similar to: In optim() function, second parameter in par() missing

Dear useRs, I have a complicated function to be optimized with optim(), and whose parameters are passed to another function within its evaluation. This function allows for the parameters to enter as arguments to various probability distribution functions. However, I am violating some scoping convention, as somewhere within the hierarchy of calls a variable is not visible. I will give a

Dear R user I follow the steps defined in Modern applied statistics page(453) to use optim. However, when I run the following code the parameters seems way off and the third parameter(p3) stayed as the initial value. below is the code: ## data da=c(418,401,416,360,411,425,537,379,484,388,486,380,394,363,405,383,392,363,398,526) ### initial values pars=c(392.25, 507.25, 0.80)

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over

As part of carrying out a complicated maximum likelihood estimation, I am trying to learn to program likelihoods in R. I started with a simple probit model but am unable to get the code to work. Any help or suggestions are most welcome. I give my code below: ************************************ mlogl <- function(mu, y, X) { n <- nrow(X) zeta <- X%*%mu llik <- 0 for (i in 1:n) { if

I believe this may be more related to analysis than it is to R, per se. Suppose I have the following function that I wish to integrate: ff <- function(x) pnorm((x - m)/sigma) * dnorm(x, observed, sigma) Then, given the parameters: mu <- 300 sigma <- 50 m <- 250 target <- 200 sigma_i <- 50 I can use the function integrate as: > integrate(ff, lower= -Inf, upper=target)

Hi May you help me correct my loop function. I want optim to estimates al_j; au_j; sigma_j; b_j by looking at 0 to 20, 21 to 40, 41 to 60 data points. The final result should have 4 columns of each of the estimates AND 4 rows of each of 0 to 20, 21 to 40, 41 to 60. ###MY code is n=20 runs=4 out=matrix(0,nrow=runs) llik = function(x) { al_j=x[1]; au_j=x[2]; sigma_j=x[3]; b_j=x[4]

I am trying to bootstrap the parameters for a model that is estimated through the optim() function and find that when I make the call to boot, it runs but returns the exact same estimate for all of the bootstrap estimates. I managed to replicate the same problem using a glm() model but was able to fix it when I made a call to the variables as data frame by their exact names. But no matter how I

I have had some weird results using the optim() function. I wrote a probit likelihood and wanted to run it with optim() with simulated data. I did not include a gradient at first and found that optim() would not even iterate using BFGS and would only occasionally work using SANN. I programmed in the gradient and it iterates fine but the estimates it returns are wrong. The simulated data work

> version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 1 minor 7.1 year 2003 month 06 day 16 language R Bug: integrate(f,lower,upper,extra_args) where f <- function(x,extra_args) { body } integrate doesn't pass the extra arguments when calling f. As a first check of this finding I integrated dnorm from

[Yes, this is related to a homework problem, but is not the problems itself.] In my mathematical statistics class, we've just learned about properties of estimators, and I can now solve manually problems like this: A sample of size n = 16 is drawn from a normal distribution where sigma = 10 but mu is unknown. If mu = 20, what is the probability that the estimator mu hat = Y bar will lie

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))

Hi I have developed the code below. I am worried that the parameters I want to be estimated are "not being found" when I ran my code. Is there a way I can code them so that R recognize that they should be estimated. This is the error I am getting. > out1=optim(llik,par=start.par) Error in pnorm(au_j, mean = b_j * R_m, sd = sigma_j) : object 'au_j' not found #Yet

I thought the curve function was a very flexible way to draw functions. So I could plot funtions like the following: # I created a function to produce functions, for instance: fp <- function(m,b) function(x) sin(x) + m*x + b # So I can produce a function like this ff <- fp(-0.08, 0.2) ff(1.5) # Is the same as executing sin(1.5) - 0.08*1.5 + 0.2 # Let's plot this

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)

Hello, First time poster here so let me know if you need any more information. I am trying to run an ordered probit with maximum likelihood model in R with a very simple model (model <- econ3 ~ partyid). Everything looks ok until i try to run the optim() command and that's when I get " Error in rep(1, nrow(x)) : invalid 'times' argument". I had to adapt the code from a 4

Here's another example of my plotmath whipping boy, the Normal distribution. A colleague asks for a Normal plotted above a series of axes that represent various other distributions (T, etc). I want to use vectors of equations in plotmath to do this, but have run into trouble. Now I've isolated the problem down to a relatively small piece of working example code (below). If you would

Dear list, i'm trying to test if a linear combination of coefficients of glm is equal to 0. For example : class 'cl' has 3 levels (1,2,3) and 'y' is a response variable. We want to test H0: mu1 + mu2 - mu3 =0 where mu1,mu2, and mu3 are the means for each level. for me, the question is how to get the covariance matrix of the estimated parameters from glm. but perhaps there

Hi, I'm trying to evaluate a character vector within pnorm. I have a vector with values and names x = c(2,3) names(x) = c("mean", "sd") so that i tried the following temp = paste(names(x), x, sep = "=") #gives #> temp #[1] "mean=2" "sd=3" #Problem is that both values 2 and 3 are taken as values for the mean argument in pnorm pnorm(0,

Hi all. I may be wrong, (and often am), but in trying to determine how to calculate the erf function, the documentation for 'pnorm' states: ## if you want the so-called 'error function' erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1 ## and the so-called 'complementary error function' erfc <- function(x) 2 * pnorm(x * sqrt(2), lower=FALSE) Should, instead, it read:

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value