Re: [R] Help with maximum likelihood estimation

[peter dalgaard]

Bricks fly fine with sufficient thrust, but you have lff with a mu argument 
that never gets used, so the negative log-likelihood is constant and mle() 
cannot minimize it.

You need to read up on the definition of (log-) likelihood and write a proper 
one for your problem.

-pd

> On 26 Jun 2019, at 14:01 , avadhoot velankar  
> wrote:
> 
> I am analyzing animal movement pattern using levy flight pattern and want
> to fit power function to observed data and estimate exponent using Maximum
> Likelihood Estimation.
> 
> I am using
> 
> lff<-function(mu){1-1/mean(log(x))}
> library(stats4)
> mle(lff, start = list(mu = 1))
> 
> where x is the observed data.
> 
> I am getting following error
> 
> Error in solve.default(oout$hessian) :
>  Lapack routine dgesv: system is exactly singular: U[1,1] = 0
> 
> This is first time i am writing function and I am as good at algebra
> as brick is good at flying.
> 
> Thank you in advance for your help.
> 
> -- 
> *Avadhoot D. Velankar*
> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd@cbs.dk  Priv: pda...@gmail.com

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[R] Help with maximum likelihood estimation

[avadhoot velankar]

I am analyzing animal movement pattern using levy flight pattern and want
to fit power function to observed data and estimate exponent using Maximum
Likelihood Estimation.

I am using

lff<-function(mu){1-1/mean(log(x))}
library(stats4)
mle(lff, start = list(mu = 1))

where x is the observed data.

I am getting following error

Error in solve.default(oout$hessian) :
  Lapack routine dgesv: system is exactly singular: U[1,1] = 0

This is first time i am writing function and I am as good at algebra
as brick is good at flying.

Thank you in advance for your help.

-- 
*Avadhoot D. Velankar*

[[alternative HTML version deleted]]

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Re: [R] help in maximum likelihood estimation

[Marc Girondot via R-help]

Le 28/03/2016 22:19, heba eldeeb via R-help a écrit :

  Dear AllI'm trying to find the maximum likelihood estimator  of a certain 
distribution using nlm command but I receive an error as:
  non-finite value supplied by 'nlm'
can't figure out what is wrong in my function
Any help?
Thank you in advance



Hi,

Whitout rproducible example, it will be impossible to help you 
efficiently. See https://www.r-project.org/posting-guide.html
Anyway, this error means that your function returns NA or error for some 
combination of parameters.


Marc

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[R] help in maximum likelihood estimation

[heba eldeeb via R-help]

 Dear AllI'm trying to find the maximum likelihood estimator  of a certain 
distribution using nlm command but I receive an error as:
 non-finite value supplied by 'nlm'
can't figure out what is wrong in my function
Any help? 
Thank you in advance



[[alternative HTML version deleted]]

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Re: [R] help in maximum likelihood

[Ravi Varadhan]

Standard error = sqrt(diag(solve(opt$hessian)))

Ravi

From: Alaa Sindi [mailto:alaasi...@icloud.com]
Sent: Wednesday, March 02, 2016 3:22 PM
To: Ravi Varadhan 
Cc: r-help@r-project.org
Subject: Re: help in maximum likelihood

Thank you very much prof. Ravi,

That was very helpful. Is there a way to get the t and p value for the 
coefficients?

Thanks
Alaa
On Mar 2, 2016, at 10:05 AM, Ravi Varadhan 
> wrote:

There is nothing wrong with the optimization.  It is a warning message.  
However, this is a good example to show that one should not simply dismiss a 
warning before understanding what it means.  The MLE parameters are also large, 
indicating that there is something funky about the model or the data or both.  
In your case, there is one major problem with the data:  for the highest dose 
(value of x), you have all subjects responding, i.e. y = n.  Even for the next 
lower dose, there is almost complete response.  Where do these data come from? 
Are they real or fake (simulated) data?

Also, take a look at the eigenvalues of the hessian at the solution.  You will 
see that there is some ill-conditioning, as the eigenvalues are widely 
separated.

x <- c(1.6907, 1.7242, 1.7552, 1.7842, 1.8113, 1.8369, 1.8610, 1.8839)
y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
n <- c(59, 60, 62, 56, 63, 59, 62, 60)

# note: there is no need to have the choose(n, y) term in the likelihood
fn <- function(p)
sum( - (y*(p[1]+p[2]*x) - n*log(1+exp(p[1]+p[2]*x))) )

out <- nlm(fn, p = c(-50,20), hessian = TRUE)

out

eigen(out$hessian)


Hope this is helpful,
Ravi



Ravi Varadhan, Ph.D. (Biostatistics), Ph.D. (Environmental Engg)
Associate Professor,  Department of Oncology
Division of Biostatistics & Bionformatics
Sidney Kimmel Comprehensive Cancer Center
Johns Hopkins University
550 N. Broadway, Suite -E
Baltimore, MD 21205
410-502-2619



[[alternative HTML version deleted]]

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Re: [R] help in maximum likelihood

[Alaa Sindi]

Thank you very much prof. Ravi,

That was very helpful. Is there a way to get the t and p value for the 
coefficients?

Thanks 
Alaa
> On Mar 2, 2016, at 10:05 AM, Ravi Varadhan  wrote:
> 
> There is nothing wrong with the optimization.  It is a warning message.  
> However, this is a good example to show that one should not simply dismiss a 
> warning before understanding what it means.  The MLE parameters are also 
> large, indicating that there is something funky about the model or the data 
> or both.  In your case, there is one major problem with the data:  for the 
> highest dose (value of x), you have all subjects responding, i.e. y = n.  
> Even for the next lower dose, there is almost complete response.  Where do 
> these data come from? Are they real or fake (simulated) data?
>  
> Also, take a look at the eigenvalues of the hessian at the solution.  You 
> will see that there is some ill-conditioning, as the eigenvalues are widely 
> separated.
>  
> x <- c(1.6907, 1.7242, 1.7552, 1.7842, 1.8113, 1.8369, 1.8610, 1.8839)
> y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
> n <- c(59, 60, 62, 56, 63, 59, 62, 60)
>  
> # note: there is no need to have the choose(n, y) term in the likelihood
> fn <- function(p)
> sum( - (y*(p[1]+p[2]*x) - n*log(1+exp(p[1]+p[2]*x))) )
>  
> out <- nlm(fn, p = c(-50,20), hessian = TRUE)
>  
> out
>  
> eigen(out$hessian)
>  
>  
> Hope this is helpful,
> Ravi
>  
>  
>  
> Ravi Varadhan, Ph.D. (Biostatistics), Ph.D. (Environmental Engg)
> Associate Professor,  Department of Oncology
> Division of Biostatistics & Bionformatics
> Sidney Kimmel Comprehensive Cancer Center
> Johns Hopkins University
> 550 N. Broadway, Suite -E
> Baltimore, MD 21205
> 410-502-2619
>  


[[alternative HTML version deleted]]

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Re: [R] help in maximum likelihood

[Ravi Varadhan]

There is nothing wrong with the optimization.  It is a warning message.  
However, this is a good example to show that one should not simply dismiss a 
warning before understanding what it means.  The MLE parameters are also large, 
indicating that there is something funky about the model or the data or both.  
In your case, there is one major problem with the data:  for the highest dose 
(value of x), you have all subjects responding, i.e. y = n.  Even for the next 
lower dose, there is almost complete response.  Where do these data come from? 
Are they real or fake (simulated) data?

Also, take a look at the eigenvalues of the hessian at the solution.  You will 
see that there is some ill-conditioning, as the eigenvalues are widely 
separated.

x <- c(1.6907, 1.7242, 1.7552, 1.7842, 1.8113, 1.8369, 1.8610, 1.8839)
y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
n <- c(59, 60, 62, 56, 63, 59, 62, 60)

# note: there is no need to have the choose(n, y) term in the likelihood
fn <- function(p)
sum( - (y*(p[1]+p[2]*x) - n*log(1+exp(p[1]+p[2]*x))) )

out <- nlm(fn, p = c(-50,20), hessian = TRUE)

out

eigen(out$hessian)


Hope this is helpful,
Ravi



Ravi Varadhan, Ph.D. (Biostatistics), Ph.D. (Environmental Engg)
Associate Professor,  Department of Oncology
Division of Biostatistics & Bionformatics
Sidney Kimmel Comprehensive Cancer Center
Johns Hopkins University
550 N. Broadway, Suite -E
Baltimore, MD 21205
410-502-2619


[[alternative HTML version deleted]]

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Re: [R] help in maximum likelihood

[ProfJCNash]

It's useful to add "print.level=2" inside your call to find that there's
essentially nothing wrong.

Rvmmin doesn't give the msg and numDeriv gives a similar (BUT NOT
EXACTLY THE SAME!) hessian estimate.

It's almost always worthwhile turning on the diagnostic printing when
doing optimization, even if you throw away the output pretty well right
away, because it can often suggest whether the results are reasonable or
rubbish.

JN


On 16-03-01 03:09 PM, Alaa Sindi wrote:
> 
> Hi all,
> 
> what is wrong with this code? I am trying to estimate the model parameters by 
> maximizing the likelihood function and I am getting this warning 
> 
> 
> Warning message:
> In nlm(fn, p = c(-50, 20), hessian = TRUE) :
>   NA/Inf replaced by maximum positive value
> 
> 
> 
> 
> x <- c(1.6907, 1.7242, 1.7552, 1.7842, 1.8113, 1.8369, 1.8610, 1.8839)
> y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
> n <- c(59, 60, 62, 56, 63, 59, 62, 60)
> fn <- function(p)
> sum( - (y*(p[1]+p[2]*x) - n*log(1+exp(p[1]+p[2]*x))
> + log(choose(n, y)) ))
> out <- nlm(fn, p = c(-50,20), hessian = TRUE)
> 
> 
> Thanks
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] help in maximum likelihood

[Alaa Sindi]

Hi all,

what is wrong with this code? I am trying to estimate the model parameters by 
maximizing the likelihood function and I am getting this warning 


Warning message:
In nlm(fn, p = c(-50, 20), hessian = TRUE) :
  NA/Inf replaced by maximum positive value




x <- c(1.6907, 1.7242, 1.7552, 1.7842, 1.8113, 1.8369, 1.8610, 1.8839)
y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
n <- c(59, 60, 62, 56, 63, 59, 62, 60)
fn <- function(p)
sum( - (y*(p[1]+p[2]*x) - n*log(1+exp(p[1]+p[2]*x))
+ log(choose(n, y)) ))
out <- nlm(fn, p = c(-50,20), hessian = TRUE)


Thanks
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