chi-Square(degrees of freedom, chi-square

Hi, I want to fit 3 beta distributions to my data which ranges between 0 and 1. What are the functions that I can easily call and specify that 3 beta distributions should be fitted? I have already looked at normalmixEM and fitdistr but they dont seem to be applicable (normalmixEM is only for fitting normal dist and fitdistr will only fit 1 distribution, not 3). Is that right? Also, my data has 26

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

Dear All, I need to fit a custom probability density (based on the symmetric beta distribution B(shape, shape), where the two parameters shape1 and shape2 are identical) to my data. The trouble is that I experience some problems also when dealing with the plain vanilla symmetric beta distribution. Please consider the code at the end of the email. In the code, dbeta1 is the density of the beta

I answer my own question: I had overlooked the fact that the normalization factor is also a function of the parameters I want to optimise, hence I should write dbeta2 <- function(x, shape){ res <- x^(shape-1)*(1-x)^(shape-1)/beta(shape, shape) return(res) } after which the results are consistent. ---------- Forwarded message ---------- From: Lorenzo Isella <lorenzo.isella

Dear R users, first I take this opportunity to greet all the R community for your continuous efforts. I wrote a function to calculate the pdf and cdf of a custom distribution (mixed gamma model). The function is the following: pmixedgamma3 <- function(y, shape1, rate1, shape2, rate2, prev) { density.function<-function(x) { shape1=shape1 rate1=rate1 shape2=shape2

Buenos dias! Os escribo para ver si me podeis ayudar con un asunto en el que me he quedado un poco encallado. Lo que tengo que hacer es sacar los percentiles (0.001, 0.005, 0.95 y 0.999) de varias distribuciones beta, concretamente 418. Cada distribucion esta definida por los parametros "shape1" y "shape2". Por lo tanto tengo una base de datos de 418 filas y en cada una de

Hi folks, I have a question about how to efficiently produce random numbers from Beta and Binomial distributions. For Beta distribution, suppose we have two shape vectors shape1 and shape2. I hope to generate a 10000 x 2 matrix X whose i th rwo is a sample from reta(2,shape1[i]mshape2[i]). Of course this can be done via loops: for(i in 1:10000) { X[i,]=rbeta(2,shape1[i],shape2[i]) } However,

I've discovered an infelicity (I guess) in qbeta(): it's not a bug, since there's a clear warning about lack of convergence of the numerical algorithm ("full precision may not have been achieved"). I can work around this, but I'm curious why it happens and whether there's a better workaround -- it doesn't seem to be in a particularly extreme corner of parameter

Hi , I am new to R. I am trying to run a simple R script as shown below: aov.R ------ data1<-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40 ,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40) matrix(data1, ncol= 4, dimnames = list(paste("subj", 1:12), c("Shape1.Color1", "Shape2.Color1", "Shape1.Color2",

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

Hallo, I need your help. I fitted my distribution of data with beta-prime, I need now to plot the Cumulative distribution. For other distribution like Gamma is easy: x <- seq (0, 100, 0.5) plot(x,pgamma(x, shape, scale), type= "l", col="red") but what about beta-prime? In R it exists only pbeta which is intended only for the beta distribution (not beta-prime) This is

Dear R-Developers, I just started learning how to use Rcpp. Earlier while using it, I encountered an error as shown below: file74d8254b96d4.cpp: In function ‘Rcpp::NumericVector foo(Rcpp::NumericVector, Rcpp::NumericVector, Rcpp::NumericVector, Rcpp::Function, Rcpp::Function)’: file74d8254b96d4.cpp:10: error: invalid operands of types ‘SEXPREC*’ and ‘R_len_t’ to binary ‘operator/’ make: ***

Hi, I noticed this problem on my home desktop running FC4 and again on my laptop running FC5. Both have previously compiled and passed make check-all on 2.3.1 svn revisions from 10 days ago or so. On both these machines, make check-all is consistently failing (4 out of 4 attempts on the FC 4 desktop and 3 out of 3 on the FC 5 laptop) in the p-r-random-tests tests. This is with both default

Hi R users: I'm trying to make a function where two of the parameters are functions, but I don't know how to put each set of parameters for each function. What am I missing? I try this code: f2<-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){ force(fun1) force(fun2) force(n) p1<-unlist(par1) p2<-unlist(par2) force(p1) force(p2)

>>>>> Ben Bolker >>>>> on Wed, 25 Mar 2020 21:09:16 -0400 writes: > I've discovered an infelicity (I guess) in qbeta(): it's not a bug, > since there's a clear warning about lack of convergence of the numerical > algorithm ("full precision may not have been achieved"). I can work > around this, but I'm