similar to: Maximizing a function - optim does not always converge

Hi All, This is not really an R question but a statistical one. If someone could either give me the brief explanation or point me to a reference that might help, I'd appreciate it. I want to estimate the mean of a log-normal distribution, given the (log scale normal) parameters mu and sigma squared (sigma2). I understood this should simply be: exp(mu + sigma2) ... but I the following code

Hi I am beginner with R I would like to compare the performance of Maximum likelihood Estimator (MLE) and Method of Moment Estimator (MME) effect probability of rejection, p. My MLE is x=rlnorm(10,meanlog = 2, sdlog =5) x xbar=mean(x) ssqrt=var(x) xbar ssqrt #MLE y=log(x) m1=mean(y) s1=var(y) m1 s1 #MME m2=log(xbar^2/sqrt(ssqrt+xbar^2)) s2 = log((ssqrt+xbar^2)/xbar^2) m2

Hello, I am facing a problem with optimization in R from 2-3 weeks. I have some Gaussian mixtures parameters and I want to find the maximum in that *Parameters are in the form * mean1 mean2 mean3 sigma1 sigma2 sigma3 c1 c2 c3 506.8644 672.8448 829.902 61.02859 9.149168 74.84682 0.1241933 0.6329082 0.2428986 I have used optima and optimx to find the

Hello, I am struggling for some time now to estimate AR(1) process for commodity price time series. I did it in STATA but cannot get a result in R. The equation I want to estimate is: p(t)=a+b*p(t-1)+error Using STATA I get 0.92 for a, and 0.73 for b. Code that I use in R is: p<-matrix(data$p) # price at time t lp<-cbind(1,data$lp) # price at time t-1

The background: I'm trying to fit a Poisson-lognormal distrbutuion to some data. This is a way of modelling species abundances: N ~ Pois(lam) log(lam) ~ N(mu, sigma2) The number of individuals are Poisson distributed with an abundance drawn from a log-normal distrbution. To fit this to data, I need to integrate out lam. In principle, I can do it this way: PLN1 <- function(lam, Count,

Hi, there I'm trying to plot what is returned from a call to tapply, and can't figure out how to do it. My guess is that it has something to do with the inclusion of row names when you ask for the values you're interested in, but if anyone has any ideas on how to get it to work, that would be stellar. Here's some example code: y1<-rnorm(40, 2) x1<-rep(1:2, each=20)

I am having issues with the following: (muhat = 1/n^2(sum of all the xi's) ) essentially if xbar = the sample mean, muhat = sample mean but square the n. Question: Use R to run a Monte Carlo simulation which compares the finite-sample performance of xbar and muhat. Specifically generate 1000 samples n=30 from a standard normal distribution. For each sample calculate xbar and muhat. I have

Hi. I am trying to analyze with BRugs the Box-Tiao variance components example in WinBUGS. The output from BRugs, mean sd MC_error val2.5pc median val97.5pc start sample sigma2.btw 681.9 1161 10.89 0.7016 253.8 4232 25001 100000 sigma2.with 4266.0 1246 4.92 2480.0000 4057.0 7262 25001 100000 doesn't match the output from WinBUGS, node mean

?optim says, in describing the control parameter, 'fnscale' An overall scaling to be applied to the value of 'fn' and 'gr' during optimization. If negative, turns the problem into a maximization problem. Optimization is performed on 'fn(par)/fnscale'. 'parscale' A vector of scaling values for the parameters.

Dear users, I want to compute r-squared values from a glm regression using a gamma distribution and an "identity" link-function, but find no such thing when using the summary() or names() function. My next guess was to calculate it by "hand", i.e. r2 = (sum((estimate - xbar)^2) /sum((x-xbar)^2)) but I am unsure if this is even allowed... Chris -- View this message in

Dear Bill, All makes perfect sense (including the late evaluation). I actually discovered the problem by looking at old code which used your proposed solution. Still I find it strange (and, hnestly, I don?t like R?s behavior in this respect), and I am wondering why u is not being copied to L just before u is assigned a new value. Of course, this would require the R interpreter to track all these

Hello, I have written a function that demonstrates the CLT by generating samples following the exponential distribution, calculating the means, plotting the histogram, and drawing the limiting normal curve as an overlay. I have the title of each histogram state the sample size and rate (1/theta) for the exponential (the output is actually 4 histograms), but I can't get the greek letter theta

Hello, One way of preventing that is to use ?force. Just put force(l) right after the commented out print and before you change 'u'. Hope this helps, Rui Barradas Citando Matthias Gondan <matthias-gondan at gmx.de>: > Dear R developers, > > sessionInfo() below > > Please have a look at the following two versions of the same function: > > 1. Intended

Another way to avoid the problem is to not redefine variables that are arguments. E.g., > Su3 <- function(u=100, l=u, mu=0.53, sigma2=4.3^2, verbose) { if (verbose) { print(c(u, l, mu)) } uNormalized <- u/sqrt(sigma2) lNormalized <- l/sqrt(sigma2) muNormalized <- mu/sqrt(sigma2) c(uNormalized, lNormalized, muNormalized) } > Su3(verbose=TRUE)

Mathias, If it's any comfort, I appreciated the example; 'expected' behaviour maybe, but a very nice example for staff/student training! S Ellison > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Matthias > Gondan > Sent: 02 September 2017 18:22 > To: r-help at r-project.org > Subject: [R] Strange lazy evaluation of

Hi everyone, I am looking to do some manual maximum likelihood estimation in R. I have done a lot of work in Stata and so I have been using output comparisons to get a handle on what is happening. I estimated a simple linear model in R with lm() and also my own maximum likelihood program. I then compared the output with Stata. Two things jumped out at me. Firstly, in Stata my coefficient

When I saw the subject of the original message on R-help, I was 95% confident that I knew the answer (before I had seen the question). This made me think that perhaps for some functions there should be a 'Troubleshooting' section in the help file. The current help file for 'integrate' does say, as Sundar points out, what the requirements are. However, I think more people would

I am new to R. I want to graph group data using a "Traditional Bar Chart with Standard Error Bar", like the kind shown here: http://samiam.colorado.edu/~mcclella/ftep/twoGroups/twoGroupGraphs.html Is there a simple way to do this? So far, I have only figured out how to plot the bars using barplot. testdata <- scan(, list(group=0,xbar=0,se=0)) 400 0.36038 0.02154 200 0.35927

Dear R-help, I'm using the 'optim' functions to minimise functions, and have read the documentation, but I'm still not sure how to determine sensible values to use for the 'fnscale' and 'parscale' options. If I have understood everything correctly, 'fnscale' should be used to scale the objective function, so that for example if the default is

Moved to R-help because there's no obvious financial content. Michael On Sat, Apr 6, 2013 at 10:56 AM, Stat Tistician <statisticiangermany at gmail.com> wrote: > Hi, > I want to implement the EM algorithm manually, with my own loops and so. > Afterwards, I want to compare it to the normalmixEM output of mixtools > package. > > Since the notation is very advanced, I