similar to: Regression coefficient constraints

Hi, I'm trying to fit the following model to data using 'nls': y = alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x) and the call I've been using is: nls(y ~ alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x), start=list(alpha_1=4, alpha_2=2, beta_1=3.5, beta_2=2.5), trace=TRUE, control=nls.control(maxiter =

Hello, I have a simple question: I want to list numbers 1:k, but if k <1, I hope nothing listed. how should we do? k=2 for (i in 1:k)  print(i) [1] 1    # <-correct [1] 2 k=0 for (i in 1:k) print(i) [1] 1      #<---- wrong [1] 0   thanks jian [[alternative HTML version deleted]]

Hi,I know only mle function but it seems that in mle one can only specify the bound of the unknowns forming the likelihood function. But I would like to specify something like, a = 2b or a <= 2b where 'a' and 'b' could be my parameters in the likelihood function. Any help would be really appreciated. Thank you!- adschai [[alternative HTML version deleted]]

Dear Group, I have a following dataset: > a A B C D 1 22 3 31 40 2 26 31 36 32 3 3 7 49 16 4 24 40 27 26 5 20 45 47 0 6 34 43 11 18 7 48 48 24 2 8 3 16 39 48 9 20 49 7 21 10 17 36 47 10 > dput(a) structure(list(A = c(22L, 26L, 3L, 24L, 20L, 34L, 48L, 3L, 20L, 17L), B = c(3L, 31L, 7L, 40L, 45L, 43L, 48L, 16L, 49L, 36L), C = c(31L, 36L, 49L, 27L, 47L, 11L, 24L,

Dear R helpers Suppose x  <- c(1:3) y  <- matrix(1:12, ncol = 3, nrow = 4) > y      [,1] [,2] [,3] [1,]    1    5    9 [2,]    2    6   10 [3,]    3    7   11 [4,]    4    8   12 I wish to multiply 1st column of y by first element of x i.e. 1, 2nd column of y by 2nd element of x i.e. 2 an so on. Thus the resultant matrix should be like > z      [,1]   [,2]    [,3] [1,]    1   

Hi, Thank you for your help. I'm trying to constrain the coefficients on a linear regression model, <-lm(...), to all equal one, except for the intercept. I've searched help(glm), help(model.fit), etc. but I can not find where to add the constraint. Your help would be greatly appreciated. Thanks. [[alternative HTML version deleted]]

Hi, I have this lame question. I want to convert a list (each with varies in length) to matrix with same row length by eliminating vectors outside the needed range. For example: l<-list(NULL) l[[1]]=1,2,3.7 l[[2]]=3,4,5,6,3 l[[3]]=4,2,5,7 l[[4]]=2,4,6,3,2 l[[5]]=3,5,7,2 #so say I want to only have 4 rows and 5 column in my matrix (or data.frame) and eliminating the 5th index value in l[[2]]

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives

I need to force a coxph() function in R to use a pre-calculated set of beta coefficients of a gene signature consisting of xx genes and the gene expression is also provided of those xx genes. If I try to use "coxph()" function in R using just the gene expression data alone, the beta coefficients and coxph$linear.predictors will change and I need to use the pre-calcuated linear predictor

Hello, I'm having trouble figuring out how to calculate a p-value for a 1-tailed test of beta_1 in a linear model fit using command lm. My model has only 1 continuous, predictor variable. I want to test the null hypothesis beta_1 is >= 0. I can calculate the p-value for a 2-tailed test using the code "2*pt(-abs(t-value), df=degrees.freedom)", where t-value and degrees.freedom

Dear All, I would appreciate any help with the following: given the vector 'x' x <- c("Ass1", "Ass.s1", "Ass2", "Ass.s2") I would like to pick up the positions where the character string contains "Ass" but does not contain "Ass.s", so for 'x' that would be positions 1 and 3. I guess this could be programmed around

Dear all I have data like this: x y [1,] 59.74889 3.1317081 [2,] 38.77629 1.7102589 [3,] NA 2.2312962 [4,] 32.35268 1.3889621 [5,] 74.01394 1.5361227 [6,] 34.82584 1.1665412 [7,] 42.72262 2.7870875 [8,] 70.54999 3.3917257 [9,] 59.37573 2.6763249 [10,] 68.87422 1.9697770 [11,] 19.00898 2.0584415 [12,] 60.27915 2.5365194 [13,] 50.76850

Dear Mr. Simon Wood, dear list members, I am trying to fit a similar model with gam from mgcv compared to what I did with BayesX, and have discovered the relatively new possibility of incorporating user-defined matrices for quadratic penalties on parametric terms using the "paraPen" argument. This was really a very good idea! However, I would like to constraint the coefficients

I would like to be able to place a normal distribution surrounding the predicted values at various places on a plot. Below is some toy code that creates a scatterplot and plots a regression line through the data. library(MASS) mu <- c(0,1) Sigma <- matrix(c(1,.8,.8,1), ncol=2) set.seed(123) x <- mvrnorm(50,mu,Sigma) plot(x) abline(lm(x[,2] ~ x[,1])) Say I want to add a normal

Greets, I'm trying to use nlminb() to estimate the parameters of a bivariate normal sample and during one of the iterations it passes a parameter vector to the likelihood function resulting in an invalid covariance matrix that causes dmvnorm() to throw an error. Thus, it seems I need to somehow communicate to nlminb() that the final three parameters in my parameter vector are used to

Hi, I'm trying to get maximum likelihood estimates of \alpha, \beta_0 and \beta_1, this can be achieved by solving the following three equations: n / \alpha + \sum\limits_{i=1}^{n} ln(\psihat(i)) - \sum\limits_{i=1}^{n} ( ln(x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} 1/(psihat(i)) - (\alpha+1) \sum\limits_{i=1}^{n} ( 1 / (x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} (

Hello, I would like to create a matrix with one of the columns named $\delta$. I have also created columns $\beta_1$ , $\beta_2$, etc. However, it seems like \d is an escape sequence which gets automatically removed. (Using these names such that they work right in xtable -> latex) colnames(simpleReg.mat) <- c("$\beta_1$","$SE(\beta_1)$", "$\beta_2$",

Hello R helpers, I am trying to do a linear OLS regression of y on two variables x1 and x2. I want to constrain the coefficients of x1 and x2 to sum up to 1. and therefore run a constrained OLS. Can anybody help with this? (I have seen some answers to similar questions but it was not clear to me what I need to do) - I have tried the lm function with offset but I must not have used it properly.

Hi, I'm quite new to R so this question will sound quite fundamental. I need to create a vector of length 160. The first element should be (1+r)^159 and each element thereafter should decrease by a factor of (1+r) until the 160th element that should be 1. Is there a function similar to seq() but increasing or decreasing by factors? I need to do this in one step i.e, not using loops. Any help

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)