[R] [Re: Significance of confidence intervals in the Non-Linear Least Squares Program.]

[glenn andrews]

Thanks for the response. I was not very clear in my original request.

What I am asking is if in a non-linear estimation problem using nls(), 
as the condition number of the Hessian matrix becomes larger, will the 
t-values of one or more of the parameters being estimated in general 
become smaller in absolute value -- that is, are low t-values a  
sign of an ill-conditioned Hessian?

Typical nls() ouput:

Formula: y ~ (a + b * log(c * x1^d + (1 - c) * x2^d))

Parameters:
 Estimate Std. Error t value Pr(>|t|) 
a  0.11918    0.07835   1.521   0.1403 
b -0.34412    0.27683  -1.243   0.2249 
c  0.33757    0.13480   2.504   0.0189 *
d -2.94165    2.25287  -1.306   0.2031 

Glenn

Prof Brian Ripley wrote:

> On Wed, 26 Mar 2008, glenn andrews wrote:
>
>> I am using the non-linear least squares routine in "R" -- nls.  I have a
>> dataset where the nls routine outputs tight confidence intervals on the
>> 2 parameters I am solving for.
>
>
> nls() does not ouptut confidence intervals, so what precisely did you do?
> I would recommend using confint().
>
> BTW, as in most things in R, nls() is 'a' non-linear least squares 
> routine: there are others in other packages.
>
>> As a check on my results, I used the Python SciPy leastsq module on the
>> same data set and it yields the same answer as "R" for the
>> coefficients.  However, what was somewhat surprising was the the
>> condition number of the covariance matrix reported by the SciPy leastsq
>> program = 379.
>>
>> Is it possible to have what appear to be tight confidence intervals that
>> are reported by nls, while in reality they mean nothing because of the
>> ill-conditioned covariance matrix?
>
>
> The covariance matrix is not relevant to profile-based confidence 
> intervals, and its condition number is scale-dependent whereas the 
> estimation process is very much less so.
>
> This is really off-topic here (it is about misunderstandings about 
> least-squares estimation), so please take it up with your statistical 
> advisor.
>