search for: w_i

...new: BEGIN OF TEXT -------------------- (A) Non-?NULL? ?weights? can be used to indicate that different observations have different variances (with the values in ?weights? being inversely proportional to the variances); (B) or equivalently, when the elements of ?weights? are positive integers w_i, that each response y_i is the mean of w_i unit-weight observations (C) (including the case that there are w_i observations equal to y_i and the data have been summarized). (D) However, in the latter case, notice that within-group variation is not used. Therefore, the sigma estimate and residua...

Dear R users, I'm trying to fit a simple random intercept model with a fixed intercept. Suppose I want to assign a weight w_i to the i-th contribute to the log-likelihood, i.e. w_i * logLik_i where logLik_i is the log-likelihood for the i-th subject. I want to maximize the likelihood for N subjects Sum_i {w_i * logLik_i} Here is a simple example to reproduce # require(nlme) > foo &l...

...am getting some weird results here and I think I am missing something. I am trying to program a function that for a set of random variables drawn from uniform distributions plots that distribution of the second order statistic of the ordered variables. (ie I have n uniform distributions on [0, w_i] for w_i different w_j and i=1..n. I want to plot the distribution of the second order statistic ie one less the maximum. I thought that the way to do this is to calculate: F= Sum over i { (1-Fi) * Product of all j different i of Fj} + Product over all i of Fi where Fi are just the respective...

...estricciones lineales, pero no consigo incluir dichas restricciones. Es decir, f<-function(w){ sd(...) # desviación típica de ciertos datos } optim(rep(1/2,8),fn = f,lower=0,upper=1,method='L-BFGS-B') # no se como incluir aquí las restricciones Las restricciones son: la suma de los w_i es 1 y todos los w_i son mayores o iguales que cero. Ya he mirado: constrOptim pero no consigo nada. Salu2, y muchas gracias. -- Víctor Nalda Castellet [[alternative HTML version deleted]]

Dear R developers, I am a little disappointed that my bug report only made it to the wishlist, with the argument: Well, it does not say it has. Only relevant to prediction intervals. predict.lm does calculate prediction intervals for linear models from weighted regression, so they should be correct, right? As far as I can see they are bound to be wrong in almost all cases, if no weights

Dear all, I am trying to use the 'glm' package as part of a semiparametric technique that involves weighting a likelihood in various ways, i.e. L(theta;data)=Sum_i=1,..,n (W_i)(log L(theta;data_i)) Where W_i can be a kernel weighting function, or W_i can be an indicator of 'non-missingness' divided by a propensity score. In a Monte Carlo exercise, the option glm(..., weights=W_i,) works very well for the Gaussian design, but if I were to change it to a Binomial...

...t; >> Arie >> >> On Sun, Oct 8, 2017 at 4:38 PM, Viechtbauer Wolfgang (SP) >> <wolfgang.viechtbauer at maastrichtuniversity.nl> wrote: >>> Ah, I think you are referring to this part from ?lm: >>> >>> "(including the case that there are w_i observations equal to y_i and the data have been summarized)" >>> >>> I see; indeed, I don't think this is what 'weights' should be used for (the other part before that is correct). Sorry, I misunderstood the point you were trying to make. >>> >>&gt...

Hello, I have a vector of samples x of length N. Associated with each sample x_i is a certain weight w_i. All the weights are in another vector w of the same length N. I have another vector of samples y of length n (small n). All these samples have equal weights 1/n. The ECDF of these samples is defined as for example at http://en.wikipedia.org/wiki/Empirical_distribution_function and I can comput...

..._LN_COBYLA (Aunque el gradiente se podría calcular, es fácil) Ahora bien, Quiero hacer una variación de este problema incluyendo que x_i pueda ser cero (no se invierte en i) o si es mayor que cero ya tenga que ser como mínimo a_i y como máximo b_i Esto lo he puesto de la siguiente manera: \sum w_i * f(x_i) donde como antes a_i < x_i < b_i \sum w_i * f(x_i) < Presupuesto Y donde ahora, w_i quiero que sea binario (entero). Por lo que he estado viendo esto es un problema de MINLP: Mixed Integer Non Linear Problem Buscando un poco y mirando en el task view de optimización de cra...

...which is dealing with a different model. The model fitted by lm is y = x\beta + \epsilon, \epsilon \sim N(0, \sigma^2) (Row vector x, column vector \beta.) If the observations are from the model, OLS is appropriate, but weighting is used in several scenarios, including: (a) case weights: w_i = 3 means `I have three observations like (y, x)' (b) inverse-variance weights, most often an indication that w_i = 1/3 means that y_i is actually the average of 3 observations at x_i. (c) multiple imputation, where a case with missing values in x is split into say 5 parts, with case weight...

...el version now has (in ?lm) Non-?NULL? ?weights? can be used to indicate that different observations have different variances (with the values in ?weights? being inversely proportional to the variances); or equivalently, when the elements of ?weights? are positive integers w_i, that each response y_i is the mean of w_i unit-weight observations (including the case that there are w_i observations equal to y_i and the data have been summarized). However, in the latter case, notice that within-group variation is not used. Therefore, the sigma estimate and...

...this is at all possible. For example we can take schools data by Gelman from R2WinBUGS package. As you can see bellow use of weights argument did not had influence on results. I do not know if my specification of weights i.e. 1 / sd^2 is ok. Under least squares one minimizes sum(e^2_i) or sum(w_i * e^2_i) with weighted LS. If I consider that \sigma_i represents uncertainty in my "data" then e'_i = e_i / \sigma_i and we minimize sum(e'^2_i) = sum((e_i / \sigma_i)^2) = sum(e_i * \sigma^-2_i). Therefore weights i.e. w_i are equal to 1 / \sigma^2_i. Can anyone help me wit...

...pars], silent = silent, par.names = names(vars), ...)) param.est = gh.fit$par In the above, vega are the parameters to estimate and pdf is the GH pdf. This seems to work well for the case where observations are equally weighted. However, I'm stuck on how to include a weighted vector (w_i) to turn this problem into a weighted ML optimization. Would you please be able to suggest a function or change in code which may allow me to do this? Thank you in advance for your time. Jason

...nately be passed. } \value{ A list with the following items: \item{AIC}{a data.frame including LL, the maximized log-likelihood; K the number of estimated parameters; N2K, number of observations/K; AIC, the Akaike index criterion; deltAIC, the difference between AIC and the lowest AIC value; w_i, the Akaike weights; AICc, the second order Akaike criterion; deltAICc, the difference between AICc and the lowest AICc value; w_ic, the AICc weights } \item{models}{the list of models} } \references{ Anderson, D.R., Link, W.A., Johnson, D.H. and Burnham, K.P. (2001). Suggestions for prese...

I tried lrm in library(Design) but there is always some error message. Is this function really doing the weighted logistic regression as maximizing the following likelihood: \sum w_i*(y_i*\beta*x_i-log(1+exp(\beta*x_i))) Does anybody know a better way to fit this kind of model in R? FYI: one example of getting error message is like: > x=runif(10,0,3) > y=c(rep(0,5),rep(1,5)) > w=rep(1/10,10) > fit=lrm(y~x,weights=w) Warning message: currently weights are ignored...

...;2.d0". How can I change it to run in R? The function can be found in page 230 from http://www.stat.wisc.edu/~mchung/teaching/stat471/stat_computing.pdf Function is as follows: gauher <- function(n) {# Gauss-Hermite: returns x,w so that #\int_-\infty^\infty exp(-x^2) f(x) dx \doteq \sum w_i f(x_i) EPS <- 3.e-14 PIM4 <- .7511255444649425D0 MAXIT <- 10 m <- trunc((n+1)/2) x <- w <- rep(-1,n) for (i in 1:m) { if (i==1) { z <- sqrt(2*n+1)-1.85575*(2*n+1)^(-.16667) } else if(i==2) { z <- z-1.14*n^.426/z } else if (i==3) { z <- 1.86*z-.86*x[1] } else if (i==4) { z...

...visualisation of a "group structure" in my dataset. So far, no problem. But i would like to estimate that density using weights according to the COS?? values that tells me if my observation is well represented on the factorial plan 1-2. I would like to use (1) instead of (2) where the w_i depends on the COS??, i hacked the MASS::kde2d function to do that. \hat{f}_n(x) = \frac{1}{n|H|} \sum_{i=1}^n w_i \times K\left[H^{-1}(x-X_i)\right] (1) \hat{f}_n(x) = \frac{1}{n|H|} \sum_{i=1}^n K\left[H^{-1}(x-X_i)\right] (2) I have uploaded to http://addictedto...

Hi list: This is my first time to post my question on the list. Thanks for your help. I am solving a quadratic programming using R. Here is my question: w = arg min 0.5*w'Mw - w'N s. t. sum(w) = 1; w>0 note: w is weight vector, each w_i must >=0, and the sum of w =1. Here is my R code: A <-matrix(c(2.26,1.26,1.12,1.12,2.27,1.13,1.12,1.13,2.2),3,3); B <- c(0.007459281,0.007448885,0.007447850); M <-nrow(A); Amat <- cbind(rep(1,M), diag(M)); bvec <- c(1,rep(0,M)) ; meq <- 1; min <- solve.QP(Dmat...

Hi, ? I have a relatively big dataset and I want to construct some histograms using the histogram function in lattice. One thing I am interested in is to look at differences between density and percent. I know I can use the hist function but it seems that this function gives sometimes some wrong answers and the density is actually a percent since it is calculated as counts in the bin divided by

...t, data=bdd2) > lmmedt9<-lme(mediane~log(0.0001+transat), random=~1|site, na.action=na.omit, data=bdd2) Using the Akaike Criterion and selMod of the package pgirmess gives the following output: > selMod(list(lmmedt1,lmmedt9)) model LL K N2K AIC deltAIC w_i AICc deltAICc w_ic 2 log(1e-04 + transat) 44.63758 4 7.5 -81.27516 0.000000 0.65 -79.67516 0.000000 0.57 1 1 43.02205 3 10.0 -80.04410 1.231069 0.35 -79.12102 0.554146 0.43 The usual conclusion would be that the two models are equivalent and to keep the null model for...