similar to: need help in using Hessian matrix

Displaying 20 results from an estimated 900 matches similar to: "need help in using Hessian matrix"

2009 Nov 01
2
intigrate function and absolute error
Hi Can we get the result of an intigration without the absolute error? for example f1<-function(x1){(1/gamma(alpha))*x1^(alpha-1)*exp(-x1)*log(x1)} I1<-integrate(f1, 0, (max(cc)-tau1+(theta2/theta1)*tau1)/theta2) I1 0.08007414 with absolute error < 7.2e-05 I need the answer 0.08007414 withou the other part(with absolute error <7.2e-05) how can we do that? thank you and take care
2008 Jul 26
4
parametric bootstrap
Hi I am trying to find a parametric bootstrap confidence interval and when I used the boot function I get zero bias and zero st.error? What could be my mistake? Thank you and take care. Laila [[alternative HTML version deleted]]
2008 Nov 26
1
Finding Stopping time
Can any one help me to solve problem in my code? I am actually trying to find the stopping index N. So first I generate random numbers from normals. There is no problem in finding the first stopping index. Now I want to find the second stopping index using obeservation starting from the one after the first stopping index. E.g. If my first stopping index was 5. I want to set 6th observation from
2023 Aug 20
2
Issues when trying to fit a nonlinear regression model
Dear Bert, Thank you so much for your kind and valuable feedback. I tried finding the starting values using the approach you mentioned, then did the following to fit the nonlinear regression model: nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x), start = list(theta1 = 0.37, theta2 = exp(-1.8), theta3 =
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Oh, sorry; I changed signs in the model, fitting theta0 + theta1*exp(theta2*x) So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 = +.055 as starting values. -- Bert On Sun, Aug 20, 2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you so much for your kind and valuable feedback. I tried finding the > starting
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Dear Bert, Thank you for your extremely valuable feedback. Now, I just want to understand why the signs for those starting values, given the following: > #Fiting intermediate model to get starting values > intermediatemod <- lm(log(y - .37) ~ x, data=mod14data2_random) > summary(intermediatemod) Call: lm(formula = log(y - 0.37) ~ x, data = mod14data2_random) Residuals: Min
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Basic algebra and exponentials/logs. I leave those details to you or another HelpeR. -- Bert On Sun, Aug 20, 2023 at 12:17?PM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you for your extremely valuable feedback. Now, I just want to > understand why the signs for those starting values, given the following: > > #Fiting intermediate model to get
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
I got starting values as follows: Noting that the minimum data value is .38, I fit the linear model log(y - .37) ~ x to get intercept = -1.8 and slope = -.055. So I used .37, exp(-1.8) and -.055 as the starting values for theta0, theta1, and theta2 in the nonlinear model. This converged without problems. Cheers, Bert On Sun, Aug 20, 2023 at 10:15?AM Paul Bernal <paulbernal07 at
2023 Aug 20
3
Issues when trying to fit a nonlinear regression model
Dear friends, This is the dataset I am currently working with: >dput(mod14data2_random) structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L, 39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4, 0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4 ), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34, 28)), row.names = c(NA, -15L), class =
2008 Apr 22
2
optimization setup
Hi, here comes my problem, say I have the following functions (example case) #------------------------------------------------------------ function1 <- function (x, theta) {a <- theta[1] ( 1 - exp(-theta[2]) ) * theta[3] ) b <- x * theta[1] / theta[3]^2 return( list( a = a, b = b )) } #----------------------------------------------------------- function2<-function (x, theta) {P
2007 Sep 12
1
enquiry
Dear R-help, I am trying to estimate a Cox model with nested effects basing on the minimization of the overall AIC; I have two frailties terms, both gamma distributed. There is a error message (theta2 argument misses) and I don?t understand why. I would like to know what I have wrong. Thank you very much for your time. fitM7 <- coxph(Surv(lifespan,censured) ~ south + frailty(id,
2008 Apr 22
4
how to convert non numeric data into numeric?
I am having the following error in my function function(theta,reqdIRR) { theta1<-theta[1] theta2<-theta[2] n<-length(reqdIRR) constant<- n*(theta1+theta2) sum1<-lapply(reqdIRR*exp(theta1),FUN = sum) sum2<-lapply(exp(theta2 - reqdIRR*exp(theta1)),FUN = sum) sum = sum1 + sum2 log.fcn = constant - as.numeric(sum) result = - log.fcn return(result) } *error :
2012 Jul 12
0
Generate random numbers with nested Archimedean Copula
Hi everybody, I try to simulate random numbers from a trivariate nested Archimedean copula. My aim is to correlate two processes with, e.g. theta2, as the so called child pair and then to correlate these two processes with a third one with theta1 (parent). This "figure" tries to capture what I am explaining theta1 theta2
2009 Oct 27
1
Poisson dpois value is too small for double precision thus corrupts loglikelihood
Hi - I have a likelihood function that involves sums of two possions: L = a*dpois(Xi,theta1)*dpois(Yi,theta2)+b*(1-c)*a*dpois(Xi,theta1+theta3)*dpois(Yi,theta2) where a,b,c,theta1,theta2,theta3 are parameters to be estimated. (Xi,Yi) are observations. However, Xi and Yi are usually big (> 20000). This causes dpois to returns 0 depending on values of theta1, theta2 and theta3. My first
2004 May 15
2
questions about optim
Hi, I am trying to do parameter estimation with optim, but I can't get it to work quite right-- I have an equation X = Y where X is a gaussian, Y is a multinomial distribution, and I am trying to estimate the probabilities of Y( the mean and sd of X are known ), Theta1, Theta2, Theta3, and Theta4; I do not know how I can specify the constraint that Theta1 + Theta2 + Theta3 + Theta4 = 1 in
2007 Jul 26
3
substituting dots in the names of the columns (sub, gsub, regexpr)
Dear R users, I have the following two problems, related to the function sub, grep, regexpr and similia. The header of the file(s) I have to import is like this. c("y (m)", "BD (g/cm3)", "PR (Mpa)", "Ks (m/s)", "SP g./g.", "P (m3/m3)", "theta1 (g/g)", "theta2 (g/g)", "AWC (g/g)") To get rid of spaces and
2008 Jun 27
3
For loop
Hi, Could you please let me know to use a list in a for loop here geneset is a loop.I am trying to match the names of the list with 1st row of the output. result<- list() for(i in 1:length(output) { result[[i]] <- geneset(which(geneset %n% output[,1])) } Kindly help me out -- View this message in context: http://www.nabble.com/For-loop-tp18163665p18163665.html Sent from the R
2011 May 23
6
Reading Data from mle into excel?
Hi there, I ran the following code: vols=read.csv(file="C:/Documents and Settings/Hugh/My Documents/PhD/Swaption vols.csv" , header=TRUE, sep=",") X<-ts(vols[,2]) #X dcOU<-function(x,t,x0,theta,log=FALSE){ Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t) Vx<-theta[3]^2*(1-exp(-2*theta[2]*t))/(2*theta[2]) dnorm(x,mean=Ex,sd=sqrt(Vx),log=log) }
2006 Oct 17
4
if statement error
Hi List, I was not able to make this work. I know it is a simple one, sorry to bother. Give me some hints pls. Thanks! Jen if(length(real.d)>=30 && length(real.b)>=30 && beta1*beta2*theta1*theta2>0 ) { r <- 1; corr <- 1; } real.d and real.b are two vectors, beta1,beta2,theta1,and theta2 are constants. The error occurred like this: Error in if
2010 Nov 12
1
Problem retrieving data from R2InBUGS
Dear list I am calling the functiton bugs() provided by R2WinBugs to performs an IRT analysis. The function returns a set of estimated parameters over n replications/iterations. For each replication, two sets of person measures (theta1 and theta2) and two sets of item difficulty parameters (diff1 and diff2) are returned. The code used to obtain these estimates is as follows: sim <-