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

From: glenn andrews <ga_at_aggies.com>
Date: Thu, 27 Mar 2008 17:29:06 -0500

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))

 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 


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.

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