Unlikely that someone will be interested in guessing. You are much more likely
to get a constructive response if you provide the minimal reproducible example
requested in the footer.
Of course, it is possible that you just might find the answer if you carefully
read the help page for nlm, since
Hi R community
I have written a loglikelihood function which I am minimizing using
nlm().
nlm() is giving me no results...I mean, I am getting initial values as
estimates. No iteration.
I have tried many initials value close to true values and far away from
tru values. But every time I
am gett
Hello,
I am trying to compute MLE for non-Gaussian AR(1). The error term follows a
difference poisson distribution. This distribution has one parameter
(vector[2]).
So in total I want to estimate two parameters: the AR(1) paramter (vector[1])
and the distribution parameter.
My function is
224-405-1425 Fax 224-405-4971
robertag...@discover.com
Steve Lianoglou
11/05/2009 04:56 PM
To
""
cc
r-help@r-project.org
Subject
Re: [R] NLM OUTPUT
Hi Bob
.com
Steve Lianoglou
11/05/2009 04:56 PM
To
""
cc
r-help@r-project.org
Subject
Re: [R] NLM OUTPUT
Hi Bob,
On Nov 5, 2009, at 3:04 PM,
wrote:
> I am missing something fundamental. I ran the fu
ax 224-405-4971
robertag...@discover.com
Steve Lianoglou
11/05/2009 04:56 PM
To
""
cc
r-help@r-project.org
Subject
Re: [R] NLM OUTPUT
Hi Bob,
On Nov 5, 2
documentation about converting such a list element to the
vector it
displays.
Are you just asking how to pull out the appropriate parts of the
returned value from the nlm function call?
Taking code from the Example section of ?nlm, run this:
R> f <- function(x, a) sum((x-a)^2)
R> r &
I am missing something fundamental. I ran the function nlm, but I don't
understand how to extract the optimal solution as a numeric vector. The
function produces it as one element of a list. I don't see anything in
the R documentation about converting such a list element to the vector it
dis
Andrew Wang wrote:
I am trying to understand NLM package, so I generated this data set consisting y and x using
y= a + b*x +c*x^2 + N(0,10), with a=3.5,b=4.5,c=5.5
Given y and x, I am trying to use NLM to have estimates of parameters a, b and c that minimize the least square error
my co
I am trying to understand NLM package, so I generated this data set consisting
y and x using
y= a + b*x +c*x^2 + N(0,10), with a=3.5,b=4.5,c=5.5
Given y and x, I am trying to use NLM to have estimates of parameters a, b and
c that minimize the least square error
my code looks like
f<- fu
I am using nlm to maximize a likelihood function. When I call the likelihood
function (garchLLH) via nlm however, nlm returns the wrong value of the
function.
When I test the likelihood function manually I get the correct answer. I'm
probably doing something really stupid, maybe someone can p
It says the value is *missing* (NA), not that the length is wrong.
My quess is that term*bexp is NaN, but you have given us no context to go
on.
On Tue, 3 Jun 2008, Redding, Matthew wrote:
Hi R-Gurus,
I've been cutting along quite nicely with nlm, until
I threw in the following condition in
Hi R-Gurus,
I've been cutting along quite nicely with nlm, until
I threw in the following condition in the function that nlm is
minimising:
if (((term*bexp) < 0.0001)) {
#warning(term*bexp, "=term*bexp",psi,"=psi")
theta<-2000
}
Now when I run this function anywhere
On May 29, 2008, at 11:54 PM, Redding, Matthew wrote:
Dear R Gurus,
I am having a little difficulty with nlm. I've searched the
archives and
found nothing that tells me why this is occuring -- though there are
some slightly similar issues.
A simple example:
lev2<-function(aaa,bbb,ccc,ddd,e
Dear R Gurus,
I am having a little difficulty with nlm. I've searched the archives and
found nothing that tells me why this is occuring -- though there are
some slightly similar issues.
A simple example:
lev2<-function(aaa,bbb,ccc,ddd,eee){
res<-aaa+bbb+ccc+ddd+eee
res
}
nlm(l
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