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