similar to: Supply linear constrain to optimizer

Many thanks to those took time replied to my question. They were very helpful and I solved my problem by reparameterization. With the help of optim() the fitting procedure is very robust and insensitive to initial starting value. Once again, many thanks. Kevin -------- Original post --------- >Dear R and S users, > >I've been working on fitting finite mixture of negative

Hi all; I recently have been used 'maxLik' function for maximizing G2StNV178 function with gradient function gradlik; for receiving this goal, I write the following program; but I have been seen an error  in calling gradient  function; The maxLik function can't enter gradlik function (definition of gradient function); I guess my mistake is in line ******** ,that the vector  ‘h’ is

Hello, The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0 and beta1 via GLM) 2) with lambda, estimate alpha via ML estimation 3) with updataed alpha, replicate 1) and 2) until alpha is converged to a value I coded 1) and 2) (it works), but faced some

Dear R users, i would like to transform the following function from R-code to C-code and call it from R in order to speed up the computation because in my other functions this function is called many times. `dgcpois` <- function(z, lambda1, lambda2) { `f1` <- function(alpha, lambda1, lambda2) return(exp(log(lambda1) * (alpha - 1) - lambda2 * lgamma(alpha))) `f2` <-

How to convert the following ms() in Splus to Optim in R? The "Calc" function is also attached. ms(~ Calc(a.init, B, v, off, d, P.a, lambda.a, P.y, lambda.y, 10^(-8), FALSE, 20, TRUE)$Bic, start = list(lambda.a = 0.5, lambda.y = 240), control = list(maxiter = 10, tol = 0.1)) Calc <- function(A.INIT., X., V., OFF., D., P1., LAMBDA1., P2., LAMBDA2., TOL., MONITOR.,

I use "while" loop but it produces an errro. I have no idea about this. Error in "[<-"(`*tmp*`, i, value = numeric(0)) : nothing to replace with The problem description is The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0

I have an error for a simple optimization problem. Is there anyone knowing about this error? lambda1=-9 lambda2=-6 L<-function(a){ s2i2f<-(exp(-lambda1*(250^a)-lambda2*(275^a-250^a)) -exp(-lambda1*(250^a)-lambda2*(300^a-250^a))) logl<-log(s2i2f) return(-logl)} optim(1,L) Error in optim(1, L) : function cannot be evaluated at initial parameters Thank you in advance -- View this

Hey all, I am trying to work under a State Space form, but I didn't get the help exactly. Have anyone eles used this functions? I was used to work with S-PLUS, but I have some codes I need to adpt. Thanks alot, Bernardo [[alternative HTML version deleted]]

Dear all, I have two functions (f1, f2) and 4 unknown parameters (p1, p2, p3, p4). Both f1 and f2 are functions of p1, p2, and p3, denoted by f1(p1, p2, p3) and f2(p1,p2,p3) respectively. The goal is to maximize f1(p1, p2, p3) subject to two constraints: (1) c = k1*p4/(k1*p4+(1-k1)*f1(p1,p2,p3)), where c and k1 are some known constants (2) p4 = f2(p1, p2, p3) In addition, each parameter

Dear R users, I am using "enet" package in R for applying "elastic net" method. In elastic net, two penalities are applied one is lambda1 for LASSO and lambda2 for ridge ( zou, 2005) penalty. But while running the analysis, I realised tht, I optimised only one lambda. ( even when I looked at the example in R, they used only one penality) So, I am

Hi, I have reconstructed ancestral character states on a phylogeny using MuSSE in the diversitree package and plotted the character state probabilities as pie charts on the nodes. I would, however, like to colour the character states of my extant species, i.e. the tip labels, the same colours as my pie charts, such that all species in state 1 are e.g. blue, species in state 2 red and species in

The following R code using different packages gives the same results for a simple logistic regression without regularization, but different results with regularization. This may just be a matter of different scaling of the regularization parameters, but if anyone familiar with these packages has insight into why the results differ, I'd appreciate hearing about it. I'm new to

Any one know is there any package or function to generate bivariate exponential distribution? I gusee there should be three parameters, two rate parameters and one correlation parameter. I just did not find any function available on R. Any suggestion is appreciated. -- View this message in context:

Hi - I have a question about the following code from Bayesian Computation with R (Jim Albert). par(mfrow=c(2,2)) m = 500 alphas = c(5, 20, 80, 400) for (j in 1:4) { mu = rgamma(m, shape=10, rate=10) lambda1 = rgamma(m, shape=alphas[j], rate=alphas[j]/mu) lambda2 = rgamma(m, shape=alphas[j], rate=alphas[j]/mu) plot(lambda1, lambda2) title(main=paste('alpha=',

On Fri, 25 Aug 2023 22:17:05 +0530 ASHLIN VARKEY <ashlinvarkey at gmail.com> wrote: > Sir, Please note that r-help is a mailing list, not a knight! ?? > I want to solve the equation Q(u)=mean, where Q(u) represents the > quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is > the quantile function of Davies (Power-pareto) distribution. Hence I > want to

Dear All The problem is about regularization methods in multiple regression when the independent variables are collinear. A modified regularization method with two tuning parameters l1 and l2 and their product l1*l2 (Lambda 1 and Lambda 2) such that l1 takes care of ridge property and l2 takes care of LASSO property is proposed The proposed method is given

Hi All, I am using the package 'penalized' to perform a multiple regression on a dataset of 33 samples and 9 explanatory variables. The analysis appears to have performed as outlined and I have ended up with 4 explanatory variables and their respective regression coefficients. What I am struggling to understand is where do I get the variance explained information from and how do I

i want to estimate parameters with maximum likelihood method with contraints (contant numbers). for example sum(Ai)=0 and sum(Bi)=0 i have done it without the constraints but i realised that i have to use the contraints. Without constraints(just a part-not complete): skellamreg_LL=function(parameters,z,design) { n=length(z); mu=parameters[1]; H=parameters[2]; Apar=parameters[3:10];

Dear R users,? ? ? ? ? ? ? ? ? ? ? ? ? I am using "Glmnet" package in R for applying "elastic net" method. In elastic net, two penalities are applied one is lambda1 for?LASSO and lambda2 for ridge ( zou, 2005) penalty.?How can I? write the code to? pre-chose the? lambda1 for?LASSO and lambda2 for ridge without using cross-validation Thanks in advance? Tayo? [[alternative

Hi all, I am attempting to fit, and test the goodness of fit of, a mixed exponential distribution to my dataset which consists of 15minute rainfall intensity data. FYI, the dataset spanning approx.2 years and 7 rainfall stations consists of some three hundred thousand 15min data records, of which some 30 thousand are non-zero rainfall amounts. Could anyone please tell me how i could do