similar to: help in M-estimator by R

You need to pay attention to the documentation more closely. If you don't know what something means, that is usually a signal that you need to study more... in this case about the difference between an input variable and a design (model) matrix. This is a concept from the standard linear algebra formulation for regression equations. (Note that I have never used RobPer, nor do I regularly

R-experts, I have fitted many different lines. The fast-tau estimator (yellow line) seems strange to me?because this yellow line is not at all in agreement with the other lines (reverse slope, I mean the yellow line has a positive slope and the other ones have negative slope). Is there something wrong in my R code ? Is it because the Y variable is 1 vector and should be a matrix ? Here is the

Dear R-experts, Here below my reproducible R code. I want to add many straight lines to a plot using "abline" The last fit (fast Tau-estimator, color yellow) will not appear on the plot. What is going wrong ? Many thanks for your reply. ########## Y=c(2,4,5,4,3,4,2,3,56,5,4,3,4,5,6,5,4,5,34,21,12,13,12,8,9,7,43,12,19,21)

On 31/03/2018 11:57 AM, varin sacha via R-help wrote: > Dear R-experts, > > Here below my reproducible R code. I want to add many straight lines to a plot using "abline" > The last fit (fast Tau-estimator, color yellow) will not appear on the plot. What is going wrong ? > Many thanks for your reply. > It's not quite reproducible: you forgot the line to create

Dear R-experts, I am trying to do cross-validation for different models using the cvTools package. I can't get the CV for the "FastTau" and "hbrfit". I guess I have to write my own functions at least for hbrfit. What is going wrong with FastTau ? Here below the reproducible example. It is a simple toy example (not my real dataset) with many warnings, what is important to

Dear R-experts, I guess I have a problem with my fast function (fast tau estimator) here below. Indeed, zero errors look highly suspicious. I guess there is a bug in my R code. How could I correct my R code ? # install.packages( "robustbase" ) # install.packages( "MASS" ) # install.packages( "quantreg" ) # install.packages( "RobPer" ) #

Hi all, I'm trying to fit a nonlinear regression by "nlrob": model3=nlrob(y~a1*x^a2,data=transient,psi=psi.bisquare, start=list(a1=0.02,a2=0.7),maxit=1000) However an error message keeps popping up saying that the function psi.bisquare doesn't exist. I also tried psi.huber, which is supposed to be the default for nlrob: model3=nlrob(y~a1*x^a2,data=transient,psi=psi.huber,

How do we get 2-tailed p-values for the rlm summary? I'm using the following: > fit <- rlm(oatRT ~ oatoacData$erp, psi=psi.bisquare, maxit=100, na.action='na.omit') > fitsum <- summary(fit, cor=F) > print(fitsum) Call: rlm(formula = oatRT ~ oatoacData$erp, psi = psi.bisquare, maxit = 100, na.action = "na.omit") Residuals: Min 1Q Median

I want to estimate the center of a distribution with lots of outliers in one tail, and thought I would use a function such as S-plus's location.m() with psi.fun=bisquare (as per MASS 3 p. 131). However, R seems not have such a function, so my questions are: 1) Is there an R equivalent to location.m()? 2) Would huber() give me results that are similar (i.e., close enough)? Thanks.

How to get robust M-estimators of multivariate scatter using Huber's psi? Which package/function should I look into? Ideally, I hope I can self-define thresholds of Huber's psi function. Thanks a lot!!! -- View this message in context: http://www.nabble.com/How-to-get-robust-M-estimator-of-multivariate-scatter-using-Huber%27s-psi--tp20585755p20585755.html Sent from the R help mailing

Hello, I'm learning about robust M-estimators right now and had settled on the "Huber Proposal 2" as implemented in MASS, but further reading made clear, that at least 2 further weighting functions (Hampel, Tukey bisquare) exist. In a post from B.D. Ripley going back to 1999 I found the following quote: >> 2) Would huber() give me results that are similar (i.e., close

Dear Group, I am having trouble with using rlm on multivariate data sets. When I call rlm I get Error in lm.wfit(x, y, w, method = "qr") : incompatible dimensions lm on the same data sets seem to work well (see code example). Am I doing something wrong? I have already browsed through the forums and google but could not find any related discussions. I use Windows XP and R

Hi, I'm doing robust regression with the following command rlm(dip~ind1+ind2-1,method="M",psi=psi,maxit=1000,acc=1e-15) now when I ask for a summary summary(rlm(dip~ind1+ind2-1,method="M",psi=psi,maxit=1000,acc=1e-15)) I get Coefficients: Value Std. Error t value ind1 -0.0377 0.0000 -24203.1415 ind2 1.0370 0.0000 668735.7195 taht is

I am looking at MM-estimates for some interlab comparison work. The usual situation in this particular context is a modest number of results from very expensive methods with abnormally well-characterised performance, so for once we have good "variance" estimates (which can differ substantially for good reason) from most labs. But there remains room for human error or unexpected chemistry

We have a choice when calculating the Huber location estimate: > set.seed(221205) > y <- 7 + 3*rt(30,1) > library(MASS) > huber(y)$mu [1] 5.9117 > coefficients(rlm(y~1)) (Intercept) 5.9204 I was surprised to get two different results. The function huber() works directly with the definition whereas rlm() uses iteratively reweighted least squares. My surprise is

----------------------------------------------------------------------------------------- Packages for the computation of optimally robust estimators ----------------------------------------------------------------------------------------- We would like to announce the availability on CRAN (with possibly a minor delay until on every mirror) of new versions of our packages for the computation of

----------------------------------------------------------------------------------------- Packages for the computation of optimally robust estimators ----------------------------------------------------------------------------------------- We would like to announce the availability on CRAN (with possibly a minor delay until on every mirror) of new versions of our packages for the computation of

Full_Name: Kirk Hampel Version: 2.4.1 OS: Windows Submission from: (NULL) (144.53.251.2) Given an AR(p) model, the last p SE's are wrong. The source of the bug is that the C code (ver 2.4.0) assumes *npsi is the length of the psi vector (which is n+p), whilst the predict.ar function in R passes out as.integer(npsi), where npsi <- n-1. Some R code following reproduces the error. Let p=4,

Hello, I'm trying to do a little rlm of some data that looks like this: UNIT COHORT perdo adjodds 1010 96 0.39890 1.06894 1010 97 0.48113 1.57500 1010 98 0.36328 1.21498 1010 99 0.44391 1.38608 It works fine like this: rlm(perdo ~ COHORT, psi=psisquare) But the problem is that I have about 100 UNITs, and I want to do a

Hello R-listers! My first post to the list is a very simple one for those who use the software continuosly. I am trying to understand the fixed-x resampling and random-x-resampling method proposed by Fox about Bootstrapping. The doubt that I have is on the side of the model run in one of the functions expressed for fixed-x resampling. What I don't understand is: X=model.matrix, and the -1