similar to: How can I estimate parameters of probability distributions?

I am trying to it a particular nonlinear model common in Soil Science to moisture release data from soil. I have written the function as shown below according to the logist example in Ch8 of Pinheiro & Bates. I am getting the following error (R version 2.1.1) *Error in qr(attr(rhs, "gradient")) : NA/NaN/Inf in foreign function call (arg 1)* Below is the function and data. /#

I have a set of data for US t-bill returns and US stock returns frm 1980-2012. I am trying to bootstrap the data and obtain the minimum variance portfolio and repeat this portfolio 1000 times. However I am unable to get the correct code function for the minimum variance portfolio. When I tried to enter Opt(OriData+1, 1, 5, 0), I get "error:subscript out of bounds" Please help!

Hi,all: I met a problem of nls. My data: x y 60 0.8 80 6.5 100 20.5 120 45.9 I want to fit exp curve of data. My code: > nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates I can't find out the reason for the error. Any suggesions are welcome. Many thanks. [[alternative HTML

Hi all, I am a bit mystified by this error message that I get when I try to apply lmer() to a simple dataset with one between factor (age) and one within factor (item): "$ operator is invalid for atomic vectors" I'll just provide the code, because I don't see where the problem is: library(lme4) options(contrasts=c("contr.helmert","contr.poly")) data =

Dear all I am so glad the R can provide the efficient calculate about eigenvector and eigenvalue. However, i have some puzzle about the procedure of eigen. Fristly, what kind of procedue does the R utilize such that the eigen are obtained? For example, A=matrix(c(1,2,4,3),2,2) we can define the eigenvalue lamda, such as det | 1-lamda 4 | =0 | 2 3-lamda | then

Hi, I have been using maxLik to do some MLE of Geometric Brownian Motion Process and everything has been going fine, but know I have tried to do it with jumps. I have create a vector of jumps and then added this into my log-likelihood equation, know I am getting a message: NA in the initial gradient My codes is hear # n<-length(combinedlr) j<-c(1,2,3,4,5,6,7,8,9,10)

Dear All, In Storey's estimator of the proportion of true nulls, the estimator depends on the tuning parameter lamda. Suppose now that an estimator of this proportion has been obtained by the qvalue package, what is the lamda that corresponds to the estimate? How to get this lamda? Thanks, -Chee [[alternative HTML version deleted]]

Hello guys.I am working on implementing the PL2 weighting scheme of the DFR framework by Gianni Amati. It uses the Poisson approximation of the Binomial as the probabilistic model (P), the Laplace law of succession to calculate the after effect of sampling or the risk gain (L) and within document frequency normalization H2(2) (as proposed by Amati in his PHD thesis). The formula for w(t,d) in

Hi I have been having issue with a ML estimator for Jump diffusion process but know I am get little error I didn't notice before like I am try to create a vector > #GBMPJ MLE Combined Ph 1 LR > # > n<-length(combinedlrph1) > j<-c(1,2,3,4,5,6,7,8,9,10) Error in c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) : unused argument(s) (3, 4, 5, 6, 7, 8, 9, 10) >

Hi I?m trying to fit a nonlinear model to a derivative of the logistic function y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls) The derivative calculated with D function is: > logis<- expression(a/(1+exp((b-x)/c))) > D(logis, "x") a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2 So I enter this expression in the nls function:

Hello. Here is my R code. I used the functional data . Now I need to use the functional data by applying the kernels instead of the xi, yi functions. Bonjour. Voici mon code en R . J'ai utiliser les donn?es fonctionnelles . Maintenant j'ai besoin d'utiliser les donn?es fonctionnelles en appliquant les noyaux ? la place des fontions xi, yi library(MASS)

Is there a likelihod function for the Weibull distribution in 'R'? I found the following reference: http://www.weibull.com/LifeDataWeb/weibull_log_likelihood_functions_and_their_partials.htm But I had a hard time understanding the parameters required Particularly 'number of groups of times-to-failure data points", "number of groups of suspension data points", and

Hm. Google tells me that kernel function is in stats package which comes with base installation and is invoked when you start R. search() [1] ".GlobalEnv" "package:stats" "package:graphics" [4] "package:grDevices" "package:utils" "package:datasets" [7] "package:methods" "Autoloads"

Good evening dear administrators, It is with pleasure that I am writing to you to ask for help to finalize my R programming algorithm. Indeed, I attach this note to my code which deals with a case of independence test statistic . My request is to introduce the kernels using the functional data for this same code that I am sending you. So I list the lines for which we need to introduce the

What do you *mean* "when you want to use the kernels". WHICH kernels? Use to do WHAT? In your browser, visit cran.r-project.org then select "Packages" from the list on the left. Then pick the alphabetic list. Now search for 'kernel'. You will find dozens of matches. On Wed, 14 Oct 2020 at 05:15, PIKAL Petr <petr.pikal at precheza.cz> wrote: > Hm. Google tells

Dear Sir/madam, I'm getting a problem with a R-code which calculate Fisher Information Matrix for Hybrid Censored Weibull Distribution. My problem is that: when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives error : Error in integrate(int2, lower = 0, upper = t) : the integral is probably

Hi folks: Does anyone know of a way to do (linear) hypothesis tests of parameters after fitting a maximum-likelihood model w/ optim? I can't seem to find anything like a Wald test whose documentation says it applies to optim output. Also, thanks again to everyone who gave me feedback on the robustness of ML estimation in R! Peter ********************************

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

Unlike L1 (lasso) regression or elastic net (mixture of L1 and L2), L2 norm regression (ridge regression) does not select variables. Selection of variables would not work properly, and it's unclear why you would want to omit "apparently" weak variables anyway. Frank maths123 wrote > I have a .txt file containing a dataset with 500 samples. There are 10 > variables. > >

I was trying to reproduce a result in a published journal, and I have come across some difficulties. I have the following equation, which is two equations combined together. http://r.789695.n4.nabble.com/file/n3006584/Screenshot.png where http://r.789695.n4.nabble.com/file/n3006584/Screenshot-1.png http://r.789695.n4.nabble.com/file/n3006584/Screenshot-2.png