similar to: Error in backSpline.npolySpline(sp) : spline must be monotone

Displaying 20 results from an estimated 200 matches similar to: "Error in backSpline.npolySpline(sp) : spline must be monotone"

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) }
2011 May 17
1
Problem with MLE
Hi there, I am trying to run the following code: > 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) + } > OU.lik<-function(theta1,theta2,theta3){ + n<-length(X) + dt<-deltat(X) +
2010 Sep 14
1
predict(backSpline(x)): losing my marbles?
I'm sure I'm doing something completely boneheaded here, but I've used this idiom (constructing an interpolation spline and using prediction from a backSpline to find an approximation profile confidence interval) many times before and haven't hit this particular problem: r2 <- c(1.04409027570601, 1.09953936543359, 1.15498845516117, 1.21043754488875,
2009 Aug 24
0
Monotone Smoothing specifically I splines
Hello I am looking for a function to create an Integrated (I) spline basis, somehting similar to the likes of 'bs' and 'ns'. I have come across the funcitons, fda::eval.monfd Values of a Monotone Functional Data Object fda::/.fd FDA internal functions fda::monfn Evaluates a monotone function fda::smooth.monotone Monotone
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
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
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
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 =
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
2008 Dec 06
0
Inversing a non-monotonic spline
I have developed a GAM model in order to predict Y using 4 X variables. 2 of these X's are factors, and 1 is a spline. Part of the data looks like: Days WRM variety PWM O_EC 31 75 1 90 234 31 79 1 78 283 31 82 1 92 281 31 84 1 96 213 31 99 2 69 247 31 100 2 77 324 31 104 2 74 259 31 118 2 81 282 31 61 3 58 478 31 98 3 83 429 31 98 3 70 379 31 156 3 87 467 31 78 4 56 283 31 97 4 67 282 31
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
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
2011 Jul 09
3
Confusing piece of R code
m0<-epxression((4*theta1*theta2-theta3^2)/(2*x*theta3^2)-0.5*theta1*x) params<-all.vars(m0) this reads all the params from m0 so theta1,2 and 3 correct? params<-params[-which(params=="x")] checks which params are multiplied by x? np<-length(params) for(i in 1:6){ esp<-get(sprintf("m%d",i-1))
2004 Jul 12
3
Smooth monotone estimation on R
Hi all, I'm looking for smooth monotone estimation packages, preferably using splines. I downloaded the 'cobs' package and intend to use it, but since it offers only quadratic splines based on L1 minimization, I'd like to compare its performance to that of a more 'mainstream' cubic-spline, L2-norm minimizing spline. Preferably a smoothing spline. Does anyone know of such
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
2004 Mar 18
1
profile error on an nls object
Hello all, This is the error message that I get. > hyp.res <- nls(log(y)~log(pdf.hyperb(theta,X)), data=dataModel, + start=list(theta=thetaE0), + trace=TRUE) 45.54325 : 0.1000000 1.3862944 -4.5577142 0.0005503 3.728302 : 0.0583857346 0.4757772859 -4.9156128701 0.0005563154 1.584317 : 0.0194149477 0.3444648833 -4.9365149150 0.0004105426 1.569333 :
2012 Mar 16
2
Elegant Code
Hi, Can anyone help to write a more elegant version of my code? I am sure this can be put into a loop but I am having trouble creating the objects b1,b2,b3,...,etc. b1 <- rigamma(50,1,1) theta1 <- rgamma(50,0.5,(1/b1)) sim1 <- rpois(50,theta1) b2 <- rigamma(50,1,1) theta2 <- rgamma(50,0.5,(1/b2)) sim2 <- rpois(50,theta2) b3 <- rigamma(50,1,1) theta3 <-
2023 Nov 07
1
non-linear regression and root finding
G'day Troels, On Mon, 6 Nov 2023 20:43:10 +0100 Troels Ring <tring at gvdnet.dk> wrote: > Thanks a lot! This was amazing. I'm not sure I see how the conditiion > pK1 < pK2 < pK3 is enforced? One way of enforcing such constraints (well, in finite computer arithemtic only "<=" can be enforced) is to rewrite the parameters as: pK1 = exp(theta1) ##
2023 Nov 07
1
non-linear regression and root finding
Thanks a lot, Berwin. Unfortunately, pK1 may well be negative and as I understand the literature it may be poorly defined as such, and also seems to be at a boundary, since when lower is set to say rep(-4,3) pK1 is returned as -4 while pK2 and pK3 are undisturbed. Perhaps the point is that pK1 is not carrying any information at the pH around 5. Fair enough, I guess. Only, I believe I need