search for: mylogfn

I am using the MASS library function fitdistr(x, dpois, list(lambda=2)) but I get Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : Function cannot be evaluated at initial parameters In addition: There were 50 or more warnings (use warnings() to see the first 50) and all the first 50 warnings say 1: non-integer x = 1.452222 etc Can anyone tell me what I am doing wrong. p.s. the data...

...erence value [1] In addition: There were 50 or more warnings (use warnings() to see the first 50) #with fitdistr() for the exponentialdistribution library(MASS) fitdistr(data2,densfun=dexp,start=list(rate=0.1),lower=6e-06,method="BFGS") #instead of a result, i get Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : ?? ?? ?? ?? non-finite finite-difference value [1] In addition: Warning messages: 1: bounds can only be used with method L-BFGS-B in: optim(start, mylogfn, x = x, hessian = TRUE, ...) 2: NaNs produced in: dexp(x, 1/rate, log) i'll be very happy for any help i c...

...le, log) 3: NaNs produced in: dgamma(x, shape, scale, log) > fitdistr(x, dgamma, list(shape = 1, rate = 0.1), lower = 0.01) shape rate 2.156761461 0.010000000 (0.277607888) (0.001433778) Warning message: bounds can only be used with method L-BFGS-B in: optim(start, mylogfn, x = x, hessian = TRUE, ...) > fitdistr(x, dgamma, list(shape = 1, rate = 0.1), lower = 0.001) shape rate 6.48686551 0.13651012 (0.89438949) (0.01956818) Warning message: bounds can only be used with method L-BFGS-B in: optim(start, mylogfn, x = x, hessian = TRUE, ......

Why the warning messages (2:4)? > x <- rexp(1000,0.2) > fitdistr(x,"exponential",list(rate=1)) rate 0.219824219 (0.006951308) Warning messages: 1: one-diml optimization by Nelder-Mead is unreliable: use optimize in: optim(start, mylogfn, x = x, hessian = TRUE, ...) 2: NaNs produced in: dexp(x, 1/rate, log) 3: NaNs produced in: dexp(x, 1/rate, log) 4: NaNs produced in: dexp(x, 1/rate, log) and with respect to this function, it is curious that only for the normal family the mle is obtained in closed form. Why not in other cases...

Dear R Users, I am trying to obtain a best-fit analytic distribution for a dataset with 11535459 entries. The data range in value from 1 to 300000000. I use: fitdistr(data, "gamma") to obtain mle's for the parameters. I get the following error: Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : non-finite finite-difference value [1] And the following warnings: NaNs produced in: dgamma(x, shape, scale, log) I have the same problem with the exponential distribution, but the lognormal and weibull distributions don't have this problem. I suspect...

Hi all, I'm using the function "fitdistr" (library MASS) to fit a distribution to given data. What I have to do further, is getting the log-Likelihood-Value from this estimation. Is there any simple possibility to realize it? Regards, Carsten

...=a, "beta", start=list(shape1=0.1,shape2=0.1))1) > shape1 shape2 0.09444627 0.12048753 (0.01120670) (0.01550129) but sometimes does not: > a <- rbeta(100,0.1,0.1) > fitdistr(x=a, "beta", start=list(shape1=0.1,shape2=0.1))1) > Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : Function cannot be evaluated at initial parameters Unfortunately, my data fall in the second case I've searched for any weird value that be present in the cases in which fitdistr exits with the error message, but could not find any. Any help? (please if...

Dear all, I get the following error message. And I cannot quite work out what is wrong. I think the optim gets infinite values. Certainly my data do not have any infinite values. How can I solve this? fitdistr(A1, "weibull") Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : non-finite value supplied by optim I am using R version 1.9.1 on RedHat Linux, Kernel 2.6.8. Ulrich

Hello, I want to estimate the parameters of a binomial distributed rv using MLE. Other distributions will follow. The equation system to solve is not very complex, but I've never done such work in R and don't have any idea how to start... The system is: (1) n*P = X (2) [sum {from j=0 to J-1} Y{j} /(n-j)] = -n * ln (1-X / n) where * only X is given (empirical mean)